Introduction to Quantum Mechanics

Contents Preface 2 1 The Wave Function 3 2 Time-Independent Schrödinger Equation 14 3 Formalism 62 4 Quantum Mec...

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Contents Preface

2

1 The Wave Function

3

2 Time-Independent Schrödinger Equation

14

3 Formalism

62

4 Quantum Mechanics in Three Dimensions

87

5 Identical Particles

132

6 Time-Independent Perturbation Theory

154

7 The Variational Principle

196

8 The WKB Approximation

219

9 Time-Dependent Perturbation Theory

236

10 The Adiabatic Approximation

254

11 Scattering

268

12 Afterword

282

Appendix Linear Algebra

283

2nd Edition – 1st Edition Problem Correlation Grid

299

2

Preface These are my own solutions to the problems in Introduction to Quantum Mechanics, 2nd ed. I have made every effort to insure that they are clear and correct, but errors are bound to occur, and for this I apologize in advance. I would like to thank the many people who pointed out mistakes in the solution manual for the first edition, and encourage anyone who finds defects in this one to alert me (griffi[email protected]). I’ll maintain a list of errata on my web page (http://academic.reed.edu/physics/faculty/griffiths.html), and incorporate corrections in the manual itself from time to time. I also thank my students at Reed and at Smith for many useful suggestions, and above all Neelaksh Sadhoo, who did most of the typesetting. At the end of the manual there is a grid that correlates the problem numbers in the second edition with those in the first edition.

David Griffiths

c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 1. THE WAVE FUNCTION

3

Chapter 1

The Wave Function Problem 1.1 (a) j2 = 212 = 441. j 2 = =

1 2 1 2 j N (j) = (14 ) + (152 ) + 3(162 ) + 2(222 ) + 2(242 ) + 5(252 ) N 14 1 6434 (196 + 225 + 768 + 968 + 1152 + 3125) = = 459.571. 14 14 j 14 15 16 22 24 25

(b)

σ2 = =

σ=

∆j = j − j 14 − 21 = −7 15 − 21 = −6 16 − 21 = −5 22 − 21 = 1 24 − 21 = 3 25 − 21 = 4

1 1 (∆j)2 N (j) = (−7)2 + (−6)2 + (−5)2 · 3 + (1)2 · 2 + (3)2 · 2 + (4)2 · 5 N 14 1 260 (49 + 36 + 75 + 2 + 18 + 80) = = 18.571. 14 14

√

18.571 = 4.309.

(c) j 2 − j2 = 459.571 − 441 = 18.571.

[Agrees with (b).]


4


Problem 1.2 (a)

h

1 1 x2 √ dx = √ 2 hx 2 h

x2 = 0

h2 σ = x − x = − 5 2

2

2

(b)

h h2 2 5/2 = x . 5 5 0

2 2h 4 2 h = h ⇒ σ = √ = 0.2981h. 3 45 3 5

√ x+ 1 1 √ 1 √

√ dx = 1 − √ (2 x) = 1 − √ x+ − x− . 2 hx 2 h h x−

x+

P =1−

x−

x+ ≡ x + σ = 0.3333h + 0.2981h = 0.6315h; √

P =1−

0.6315 +

√

x− ≡ x − σ = 0.3333h − 0.2981h = 0.0352h.

0.0352 = 0.393.

Problem 1.3 (a)

∞

Ae−λ(x−a) dx. 2

1=

Let u ≡ x − a, du = dx, u : −∞ → ∞.

−∞

∞

−λu2

1=A

e

du = A

−∞

(b)

x = A

∞

π λ

⇒ A=

xe−λ(x−a) dx = A 2

−∞ ∞

=A

ue

−λu

2

x = A

∞

2

−∞ ∞

−λu2

e

π du = A 0 + a = a. λ

x2 e−λ(x−a) dx 2

−∞ ∞

=A

2 −λu2

u e

∞

du + 2a

−∞

1 =A 2λ

(u + a)e−λu du

−∞

∞

du + a

−∞ 2

λ . π

π + 0 + a2 λ

σ 2 = x2 − x2 = a2 +

π λ

ue

−λu2

1 1 − a2 = ; 2λ 2λ

∞

du + a

−∞

= a2 +

2

−λu2

e

du

−∞

1 . 2λ 1 σ=√ . 2λ



5

(c) ρ(x) A

x

a

Problem 1.4 (a) |A|2 1= 2 a

a

|A|2

b

(b − x) dx = |A| 2 (b − a) a

3 b−a 2 a 2b = |A| + = |A| ⇒ A= . 3 3 3 b 2

x dx +

2

2

0

(b)

1 a2

a b (b − x)3 x3 1 − + 3 0 (b − a)2 3 a

Ψ A

a

b

x

(c) At x = a. (d)

a

|Ψ|2 dx =

P = 0

|A|2 a2

a

x2 dx = |A|2 0

a a = . 3 b

P = 1 if b = a, P = 1/2 if b = 2a.

(e) a b 1 1 2 x|Ψ|2 dx = |A|2 2 x3 dx + x(b − x) dx a 0 (b − a)2 a 4 a b 2 3 1 x x3 x4 1 2x = b − 2b + + b a2 4 0 (b − a)2 2 3 4 a 2 3 = a (b − a)2 + 2b4 − 8b4 /3 + b4 − 2a2 b2 + 8a3 b/3 − a4 2 4b(b − a) 4 3 b 2a + b 2 3 1 2 2 = (b3 − 3a2 b + 2a3 ) = −a b + a b = . 2 2 4b(b − a) 3 3 4(b − a) 4

x =


6


Problem 1.5 (a)

∞

|Ψ| dx = 2|A| 2

1=

2

−2λx

e

2

dx = 2|A|

0

∞ e−2λx |A|2 = ; −2λ 0 λ

A=

√

λ.

(b) x =

x|Ψ| dx = |A| 2

x = 2|A| 2

2

2

∞

∞

xe−2λ|x| dx = 0.

[Odd integrand.]

−∞

2 −2λx

x e 0

1 2 = dx = 2λ . 3 (2λ) 2λ2

(c) σ 2 = x2 − x2 =

1 ; 2λ2

√

1 σ=√ . 2λ

|Ψ(±σ)|2 = |A|2 e−2λσ = λe−2λ/

λ

2λ

√

= λe−

2

= 0.2431λ.

|Ψ| 2

.24λ

−σ

+σ

x

Probability outside:

∞

2

|Ψ|2 dx = 2|A|2

σ

∞

e−2λx dx = 2λ

σ

∞ √ e−2λx = e−2λσ = e− 2 = 0.2431. −2λ σ

Problem 1.6 For integration by parts, the differentiation has to be with respect to the integration variable – in this case the differentiation is with respect to t, but the integration variable is x. It’s true that ∂ ∂x 2 ∂ ∂ (x|Ψ|2 ) = |Ψ| + x |Ψ|2 = x |Ψ|2 , ∂t ∂t ∂t ∂t but this does not allow us to perform the integration: a

b

∂ x |Ψ|2 dx = ∂t

a

b

b ∂ (x|Ψ|2 )dx = (x|Ψ|2 )a . ∂t



Problem 1.7 From Eq. 1.33, ∂ ∂t

dp dt

∗ ∂Ψ

Ψ

∂x

= −i

∂ ∂t

7

Ψ∗ ∂Ψ ∂x dx. But, noting that

∂2Ψ ∂x∂t

=

∂2Ψ ∂t∂x

and using Eqs. 1.23-1.24:

∂Ψ∗ ∂Ψ ∂Ψ i ∂ 2 Ψ∗ i ∂ 2 Ψ i i ∗ ∂ ∗ ∂Ψ ∗ ∂ = +Ψ = − +Ψ + VΨ − VΨ ∂t ∂x ∂x ∂t 2m ∂x2 ∂x ∂x 2m ∂x2

3 2 ∗ i ∂ Ψ ∂ Ψ ∂Ψ ∂Ψ ∂ i = Ψ∗ 3 − + V Ψ∗ − Ψ∗ (V Ψ) 2 2m ∂x ∂x ∂x ∂x ∂x

The first term integrates to zero, using integration by parts twice, and the second term can be simplified to ∗ ∂Ψ ∗ ∂V 2 ∂V V Ψ∗ ∂Ψ ∂x − Ψ V ∂x − Ψ ∂x Ψ = −|Ψ| ∂x . So dp = −i dt

∂V i ∂V −|Ψ|2 dx = − . ∂x ∂x

QED

Problem 1.8 ∂ Ψ Suppose Ψ satisfies the Schr¨ odinger equation without V0 : i ∂Ψ ∂t = − 2m ∂x2 + V Ψ. We want to find the solution 2 2 ∂ Ψ0 0 Ψ0 with V0 : i ∂Ψ ∂t = − 2m ∂x2 + (V + V0 )Ψ0 . 2

2

Claim: Ψ0 = Ψe−iV0 t/ .

2 2

∂Ψ −iV0 t/ ∂ Ψ −iV0 t/ 0 Proof: i ∂Ψ + iΨ − iV0 e−iV0 t/ = − 2m + V0 Ψe−iV0 t/ ∂t = i ∂t e ∂x2 + V Ψ e 2

∂ Ψ0 = − 2m ∂x2 + (V + V0 )Ψ0 . 2

QED

This has no effect on the expectation value of a dynamical variable, since the extra phase factor, being independent of x, cancels out in Eq. 1.36.

Problem 1.9 (a) 2

∞

1 = 2|A|

−2amx2 /

e

21

dx = 2|A|

0

2

π = |A|2 (2am/)

π ; 2am

A=

2am π

1/4 .

(b) ∂Ψ = −iaΨ; ∂t

∂Ψ 2amx =− Ψ; ∂x

∂2Ψ 2am =− ∂x2

∂Ψ Ψ+x ∂x

2am =−

2amx2 1−

Ψ.

∂ Ψ Plug these into the Schr¨ odinger equation, i ∂Ψ ∂t = − 2m ∂x2 + V Ψ: 2

2

2 2am 2amx2 V Ψ = i(−ia)Ψ + − 1− Ψ 2m 2amx2 = a − a 1 − Ψ = 2a2 mx2 Ψ, so

V (x) = 2ma2 x2 .


8

CHAPTER 1. THE WAVE FUNCTION (c) x =

∞

x|Ψ|2 dx = 0.

[Odd integrand.]

−∞

∞

x = 2|A| 2

2

2 −2amx2 /

x e 0

p = m

1 dx = 2|A| 2 2 (2am/)

2

π = . 2am 4am

dx = 0. dt

2 ∂ ∂2Ψ 2 Ψ∗ 2 dx p = Ψ Ψdx = − i ∂x ∂x 2 2am 2amx 2am 2 ∗ 2 2 2 = − Ψ − 1− Ψ dx = 2am |Ψ| dx − x |Ψ| dx 2am 2 2am 1 = 2am 1 − x = 2am 1 − = 2am = am. 4am 2

∗

2

(d) σx2

=⇒ σx = = x − x = 4am 2

σx σp =

2

4am

√

; 4am

σp2 = p2 − p2 = am =⇒ σp =

√

am.

am = 2 . This is (just barely) consistent with the uncertainty principle.

Problem 1.10 From Math Tables: π = 3.141592653589793238462643 · · · (a)

P (0) = 0 P (5) = 3/25

P (1) = 2/25 P (6) = 3/25

In general, P (j) =

=

1 25 [0

(c) j 2 = =

1 25 [0

P (3) = 5/25 P (8) = 2/25

1 25 [0

Median: 13 are ≤ 4, 12 are ≥ 5, so median is 4. · 0 + 1 · 2 + 2 · 3 + 3 · 5 + 4 · 3 + 5 · 3 + 6 · 3 + 7 · 1 + 8 · 2 + 9 · 3]

+ 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] = 1 25 [0

P (4) = 3/25 P (9) = 3/25

N (j) N .

(b) Most probable: 3. Average: j =

P (2) = 3/25 P (7) = 1/25

118 25

= 4.72.

+ 12 · 2 + 22 · 3 + 32 · 5 + 42 · 3 + 52 · 3 + 62 · 3 + 72 · 1 + 82 · 2 + 92 · 3]

+ 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] =

σ 2 = j 2 − j2 = 28.4 − 4.722 = 28.4 − 22.2784 = 6.1216;

710 25

= 28.4. √ σ = 6.1216 = 2.474.



9

Problem 1.11 (a) Constant for 0 ≤ θ ≤ π, otherwise zero. In view of Eq. 1.16, the constant is 1/π. 1/π, if 0 ≤ θ ≤ π, ρ(θ) = 0, otherwise.

ρ(θ) 1/π −π/2

π

0

θ 3π/2

(b) θ =

θρ(θ) dθ =

1 θ = π

π

1 π

1 θ dθ = π

2

π

θdθ = 0

2

0

σ 2 = θ2 − θ2 =

1 π

π π θ2 = 2 0 2

[of course].

π θ3 π2 . = 3 0 3

π2 π2 π2 − = ; 3 4 12

π σ= √ . 2 3

(c) 1 sin θ = π

cos θ =

1 π

π

sin θ dθ =

2 1 1 π (− cos θ)|0 = (1 − (−1)) = . π π π

cos θ dθ =

1 π (sin θ)|0 = 0. π

0

1 cos θ = π

π

0

2

π

1 cos θ dθ = π

2

0

π

(1/2)dθ = 0

1 . 2

[Because sin2 θ + cos2 θ = 1, and the integrals of sin2 and cos2 are equal (over suitable intervals), one can replace them by 1/2 in such cases.]

Problem 1.12 (a) x = r cos θ ⇒ dx = −r sin θ dθ. The probability that the needle lies in range dθ is ρ(θ)dθ = probability that it’s in the range dx is ρ(x)dx =

1 π dθ,

so the

1 dx dx 1 dx = √ . = 2 π r sin θ π r 1 − (x/r) π r2 − x2


10

CHAPTER 1. THE WAVE FUNCTION ρ(x)

-2r

√ 1 , π r 2 −x2

∴ ρ(x) =

Total:

0,

r

√ 1 dx −r π r 2 −x2

(b) x =

1 π

2 π

x2 =

r

−r

0

x√

r

√

=

r2

2 π

-r

r

if − r < x < r, otherwise.

r 0

√ 1 dx r 2 −x2

1 dx = 0 − x2

2 π

=

x

2r

[Note: We want the magnitude of dx here.] sin−1

x r r 0

=

2 π

sin−1 (1) =

2 π

·

π 2

= 1.

[odd integrand, even interval].

x r 2 2 x2 2 r2 x 2 = 2 r sin−1 (1) = r . dx = r − x2 + − sin−1 π 2 2 r π 2 2 r2 − x2 0

√ σ 2 = x2 − x2 = r2 /2 =⇒ σ = r/ 2. To get x and x2 from Problem 1.11(c), use x = r cos θ, so x = rcos θ = 0, x2 = r2 cos2 θ = r2 /2.

Problem 1.13 Suppose the eye end lands a distance y up from a line (0 ≤ y < l), and let x be the projection along that same direction (−l ≤ x < l). The needle crosses the line above if y + x ≥ l (i.e. x ≥ l − y), and it crosses the line below if y + x < 0 (i.e. x < −y). So for a given value of y, the probability of crossing (using Problem 1.12) is

−y

P (y) = −l

1 = π

−1

sin

l

1 ρ(x)dx + ρ(x)dx = π l−y

−y

−l

1 √ dx + 2 l − x2

l

l−y

1 √ dx 2 l − x2

x −y l 1 −1 x = − sin−1 (y/l) + 2 sin−1 (1) − sin−1 (1 − y/l) + sin l −l l l−y π

sin−1 (y/l) sin−1 (1 − y/l) − . π π Now, all values of y are equally likely, so ρ(y) = 1/l, and hence the probability of crossing is 1 l l−y 1 l −1 y −1 P = π − sin π − 2 sin−1 (y/l) dy − sin dy = πl 0 l l πl 0 =1−

=

l 1 2 2 2 πl − 2 y sin−1 (y/l) + l 1 − (y/l)2 = 1 − [l sin−1 (1) − l] = 1 − 1 + = . πl πl π π 0



11

Problem 1.14 (a) Pab (t) =

b a

|Ψ(x, t)2 dx,

so

dPab dt

=

b

∂ |Ψ|2 dx. a ∂t

But (Eq. 1.25):

∂|Ψ|2 ∂ i ∂Ψ∗ ∂ ∗ ∂Ψ = Ψ − Ψ = − J(x, t). ∂t ∂x 2m ∂x ∂x ∂t

∴

dPab =− dt

b

a

∂ b J(x, t)dx = − [J(x, t)]|a = J(a, t) − J(b, t). ∂x

QED

Probability is dimensionless, so J has the dimensions 1/time, and units seconds−1 . (b) Here Ψ(x, t) = f (x)e−iat , where f (x) ≡ Ae−amx

2

/

∗

df −iat df iat , so Ψ ∂Ψ = f dx , ∂x = f e dx e

df and Ψ∗ ∂Ψ ∂x = f dx too, so J(x, t) = 0.

Problem 1.15 (a) Eq. 1.24 now reads

∂Ψ∗ ∂t

2

∗

i ∂ Ψ i ∗ ∗ = − 2m ∂x2 + V Ψ , and Eq. 1.25 picks up an extra term:

∂ i i 2Γ 2 |Ψ|2 = · · · + |Ψ|2 (V ∗ − V ) = · · · + |Ψ|2 (V0 + iΓ − V0 + iΓ) = · · · − |Ψ| , ∂t 2Γ ∞ 2Γ 2 and Eq. 1.27 becomes dP QED dt = − −∞ |Ψ| dx = − P . (b) dP 2Γ 2Γ = − dt =⇒ ln P = − t + constant =⇒ P (t) = P (0)e−2Γt/ , so τ = . P 2Γ

Problem 1.16 Use Eqs. [1.23] and [1.24], and integration by parts: d dt

∞

−∞

Ψ∗1 Ψ2

∞ ∗ ∂Ψ1 ∂ ∗ ∗ ∂Ψ2 (Ψ1 Ψ2 ) dx = Ψ2 + Ψ1 dx ∂t ∂t −∞ ∂t −∞ ∞ −i ∂ 2 Ψ∗1 i ∂ 2 Ψ2 i i ∗ ∗ Ψ dx + + Ψ − V Ψ V Ψ 2 2 1 1 2m ∂x2 2m ∂x2 −∞ ∞ 2 ∗ 2 i ∂ Ψ1 ∗ ∂ Ψ2 − dx Ψ − Ψ 2 1 2m −∞ ∂x2 ∂x2 ∞ ∞ ∞ ∞ i ∂Ψ∗1 ∂Ψ∗1 ∂Ψ2 ∂Ψ∗1 ∂Ψ2 ∗ ∂Ψ2 − − + Ψ2 dx − Ψ1 dx = 0. QED 2m ∂x ∂x −∞ −∞ ∂x ∂x −∞ ∂x ∂x −∞

dx = = = =

∞


12


Problem 1.17 (a)

a 3 x5 4 2x 1 = |A| a − x dx = 2|A| a − 2a x + x dx = 2|A| a x − 2a + 3 5 0 −a 0 15 2 1 16 5 2 = 2|A|2 a5 1 − + . = a |A| , so A = 3 5 15 16a5

a

2

2 2

2

2

a

4

2 2

4

2

(b) x =

a

x|Ψ|2 dx = 0.

(Odd integrand.)

−a

(c) p =

2 A i

a

−a

a2 − x2

d 2

a − x2 dx = 0. dx

(Odd integrand.)

−2x

Since we only know x at t = 0 we cannot calculate dx/dt directly. (d) x2 = A2

a

−a

2 x2 a2 − x2 dx = 2A2

0

a

a 3 5 15 x7 4x 2x =2 a − 2a + = 16a5 3 5 7 0 ✚ 2 35 − 42 + 15 15a a2 8 ✚ = = · = 8 8 7 3·✁ 5·7 ✁

a4 x2 − 2a2 x4 + x6 dx

15 7 1 2 1 a − + 8a5 3 5 7 a2 . 7

(e) p2 = −A2 2

=4·

a

−a

15 2 16a5

a2 − x2

d2 2

a − x2 dx = 2A2 2 2 2 dx

a

a2 − x2 dx

0

−2

a2 x −

a 5 2 152 2 x3 152 3 a3 a = = − · = . 5 2 3 0 4a 3 4a 3 2 a2

(f ) σx =

x2 − x2 =

1 2 a a = √ . 7 7

(g) σp =

p2

−

p2

=

5 2 = 2 a2

5 . 2a



13

(h) a σx σp = √ · 7

5 = 2a

5 = 14

10 > . 7 2 2

Problem 1.18 √

h h2 >d ⇒ T < . 3mkB d2 3mkB T

(a) Electrons (m = 9.1 × 10−31 kg): T
0

(d) x =

∞

−∞

x = 2

x|Ψ|2 dx = 0 (odd integrand); p = m

2 w π

∞

2 −2w2 x2

x e −∞

dx =

2 1 w π 4w2

dx = 0. dt

π 1 = . 2w2 4w2

p = − 2

2

∞

−∞

Ψ∗

d2 Ψ dx. dx2


¨ CHAPTER 2. THE TIME-INDEPENDENT SCHRODINGER EQUATION

Write Ψ = Be−bx , where B ≡ 2

2a π

1/4 √

29

1 a . and b ≡ 1 + iθ 1 + iθ

2 d2 Ψ d −bx2 = −2bB(1 − 2bx2 )e−bx . = B −2bxe 2 dx dx Ψ∗

∗ 2 d2 Ψ a a 2a = −2b|B|2 (1 − 2bx2 )e−(b+b )x ; b + b∗ = = 2w2 . + = 2 dx 1 + iθ 1 − iθ 1 + θ2

|B| = 2

2a 1 √ = π 1 + θ2

p = 2b 2

2

= 2b2

But 1 −

2 w π 2 w π

∞

−∞

b =1− 2w2

p2 = 2b2

2 d2 Ψ w. So Ψ∗ 2 = −2b π dx

(1 − 2bx2 )e−2w

π 1 − 2b 2 2w2 4w

a 1 + iθ

a = 2 a. 2b

2

x2

σx =

1 ; 2w

2 2 2 w(1 − 2bx2 )e−2w x . π

dx

π 2w2

1 + θ2 2a

b = 2b2 1 − . 2w2

=1−

(1 − iθ) 1 + iθ a = = , so 2 2 2b

√ σp = a.

(e) σx σp =

1 √ 1 + θ2 = 1 + (2at/m)2 ≥ . a= 2w 2 2 2

Closest at t = 0, at which time it is right at the uncertainty limit.

Problem 2.23 (a) (−2)3 − 3(−2)2 + 2(−2) − 1 = −8 − 12 − 4 − 1 = −25. (b) cos(3π) + 2 = −1 + 2 = 1. (c) 0 (x = 2 is outside the domain of integration).



30

Problem 2.24

If c > 0, y : −∞ → ∞. If c < 0, y : ∞ → −∞.  1 ∞ 1 ∞  c −∞ f (y/c)δ(y)dy = c f (0) (c > 0); or f (x)δ(cx)dx =  1 −∞ 1 ∞ 1 −∞ c ∞ f (y/c)δ(y)dy = − c −∞ f (y/c)δ(y)dy = − c f (0) (c < 0).

1 (a) Let y ≡ cx, so dx = dy. c

In either case, (b)

∞

f (x) −∞

∞

1 f (x)δ(cx)dx = f (0) = |c| −∞

∞

f (x) −∞

1 1 δ(x)dx. So δ(cx) = δ(x). |c| |c|

∞ ∞ df dθ − dx = f θ θdx (integration by parts) dx dx −∞ −∞ ∞ ∞ df = f (∞) − f (x)δ(x)dx. dx = f (∞) − f (∞) + f (0) = f (0) = dx 0 −∞

So dθ/dx = δ(x). [Makes sense: The θ function is constant (so derivative is zero) except at x = 0, where the derivative is infinite.]

Problem 2.25

√ ψ(x) =

mα −mα|x|/2 = e

√

mα

2 e−mαx/ , (x ≥ 0), 2 emαx/ , (x ≤ 0).

x = 0 (odd integrand). 2 3 2 mα 2mα 4 2 x2 = x2 |ψ|2 dx = 2 2 x2 e−2mαx/ dx = 2 2 = ; σx = √ . 2 2 2mα 2m α 2mα −∞ 0   mα −mαx/2 3 √ , (x ≥ 0)  √ − e 2 2 dψ mα  2 mα −θ(x)e−mαx/ + θ(−x)emαx/ . = =  dx  mα mαx/2 , (x ≤ 0) 2 e

∞

√

3

∞

mα mα −mαx/2 mαx/2 mαx/2 θ(x)e − δ(−x)e + θ(−x)e 2 2 √ 3 2 mα mα = −2δ(x) + 2 e−mα|x|/ .

d2 ψ = dx2

mα

−δ(x)e−mαx/ + 2

In the last step I used the fact that δ(−x) = δ(x) (Eq. 2.142), f (x)δ(x) = f (0)δ(x) (Eq. 2.112), and θ(−x) + θ(x) = 1 (Eq. 2.143). Since dψ/dx is an odd function, p = 0. √ 3 ∞ √ ∞ 2 2 mα mα mα d2 ψ p2 = −2 ψ 2 dx = −2 e−mα|x|/ −2δ(x) + 2 e−mα|x|/ dx dx −∞ −∞

∞ mα 2 2 2 2 mα mα mα mα 2 = 2−2 2 1− 2 e−2mαx/ dx = 2 . = 2mα 0 Evidently mα 2 mα √ σp = , so σx σp = √ = 2 > . 2 2 2mα c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


31

Problem 2.26 1 Put f (x) = δ(x) into Eq. 2.102: F (k) = √ 2π

1 ∴ f (x) = δ(x) = √ 2π

∞

−∞

1 1 √ eikx dk = 2π 2π

∞

1 δ(x)e−ikx dx = √ . 2π −∞

∞

eikx dk. QED

−∞

Problem 2.27 V(x)

(a) -a

a x

(b) From Problem 2.1(c) the solutions are even or odd. Look first for even solutions:  −κx (x < a),  Ae ψ(x) = B(eκx + e−κx ) (−a < x < a),  κx (x < −a). Ae Continuity at a : Ae−κa = B(eκa + e−κa ), or A = B(e2κa + 1).

Discontinuous derivative at a, ∆

dψ 2mα = − 2 ψ(a) : dx

2mα −κa 2mα Ae ⇒ A + B(e2κa − 1) = 2 A; or 2 κ 2mα 2mα 2mα 2mα 2κa 2κa 2κa − 1 = B(e − 1 ⇒ e − 1 + 2 − 1. B(e2κa − 1) = A + 1) − 1 = e 2 κ 2 κ 2 κ κ −κAe−κa − B(κeκa − κe−κa ) = −

1=

2mα 2mα −2κa 2 κ 2 κ −2κa −2κa − 1 + e = 1 + e − 1. ; , or e = 2 κ 2 κ mα mα

This is a transcendental equation for κ (and hence for E). I’ll solve it graphically: Let z ≡ 2κa, c ≡ so e−z = cz − 1. Plot both sides and look for intersections: 1

2 2amα ,

cz-1

e -z 1/c

z



32

From the graph, noting that c and z are both positive, we see that there is one (and only one) solution 2 z2 (for even ψ). If α = 2ma , so c = 1, the calculator gives z = 1.278, so κ2 = − 2mE = (2a) 2 ⇒ E = 2 (1.278)2 2 2 − 8 ma2 = −0.204 ma2 . Now look for odd solutions:  −κx (x < a),  Ae ψ(x) = B(eκx − e−κx ) (−a < x < a),  −Aeκx (x < −a). Continuity at a : Ae−κa = B(eκa − e−κa ), or A = B(e2κa − 1). 2mα −κa Ae ⇒ B(e2κa + 1) = A 2 2mα 2mα 2mα 2κa 2κa 2κa e + 1 = (e − 1) −1 =e − 1 − 2 + 1, 2 κ 2 κ κ

Discontinuity in ψ : −κAe−κa − B(κeκa + κe−κa ) = −

1=

2mα − 1 , 2 κ

2mα 2 κ 2mα −2κa 2 κ −2κa −2κa ; , e = 1 − − 1 − e = 1 − e , or e−z = 1 − cz. 2 κ 2 κ mα mα 1

z

1/c

1/c

This time there may or may not be a solution. Both graphs have their y-intercepts at 1, but if c is too large (α too small), there may be no intersection (solid line), whereas if c is smaller (dashed line) there will be. (Note that z = 0 ⇒ κ = 0 is not a solution, since ψ is then non-normalizable.) The slope of e−z (at z = 0) is −1; the slope of (1 − cz) is −c. So there is an odd solution ⇔ c < 1, or α > 2 /2ma. Conclusion:

One bound state if α ≤ 2 /2ma; two if α > 2 /2ma. ψ

ψ -a

-a

a Even

2 1 α= ⇒c= . ma 2

a

x

x

Odd

Even: e−z = 12 z − 1 ⇒ z = 2.21772, Odd: e−z = 1 − 12 z ⇒ z = 1.59362.

E = −0.615(2 /ma2 ); E = −0.317(2 /ma2 ). α=

2 ⇒ c = 2. Only even: e−z = 2z − 1 ⇒ z = 0.738835; 4ma

E = −0.0682(2 /ma2 ).



33

Problem 2.28  ikx   Ae + Be−ikx (x < −a)  ψ = Ceikx + De−ikx (−a < x < a) . Impose boundary conditions:  ikx  Fe (x > a) (1) Continuity at −a : Aeika + Beika = Ce−ika + Deika ⇒ βA + B = βC + D, where β ≡ e−2ika . (2) Continuity at +a : Ceika + De−ika = F eika ⇒ F = C + βD. −ika (3) Discontinuity in ψ at −a : ik(Ce−ika − Deika ) − ik(Ae−ika − Beika ) = − 2mα + Beika ) 2 (Ae 2 ⇒ βC − D = β(γ + 1)A + B(γ − 1), where γ ≡ i2mα/ k. ika (4) Discontinuity in ψ at +a : ikF eika − ik(Ceika − De−ika ) = − 2mα ) 2 (F e ⇒ C − βD = (1 − γ)F. To solve for C and D,

add (2) and (4) : 2C = F + (1 − γ)F ⇒ 2C = (2 − γ)F. subtract (2) and (4) : 2βD = F − (1 − γ)F ⇒ 2D = (γ/β)F.

add (1) and (3) : 2βC = βA + B + β(γ + 1)A + B(γ − 1) ⇒ 2C = (γ + 2)A + (γ/β)B. subtract (1) and (3) : 2D = βA + B − β(γ + 1)A − B(γ − 1) ⇒ 2D = −γβA + (2 − γ)B. Equate the two expressions for 2C : (2 − γ)F = (γ + 2)A + (γ/β)B. Equate the two expressions for 2D : (γ/β)F = −γβA + (2 − γ)B.

Solve these for F and B, in terms of A. Multiply the first by β(2 − γ), the second by γ, and subtract: 2 β(2 − γ)2 F = β(4 − γ 2 )A + γ(2 − γ)B ; (γ /β)F = −βγ 2 A + γ(2 − γ)B . 4 F ⇒ β(2 − γ)2 − γ 2 /β F = β 4 − γ 2 + γ A = 4βA ⇒ = . 2 A (2 − γ) − γ 2 /β 2

Let g ≡ i/γ =

2 k i 4g 2 F ; φ ≡ 4ka, so γ = , β 2 = e−iφ . Then: = . 2mα g A (2g − i)2 + eiφ

Denominator: 4g 2 − 4ig − 1 + cos φ + i sin φ = (4g 2 − 1 + cos φ) + i(sin φ − 4g). |Denominator|2 = (4g 2 − 1 + cos φ)2 + (sin φ − 4g)2 = 16g 4 + 1 + cos2 φ − 8g 2 − 2 cos φ + 8g 2 cos φ + sin2 φ − 8g sin φ + 16g 2 = 16g 4 + 8g 2 + 2 + (8g 2 − 2) cos φ − 8g sin φ. 2 F 2 k 8g 4 T = = , where g ≡ and φ ≡ 4ka. 4 2 2 A (8g + 4g + 1) + (4g − 1) cos φ − 4g sin φ 2mα



34

Problem 2.29   −κx (x > a)   Fe In place of Eq. 2.151, we have: ψ(x) = D sin(lx) (0 < x < a) .   −ψ(−x) (x < 0) Continuity of ψ : F e−κa = D sin(la); continuity of ψ : −F κe−κa = Dl cos(la). Divide: − κ = l cot(la), or − κa = la cot(la) ⇒

z02 − z 2 = −z cot z, or − cot z = (z0 /z)2 − 1.

Wide, deep well: Intersections are at π, 2π, 3π, etc. Same as Eq. 2.157, but now for n even. This fills in the rest of the states for the infinite square well. Shallow, narrow well: If z0 < π/2, there is no odd bound state. The corresponding condition on V0 is V0
a) Fe

√ k=

2mE ; κ=

2m(V0 − E) .

(1) Continuity of ψ at −a: Ae−ika + Beika = Ce−κa + Deκa . (2) Continuity of ψ at −a: ik(Ae−ika − Beika ) = κ(Ce−κa − Deκa ). κ −κa κ κa ⇒ 2Ae−ika = 1 − i Ce De . + 1+i k k (3) Continuity of ψ at +a: Ceκa + De−κa = F eika . (4) Continuity of ψ at +a: κ(Ceκa − De−κa ) = ikF eika . c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

¨ CHAPTER 2. THE TIME-INDEPENDENT SCHRODINGER EQUATION ⇒ 2Ce

κa

−ika

2Ae

T

T

=

ik 1+ κ

ika

Fe

;

−κa

2De

=

ik 1− κ

37

F eika .

−2κa e2κa iκ ik iκ ik ika e = 1− 1+ Fe + 1+ 1− F eika k κ 2 k κ 2

F eika κ k κ k = 1+i − + 1 e−2κa + 1 + i − + 1 e2κa 2 κ k k κ

2 2

−2κa F eika − k ) (κ = 2 e e2κa − e−2κa . + e2κa + i 2 kκ x −x e −e ex + e−x But sinh x ≡ , cosh x ≡ , so 2 2

F eika (κ2 − k 2 ) = 4 cosh(2κa) + i 2 sinh(2κa) 2 kκ

(κ2 − k 2 ) ika = 2F e cosh(2κa) + i sinh(2κa) . 2kκ

−1

2 A (κ2 − k 2 )2 = = cosh2 (2κa) + sinh2 (2κa). But cosh2 = 1 + sinh2 , so F (2κk)2

−1

2a (κ2 − k 2 )2 V02 2 2 sinh (2κa) = 1 + =1+ 1+ 2m(V0 − E) , sinh (2κk)2 4E(V0 − E)

where =

4κ k + k + κ − 2κ k (κ + k ) = = (2κk)2 (2κk)2 2 2

4

4

2 2

2

2 2

2mE 2

+

2m(V0 −E) 2

2

2m(V0 −E) 4 2mE 2 2

=

V02 . 4E(V0 − E)

(You can also get this from Eq. 2.169 by switching the sign of V0 and using sin(iθ) = i sinh θ.)  ikx   Ae + Be−ikx (x < −a)  (−a < x < a) E = V0 . ψ = C + Dx  ikx  Fe (x > a)

(In central region −

2 d 2 ψ d2 ψ + V0 ψ = Eψ ⇒ = 0, so ψ = C + Dx.) 2 2m dx dx2

(1) Continuous ψ at −a : Ae−ika + Beika = C − Da. (2) Continuous ψ at +a : F eika = C + Da. ⇒ (2.5) 2Da = F eika − Ae−ika − Beika .

(3) Continuous ψ at −a : ik Ae−ika − Beika = D. (4) Continuous ψ at +a : ikF eika = D. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


38 ⇒ (4.5) Ae−2ika − B = F.

Use (4) to eliminate D in (2.5): Ae−2ika + B = F − 2aikF = (1 − 2iak)F , and add to (4.5): 2 A 2mE −2ika −1 2Ae = 2F (1 − ika), so T = = 1 + (ka)2 = 1 + 2 a2 . F (You can also get this from Eq. 2.169 by changing the sign of V0 and taking the limit E → V0 , using sin 9 ∼ = 9.) This case is identical to the one in the book, only with V0 → −V0 . So

E > V0 . T −1 = 1 +

V02 sin2 4E(E − V0 )

2a 2m(E − V0 ) .

Problem 2.34 (a) ψ=

Aeikx + Be−ikx (x < 0) F e−κx (x > 0)

√

where k =

2mE ; κ=

2m(V0 − E) .

(1) Continuity of ψ : A + B = F. (2) Continuity of ψ : ik(A − B) = −κF. ik ik ik ⇒ A + B = − (A − B) ⇒ A 1 + = −B 1 − . κ κ κ 2 B |(1 + ik/κ)|2 1 + (k/κ)2 R = = = = 1. A |(1 − ik/κ)|2 1 + (k/κ)2 Although the wave function penetrates into the barrier, it is eventually all reflected. (b) ψ=

Aeikx + Be−ikx (x < 0) F eilx (x > 0)

√

where k =

2mE ; l=

2m(E − V0 ) .

(1) Continuity of ψ : A + B = F. (2) Continuity of ψ : ik(A − B) = ilF. ⇒A+B =

k k k (A − B); A 1 − = −B 1 + . l l l

2 B (1 − k/l)2 (k − l)2 (k − l)4 R = = = = 2 . 2 2 A (1 + k/l) (k + l) (k − l2 )2 2m Now k − l = 2 (E − E + V0 ) = √ √ ( E − E − V0 )4 R= . V02 2

2

2m 2

√

V0 ; k − l =

2m √ [ E − E − V0 ], so



39

(c)

vi

vi dt

vt dt

vt

From the diagram, T = Pt /Pi = |F |2 vt /|A|2 vi , where Pi is the probability of finding the incident particle in the box corresponding to the time interval dt, and Pt is the probability of finding the transmitted particle in the associated box to the right of the barrier. 2 √ vt E − V0 E − V0 F √ But = . Alternatively, from Problem 2.19: (from Eq. 2.98). So T = vi E A E k 2 Ji = |A| ; m

l Jt = |F |2 ; m

2 2 F l F Jt E − V0 T = = = . Ji A k A E

For E < V0 , of course, T = 0. (d) k

For E > V0 , F = A + B = A +

A kl l

−1

2k/l 2k

= A k

= A. k +l +1 + 1 l

√ √ √ 2 √ 2 F l l 4 E E − V0 ( E − E − V0 )2 4kl(k − l)2 2k 4kl T = = = . = = A k k+l k (k + l)2 (k 2 − l2 )2 V02

T +R=

4kl (k − l)2 4kl + k 2 − 2kl + l2 k 2 + 2kl + l2 (k + l)2 + = = = = 1. 2 2 2 2 (k + l) (k + l) (k + l) (k + l) (k + l)2

Problem 2.35 (a) ψ(x) =

Aeikx + Be−ikx (x < 0) F eilx (x > 0)

√

where k ≡

Continuity of ψ ⇒ A + B = F Continuity of ψ ⇒ ik(A − B) = ilF k A + B = (A − B); l

k A 1− l

2mE , l≡

2m(E + V0 ) .

=⇒

k = −B 1 + l

;

B =− A

1 − k/l 1 + k/l

.



40

√ (2 2 2 ' √ B E + V0 − E l−k √ R= = = √ A l+k E+V + E '

=

1 + V0 /E − 1 1 + V0 /E + 1

(2

0

2 2 √ 1+3−1 1 2−1 = √ = = . 2 + 1 9 1+3+1

(b) The cliff is two-dimensional, and even if we pretend the car drops straight down, the potential as a function of distance along the (crooked, but now one-dimensional) path is −mgx (with x the vertical coordinate), as shown. V(x) x -V0

(c) Here V0 /E = 12/4 = 3, the same as in part (a), so R = 1/9, and hence T = 8/9 = 0.8889.

Problem 2.36 Start with Eq. 2.22: ψ(x) = A sin kx + B cos kx. This time the boundary conditions are ψ(a) = ψ(−a) = 0: A sin ka + B cos ka = 0; −A sin ka + B cos ka = 0. Subtract : A sin ka = 0 ⇒ ka = jπ or A = 0, Add : B cos ka = 0 ⇒ ka = (j − 12 )π or B = 0, (where j = 1, 2, 3, . . . ). If B = 0 (so A = 0), k = jπ/a. In this case let n ≡ 2j (so n is an even √ integer); then k = nπ/2a, a ψ = A sin(nπx/2a). Normalizing: 1 = |A|2 −a sin2 (nπx/2a) dx = |A|2 /2 ⇒ A = 2. If A = 0 (so B = 0), k = (j − 12 )π/a. In this case let n ≡ 2j − 1 (n is an odd integer); again k = nπ/2a, √ a ψ = B cos(nπx/2a). Normalizing: 1 = |B|2 −a cos2 (nπx/2a)dx = |a|2 /2 ⇒ B = 2. k n π In either case Eq. 2.21 yields E = 2m = 2m(2a) 2 (in agreement with Eq. 2.27 for a well of width 2a). The substitution x → (x + a)/2 takes Eq. 2.28 to  nπx

2 n/2  (n even), (−1)  a sin 2a  2 2 nπ (x + a) nπx nπ sin = sin + =  a a 2 a 2a 2  (−1)(n−1)/2 2 cos nπx (n odd). a 2a 2 2

2

2 2

So (apart from normalization) we recover the results above. The graphs are the same as Figure 2.2, except that some are upside down (different normalization).



cos(πx/2a)

sin(2πx/2a)

41

cos(3πx/2a)

Problem 2.37 Use the trig identity sin 3θ = 3 sin θ − 4 sin3 θ to write 3

sin

πx a

3 = sin 4

πx a

1 − sin 4

9 1 Normalize using Eq. 2.38: |A| + 2 16 16 2a

So Ψ(x, 0) =

√1 10

a 3 3πx 1 . So (Eq. 2.28): Ψ(x, 0) = A ψ1 (x) − ψ3 (x) . a 2 4 4

=

5 4 a|A|2 = 1 ⇒ A = √ . 16 5a

[3ψ1 (x) − ψ3 (x)] , and hence (Eq. 2.17) 1 Ψ(x, t) = √ 3ψ1 (x)e−iE1 t/ − ψ3 (x)e−iE3 t/ . 10

|Ψ(x, t)|2 = x =

a

x|Ψ(x, t)|2 dx = 0

1 E3 − E1 9ψ12 + ψ32 − 6ψ1 ψ3 cos t ; so 10

9 1 3 x1 + x3 − cos 10 10 5

E3 − E1 t

a

xψ1 (x)ψ3 (x)dx, 0

where xn = a/2 is the expectation value of x in the nth stationary state. The remaining integral is πx 2πx 3πx 1 a 4πx x cos sin dx = − cos dx a a a 0 a a 0 2 2 a 1 a xa 2πx a xa 4πx 2πx 4πx = + sin − − sin cos cos = 0. a 2π a 2π a 4π a 4π a

2 a

a

x sin

0

Evidently then, 9 a a 1 a x = + = . 10 2 10 2 2



42

Problem 2.38

π nπ 2 2 n2 π 2 2 (a) New allowed energies: En = ; Ψ(x, 0) = sin x , ψn (x) = sin x . 2 2m(2a) a a 2a 2a √ a+ √ a n nπ πx n πx , π 2 2 cn = cos x sin x dx = −1 − cos +1 dx. sin a 0 a 2a 2a 0 2 a 2 a n

πx

πx a n sin 2 − 1 a sin 2 + 1 a 1

π

π n n =√ − (for n = 2) − 1 + 1 2a 2 a 2 a 0

sin n2 + 1 π sin n2 + 1 π sin n2 − 1 π 1 1 1

− n

n n n √ =√ − = 2π 2π 2 −1 2 +1 2 −1 2 +1

√ 0, √ if n is even 4 2 sin n2 + 1 π = = . 4 2 ± π(n π (n2 − 4) 2 −4) , if n is odd √ a √ a 2 2 1 1 2 π c2 = sin x dx = dx = √ . So the probability of getting En is a 0 a a 0 2 2  if n = 2  Pn = |cn |2 = π2 (n32 . 2 −4)2 , if n is odd   0, otherwise 1  2,

Most probable: E2 =

π 2 2 (same as before). 2ma2

Probability: P2 = 1/2.

π 2 2 32 , with probability P1 = = 0.36025. 8ma2 9π 2 π

2 d2 a (c) H = Ψ∗ HΨ dx = a2 0 sin πa x − 2m dx2 sin a x dx, but this is exactly the same as before the wall

(b) Next most probable: E1 =

moved – for which we know the answer:

π 2 2 . 2ma2

Problem 2.39 (a) According to Eq. 2.36, the most general solution to the time-dependent Schr¨ odinger equation for the infinite square well is ∞ 2 2 2 Ψ(x, t) = cn ψn (x)e−i(n π /2ma )t . n=1 2 2 2 2 2 2 2 n π n π 4ma T = = 2πn2 , so e−i(n π /2ma )(t+T ) = e−i(n π /2ma )t e−i2πn , and since n2 is 2 π 2ma2 2ma 2 an integer, e−i2πn = 1. Therefore Ψ(x, t + T ) = Ψ(x, t). QED

2 2

2 2

2

Now

(b) The classical revival time is the time it takes the particle to go down and back: Tc = 2a/v, with the velocity given by 1 2E 2m 2 E = mv ⇒ v = ⇒ Tc = a . 2 m E c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


43

(c) The two revival times are equal if 4ma2 2m =a , π E

or

E=

π 2 2 E1 = . 8ma2 4

Problem 2.40 (a) Let V0 ≡ 322 /ma2 . This is just like the odd bound states for the finite square well, since they are the ones that go to zero at the origin. Referring to the solution to Problem 2.29, the wave function is D sin lx, l ≡ 2m(E + V0 )/ (0 < x < a), √ ψ(x) = F e−κx , κ ≡ −2mE/ (x > a), and the boundary conditions at x = a yield − cot z =

(z0 /z)2 − 1

2m(322 /ma2 ) 2mV0 z0 = a= a = 8. Referring to the figure (Problem 2.29), and noting that (5/2)π = 7.85 < z0 < 3π = 9.42, we see that there are three bound states. with

√

(b) Let

a

x 1 1 2 a I1 ≡ |ψ| dx = |D| sin lx dx = |D| − sin lx cos lx = |D| − sin lz cos la ; 2 2l 2 2l 0 0 0 −2κx ∞ ∞ ∞ −2κa e = |F |2 e I2 ≡ |ψ|2 dx = |F |2 e−2κx dx = |F |2 − . 2κ 2κ a a a

a

2

2

a

2

2

2

But continuity at x = a ⇒ F e−κa = D sin la, so I2 = |D|2 sin2κla . Normalizing:

a κ 1 sin2 la 1 1 = I1 + I2 = |D| − sin la cos la + = |D|2 κa − sin la cos la + sin2 la 2 2l 2κ 2κ l But (referring again to Problem 2.29) κ/l = − cot la, so 1 (1 + κa) = |D|2 κa + cot la sin la cos la + sin2 la = |D|2 . 2κ 2κ 2

So |D|2 = 2κ/(1 + κa), and the probability of finding the particle outside the well is 2κ sin2 la sin2 la = . 1 + κa 2κ 1 + κa We can express this interms of z ≡ la and z0 : κa = z02 − z 2 (page 80), 2 1 z2 1 z sin2 la = sin2 z = ⇒P = 2 . = = 2 2 1 + (z0 /z) − 1 z0 1 + cot z z0 (1 + z02 − z 2 ) P = I2 =

So far, this iscorrect for any bound state. In the present case z0 = 8 and z is the third solution to − cot z = (8/z)2 − 1, which occurs somewhere in the interval 7.85 < z < 8. Mathematica gives z = 7.9573 and P = 0.54204. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


44

Problem 2.41 (a) In the standard notation ξ ≡

mω/ x, α ≡ (mω/π)1/4 ,

Ψ(x, 0) = A(1 − 2ξ)2 e−ξ

2

/2

= A(1 − 4ξ + 4ξ 2 )e−ξ

2

/2

.

It can be expressed as a linear combination of the first three stationary states (Eq. 2.59 and 2.62, and Problem 2.10): √ 2 2 2 α ψ0 (x) = αe−ξ /2 , ψ1 (x) = 2 αξe−ξ /2 , ψ2 (x) = √ (2ξ 2 − 1)e−ξ /2 . 2 √ √ 2 2 1 −ξ /2 So Ψ(x, 0) = c0 ψ0 + c1 ψ1 + c2 ψ2 = α(c0 + 2ξc1 + 2ξ c2 − √2 c2 )e with (equating like powers)  √ √  ⇒ c2 = 2 2A/α, α√2c2 = 4A √ α 2c1 = −4A ⇒ c1 = −2 2A/α,  √ √  α(c0 − c2 / 2) = A ⇒ c0 = (A/α) + c2 / 2 = (1 + 2)A/α = 3A/α. Normalizing: 1 = |c0 |2 + |c1 |2 + |c2 |2 = (8 + 8 + 9)(A/α)2 = 25(A/α)2 ⇒ A = α/5. √ √ 3 2 2 2 2 c0 = , c1 = − , c2 = . 5 5 5 73 1 9 1 8 3 8 5 ω 2 H = |cn | (n + )ω = ω + ω + ω = (9 + 24 + 40) = ω. 2 25 2 25 2 25 2 50 50 (b)

√ √ √ √ 3 2 2 2 2 −iωt/2 2 2 −3iωt/2 2 2 −5iωt/2 −iωt/2 3 −iωt −2iωt Ψ(x, t) = ψ0 e . ψ1 e ψ2 e ψ0 − ψ1 e ψ2 e − + =e + 5 5 5 5 5 5 To change the sign of the middle term we need e−iωT = −1 (then e−2iωT = 1); evidently ωT = π, or T = π/ω.



45

Problem 2.42 Everything in Section 2.3.2 still applies, except that there is an additional boundary condition: ψ(0) = 0. This eliminates all the even solutions (n = 0, 2, 4, . . . ), leaving only the odd solutions. So En =

1 n+ 2

Problem 2.43 (a) Normalization is the same as before: A =

ω, n = 1, 3, 5, . . . .

2a 1/4 π

.

(b) Equation 2.103 says 1 φ(k) = √ 2π

2a π

1/4

∞

e−ax eilx e−ikx dx 2

−∞

1 1 Ψ(x, t) = √ 2π (2πa)1/4

∞

[same as before, only k → k − l] =

2 1 e−(k−l) /4a . (2πa)1/4

i(kx−k t/2m) e−(k−l) /4a e dk −∞ 1 l t − k2 −(ix+ k +i 2a ) ] e−l2 /4a e [( 4a 2m ) 2 2 1 π 1

e(ix+l/2a) /[4(1/4a+it/2m)] =√ e−l /4a 1 t + i 2π (2πa)1/4 4a 2m

=

2a π

1/4

2

2 2 1 e−l /4a ea(ix+l/2a) /(1+2iat/m) . 1 + 2iat/m

(c) Let θ ≡ 2at/m, as before: |Ψ|2 = square brackets:

2

a 2 2a 1 √ e−l /2a e π 1 + θ2

(ix+l/2a)2 (1+iθ)

+

(−ix+l/2a)2 (1−iθ)

. Expand the term in

2 2 1 l l []= (1 − iθ) ix + + (1 + iθ) −ix + 1 + θ2 2a 2a 1 ixl ixl l2 l2 2 2 = −x + −x + − + + 1 + θ2 a 4a2 a 4a2 l2 l2 ixl ixl +iθ x2 − − 2 + iθ −x2 − + 2 a 4a a 4a

2 1 l xl xl θ2 l2 θ 2 l2 l2 1 2 2 = −2x −2x + + 2θ + 2θ + + = − 1 + θ2 2a2 a 1 + θ2 a 2a2 2a2 2a2 2 −2 θl l2 = x − + . 1 + θ2 2a 2a2 2 2a 2 a 2 −2w2 (x−θl/2a)2 2 −l2 /2a − 1+θ2 (x−θl/2a) l2 /2a e e e = , |Ψ(x, t)| = we 2 π 1+θ π c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


46

θl l = x− m where w ≡ a/(1 + θ2 ). The result is the same as before, except x → x − 2a t , so |Ψ|2 has the same (flattening Gaussian) shape – only this time the center moves at constant speed v = l/m. (d) x =

∞

x|Ψ(x, t)|2 dx. Let y ≡ x − θl/2a = x − vt, so x = y + vt. 2 −2w2 y2 = (y + vt) dy = vt. we π −∞ −∞ ∞

(The first integral is trivially zero; the second is 1 by normalization.) =

l t; m

x = 2

p = m

∞

2

(y + vt) −∞

1 x = + 4w2

2

d2 Ψ = dx2 p2 = = =

= = =

σx2

lt m

2 −2w2 y2 1 dy = + 0 + (vt)2 (the first integral is same as before). we π 4w2

2

.

p = − 2

2

∞

Ψ∗

−∞

d2 Ψ dx. dx2

l 2ia ix + 2a 1 dΨ −l2 /4a a(ix+l/2a)2 /(1+iθ) √ = Ψ; e e , so dx (1 + iθ) 1 + iθ

2 2i2 a −4a2 (ix + l/2a) 2ia(ix + l/2a) dΨ 2a + Ψ= Ψ. − 1 + iθ dx 1 + iθ (1 + iθ)2 1 + iθ 2 ∞ l 4a2 2 (1 + iθ) ix + + |Ψ|2 dx (1 + iθ)2 −∞ 2a 2a 2 ∞ 4a2 2 il (1 + iθ) − y + vt − + |Ψ|2 dy (1 + iθ)2 −∞ 2a 2a ∞ ∞ 4a2 2 il 2 2 − y |Ψ| dy − 2 vt − y|Ψ|2 dy (1 + iθ)2 2a −∞ −∞ 2 ∞ il (1 + iθ) 2 + − vt − + |Ψ| dy 2a 2a −∞ 2 4a2 2 1 il (1 + iθ) − 2 + 0 − vt − + (1 + iθ)2 4w 2a 2a

2 4a2 2 1 + θ2 −il (1 + iθ) − − (1 + iθ) + (1 + iθ)2 4a 2a 2a

a l2 a2 l2 −(1 − iθ) + (1 + iθ) + 2 = (1 + iθ) 1 + = 2 (a + l2 ). 1 + iθ a 1 + iθ a

Ψ=

dx = l. dt

2a π

1/4

1 = x − x = + 4w2 2

2

lt m

2 −

lt m

2 =

1 1 ⇒ σx = ; 4w2 2w



47

√ σp2 = p2 − p2 = 2 a + 2 l2 − 2 l2 = 2 a, so σp = a. (e) σx and σp are same as before, so the uncertainty principle still holds.

Problem 2.44

√ Equation 2.22 ⇒ ψ(x) = A sin kx + B cos kx, 0 ≤ x ≤ a, with k = 2mE/2 . Even solutions: ψ(x) = ψ(−x) = A sin(−kx) + B cos(−kx) = −A sin kx + B cos kx (−a ≤ x ≤ 0).   ψ continuous at 0 : B = B (no new condition). Boundary ψ discontinuous (Eq. 2.125 with sign of α switched): Ak + Ak = 2mα 2 B ⇒ B = conditions  2 k 2 k ψ → 0 at x = a : A sin(ka) + mα A cos(ka) = 0 ⇒ tan(ka) = − mα .

2 k mα A.

2 k ψ(x) = A sin kx + cos kx (0 ≤ x ≤ a); ψ(−x) = ψ(x). mα π

2π

3π

ka

-h2k mα

tan(ka)

From the graph, the allowed energies are slightly above ka =

nπ (n = 1, 3, 5, . . . ) so 2

En

n2 π 2 2 (n = 1, 3, 5, . . . ). 2m(2a)2

These energies are somewhat higher than the corresponding energies for the infinite square well (Eq. 2.27, with a → 2a). As α → 0, the straight line (−2 k/mα) gets steeper and steeper, and the intersections get closer to nπ/2; the energies then reduce to those of the ordinary infinite well. As α → ∞, the straight line approaches 2 2 2 π horizontal, and the intersections are at nπ (n = 1, 2, 3, . . . ), so En → n2ma – these are the allowed energies for 2 the infinite square well of width a. At this point the barrier is impenetrable, and we have two isolated infinite square wells. Odd solutions: ψ(x) = −ψ(−x) = −A sin(−kx) − B cos(−kx) = A sin(kx) − B cos(kx) (−a ≤ x ≤ 0).   ψ continuous at 0 : B = −B ⇒ B = 0. ψ discontinuous: Ak − Ak = 2mα Boundary conditions 2 (0) (no new condition).  ψ(a) = 0 ⇒ A sin(ka) = 0 ⇒ ka = nπ 2 (n = 2, 4, 6, . . . ). ψ(x) = A sin(kx), (−a < x < a);

En =

n2 π 2 2 (n = 2, 4, 6, . . . ). 2m(2a)2

These are the exact (even n) energies (and wave functions) for the infinite square well (of width 2a). The point is that the odd solutions (even n) are zero at the origin, so they never “feel” the delta function at all.



48

Problem 2.45

 d2 ψ1 2 d 2 ψ 1 2  ψ2 − + V ψ1 = Eψ1 ⇒ − + V ψ1 ψ2 = Eψ1 ψ2  

 2m dx2 2m dx2 2 d2 ψ1 d2 ψ2 ⇒− = 0. − ψ1 ψ2  2m dx2 dx2  2 d 2 ψ 2 2 d2 ψ2  − + V ψ2 = Eψ2 ⇒ − + V ψ1 ψ2 = Eψ1 ψ2  ψ1 2m dx2 2m dx2

d dψ1 dψ2 d2 ψ1 d2 ψ2 d2 ψ1 d2 ψ2 dψ1 dψ2 dψ2 dψ1 But − = ψ2 − ψ1 . Since this is ψ2 − ψ1 = + ψ2 − ψ1 2 2 2 dx dx dx dx dx dx dx dx dx dx dx2 dψ1 dψ2 zero, it follows that ψ2 − ψ1 = K (a constant). But ψ → 0 at ∞ so the constant must be zero. Thus dx dx dψ1 dψ2 1 dψ1 1 dψ2 ψ2 = ψ1 , or = , so ln ψ1 = ln ψ2 + constant, or ψ1 = (constant)ψ2 . QED dx dx ψ1 dx ψ2 dx

Problem 2.46 −

2 d 2 ψ d2 ψ = Eψ (where x is measured around the circumference), or = −k 2 ψ, with k ≡ 2 2m dx dx2

√

2mE , so

ψ(x) = Aeikx + Be−ikx . But ψ(x + L) = ψ(x), since x + L is the same point as x, so Aeikx eikL + Be−ikx e−ikL = Aeikx + Be−ikx , and this is true for all x. In particular, for x = 0 : (1) AeikL + Be−ikL = A + B. And for x =

π : 2k

Aeiπ/2 eikL + Be−iπ/2 e−ikL = Aeiπ/2 + Be−iπ/2 , or iAeikL − iBe−ikL = iA − iB, so (2) AeikL − Be−ikL = A − B. Add (1) and (2): 2AeikL = 2A. Either A = 0, or else eikL = 1, in which case kL = 2nπ (n = 0, ±1, ±2, . . . ). But if A = 0, then Be−ikL = B, leading to the same conclusion. So for every positive n there are two solutions: ψn+ (x) = Aei(2nπx/L) and L ψn− (x) = Be−i(2nπx/L) (n = 0 is ok too, but in that case there is just one solution). Normalizing: 0 |ψ± |2 dx = √ 1 ⇒ A = B = 1/ L. Any other solution (with the same energy) is a linear combination of these. 1 2n2 π 2 2 ψn± (x) = √ e±i(2nπx/L) ; En = (n = 0, 1, 2, 3, . . . ). mL2 L The theorem fails because here ψ does not go to zero at ∞; x is restricted to a finite range, and we are unable to determine the constant K (in Problem 2.45).

Problem 2.47 (a) (i) b = 0 ⇒ ordinary finite square well. Exponential decay outside; sinusoidal inside (cos for ψ1 , sin for ψ2 ). No nodes for ψ1 , one node for ψ2 . (ii) Ground state is even. Exponential decay outside, sinusoidal inside the wells, hyperbolic cosine in barrier. First excited state is odd – hyperbolic sine in barrier. No nodes for ψ1 , one node for ψ2 . c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


ψ1

49

ψ2 -a

-a

a

a

x

ψ1

x

ψ2

-(b/2+a) -b/2 -(b/2+a) -b/2

b/2 b/2+a

x

b/2 b/2+a

x

(iii) For b a, same as (ii), but wave function very small in barrier region. Essentially two isolated finite square wells; ψ1 and ψ2 are degenerate (in energy); they are even and odd linear combinations of the ground states of the two separate wells. ψ1

ψ

2

b/2

-b/2 -(b/2+a)

b/2+a x

-(b/2+a) -b/2

b/2 b/2+a x

(b) From Eq. 2.157 we know that for b = 0 the energies fall slightly below π 2 2 h E1 + V0 ≈ 2m(2a) π 2 2 2 = 4 . where h ≡ 2 2 4π 2ma2 E2 + V0 ≈ 2m(2a) 2 = h For b a, the width of each (isolated) well is a, so E1 + V 0 ≈ E 2 + V 0 ≈

π 2 2 = h (again, slightly below this). 2ma2 2

Hence the graph (next page). [Incidentally, within each well, ddxψ2 = − 2m 2 (V0 + E)ψ, so the more curved the wave function, the higher the energy. This is consistent with the graphs above.] (c) In the (even) ground state the energy is lowest in configuration (i), with b → 0, so the electron tends to draw the nuclei together, promoting bonding of the atoms. In the (odd) first excited state, by contrast, the electron drives the nuclei apart. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


50

E+V0 h

E 2 +V0

E 1+V0 h/4 b

Problem 2.48 (a) √ √ dΨ 2 3 a 2 3 1, (0 < x < a/2) = √ 1 − 2θ x − = √ · . −1, (a/2 < x < a) dx 2 a a a a (b) √ √ d2 Ψ 2 3 4 3 a a √ √ = − − 2δ x − δ x − = . dx2 2 2 a a a a (c) H = −

2 2m

√ √ 62 a 2 · 3 · 2 4 3 2 32 ∗ a √ Ψ = . Ψ∗ δ x − − √ dx = = 2 2 m·a·a ma2 a a ma a √ 3/a

Problem 2.49 (a)

i ∂Ψ mω a2 = − −2iωe−2iωt + − 2ax(−iω)e−iωt Ψ, so ∂t 2 2 m

∂Ψ 1 1 2 2 −2iωt 2 −iωt i Ψ. + ω + maxω e = − ma ω e ∂t 2 2

∂Ψ mω mω = − 2x − 2ae−iωt Ψ = − x − ae−iωt Ψ; ∂x 2

∂2Ψ mω mω mω 2 mω −iωt ∂Ψ −iωt 2 Ψ. x − ae Ψ− x − ae = − + =− ∂x2 ∂x c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


51

1 2 ∂ 2 Ψ 1 2 mω mω 2 2 2 −iωt 2 Ψ + mω 2 x2 Ψ − + mω x Ψ = − x − ae − + 2m ∂x2 2 2m 2

2

1 1 1 2 −iωt 2 −2iωt 2 2 = +a e + mω x Ψ ω − mω x − 2axe 2 2 2

1 1 = ω + maxω 2 e−iωt − mω 2 a2 e−2iωt Ψ 2 2 ∂Ψ = i (comparing second line above). ∂t (b)

2 2 iωt −iωt mω − mω x2 + a2 (1+e2iωt )− it + x2 + a2 (1+e−2iωt )+ it m −2axe m −2axe e 2 π 2 2 2 mω − mω = e 2 [2x +a +a cos(2ωt)−4ax cos(ωt)] . But a2 [1 + cos(2ωt)] = 2a2 cos2 ωt, so π mω − mω mω − mω (x−a cos ωt)2 x2 −2ax cos(ωt)+a2 cos2 (ωt)] [ = . = e e π π

|Ψ| = 2

The wave packet is a Gaussian of fixed shape, whose center oscillates back and forth sinusoidally, with amplitude a and angular frequency ω. (c) Note that this wave function is correctly normalized (compare Eq. 2.59). Let y ≡ x − a cos ωt : x = x|Ψ|2 dx = (y + a cos ωt)|Ψ|2 dy = 0 + a cos ωt |Ψ|2 dy = a cos ωt.

p = m

−

dx = −maω sin ωt. dt

dp = −maω 2 cos ωt. dt

V =

1 dV mω 2 x2 =⇒ = mω 2 x. 2 dx

dV dp = −mω 2 x = −mω 2 a cos ωt = , so Ehrenfest’s theorem is satisfied. dx dt

Problem 2.50 (a)

(E + 12 mv 2 ) ∂Ψ mα ∂ = − 2 |x − vt| − i Ψ; ∂t ∂t

∂ |x − vt| = ∂t

−v, if x − vt > 0 v, if x − vt < 0

.

We can write this in terms of the θ-function (Eq. 2.143): ∂ 1, if z > 0 2θ(z) − 1 = , so |x − vt| = −v[2θ(x − vt) − 1]. −1, if z < 0 ∂t ∂Ψ i = ∂t

mαv 1 2 i [2θ(x − vt) − 1] + E + mv Ψ. 2

[]



52

∂Ψ mα ∂ imv = − 2 |x − vt| + Ψ ∂x ∂x ∂ |x − vt| = {1, if x > vt; −1, if x < vt} = 2θ(x − vt) − 1. ∂x mα imv = − 2 [2θ(x − vt) − 1] + Ψ. ∂2Ψ = ∂x2

−

mα imv [2θ(x − vt) − 1] + 2

But (from Problem 2.24(b)) −

∂ ∂x θ(x

2 Ψ−

2mα ∂ θ(x − vt) Ψ. 2 ∂x

− vt) = δ(x − vt), so

2 ∂ 2 Ψ − αδ(x − vt)Ψ 2m'∂x2 ( 2 2 imv mα = − + αδ(x − vt) − αδ(x − vt) Ψ − 2 [2θ(x − vt) − 1] + 2m 2 m2 α 2 m2 v 2 mv mα 2 =− [2θ(x − vt) − 1] − − 2i [2θ(x − vt) − 1] Ψ 2m 4 2 2 1 mα2 1 mvα ∂Ψ = − 2 + mv 2 + i [2θ(x − vt) − 1] Ψ = i (compare []). 2 2 ∂t

(b) |Ψ|2 =

mα −2mα|y|/2 e (y ≡ x − vt). 2

Check normalization: 2

∞

mα 2

∞

e−2mαy/ dy =

0

2

2mα 2 = 1. 2 2mα

∂Ψ , which we calculated above []. ∂t −∞ imαv 1 1 2 = [2θ(y) − 1] + E + mv |Ψ|2 dy = E + mv 2 . 2 2

H =

Ψ∗ HΨdx. But HΨ = i

(Note that [2θ(y) − 1] is an odd function of y.) Interpretation: The wave packet is dragged along (at speed v) with the delta-function. The total energy is the energy it would have in a stationary delta-function (E), plus kinetic energy due to the motion ( 12 mv 2 ).

Problem 2.51 (a) Figure at top of next page. (b)

dψ0 = −Aa sech(ax) tanh(ax); dx

d 2 ψ0 = −Aa2 − sech(ax) tanh2 (ax) + sech(ax) sech2 (ax) . 2 dx



53

V(x)

x

2 d 2 ψ0 2 a2 − sech2 (ax)ψ0 2m dx2 m 2 a2 2 Aa2 − sech(ax) tanh2 (ax) + sech3 (ax) − A sech3 (ax) = 2m m 2 a2 A = − sech(ax) tanh2 (ax) + sech3 (ax) − 2 sech3 (ax) 2m 2 a2 =− A sech(ax) tanh2 (ax) + sech2 (ax) . 2m 1 sinh2 θ + 1 sinh2 θ But (tanh2 θ + sech2 θ) = + = = 1, so 2 2 cosh θ cosh θ cosh2 θ

Hψ0 = −

=−

2 a2 ψ0 . QED 2m

1 = |A|

2 a2 . 2m

∞ 1 a 2 2 tanh(ax) . sech (ax)dx = |A| = |A| =⇒ A = a a 2 −∞ −∞

2

Evidently E = −

∞

2

2

ψ(x)

x

(c) dψk A = (ik − a tanh ax)ik − a2 sech2 ax eikx . dx ik + a

. d2 ψk A - = ik (ik − a tanh ax)ik − a2 sech2 ax − a2 ik sech2 ax + 2a3 sech2 ax tanh ax eikx . 2 dx ik + a c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


54

2 2 a2 2 d 2 ψ k A − ik 2 2 2 − + V ψ = − iak tanh ax − a sech ax + −k ik sech2 ax k 2m dx2 ik + a 2m 2m 2 a3 2 a2 2 2 − sech ax tanh ax − sech ax(ik − a tanh ax) eikx m m Aeikx 2 3 = ik − ak 2 tanh ax + ia2 k sech2 ax + ia2 k sech2 ax ik + a 2m

−2a3 sech2 ax tanh ax − 2ia2 k sech2 ax + 2a3 sech2 ax tanh ax =

Aeikx 2 2 2 k 2 k (ik − a tanh ax) = ψk = Eψk . QED ik + a 2m 2m

As x → +∞, tanh ax → +1, so ψk (x) → A

R = 0.

ik − a ik + a

eikx , which represents a transmitted wave.

ik − a 2 ik − a = −ik − a T = = 1. ik + a −ik + a ik + a

Problem 2.52 (a) (1) From Eq. 2.133: F + G = A + B. (2) From Eq. 2.135: F − G = (1 + 2iβ)A − (1 − 2iβ)B, where β = mα/2 k. 1 Subtract: 2G = −2iβA + 2(1 − iβ)B ⇒ B = (iβA + G). Multiply (1) by (1 − 2iβ) and add: 1 − iβ 1 1 iβ 1 2(1 − iβ)F − 2iβG = 2A ⇒ F = (A + iβG). S = . 1 − iβ 1 − iβ 1 iβ (b) For an even potential, V (−x) = V (x), scattering from the right is the same as scattering from the left, with x ↔ −x, A ↔ G, B ↔ F (see Fig. 2.22): F = S11 G + S12 A, B = S21 G + S22 A. So S11 = S22 , S21 = S12 . (Note that the delta-well S matrix in (a) has this property.) In the case of the finite square well, Eqs. 2.167 and 2.168 give S21 =

cos 2la − i

(k2 +l2 ) 2kl

−k i (l 2kl 2

e−2ika sin 2la

;

S11 =

2

)

sin 2la e−2ika 2

2

+l ) cos 2la − i (k 2kl sin 2la

. So

' 2 2 ( −k ) sin 2la 1 i (l 2kl S= . 2 2 +l2 ) −k2 ) 1 i (l 2kl sin 2la cos 2la − i (k 2kl sin 2la e−2ika

Problem 2.53 (a) B = S11 A + S12 G ⇒ G =

1 S11 1 (B − S11 A) = M21 A + M22 B ⇒ M21 = − , M22 = . S12 S12 S12



F = S21 A + S22 B = S21 A +

⇒ M11 = −

S22 (S11 S22 − S12 S21 ) S22 (B − S11 A) = − A+ B = M11 A + M12 B. S12 S12 S12

det S S22 , M12 = . S12 S12

G = M21 A + M22 B ⇒ B =

M=

1 S12

− det(S) S22 . −S11 1

Conversely:

1 M21 1 (G − M21 A) = S11 A + S12 G ⇒ S11 = − ; S12 = . M22 M22 M22

F = M11 A + M12 B = M11 A +

⇒ S21

55

M12 (M11 M22 − M12 M21 ) M12 (G − M21 A) = A+ G = S21 A + S22 G. M22 M22 M22

det M M12 = ; S22 = . M22 M22

1 S= M22

−M21 1 . det(M) M12

[It happens that the time-reversal invariance of the Schr¨ odinger equation, plus conservation of probability, ∗ ∗ requires M22 = M11 , M21 = M12 , and det(M) = 1, but I won’t use this here. See Merzbacher’s Quantum Mechanics. Similarly, for even potentials S11 = S22 , S12 = S21 (Problem 2.52).] 2 2 M21 2 1 , Tl = |S21 |2 = det(M) , Rr = |S22 |2 = M12 , Tr = |S12 |2 = Rl = |S11 |2 = . M22 M22 M22 |M22 |2 (b)

A

C

F

B

D

G M2

M1

x

F C C A F A A = M2 , = M1 , so = M 2 M1 =M , with M = M2 M1 . QED G D D B G B B (c) ψ(x) =

Aeikx + Be−ikx (x < a) F eikx + Ge−ikx (x > a)

.

−ika Continuity of ψ : Ae ika + Be−ika = F eika + Geika

ika

2mα ika −ika − ik Ae − Be−ika = − 2mα Ae + Be−ika . Discontinuity of ψ : ik F e − Ge 2 ψ(a) = − 2

(1) F e2ika + G = Ae2ika + B.

2ika

(2) F e2ika − G = Ae2ika − B + i 2mα +B . 2 k Ae c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


56 Add (1) and (2): 2F e2ika = 2Ae2ika + i

mα 2mα 2ika mα Ae A + i 2 e−2ika B = M11 A + M12 B. + B ⇒ F = 1 + i 2 2 k k k

So M11 = (1 + iβ); M12 = iβe−2ika ; β ≡

mα . 2 k

Subtract (2) from (1): 2G = 2B − 2iβe2ika A − 2iβB ⇒ G = (1 − iβ)B − iβe2ika A = M21 A + M22 B. So M21 = −iβe

2ika

; M22 = (1 − iβ).

M=

(1 + iβ) iβe−2ika . −iβe2ika (1 − iβ)

(d) M1 =

(1 + iβ) iβe−2ika (1 + iβ) iβe2ika ; to get M2 , just switch the sign of a: M2 = . −iβe2ika (1 − iβ) −iβe−2ika (1 − iβ)

M = M2 M1 =

T = Tl = Tr =

[1 + 2iβ + β 2 (e4ika − 1)] 2iβ[cos 2ka + β sin 2ka] . −2iβ[cos 2ka + β sin 2ka] [1 − 2iβ + β 2 (e−4ika − 1)]

1 ⇒ |M22 |2

T −1 = [1 + 2iβ + β 2 (e4ika − 1)][1 − 2iβ + β 2 (e−4ika − 1)] = 1 − 2iβ + β 2 e−4ika − β 2 + 2iβ + 4β 2 + 2iβ 3 e−4ika − 2iβ 3 + β 2 e4ika − β 2 − 2iβ 3 e4ika + 2iβ 3 + β 4 (1 − e4ika − e−4ika + 1) = 1 + 2β 2 + β 2 (e4ika + e−4ika ) − 2iβ 3 (e4ika − e−4ika ) + 2β 4 − β 4 (e4ika + e−4ika ) = 1 + 2β 2 + 2β 2 cos 4ka − 2iβ 3 2i sin 4ka + 2β 4 − 2β 4 cos 4ka = 1 + 2β 2 (1 + cos 4ka) + 4β 3 sin 4ka + 2β 4 (1 − cos 4ka) = 1 + 4β 2 cos2 2ka + 8β 3 sin 2ka cos 2ka + 4β 4 sin2 2ka

T =

1 1 + 4β 2 (cos 2ka + β sin 2ka)2

Problem 2.54 I’ll just show the first two graphs, and the last two. Evidently K lies between 0.9999 and 1.0001.


57

Problem 2.55 The correct values (in Eq. 2.72) are K = 2n + 1 (corresponding to En = (n + 12 )ω). I’ll start by “guessing” 2.9, 4.9, and 6.9, and tweaking the number until I’ve got 5 reliable significant digits. The results (see below) are 3.0000, 5.0000, 7.0000. (The actual energies are these numbers multiplied by 12 ω.)


58


59


60

Problem 2.56 The Schr¨ odinger equation says − 2m ψ = Eψ, or, with the correct energies (Eq. 2.27) and a = 1, ψ + (nπ)2 ψ = 0. I’ll start with a “guess” using 9 in place of π 2 (that is, I’ll use 9 for the ground state, 36 for the first excited state, 81 for the next, and finally 144). Then I’ll tweak the parameter until the graph crosses the axis right at x = 1. The results (see below) are, to five significant digits: 9.8696, 39.478, 88.826, 157.91. (The actual energies are these numbers multiplied by 2 /2ma2 .) 2


61


62

CHAPTER 3. FORMALISM

Chapter 3

Formalism Problem 3.1 (a) All conditions are trivial except Eq. A.1: we need to show that the sum of two square-integrable functions is itself square-integrable. Let h(x) = f (x) + g(x), so that |h|2 = (f + g)∗ (f + g) = |f |2 + |g|2 + f ∗ g + g ∗ f and hence ∗ 2 2 2 ∗ ∗ f g dx . |h| dx = |f | dx + |g| dx + f g dx + If f (x) and g(x) are square-integrable, then the first two terms are finite, and (by Eq. 3.7) so too are the last two. So |h|2 dx is finite. QED The set of all normalized functions is certainly not a vector space: it doesn’t include 0, and the sum of two normalized functions is not (in general) normalized—in fact, if f (x) is normalized, then the square integral of 2f (x) is 4. (b) Equation A.19 is trivial:

b

g|f =

∗

'

g(x) f (x) dx = a

b

(∗ ∗

f (x) g(x) dx

= f |g∗ .

a

Equation A.20 holds (see Eq. 3.9) subject to the understanding in footnote 6. As for Eq. A.21, this is pretty obvious: f |(b|g + c|h) = f (x)∗ (bg(x) + ch(x)) dx = b f ∗ g dx + c f ∗ h dx = bf |g + cf |h.

Problem 3.2 (a)

f |f =

1

x2ν dx = 0

1

1 1 x2ν+1 = 1 − 02ν+1 . 2ν + 1 2ν + 1 0

Now 02ν+1 is finite (in fact, zero) provided (2ν + 1) > 0, which is to say, ν > − 12 . If (2ν + 1) < 0 the integral definitely blows up. As for the critical case ν = − 12 , this must be handled separately: 1 1 f |f = x−1 dx = ln x0 = ln 1 − ln 0 = 0 + ∞. 0



63

So f (x) is in Hilbert space only for ν strictly greater than -1/2. (b) For ν = 1/2, we know from (a) that f (x) is in Hilbert space: yes. Since xf = x3/2 , we know from (a) that it is in Hilbert space: yes. For df /dx = 12 x−1/2 , we know from (a) that it is not in Hilbert space: no. [Moral: Simple operations, such as differenting (or multiplying by 1/x), can carry a function out of Hilbert space.]

Problem 3.3 ˆ = Qh|h ˆ Suppose h|Qh for all functions h(x). Let h(x) = f (x) + cg(x) for some arbitrary constant c. Then ˆ = (f + cg)|Q(f ˆ + cg) = f |Qf ˆ + cf |Qg ˆ + c∗ g|Qf ˆ + |c|2 g|Qg; ˆ h|Qh ˆ ˆ + cg)|(f + cg) = Qf ˆ |f + cQf ˆ |g + c∗ Qg|f ˆ + |c|2 Qg|g. ˆ Qh|h = Q(f ˆ = Qf ˆ |f and g|Qg ˆ = Qg|g ˆ Equating the two and noting that f |Qf leaves ˆ + c∗ g|Qf ˆ = cQf ˆ |g + c∗ Qg|f ˆ . cf |Qg In particlar, choosing c = 1: whereas if c = i: Adding the last two equations:

ˆ + g|Qf ˆ = Qf ˆ |g + Qg|f ˆ , f |Qg ˆ − g|Qf ˆ = Qf ˆ |g − Qg|f ˆ . f |Qg ˆ = Qf ˆ |g. QED f |Qg

Problem 3.4 ˆ + K)g ˆ ˆ + f |Kg ˆ = Hf ˆ |g + Kf ˆ |g = (H ˆ + K)f ˆ |g. (a) f |(H = f |Hg ˆ |g. Hermitian ⇔ α is real. ˆ = αf |Qg; ˆ ˆ |g = α∗ Qf (b) f |αQg αQf ˆ Kg ˆ = Hf ˆ |Kg ˆ = K ˆ Hf ˆ |g, so H ˆK ˆ is hermitian ⇔ H ˆK ˆ =K ˆ H, ˆ or [H, ˆ K] ˆ = 0. (c) f |H (d) f |ˆ xg = f ∗ (xg) dx = (xf )∗ g dx = ˆ xf |g. 2 2 2 2 ∗d g ˆ = f ∗ − d + V g dx = − f |Hg dx + f ∗ V g dx. f 2m dx2 2m dx2 Integrating by parts (twice): ∞ ∞ ∞ ∞ ∞ ∗ ∞ 2 ∗ 2 df dg d f df ∗ ∗d g ∗ dg ∗ dg f dx = f − − + g dx. dx = f g 2 2 dx dx dx dx dx dx −∞ −∞ −∞ dx −∞ −∞ −∞ But for functions f (x) and g(x) in Hilbert space the boundary terms vanish, so ∞ ∞ 2 ∗ 2 d f ∗d g f dx = g dx, and hence (assuming that V (x) is real): 2 2 dx −∞ −∞ dx ∗ ∞ 2 d 2 f ˆ = ˆ |g. f |Hg − + V f g dx = Hf 2m dx2 −∞


64


Problem 3.5

f ∗ (xg) dx = (xf )∗ g dx = xf |g, so x† = x. f |ig = f ∗ (ig) dx = (−if )∗ g dx = −if |g, so i† = −i.

(a) f |xg =

dg f | = dx

∞

−∞

∞ f − dx = f g dx ∗ dg

∗

−∞

∞

−∞

df dx

∗ g dx = −xf |g, so

d dx

† =−

d . dx

1 1 (−ip + mωx). But p and x are hermitian, and i† = −i, so (a+ )† = √ (ip + mωx), or 2mω 2mω (a+ )† = (a− ).

(b) a+ = √

ˆ † f |g = (Q ˆ†. ˆ R)g ˆ ˆ † f |Rg ˆ = R ˆ†Q ˆ R) ˆ † f |g, so (Q ˆ R) ˆ †=R ˆ†Q (c) f |(Q = Q

Problem 3.6

2π 2π ∗ 2π 2π 2π 2 ∗ d2 g df dg d f df ∗ ∗ dg ∗ dg dφ = f g dφ = f − − g dφ. + 2 dφ dφ dφ dφ dφ dφ dφ2 0 0 0 0 0 0 ˆ = As in Example 3.1, for periodic functions (Eq. 3.26) the boundary terms vanish, and we conclude that f |Qg ˆ ˆ Qf |g, so Q is hermitian: yes.

ˆ = f |Qg

2π

f∗

2 √ ˆ = qf ⇒ d f = qf ⇒ f± (φ) = Ae± qφ . Qf 2 dφ √ √ The periodicity condition (Eq. 3.26) requires that q(2π) = 2nπi, or q = in, so the eigenvalues are

q = −n2 , (n = 0, 1, 2, . . . ). The spectrum is doubly degenerate; for a given n there are two eigenfunctions (the plus sign or the minus sign, in the exponent), except for the special case n = 0, which is not degenerate.

Problem 3.7 ˆ = qf and Qg ˆ = qg. Let h(x) = af (x) + bg(x), for arbitrary constants a and b. Then (a) Suppose Qf ˆ = Q(af ˆ ˆ ) + b(Qg) ˆ = a(qf ) + b(qg) = q(af + bg) = qh. Qh + bg) = a(Qf

(b)

d2 f d −x

d2 d x d2 g d2 = e−x = g. = 2 (ex ) = = 2 e−x = (e ) = ex = f, −e 2 2 dx dx dx dx dx dx So both of them are eigenfunctions, with the same eigenvalue 1. The simplest orthogonal linear combinations are sinh x =

1 1 x e − e−x = (f − g) 2 2

and

cosh x =

1 1 x e + e−x = (f + g). 2 2

(They are clearly orthogonal, since sinh x is odd while cosh x is even.)



65

Problem 3.8 (a) The eigenvalues (Eq. 3.29) are 0, ±1, ±2, . . . , which are obviously real. For any two eigenfunctions, f = Aq e−iqφ and g = Aq e−iq φ (Eq. 3.28), we have 2π 2π i(q−q )φ A∗q Aq i(q−q )2π e f |g = A∗q Aq eiqφ e−iq φ dφ = A∗q Aq = − 1 . e i(q − q ) i(q − q ) 0 0

i(q−q )2π

But q and q are integers, so e nonzero).

= 1, and hence f |g = 0 (provided q = q , so the denominator is

(b) In Problem 3.6 the eigenvalues are q = −n2 , with n = 0, 1, 2, . . . , which are obviously real. two eigenfunctions, f = Aq e±inφ and g = Aq e±in φ , we have 2π 2π A∗q Aq ±i(n −n)2π e±i(n −n)φ ∗ ∓inφ ±in φ ∗ f |g = Aq Aq e e dφ = Aq Aq = − 1 =0 e ±i(n − n) ±i(n − n) 0

For any

0

(provided n = n ). But notice that for each eigenvalue (i.e. each value of n) there are two eigenfunctions (one with the plus sign and one with the minus sign), and these are not orthogonal to one another.

Problem 3.9 (a) Infinite square well (Eq. 2.19). (b) Delta-function barrier (Fig. 2.16), or the finite rectangular barrier (Prob. 2.33). (c) Delta-function well (Eq. 2.114), or the finite square well (Eq. 2.145) or the sech2 potential (Prob. 2.51).

Problem 3.10 From Eq. 2.28, with n = 1: d pˆ ψ1 (x) = i dx

π 2 π π π π 2 sin x = cos x = −i cot x ψ1 (x). a a i aa a a a

Since pˆ ψ1 is not a (constant) multiple of ψ1 , ψ1 is not an eigenfunction of pˆ: no. It’s true that the magnitude √ of the momentum, 2mE1 = π/a, is determinate, but the particle is just as likely to be found traveling to the left (negative momentum) as to the right (positive momentum).

Problem 3.11 Ψ0 (x, t) =

mω π

1/4

2 − mω −iωt/2 2 x

e

e

;

1/4 ∞ mω 2 1 mω −iω/2 Φ(p, t) = √ e e−ipx/ e− 2 x dx. 2π π −∞

From Problem 2.22(b): 1/4 2 1 2π −p2 /2mω 1 mω −iωt/2 Φ(p, t) = √ e = e−p /2mω e−iωt/2 . e 1/4 mω (πmω) 2π π c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

66

CHAPTER 3. FORMALISM |Φ(p, t)|2 = √

√ 2 1 p2 1 e−p /mω . Maximum classical momentum: = E = ω =⇒ p = mω. 2m 2 πmω

So the probability it’s outside classical range is:

√ − mω

P = −∞ √

mω

|Φ| dp + 2

∞

√

√

|Φ|2 dp.

Now

0

mω

√mω 2 1 |Φ| dp = √ e−p /mω dp. πmω 0 √2 √ 2 1 =√ e−z /2 dz = F ( 2) − 2π 0 2

0

mω

|Φ| dp = 1 − 2 2

Let z ≡

2 p, so dp = mω

mω dz. 2

1 , in CRC Table notation. 2

√ √ √ 1 P = 1 − 2 (F ( 2) − = 1 − 2F ( 2) + 1 = 2 1 − F ( 2) = 0.157. 2 To two digits: 0.16 (compare Prob. 2.15).

Problem 3.12

∞ 1 From Eq. 3.55: Ψ(x, t) = √ eipx/ Φ(p, t)dp. 2π −∞

1 1 ∗ −ip x/ ∗ +ipx/ √ x = Ψ xΨdx = Φ (p , t)dp x √ Φ(p, t)dp dx. e e 2π 2π But xeipx/ = −i x

d ipx/

, so (integrating by parts): e dp

ipx/

e

Φ dp =

1 So x = 2π

d ipx/ )Φ dp = e i dp

−ip x/

e

∗

ipx/

e

ipx/

Φ (p , t)e

∂ − Φ(p, t) dp. i ∂p

∂ − Φ(p, t) dp dp dx. i ∂p

Do the x integral first, letting y ≡ x/: 1 1 e−ip x/ eipx/ dx = ei(p−p )y dy = δ(p − p ), (Eq. 2.144), so 2π 2π x =

∂ ∂ Φ∗ (p , t)δ(p − p ) − Φ(p, t) dp dp = Φ∗ (p, t) − Φ(p, t) dp. i ∂p i ∂p

QED



67

Problem 3.13 (a) [AB, C] = ABC − CAB = ABC − ACB + ACB − CAB = A[B, C] + [A, C]B. (b) Introducing a test function g(x), as in Eq. 2.50: d n n n dg n dg n−1 n dg [x , p]g = x g+x − (x g) = x − nx = inxn−1 g. i dx i dx i dx i dx So, dropping the test function, [xn , p] = inxn−1 . dg d dg df dg df df (c) [f, p]g = f − (f g) = f − g+f = i g ⇒ [f, p] = i . i dx i dx i dx i dx dx dx dx

Problem 3.14

p2 1 x, +V = x, p2 + [x, V ]; 2m 2m

x, p2 = xp2 − p2 x = xp2 − pxp + pxp − p2 x = [x, p]p + p[x, p].

p2 1 ip so x, +V = 2ip = . 2m 2m m The generalized uncertainty principle (Eq. 3.62) says, in this case, 2 2 1 i 2 σx2 σH ≥ p = p ⇒ σx σH ≥ |p|. QED 2i m 2m 2m

Using Eq. 2.51: x, p2 = ip + pi = 2ip. And [x, V ] = 0,

For stationary states σH = 0 and p = 0, so it just says 0 ≥ 0.

Problem 3.15 ˆ n = µn fn (that is: fn (x) is an eigenfunction both of Pˆ and of Q), ˆ and the set {fn } Suppose Pˆ fn = λn fn and Qf / is complete, so that any function f (x) (in Hilbert space) can be expressed as a linear combination: f = cn fn . Then ˆ = (Pˆ Q ˆ−Q ˆ Pˆ ) ˆ [Pˆ , Q]f cn fn = Pˆ cn λn fn = cn µn λn fn − cn µn fn − Q cn λn µn fn = 0. ˆ = 0. Since this is true for any function f , it follows that [Pˆ , Q]

Problem 3.16 dΨ i a = (iax − iax + p)Ψ = dx dΨ a = Ψ

− x + x +

Let constant = −

i −x + x + p Ψ. a

ip a x2 ip dx ⇒ ln Ψ = − + xx + x + constant. a 2 a

x2 a a ip + B (B a new constant). Then ln Ψ = − (x − x)2 + x + B. 2 2

Ψ = e− 2 (x−x) eipx/ eB = Ae−a(x−x) a

2

2

/2 ipx/

e

, where A ≡ eB .


68


Problem 3.17 (a) 1 commutes with everything, so

d Ψ|Ψ = 0 (this is the conservation of normalization, which we origidt

nally proved in Eq. 1.27). d H = 0 (assuming H has no explicit time dt dependence); this is conservation of energy, in the sense of the comment following Eq. 2.40. ip p dx i ip (c) [H, x] = − (Eq. 1.33). (see Problem 3.14). So = − = m dt m m

(b) Anything commutes with itself, so [H, H] = 0, and hence

0

0 1 1 p2 ∂V dV dp i ∂V (d) [H, p] = + V, p = [V, p] = i (Problem 3.13(c)). So = i = − . 2m dx dt ∂x ∂x This is Ehrenfest’s theorem (Eq. 1.38).

Problem 3.18

1 1 Ψ(x, t) = √ ψ1 e−iE1 t/ + ψ2 e−E2 t/ . H 2 Ψ = √ (H 2 ψ1 )e−E1 t/ + (H 2 ψ2 )e−iEn t/ . 2 2 Hψ1 = E1 ψ1 ⇒ H 2 ψ1 = E1 Hψ1 = E12 ψ1 , and H 2 ψ2 = E22 ψ2 , so

1 ψ1 e−iE1 t/ + ψ2 e−iE2 t/ | E12 ψ1 e−iE1 t/ + E22 ψ2 e−iE2 t/ 2 1 = ψ1 |ψ1 eiE1 t/ E12 e−iE1 t/ + ψ1 |ψ2 eiE1 t/ E22 e−iE2 t/ 2

1

+ ψ2 |ψ1 eiE2 t/ E12 e−iE1 t/ + ψ2 |ψ2 eiE2 t/ E22 e−iE2 t/ = E12 + E22 . 2

H 2 =

Similarly, H = 12 (E1 + E2 )

(Problem 2.5(e)).

1

1 2 1 E1 + E22 − (E1 + E2 )2 = 2E12 + 2E22 − E22 − E12 − 2E1 E2 − E22 2 4 4

1 2 1 1 = E1 − 2E1 E2 + E22 = (E2 − E1 )2 . σH = (E2 − E1 ). 4 4 2

2 σH = H 2 − H2 =

1 ψ1 |x2 |ψ1 + ψ2 |x2 |ψ2 + ψ1 |x2 |ψ2 ei(E1 −E2 )t/ + ψ2 |x2 |ψ1 ei(E2 −E1 )t/ . 2 mπ 1 a 2 n+m n−m 2 a 2 nπ ψn |x2 |ψm = x sin x dx = πx − cos πx dx. x sin x cos a 0 a a a 0 a a

x2 =

a 2

Now

x cos 0

2 3 2 a

kπx 2a x a k k k πx dx = πx + πx cos − 2 sin a k2 π2 a kπ a a 0

2a3 2a3 = 2 2 cos(kπ) = 2 2 (−1)k k π k π

(for k = nonzero integer).



69

2a2 4nm 2a2 (−1)n−m (−1)n+m = ∴ ψn |x |ψm = 2 − (−1)n+m 2 . π (n − m)2 (n + m)2 π2 (n − m2 )2 2

16a2 1 2 2 1 So ψ1 |x |ψ2 = ψ2 |x |ψ1 = − 2 . Meanwhile, from Problem 2.4, ψn |x |ψn = a . − 9π 3 2(nπ)2

1 16a2 i(E2 −E1 )t/ 1 2 1 1 1 Thus x2 = e + e−i(E2 −E1 )t/ . a − 2 + a2 − 2 − 2 3 2π 3 8π 9π 2 E −E 2 cos( 2 1 t) E2 − E 1 (4 − 1)π 2 2 3π 2 = = = 3ω [in the notation of Problem 2.5(b)]. 2 2ma 2ma2

a2 2 32 a 5 32 2 x = − 2 cos(3ωt) . From Problem 2.5(c), x = − 1 − 2 cos(3ωt) . 2 3 8π 2 9π 2 9π 2

2 a 4 64 64 32 5 2 2 2 2 So σx = x − x = − 2 cos(3ωt) − 1 + 2 cos(3ωt) − cos (3ωt) . − 4 3 4π 2 9π 9π 9π 2 2

dx a2 1 32 5 8 2 σx2 = cos (3ωt) . And, from Problem 2.5(d): − 2− = sin(3ωt). 2 4 3 4π 9π dt 3ma 2 2 dx 2 2 Meanwhile, the energy-time uncertainty principle (Eq. 3.72) says σH σx ≥ . Here 4 dt 2 2 2

2 1 1 32 32 1 5 5 2 2 2a 2 2 3 2 σH σx = (3ω) − cos (3ωt) = (ωa) − cos (3ωt) . − − 4 4 3 4π 2 9π 2 4 3 4π 2 9π 2 2 2 2 2 dx 2aω 8 8 2 = . = sin (3ωt) = (ωa)2 sin2 (3ωt), since 2 2 4 dt 2 3ma 3π ma π So the uncertainty principle holds if 2 2 2

3 1 32 8 5 2 cos (3ωt) ≥ sin2 (3ωt), − 2− 4 3 4π 9π 2 3π 2 which is to say, if 2 2 2 1 5 32 4 8 32 2 2 − ≥ cos (3ωt) + sin (3ωt) = . 3 4π 2 9π 2 3 3π 2 9π 2 32 2 1 5 Evaluating both sides: − 2 = 0.20668; = 0.12978. So it holds. (Whew!) 3 4π 9π 2 2

2

Problem 3.19 From Problem 2.43, we have: x =

l 1 2at 1 2 dx 1 + θ2 l 1 2 = t, so = , σx2 = , where θ = ; H = p = (a + l2 ). 2 m dt m 4w 4a m 2m 2m

p2 We need H 2 (to get σH ). Now, H = , so 2m ∞ ∞ 1 1 1 2 4 4 2 H = p = p |Φ(p, t)| dp, where (Eq. 3.54): Φ(p, t) = √ e−ipx/ Ψ(x, t) dx. 4m2 4m2 −∞ 2π −∞ c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

70


From Problem 2.43: Ψ(x, t) =

2a π

1/4 √

2 l l2 1 e− 4a ea(ix+ 2a ) /(1+iθ) . 1 + iθ

1/4 ∞ 2 2 l 1 1 il 2a √ e−l /4a e−ipx/ ea(ix+ 2a ) /(1+iθ) dx. Let y ≡ x − . π 2a 1 + iθ 2π −∞ 1/4 ∞ 2 2 1 1 2a √ =√ e−l /4a epl/2a e−ipy/ e−ay /(1+iθ) dy. 1 + iθ 2π π −∞ [See Prob. 2.22(a) for the integral.] 1/4 1 1 π(1 + iθ) − p2 (1+iθ) 2a −l2 /4a pl/2a √ =√ e e 4a2 e a 1 + iθ 2π π 1/4 p2 (1+iθ) pl l2 1 1 =√ e− 4a e 2a e− 4a2 . 2aπ

So Φ(p, t) = √

|Φ(p, t)| = √ 2

1 1 1 −l2 /2a pl/a −p2 /2a2 1 e e = √ e 2a e 2aπ 2aπ

2

p l2 − 2pl + 2

2 1 = √ e−(l−p/) /2a . 2aπ

∞ 2 1 p p4 = √ p4 e−(l−p/) /2a dp. Let − l ≡ z, so p = (z + l). 2aπ −∞ ∞ 2 1 = √ (z + l)4 e−z /2a dz. Only even powers of z survive: 5 2aπ −∞

∞ √ 4

2 4 (2a) √ 4 3(2a)2 √ =√ 2aπ + 6l2 2aπ + l4 2aπ z + 6z 2 l2 + l4 e−z /2a dz = √ 4 2 2aπ −∞ 2aπ 4

2 = 4 3a2 + 6al2 + l4 . ∴ H 2 = 3a + 6al2 + l4 . 4m2 2 σH = H 2 − H2 =

2 2 σH σx

4 a 2 2 4 2 2 4 2 2 4 2 + 6al + l − a − 2al − l + 4al 3a = 2a = a + 2l2 . 4m2 4m2 2m2

2 2

4 l 2 a 2at 2at 4 a 2 1 = 1+ 2 1+ a + 2l 1+ = 2m2 4a m 4m2 2l m 2 2 4 l 2 2 l 2 dx ≥ = = , so it works. 2 4m 4 m 4 dt

Problem 3.20

dx dx For Q = x, Eq. 3.72 says σH σx ≥ , so σx σH ≥ |p|, which is the Griffiths . But p = m 2 dt dt 2m uncertainty principle of Problem 3.14.

Problem 3.21 P 2 |β = P (P |β) = P (α|β|α) = α|β(P |α) = α|β α|α |α = α|β|α = P |β. 1 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


71

Since P 2 |β = P |β for any vector |β, P 2 = P . QED [Note: To say two operators are equal means that they have the same effect on all vectors.] If |γ is an eigenvector of Pˆ with eigenvalue λ, then Pˆ |γ = λ|γ, and it follows that Pˆ 2 |γ = λPˆ |γ = λ2 |γ. But Pˆ 2 = Pˆ , and |γ = 0, so λ2 = λ, and hence the eigenvalues of Pˆ are 0 and 1. Any (complex) multiple of |α is an eigenvector of Pˆ , with eigenvalue 1; any vector orthogonal to |α is an eigenvector of Pˆ , with eigenvalue 0.

Problem 3.22 (a) α| = −i1| − 22| + i3|;

β| = −i1| + 23|.

(b) α|β = (−i1| − 22| + i3|) (i|1 + 2|3) = (−i)(i)1|1 + (i)(2)3|3 = 1 + 2i. β|α = (−i1| + 23|) (i|1 − 2|2 − i|3) = (−i)(i)1|1 + (2)(−i)3|3 = 1 − 2i = α|β∗ . (c) A11 = 1|αβ|1 = (i)(−i) = 1; A12 = 1|αβ|2 A21 = 2|αβ|1 = (−2)(−i) = 2i; A22 = 2|αβ|2 A31 = 3|αβ|1 = (−i)(−i) = −1; A32 = 3|αβ|2   1 0 2i A =  2i 0 −4  . No, −1 0 −2i

= (i)(0) = 0; A13 = 1|αβ|3 = (i)(2) = 2i; = (−2)(0) = 0; A23 = 2|αβ|3 = (−2)(2) = −4; = (−i)(0) = 0; A33 = 3|αβ|3 = (−i)(2) = −2i. it’s not hermitian.

Problem 3.23 Write the eigenvector as |ψ = c1 |1 + c2 |2, and call the eigenvalue E. The eigenvalue equation is ˆ H|ψ = 9 (|11| − |22| + |12| + |21|) (c1 |1 + c2 |2) = 9 (c1 |1 + c1 |2 − c2 |2 + c2 |1) = 9 [(c1 + c2 )|1 + (c1 − c2 )|2] = E|ψ = E(c1 |1 + c2 |2). E E 9(c1 + c2 ) = Ec1 ⇒ c2 = − 1 c1 ; 9(c1 − c2 ) = Ec2 ⇒ c1 = + 1 c2 . 9 9 2 √ E E E c2 = −1 + 1 c2 ⇒ − 1 = 1 ⇒ E = ± 2 9. 9 9 9 √ √ The eigenvectors are: c2 = (± 2 − 1)c1 ⇒ |ψ± = c1 |1 + (± 2 − 1)|2 . 1 1 The Hamiltonian matrix is H=9 . 1 −1

Problem 3.24 |α =

n

ˆ cn |en ⇒ Q|α =

n

ˆ n = cn Q|e

n

en |αqn |en =

' n

( ˆ= qn |en en | |α ⇒ Q

qn |en en |.

n


72


Problem 3.25

e 1 |e2

|e2 = x;

|e3 = x ; 2

1

1 1 dx = 2. So |e 1 = √ . 2 −1

|e1 = 1; e1 |e1 =

1 =√ 2

e 1 |e3

1 3 x3 2 x dx = 0; e2 |e2 = x dx = = 3 . So |e2 = 2 x. 3 −1 −1 −1

1 =√ 2

1

1

2

1

1 2 x dx = √ ; e 2 |e3 = 23 −1 2

1 2 x3 dx = 0. 3 −1

1 2 1 So (Problem A.4): |e 3 = |e3 − √ |e 1 = x2 − . 3 23 e 3 |e 3 = |e 3

=

x2 −

1 3

2

dx =

1 2 x3 x 8 x5 2 4 2 − · + . So = − + = 5 3 3 9 −1 5 9 9 45

45 2 1 5 3 2 1 x − = x − . 8 3 2 2 2

e 1 |e4

|e4 = x .

1 =√ 2

1 3

x dx = 0; −1

e 2 |e4

1 3 3 2 4 = x dx = · ; 2 −1 2 5

1 3 5 1 3 5 3 2 3 3 3 = x − x dx = 0. |e4 = |e4 − e2 |e4 |e2 = x − x = x3 − x. 2 −1 2 2 2 5 2 5

e 4 |e 4 =

|e 4

−1

3

e 3 |e4

1

5 = 2

1

−1

7 2

1 3 x 2 12 18 2 · 3 x5 9 x3 8 x3 − x dx = = − − + + = . 5 7 5 5 25 3 −1 7 25 75 7 · 25

7 3 3 7 5 3 3 x − x = x − x . 2 5 2 2 2

Problem 3.26 ∗ ˆ = Q ˆ † ψ|ψ = −Qψ|ψ ˆ ˆ (a) Q = ψ|Qψ = −(ψ|Qψ) = −Q∗ , so Q is imaginary.

ˆ † Pˆ † , so if Pˆ = Pˆ † and Q ˆ=Q ˆ † then ˆ †=Q (b) From Problem 3.5(c) we know that (Pˆ Q) ˆ † = (Pˆ Q ˆ−Q ˆ Pˆ )† = Q ˆ † Pˆ † − Pˆ † Q ˆ Pˆ − Pˆ Q ˆ = −[Pˆ , Q]. ˆ ˆ† = Q [Pˆ , Q] ˆ = −Q ˆ † , then [Pˆ , Q] ˆ†=Q ˆ † Pˆ † − Pˆ † Q ˆ ˆ = −[Pˆ , Q]. ˆ ˆ † = (−Q)(− Pˆ ) − (−Pˆ )(−Q) If Pˆ = −Pˆ † and Q So in either case the commutator is antihermitian.



73

Problem 3.27 (a) ψ1 . (b) b1 (with probability 9/25) or b2 (with probability 16/25). (c) Right after the measurement of B: • With probability 9/25 the particle is in state φ1 = (3ψ1 + 4ψ2 )/5; in that case the probability of getting a1 is 9/25. • With probability 16/25 the particle is in state φ2 = (4ψ1 − 3ψ2 )/5; in that case the probability of getting a1 is 16/25. So the total probability of getting a1 is

337 16 16 9 9 · + · = = 0.5392. 25 25 25 25 625

[Note: The measurment of B (even if we don’t know the outcome of that measurement) collapses the wave function, and thereby alters the probabilities for the second measurment of A. If the graduate student inadvertantly neglected to measure B, the second measurement of A would be certain to reproduce the result a1 .]

Problem 3.28

nπ 2 sin x e−iEn t/ , a a

n2 π 2 2 . 2ma2 ∞ nπ 2 −iEn t/ a −ipx/ 1 1 −ipx/ Φn (p, t) = √ e Ψn (x, t) dx = √ e sin e x dx a 2π −∞ 2π a 0 1 1 a i(nπ/a−p/)x =√ e − ei(−nπ/a−p/)x dx e−iEn t/ 2i 0 πa

a 1 1 ei(nπ/a−p/)x ei(−nπ/a−p/)x =√ e−iEn t/ − 2i i(nπ/a − p/) i(−nπ/a − p/) 0 πa

−1 −iEn t/ ei(nπ−pa/) − 1 e−i(nπ+pa/) − 1 = √ e + (nπ/a − p/) (nπ/a + p/) 2 πa

n −ipa/ −1 −1 −iEn t/ (−1) e (−1)n e−ipa/ − 1 = √ e a+ a (nπ − ap/) (nπ + ap/) 2 πa a −iEn t/ 2nπ 1 n −ipa/ (−1) =− e e − 1 2 π (nπ)2 − (ap/)2 ne−iEn t/ aπ = 1 − (−1)n e−ipa/ . 2 2 (nπ) − (ap/)

Ψn (x, t) =

with En =

Noting that n −ipa/

1 − (−1) e

−ipa/2

=e

ipa/2 e − (−1)n e−ipa/2 = 2e−ipa/2

cos(pa/2) i sin(pa/2)

(n odd), (n even),


74


we have |Φ1 (p, t)|2 =

4πa cos2 (pa/2) , [π 2 − (pa/)2 ]2

|Φ2 (p, t)|2 =

16πa sin2 (pa/2) . [(2π)2 − (pa/)2 ]2

Mathematica has no trouble with the points p = ±nπ/a, where the denominator vanishes. The reason is that the numerator is also zero there, and the function as a whole is finite—in fact, the graphs show no interesting behavior at these points. |Φ1 | 2

|Φ 2| 2

p

p2 =

∞

−∞

4n2 = a2

p2 |Φn (p, t)|2 dp =

∞

−∞

4n2 πa

∞

p

p2 2

2 2 −∞ [(nπ) − (ap/) ] 2 2 x 4n Tn (x) dx = 2 In , 2 2 (1 − x ) a

where

Tn (x) ≡

cos2 (nπx/2), sin2 (nπx/2),

cos2 (pa/2) sin2 (pa/2)

if n is odd, if n is even.

[let x ≡

dp

ap ] nπ

The integral can be evaluated by partial fractions:

x2 1 1 1 1 1 − ⇒ = + + (x2 − 1)2 4 (x − 1)2 (x + 1)2 (x − 1) (x + 1) 1 In = 4

∞

−∞

1 Tn (x) dx + (x − 1)2

∞

−∞

1 Tn (x) dx + (x + 1)2

∞

−∞

1 Tn (x) dx − (x − 1)

∞

−∞

1 Tn (x) dx . (x + 1)

For odd n:

∞

−∞

1 2 nπx cos dx = (x ± 1)k 2

1 2 nπx sin dx = (x ± 1)k 2

∞

−∞

∞ 1 1 2 nπy 2 nπ cos sin (y ∓ 1) dy = dy. k yk 2 2 −∞ y

For even n:

∞

−∞

In either case, then, In =

1 2

∞

−∞

∞

−∞

∞ 1 1 2 nπ 2 nπy sin sin (y ∓ 1) dy = dy. k yk 2 2 −∞ y

1 nπ 2 nπy sin dy = 2 y 2 4

Therefore 4n2 4n2 nπ 2 p = 2 In = 2 = a a 4 2

nπ a

∞

−∞

sin2 u nπ 2 du = . u2 4

2 (same as Problem 2.4).



75

Problem 3.29

nλ 1 e Ψ(x, 0) dx = √ ei(2π/λ−p/)x dx 2 nπλ −nλ −∞ nλ 1 1 ei(2π/λ−p/)x ei2πn e−ipnλ/) − e−i2πn eipnλ/) = √ = √ i(2π/λ − p/) 2 nπλ i(2π/λ − p/) −nλ 2 nπλ λ sin(npλ/) = . nπ (pλ − 2π)

1 Φ(p, 0) = √ 2π

∞

−ipx/

|Ψ(x, 0)|2 =

1 2nλ

(−nλ < x < nλ); |Ψ| 2

λ sin2 (npλ/) . nπ (pλ − 2π)2

|Φ| 2

x

nλ

-nλ

|Φ(p, 0)|2 =

2πh/λ

p

The width of the |Ψ|2 graph is wx = 2nλ. The |Φ|2 graph is a maximum at 2π/λ, and goes to zero on either 2π 1 2π side at 1± , so wp = . As n → ∞, wx → ∞ and wp → 0; in this limit the particle has a λ 2n nλ well-defined momentum, but a completely indeterminate position. In general, 2π = 4π > /2, nλ so the uncertainty principle is satisfied (using the widths as a measure of uncertainty). If we try to check the uncertainty principle more rigorously, using standard deviation as the measure, we get an uninformative result, because λ ∞ 2 sin2 (npλ/) 2 p = p dp = ∞. nπ −∞ (pλ − 2π)2 wx wp = (2nλ)

(At large |p| the integrand is approximately (1/λ2 ) sin2 (npλ/), so the integral blows up.) Meanwhile p is zero, so σp = ∞, and the uncertainty principle tells us nothing. The source of the problem is the discontinuity in Ψ at the end points; here pˆ Ψ = −i dΨ/dx picks up a delta function, and Ψ|ˆ p2 Ψ = ˆ p Ψ|ˆ p Ψ → ∞ because the integral of the square of the delta function blows up. In general, if you want σp to be finite, you cannot allow discontinuities in Ψ.

Problem 3.30 (a) 1 = |A|

2

=

∞

−∞

1 dx = 2|A|2 (x2 + a2 )2

1 π 1 |A|2 tan−1 (∞) = 3 |A|2 2 a a 2a

∞ 1 1 x 2 1 −1 x dx = 2|A| + tan (x2 + a2 )2 2a2 x2 + a2 a a 0 0 2a ⇒ A=a . π

∞


76


(b) x = A

∞

2

−∞ ∞

x2 = 2A2

(a2

0

x dx = 0. + x2 )2

x2 x2 a √ dx. [Let y ≡ , x = a y, dx = √ dy.] 2 2 2 2 (a + x ) a 2 y √ √ 1/2 2 2 2a ( π/2)( π) y 2a Γ(3/2)Γ(1/2) = = a2 . dy = (1 + y)2 π Γ(2) π 1

∞ 2a π 0 σx = x2 − x2 = a. 2

=

(c)

px px 1 − i sin , and sine is odd.] dx. [But e−ipx/ = cos 2 +a −∞ ∞ 2A 2A π −|p|a/ cos(px/) a −|p|a/ = √ dx = √ . = e e 2 + a2 x 2a 2π 0 2π

Φ(p, 0) = √

A 2π

∞

−∞

∞

e−ipx/

x2

|Φ(p, 0)|2 dp =

a

∞

e−2|p|a/ dp =

−∞

2a

∞ e−2pa/ = 1. −2a/ 0

(d) a p =

a p = 2

∞

2

(e)

pe−2|p|a/ dp = 0.

−∞ ∞

2 −2pa/

p e 0

2a 2 dp =

2a

3 =

2 . 2a2

σp =

p2 − p2 = √ . 2a

√ σx σp = a √ = 2 > . 2 2 2a

Problem 3.31 d i Equation 3.71 ⇒ xp = [H, xp]; Eq. 3.64 ⇒ [H, xp] = [H, x]p + x[H, p]; Problem 3.14 ⇒ [H, x] = dt ip dV − ; Problem 3.17(d) ⇒ [H, p] = i . So m dx

d i i p2 dV dV dV xp = − p2 + ix = 2 − x = 2T − x . QED dt m dx 2m dx dx In a stationary state all expectation values (at least, for operators that do not depend explicitly on t) are time-independent (see item 1 on p. 26), so dxp/dt = 0, and we are left with Eq. 3.97. For the harmonic oscillator: 1 dV dV V = mω 2 x2 ⇒ = mω 2 x ⇒ x = mω 2 x2 = 2V ⇒ 2T = 2V ⇒ T = V . QED 2 dx dx 1 3 In Problem 2.11(c) we found that T = V 1 = 4 ω (for n2 = 0); T 1 = V = 4 ω (for1 n =21).2 1 1 In Problem 2.12 we found that T = 2 n+ 2 ω, while x = (n+ 2 )/mω, so V = 2 mω x = 2 (n+ 12 )ω, and hence T = V for all stationary states.



77

Problem 3.32

1 Ψ(x, t) = √ ψ1 e−iE1 t/ + ψ2 e−iE2 t/ ; 2

Ψ(x, t)|Ψ(x, 0) = 0 ⇒

1 iE1 t/ ψ1 |ψ1 + eiE1 t/ ψ1 |ψ2 + eiE2 t/ ψ2 |ψ1 + eiE2 t/ ψ2 |ψ2 e 2 =

1 iE1 t/ + eiE2 t/ = 0, or eiE2 t/ = −eiE1 t/ , so ei(E2 −E1 )t/ = −1 = eiπ . e 2

Thus (E2 − E1 )t/ = π (orthogonality also at 3π, 5π, etc., but this is the first occurrence). ∴ ∆t ≡

t 1 . But ∆E = σH = (E2 − E1 ) (Problem 3.18). So ∆t ∆E = . = π E2 − E1 2 2

Problem 3.33

√ mω a+ |n = √n + 1 |n + 1, Equation 2.69: x = (a+ + a− ), p = i (a+ − a− ); Eq. 2.66 : a− |n = n |n − 1. 2mω 2 √ √ n|x|n = n|(a+ + a− )|n = n + 1 n|n + 1 + n n|n − 1 2mω 2mω √ √

√ √ = n + 1 δn,n +1 + n δn,n −1 = n δn ,n−1 + n δn,n −1 . 2mω 2mω √

mω √ n|p|n = i n δn ,n−1 − n δn,n −1 . 2 Noting that n and n run from zero to infinity, the matrices are: √   1 √0 0 0 0 √0  1 0 2 √0 0 0   √ .   0 mω 2 √0 3 √0 0 :  X= ; P = i 2mω  2 3 0 4 0 0 0    0 0 0 √4 0 √5  ··· Squaring these matrices: √  1·2 √0 0 0 1 0  0 2 · 3 0 0 3 0 √ √   1·2 0 X2 = 5 0 3 · 4 √0 2mω   0 √2 · 3 0 7 0 4·5 ··· 

−1 0  0 −3 √ mω   1·2 0 P2 = −  √ 2  2·3 0

√

√ 0 0 0 0 √0 − 1 √  1 0 − 2 0 0 0  √ √  0 2 0 − 3 0 0  √ √  0 3 0 − 4 0 0  √ √  0 0 0 4 − 5 ··· 

  : .    .

 

. 

: ; 

1·2 √0 0 0 2·3 √0 0 0 −5 0 3·4 √0 0 −7 0 4·5 ···

 

. 

: . 


78


So the Hamiltonian, in matrix form, is 1 2 mω 2 2 P + X 2m 2 √   1·2 √0 0 0 −1 0  0  −3 0 2·3 √0 0 √ ω  .  1·2 0  =− −5 0 3 · 4 0 :  √ 4   0 √2 · 3 0 −7 0 4·5  ···  √   1 1·2 √0 0 0 1 0    0 2·3 √0 0 3 0 0  √ ω  ω . 0  1·2 0 + 5 0 3 · 4 √ 0 : = 2  √ 0 4   0  2·3 0 7 0 4·5  ···

H=

0 3 0 0

0 0 5 0



0 0 0 7 ..

   .   .

It’s plainly diagonal, and the nonzero elements are Hnn = n + 12 ω, as they should be.

Problem 3.34 √ √ Evidently Ψ(x, t) = c0 ψ0 (x)e−iE0 t/ + c1 ψ1 (x)e−iE1 t/ , with |c0 |2 = |c1 |2 = 1/2, so c0 = eiθ0 / 2, c1 = eiθ1 / 2, for some real θ0 , θ1 . p = |c0 |2 ψ0 |pψ0 + |c1 |2 ψ1 |pψ1 + c∗0 c1 ei(E0 −E1 )t/ ψ0 |pψ1 + c∗1 c0 ei(E1 −E0 )t/ ψ1 |pψ0 . But E1 − E0 = ( 32 ω) − ( 12 ω) = ω, and (Problem 2.11) ψ0 |pψ0 = ψ1 |pψ1 = 0, while (Eqs. 2.69 and 2.66) ψ0 |pψ1 = i

√ √ mω mω mω mω ψ0 |(a+ − a− )ψ1 = i ψ0 | 2ψ2 − ψ0 | 1ψ0 = −i ; ψ1 |pψ0 = i . 2 2 2 2

' ' ( ( 1 −iθ0 1 iθ1 −iωt 1 −iθ1 1 iθ0 iωt mω mω √ e e √ e e p = √ e −i i +√ e 2 2 2 2 2 2 i mω mω = −e−i(ωt−θ1 +θ0 ) + ei(ωt−θ1 +θ0 ) = − sin(ωt + θ0 − θ1 ). 2 2 2 The maximum is mω/2; it occurs at t = 0 ⇔ sin(θ0 − θ1 ) = −1, or θ1 = θ0 + π/2. We might as well pick θ0 = 0, θ1 = π/2; then

1 1 Ψ(x, t) = √ ψ0 e−iωt/2 + ψ1 eiπ/2 e−3iωt/2 = √ e−iωt/2 ψ0 + iψ1 e−iωt . 2 2



Problem 3.35

(a) x = α|xα = 2mω = 2mω

x2 =

α|(a+ + a− )α = 2mω

79

(a− α|α + α|a− α) = 2mω

(α + α∗ ). 2mω

a2+ + a+ a− + a− a+ + a2− . But a− a+ = [a− , a+ ] + a+ a− = 1 + a+ a− (Eq. 2.55).

a2+ + 2a+ a− + 1 + a2− .

2 α| a2+ + 2a+ a− + 1 + a2− α = a− α|α + 2a− α|a− α + α|α + α|a2− α 2mω 2mω = (α∗ )2 + 2(α∗ )α + 1 + α2 = 1 + (α + α∗ )2 . 2mω 2mω

x2 =

mω mω mω α|(a+ − a− )α = i (a− α|α − α|a− α) = −i (α − α∗ ). p = α|pα = i 2 2 2

mω 2 mω 2 a+ − a+ a− − a− a+ + a2− = − a+ − 2a+ a− − 1 + a2− . p2 = − 2 2

mω mω 2 α| a2+ − 2a+ a− − 1 + a2− α = − a− α|α − 2a− α|a− α − α|α + α|a2− α 2 2 mω mω =− (α∗ )2 − 2(α∗ )α − 1 + α2 = 1 − (α − α∗ )2 . 2 2

p2 = −

(b) 1 + (α + α∗ )2 − (α + α∗ )2 = ; 2mω 2mω mω mω mω 2 2 2 ∗ 2 ∗ 2 σp = p − p = 1 − (α − α ) + (α − α ) = . σx σ p = = . QED 2 2 2mω 2 2

σx2 = x2 − x2 =

(c) Using Eq. 2.67 for ψn : 1 1 1 αn cn = ψn |α = √ (a+ )n ψ0 |α = √ ψ0 |(a− )n α = √ αn ψ0 |α = √ c0 . n! n! n! n! (d) 1 =

∞

|cn |2 = |c0 |2

n=0

(e) |α(t) =

∞ n=0

∞ 2 |α|2n = |c0 |2 e|α| n! n=0

−iEn t/

cn e

⇒

c0 = e−|α|

2

/2

.

n ∞ ∞ αe−iωt 2 αn −|α|2 /2 −i(n+ 1 )ωt −iωt/2 2 √ e √ |n = e |n = e e−|α| /2 |n. n! n! n=0 n=0

Apart form the overall phase factor e−iωt/2 (which doesn’t affect its status as an eigenfunction of a− , or its eigenvalue), |α(t) is the same as |α, but with eigenvalue α(t) = e−iωt α. (f ) Equation 2.58 says a− |ψ0 = 0, so yes, it is a coherent state, with eigenvalue α = 0.


80


Problem 3.36

(a) Equation 3.60 becomes |z|2 = [Re(z)]2 + [Im(z)]2 =

2

2 1 1 (z + z ∗ ) + (z − z ∗ ) ; Eq. 3.61 generalizes to 2 2i

2 2 σA σB

2

2 1 1 ≥ (f |g + g|f ) + (f |g − g|f ) . 2 2i

ˆ B] ˆ (p. 111), and, by the same argument, But f |g − g|f = [A, ˆ − AB + B ˆ A ˆ − AB = AˆB ˆ +B ˆ Aˆ − 2AB = D. f |g + g|f = AˆB 2 2 So σA σB ≥

1 D2 + C2 . 4

ˆ = A, ˆ (b) If B 2 2 σA σA

≥

then Cˆ = 0,

4 (1/4)4σA

=

4 σA ,

ˆ = 2 Aˆ2 − A2 ; D

2 D = 2 Aˆ2 − A2 = 2σA .

So Eq. 3.99 says

which is true, but not very informative.

Problem 3.37 First find the eigenvalues and eigenvectors of the Hamiltonian. The characteristic equation says (a − E) 0 b 0 (c − E) 0 = (a − E)(c − E)(a − E) − b2 (c − E) = (c − E) (a − E)2 − b2 = 0, b 0 (a − E) Either E = c, or else (a − E)2 = b2 ⇒ E = a ± b. So the eigenvalues are E1 = c,

E2 = a + b,

E3 = a − b.

To find the corresponding eigenvectors, write      a0 b α α  0 c 0   β  = En  β  . b 0a γ γ (1)  aα + bγ = cα ⇒ (a − c)α + bγ = 0;  cβ = cβ (redundant) ; ⇒ (a − c)2 − b2 α = 0.  bα + aγ = cγ ⇒ (a − c)γ + bα = 0. So (excluding the degenerate case a − c = ±b) α = 0, and hence also γ = 0. (2) aα + bγ = (a + b)α ⇒ α − γ = 0; cβ = (a + b)β ⇒ β = 0; bα + aγ = (a + b)γ (redundant). So α = γ and β = 0. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


81

(3) aα + bγ = (a − b)α ⇒ α + γ = 0; cβ = (a − b)β ⇒ β = 0; bα + aγ = (a − b)γ (redundant). So α = −γ and β = 0. Conclusion: The (normalized) eigenvectors of H are     0 1 1 |s1 = 1 , |s2 = √ 0 , 2 1 0 (a) Here |S(0) = |s1 , so



 1 1 |s3 = √  0  . 2 −1

  0 |S(t) = e−iE1 t/ |s1 = e−ict/ 1 . 0

(b) 1 |S(0) = √ (|s2 + |s3 ) . 2      1 1 1 −iE2 t/ 1 1 1 |S(t) = √ e |s2 + e−iE3 t/ |s3 = √ e−i(a+b)t/ √ 0 + e−i(a−b)t/ √  0  2 2 2 1 2 −1 

   e−ibt/ + eibt/ cos(bt/) 1  = e−iat/  . 0 0 = e−iat/  2 −ibt/ ibt/ −i sin(bt/) e −e

Problem 3.38 (a) H:

E1 = ω, E2 = E3 = 2ω;

A:

      1 0 0 |h1 = 0 , |h2 = 1 , |h3 = 0 . 0 0 1

−a λ 0 λ −a = a2 (2λ − a) − (2λ − a)λ2 = 0 ⇒ a1 = 2λ, a2 = λ, a3 = −λ. 0 0 0 (2λ − a)      010 α α  λβ = aα λα = aβ λ 1 0 0 β  = a β  ⇒  002 γ γ 2λγ = aγ 


82

CHAPTER 3. FORMALISM (1)  λβ = 2λα ⇒ β = 2α,  λα = 2λβ ⇒ α = 2β,  2λγ = 2λγ;

α = β = 0;

  0 |a1 = 0 . 1

(2)  λβ = λα ⇒ β = α,  λα = λβ ⇒ α = β,  2λγ = λγ; ⇒ γ = 0.

  1 1   1 . |a2 = √ 2 0

(3)  λβ = −λα ⇒ β = −α,  λα = −λβ ⇒ α = −β,  2λγ = −λγ; ⇒ γ = 0.

  1 1   −1 . |a3 = √ 2 0

B: (2µ − b) 0 0 0 −b µ = b2 (2µ − b) − (2µ − b)µ2 = 0 ⇒ b1 = 2µ, b2 = µ, b3 = −µ. 0 µ −b      200 α α  2µα = bα µγ = bβ µ 0 0 1 β  = b β  ⇒  γ µβ = bγ 010 γ 

(1)  2µα = 2µα,  µγ = 2µβ ⇒ γ = 2β,  µβ = 2µγ ⇒ β = 2γ;

β = γ = 0;

  1 |b1 = 0 . 0

(2)  2µα = µα ⇒ α = 0,  µγ = µβ ⇒ γ = β,  µβ = µγ; ⇒ β = γ.

  0 1 |b2 = √ 1 . 2 1

(3)  2µα = −µα ⇒ α = 0,  µγ = −µβ ⇒ γ = −β,  µβ = −µγ; ⇒ β = −γ.

  0 1   1 . |b3 = √ 2 −1



83

(b)    1 0 0 c1

H = S(0)|H|S(0) = ω c∗1 c∗2 c∗3 0 2 0 c2  = ω |c1 |2 + 2|c2 |2 + 2|c3 |2 . 0 0 2 c3    c1

0 1 0

A = S(0)|A|S(0) = λ c∗1 c∗2 c∗3 1 0 0 c2  = λ c∗1 c2 + c∗2 c1 + 2|c3 |2 . 0 0 2 c3    c1

2 0 0

B = S(0)|B|S(0) = µ c∗1 c∗2 c∗3 0 0 1 c2  = µ 2|c1 |2 + c∗2 c3 + c∗3 c2 . 0 1 0 c3

(c) |S(0) = c1 |h1 + c2 |h2 + c3 |h3 −iE1 t/

⇒

|h1 + c2 e |h2 + c3 e−iE3 t/ |h3 = c1 e−iωt |h1 + c2 e−2iωt |h2 + c3 e−2iωt |h3        iωt  1 0 0 c1 e = e−2iωt c1 eiωt 0 + c2 1 + c3 0 = e−2iωt  c2  . 0 0 1 c3

|S(t) = c1 e

−iE2 t/



H: h1 = ω, probability |c1 |2 ; A: a1 = 2λ,

a2 = λ,

h2 = h3 = 2ω, probability (|c2 |2 + |c3 |2 ).  iωt  c1 e

a1 |S(t) = e−2iωt 0 0 1  c2  = e−2iωt c3 ⇒ probability |c3 |2 . c3  iωt  c1 e

1 1 a2 |S(t) = e−2iωt √ 1 1 0  c2  = √ e−2iωt c1 eiωt + c2 ⇒ 2 2 c3

1 ∗ −iωt 1 2 c e |c1 | + |c2 |2 + c∗1 c2 e−iωt + c∗2 c1 eiωt . + c∗2 c1 eiωt + c2 = 2 1 2  iωt  c1 e

1 1 a3 |S(t) = e−2iωt √ 1 −1 0  c2  = √ e−2iωt c1 eiωt − c2 ⇒ 2 2 c3

probability = a3 = −λ,

1 ∗ −iωt 1 2 − c∗2 c1 eiωt − c2 = c1 e |c1 | + |c2 |2 − c∗1 c2 e−iωt − c∗2 c1 eiωt . 2 2 Note that the sum of the probabilities is 1.  iωt  c1 e

B: b1 = 2µ, b1 |S(t) = e−2iωt 1 0 0  c2  = e−2iωt c1 ⇒ probability |c1 |2 . c3  iωt  c1 e

1 1 b2 = µ, b2 |S(t) = e−2iωt √ 0 1 1  c2  = √ e−2iωt (c2 + c3 ) ⇒ 2 2 c probability =

3

probability =

1 ∗ 1 2 (c + c∗2 ) (c1 + c2 ) = |c1 | + |c2 |2 + c∗1 c2 + c∗2 c1 . 2 1 2


84


b3 = −µ,

 iωt  c1 e

1 1 b3 |S(t) = e−2iωt √ 0 1 −1  c2  = √ e−2iωt (c2 − c3 ) ⇒ 2 2 c 3

1 ∗ 1 2 (c − c∗3 ) (c2 − c3 ) = |c2 | + |c3 |2 − c∗2 c3 − c∗3 c2 . 2 2 2 Again, the sum of the probabilities is 1. probability =

Problem 3.39 (a) Expanding in a Taylor series: f (x + x0 ) =

n ∞ 1 n d x0 f (x). n! dx n=0

n ∞ d 1 n ip d ip But p = f (x) = eipx0 / f (x). , so = . Therefore f (x + x0 ) = x i dx dx n! 0 n=0 (b)

n ∞ 1 n ∂ Ψ(x, t + t0 ) = Ψ(x, t); t n! 0 ∂t n=0

i

∂Ψ = HΨ. ∂t

∂ [Note: It is emphatically not the case that i ∂t = H. These two operators have the same effect only when (as here) they are acting on solutions to the (time-dependent) Schr¨ odinger equation.] Also, 2 ∂ ∂Ψ ∂ i = H 2 Ψ, Ψ = i (HΨ) = H i ∂t ∂t ∂t

provided H is not explicitly dependent on t. And so on. So n ∞ i 1 n Ψ(x, t + t0 ) = t0 − H Ψ = e−iHt0 / Ψ(x, t). n! n=0 (c) Qt+t0 = Ψ(x, t + t0 )|Q(x, p, t + t0 )|Ψ(x, t + t0 ).

† But Ψ(x, t + t0 ) = e−iHt0 / Ψ(x, t), so, using the hermiticity of H to write e−iHt0 / = eiHt0 / : Qt+t0 = Ψ(x, t)|eiHt0 / Q(x, p, t + t0 )e−iHt0 / |Ψ(x, t). If t0 = dt is very small, expanding to first order, we have:

iH ∂Q iH dQ Qt + dt = Ψ(x, t)| 1 + dt Q(x, p, t) + dt 1 − dt |Ψ(x, t) dt ∂t

iH iH ∂Q i ∂Q = Q(x, p, t) + dt Q − Q dt + dt = Q + [H, Q]dt + dt ∂t ∂t i ∂Q = Qt + [H, Q]dt + dt. ∂t c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 3. FORMALISM ∴

85

dQ i ∂Q = [H, Q] + . QED dt ∂t

Problem 3.40 (a) For the free particle, V (x) = 0, so the time-dependent Schr¨ odinger equation reads ∞ 2 2 ∂Ψ 1 ∂ Ψ i . Ψ(x, t) = √ eipx/ Φ(p, t) dp ⇒ =− 2 ∂t 2m ∂x 2π −∞ ∞ ∞ 2 ∂Ψ p ∂Φ 1 1 ∂2Ψ √ − 2 eipx/ Φ dp. So eipx/ = =√ dp, 2 ∂t ∂t ∂x 2π −∞ 2π −∞ 2

∞ ∞ 1 ∂Φ p 1 √ dp = √ Φ dp. eipx/ i eipx/ ∂t 2m 2π −∞ 2π −∞ But two functions with the same Fourier transform are equal (as you can easily prove using Plancherel’s theorem), so i

p2 ∂Φ = Φ. ∂t 2m

1 ip2 dΦ = − dt Φ 2m

⇒

Φ(p, t) = e−ip

2

t/2m

Φ(p, 0).

(b) −ax2 ilx

Ψ(x, 0) = Ae Φ(p, 0) = √

e

1 2π

Φ(p, t) =

,

2a π

A=

1/4

∞

2a π

1/4 (Problem2.43(a)).

e−ipx/ e−ax eilx dx = 2

−∞

2 2 1 e−(l−p/) /4a e−ip t/2m ; 2 1/4 (2πa )

2 1 e−(l−p/) /4a (2πa2 )1/4

|Φ(p, t)|2 = √

(Problem2.43(b)).

2 1 e−(l−p/) /2a . 2πa

(c)

∞ 2 1 pe−(l−p/) /2a dp 2πa −∞ −∞ [Let y ≡ (p/) − l, so p = (y + l) and dp = dy.] ∞ 2 √ = (y + l)e−y /2a dy [but the first term is odd] 2πa −∞ ∞ 2l πa 2l −y 2 /2a = √ = l [as in Problem 2.43(d)]. e dy = √ 2 2πa 0 2πa

p =

∞

p|Φ(p, t)|2 dp = √

∞ ∞ 2 1 2 2 −(l−p/)2 /2a p = p |Φ(p, t)| dp = √ p e dp = √ (y 2 + 2yl + l2 )e−y /2a dy 2πa −∞ 2πa −∞ −∞ ∞

∞ 2 2 22 = √ y 2 e−y /2a dy + l2 e−y /2a dy 2πa 0 0 3 √ 22 a πa = √ 2 π = (a + l2 )2 [as in Problem 2.43(d)]. + l2 2 2 2πa 2

∞

2

2


86

CHAPTER 3. FORMALISM p2 2 2 1 2 a 1 2 2 a 1 2 ; H = p = (l +a) = p2 + . But H0 = p 0 = (Problem 2.22(d)). 2m 2m 2m 2m 2m 2m 2m 1 So H = p2 + H0 . QED Comment: The energy of the traveling gaussian is the energy of the 2m same gaussian at rest, plus the kinetic energy (p2 /2m) associated with the motion of the wave packet as a whole.

(d) H =


CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS

87

Chapter 4

Quantum Mechanics in Three Dimensions Problem 4.1 (a) [x, y] = xy − yx = 0, etc., so [ri , rj ] = 0. ∂ [px , py ]f = i ∂x

∂f i ∂y

∂ − i ∂y

∂f i ∂x

= −

2

∂2f ∂2f − ∂x∂y ∂y∂x

=0

(by the equality of cross-derivatives), so [pi , pj ] = 0. [x, px ]f = i

∂f ∂ ∂f ∂f x − (xf ) = x −x − f = if, ∂x ∂x i ∂x ∂x

so [x, px ] = i (likewise [y, py ] = i and [z, pz ] = i). [y, px ]f = i

∂f ∂ ∂f ∂f ∂y y − (yf ) = y −y = 0 (since = 0). So [y, px ] = 0, ∂x ∂x i ∂x ∂y ∂x

and same goes for the other “mixed” commutators. Thus [ri , pj ] = −[pj , ri ] = iδij . (b) The derivation of Eq. 3.71 (page 115) is identical in three dimensions, so

i dx = [H, x]; dt

p2 1 2 1 2 + V, x = [px + p2y + p2z , x] = [p , x] 2m 2m 2m x 1 1 = (px [px , x] + [px , x]px ) = [(−i)px + (−i)px ] = −i px . 2m 2m m

[H, x] =

dx i ∴ = dt

dr 1 1 −i px = px . The same goes for y and z, so: = p. m m dt m


88


2 dpx p ∂V i = [H, px ]; [H, px ] = + V, px = [V, px ] = i (Eq. 3.65) dt 2m ∂x 0 1 0 1 i dp ∂V ∂V = (i) = − . Same for y and z, so: = −∇V . ∂x ∂x dt

(c) From Eq. 3.62: σx σpx

1 1 ≥ [x, px ] = i = . Generally, σri σpj ≥ δij . 2i 2i 2 2

Problem 4.2 2 ∂ 2 ψ ∂ 2 ψ ∂ 2 ψ = Eψ (inside the box). Separable solutions: ψ(x, y, z) = + + 2m ∂x2 ∂y 2 ∂z 2 X(x)Y (y)Z(z). Put this in, and divide by XY Z:

(a) Equation 4.8 ⇒ −

1 d2 X 1 d2 X 1 d2 Z 2m + + = − 2 E. 2 2 X dx Y dy Z dz 2 The three terms on the left are functions of x, y, and z, respectively, so each must be a constant. Call the separation constants kx2 , ky2 , and kz2 (as we’ll soon seen, they must be positive). d2 X = −kx2 X; dx2

d2 Y = −ky2 Y ; dy 2

d2 Z = −kz2 Z, dz 2

with E =

2 2 (k + ky2 + kz2 ). 2m x

Solution: X(x) = Ax sin kx x + Bx cos kx x;

Y (y) = Ay sin ky y + By cos ky y;

Z(z) = Az sin kz z + Bz cos kz z.

But X(0) = 0, so Bx = 0; Y (0) = 0, so By = 0; Z(0) = 0, so Bz = 0. And X(a) = 0 ⇒ sin(kx a) = 0 ⇒ kx = nx π/a (nx = 1, 2, 3, . . . ). [As before (page 31), nx = 0, and negative values are redundant.] Likewise ky = ny π/a and kz = nz π/a. So

ψ(x, y, z) = Ax Ay Az sin

n π n π n π x y z x sin y sin z , a a a

E=

We might as well normalize X, Y, and Z separately: Ax = Ay = Az = 3/2 n π n π n π 2 x y z ψ(x, y, z) = sin x sin y sin z ; a a a a

E=

2 π 2 2 (n + n2y + n2z ). 2m a2 x

2/a. Conclusion:

π 2 2 2 (n + n2y + n2z ); 2ma2 x

nx , ny , nz = 1, 2, 3, . . .



89

(b) nx ny nz (n2x + n2y + n2z ) 1 1 1 3 Energy 1 1 2 1 2 1 2 1 1

6 6 6

1 2 2 2 1 2 2 2 1

9 9 9

1 1 3 1 3 1 3 1 1

11 11 11

2 2 2

12

1 1 2 2 3 3

2 3 1 3 1 2

3 2 3 1 2 1

14 14 14 14 14 14

Degeneracy

E1 = 3

π 2 2 ; 2ma2

d=1

E2 = 6

π 2 2 ; 2ma2

d = 3.

E3 = 9

π 2 2 ; 2ma2

d = 3.

E4 = 11

π 2 2 ; 2ma2

d = 3.

E5 = 12

π 2 2 ; 2ma2

d = 1.

E6 = 14

π 2 2 ; 2ma2

d = 6.

(c) The next combinations are: E7 (322), E8 (411), E9 (331), E10 (421), E11 (332), E12 (422), E13 (431), and E14 (333 and 511). The degeneracy of E14 is 4. Simple combinatorics accounts for degeneracies of 1 (nx = ny = nz ), 3 (two the same, one different), or 6 (all three different). But in the case of E14 there is a numerical “accident”: 32 + 32 + 32 = 27, but 52 + 12 + 12 is also 27, so the degeneracy is greater than combinatorial reasoning alone would suggest.

Problem 4.3 1 Eq. 4.32 ⇒ Y00 = √ P00 (cos θ); Eq. 4.27 ⇒ P00 (x) = P0 (x); Eq. 4.28 ⇒ P0 (x) = 1. 4π Y21 = −

P2 (x) =

P21 (x)

=

5 1 iφ 1 e P2 (cos θ); 4π 3 · 2

1 4·2

d dx

2

2

x2 − 1

P21 (x) =

=

1 − x2

1 Y00 = √ . 4π

d P2 (x); dx

1 2 1 2

1 d 2 2(x − 1)2x = x − 1 + x(2x) = 3x − 1 ; 8 dx 2 2

d 15 iφ 1 3 2 1 1 1 − x2 x − = 1 − x2 3x; P2 (cos θ) = 3 cos θ sin θ. Y2 = − e sin θ cos θ. dx 2 2 8π


90

CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS |Y00 |2

Normalization:

|Y21 |2 sin θ dθ dφ =

1 sin θ dθ dφ = 4π

15 8π

π

2π

sin θ dθ

dφ =

0

0

π

2π

sin2 θ cos2 θ sin θ dθ 0

dφ = 0

15 4

1 (2)(2π) = 1. 4π

π

cos2 θ(1 − cos2 θ) sin θ dθ 0

π

15 15 2 2 cos3 θ cos5 θ 5 3 = − + − = − =1 4 3 5 4 3 5 2 2 0

=

Orthogonality:

∗ Y00 Y21

1 sin θ dθ dφ = − √ 4π

15 8π

π

sin θ cos θ sin θ dθ

0

(sin3

0

θ)/3|π 0 =0

eiφ dφ = 0.

2π

(eiφ )/i|2π 0 =0

Problem 4.4 A 1 1 A dΘ A = sec2 (θ/2) = = . Therefore dθ tan(θ/2) 2 2 sin(θ/2) cos(θ/2) sin θ d With l = m = 0, Eq. 4.25 reads: dθ

dΘ sin θ dθ

d dθ

sin θ

dΘ dθ

=

d (A) = 0. dθ

= 0. So A ln[tan(θ/2)] does satisfy Eq. 4.25. However,

π Θ(0) = A ln(0) = A(−∞); Θ(π) = A ln tan = A ln(∞) = A(∞). Θ blows up at θ = 0 and at θ = π. 2

Problem 4.5 Yll

(2l + 1) 1 ilφ l e Pl (cos θ). 4π (2l)!

l

= (−1)

Pl (x) =

1 2l l!

d dx

Pll (x) = (1 − x2 )l/2

l (x2 − 1)l ,

so Pll (x) =

1 (1 − x2 )l/2 2l l!

d dx

d dx

l Pl (x).

2l (x2 − 1)l .

Now (x2 − 1)l = x2l + · · · , where all the other terms involve powers of x less than 2l, and hence give zero when differentiated 2l times. So 2l n d 1 d (2l)! Pll (x) = l (1 − x2 )l/2 x2l . But xn = n!, so Pll = l (1 − x2 )l/2 . 2 l! dx dx 2 l! ∴

Yll

l

= (−1)

(2l + 1) ilφ (2l)! 1 e (sin θ)l = l 4π(2l)! 2 l! l!

(2l + 1)! 4π

1 − eiφ sin θ 2

l .


CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS Y32

=

7 1 2iφ 2 · e P3 (cos θ); 4π 5!

P32 (x) = (1 − x2 )

d dx

91

2 P3 (x);

P3 (x) =

1 8 · 3!

d dx

3 (x2 − 1)3 .

2 1 d 2 d 1 6x(x2 − 1)2 = (x − 1)2 + 4x2 (x2 − 1) 8 · 3 · 2 dx 8 dx

1 3 1 3

1 = 4x(x2 − 1) + 8x(x2 − 1) + 4x2 · 2x = x − x + 2x3 − 2x + 2x3 = 5x − 3x . 8 2 2

P3 =

P32 (x)

1 1 − x2 = 2

Y32

=

d dx

2

d

1

1 5x3 − 3x = 1 − x2 15x2 − 3 = (1 − x2 )30x = 15x(1 − x2 ). 2 dx 2

7 1 1 15e2iφ cos θ sin2 θ = 4π 5! 4

Check that

Yll

1 l!

satisfies Eq. 4.18: Let

∂Yll = Aeilφ l(sin θ)l−1 cos θ; ∂θ

(2l + 1)! 4π

1 − 2

l ≡ A, so

Yll = A(eiφ sin θ)l .

∂Yll = l cos θYll ; ∂θ

∂ 2 Yll ∂Y l = −l2 Yll . = l cos θ sin θ l − l sin2 θYll = l2 cos2 θ − l sin2 θ Yll . ∂θ ∂φ2 So the left side of Eq. 4.18 is l2 (1 − sin2 θ) − l sin2 θ − l2 Yll = −l(l + 1) sin2 θ Yll , which matches the right side. 1 105 Check that Y32 satisfies Eq. 4.18: Let B ≡ , so Y32 = Be2iφ sin2 θ cos θ. 4 2π

∂Y32 ∂Y32 ∂ ∂ 3 2iφ 2 = Be sin θ = Be2iφ sin θ 2 sin2 θ cos2 θ − sin4 θ 2 sin θ cos θ − sin θ ; sin θ ∂θ ∂θ ∂θ ∂θ sin θ

∂ ∂θ

sin θ

105 2iφ 2 e sin θ cos θ. 2π

sin θ

∂Yll ∂θ

= Be2iφ sin θ 4 sin θ cos3 θ − 4 sin3 θ cos θ − 4 sin3 θ cos θ = 4Be2iφ sin2 θ cos θ cos2 θ − 2 sin2 θ

= 4(cos2 θ − 2 sin2 θ)Y32 .

∂ 2 Y32 = −4Y32 . So the left side of Eq. 4.18 is ∂φ2

4(cos2 θ − 2 sin2 θ − 1)Y32 = 4(−3 sin2 θ)Y32 = −l(l + 1) sin2 θ Y32 , where l = 3, so it fits the right side of Eq. 4.18.

Problem 4.6

1

1 1 Pl (x)Pl (x)dx = l l 2 l! 2 l ! −1

1

−1

d dx

l (x − 1) 2

l

d dx

l

l

(x − 1) 2

dx.


92


If l = l , we may as well let l be the larger of the two (l > l ). Integrate by parts, pulling successively each derivative off the first term onto the second: 1 1 l−1 l d d l l 2 l 2 l 2 l!2 l ! Pl (x)Pl (x)dx = (x − 1) (x − 1) dx dx −1 −1 1 l−1 l +1 d d 2 l 2 l − (x − 1) (x − 1) dx dx dx −1 l +l 1 d = . . . (boundary terms) . . . + (−1)l (x2 − 1)l (x2 − 1)l dx. dx −1 l +l

But (d/dx) (x2 − 1)l = 0, because (x2 − 1)l is a polynomial whose highest power is 2l , so more than 2l derivatives will kill it, and l + l > 2l . Now, the boundary terms are of the form:

d dx

l−n (x − 1) 2

l

d dx

l +n−1

+1 (x − 1) , n = 1, 2, 3, . . . , l. l

2

−1

Look at the first term: (x2 − 1)l = (x2 − 1)(x2 − 1) . . . (x2 − 1); l factors. So 0, 1, 2, . . . , l − 1 derivatives will still leave at least one overall factor of (x2 − 1). [Zero derivatives leaves l factors; one derivative leaves l − 1 : d/dx(x2 −1)l = 2lx(x2 −1)l−1 ; two derivatives leaves l−2 : d2 /dx2 (x2 −1)l = 2l(x2 −1)l−1 +2l(l−1)2x2 (x2 −1)l−2 , 1 and so on.] So the boundary terms are all zero, and hence −1 Pl (x)Pl (x)dx = 0. This leaves only the case l = l . Again the boundary terms vanish, but this time the remaining integral does not: 2l 1 1 d (2l l!)2 [Pl (x)]2 dx = (−1)l (x2 − 1)l (x2 − 1)l dx dx −1 −1

(d/dx)2l (x2l )=(2l)!

= (−1)l (2l)! −1

Let x ≡ cos θ, so dx = − sin θ dθ,

1

(x2 − 1)l dx = 2(2l)!

(sin θ)2l (− sin θ)dθ =

π/2

1

(1 − x2 )l dx. 0

(1 − x2 ) = sin2 θ,

0

(1 − x2 )l dx = 0

1

θ : π/2 → 0. Then

π/2

(sin θ)2l+1 dθ 0

(2)(4) · · · (2l) (2l l!)2 (2l l!)2 = = = . (1)(3)(5) · · · (2l + 1) 1 · 2 · 3 · · · · (2l + 1) (2l + 1)! ∴

1

−1

[Pl (x)]2 dx =

1 (2l l!)2 2 2(2l)! = . So l 2 (2 l!) (2l + 1)! 2l + 1

1

−1

Pl (x)Pl (x)dx =

2 δll . QED 2l + 1

Problem 4.7 (a) cos x sin x 1 d cos x n1 (x) = −(−x) =− 2 − . x dx x x x c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


93

2 1 d 1 d cos x 1 d cos x n2 (x) = −(−x)2 = −x2 x dx x x dx x dx x d 1 −x sin x − cos x d sin x cos x = −x =x + 3 · dx x x2 dx x2 x 2 x cos x − 2x sin x −x3 sin x − 3x2 cos x =x + x4 x6 cos x sin x sin x 3 cos x 3 3 1 = −2 2 − 2 − cos x − 2 sin x. = − 3− 3 x x x x x x x (b) Letting sin x ≈ x and cos x ≈ 1, and keeping only the lowest power of x: n1 (x) ≈ −

1 1 1 + x ≈ − 2. 2 x x x

As x → 0, this blows up.

3 3 1 3 n2 (x) ≈ − 3 − − 2 x ≈ − 3 , which again blows up at the origin. x x x x

Problem 4.8 (a) u = Arj1 (kr) = A

sin(kr) cos(kr) A sin(kr) − = − cos(kr) . k2 r k k (kr)

du cos(kr) sin(kr) A k 2 r cos(kr) − k sin(kr) + k sin(kr) = A + sin(kr) . = − dr k (kr)2 kr (kr)2 2

d2 u −k r sin(kr) − k cos(kr) k 3 r2 cos(kr) − 2k 2 r sin(kr) = A − + k cos(kr) dr2 (kr)2 (kr)4

sin(kr) cos(kr) cos(kr) sin(kr) = Ak − − + 2 + cos(kr) − (kr) (kr)2 (kr)2 (kr)3

2 2 1 = Ak 1 − cos(kr) + − sin(kr) . (kr)2 (kr)3 (kr)

With V = 0 and l = 1, Eq. 4.37 reads: Ak 1 −

2 (kr)2

cos(kr) +

d2 u 2 2mE − 2 u = − 2 u = −k 2 u. In this case the left side is dr2 r

2 1 − (kr)3 (kr)

sin(kr) = Ak cos(kr) − = −k 2 u. kr

2 sin(kr) − (kr)2

sin(kr) − cos(kr) (kr)

So this u does satisfy Eq. 4.37.


94


z cos z (b) Equation 4.48 ⇒ j1 (z) = 0, where z = ka. Thus sin = 0, or tan z = z. For high z (large n, z2 − z if n = 1, 2, 3, . . . counts the allowed energies in increasing order), the intersections occur slightly below z = (n + 12 )π.

2 k 2 2 z 2 2 π 2 ∴E= = = 2m 2ma2 2ma2

1 n+ 2

2 . QED

z

tan z

π/2

Problem 4.9 For r ≤ a, u(r) = A sin(kr), with k ≡ state) E < 0 ⇒:

3π/2

5π/2

z

2m(E + V0 )/. For r ≥ a, Eq. 4.37 with l = 0, V = 0, and (for a bound

√ d2 u 2m = − 2 Eu = κ2 u, with κ ≡ −2mE/ ⇒ u(r) = Ceκr + De−κr . 2 dr κr But the Ce term blows up as r → ∞, so u(r) = De−κr . 1 1 κ Continuity of u at r = a : A sin(ka) = De−κa divide: tan(ka) = − , or − cot ka = . Continuity of u at r = a : Ak cos(ka) = −Dκe−κa k κ k √ 2mV0 a2 /2 − z 2 κ 2mV0 Let ka ≡ z; = . Let z0 ≡ a. − cot z = (z0 /z)2 − 1. This is exactly the k z same transcendental equation we encountered in Problem 2.29—see graph there. There is no solution if z0 < π/2, which is to say, if 2mV0 a2 /2 < π 2 /4, or V0 a2 < π 2 2 /8m. Otherwise, the ground state energy occurs somewhere between z = π/2 and z = π: E + V0 =

2 π 2 2 k 2 a2 2 2 2 π 2 = z , so < (E0 + V0 ) < 2 2 2 2ma 2ma 8ma 2ma2

Problem 4.10 R30 (n = 3, l = 0) : Eq. 4.62 ⇒ v(ρ) = Eq. 4.76 ⇒ c1 =

/

j=0 cj ρ

2(1 − 3) c0 = −2c0 ; (1)(2)

r Eq. 4.73 ⇒ ρ = ; 3a

Eq. 4.75 ⇒ R30

j

c2 =

(precise value depends on V0 ).

. 2(2 − 3) 1 2 c1 = − c1 = c0 ; (2)(3) 3 3

c3 =

2(3 − 3) c2 = 0. (3)(4)

r 2 r 2 1 −ρ 1 r −r/3a c0 − 2c0 e + c0 = ρe v(ρ) = r r 3a 3a 3 3a



R30 =

c 2 r 2 r 2 −r/3a 0 e 1− + . 3a 3 a 27 a 2(2 − 3) 1 c0 = − c0 ; (1)(4) 2

R31 (n = 3, l = 1) : c1 =

R31

95

c2 =

2(3 − 3) c1 = 0. (2)(5)

c 1 r −r/3a 1 r 2 −r/3a 1 r 0 c0 − c0 r 1− = e . = e r 3a 2 3a 9a2 6 a 2(3 − 3) c0 = 0. (1)(6)

R32 (n = 3, l = 2) : c1 =

R32 =

c 1 r 3 −r/3a 0 r2 e−r/3a . e (c0 ) = r 3a 27a3

Problem 4.11 (a)

∞

Eq. 4.31 ⇒

c 0

|R|2 r2 dr = 1. Eq. 4.82 ⇒ R20 =

2a

0

1=

c 2 0

2a

∞

a

0

∴ c0 =

3

2 . a

z 2 −z 2 c2 a 1− e z dz = 0 2 4

∞

r −r/2a . e 2a

Let z ≡

r . a

1 4 −z c20 a 24 a z − z + z e dz = 2−6+ = c20 . 4 4 4 2 2

0

1−

3

1 Eq. 4.15 ⇒ ψ200 = R20 Y00 . Table 4.3 ⇒ Y00 = √ . 4π

1 ∴ ψ200 = √ 4π

2 1 r −r/2a r −r/2a 1 1 1− e 1− e ⇒ ψ200 = √ . a 2a 2a 2a 2πa 2a

(b) R21

R21

c0 = 2 re−r/2a ; 4a

1 1 =√ re−r/2a ; 6a 2a2

ψ210

c 2 ∞ 3 2 2 c20 a 0 5 4 −z 24 = ac0 , so c0 = . 1= a z e dz = 4a2 16 2 3a 0

ψ21±1

1 1 =√ re−r/2a 6a 2a2

1 1 =√ re−r/2a 6a 2a2

'

' ∓

3 cos θ 4π

3 sin θe±iφ 8π

( = √

(

1 1 re−r/2a sin θe±iφ ; = ∓√ πa 8a2

1 1 re−r/2a cos θ. 2πa 4a2


96


Problem 4.12 (a) L0 = ex e−x = 1.

2

L1 = ex

d −x

e x = ex e−x − e−x x = 1 − x. dx

d e−x x2 = ex 2xe−x − e−x x2 dx −x

x −x −x 2 = e 2e − 2xe + e x − 2xe−x = 2 − 4x + x2 .

L2 = ex

d dx

2 3 −x 3

−x 3

d d e x = ex −e x + 3x2 e−x dx dx

−x 3 x d 2 −x =e e x − 3x e − 3x2 e−x + 6xe−x dx

= ex −e−x x3 + 3x2 e−x + 6x2 e−x − 12xe−x − 6xe−x + 6e−x

L3 = ex

= 6 − 18x + 9x2 − x3 . (b) v(ρ) = L52 (2ρ);

L52 (x) = L57−5 (x) = (−1)5

d dx

5 L7 (x).

6 7 7 −x

6 −x

d d x L7 (x) = e x e =e 7x e − x7 e−x dx dx 5

d x =e 42x5 e−x − 7x6 e−x − 7x6 e−x + x7 e−x dx 4

d = ex 210x4 e−x − 42x5 e−x − 84x5 e−x + 14x6 e−x + 7x6 e−x − x7 e−x dx 3 d = ex 840x3 e−x − (210 + 630)x4 e−x dx

x

+ (126 + 126)x5 e−x − (21 + 7)x6 e−x + x7 e−x 2 d x =e 2520x2 e−x − (840 + 3360)x3 e−x dx

+(840 + 1260)x4 e−x − (252 + 168)x5 e−x + (28 + 7)x6 e−x − x7 e−x d = ex 5040xe−x − (2520 + 12600)x2 e−x + (4200 + 8400)x3 e−x dx

− (2100 + 2100)x4 e−x + (420 + 210)x5 e−x − (35 + 7)x6 e−x + x7 e−x x = e 5040e−x − (5040 + 30240)xe−x + (15120 + 37800)x2 e−x − (12600 + 8400 + 8400)x3 e−x + (2100 + 2100 + 3150)x4 e−x

− (630 + 252)x5 e−x + (42 + 7)x6 e−x − x7 e−x = 5040 − 35280x + 52920x2 − 29400x3 + 7350x4 − 882x5 + 49x6 − x7 .


CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS L52 = −

d dx

5

−882x5 + 49x6 − x7

97

= − −882(5 · 4 · 3 · 2) + 49(6 · 5 · 4 · 3 · 2)x − 7 · 6 · 5 · 4 · 3x2 = 60 (882 × 2) − (49 × 12)x + 42x2 = 2520(42 − 14x + x2 ).

v(ρ) = 2520(42 − 28ρ + 4ρ2 ) = 5040 21 − 14ρ + 2ρ2 . (c) Eq. 4.62 ⇒ v(ρ) =

∞

cj ρj . Eq. 4.76 ⇒ c1 =

j=0

c2 =

2(4 − 5) 1 2 c1 = − c1 = c0 ; (2)(7) 7 21

c3 =

2(3 − 5) 2 c0 = − c0 . (1)(6) 3

2(5 − 5) c2 = 0. (3)(8)

c0 2 2 v(ρ) = c0 − c0 ρ + c0 ρ2 = 21 − 14ρ + 2ρ2 . 3 21 21

Problem 4.13 (a) ψ=√

r =

1

e−r/a ,

πa3

4 a3

∞

so rn =

r3 e−2r/a dr =

0

1 πa3

4π rn e−2r/a r2 sin θ dr dθ dφ = πa3

3 4 a 4 3! = a; 3 a 2 2

r2 =

4 a3

∞

∞

rn+2 e−2r/a dr.

0

r4 e−2r/a dr =

0

4 a 5 4! = 3a2 . a3 2

(b) x = 0;

x2 =

1 2 r = a2 . 3

(c) 1 1 re−r/2a sin θeiφ ψ211 = R21 Y11 = − √ πa 8a2

(Problem 4.11(b)).

1 1 2 2 −r/a x = r e sin θ r2 sin2 θ cos2 φ r2 sin θ dr dθ dφ 2 2 πa (8a ) ∞ π 2π 1 5 6 −r/a = r e dr sin θ dθ cos2 φ dφ 64πa5 0 0 0 1 7

2·4 1 = 6!a 2 · 2π = 12a2 . 64πa5 1·3·5 2 2


98


Problem 4.14 4 4 P = |ψ|2 4πr2 dr = 3 e−2r/a r2 dr = p(r) dr; p(r) = 3 r2 e−2r/a . a a dp 2 −2r/a 8r −2r/a r 4 −2r/a 2 = 3e 1− +r − e = 3 2re = 0 ⇒ r = a. dr a a a a

ψ=√

1

πa3

e−r/a ;

Problem 4.15 1 1 (a) Ψ(r, t) = √ ψ2 1 1 e−iE2 t/ + ψ2 1 −1 e−iE2 t/ = √ (ψ2 1 1 + ψ2 1 −1 ) e−iE2 t/ ; 2 2

E2 =

E1 2 . =− 4 8ma2

From Problem 4.11(b):

i 1 1 ψ2 1 1 + ψ2 1 −1 = − √ re−r/2a sin θ eiφ − e−iφ = − √ re−r/2a sin θ sin φ. 2 πa 8a πa 4a2 Ψ(r, t) = − √

i re−r/2a sin θ sin φ e−iE2 t/ . 2πa 4a2

(b) 1 e2 1 1 e2 |Ψ|2 − r2 e−r/a sin2 θ sin2 φ d3 r = − r2 sin θ dr dθ dφ 4 4π90 r (2πa)(16a ) 4π90 r ∞ π 2π 4 4 1 2 2 3 2 3 −r/a = (π) − r e dr sin θ dθ sin φ dφ = − 3!a 32πa5 ma2 32πma6 3 0 0 0

V =

= −

2 1 1 = E1 = (−13.6eV) = −6.8eV 4ma2 2 2

(independent of t).

Problem 4.16 En (Z) = Z 2 En ;

E1 (Z) = Z 2 E1 ;

a(Z) = a/Z;

R(Z) = Z 2 R.

Lyman lines range from from ni = 2 to ni = ∞ (with nf = 1);the wavelengths range 3 1 1 1 4 1 1 = R ⇒ λ2 = down to = R ⇒ λ1 = . =R 1− =R 1− λ2 4 4 3R λ1 ∞ R 1 1 1 −8 For Z = 2 : λ1 = = = 2.28 × 10 m to λ2 = = 3.04 × 10−8 m, ultraviolet. 4R 4(1.097 × 107 ) 3R 1 4 For Z = 3 : λ1 = = 1.01 × 10−8 m to λ2 = = 1.35 × 10−8 m, also ultraviolet. 9R 27R

Problem 4.17 (a) V (r) = −G

Mm . r

So

e2 → GM m translates hydrogen results to the gravitational analogs. 4π90


CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS (b) Equation 4.72: a = =

(6.6726 × 10−11

4π90 e2

2 , m

so

ag =

99

2 GM m2

(1.0546 × 10−34 Js)2 = 2.34 × 10−138 m. m3 /kg · s2 )(1.9892 × 1030 kg)(5.98 × 1024 kg)2

(c) Equation 4.70 ⇒ En = −

1 m (GM m)2 2 . 2 2 n

1 Mm Mm mv 2 1 GM m mv 2 − G . But G 2 = ⇒ mv 2 = , so 2 ro ro ro 2 2ro 1 m ro GM m GM m2 ro 2 2 Ec = − =− (GM m) ⇒ n = r = ⇒ n = . o 2ro 22 n2 2 ag ag 1.496 × 1011 11 ro = earth-sun distance = 1.496 × 10 m ⇒ n = = 2.53 × 1074 . 2.34 × 10−138 Ec =

(d)

∆E = − So

G 2 M 2 m3 22

1 1 . − (n + 1)2 n2

1 1 1 = 2 ≈ 2 2 2 (n + 1) n (1 + 1/n) n

1 1 2 1 2 ≈ 1 − − − 1 = − 3; 2 2 2 (n + 1) n n n n

∆E =

∆E =

1−

2 n

.

G 2 M 2 m3 . 2 n3

(6.67 × 10−11 )2 (1.99 × 1030 )2 (5.98 × 1024 )3 = 2.09 × 10−41 J. (1.055 × 10−34 )2 (2.53×74 )3

Ep = ∆E = hν =

hc . λ

λ = (3 × 108 )(6.63 × 10−34 )/(2.09 × 10−41 ) = 9.52 × 1015 m. But 1 ly = 9.46 × 1015 m. Is it a coincidence that λ ≈ 1 ly? No: From part (c), n2 = GM m2 ro /2 , so ' ( 3/2 GM m2 ro ro3 ch 2π3 2 n 3 λ= = c 2π = c2π 2 2 3 = c 2 2 3 . ∆E G M m G M m 2 GM But (from (c)) v = GM/ro = 2πro /T , where T is the period of the orbit (in this case one year), so T = 2π ro3 /GM , and hence λ = cT (one light year). [Incidentally, the same goes for hydrogen: The wavelength of the photon emitted in a transition from a highly excited state to the next lower one is equal to the distance light would travel in one orbital period.]

Problem 4.18 f |L± g = f |Lx g ± if |Ly g = Lx f |g ± iLy f |g = (Lx ∓ iLy )f |g = L∓ f |g, so (L± )† = L∓ . Now, using Eq. 4.112, in the form L∓ L± = L2 − L2z ∓ Lz : flm |L∓ L± flm = flm |(L2 − L2z ∓ Lz )flm = flm | 2 l(l + 1) − 2 m2 ∓ 2 m flm = 2 [l(l + 1) − m(m ± 1)] flm |flm = 2 [l(l + 1) − m(m ± 1)] m±1 m±1 2 m±1 m±1 2 |Am = |Am |fl = |Am = L± flm |L± flm = Am l fl l fl l | fl l | .


100


Conclusion:

Am l =

l(l + 1) − m(m ± 1).

Problem 4.19 (a) [Lz , x] = [xpy − ypx , x] = [xpy , x] − [ypx , x] = 0 − y[px , x] = iy. [Lz , y] = [xpy − ypx , y] = [xpy , y] − [ypx , y] = x[py , y] − 0 = −ix. [Lz , z] = [xpy − ypx , z] = [xpy , z] − [ypx , z] = 0 − 0 = 0. [Lz , px ] = [xpy − ypx , px ] = [xpy , px ] − [ypx , px ] = py [x, px ] − 0 = ipy . [Lz , py ] = [xpy − ypx , py ] = [xpy , py ] − [ypx , py ] = 0 − px [y, py ] = −ipx . [Lz , pz ] = [xpy − ypx , pz ] = [xpy , pz ] − [ypx , pz ] = 0 − 0 = 0. (b) [Lz , Lx ] = [Lz , ypz − zpy ] = [Lz , ypz ] − [Lz , zpy ] = [Lz , y]pz − [Lz , py ]z = −ixpz + ipx z = i(zpx − xpz ) = iLy .

(So, by cyclic permutation of the indices, [Lx , Ly ] = iLz .) (c) [Lz , r2 ] = [Lz , x2 ] + [Lz , y 2 ] + [Lz , z 2 ] = [Lz , x]x + x[Lz , x] + [Lz , y]y + y[Lz , y] + 0 = iyx + xiy + (−ix)y + y(−ix) = 0.

[Lz , p2 ] = [Lz , p2x ] + [Lz , p2y ] + [Lz , p2z ] = [Lz , px ]px + px [Lz , px ] + [Lz , py ]py + py [Lz , py ] + 0 = ipy px + px ipy + (−ipx )py + py (−ipx ) = 0. (d) It follows from (c) that all three components of L commute with r2 and p2 , and hence with the whole √ 2 2 Hamiltonian, since H = p /2m + V ( r ). QED



101

Problem 4.20 (a) dLx i 1 2 = [H, Lx ]. [H, Lx ] = [p , Lx ] + [V, Lx ]. dt 2m The first term is zero (Problem 4.19(c)); the second would be too if V were a function only of r = |r|, but in general Equation 3.71 ⇒

[H, Lx ] = [V, ypz − zpy ] = y[V, pz ] − z[V, py ]. Now (Problem 3.13(c)): [V, pz ] = i Thus

∂V ∂V ∂V ∂V and [V, py ] = i . So [H, Lx ] = yi − zi = i[r × (∇V )]x . ∂z ∂y ∂z ∂y

dLx = −[r × (∇V )]x , and the same goes for the other two components: dt dL = [r × (−∇V )] = N. QED dt

(b) If V (r) = V (r), then ∇V =

∂V dL rˆ, and r × rˆ = 0, so = 0. QED ∂r dt

Problem 4.21 (a)

∂ ∂f ∂ ∂f + i cot θ e−iφ − i cot θ ∂θ ∂φ ∂θ ∂φ 2 2 ∂ f 2 iφ −iφ ∂ f 2 ∂f e − e + cot θ − i − csc θ ∂θ2 ∂φ ∂θ ∂φ 2 ∂f ∂ f ∂f ∂2f −iφ −iφ +i cot θ −ie − i cot θ +e − i cot θ 2 ∂θ ∂φ ∂φ ∂θ ∂φ 2 2 2 ∂ f ∂ f ∂f ∂2f 2 2 ∂f 2 ∂f 2 ∂ f − − i cot θ + cot θ − i cot θ + i cot θ + cot θ 2 + i csc θ ∂θ2 ∂φ ∂θ ∂φ ∂θ ∂φ ∂φ ∂θ ∂φ 2

2 ∂ ∂ ∂ ∂ −2 + cot θ + cot2 θ 2 + i(csc2 θ − cot2 θ) f, so ∂θ2 ∂θ ∂φ ∂φ 2 ∂ ∂ ∂2 ∂ 2 −2 + cot θ θ +i + cot . QED 2 2 ∂θ ∂θ ∂φ ∂φ

L+ L− f = −2 eiφ =

= = L+ L− =

∂ , Eq. 4.112 ⇒ L2 = L+ L− + L2z − Lz , so, using (a): i ∂φ 2 2 ∂ ∂ ∂2 ∂ ∂ 2 2 2 2 ∂ L = − + cot θ 2 + i − + cot θ − ∂θ2 ∂θ ∂φ ∂φ ∂φ2 i ∂φ 2 2 2 ∂ ∂ ∂ ∂ ∂ ∂ ∂ 1 ∂2 2 2 2 = − + cot θ + cot θ + (cot θ + 1) 2 + i −i = − + ∂θ2 ∂θ ∂φ ∂φ ∂φ ∂θ2 ∂θ sin2 θ ∂φ2

1 ∂ ∂ 1 ∂2 = −2 . QED sin θ + sin θ ∂θ ∂θ sin2 θ ∂φ2

(b) Equation 4.129 ⇒ Lz =


102


Problem 4.22 (a) L+ Yll = 0 (top of the ladder). (b) Lz Yll = lYll ⇒

∂ l Y = lYll , i ∂φ l

∂Yll = ilYll , and hence Yll = f (θ)eilφ . ∂φ

so

[Note: f (θ) is the “constant” here—it’s constant with respect to φ . . . but still can depend on θ.] ∂ df ilφ ∂ l iφ L+ Yl = 0 ⇒ e + i cot θ f (θ)eilφ = 0, or e + if cot θ il eilφ = 0, so ∂θ ∂φ dθ

df df = l cot θf ⇒ = l cot θdθ ⇒ dθ f ln f = ln(sinl θ) + K ⇒ ln

f sinl θ

df =l f

=K⇒

cos θ dθ ⇒ ln f = l ln(sin θ) + constant. sin θ

f = constant ⇒ f (θ) = A sinl θ. sinl θ

Yll (θ, φ) = A(eiφ sin θ)l . (c) 1=A

2l

2

sin θ sin θ dθ dφ = 2πA

π

sin(2l+1) θ dθ = 2πA2 2

2 0

= 4πA2

(2 · 4 · 6 · · · · · 2l)2 (2l l!)2 = 4πA2 , 1 · 2 · 3 · 4 · 5 · · · · · (2l + 1) (2l + 1)!

so

(2 · 4 · 6 · · · · · (2l)) 1 · 3 · 5 · · · · · (2l + 1) (2l + 1)! 1 A = l+1 , 2 l! π

the same as Problem 4.5, except for an overall factor of (−1)l , which is arbitrary anyway.

Problem 4.23 ∂ ∂ 15 iφ + i cot θ − sin θ cos θe = e ∂θ ∂θ 8π

15 iφ iφ cos θ 2 2 iφ =− e (cos θ − sin θ) + i e sin θ cos θ ie 8π sin θ

2 15 2iφ 2 15 iφ =− e e sin θ cos θ − sin2 θ − cos2 θ = 8π 8π √

2 1 15 iφ e sin θ . = 2 · 3 − 1 · 2 Y22 = 2Y22 . ∴ Y22 = 4 2π

L+ Y21

iφ



103

Problem 4.24 (a)

H=2

1 mv 2 2

= mv 2 ;

a |L| = 2 mv = amv, 2

so L2 = a2 m2 v 2 ,

and hence H =

L2 . ma2

But we know the eigenvalues of L2 : 2 l(l + 1); or, since we usually label energies with n: En =

2 n(n + 1) (n = 0, 1, 2, . . . ). ma2

(b) ψnm (θ, φ) = Ynm (θ, φ), the ordinary spherical harmonics. The degeneracy of the nth energy level is the number of m-values for given n: 2n + 1.

Problem 4.25 (1.6 × 10−19 )2 = 2.81 × 10−15 m. 4π(8.85 × 10−12 )(9.11 × 10−31 )(3.0 × 108 )2 1 2 2 v 2 L = = Iω = mr = mrv so 2 5 r 5

rc =

v=

5 (5)(1.055 × 10−34 ) = = 5.15 × 1010 m/s. 4mr (4)(9.11 × 10−31 )(2.81 × 10−15 )

Since the speed of light is 3 × 108 m/s, a point on the equator would be going more than 100 times the speed of light. Nope : This doesn’t look like a very realistic model for spin.

Problem 4.26 (a)

2 0 1 0 −i 0 −i 0 1 [Sx , Sy ] = Sx Sy − Sy Sx = − 1 0 i 0 i 0 1 0 4 2 2 2i 0 1 0 i 0 −i 0 = − = = i = iSz . 0 −i 0 i 4 4 0 −2i 2 0 −1

(b)

σx σ x = σx σy = σy σ x =

10 01

= 1 = σy σy = σz σz ,

i 0 = iσz ; 0 −i

σy σz =

−i 0 0 i

so σj σj = 1 for j = x, y, or z.

0 i i 0

= −iσz ;

σz σy =

= iσx ;

σz σx =

0 −i = −iσx ; −i 0

0 1 −1 0

σx σz =

= iσy ; 0 −1 1 0

= −iσy .

Equation 4.153 packages all this in a single formula.


104


Problem 4.27 (a) χ† χ = |A|2 (9 + 16) = 25|A|2 = 1 ⇒ A = 1/5. (b) Sx = χ† Sx χ =

0 1

4 1 3i −3i 4 −3i 4 = = (12i + 12i) = 0. 1 0 4 3i 25 2 50 50

Sy = χ† Sy χ =

0 −i

−4i 1 12 3i −3i 4 −3i 4 = = (−12 − 12) = − . i 0 4 −3 25 2 50 50 25

1 0

3i 1 7 3i −3i 4 −3i 4 = = (9 − 16) = − . Sz = χ Sz χ = 0 −1 4 −4 25 2 50 50 50 †

(c) Sx2 = Sy2 = Sz2 =

2 2 (always, for spin 1/2), so σS2 x = Sx2 − Sx 2 = − 0, σSx = . 4 4 2

σS2 y = Sy2 − Sy 2 =

− 4

σS2 z

=

Sz2

2 − Sz = − 4 2

12 25

2

7 50

2 =

2 7 49 2 (625 − 576) = , σSy = . 2500 2500 50

2 2 =

2 12 576 2 (625 − 49) = , σSz = . 2500 2500 25

(d) σSx σSy =

7 ? 7 · ≥ |Sz | = · 2 50 2 2 50

σSy σSz =

7 12 ? · ≥ |Sx | = 0 50 25 2

σSz σSx =

12 12 ? · ≥ |Sy | = · 25 2 2 2 25

(right at the uncertainty limit).

(trivial).

(right at the uncertainty limit).



105

Problem 4.28 ∗ ∗ b ∗ ∗ 0 1 a a b a b = = (a∗ b + b∗ a) = Re(ab∗ ). Sx = 1 0 b a 2 2 2 Sy = =

Sz =

∗ ∗ 0 −i ∗ ∗ −ib a a b a b = i 0 b ia 2 2 (−ia∗ b + iab∗ ) = i(ab∗ − a∗ b) = − Im(ab∗ ). 2 2 ∗ ∗ 1 0 ∗ ∗ a a a b a b = = (a∗ a − b∗ b) = (|a|2 − |b|2 ). 0 −1 b −b 2 2 2 2

S2x

2 = 4

S2z

2 = 4

01 10

2 1 0 2 01 = = ; 10 4 01 4

1 0 0 −1

1 0 0 −1

Sx2 + Sy2 + Sz2 =

=

2 ; 4

S2y

2 = 4

2 0 −i 0 −i = ; i 0 i 0 4

so Sx2 = Sy2 = Sz2 =

2 . 4

3 2 ? 1 1 3 = s(s + 1)2 = ( + 1)2 = 2 = S 2 . 4 2 2 4

Problem 4.29 (a) Sy = 2

2

0 −i ; i 0

−λ −i/2 2 2 i/2 −λ = λ − 4 ⇒ λ = ± 2 (of course).

α 0 −i α =± ⇒ −iβ = ±α; i 0 β 2 β

(y) χ+

1 =√ 2

1 ; i

(y) χ−

1 =√ 2

1 |α|2 + |β|2 = 1 ⇒ |α|2 + |α|2 = 1 ⇒ α = √ . 2

1 . −i

(b) †

a 1 1 (y) c+ = χ+ χ = √ 1 −i = √ (a − ib); b 2 2 † 1 1 a (y) = √ (a + ib); c− = χ− χ= √ 1 i b 2 2

1 + , with probability |a − ib|2 . 2 2 1 − , with probability |a + ib|2 . 2 2

1 ∗ [(a + ib∗ )(a − ib) + (a∗ − ib∗ )(a + ib)] 2 1 2 = |a| − ia∗ b + iab∗ + |b|2 + |a|2 + ia∗ b − iab∗ + |b|2 = |a|2 + |b|2 = 1. 2

P+ + P − =


106

(c)

CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 2 , with probability 1. 4

Problem 4.30 Sr = S · rˆ = Sx sin θ cos φ + Sy sin θ sin φ + Sz cos θ 0 sin θ cos φ 0 −i sin θ sin φ cos θ 0 = + + sin θ cos φ 0 i sin θ sin φ 0 0 − cos θ 2 cos θ sin θ(cos φ − i sin φ) cos θ e−iφ sin θ = . = − cos θ 2 sin θ(cos φ + i sin φ) 2 eiφ sin θ − cos θ ( cos θ − λ) e−iφ sin θ 2 2 2 2 2 2 θ + λ − cos sin2 θ = 0 ⇒ = − eiφ sin θ (− cos θ − λ) 4 4 2 2

λ2 =

2

2 2 (sin2 θ + cos2 θ) = ⇒ λ = ± (of course). 4 4 2

cos θ e−iφ sin θ iφ e sin θ − cos θ

α α ⇒ α cos θ + βe−iφ sin θ = ±α; =± β 2 β

β = eiφ

(±1 − cos θ) α. sin θ

sin(θ/2) α. Normalizing: cos(θ/2) 1 sin2 (θ/2) 2 θ θ cos(θ/2) (r) iφ 1 = |α|2 +|β|2 = |α2 |+ 2 sin χ = . |α| = |α|2 ⇒ α = cos , β = e , + eiφ sin(θ/2) cos (θ/2) cos2 (θ/2) 2 2

Upper sign: Use 1 − cos θ = 2 sin2 θ2 , sin θ = 2 sin θ2 cos θ2 . Then β = eiφ

θ cos(θ/2) 1 cos2 (θ/2) 2 Lower sign: Use 1 + cos θ = 2 cos2 , β = −eiφ α; 1 = |α|2 + |α| = |α|2 2 . 2 sin(θ/2) sin2 (θ/2) sin (θ/2) −iφ sin(θ/2) e (r) −iφ Pick α = e . sin(θ/2); then β = − cos(θ/2), and χ− = − cos(θ/2)

Problem 4.31       1 0 0 There are three states: χ+ = 0 , χ0 = 1 , χ− = 0 . 0 0 1 

Sz χ+ = χ+ , Sz χ0 = 0, Sz χ− = −χ− ,

 1 0 0 ⇒ Sz = 0 0 0  . 0 0 −1

From Eq. 4.136:



107

    √ √ 0 1 0 0 0 0 √ √ S+ χ+ = 0,√ S+ χ0 = √2χ+ , S+ χ− = 2χ0 ⇒ S+ = 2 0 0 1 , S− = 2 1 0 0 . S− χ+ = 2χ0 , S− χ0 = 2χ− , S− χ− = 0 0 0 0 0 1 0   010  1 1 0 1 , Sx = (S+ + S− ) = √ 2 2 010

  0 −1 0 i  1 1 0 −1 . Sy = (S+ − S− ) = √ 2i 2 0 1 0

Problem 4.32 (a) Using Eqs. 4.151 and 4.163: 1 1 cos α2 eiγB0 t/2 α −iγB0 t/2 α iγB0 t/2 (x) (x)† √ c+ = χ+ χ = √ 1 1 = . + sin e e cos sin α2 e−iγB0 t/2 2 2 2 2 α α 1 α α (x) (x) P+ (t) = |c+ |2 = cos e−iγB0 t/2 + sin eiγB0 t/2 cos eiγB0 t/2 + sin e−iγB0 t/2 2 2 2 2 2

1 2α α α iγB0 t 2 α −iγB0 t = cos + sin + sin cos e +e 2 2 2 2 2 1 1 α α = 1 + 2 sin cos cos(γB0 t) = [1 + sin α cos(γB0 t)] . 2 2 2 2 1 1 (y) (b) From Problem 4.29(a): χ+ = √ . 2 i

cos α eiγB0 t/2 1 1 α α (y) (y)† 2 c+ = χ+ χ = √ 1 −i = √ cos eiγB0 t/2 − i sin e−iγB0 t/2 ; α iγB0 t/2 sin 2 e 2 2 2 2 α 1 α α α (y) (y) P+ (t) = |c+ |2 = cos e−iγB0 t/2 + i sin eiγB0 t/2 cos eiγB0 t/2 − i sin e−iγB0 t/2 2 2 2 2 2

1 2α α α iγB0 t 2 α −iγB0 t = −e cos + sin + i sin cos e 2 2 2 2 2 1 1 α α = 1 − 2 sin cos sin(γB0 t) = [1 − sin α sin(γB0 t)] . 2 2 2 2 (c) (z)

χ+ =

1 ; 0

(z) c+ = 1 0

cos α2 eiγB0 t/2 sin α2 e−iγB0 t/2

= cos

α iγB0 t/2 ; e 2

(z)

(z)

P+ (t) = |c+ |2 = cos2

α . 2

Problem 4.33 (a) γB0 1 0 H = −γB · S = −γB0 cos ωt Sz = − cos ωt . 0 −1 2 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

108


(b) χ(t) =

i

1 α(t) , with α(0) = β(0) = √ . β(t) 2

∂χ γB0 γB0 α˙ 1 0 α α = i ˙ = Hχ = − cos ωt =− cos ωt . 0 −1 β −β β ∂t 2 2

α˙ = i

γB0 2

cos ωt α ⇒

α(t) = Aei(γB0 /2ω) sin ωt ; β˙ = −i

γB0 2

dα =i α

γB0 2

cos ωt dt ⇒ ln α =

1 α(0) = A = √ , 2

iγB0 sin ωt + constant. 2 ω

1 so α(t) = √ ei(γB0 /2ω) sin ωt . 2

1 cos ωt β ⇒ β(t) = √ e−i(γB0 /2ω) sin ωt . 2

1 χ(t) = √ 2

ei(γB0 /2ω) sin ωt . e−i(γB0 /2ω) sin ωt

(c) (x)

i(γB /2ω) sin ωt 0 1 i(γB0 /2ω) sin ωt 1 e − e−i(γB0 /2ω) sin ωt (1 − 1) −i(γB0 /2ω) sin ωt = e e 2 2

γB0 γB0 (x) (x) = i sin sin ωt . P− (t) = |c− |2 = sin2 sin ωt . 2ω 2ω (x)†

c− = χ− χ =

(d) The argument of sin2 must reach π/2 (so P = 1) ⇒

γB0 πω π = , or B0 = . 2ω 2 γ

Problem 4.34 (a) 1 (1) (2) 1 S− |1 0 = (S− + S− ) √ (↑↓ + ↓↑) = √ [(S− ↑) ↓ +(S− ↓) ↑ + ↑ (S− ↓)+ ↓ (S− ↑)] . 2 2 √ √ 1 But S− ↑= ↓, S− ↓= 0 (Eq. 4.143), so S− |10 = √ [ ↓↓ +0 + 0 + ↓↓] = 2 ↓↓= 2|1 − 1. 2 (b) 1 (1) (2) 1 S± |0 0 = (S± + S± ) √ (↑↓ − ↓↑) = √ [(S± ↑) ↓ −(S± ↓) ↑ + ↑ (S± ↓)− ↓ (S± ↑)] . 2 2 1 1 S+ |0 0 = √ (0 − ↑↑ + ↑↑ −0) = 0; S− |0 0 = √ ( ↓↓ −0 + 0 − ↓↓) = 0. 2 2 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


109

(c) S 2 |1 1 = (S (1) )2 + (S (2) )2 + 2S(1) · S(2) ↑↑ = (S 2 ↑) ↑ + ↑ (S 2 ↑) + 2 [(Sx ↑)(Sx ↑) + (Sy ↑)(Sy ↑) + (Sz ↑)(Sz ↑)]

3 3 i i = 2 ↑↑ + 2 ↑↑ +2 ↓ ↓+ ↓ ↓+ ↑ ↑ 4 4 2 2 2 2 2 2 2 3 = 2 ↑↑ +2 ↑↑ = 22 ↑↑= 22 |1 1 = (1)(1 + 1)2 |1 1, as it should be. 2 4 S 2 |1 − 1 = (S (1) )2 + (S (2) )2 + 2S(1) · S(2) ↓↓ 32 32 ↓↓ + ↓↓ +2 [(Sx ↓)(Sx ↓) + (Sy ↓)(Sy ↓) + (Sz ↓)(Sz ↓)] 4 4 3 2 i i = ↓↓ +2 ↑ ↑ + − ↑ − ↑ + − ↓ − ↓ 2 2 2 2 2 2 2 3 2 2 = ↓↓ +2 ↓↓= 22 ↓↓= 22 |1 − 1. 2 4

=

Problem 4.35 (a) 1/2 and 1/2 gives 1 or zero; 1/2 and 1 gives 3/2 or 1/2; 1/2 and 0 gives 1/2 only. So baryons can have spin 3/2 or spin 1/2 (and the latter can be acheived in two distinct ways). [Incidentally, the lightest baryons do carry spin 1/2 (proton, neutron, etc.) or 3/2 (∆, Ω− , etc.); heavier baryons can have higher total spin, but this is because the quarks have orbital angular momentum as well.] (b) 1/2 and 1/2 gives spin 1 or spin 0. [Again, these are the observed spins for the lightest mesons: π’s and K’s have spin 0, ρ’s and ω’s have spin 1.]

Problem 4.36 (a) From the 2 × 1 Clebsch-Gordan table we get |3 1 =

1 |2 2|1 − 1 + 15

8 |2 1|1 0 + 15

6 |2 0|1 1, 15

so you might get 2 (probability 1/15), (probability 8/15), or (probability 6/15). (b) From the 1× 12 table: |1 0| 12 − 12 =

2 3 3 |2

− 12 +

15/42 and 3/42 , respectively. Thus you get

1 1 3 |2

− 12 . So the total is 3/2 or 1/2, with l(l+1)2 =

15 2 3 (probability 2/3), or 2 (probability 1/3). 4 4


110


Problem 4.37 (1)

2

2

(1)

(1)

(1)

Using Eq. 4.179: [S 2 , Sz ] = [S (1) , Sz ] + [S (2) , Sz ] + 2[S(1) · S(2) , Sz ]. But [S 2 , Sz ] = 0 (Eq. 4.102), and anything with superscript (2) commutes with anything with superscript (1). So + , [S 2 , Sz(1) ] = 2 Sx(2) [Sx(1) , Sz(1) ] + Sy(2) [Sy(1) , Sz(1) ] + Sz(2) [Sz(1) , Sz(1) ] + , = 2 −iSy(1) Sx(2) + iSx(1) Sy(2) = 2i(S(1) × S(2) )z. [S 2 , Sz(1) ] = 2i(Sx(1) Sy(2) − Sy(1) Sx(2) ), and [S 2 , S(1) ] = 2i(S(1) × S(2) ). Note that [S 2 , S(2) ] = 2i(S(2) × S(1) ) = −2i(S(1) × S(2) ), so [S 2 , (S(1) + S(2) )] = 0.]

Problem 4.38 (a) −

2 2m

∂2ψ ∂2ψ ∂2ψ + + ∂x2 ∂y 2 ∂z 2

1 + mω 2 x2 + y 2 + z 2 ψ = Eψ. 2

Let ψ(x, y, z) = X(x)Y (y)Z(z); plug it in, divide by XY Z, and collect terms: −

2 1 d 2 X 1 + mω 2 x2 2m X dx2 2

2 1 d 2 Y 2 1 d2 Z 1 1 2 2 2 2 mω mω + − + − = E. + y + z 2m Y dy 2 2 2m Z dz 2 2

The first term is a function only of x, the second only of y, and the third only of z. So each is a constant (call the constants Ex , Ey , Ez , with Ex + Ey + Ez = E). Thus: −

2 d 2 X 1 2 d 2 Y 1 2 d 2 Z 1 2 2 2 2 + x X = E X; − + y Y = E Y ; − + mω 2 z 2 Z = Ez Z. mω mω x y 2 2 2 2m dx 2 2m dy 2 2m dz 2

Each of these is simply the one-dimensional harmonic oscillator (Eq. 2.44). We know the allowed energies (Eq. 2.61): Ex = (nx + 12 )ω; Ey = (ny + 12 )ω; Ez = (nz + 12 )ω; where nx , ny , nz = 0, 1, 2, 3, . . . . So E = (nx + ny + ny + 32 )ω = (n + 32 )ω,

with n ≡ nx + ny + nz .

(b) The question is: “How many ways can we add three non-negative integers to get sum n?” If nx = n, then ny = nz = 0; one way. If nx = n − 1, then ny = 0, nz = 1, or else ny = 1, nz = 0; two ways. If nx = n − 2, then ny = 0, nz = 2, or ny = 1, nz = 1, or ny = 2, nz = 0; three ways. And so on. Evidently d(n) = 1 + 2 + 3 + · · · + (n + 1) =

(n + 1)(n + 2) . 2



111

Problem 4.39

2 d 2 u 1 2 l(l + 1) 2 2 u = Eu. Eq. 4.37: − + mω r + 2m dr2 2 2m r2 Following Eq. 2.71, let ξ ≡

1 2 mω l(l + 1) mω 2 mω d2 u 2 2 u = Eu, + + r. Then − mω ξ 2m dξ 2 2 mω 2m ξ2

d2 u l(l + 1) 2E 2 or = ξ + − K u, where K ≡ (as in Eq. 2.73). 2 2 dξ ξ ω

At large ξ,

At small ξ,

2 d2 u ≈ ξ 2 u, and u ∼ ( )e−ξ /2 (see Eq. 2.77). dξ 2

d2 u l(l + 1) ≈ u, and u ∼ ( )ξ l+1 (see Eq. 4.59). dξ 2 ξ2

So let u(ξ) ≡ ξ l+1 e−ξ

2

/2

v(ξ).

[This defines the new function v(ξ).]

2 2 2 du = (l + 1)ξ l e−ξ /2 v − ξ l+2 e−ξ /2 v + ξ l+1 e−ξ /2 v . dξ 2 2 2 2 d2 u = l(l + 1)ξ l−1 e−ξ /2 v − (l + 1)ξ l+1 e−ξ /2 v + (l + 1)ξ l e−ξ /2 v − (l + 2)ξ l+1 e−ξ /2 v 2 dξ

+ ξ l+3 e−ξ /2 v − ξ l+2 e−ξ /2 v + (l + 1)ξ l e−ξ /2 v − ξ l+2 e−ξ /2 v + ξ l+1 e−ξ /2 v ✭2✭ 2 2✘ 2 ✘ −ξ /2 ✭✭ ✭l−1 ✘e✘ =✭ l(l✭ +✭ 1)ξ e−ξ /2 v − (2l + 3)ξ l+1 e−ξ /2 v + ✘ ξ l+3 v + 2(l + 1)ξ l e−ξ /2 v ✭2✭ 2 2 2✘ 2 ✘ −ξ /2 ✭✭ ✭l−1 ✘e✘ − 2ξ l+2 e−ξ /2 v + ξ l+1 e−ξ /2 v = ✘ ξ l+3 v +✭ l(l✭ +✭ 1)ξ e−ξ /2 v − Kξ l+1 e−ξ /2 v. 2

2

2

Cancelling the indicated terms, and dividing off ξ l+1 e−ξ l+1 v + 2v − ξ + (K − 2l − 3) v = 0. ξ

Let v(ξ) ≡

∞

aj ξ j , so v =

j=0

∞

∞

jaj ξ j−1 ;

v =

j=0

j(j − 1)aj ξ j−2 + 2(l + 2)

j=2

∞

2

2

/2

∞

2

, we have:

j(j − 1)aj ξ j−2 . Then

j=2

jaj ξ j−2 − 2

j=1

∞

jaj ξ j + (K − 2l − 3)

j=1

∞

aj ξ j = 0.

j=0

In the first two sums, let j → j + 2 (rename the dummy index): ∞ j=0

j

(j + 2)(j + 1)aj+2 ξ + 2(l + 1)

∞ j=0

(j + 2)aj+2 ξ − 2 j

∞ j=0

jaj ξ + (K − 2l − 3) j

∞

aj ξ j = 0.

j=0


112


Note: the second sum should start at j = −1; to eliminate this term (there is no compensating one in ξ −1 ) we must take a1 = 0. Combining the terms: ∞

[(j + 2)(j + 2l + 3)aj+2 + (K − 2j − 2l − 3)aj ] = 0, so aj+2 =

j=0

(2j + 2l + 3 − K) aj . (j + 2)(j + 2l + 3)

Since a1 = 0, this gives us a single sequence: a0 , a2 , a4 , . . . . But the series must terminate (else we get the wrong behavior as ξ → ∞), so there occurs some maximal (even) number jmax such that ajmax+2 = 0. Thus 1 3 K = 2jmax + 2l + 3. But E = ωK, so E = jmax + l + ω. Or, letting jmax + l ≡ n, 2 2 En = (n + 32 )ω, and n can be any nonnegative integer. [Incidentally, we can also determine the degeneracy of En . Suppose n is even; then (since jmax is even) l = 0, 2, 4, . . . , n. For each l there are (2l + 1) values for m. So n

d(n) =

(2l + 1). Let j = l/2;

l=0,2,4,... ( n2 )( n2 +

=4

then d(n) =

n/2

(4j + 1) = 4

j=0

1)

2

+ ( n2 + 1) = ( n2 + 1)(n + 1) =

n/2

j+

j=0

n/2

1

j=0

(n + 1)(n + 2) , as before (Problem 4.38(b)).] 2

Problem 4.40 (a) d i r · p = [H, r · p]. dt [H, r · p] =

3

[H, ri pi ] =

i=1

[p2 , ri ] =

3

([H, ri ]pi + ri [H, pi ]) =

i=1

[pj pj , ri ] =

j=1

[V, pi ] = i

3

3 j=1

(pj [pj , ri ] + [pj , ri ]pj ) =

3 1 2 [p , ri ]pi + ri [V, pi ] . 2m i=1

3

[pj (−iδij ) + (−iδij )pj ] = −2ipi .

j=1

3 1 ∂V ∂V (Problem 3.13(c)). [H, r · p] = (−2i)pi pi + ri i ∂ri 2m ∂ri i=1

2 p = i − + r · ∇V . m For stationary states

d p2 r · p = − r · ∇V = 2T − r · ∇V . dt m

d r · p = 0, so 2T = r · ∇V . QED dt

(b) V (r) = −

e2 1 e2 1 e2 1 r ˆ ⇒ r · ∇V = ⇒ ∇V = = −V. 4π90 r 4π90 r2 4π90 r

So 2T = −V .

But T = V = En , so T − 2T = En , or T = −En ; V = 2En . QED c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


113

(c) V =

1 mω 2 r2 ⇒ ∇V = mω 2 r rˆ ⇒ r · ∇V = mω 2 r2 = 2V. 2

But T + V = En , so T = V =

So 2T = 2V , or T = V .

1 En . QED 2

Problem 4.41 i ∇Ψ · ∇Ψ∗ + Ψ(∇2 Ψ∗ ) − ∇Ψ∗ · ∇Ψ − Ψ∗ (∇2 Ψ) = 2m ∂Ψ 2 2 But the Schr¨ odinger equation says i =− ∇ Ψ + V Ψ, ∂t 2m ∂Ψ 2m 2m ∂Ψ∗ ∇2 Ψ = 2 V Ψ − i , ∇2 Ψ∗ = 2 V Ψ∗ + i . ∂t ∂t

(a) ∇ · J =

i Ψ(∇2 Ψ∗ ) − Ψ∗ (∇2 Ψ) . 2m so Therefore

i 2m ∂Ψ ∂Ψ∗ ∗ ∗ − Ψ Ψ V Ψ V Ψ − i + i 2m 2 ∂t ∂t ∗ i ∂Ψ ∂Ψ ∂ ∂ = i Ψ + Ψ∗ = − (Ψ∗ Ψ) = − |Ψ|2 . ∂t ∂t ∂t ∂t

∇·J =

1 1 Ψ2 1 1 = − √ re−r/2a sin θeiφ e−iE2 t/ . πa 8a2 ∂Ψ 1 ∂Ψ ˆ 1 ∂Ψ ˆ ∇Ψ = rˆ + θ+ φ, so ∂r r ∂θ r sin θ ∂φ

(b) From Problem 4.11(b),

In spherical coordinates,

r −r/2a 1 1 1 ∇Ψ2 1 1 = − √ 1− sin θeiφ e−iE2 t/ rˆ + re−r/2a cos θeiφ e−iE2 t/ θˆ e 2 2a r πa 8a

1 i ˆ 1 r + re−r/2a sin θ ieiφ e−iE2 t/ φˆ = 1 − rˆ + cot θ θˆ + φ Ψ2 1 1 . r sin θ 2a sin θ r Therefore J = =

i r i ˆ r i ˆ 1 1− rˆ + cot θ θˆ − φ− 1− rˆ − cot θ θˆ − φ |Ψ2 1 1 |2 2m 2a sin θ 2a sin θ r i (−2i) 1 1 r2 e−r/a sin2 θ ˆ ˆ re−r/a sin θ φ. |Ψ2 1 1 |2 φˆ = φ= 4 2m r sin θ m πa 64a r sin θ 64πma5

2 −r/a ˆ , while rˆ × φˆ = −θˆ and zˆ · θˆ = − sin θ, so r e sin θ r ˆ × φ 64πma5 r × Jz = r2 e−r/a sin2 θ, and hence 64πma5 2 2 −r/a r Lz = m e sin θ r2 sin θ dr dθ dφ 64πma5 ∞ π 2π 5 4 3 4 −r/a = 4!a (2π) = , r e dr sin θ dθ dφ = 64πa5 0 64πa5 3 0 0

(c) Now r × J =


114

CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS as it should be, since (Eq. 4.133) Lz = m, and m = 1 for this state.

Problem 4.42 (a) ψ=√

1

−r/a

πa3

e

1 1 √ ⇒ φ(p) = 3/2 (2π) πa3

e−ip·r/ e−r/a r2 sin θ dr dθ dφ.

With axes as suggested, p · r = pr cos θ. Doing the (trivial) φ integral: π

∞ 2π 1 2 −r/a −ipr cos θ/ √ φ(p) = r e e sin θ dθ dr. (2πa)3/2 π 0 0

π

2 pr −ipr cos θ/ π ipr/ − e−ipr/ = e e sin . = ipr ipr pr 0

e−ipr cos θ/ sin θ dθ =

0

1 √

2 1 φ(p) = 3/2 p (a) π 2

∞

−r/a

re

sin

pr

0

1 dr = 2i

=

re−r/a sin

pr

0

=

φ(p) =

∞

∞

−r/a ipr/

re

dr.

e

∞

dr −

0

−r/a −ipr/

re 0

1 1 1 − 2 2i (1/a − ip/)2 (1/a + ip/) (2p/)a3

e

=

1 2i

dr (2ip/a)2 2

2

2

(1/a) + (p/)

2.

[1 + (ap/)2 ]

2 1 1 2pa3 1 1 = a3/2 πp [1 + (ap/)2 ]2 π

2a

3/2

1

2.

[1 + (ap/)2 ]

(b)

|φ|2 d3 p = 4π

∞

p2 |φ|2 dp = 4π

0

∞

From math tables: 0

0

∞

p2

4 dp = [1 + (ap/)2 ]

1 π2

2a

3 0

∞

p2 4

[1 + (ap/)2 ]

x2 π −5/2 dx = , m (m + x2 )4 32

dp.

so

−5 3 3 8 π π 32 a 3 π = ; |φ|2 d3 p = = 1. a 32 a 32 a π 32 a



115

(c) p2 = 0

∞

p2 |φ|2 d3 p =

1 π2

2a

3

∞

4π 0

p4

4 dp.

[1 + (ap/)2 ]

π x4 4 m−3/2 . So p2 = dx = 2 4 [m + x ] 32 π

2a

From math tables:

3 8 −3 π 2 = 2. a 32 a a

(d) T =

1 2 1 2 2 m2 = p = 2 2m 2m a 2m 4

e2 4π90

2 =

m 22

e2 4π90

2 = −E1 ,

which is consistent with Eq. 4.191.

Problem 4.43 (a) From Tables 4.3 and 4.7, ψ321 =

R32 Y21

15 1 1 r 2 −r/3a 1 iφ − = −√ = e r2 e−r/3a sin θ cos θeiφ . sin θ cos θe 3/2 a 8π π 81a7/2 81 30 a 4 √

(b) |ψ| d r = 2 3

= = =

1 1 2 4 −2r/3a 2 r e sin θ cos θ r2 sin θ dr dθ dφ π (81)2 a7 ∞ π 1 6 −2r/3a 2π r e dr (1 − cos2 θ) cos2 θ sin θ dθ π(81)2 a7 0 0

π 7 2 3a cos3 θ cos5 θ 6! − + (81)2 a7 2 3 5 0

7 7 2 3 a 2 2 3·5 4 6·5·4·3·2 7 − = · = 1. 38 a7 2 3 5 4 15

(c)

2 ∞ 4 1 1 rs+6 e−2r/3a dr 81 30 a7 0 0 s+7 5 3 8 1 3a 3a (s + 6)! 3a = (s + 6)! = (s + 6)! = . 15(81)2 a7 2 2 720 6! 2

rs =

∞

rs |R32 |2 r2 dr =

Finite for s > −7 .


116


Problem 4.44 (a) From Tables 4.3 and 4.7, ψ433 = (b)

R43 Y33

1 1 r 3 −r/4a √ = e 768 35 a3/2 a

' −

35 sin3 θ cos θe3iφ 64π

( = −

1 √ r3 e−r/4a sin3 θe3iφ . 6144 πa9/2

1 6 6 −r/2a r = r|ψ| d r = r r e sin θ r2 sin θ dr dθ dφ (6144)2 πa9 ∞ π 2π 1 7 9 −r/2a = r e dr sin θ dθ dφ (6144)2 πa9 0 0 0 1 2·4·6 = 9!(2a)10 2 (2π) = 18a. 2 9 (6144) πa 3·5·7 2 3

(c) Using Eq. 4.133: L2x + L2y = L2 − L2z = 4(5)2 − (3)2 = 112 , with probability 1.

Problem 4.45 (a) b b a 2 −2r/a a3 −2r/a 2r 4π 4 −2r/a 2 − − 1 P = |ψ| d r = e r dr = 3 − r e + e 3 πa 0 a 2 4 a 0 b 2 2 2r 2r 2b b =− 1+ + 2 e−2r/a = 1 − 1 + + 2 2 e−2b/a . a a a a 0

2

3

(b)

1 2 −5 P =1− 1+9+ 9 e ≈1− 1+9+ 2 92 93 93 92 ≈1−1+9− + − 9 + 92 − − 2 6 2 2 3 3 1 2b 4 b = = . 6 a 3 a

92 1 2 93 1−9+ 9 − 2 2 3! 1 93 1 1 3 + =9 − + 2 6 2 2

(c) |ψ(0)|2 =

1 1 4 4 ⇒ P ≈ πb3 3 = πa3 3 πa 3

3 b . a

(d) 4 P = 3

10−15 0.5 × 10−10

3 =

3 4 4 32 2 × 10−5 = · 8 × 10−15 = × 10−15 = 1.07 × 10−14 . 3 3 3



117

Problem 4.46 (a) Equation 4.75 ⇒ Rn(n−1) = So

1 n −ρ ρ e v(ρ), where r

v(ρ) = c0 , and hence

∞

1=

r ; na

Eq. 4.76 ⇒ c1 =

Rn(n−1) = Nn rn−1 e−r/na , where

∞

|R| r dr = (Nn ) 2 2

ρ≡

2

0

2n −2r/na

r e 0

2(n − n) c0 = 0. (1)(2n)

c0 . (na)n

Nn ≡

2n+1 n 2 na 2 dr = (Nn ) (2n)! ; Nn = . 2 na na(2n)! 2

(b) r = l

∞

|R| r

2 l+2

0

r =

r =

∞

r2n+l e−2r/na dr.

0

2 na

2

dr =

Nn2

2n+1

2 na

2n+2 1 1 na = n+ (2n + 1)! na. (2n)! 2 2

2n+1

2 2n+3 1 1 na na = (2n + 2)(2n + 1) = n+ (2n + 2)! (n + 1)(na)2 . (2n)! 2 2 2

(c) σr2

2 1 1 2 2 = r − r = n+ (na) (n + 1)(na) − n + 2 2 1 1 r 1 = σr = √ n+ (na)2 = r2 ; . 2 2 2(n + 1/2) 2n + 1 2

2

R10

a

Maxima occur at:

R 32

r

6a

R 26 25

r

650a

r

dRn,n−1 1 n−1 −r/na e = 0 ⇒ r = na(n − 1). = 0 ⇒ (n − 1)rn−2 e−r/na − r dr na

Problem 4.47 Here are a couple of examples: {32, 28} and {224,56}; {221, 119} and {119, 91}. For further discussion see D. Wyss and W. Wyss, Foundations of Physics 23, 465 (1993).


118


Problem 4.48 (a) Using Eqs. 3.64 and 4.122: [A, B] = [x2 , Lz ] = x[x, Lz ] + [x, Lz ]x = x(−iy) + (−iy)x = −2ixy.

2 1 2 2 Equation 3.62 ⇒ σA σB ≥ (−2i)xy = 2 xy2 ⇒ σA σB ≥ |xy|. 2i (b) Equation 4.113 ⇒ B = Lz = m;

B 2 = L2z = m2 2 ;

σB = m2 2 − m2 2 = 0.

so

(c) Since the left side of the uncertainty principle is zero, the right side must also be: xy = 0, for eigenstates of Lz .

Problem 4.49 (a) 1 = |A|2 (1 + 4 + 4) = 9|A|2 ; (b)

A = 1/3.

5 4 5 4 , with probability ; − , with probability . Sz = + 2 9 2 9 92 9

−

2

=

. 18

(c) From Eq. 4.151, (x) c+ (x)

c−

13 1 1 1 − 2i 9+4 1 3 − 2i (x) = . = χ= √ 1 1 = √ (1 − 2i + 2) = √ ; |c+ |2 = 2 3 2 9 · 2 18 3 2 3 2 †

1 − 2i 1 1 1+4 1 1 + 2i 5 (x) (x) = χ− χ = √ 1 −1 = √ (1 − 2i − 2) = − √ ; |c− |2 = = . 2 3 2 9 · 2 18 3 2 3 2

(x) χ+

†

13 5 , with probability ; − , with probability . 2 18 2 18

13 5 Sx = + 18 2 18

− 2

=

2 . 9

(d) From Problem 4.29(a), †

1 − 2i 17 1 1 1 + 16 1 1 − 4i (y) (y) (y) c+ = χ+ = . χ = √ 1 −i = √ (1 − 2i − 2i) = √ ; |c+ |2 = 2 3 2 9·2 18 3 2 3 2 † 1 1 1 1 − 2i 1 1 1 (y) (y) (y) c− = χ− = . χ= √ 1 i = √ (1 − 2i + 2i) = √ ; |c− |2 = 2 3 2 9·2 18 3 2 3 2 17 1 , with probability ; − , with probability . 2 18 2 18

1 17 + Sy = 18 2 18

− 2

=

4 . 9

Problem 4.50 (1)

(1)

(2)

We may as well choose axes so that a ˆ lies along the z axis and ˆb is in the xz plane. Then Sa = Sz , and Sb (2) (2) (1) (2) cos θ Sz + sin θ Sx . 0 0|Sa Sb |0 0 is to be calculated.

=

1 (2) Sa(1) Sb |0 0 = √ Sz(1) (cos θ Sz(2) + sin θ Sx(2) ) (↑↓ − ↓↑) 2 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


119

1 = √ [(Sz ↑)(cos θ Sz ↓ + sin θ Sx ↓) − (Sz ↓)(cos θ Sz ↑ + sin θ Sx ↑)] 2 1 =√ 2

=

cos θ − ↓ + sin θ ↑ − − ↓ cos θ ↑ + sin θ ↓ (using Eq. 4.145) 2 2 2 2 2

↑ 2

2 1 2 1 1 cos θ √ (− ↑↓ + ↓↑) + sin θ √ (↑↑ + ↓↓) = − cos θ|0 0 + sin θ √ (|1 1 + |1 − 1) . 4 4 2 2 2 (2)

(2)

so Sa(1) Sb = 0 0|Sa(1) Sb |0 0 =

2 1 2 0 0| − cos θ|0 0 + sin θ √ (|1 1 + |1 − 1) = − cos θ0 0|0 0 4 4 2 (2)

(by orthogonality), and hence Sa(1) Sb = −

2 cos θ. 4

QED

Problem 4.51 (a) First note from Eqs. 4.136 and 4.144 that 1 [S+ |s m + S− |s m] 2 = s(s + 1) − m(m + 1)|s m + 1 + s(s + 1) − m(m − 1)|s m − 1 2

Sx |s m =

1 [S+ |s m − S− |s m] 2i = s(s + 1) − m(m + 1)|s m + 1 − s(s + 1) − m(m − 1)|s m − 1 2i

Sy |s m =

Now, using Eqs. 4.179 and 4.147:

(1) (2) (1) (2) (1) (2) 1 1 1 1 1 1 2 (1) 2 (2) 2 S |s m = (S ) + (S ) + 2(Sx Sx + Sy Sy + Sz Sz ) A| 2 2 |S2 m − 2 + B| 2 − 2 |s2 m + 2

= A S 2 | 12 12 |s2 m − 12 + | 12 21 S 2 |s2 m − 12

1 1

1 1

1 1 1 1 1 +2 Sx | 2 2 Sx |s2 m − 2 + Sy | 2 2 Sy |s2 m − 2 + Sz | 2 2 Sz |s2 m − 2

+B S 2 | 12 − 12 |s2 m + 12 + | 12 − 12 S 2 |s2 m + 12

+2 Sx | 12 − 12 Sx |s2 m + 12 + Sy | 12 − 12 Sy |s2 m + 12 + Sz | 12 − 12 Sz |s2 m + 12 = A 34 2 | 12 21 |s2 m − 12 + 2 s2 (s2 + 1)| 12 12 |s2 m − 12 1 1 +2 2 | 2 − 2 2 s2 (s2 + 1) − (m − 12 )(m + 12 )|s2 m + 12 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

120

CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS 1 3 3 + s2 (s2 + 1) − (m − 2 )(m − 2 )|s2 m − 2 i 1 1 + 2 | 2 − 2 2i s2 (s2 + 1) − (m − 12 )(m + 12 )|s2 m + 12 − s2 (s2 + 1) − (m − 12 )(m − 32 )|s2 m − 32 + 2 | 12

1 2 (m

− 12 )|s2 m − 12

+B

3 2 1 4 |2

+2 2 | 12

− 12 |s2 m + 12 + 2 s2 (s2 + 1)| 12 − 12 |s2 m + 12

1 22

+ −i | 12 2

1 2 2i

s2 (s2 + 1) − (m +

1 2 )(m

+

3 2 )|s2

m+

3 2

+

s2 (s2 + 1) − (m +

1 2 )(m

−

1 2 )|s2

m−

1 2

s2 (s2 + 1) − (m + 12 )(m + 32 )|s2 m + 32

− 1 1 1 1 1 1 1 − s2 (s2 + 1) − (m + 2 )(m − 2 )|s2 m − 2 + 2 | 2 − 2 (m + 2 )|s2 m + 2

3 1 1 2 = A 4 + s2 (s2 + 1) + m − 2 + B s2 (s2 + 1) − m + 4 | 12 2

1 2 |s2

m − 12

+2 B 34 + s2 (s2 + 1) − m − 12 + A s2 (s2 + 1) − m2 + 14 | 12 − 12 |s2 m + 12 = s(s + 1)|s m = s(s + 1) A| 12 2

2

1 2 |s2

m−

1 2

+

B| 12

−

1 2 |s2

m+

.

1 2

   A s2 (s2 + 1) + 1 + m +B s2 (s2 + 1) − m2 + 1 = s(s + 1)A,  4 4  B s2 (s2 + 1) + 1 − m +A s2 (s2 + 1) − m2 + 1 = s(s + 1)B,  4 4

or

   A s2 (s2 + 1) − s(s + 1) + 1 + m +B s2 (s2 + 1) − m2 + 1 = 0,  A(a + m) +Bb = 0 4 4 or , B(a − m) +Ab = 0  B s2 (s2 + 1) − s(s + 1) + 1 − m +A s2 (s2 + 1) − m2 + 1 = 0,  4 4 where a ≡ s2 (s2 + 1) − s(s + 1) + 14 , b ≡ s2 (s2 + 1) − m2 + 14 . Multiply by (a − b) and b, then subtract: A(a2 − m2 ) + Bb(a − m) = 0; Bb(a − m) + Ab2 = 0 ⇒ A(a2 − m2 − b2 ) = 0 ⇒ a2 − b2 = m2 , or: 2 s2 (s2 + 1) − s(s + 1) + 14 − s2 (s2 + 1) + m2 − 14 = m2 , 2

2 s2 (s2 + 1) − s(s + 1) + 14 = s22 + s2 + 14 = s2 + 12 , so

s2 (s2 + 1) − s(s + 1) + 14 = ± s2 + 12 ; s(s + 1) = s2 (s2 + 1) ∓ s2 + 12 + 14 . Add

1 4

to both sides:

s2 + s +

So

1 4

   s+

1 2

  s+

1 2

= s+

1 2 2

= s2 (s2 + 1) ∓ s2 +

1 2

+

1 2

 2  s2 + s2 − s2 − =



s22 + s2 + s2 +  s2 − 12  1  = ±s2 ⇒ s = ±s2 − 2 =  −s2− 12 . 1 s2 + 2   = ±(s2 + 1) ⇒ s = ±(s2 + 1) − 12 = 3  −s2 − 2

1 2

+

1 2

= s22

1 2

+

1 2

= (s2 + 1)2

  

.


CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS But s ≥ 0, so the possibilities are s = s2 ± 1/2.

121

Then:

1 1 1 a = s22 + s2 − s2 ± s2 ± + 1 + 2 2 4 1 1 1 1 1 1 1 2 2 = s2 + s2 − s2 ∓ s2 − s2 ∓ s2 − ∓ + = ∓s2 ∓ = ∓ s2 + . 2 2 4 2 4 2 2 2 1 1 1 1 2 b= s2 + s2 + s2 + s2 + + m − m2 = − m2 = s2 + − m . 4 2 2 2

∴ A ∓ s2 + 12 + m = ∓A s2 + 12 ∓ m = −Bb = −B s2 + 12 + m s2 + ⇒ A s2 + 12 ∓ m = ±B s2 + 12 ± m. But |A|2 + |B|2 = 1, so |A| + |A| 2

2

s2 + s2 +

1 2 1 2

∓m ±m

⇒ A=

1 2

−m

|A|2 1 1 (2s2 + 1) = s2 + ± m + s2 + ∓ m = |A|2 . 1 2 2 (s2 + 2 ± m) (s2 + 12 ± m)

s2 ± m + . 2s2 + 1 1 2

s2 + B = ±A s2 +

1 2

∓m

1 2

±m

= ±

s2 ∓ m + 12 . 2s2 + 1

(b) Here are four examples: (i) From the 1/2 × 1/2 table (s2 = 1/2), pick s = 1 (upper signs), m = 0. Then 1 1 1 1 2 +0+ 2 2 −0+ 2 √1 ; A= = B = = √12 . 1+1 1+1 2 (ii) From the 1 × 1/2 table (s2 = 1), pick s = 3/2 (upper signs), m = 1/2. Then 1 1 1 1 1+ 2 + 2 1− 2 + 2 2 A= = ; B = = √13 . 2+1 3 2+1 (iii) From the 3/2 × 1/2 table (s2 = 3/2), pick s = 1 (lower signs), m = −1. Then 3 3 √ 1 1 3 2 +1+ 2 2 −1+ 2 A= = ; B = − = − 12 . 3+1 2 3+1 (iv) From the 2 × 1/2 table (s2 = 2), pick s = 3/2 (lower signs), m = 1/2. Then 1 1 1 1 2− 2 + 2 2+ 2 + 2 2 3 A= = ; B = − = − 4+1 5 4+1 5. These all check with the values on Table 4.8, except that the signs (which are conventional) are reversed in (iii) and (iv). Normalization does not determine the sign of A (nor, therefore, of B).

Problem 4.52 | 32

        1 0 0 0         0 1 0 3 −1 3 −3 3      0 | 32 21 =  2 = 0 ; 0 ; | 2 2 = 1 ; | 2 2 = 0 . 0 0 0 1  √ 3 1  S+ | 32 32 = 0, S+ | 32 21 = 3| 32 32 , S+ | 32 −1 2 = 2| 2 2 , 

S− | 32

3 2

=

√

3| 32

1 2 ,

S− | 32

1 2

= 2| 32

−1 2 ,

S− | 32

−1 2

=

√

Equation 4.136 ⇒

S+ | 32 3| 32

−3 2

−3 2 ,

=

√

3| 32

S− | 32

−1 2 ;

−3 2

= 0.


122

CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS  √ 0 3 0 0 So: S+ =  0 0 0 0 −λ √ 3 0 0

√

3 −λ 2 0

0 2 −λ √ 3

0 2 0 0

  0 √0   3 0 √  ; S− =   0 3 0 0

0 0 2 0

0 0 √0 3

  0 √0  3 1 0   ; Sx = (S+ + S− ) = 0 2 2 0 0 0

√

3 0 2 0

0 2 √0 3

 0  √0  . 3 0

√ 0 −λ 2 0 √ √ 3 2 √0 0 √ = −λ 2 −λ 3 − 3 0 −λ 3 √ 3 0 √3 −λ 0 3 −λ −λ √ √ √ = −λ −λ3 + 3λ + 4λ − 3 3λ2 − 3 3 = λ4 − 7λ2 − 3λ2 + 9 = 0,

or λ4 − 10λ2 + 9 = 0;

(λ2 − 9)(λ2 − 1) = 0;

λ = ±3, ±1. So the eigenvalues of Sx are

3 1 2 , 2 ,

− 12 , − 32 .

Problem 4.53 From Eq. 4.135, Sz |s m = m|s m. Since s is fixed, here, let’s just identify the states by the value of m (which runs from −s to +s). The matrix elements of Sz are Snm = n|Sz |m = mn|m = mδn m . It’s a diagonal matrix, with elements m, ranging from m = s in right corner:  s 0 0 ··· 0 s − 1 0 · · ·   Sz = 0 0 s − 2 · · ·  .. .. .. . . . . . . 0

0

0

the upper left corner to m = −s in the lower 0 0 0 .. .



   .   · · · −s

From Eq. 4.136, S± |s m =

s(s + 1) − m(m ± 1) |s (m ± 1) =

(S+ )nm = n|S+ |m =

(s ∓ m)(s ± m + 1) |s (m ± 1).

(s − m)(s + m + 1) n|m + 1 = bm+1 δn (m+1) = bn δn (m+1) .

All nonzero elements have row index (n) one greater than the column index (m), so they are on the diagonal just above the main diagonal (note that the indices go down, here: s, s − 1, s − 2 . . . , −s): 

0 0  0  S+ =  .  ..  0 0 Similarly (S− )nm = n|S− |m =

0 bs 0 0 bs−1 0 0 0 bs−2 .. .. .. . . . 0 0 0 0 0 0

··· ··· ··· .. .

0 0 0 .. .



    .   · · · b−s+1  ··· 0

(s + m)(s − m + 1) n|m − 1 = bm δn (m−1) .



123

This time the nonzero elements are on the diagonal just below  0 0 0 ··· b s 0 0 · · ·   S− =  0 bs−1 0 · · ·  .. .. .. . . . . . . 0

0

the main diagonal:  0 0 0 0  0 0 . .. ..  . . 0 · · · b−s+1 0

1 To construct Sx = 12 (S+ + S− ) and Sy = 2i (S+ − S− ), simply add and subtract the matrices S+ and S− :     0 bs 0 bs 0 0 ··· 0 0 0 0 ··· 0 0 bs 0 bs−1 0 · · · −bs 0 0  0 bs−1 0 ··· 0 0       0 bs−1 0 bs−2 · · ·   0 −bs−1 0 0  0 bs−2 · · · 0 0       0 0 0  0 0  0 −bs−2 0 · · · Sx =  0 0 bs−2 0 · · ·   ; Sy = . 2  .. .. 2i  .. .. .. . . .. ..  .. .. .. . . .. ..  . .   . . . . . .  . . . . .    .  0 0   0 0 ··· 0 b−s+1 0 0 0 0 ··· 0 b−s+1  0 0 0 0 · · · b−s+1 0 0 0 0 0 · · · −b−s+1 0

Problem 4.54 L+ Ylm =

l(l + 1) − m(m + 1)Ylm±1 (Eqs. 4.120 and 121). Equation 4.130 ⇒ ∂ ∂ iφ e + i cot θ Blm eimφ Plm (cos θ) = l(l + 1) − m(m + 1)Blm+1 ei(m+1)φ Plm+1 (cos θ). ∂θ ∂φ d Blm − m cot θ Plm (cos θ) = l(l + 1) − m(m + 1)Blm+1 Plm+1 (cos θ). dθ cos θ d d x dx d d Let x ≡ cos θ; cot θ = ; =√ = = − sin θ = − 1 − x2 . 2 sin θ dθ dx dx dx 1 − x dθ

d 1 dP m x Plm (x) = −Blm √ (1 − x2 ) l + mxPlm = −Blm Plm+1 Blm − 1 − x2 − m√ dx dx 1 − x2 1 − x2 =

l(l + 1) − m(m + 1)Blm+1 Plm+1 (x).

⇒

Blm+1 =

−1 l(l + 1) − m(m + 1)

Blm .

Now l(l + 1) − m(m + 1) = (l − m)(l + m + 1), so Blm+1 = √

−1 −1 √ Blm ⇒ Bl1 = √ √ Bl0 ; l−m l+1+m l l+1 Bl3 = √

Bl2 = √

−1 1 √ Bl1 = Bl0 ; l−1 l+2 l(l − 1) (l + 1)(l + 2)

−1 −1 √ Bl0 , Bl2 = l−2 l+3 (l + 3)(l + 2)(l + 1)l(l − 1)(l − 2)

etc.

Evidently there is an overall sign factor (−1)m , and inside the square root the quantity is [(l + m)!/(l − m)!]. (l − m)! Thus: Blm = (−1)m C(l) (where C(l) ≡ Bl0 ), for m ≥ 0. For m < 0, we have (l + m)! −Bl0

Bl−1 =

(l + 1)l

; Bl−2 =

−1 (l + 2)(l − 1)

Bl−1 =

1 (l + 2)(l + 1)l(l − 1)

Bl0 ,

etc.


124


Thus Bl−m = Blm , so in general: Blm = (−1)m (l−|m|)! Now, Problem 4.22 says: (l+|m|! C(l). (2l + 1)! iφ 1 Yll = l (e sin θ)l = Bll eilφ Pll (cos θ). But 2 l! π l l 2l d 1 d (1 − x2 )l/2 d (2l)! l 2 l/2 2 l Pl (x) = (1 − x ) (x − 1) = (x2l − . . . ) = l (1 − x2 )l/2 , l l dx 2 l! dx 2 l! dx 2 l! (2l)!

so

(2l)! Pll (cos θ) = l (sin θ)l . Therefore 2 l! 1 (2l + 1)! iφ (2l + 1)! (2l + 1) 1 l l ilφ (2l)! l l (e sin θ) = Bl e (sin θ) ⇒ Bl = = . l l 2 l! π 2 l! (2l)! π π(2l)! 1 C(l), (2l)!

But Bll = (−1)l

so C(l) = (−1)l

2l + 1 , π

and hence

Blm = (−1)l+m

(2l + 1) (l − |m|)! . π (l + |m|)!

This agrees with Eq. 4.32 except for the overall sign, which of course is purely conventional.

Problem 4.55 (a) For both terms, l = 1, so (b) 0, P =

2 (1)(2) = 22 , P = 1.

1 2 , or , P = . 3 3

(c)

3 2 , P = 1. 4

(d)

1 2 , P = , or − , P = . 2 3 2 3

(e) From the 1 ×

1 2

Clebsch-Gordan table (or Problem 4.51): 2 1 −1 2 3 1 2 √1 3 1 2 1 √1 | 1 1 |1 0 + √1 √1 | 1 1 + | |1 1 = | − | + 3 2 2 3 2 2 3 3|2 3 2 2 3 3 2 2 3 2 2 √

15 2 3 8 1 = 2 32 | 32 12 + 13 | 12 12 . So s = 32 or 12 . , P = , or 2 , P = . 4 9 4 9

(f )

1 2

1 , P = 1. 2

(g)

|ψ| = |R21 | 2

2

√

1 02 † 2 12 † 2 0∗ 1 † † 1∗ 0 |Y | (χ χ+ ) + Y1 Y1 (χ+ χ− ) +Y1 Y1 (χ− χ+ ) + |Y1 | (χ− χ− ) 3 1 + 3 3 1

0

0

02

1 1 1 r2 −r/a 3 1 3 2 2 1 2 2 = |R21 | |Y1 | + 2|Y1 | = · · e · cos θ + 2 sin θ 3 3 24 a3 a2 4π 8π =

1

[Tables 4.3, 4.7]

1 1 3 r2 e−r/a · r2 e−r/a . (cos2 θ + sin2 θ) = 3 · 24 · a5 4π 96πa5



125

(h) 1 |R21 |2 3

|Y10 |2 sin2 θ dθ dφ =

1 2 −r/a 1 1 1 2 −r/a r e = r e . |R21 |2 = · 3 3 24a3 72a5

Problem 4.56 (a) Equation 4.129 says Lz =

∂ i ∂φ ,

so this problem is identical to Problem 3.39, with pˆ → Lz and x → φ.

(b) First note that if M is a matrix such that M2 = 1, then 1 1 1 1 eiMφ = 1 + iMφ + (iMφ)2 + (iMφ)3 + · · · = 1 + iMφ − φ2 − iM φ3 + · · · 2 3! 2 3! 1 1 1 1 = (1 − φ2 + φ4 − · · · ) + iM(φ − φ3 + φ5 − · · · ) = cos φ + iM sin φ. 2 4! 3! 5! 0 1 π π iπσx /2 2 So R = e = cos 2 + iσx sin 2 (because σx = 1 – see Problem 4.26) = iσx = i . 1 0 01 1 0 Thus Rχ+ = i =i = iχ− ; it converts “spin up” into “spin down” (with a factor of i). 10 0 1 (c) iπσy /4

R=e

1 Rχ+ = √ 2

π π 1 1 = cos + iσy sin = √ (1 + iσy ) = √ 4 4 2 2

1 1 0 0 −i 1 1 +i = √ . 0 1 i 0 2 −1 1

1 1 1 1 (x) =√ = √ (χ+ − χ− ) = χ− (Eq. 4.151). 0 −1 2 2

1 1 −1 1

What had been spin up along z is now spin down along x (see figure). z

y y' x z' x'

(d) R = eiπσz = cos π + iσz sin π = −1; rotation by 360◦ changes the sign of the spinor. But since the sign of χ is arbitrary, it doesn’t matter. (e) (σ · n ˆ )2 = (σx nx + σy ny + σz nz )(σx nx + σy ny + σz nz ) = σx2 n2x + σy2 n2y + σz2 n2z + nx ny (σx σy + σy σx ) + nx nz (σx σz + σz σx ) + ny nz (σy σz − σz σy ). c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

126

CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS But σx2 = σy2 = σz2 = 1, and σx σy + σy σx = σx σz + σz σx = σy σz + σz σy = 0 (Problem 4.26), so (σ · n ˆ )2 = n2x + n2y + n2z = 1. So ei(σ·ˆn)φ/2 = cos

φ φ + i(σ · n ˆ ) sin . QED 2 2

Problem 4.57 (a)

1 x + a2 / py , 2

1 [p1 , p2 ] = px − /a2 y, 2

1 [q1 , p1 ] = x + a2 / py , 2

1 [q2 , p2 ] = x − a2 / py , 2 [q1 , q2 ] =

x − a2 / py = 0, because [x, py ] = [x, x] = [py , py ] = 0.

px + /a2 y = 0, because [y, px ] = [y, y] = [px , px ] = 0.

1 1 px − /a2 y = ([x, px ] − [py , y]) = [i − (−i)] = i. 2 2

1 2 px + /a y = ([x, px ] − [py , y]) = i. 2

[See Eq. 4.10 for the canonical commutators.] (b) 1 q12 − q22 = x2 + 2 1 2 2 2 p1 − p2 = p − 2 x

So

a2 (xpy + py x) + (px y + ypx ) + a2

a2 a2

2

a2 p2y − x2 + (xpy + py x) −

2 y − 2

p2x

− 2 (px y + ypx ) − a

a2

a2

2

p2y =

2 y

2

2a xpy .

=−

2 ypx . a2

2 a2 2 (q1 − q22 ) + (p − p22 ) = xpy − ypx = Lz . 2 2a 2 1

(c) H=

1 2 1 a2 2 p + mω 2 x2 = p + 2 x2 = H(x, p). 2m 2 2 2a

Then H(q1 , p1 ) =

a2 2 2 p + q ≡ H1 , 2 1 2a2 1

H(q2 , p2 ) =

a2 2 2 p + q ≡ H2 ; 2 2 2a2 2

Lz = H1 − H2 .

(d) The eigenvalues of H1 are (n1 + 12 ), and those of H2 are (n2 + 12 ), so the eigenvalues of Lz are (n1 + 12 ) − (n2 + 12 ) = (n1 − n2 ) = m, and m is an integer, because n1 and n2 are.



127

Problem 4.58

a From Problem 4.28 we know that in the generic state χ = (with |a|2 + |b|2 = 1), b Sz =

2 |a| − |b|2 , 2

Sx = Re(ab∗ ),

Sy = −Im(ab∗ );

Sx2 = Sy2 =

2 . 4

Writing a = |a|eiφa , b = |b|eiφb , we have ab∗ = |a||b|ei(φa −φb ) = |a||b|eiθ , where θ ≡ φa − φb is the phase difference between a and b. Then Sx = Re(|a||b|eiθ ) = |a||b| cos θ, σS2 x = Sx2 − Sx 2 = We want σS2 x σS2 y =

Sy = −Im(|a||b|eiθ ) = −|a||b| sin θ.

2 − 2 |a|2 |b|2 cos2 θ; 4

σS2 y = Sy2 − Sy 2 =

2 − 2 |a|2 |b|2 sin2 θ. 4

2 Sz 2 , or 4

2

2

2 2 2 2 1 − 4|a|2 |b|2 cos2 θ 1 − 4|a|2 |b|2 sin2 θ = |a| − |b|2 . 4 4 4 4

2 2 2 2 2 4 4 2 1 − 4|a| |b| cos θ + sin θ + 16|a| |b| sin θ cos θ = |a|4 − 2|a|2 |b|2 + |b|4 .

2 1 + 16|a|4 |b|4 sin2 θ cos2 θ = |a|4 + 2|a|2 |b|2 + |b|4 = |a|2 + |b|2 = 1 ⇒ |a|2 |b|2 sin θ cos θ = 0. So either θ = 0 or π, in which case a and b are relatively real, or else θ = ±π/2, in which case a and b are relatively imaginary (these two options subsume trivially the solutions a = 0 and b = 0).

Problem 4.59 (a) Start with Eq. 3.71:

H=

dr i = [H, r]. dt

1 1 2 (p − qA) · (p − qA) + qϕ = p − q(p · A + A · p) + q 2 A2 + qϕ. 2m 2m

[H, x] =

1 2 q [p , x] − [(p · A + A · p), x]. 2m 2m

[p2 , x] = [(p2x + p2y + p2z ), x] = [p2x , x] = px [px , x] + [px , x]px = px (−i) + (−i)px = −2ipx . [p · A, x] = [(px Ax + py Ay + pz Az ), x] = [px Ax , x] = px [Ax , x] + [px , x]Ax = −iAx . [A · p, x] = [(Ax px + Ay py + Az pz ), x] = [Ax px , x] = Ax [px , x] + [Ax , x]px = −iAx . [H, x] =

1 q i (−2ipx ) − (−2iAx ) = − (px − qAx ); 2m 2m m

[H, r] = −

i (p − qA). m

dr 1 = (p − qA). QED dt m c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

128


(b) We define the operator v ≡

H=

1 (p − qA); m

dv i ∂v = [H, v] + ; dt ∂t

1 m mv 2 + qϕ ⇒ [H, v] = [v 2 , v] + q[ϕ, v]; 2 2

[ϕ, px ] = i

∂ϕ ∂x

[ϕ, v] =

∂v q ∂A =− . ∂t m ∂t

1 [ϕ, p]. m

(Eq. 3.65), so [ϕ, p] = i∇ϕ, and [ϕ, v] =

i ∇ϕ. m

[v 2 , vx ] = [(vx2 + vy2 + vz2 ), vx ] = [vy2 , vx ] + [vz2 , vx ] = vy [vy , vx ] + [vy , vx ]vy + vz [vz , vx ] + [vz , vx ]vz . 1 q [(py − qAy ), (px − qAx )] = − 2 ([Ay , px ] + [py , Ax ]) 2 m m q iq ∂Ay ∂Ax iq = − 2 i − i = − 2 (∇ × A)z = − 2 Bz . m ∂x ∂y m m 1 q [vz , vx ] = 2 [(pz − qAz ), (px − qAx )] = − 2 ([Az , px ] + [pz , Ax ]) m m q iq ∂Ax iq ∂Az = − 2 i − i = 2 (∇ × A)y = 2 By . m ∂x ∂y m m

[vy , vx ] =

∴ [v 2 , vx ] =

[v 2 , v] =

iq iq (−vy Bz − Bz vy + vz By + By vz ) = 2 [−(v × B)x + (B × v)x ] . m2 m

iq [(B × v) − (v × B)] . Putting all this together: m2

dv i = dt

0

qi m iq (B × v − v × B) + ∇ϕ 2 m2 m

1 −

q ∂A . m ∂t

0 1 dv q ∂A q [] m = (v × B) − (B × v) + q −∇ϕ − = (v × B − B × v) + qE. Or, since dt 2 dt 2 v×B−B×v =

1 1 q [(p − qA) × B − B × (p − qA)] = [p × B − B × p] − [A × B − B × A] . m m m

[Note: p does not commute with B, so the order does matter in the first term. But A commutes with B, so B × A = −A × B in the second.] m

dv q q2 = qE + p × B − B × p − A × B. QED dt 2m m

(c) Go back to Eq. , and use E = E, v × B = v × B; B × v = B × v = −v × B. Then m

dv = qv × B + qE. QED dt



129

Problem 4.60 (a) ˆ E = −∇ϕ = −2Kz k.

î ˆj kˆ ˆ B = ∇ × A = ∂/∂x ∂/∂y ∂/∂z = B0 k. −B0 y/2 B0 x/2 0

(b) For time-independent potentials Eq. 4.205 separates in the usual way: 1 ∇ − qA · ∇ − qA ψ + qϕψ = Eψ, 2m i i −

2 2 q2 2 iq ∇ ψ+ [∇ · (Aψ) + A · (∇ψ)] + A + qϕψ = Eψ. 2m 2m 2m 2 2 iq − ∇ ψ+ [2A · (∇ψ) + ∇ · (Aψ)] + 2m 2m

or

But ∇ · (Aψ) = (∇ · A)ψ + A · (∇ψ),

so

q2 2 A + qϕ ψ = Eψ. 2m

This is the time-independent Schr¨ odinger equation for electrodynamics. In the present case

B0 ∂ψ ∂ψ B2 ∇ · A = 0, A · (∇ψ) = x −y , A2 = 0 x2 + y 2 , ϕ = Kz 2 . 2 ∂y ∂x 4

qB0 q 2 B02 2 ∂ ∂ 2 2 2 2 But Lz = x −y , so − ∇ ψ− Lz ψ + x + y + qKz ψ = Eψ. i ∂y ∂x 2m 2m 8m Since Lz commutes with H, we may as well pick simultaneous eigenfunctions of both: Lz ψ = mψ, ¯ where m ¯ = 0, ±1, ±2, . . . (with the overbar to distinguish the magnetic quantum number from the mass). Then

2 2 (qB0 )2 2 qB0 − ∇ + x + y 2 + qKz 2 ψ = E + m ¯ ψ. 2m 8m 2m Now let ω1 ≡ qB0 /m, ω2 ≡ 2Kq/m, and use cylindrical coordinates (r, φ, z):

2

1 2 1 ∂ 1 1 ∂ψ 1 ∂2ψ ∂2ψ 2 2 2 2 − + mω1 x + y + mω2 z ψ = E + mω r + 2 ¯ 1 ψ. + 2m r ∂r ∂r r ∂φ2 ∂z 2 8 2 2 ∂2ψ ∂ 1 1 2 2 , so = − 2 L2z ψ = − 2 m ¯ ψ = −m ¯ 2 ψ. i ∂φ ∂φ2 R(r)Φ(φ)Z(z) :

2 d2 Z 1 d dR m ¯2 1 − ΦZ r − 2 RΦZ + RΦ 2 + mω12 r2 + 2m r dr dr r dz 8 But Lz =

Use separation of variables: ψ(r, φ, z) = 1 1 2 2 mω2 z RΦZ = E + mω ¯ 1 RΦZ. 2 2

Divide by RΦZ and collect terms:

2 1 2 1 d 2 Z 1 1 1 d dR m ¯2 2 2 − = E + + z r − 2 + mω12 r2 + − mω mω ¯ 1 . 2 2m rR dr dr r 8 2m Z dz 2 2 2 The first term depends only on r, the second only on z, so they’re both constants; call them Er and Ez :

2 1 d dR m ¯2 1 2 d 2 Z 1 1 − r − 2 R + mω12 r2 R = Er R; − ¯ 1. + mω22 z 2 Z = Ez Z; E = Er +Ez − mω 2m r dr dr r 8 2m dz 2 2 2 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

130

CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS The z equation is a one-dimensional harmonic oscillator, and we can read off immediately that Ez = (n2 + 1/2)ω2 , with √n2 = 0, 1, 2, . . . . The r equation is actually a two-dimensional harmonic oscillator; to get Er , let u(r) ≡ r R, and follow the method of Sections 4.1.3 and 4.2.1: √ u dR √ u u dR u d u dR R= √ , = √ − 3/2 , r = ru − √ , r = r u + 3/2 , dr dr dr dr r r 2r 2 r 4r 1 d u u 2 u 1 u 1 m ¯2 u 1 dR u u √ + √ − 2 √ + mω12 r2 √ = Er √ r = √ + 5/2 ; − 2 r dr dr 2m r 8 r 4r r 4r r r r r 2 2 u 1 1 d u − + mω12 r2 u = Er u. + −m ¯2 2 2 2m dr 4 r 8 This is identical to the equation we encountered in Problem 4.39 (the three-dimentional harmonic oscillator), only with ω → ω1 /2, E → Er , and l(l + 1) → m ¯ 2 − 1/4, which is to say, l2 + l + 1/4 = m ¯ 2 , or 2 2 2 (l + 1/2) = m ¯ , or l = |m| ¯ − 1/2. [Our present equation depends only on m ¯ , and hence is the same for either sign, but the solution to Problem 4.39 assumed l + 1/2 ≥ 0 (else u is not normalizable), so we need |m| here.] Quoting 4.39: E = (jmax + l + 3/2)ω → Er = (jmax + |m| ¯ + 1)ω1 /2,

where

jmax = 0, 2, 4, . . . .

E = jmax + |m| ¯ + 1)ω1 /2 + (n2 + 1/2)ω2 − mω ¯ 1 /2 = (n1 + 12 )ω1 + (n2 + 12 )ω2 , where n1 = 0, 1, 2, . . . (if m ¯ ≥ 0, then n1 = jmax /2; if m ¯ < 0, then n1 = jmax /2 − m). ¯

Problem 4.61 (a) B = ∇ × A = ∇ × A + ∇ × (∇λ) = ∇ × A = B. [∇ × ∇λ = 0, by equality of cross-derivatives: (∇ × ∇λ)x = ∂A E = −∇ϕ − = −∇ϕ + ∇ ∂t

[Again: ∇

∂Λ ∂t

=

∂Λ ∂t

−

∂ ∂y

∂λ ∂z

−

∂ ∂z

∂λ ∂y

= 0, etc.]

∂A ∂ ∂A − (∇Λ) = −∇ϕ − = E. ∂t ∂t ∂t

∂ (∇Λ) by the equality of cross-derivatives.] ∂t

(b)

∇ − qA − q(∇Λ) eiqΛ/ Ψ = q(∇Λ)eiqΛ/ Ψ + eiqΛ/ ∇Ψ − qAeiqΛ/ Ψ − q(∇Λ)eiqΛ/ Ψ i i = eiqΛ/ ∇Ψ − qAeiqΛ/ Ψ. i ∇ − qA − q(∇Λ) i

2

iqΛ/

e

Ψ=

∇ − qA − q(∇Λ) i

iqΛ/ iqΛ/ e ∇Ψ − qAe Ψ i



131

iq q iqΛ/ iqΛ/ 2 = − +e ∇ Ψ − (∇ · A)eiqΛ/ Ψ − q 2 (A · ∇Λ)eiqΛ/ Ψ (∇Λ · ∇Ψ)e i q q − eiqΛ/ A · (∇Ψ) − eiqΛ/ (A · ∇Ψ) + q 2 A2 eiqΛ/ Ψ i i q iqΛ/ − e (∇Λ · ∇Ψ) + q 2 (A · ∇Λ)eiqΛ/ Ψ i - = eiqΛ/ −2 ∇2 Ψ + iq(∇ · A)Ψ + 2iq(A · ∇Ψ) + q 2 A2 Ψ . −iq(∇Λ) · (∇Ψ) − q 2 (A · ∇Λ)Ψ + iq(∇Λ) · (∇Ψ) + q 2 (A · ∇Λ)Ψ 2 = eiqΛ/ ∇ − qA Ψ . i 2

So:

1 2m

∇ − qA i

[using Eq. 4.205]

2 + qϕ

iqΛ/

Ψ =e

1 2m

∇ − qA i

2

∂Λ + qϕ − q Ψ ∂t

∂Λ ∂ iqΛ/ ∂Ψ ∂Ψ = eiqΛ/ i −q Ψ = i e . QED Ψ = i ∂t ∂t ∂t ∂t


132

CHAPTER 5. IDENTICAL PARTICLES

Chapter 5

Identical Particles Problem 5.1 (a) (m1 + m2 )R = m1 r1 + m2 r2 = m1 r1 + m2 (r1 − r) = (m1 + m2 )r1 − m2 r

r1 = R +

⇒

m2 µ r=R+ r. m1 + m2 m1

(m1 + m2 )R = m1 (r2 + r) + m2 r2 = (m1 + m2 )r2 + m1 r ⇒ r2 = R −

m1 µ r=R− r. m1 + m2 m2

Let R = (X, Y, Z), r = (x, y, z). ∂ ∂X ∂ ∂x ∂ = + ∂x1 ∂x1 ∂X ∂x1 ∂x m1 ∂ µ ∂ = + (1) = (∇R )x + (∇r )x , m1 + m2 ∂X ∂x m2 ∂ ∂X ∂ ∂x ∂ (∇2 )x = = + ∂x ∂x2 ∂X ∂x2 ∂x 2 ∂ m2 ∂ µ = (∇R )x − (∇r )x , − (1) = m1 + m2 ∂X ∂x m1 (∇1 )x =

so

∇1 =

µ ∇R + ∇ r . m2

so

∇2 =

µ ∇ R − ∇r . m1

(b)

µ ∇21 ψ = ∇1 · (∇1 ψ) = ∇1 · ∇R ψ + ∇ r ψ m2 µ µ µ = ∇R · ∇R ψ + ∇r ψ + ∇r · ∇R ψ + ∇r ψ m2 m2 m2 2 µ µ = ∇2R ψ + 2 (∇r · ∇R )ψ + ∇2r ψ. m2 m2 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 5. IDENTICAL PARTICLES Likewise,

∇22 ψ =

µ m1

133

2 ∇2R ψ − 2

µ (∇r · ∇R ) + ∇2r ψ. m1

2 2 2 2 ∴ Hψ = − ∇1 ψ − ∇ ψ + V (r1 , r2 )ψ 2m1 2m2 2 2 2µ 1 2 µ2 2µ 1 2 µ2 2 2 =− ψ ∇ + ∇ · ∇ + ∇ + ∇ − ∇ · ∇ + ∇ r R r R 2 m1 m22 R m1 m2 m1 r m2 m21 R m2 m1 m2 r 2 1 1 1 1 µ2 2 + V (r)ψ = − ∇R + ∇2r ψ + V (r)ψ = Eψ. + + 2 m1 m 2 m2 m1 m1 m2 But

−

1 1 + m1 m2

=

m 1 + m2 µ2 1 = , so m 1 m2 µ m1 m 2

2 2 2 ∇2R ψ − ∇ ψ + V (r)ψ = Eψ. 2(m1 + m2 ) 2µ r

1 1 + m2 m1

=

µ 1 m 1 m2 + = . m1 m2 m1 m2 (m1 + m2 ) m1 + m2

(c) Put in ψ = ψr (r)ψR (R), and divide by ψr ψR :

2 2 1 2 1 2 − ∇R ψR + − ∇r ψr + V (r) = E. 2(m1 + m2 ) ψR 2µ ψr The first term depends only on R, the second only on r, so each must be a constant; call them ER and Er , respectively. Then: −

2 ∇ 2 ψ R = ER ψ R ; 2(m1 + m2 )

−

2 2 ∇ ψr + V (r)ψr = Er ψr , 2µ

with

ER + Er = E.

Problem 5.2 ∆E1 ∆m m−µ m(m + M ) M m = = = − = . E1 µ µ mM M M The fractional error is the ratio of the electron mass to the proton mass: 9.109 × 10−31 kg = 5.44 × 10−4 . The percent error is 0.054% (pretty small). 1.673 × 10−27 kg

(a) From Eq. 4.77, E1 is proportional to mass, so

∆(1/λ) ∆R ∆µ (1/λ2 )∆λ ∆λ = = =− =− . (1/λ) R µ (1/λ) λ But µ = mM/(m + M ), where m = electron mass, and M =

(b) From Eq. 4.94, R is proportional to m, so So (in magnitude) ∆λ/λ = ∆µ/µ. nuclear mass.

∆µ =

=

m(2mp ) mmp mmp (2m + 2mp − m − 2mp ) − = m + 2mp m + mp (m + mp )(m + 2mp ) m2 mp mµ . = (m + mp )(m + 2mp ) m + 2mp


134


∆λ m m ∆µ m ≈ , so ∆λ = λh , where λh is the hydrogen wavelength. = = λ µ m + 2mp 2mp 2mp

1 =R λ

∴ ∆λ =

(c) µ =

1 1 − 4 9

5 36 36 R⇒λ= = m = 6.563 × 10−7 m. 36 5R 5(1.097 × 107 )

9.109 × 10−31 (6.563 × 10−7 )m = 1.79 × 10−10 m. 2(1.673 × 10−27 )

mm m = , so the energy is half what it would be for hydrogen: (13.6/2)eV = 6.8 eV. m+m 2

mp m µ m p mµ mp + me mµ (mp + me ) ; R ∝ µ, so R is changed by a factor · = , as compared mp + m µ mp + mµ mp me me (mp + mµ ) with hydrogen. For hydrogen, 1/λ = R(1−1/4) = 34 R ⇒ λ = 4/3R = 4/3(1.097×107 ) m = 1.215×10−7 m, and λ ∝ 1/R, so for muonic hydrogen the Lyman-alpha line is at

(d) µ =

λ=

me (mp + mµ ) 1 (1.673 × 10−27 + 206.77 × 9.109 × 10−31 ) (1.215 × 10−7 m) = (1.215 × 10−7 m) mµ (mp + me ) 206.77 (1.673 × 10−27 + 9.109 × 10−31 )

= 6.54 × 10−10 m.

Problem 5.3 The energy of the emitted photon, in a transition from vibrational state ni to state nf , is Ep = (ni + 12 )ω − (nf + 12 )ω = nω, (where n ≡ ni − nf ). The frequency of the photon is Ep k nω n ν= = = . The splitting of this line is given by h 2π 2π µ n√ 1 n k ∆µ 1 ∆µ 1 ∆ν = = ν . k − 3/2 ∆µ = 2π 2 2π µ µ 2 µ 2µ Now µ=

mh mc = mh + mc

1 mc

1 +

1 mh

∆ν =

⇒ ∆µ =

−1 1 mc

+

1 mh

2

−

1 ∆mc m2c

=

µ2 ∆mc . m2c

1 µ∆mc 1 (∆mc /mc ) . = ν ν 2 m2c 2 1 + mc mh

Using the average value (36) for mc , we have ∆mc /mc = 2/36, and mc /mh = 36/1, so ∆ν =

1 (1/18) 1 ν= ν = 7.51 × 10−4 ν. 2 (1 + 36) (36)(37)



135

Problem 5.4 (a) |ψ± |2 d3 r1 d3 r2

1=

∗

= |A|2

[ψa (r1 )ψb (r2 ) ± ψb (r1 )ψa (r2 )] [ψa (r1 )ψb (r2 ) ± ψb (r1 )ψa (r2 )] d3 r1 d3 r2

±

|ψa (r1 )|2 d3 r1

= |A|2

ψb (r1 )∗ ψa (r1 )d3 r1

|ψb (r2 )|2 d3 r2 ±

ψa (r1 )∗ ψb (r1 )d3 r1

ψa (r2 )∗ ψb (r2 )d3 r2 +

ψb (r2 )∗ ψa (r2 )d3 r2

|ψb (r1 )|2 d3 r1

|ψa (r2 )|2 d3 r2

√ = |A|2 (1 · 1 ± 0 · 0 ± 0 · 0 + 1 · 1) = 2|A|2 =⇒ A = 1/ 2. (b) 1 = |A|

2

∗

[2ψa (r1 )ψa (r2 )] [2ψa (r1 )ψa (r2 )] d3 r1 d3 r2

= 4|A|2

|ψa (r1 )|2 d3 r1

|ψa (r2 )|2 d3 r2 = 4|A|2 .

A = 1/2.

Problem 5.5 (a) −

2 ∂ 2 ψ 2 ∂ 2 ψ − = Eψ 2 2m ∂x1 2m ∂x22

(for 0 ≤ x1 , x2 ≤ a, otherwise ψ = 0).

√ πx 2 πx1 2πx2 2πx1 2 ψ= sin sin − sin sin a a a a a √ 2 πx πx d2 ψ 2 2πx1 π 2 2πx2 2π 1 2 = sin sin − sin + sin dx21 a a a a a a a √ 2 πx πx d2 ψ 2π 2πx2 π 2 2 2πx1 1 2 − sin + sin = sin sin dx22 a a a a a a a c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

136


−

d2 ψ d2 ψ + 2 dx21 dx2 2 2m

=−

d2 ψ d2 ψ + 2 dx21 dx2

π 2 a

=

+

2π a

2 ψ = −5

5π 2 2 ψ = Eψ, 2ma2

π2 ψ, a2

with E =

5π 2 2 = 5K. 2ma2

(b) Distinguishable: ψ22 = (2/a) sin (2πx1 /a) sin (2πx2 /a) , with E22 = 8K (nondegenerate). ψ13 = (2/a) sin (πx1 /a) sin (3πx2 /a) , with E13 = E31 = 10K (doubly degenerate). ψ31 = (2/a) sin (3πx1 /a) sin (πx2 /a) Identical Bosons: ψ22 = (2/a) sin (2πx1 /a) sin (2πx2 /a), E22 = 8K (nondegenerate). √ ψ13 = ( 2/a) [sin (πx1 /a) sin (3πx2 /a) + sin (3πx1 /a) sin (πx2 /a)], E13 = 10K Identical Fermions: √

2

1

ψ13 = ( 2/a) sin πxa 1 sin 3πx − sin 3πx sin πxa 2 , E13 = 10K a a

(nondegenerate).

√

2

1

2 1 ψ23 = ( 2/a) sin 2πx sin 3πx − sin 3πx sin 2πx , E23 = 13K a a a a

Problem 5.6

(a) Use Eq. 5.19 and Problem 2.4, with xn = a/2 and x2 n = a2

1 3

−

(nondegenerate).

1 2(nπ)2

(nondegenerate).

.

1 1 1 1 + 2 − 6 2π 2 n2 m

nπ

a (m−n)π 1 a (b) xmn = a2 0 x sin mπ x − cos (m+n)π x dx a x sin a x dx = a 0 x cos a a 2 (m−n)π (m−n)π 1 a ax =a sin cos x + x (m−n)π a (m−n)π a 2 a (m+n)π (m+n)π a ax − (m+n)π cos x − (m+n)π sin x a a 0

2 2 1 a a =a (cos[(m − n)π] − 1) − (cos[(m + n)π] − 1) . (m−n)π (m+n)π (x1 − x2 )2 = a2

1 3

−

1 2(nπ)2

+ a2

1 3

−

1 2(mπ)2

−2·

a 2

·

a 2

= a2

.

But cos[(m ± n)π] = (−1)m+n , so

xmn

a = 2 (−1)m+n − 1 π

1 1 − 2 (m − n) (m + n)2

So Eq. 5.21 ⇒ (x1 − x2 ) = a 2

2

1 1 − 6 2π 2

=

1 1 + 2 n2 m

a(−8mn) π 2 (m2 −n2 )2 ,

0,

−

if m and n have opposite parity, if m and n have same parity.

128a2 m2 n2 . π 4 (m2 − n2 )4

(The last term is present only when m, n have opposite parity.) c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 5. IDENTICAL PARTICLES (c) Here Eq. 5.21 ⇒ (x1 − x2 )2 = a2

137 1 1 − 6 2π 2

1 1 + 2 n2 m

+

128a2 m2 n2 . π 4 (m2 − n2 )4

(Again, the last term is present only when m, n have opposite parity.)

Problem 5.7 (a) ψ(x1 , x2 , x3 ) = ψa (x1 )ψb (x2 )ψc (x3 ).

(b) ψ(x1 , x2 , x3 ) =

(c) ψ(x1 , x2 , x3 ) =

√1 [ψa (x1 )ψb (x2 )ψc (x3 ) 6

+ ψa (x1 )ψc (x2 )ψb (x3 ) + ψb (x1 )ψa (x2 )ψc (x3 ) . +ψb (x1 )ψc (x2 )ψa (x3 ) + ψc (x1 )ψb (x2 )ψa (x3 ) + ψc (x1 )ψa (x2 )ψb (x3 )] − ψa (x1 )ψc (x2 )ψb (x3 ) − ψb (x1 )ψa (x2 )ψc (x3 ) . +ψb (x1 )ψc (x2 )ψa (x3 ) − ψc (x1 )ψb (x2 )ψa (x3 ) + ψc (x1 )ψa (x2 )ψb (x3 )]

√1 [ψa (x1 )ψb (x2 )ψc (x3 ) 6

Problem 5.8 ψ = A ψ(r1 , r2 , r3 , . . . , rZ ) ± ψ(r2 , r1 , r3 , . . . , rZ ) + ψ(r2 , r3 , r1 , . . . , rZ ) + etc. , where “etc.” runs over all permutations of the arguments r1 , r2 , . . . , rZ , with a + sign for all even permutations (even number of transpositions ri ↔ rj , starting from r1 , r2 , . . . , rZ ), and ± for all odd permutations (+ √ for bosons, − for fermions). At the end of the process, normalize the result to determine A. (Typically A = 1/ Z!, but this may not be right if the starting function is already symmetric under some interchanges.)

Problem 5.9 (a) The energy of each electron is E = Z 2 E1 /n2 = 4E1 /4 = E1 = −13.6eV, so the total initial energy is 2 × (−13.6) eV= −27.2 eV. One electron drops to the ground state Z 2 E1 /1 = 4E1 , so the other is left with 2E1 − 4E1 = −2E1 = 27.2 eV. (b) He+ has one electron; it’s a hydrogenic ion (Problem 4.16) with Z = 2, so the spectrum is 1/λ = 4R 1/n2f − 1/n2i , where R is the hydrogen Rydberg constant, and ni , nf are the initial and final quantum numbers (1, 2, 3, . . . ).

Problem 5.10 (a) The ground state (Eq. 5.30) is spatially symmetric, so it goes with the symmetric (triplet) spin configuration. Thus the ground state is orthohelium, and it is triply degerate. The excited states (Eq. 5.32) come in ortho (triplet) and para (singlet) form; since the former go with the symmetric spatial wave function, the orthohelium states are higher in energy than the corresponding (nondegenerate) para states. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

138


(b) The ground state (Eq. 5.30) and all excited states (Eq. 5.32) come in both ortho and para form. All are quadruply degenerate (or at any rate we have no way a priori of knowing whether ortho or para are higher in energy, since we don’t know which goes with the symmetric spatial configuration).

Problem 5.11 (a) 0

1 |r1 − r2 |

1 =

∞

= 2π

8 πa3

2

1 = r1 r 2

r12

+

r22

= 4πe

1 r1

r1

0

r12 + r22 − 2r1 r2 cos θ2

∞

r1

= 4π

1 r1

dθ2 r22 dr2

π

1 2 2 2 2 − 2r1 r2 cos θ2 = r1 + r2 + 2r1 r2 − r1 + r2 − 2r1 r2 r1 r 2 0

r1

r22 e−4r2 /a dr2

2/r1 (r2 < r1 ) 2/r2 (r2 > r1 )

r22 e−4r2 /a dr2

∞

+

0

0

=−

sin θ2

1 [(r1 + r2 ) − |r1 − r2 |] = r1 r2

−4r1 /a

d r2 d3 r1

π

e 0

=

3

r12 + r22 − 2r1 r2 cos θ2

−4(r1 +r2 )/a

e−4(r1 +r2 )/a

−4r2 /a

r2 e

dr2 .

r1

r1 a 2 −4r2 /a a a 2 −4r2 /a 4r2 1 − r2 e − − 1 = + e r1 4 2 4 a 0

a ar1 −4r1 /a a2 −4r1 /a a2 e . r12 e−4r1 /a + + e − 4r1 2 8 8

r2 e−4r2 /a dr2 =

a 2 4

e−4r2 /a

−

∞ ar1 −4r1 /a a2 −4r1 /a 4r2 + e . − 1 = e a 4 16 r1

a3 −4r1 /a a2 a3 a2 −8r1 /a ar1 ar1 − − + e e + − + 32r1 a 8 32r1 4 16



=

0

πa2 8

32 = 4 a

32 = 4 a

a a −4r1 /a e−8r1 /a . e − 2+ r1 r1

1

1 |r1 − r2 | a

8 = · 4π πa4

∞

−4r1 /a

r1 e

∞

a·

0

a −4r1 /a a −8r1 /a e r12 dr1 e − 2+ r1 r1

dr1 − 2

0

139

∞

r12 e−8r1 /a dr1

−a

0

a 2 4

−2·2

a 3 8

−a·

∞

−8r1 /a

r1 e

dr1

0

a 2 8

32 = a

1 1 1 − − 16 128 64

=

5 . 4a

(b) Vee ≈

e2 4π90

0

1 |r1 − r2 |

1 =

5 e2 1 5m = 4 4π90 a 4 2

e2 4π90

2 =

5 5 (−E1 ) = (13.6 eV) = 34 eV. 2 2

E0 + Vee = (−109 + 34)eV = −75 eV, which is pretty close to the experimental value (−79 eV).

Problem 5.12 (a) Hydrogen: (1s); helium: (1s)2 ; lithium: (1s)2 (2s); beryllium: (1s)2 (2s)2 ; boron: (1s)2 (2s)2 (2p); carbon: (1s)2 (2s)2 (2p)2 ; nitrogen: (1s)2 (2s)2 (2p)3 ; oxygen: (1s)2 (2s)2 (2p)4 ; fluorine: (1s)2 (2s)2 (2p)5 ; neon: (1s)2 (2s)2 (2p)6 . These values agree with those in Table 5.1—no surprises so far. (b) Hydrogen: 2 S1/2 ; helium: 1 S0 ; lithium: 2 S1/2 ; beryllium 1 S0 . (These four are unambiguous, because the orbital angular momentum is zero in all cases.) For boron, the spin (1/2) and orbital (1) angular momenta could add to give 3/2 or 1/2, so the possibilities are 2 P3/2 or 2 P1/2 . For carbon, the two p electrons could combine for orbital angular momentum 2, 1, or 0, and the spins could add to 1 or 0: 1 S0 , 3 S1 , 1 P1 , 3 P2 , 3 P1 , 3 P0 , 1 D2 , 3 D3 , 3 D2 , 3 D1 . For nitrogen, the 3 p electrons can add to orbital angular momentum 3, 2, 1, or 0, and the spins to 3/2 or 1/2: 2 4

S1/2 , 4 S3/2 , 2 P1/2 , 2 P3/2 , 4 P1/2 , 4 P3/2 , 4 P5/2 , 2 D3/2 , 2 D5/2 , D1/2 , 4 D3/2 , 4 D5/2 , 4 D7/2 , 2 F5/2 , 2 F3/2 , 4 F3/2 , 4 F5/2 , 4 F7/2 , 4 F9/2 .


140


Problem 5.13 (a) Orthohelium should have lower energy than parahelium, for corresponding states (which is true). (b) Hund’s first rule says S = 1 for the ground state of carbon. But this (the triplet) is symmetric, so the orbital state will have to be antisymmetric. Hund’s second rule favors L = 2, but this is symmetric, as you can see most easily by going to the “top of the ladder”: |2 2 = |1 11 ||1 12 . So the ground state of carbon will be S = 1, L = 1. This leaves three possibilities: 3 P2 , 3 P1 , and 3 P0 . (c) For boron there is only one electron in the 2p subshell (which can accommodate a total of 6), so Hund’s third rule says the ground state will have J = |L − S|. We found in Problem 5.12(b) that L = 1 and S = 1/2, so J = 1/2, and the configuration is

2

P1/2 .

(d) For carbon we know that S = 1 and L = 1, and there are only two electrons in the outer subshell, so Hund’s third rule says J = 0, and the ground state configuration must be

3

P0 .

For nitrogen Hund’s first rule says S = 3/2, which is symmetric (the top of the ladder is | 32 32 = Hund’s second rule favors L = 3, but this is also symmetric. In fact, the only antisymmetric orbital configuration here is L = 0. [You can check this directly by working out the Clebsch-Gordan coefficients, but it’s easier to reason as follows: Suppose the three outer electrons are in the “top of the ladder” spin state, so each one has spin up (| 12 12 ); then (since the spin states are all the same) the orbital states have to be different: |1 1, |1 0, and |1−1. In particular, the total z-component of orbital angular momentum has to be zero. But the only configuration that restricts Lz to zero is L = 0.] The outer subshell is exactly half filled (three electrons with n = 2, l = 1), so Hund’s third rule says | 12 12 1 | 12 21 2 | 12 12 3 ).

J = |L − S| = |0 − 32 | = 3/2. Conclusion: The ground state of nitrogen is

4

S3/2 . (Table 5.1 confirms

this.)

Problem 5.14 S = 2; L = 6; J = 8. (1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)10 (4s)2 (4p)6 (4d)10 (5s)2 (5p)6 (4f )10 (6s)2 . definite (36 electrons) likely (30 electrons)

Problem 5.15 Divide Eq. 5.45 by Eq. 5.43, using Eq. 5.42: Etot /N q 3 2 (3π 2 N q)5/3 1 2m = = . 2 2/3 2 2 2/3 EF N q (3π N q/V ) 5 10π mV

Problem 5.16 N atoms moles gm NA 2 Nq (3ρπ 2 )2/3 . ρ = = = × × = · d, where NA is Avogadro’s 2m V V mole gm volume M 23 number (6.02 × 10 ), M = atomic mass = 63.5 gm/mol, d = density = 8.96 gm/cm3 . (6.02 × 1023 )(8.96 gm/cm3 ) ρ= = 8.49 × 1022 /cm3 = 8.49 × 1028 /m3 . (63.5 gm) (1.055 × 10−34 J · s)(6.58 × 10−16 eV · s) EF = (3π 2 8.49 × 1028 /m3 )2/3 = 7.04 eV. (2)(9.109 × 10−31 kg)

(a) EF =



141

(b) 7.04 eV =

v2 14.08 v 1 = 2.76 × 10−5 ⇒ = 5.25 × 10−3 , (0.511 × 106 eV/c2 )v 2 ⇒ 2 = 2 c .511 × 106 c

so it’s nonrelativistic.

v = (5.25 × 10−3 ) × (3 × 108 ) = 1.57 ×106 m/s.

(c) T =

7.04 eV = 8.17 × 104 K. 8.62 × 10−5 eV/K

P =

(3π 2 )2/3 2 5/3 (3π 2 )2/3 (1.055 × 10−34 )2 2 ρ (8.49 × 1028 )5/3 N/m = 3.84 × 1010 N/m2 . = 5m 5(9.109 × 10−31 )

(d)

Problem 5.17 (3π 2 )2/3 2 P = 5m

Nq V

5/3 = AV

−5/3

dP = −V A ⇒ B = −V dV

−5 3

V −5/3−1 =

5 5 AV −5/3 = P. 3 3

For copper, B = 53 (3.84 × 1010 N/m2 ) = 6.4 × 1010 N/m2 .

Problem 5.18 (a) Equations 5.59 and 5.63 ⇒ ψ = A sin kx + B cos kx;

A sin ka = eiKa − cos ka B. So

iKa A sin ka A cos kx = iKa e sin kx − sin kx cos ka + cos kx sin ka (eiKa − cos ka) (e − cos ka) . AeiKa = C sin kx + e−iKa sin[k(a − x)] , where C ≡ iKa . e − cos ka

ψ = A sin kx +

(b) If z = ka = jπ, then sin ka = 0, Eq. 5.64 ⇒ cos Ka = cos ka = (−1)j ⇒ sin Ka = 0, so eiKa = cos Ka + i sin Ka = (−1)j , and the constant C involves division by zero. In this case we must go back to Eq. 5.63, which is a tautology (0=0) yielding no constraint on A or B, Eq. 5.61 holds automatically, and Eq. 5.62 gives 2mα kA − (−1)j k A(−1)j − 0 = 2 B ⇒ B = 0. So ψ = A sin kx. Here ψ is zero at each delta spike, so the wave function never “feels” the potential at all.

Problem 5.19 We’re looking for a solution to Eq. 5.66 with β = 10 and z π: f (z) = cos z + 10 Mathematica gives z = 2.62768.

So

E=

sin z = 1. z

2 k 2 2 z 2 (2.62768)2 z2 α = = eV = 0.345 eV. = 2 2m 2ma 2β a 20


142


Problem 5.20 Positive-energy solutions. These are the same as before, except that α (and hence also β) is now a negative number. Negative-energy solutions. On 0 < x < a we have √ d2 ψ −2mE 2 = κ ψ, where κ ≡ ⇒ ψ(x) = A sinh kx + B cosh kx. dx2 According to Bloch’s theorem the solution on −a < x < 0 is ψ(x) = e−iKa [A sinh κ(x + a) + B cosh κ(x + a)] . Continuity at x = 0 ⇒

or A sinh κa = B eiKa − cosh κa .

B = e−iKa [A sinh κa + B cosh κa] , The discontinuity in ψ (Eq. 2.125) ⇒ κA−e−iKa κ [A cosh κa + B sinh κa] =

(1)

2mα 2mα −iKa −iKa B, or A 1 − e cosh κa = B sinh κa . (2) + e 2 2 κ

Plugging (1) into (2) and cancelling B:

2mα eiKa − cosh κa 1 − e−iKa cosh κa = 2 sinh κa + e−iKa sinh2 κa. κ 2mα sinh κa. 2 κ mα cos Ka = cosh κa + 2 sinh κa. κ

eiKa − 2 cosh κa + e−iKa cosh2 κa − e−iKa sinh2 κa = eiKa + e−iKa = 2 cosh κa +

2ma sinh κa, 2 κ

This is the analog to Eq. 5.64. As before, we let β ≡ mαa/2 (but remember it’s now a negative number), and this time we define z ≡ −κa, extending Eq. 5.65 to negative z, where it represents negative-energy solutions. In this region we define sinh z f (z) = cosh z + β . (3) z In the Figure I have plotted f (z) for β = −1.5, using Eq. 5.66 for postive z and (3) for negative z. As before, allowed energies are restricted to the range −1 ≤ f (z) ≤ 1, and occur at intersections of f (z) with the N horizontal lines cos Ka = cos(2πn/N a), with n = 0, 1, 2 . . . N − 1. Evidently the first band (partly negative, and partly positive) contains N states, as do all the higher bands.

1

0

-1 0

π

2π

3π

4π



143

Problem 5.21 2πn n Equation 5.56 says K = ⇒ Ka = 2π ; at the bottom of page 227 we found that n = 0, 1, 2, . . . , N − 1. Na N Each value of n corresponds to a distinct state. To find the allowed energies we draw N horizontal lines on Figure 5.6, at heights cos Ka = cos(2πn/N ), and look for intersections with f (z). The point is that almost all of these lines come in pairs—two different n’s yielding the same value of cos Ka: N = 1 ⇒ n = 0 ⇒ cos Ka = 1. Nondegenerate. N = 2 ⇒ n = 0, 1 ⇒ cos Ka = 1, −1. Nondegenerate. N = 3 ⇒ n = 0, 1, 2 ⇒ cos Ka = 1, − 12 , − 12 . The first is nondegenerate, the other two are degenerate. N = 4 ⇒ n = 0, 1, 2, 3 ⇒ cos Ka = 1, 0, −1, 0. Two are nondegenerate, the others are degenerate. Evidently they are doubly degenerate (two different n’s give same cos Ka) except when cos Ka = ±1, i.e., at the top or bottom of a band. The Bloch factors eiKa lie at equal angles in the complex plane, starting with 1 (see Figure, drawn for the case N = 8); by symmetry, there is always one with negative imaginary part symmetrically opposite each one with positive imaginary part; these two have the same real part (cos Ka). Only points which fall on the real axis have no twins.

sin(Ka) n=2 n=3

n=1 n=0

n=4 n=5

cos(Ka)

n=7 n=6

Problem 5.22 (a) ' (3 5πxA 2 7πxB 17πxC 5πxA 17πxB 7πxC sin sin sin − sin sin sin a a a a a a a 7πxA 17πxB 5πxC 7πxA 5πxB 17πxC + sin sin sin − sin sin sin a a a a a a 17πxA 5πxB 7πxC 17πxA 7πxB 5πxC + sin sin sin − sin sin sin . a a a a a a

1 ψ(xA , xB , xC ) = √ 6

(b) (i)

' (3 11πxB 11πxC 2 11πxA ψ= sin sin . sin a a a a


144

CHAPTER 5. IDENTICAL PARTICLES (ii) ' (3 πx πx 2 19πxC A B sin sin sin a a a a πx πx πx 19πxB πxC 19πxA A B C + sin sin sin + sin sin sin . a a a a a a

1 ψ = √ 3

(iii) ' (3 5πxA 2 7πxB 17πxC 5πxA 17πxB 7πxC sin sin sin + sin sin sin a a a a a a a 7πxA 17πxB 5πxC 7πxA 5πxB 17πxC + sin sin sin + sin sin sin a a a a a a 17πxA 5πxB 7πxC 17πxA 7πxB 5πxC + sin sin sin + sin sin sin . a a a a a a

1 ψ = √ 6

Problem 5.23 (a) En1 n2 n3 = (n1 + n2 + n3 + 32 )ω = 92 ω ⇒ n1 + n2 + n3 = 3. (n1 , n2 , n3 = 0, 1, 2, 3 . . . ). n1 0 0 3 0 0 1 1 2 2 1

State n2 n3 0 3 3 0 0 0 1 2 2 1 0 2 2 0 0 1 1 0 1 1

Configuration (N0 , N1 , N2 . . . )

# of States

(2,0,0,1,0,0 . . . )

3

(1,1,1,0,0,0 . . . )

6

(0,3,0,0,0 . . . )

1

Most probable configuration:

Possible single-particle energies: E0 = ω/2 : E1 = 3ω/2 : E2 = 5ω/2 : E3 = 7ω/2 :

P0 P1 P2 P3

= 12/30 = 4/10. = 9/30 = 3/10. = 6/30 = 2/10. = 3/30 = 1/10.

(1,1,1,0,0,0 . . . ).

Most probable single-particle energy:

E0 = 12 ω.

(b) For identical fermions the only configuration is (1,1,1,0,0,0 . . . ) (one state), so this is also the most probable configuration. The possible one-particle energies are E0 (P0 = 1/3),

E1 (P1 = 1/3),

E2 (P2 = 1/3),

and they are all equally likely, so it’s a 3-way tie for the most probable energy. (c) For identical bosons all three configurations are possible, and there is one state for each. Possible oneparticle energies: E0 (P0 = 1/3), E1 (P1 = 4/9), E2 (P2 = 1/9), E3 (P3 = 1/9). Most probable energy: E1 .



145

Problem 5.24

 ∞ < 1   (distinguishable), Eq. 5.74 ⇒ Q = 6    N n!  n=1 ∞ < Here N = 3, and dn = 1 for all states, so: 1  Eq. 5.75 ⇒ Q = (fermions),   N !(1 − Nn )!  n  n=1  Eq. 5.77 ⇒ Q = 1 (bosons). (In the products, most factors are 1/0! or1/1!, both of which are 1, so I won’t write them.) 1   = 1 (distinguishable), Q=6×   3! 1 1 Configuration 1 (N11 = 3, others 0): Q= (fermions), × = 0   3! (−2)!   Q= 1 (bosons).  1 1   (distinguishable), Q=6× × = 3   1! 2! 1 1 Configuration 2 (N5 = 1, N13 = 2): Q= × = 0 (fermions),   1!0! 2!(−1)!   Q= 1 (bosons).  1 1  Q = 6× × (distinguishable), = 3   2! 1! 1 1 Configuration 3 (N1 = 2, N19 = 1): Q= × = 0 (fermions),   2!(−1)! 1!0!   Q= 1 (bosons).  1 1 1     Q = 6 × 1! × 1! × 1! = 6 (distinguishable), 1 1 1 Configuration 4 (N5 = N7 = N17 = 1): Q= × × = 1 (fermions),   1!0! 1!0! 1!0!   Q= 1 (bosons). All of these agree with what we got “by hand” at the top of page 231.

Problem 5.25 N = 1 : - can put the ball in any of d baskets, so d ways.   - could put both balls in any of the d baskets : d ways, or N = 2 : - could put one in one basket (d ways), the other in another(d − 1) ways—but it  doesn’t matter which is which, so divide by 2. Total: d + 12 d(d − 1) = 12 d(2 + d − 1) = 12 d(d + 1) ways.   - could put all three in one basket : d ways, or N = 3 : - 2 in one basket, one in another : d(d − 1) ways, or  - 1 each in 3 baskets : d(d − 1)(d − 2)/3! ways. Total: d + d(d − 1) + d(d − 1)(d − 2)/6 = 16 d(6 + 6d − 6 + d2 − 3d + 2) = 16 d(d2 + 3d + 2) c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

146

CHAPTER 5. IDENTICAL PARTICLES d(d + 1)(d + 2) ways. 6

=       N =4:      -

all in one basket: d ways, or 3 in one basket, 1 in another: d(d − 1) ways, or 2 in one basket, 2 in another: d(d − 1)/2 ways, or 2 in one basket, one each in others: d(d − 1)(d − 2)/2, or all in different baskets: d(d − 1)(d − 2)(d − 3)/4!

Total: d + d(d − 1) + d(d − 1)/2 + d(d − 1)(d − 2)/2 + d(d − 1)(d − 2)(d − 3)/24 1 = 24 (24 + 24d − 24 + 12d − 12 + 12d2 − 36d + 24 + d3 − 6d2 + 11d − 6) =

1 3 24 d(d

+ 6d2 + 11d + 6) =

d(d + 1)(d + 2)(d + 3) ways. 24

d(d + 1)(d + 2) · · · (d + N − 1) d+N −1 (d + N − 1)! The general formula seems to be f (N, d) = . = = N N! N !(d − 1)! Proof: How many ways to put N identical balls in d baskets? Call it f (N, d). - Could put all of them in the first basket: 1 way. - Could put all but one in the first basket; there remains 1 ball for d − 1 baskets: f (1, d − 1) ways. - Could put all but two in the first basket; there remain 2 for d − 1 baskets: f (2, d − 1) ways. .. . - Could put zero in the first basket, leaving N for d − 1 baskets: f (N, d − 1) ways. /N Thus: f (N, d) = f (0, d−1)+f (1, d−1)+f (2, d−1)+· · ·+f (N, d−1) = j=0 f (j, d−1) (where f (0, d) ≡ 1). /N −1 It follows that f (N, d) = j=0 f (j, d − 1) + f (N, d − 1) = f (N − 1, d) + f (N, d − 1). Use this recursion relation to confirm the conjectured formula by induction:

d+N −1 N

?

=

d+N −2 d+N −2 (d + N − 2)! (d + N − 2)! + = + N −1 N (N − 1)!(d − 1)! N !(d − 2)!

(d + N − 2)! (d + N − 1)! = (N + d − 1) = = N !(d − 1)! N !(d − 1)! It works for N = 0 :

d−1

0

= 1, and for d = 1 :

N

N

d+N −1 . d−1

= 1 (which is obviously correct for just one basket). QED

Problem 5.26 A(x, y) = (2x)(2y) = 4xy; maximize, subject to the constraint (x/a)2 + (y/b)2 = 1.

b

(x,y) a



147

G(x, y, λ) ≡ 4xy + λ (x/a)2 + (y/b)2 − 1 . ∂G 2λ 2λy = 4x + 2 = 0 ⇒ 4x = − 2 ∂y b b

λx − 2 2a

⇒ 4x =

∂G λx 2λx = 4y + 2 = 0 ⇒ y = − 2 . ∂x a 2a λ2 x ⇒ x = 0 (minimum), or else λ = ±2ab. a2 b2

2abx b = ∓ x. We may as well pick x and y positive, (as in the figure); then y = (b/a)x (and 2a2 a √ ∂G x 2 y 2 x2 b2 x2 2 λ = −2ab). =0⇒ + = 1 (of course), so 2 + 2 2 = 1, or 2 x2 = 1, or x = a/ 2, and hence ∂λ a b a a b a √ √ a b y = ba/(a 2) ⇒ y = b/ 2. A = 4 √ √ = 2ab. 2 2 So y = ∓

Problem 5.27 (a) ln(10!) = ln(3628800) = 15.1044; 10 ln(10) − 10 = 23.026 − 10 = 13.0259; 2.0785; 2.0785/15.1044 = 0.1376, or 14% .

(b) The percent error is:

ln(z!) − z ln(z) + z × 100. ln(z!)

z 20 100 50 90 85 89

15.1044 − 13.0259 =

% 5.7 0.89 1.9 0.996 1.06 1.009

Since my calculator cannot compute factorials greater than 69! I used Mathematica to construct the table. Evidently, the smallest integer for which the error is < 1% is 90.

Problem 5.28

∞ V Equation 5.108 ⇒ N = k 2 n(9) dk, where n(9) is given (as T → 0) by Eq. 5.104. 2π 2 0 √ kmax 3 2 V 2mEF V kmax 2 kmax 2 So N = k dk = , where kmax is given by = µ(0) = EF ⇒ kmax = . 2π 2 0 2π 2 3 2m V N= (2mEF )3/2 . Compare Eq. 5.43, which says 6π 2 3 2/3 Nq Nq 2 (2mEF )3/2 V EF = , or = 3π 2 (2mEF )3/2 . 3π 2 , or N = 2m V 3 V 3π 2 q3 Here q = 1, and Eq. 5.108 needs an extra factor of 2 on the right, to account for spin, so the two formulas agree. kmax 5 V V 2 V 2 kmax Equation 5.109 ⇒ Etot = ⇒ Etot = k 4 dk = (2mEF )5/2 . 2 2 4π m 0 4π m 5 20π 2 m3 V 2 5 Compare Eq. 5.45, which says Etot = . Again, Eq. 5.109 for electrons has an extra factor of 2, so k 10π 2 m max the two agree.


148


Problem 5.29 (a) Equation 5.103, n(9) > 0 ⇒ allowed energies 9.

1 e(5−µ)/kB T

−1

> 0 ⇒ e(5−µ)/kB T > 1 ⇒

(9 − µ) > 0 ⇒ 9 > µ(T ), for all kB T

2 2 k → 0 (as k → 0, in the continuum limit), so µ(T ) is always negative. 2m 2 π 2 1 1 1 (Technically, the lowest energy is , but we take the dimensions lx ly lz to be very large + + 2m lx2 ly2 lz2 ∞ 1 k2 in the continuum limit.) Equation 5.108 ⇒ N/V = dk. The integrand is 2 2 2π 2 0 e( k /2m−µ)/kB T − 1 always positive, and the only T dependence is in µ(T ) and kB T . So, as T decreases, (2 k 2 /2m) − µ(T ) must also decrease, and hence −µ(T ) decreases, or µ(T ) increases (always negative). √ √ ∞ N k2 2mkB T 1/2 2mkB T 1 −1/2 2 k 2 1 (c) ; dk = dx. dk. Let x ≡ = , so k = x x V 2π 2 0 e2 k2 /2mkB T − 1 2mkB T 2 3/2 ∞ 1/2 ∞ 3/2−1 N 1 2mkB T 1 x x = dx, where dx = Γ(3/2)ζ(3/2). V 2π 2 2 2 0 ex − 1 ex − 1 0 2/3 3/2 √ mkB T N N 2π2 Now Γ(3/2) = π/2; ζ(3/2) = 2.61238, so = 2.612 ; T = . c V 2π2 mkB 2.612V

(b) For a free particle gas, E =

(d) 3

N mass/volume 0.15 × 103 kg/m = = = 2.2 × 1028 /m3 . V mass/atom 4(1.67 × 10−27 kg)

Tc =

2π(1.05 × 10−34 J · s)2 4(1.67 × 10−27 kg)(1.38 × 10−23 J/K)

2.2 × 1028 2.61 m3

2/3 = 3.1 K.

Problem 5.30 (a) ω = 2πν =

2πc , λ

ρ(ω)|dω| = 8π

so

dω = −

2πc dλ, λ2

and

ρ(ω) =

(2πc)3 . π 2 c3 λ3 (e2πc/kB T λ − 1)

2πc 16π 2 c − = ρ(λ) dλ ⇒ ρ(λ) = . dλ 2 3 2πc/k T λ 5 2πc/k B B T λ − 1) λ λ (e − 1) λ (e 1

(For density, we want only the size of the interval, not its sign.) (b) To maximize, set dρ/dλ = 0:

−5

e2πc/kB T λ (2πc/kB T ) 0 = 16π c 6 2πc/k T λ − B λ (e − 1) λ5 (e2πc/kB T λ − 1)2 2

1 − 2 λ



149 5 4

5-x

3 2

5e -x

1

1

2

5

4

3

⇒ 5(e

2πc/kB T λ

− 1) = e

2πc/kB T λ

2πc kB T λ

.

Let x ≡ 2πc/kB T λ; then 5(ex − 1) = xex ; or 5(1 − e−x ) = x, or 5e−x = 5 − x. From the graph, the solution occurs slightly below x = 5. Mathematica says x = 4.966, so λmax =

2πc 1 (4.966)kB T

=

(6.626 × 10−34 J · s)(2.998 × 108 m/s) 1 = (4.966)(1.3807 × 10−23 J/K) T

2.897 × 10−3 m·K/T.

Problem 5.31 From Eq. 5.113: ∞ ∞ E ω3 ρ(ω) dω = 2 3 dω. = V π c 0 (eω/kB T − 1) 0 E = 2 3 V π c =

kB T

4

∞

0

Let x ≡

ω . kB T

Then

(kB T )4 x3 (kB T )4 π4 dx = 2 3 3 Γ(4)ζ(4) = 2 3 3 · 6 · = x e −1 π c π c 90

4 π 2 kB 3 15c 3

T4

J π 2 (1.3807 × 10−23 J/K)4 T 4 = 7.566 × 10−16 3 4 T 4 . QED 15(2.998 × 108 m/s)3 (1.0546 × 10−34 J · s)3 m K

Problem 5.32 From Problem 2.11(a), x0 = x1 = 0;

x2 0 =

; 2mω

x2 1 =

3 . 2mω

From Eq. 3.98, x01 =

∞

−∞

xψ0 (x)ψ1 (x) dx = 0|x|1 =

√ √ . 1 δ0 0 + 0 δ1 −1 = 2mω 2mω

(a) Equation 5.19 ⇒ (x1 − x2 )2 d =

3 2 + −0= . 2mω 2mω mω

(b) Equation 5.21 ⇒ (x1 − x2 )2 + =

2 −2 = . mω 2mω mω


150


(c) Equation 5.21 ⇒ (x1 − x2 )2 − =

3 2 +2 = . mω 2mω mω

Problem 5.33 (a) Each particle has 3 possible states: 3 × 3 × 3 = 27. (b) All in same state: aaa, bbb, ccc ⇒ 3. 2 in one state: aab, aac, bba, bbc, cca, ccb ⇒ 6 (each symmetrized). 3 different states: abc (symmetrized) ⇒ 1. Total: 10. (c) Only abc (antisymmetrized) =⇒ 1.

Problem 5.34

( ' n2y 2 k 2 π 2 2 n2x πnx πny = . Each state is represented by an Equation 5.39 ⇒ Enx ny = + , , with k = 2m lx2 ly2 2m lx ly intersection on a grid in “k-space”—this time a plane—and each state occupies an area π 2 /lx ly = π 2 /A (where A ≡ lx ly is the area of the well). Two electrons per state means 1 2 Nq πkF = 4 2

π2 A

, or kF =

Nq 2π A

1/2 = (2πσ)1/2 ,

where σ ≡ N q/A is the number of free electrons per unit area. ∴ EF =

π2 σ 2 kF2 2 = 2πσ = . 2m 2m m

ky

kx



151

Problem 5.35 (a) V =

4 3 πR , 3

so E =

2 (3π 2 N q)5/3 10π 2 m

4 3 πR 3

−2/3

22 15πmR2

=

9 πN q 4

5/3 .

(b) Imagine building up a sphere by layers. When it has reached mass m, and radius r, the work necessary to bring in the next increment dm is: dW = −(Gm/r) dm. In terms of the mass density ρ, m = 43 πr3 ρ, and dm = 4πr2 drρ, where dr is the resulting increase in radius. Thus: 4 dr 16π 2 2 4 dW = −G πr3 ρ 4πr2 ρ =− ρ Gr dr, 3 r 3 and the total energy of a sphere of radius R is therefore R NM 16π 2 2 16π 2 ρ2 R5 Egrav = − ρ G G. But ρ = r4 dr = − , so 3 15 4/3πR3 0

Egrav = −

16π 2 R5 9N 2 M 2 3 N 2M 2 = − G G . 15 16π 2 R6 5 R

(c) Etot

A B = 2− , R R

22 where A ≡ 15πm

9 πN q 4

dEtot 2A B = − 3 + 2 = 0 ⇒ 2A = BR, dR R R R= R=

4 9π

9π 4

9π 4

5/3

N 5/3 N2

2/3 (6.673 ×

10−11

so

5/3

R=

2 q 5/3 = GmM 2

4 2A = B 15πm

9π 4

3 GN 2 M 2 . 5

B≡

and

2/3

9 πN q 4

5/3

5 . 3GN 2 M 2

2 q 5/3 . GmM 2 N 1/3

(1.055 × 10−34 J · s)2 (1/2)5/3 N −1/3 Nm2 /kg2 )(9.109 × 10−31 kg)(1.674 × 10−27 kg)2

= (7.58 × 1025 m)N −1/3 . (d) Mass of sun:

1.989 × 1030 kg,

so N =

1.989 × 1030 = 1.188 × 1057 ; N −1/3 = 9.44 × 10−20 . 1.674 × 10−27

R = (7.58 × 1025 )(9.44 × 10−20 ) m = 7.16 × 106 m (slightly larger than the earth). (e) 2 From Eq. 5.43: EF = 2m

Nq 3π 4/3πR3 2

2/3

2 = 2mR2

9π Nq 4

2/3 .

Numerically:


152


EF =

2/3 (1.055 × 10−34 J · s)2 9π 57 1 ) = 3.102 × 10−14 J, (1.188 × 10 2(9.109 × 10−31 kg)(7.16 × 106 m)2 4 2

or, in electron volts: EF =

3.102 × 10−14 eV = 1.94 × 105 eV. 1.602 × 10−19

Erest = mc2 = 5.11 × 105 eV, so the Fermi energy (which is the energy of the most energetic electrons) is comparable to the rest energy, so they are getting fairly relativistic.

Problem 5.36 (a) V 2 cV k dk ⇒ Etot = 2 2 π π

dE = (ck)

So Etot =

kF

k 3 dk =

0

cV 4 k ; 4π 2 F

kF =

3π 2 N q V

4/3

1/3 .

c (3π 2 N q)4/3 V −1/3 . 4π 2

(b) 4 c V = πR3 ⇒ Edeg = (3π 2 N q)4/3 3 4π 2 R

4π 3

−1/3

c = 3πR

9 πN q 4

Adding in the gravitational energy, from Problem 5.35(b), 4/3 A B 3 c 9 Etot = and B ≡ GN 2 M 2 . − , where A ≡ πN q R R 3π 4 5

.

dEtot (A − B) = 0 ⇒ A = B, =− dR R2

but there is no special value of R for which Etot is minimal. Critical value: A = B(Etot = 0) ⇒ 4/3 c 9 3 = GN 2 M 2 , or πN q 3π 4 5

Nc =

15 √ 5π 16

= 2.04 × 1057 .

c G

3/2

q2 15 √ = 5π 3 M 16

1.055 × 10−34 J · s × 2.998 × 108 m/s 6.673 × 10−11 N · m2 /kg2

3/2

(1/2)2 (1.674 × 10−27 kg)3

(About twice the value for the sun—Problem 5.35(d).)

(c) Same as Problem 5.35(c), with m → M and q → 1, so multiply old answer by (2)5/3 m/M : R = 25/3

(9.109 × 10−31 ) (7.58 × 1025 m)N −1/3 = (1.31 × 1023 m)N −1/3 . Using N = 1.188 × 1057 , (1.674 × 10−27 )

R = (1.31 × 1023 m)(9.44 × 10−20 ) = 12.4 km. To get EF , use Problem 5.35(e) with q = 1, the new R, and the neutron mass in place of m: 2 7.16 × 106 9.11 × 10−31 EF = 22/3 (1.94 × 105 eV) = 5.60 × 107 eV = 56.0 MeV. 1.24 × 104 1.67 × 10−27 The rest energy of a neutron is 940 MeV, so a neutron star is reasonably nonrelativistic.



153

Problem 5.37 (a) From Problem 4.38: En = (n + 32 )ω, with n = 0, 1, 2, . . . ; dn = 12 (n + 1)(n + 2). From Eq. 5.103, n(9) = e−(5−µ)/kB T , so Nn = 12 (n + 1)(n + 2)e(µ− 2 ω)/kB T e−nω/kB T . ∞ ∞ 3 1 N= Nn = e(µ− 2 ω)/kB T (n + 1)(n + 2)xn , where x ≡ e−ω/kB T . Now 2 n=0 n=0 3

∞ ∞ ∞ ∞ 1 x d 1 x n n+1 x ⇒ x ⇒ (n + 1)xn ⇒ = (n + 1)xn . = = = 2 1 − x n=0 1 − x n=0 dx 1 − x (1 − x) n=0 n=0 ∞ ∞ x2 d 2x x2 n+2 = = (n + 1)x , and hence (n + 1)(n + 2)xn+1 = . 2 2 (1 − x) dx (1 − x) (1 − x)3 n=0 n=0 ∞

(n + 1)(n + 2)xn =

n=0

2 . (1 − x)3

So N = eµ/kB T e− 2 ω/kB T

eµ/kB T = N (1 − e−ω/kB T )3 e 2 ω/kB T ; 3

E=

∞ n=0

N n En =

1

3

(1 −

e−ω/kB T )3

.

µ = kB T ln N + 3 ln(1 − e−ω/kB T ) + 32 ω/kB T .

∞ 3 1 (n + 3/2)(n + 1)(n + 2)xn . ωe(µ− 2 ω)/kB T 2 n=0

From above,

∞ ∞ 2x3/2 d 2x3/2 n+3/2 = = (n + 1)(n + 2)x ⇒ (n + 3/2)(n + 1)(n + 2)xn+1/2 , or 3 (1 − x)3 dx (1 − x) n=0 n=0 ∞

d (n + 3/2)(n + 1)(n + 2)xn = 1/2 dx x n=0 E=

1

2x3/2 (1 − x)3

=

2 x1/2

3 1/2 2x

3x3/2 + (1 − x)3 (1 − x)4

3 3 1 3(1 + e−ω/kB T ) ωe(µ− 2 ω)/kB T . But e(µ− 2 ω)/kB T = N (1 − e−ω/kB T )3 , 2 (1 − e−ω/kB T )4

=

3(1 + x) . (1 − x)4

so

1 + e−ω/kB T 3

. E = N ω 2 1 − e−ω/kB T (b) kB T > ω (high temperature) ⇒ e−ω/kB T ≈ 1 − (ω/kB T ), so E ≈ 3N kB T (µ ≈ kB T [ln N + 3 ln (ω/kB T )]) . The equipartition theorem says E = N # 12 kB T , where # is the number of degrees of freedom for each particle. In this case #/2 = 3, or # = 6 (3 kinetic, 3 potential, for each particle—one of each for each direction in space).


154

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY

Chapter 6

Time-Independent Perturbation Theory Problem 6.1 (a)

a nπ nπ 2 2 a sin2 sin x , so En1 = ψn0 |H |ψn0 = α x δ x− dx. a a a 0 a 2

ψn0 (x)

=

En1 =

nπ a 2α nπ 2α sin2 = sin2 = a a 2 a 2

0, if n is even, 2α/a, if n is odd.

For even n the wave function is zero at the location of the perturbation (x = a/2), so it never “feels” H . (b) Here n = 1, so we need 0 ψm |H |ψ10 =

2α a

sin

mπ π mπ π 2α mπ a 2α x δ x− sin x dx = sin sin = sin . a 2 a a 2 2 a 2

This is zero for even m, so the first three nonzero terms will be m = 3, m = 5, and m = 7. Meanwhile, π 2 2 0 E10 − Em = (1 − m2 ), so 2ma2

(2α/a) sin(mπ/2) 0 2α 2ma2 −1 0 1 −1 0 0 ψ = + + + . . . ψ ψ ψ m 0 E10 − Em a π 2 2 1 − 9 3 1 − 25 5 1 − 49 7 m=3,5,7,...

4maα 2 1 3π 1 5π 1 7π = 2 2 sin x − sin x + sin x + ... π a 8 a 24 a 48 a

3π 1 5π 1 7π mα a = 2 2 sin x − sin x + sin x + ... . π 2 a 3 a 6 a

ψ11 =



155

Problem 6.2

√ (a) En = (n + 12 )ω , where ω ≡ k(1 + 9)/m = ω 1 + 9 = ω(1 + 12 9 − 18 92 + √ En = (n + 12 )ω 1 + 9 = (n + 12 )ω(1 + 12 9 − 18 92 + · · · ).

1 3 16 9

· · · ),

so

(b) H = 12 k x2 − 12 kx2 = 12 kx2 (1 + 9 − 1) = 9( 12 kx2 ) = 9V , where V is the unperturbed potential energy. So En1 = ψn0 |H |ψn0 = 9n|V |n, with n|V |n the expectation value of the (unperturbed) potential energy in the nth unperturbed state. This is most easily obtained from the virial theorem (Problem 3.31), but it can also be derived algebraically. In this case the virial theorem says T = V . But T + V = En . So V = 12 En0 = 12 (n + 12 )ω;

En1 = 25 (n + 12 )ω, which is precisely the 91 term in the power series from

part (a).

Problem 6.3 (a) In terms of the one-particle states (Eq. 2.28) and energies (Eq. 2.27): πx πx 2 π 2 2 1 2 Ground state: ψ10 (x1 , x2 ) = ψ1 (x1 )ψ1 (x2 ) = sin sin ; E10 = 2E1 = . a a a ma2 First excited state: ψ20 (x1 , x2 ) =

√1 2

[ψ1 (x1 )ψ2 (x2 ) + ψ2 (x1 )ψ1 (x2 )]

√ πx 2 πx1 2πx2 2πx1 2 = sin sin + sin sin ; a a a a a (b) E11

=

=−

ψ10 |H |ψ10

4V0 a

a

sin4

5 π 2 2 . 2 ma2

2 a a πx πx 2 1 2 = (−aV0 ) sin2 sin2 δ(x2 − x2 ) dx1 dx2 a a a 0 0

πx

0

E20 = E1 + E2 =

a

dx = −

4V0 a a π

π

sin4 y dy = − 0

4V0 3π 3 · = − V0 . π 8 2

E21 = ψ20 |H |ψ20 = (−aV0 )

2 a2

πx 2 πx1 2πx2 2πx1 2 sin δ(x1 − x2 ) dx1 dx2 sin + sin sin a a a a 0 a

πx 2 πx 2πx 2πx sin sin + sin sin dx a a a a 0 8V0 a 2 πx 2 2πx 8V0 a π 2 =− sin dx = − · sin sin y sin2 (2y) dy a 0 a a a π 0 π 8V0 32V0 π =− sin2 y sin2 y cos2 y dy = − (sin4 y − sin6 y) dy ·4 π π 0 0 32V0 3π 5π =− − = −2V0 . π 8 16

2V0 =− a

a


156


Problem 6.4 (a) 0 ψm |H|ψn0

2 = α a

a

sin 0

nπ mπ nπ mπ a 2α x δ x− sin x dx = sin sin , a 2 a a 2 2

which is zero unless both m and n are odd—in which case it is ±2α/a. So Eq. 6.15 says

En2 =

m=n, odd

En2

   =

  2m

2α a

2α π

2 (En0

2

1 . 0 ) − Em

0, m=n, odd

if n is even; 1 , if n is odd. (n2 − m2 )

1 1 1 1 = − . To sum the series, note that 2 (n − m2 ) 2n m + n m − n 1 1 1 for n = 1: = − 2 3,5,7,... m + 1 m − 1 1 = 2

1 1 1 1 1 1 1 + + + ··· − − − − ··· 4 6 8 2 4 6 8

for n = 3:

=

1 6

=

π 2 2 2 n , so 2ma2

But Eq. 2.27 says En0 =

1 = 2

1 − 2

Thus,

1 =− ; 4

1 1 1 − 6 1,5,7,... m + 3 m − 3

1 1 1 1 1 1 1 1 1 + + + ··· + − − − − − ··· 4 8 10 2 2 4 6 8 10

=

1 6

−

1 6

=−

1 . 36

In general, there is perfect cancellation except for the “missing” term 1/2n in the first sum, so the total 1 1 1 0, if n is even; 2 is − =− . Therefore: En = 2 −2m (α/πn) , if n is odd. 2n 2n (2n)2 (b) H =

1 9kx2 ; 2

m|x2 |n =

=

0 ψm |H |ψn0 =

1 9km|x2 |n. Using Eqs. 2.66 and 2.69: 2

m|(a2+ + a+ a− + a− a+ + a2− )|n 2mω

(n + 1)(n + 2)m|n + 2 + nm|n + (n + 1)m|n + n(n − 1)m|n − 2 . 2mω


CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY So, for m = n,

En2 =

=

0 ψm |H |ψn0

=

157

1 (n + 1)(n + 2) δm,n+2 + n(n − 1) δm,n−2 . k9 2 2mω

2 2 (n + 1)(n + 2) δ + n(n − 1) δ m,n+2 m,n−2 9ω 4

(n + 12 )ω − (m + 12 )ω

m=n

92 ω [(n + 1)(n + 2) δm,n+2 + n(n − 1) δm,n−2 ] 16 (n − m) m=n

=

92 ω (n + 1)(n + 2) n(n − 1) 92 ω 1 1 + = − (n + 1)(n + 2) + n(n − 1) 16 n − (n + 2) n − (n − 2) 16 2 2

=

92 ω 92 ω 2 1 1 −n − 3n − 2 + n2 − n = (−4n − 2) = −92 ω n + 32 32 8 2

(which agrees with the 92 term in the exact solution—Problem 6.2(a)).

Problem 6.5 (a) En1 = ψn0 |H |ψn0 = −qEn|x|n = 0 (Problem 2.12). From Eq. 6.15 and Problem 3.33: En2 = (qE)2

|m|x|n|2 (n − m)ω

m=n

=

√ √ (qE)2 [ n + 1 δm,n+1 + n, δm,n−1 ]2 (qE)2 [(n + 1) δm,n+1 + n δm,n−1 ] = ω 2mω (n − m) 2mω 2 (n − m) m=n

=

(b) −

(qE)2 n (qE)2 (n + 1) (qE)2 + = [−(n + 1) + n] = − . 2 2 2mω n − (n + 1) n − (n − 1) 2mω 2mω 2

2 d 2 ψ + 2m dx2

=

m=n

1 mω 2 x2 − qEx ψ = Eψ. 2

1 mω 2 x2 − qEx 2

=

With the suggested change of variables,

2 1 qE qE mω 2 x + − qE x + 2 mω 2 mω 2

1 qE (qE)2 1 (qE)2 1 1 (qE)2 2 2 + mω 2 2 4 − qEx − = mω 2 x − . mω 2 x + mω 2 x 2 2 2 mω 2 m ω mω 2 2 mω 2


158

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY So the Schr¨ odinger equation says

2 d 2 ψ 1 1 (qE)2 2 2 − ψ, + x ψ = E + mω 2m dx 2 2 2 mω 2 which is the Schr¨ odinger equation for a simple harmonic oscillator, in the variable x . The constant on the right must therefore be (n + 12 )ω, and we conclude that 1 1 (qE)2 En = (n + )ω − . 2 2 mω 2 The subtracted term is exactly what we got in part (a) using perturbation theory. Evidently all the higher corrections (like the first-order correction) are zero, in this case.

Problem 6.6 (a) 0 0 ψ+ |ψ− = (α+ ψa0 + β+ ψb0 )|(α− ψa0 + β− ψb0 ) ∗ ∗ ∗ ∗ = α+ α− ψa0 |ψa0 + α+ β− ψa0 |ψb0 + β+ α− ψb0 |ψa0 + β+ β− ψb0 |ψb0 ∗ ∗ 1 = α+ α− + β+ β− . But Eq. 6.22 ⇒ β± = α± (E± − Waa )/Wab ,

so

1 − Waa ) (E 1 − Waa )(E− α∗ α− 0 0 ∗ 1 1 ψ+ = + 2 |Wab |2 + (E+ |ψ− = α+ α− 1 + + − Waa )(E− − Waa ) . ∗ Wab Wab |Wab | The term in square brackets is: 1 1 1 1 2 1 [ ] = E+ E− − Waa (E+ + E− ) + |Wab |2 + Waa . But Eq. 6.27 ⇒ E± = 12 [(Waa + Wbb ) ± 1 1 shorthand for the square root term. So E+ + E− = Waa + Wbb , and 1 1 E+ E− =

], where

√

is

√ 1 1 (Waa + Wbb )2 − ( )2 = (Waa + Wbb )2 − (Waa − Wbb )2 − 4|Wab |2 = Waa Wbb − |Wab |2 . 4 4

2 [ ] = Waa Wbb − |Wab |2 − Waa (Waa + Wbb ) + |Wab |2 + Waa = 0,

Thus

√

so

0 0 ψ+ |ψ− = 0. QED

(b) 0 0 ∗ ∗ ∗ ∗ ψ+ |H |ψ− = α+ α− ψa0 |H |ψa0 + α+ β− ψa0 |H |ψb0 + β+ α− ψb0 |H |ψa0 + β+ β− ψb0 |H |ψb0 ∗ ∗ ∗ ∗ = α+ α− Waa + α+ β− Wab + β+ α− Wba + β+ β− Wbb

=

∗ α+ α−

1 1 1 1 (E− − Waa ) (E+ − Waa ) (E+ − Waa ) (E− − Waa ) + Wba + Wbb Waa + Wab ∗ ∗ Wab Wab Wab Wab

=

∗ α+ α−

1 1 (E+ − Waa )(E− − Waa ) 1 1 Waa + E− − Waa + E+ − Waa + Wbb . |Wab |2

But we know from (a) that

1 1 (E+ − Waa )(E− − Waa ) = −1, so |Wab |2

0 0 ∗ 1 1 ψ+ |H |ψ− = α+ α− [E− + E+ − Waa − Wbb ] = 0. QED c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


159

(c) 0 0 ∗ ∗ ∗ ∗ ψ± |H |ψ± = α± α± ψa0 |H |ψa0 + α± β± ψa0 |H |ψb0 + β± α± ψb0 |H |ψa0 + β± β± ψb0 |H |ψb0

1 1 (E± − Waa ) (E± − Wbb ) 2 = |α± | Waa + Wab + |β± | Wba + Wbb Wab Wba 2

(this time I used Eq. 6.24 to express α in terms of β, in the third term).

1 0 0 1 1 1 ∴ ψ± |H |ψ± = |α± |2 (E± ) + |β± |2 (E± ) = |α± |2 + |β± |2 E± = E± . QED

Problem 6.7 (a) See Problem 2.46. (b) With a → n, b → −n, we have: Waa = Wbb = −

Wab = −

V0 L

V0 L

L/2

L/2

e−x

2

/a2

−L/2

e−x

2

V0 L

dx ≈ −

/a2 −4πnix/L

e

−L/2

dx ≈ −

∞

e−x

2

/a2

−∞

V0 L

∞

e−(x

dx = −

2

V0 √ a π. L

/a2 +4πnix/L)

−∞

dx = −

V0 √ −(2πna/L)2 . a πe L

(We did this integral in Problem 2.22.) In this case Waa = Wbb , and Wab is real, so Eq. 6.26 ⇒ √ V0 a 2 1 1 ∓ e−(2πna/L) . or E± = − π L

1 E± = Waa ± |Wab |,

√ 2 1 − Waa ) (E− ± π(V0 a/L)e−(2πna/L) √ = ∓α. Evidently, the “good” linear (c) Equation 6.22 ⇒ β = α =α Wab − π(V0 a/L)e−(2πna/L)2 combinations are: 2 2πnx 1 1 i2πnx/L −i2πnx/L ψ+ = αψn − αψ−n = √ √ e = i sin −e and L L 2 L ψ− = αψn + αψ−n =

1 E+

2 cos L

2 = ψ+ |H |ψ+ = (−V0 ) L

1 E− = ψ− |H |ψ− =

2 (−V0 ) L

2πnx . L

L/2

−x2 /a2

e

Using Eq. 6.9, we have : 2

sin

−L/2

L/2

−L/2

e−x

2

/a2

2πnx L

cos2

2πnx L

dx, dx.


160

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY sin2 θ = (1 − cos 2θ)/2, and cos2 θ = (1 + cos 2θ)/2, so ∞ ∞ 2 2 2 2 V0 ∞ −x2 /a2 4πnx V0 4πnx 1 E± ≈− e e−x /a dx ∓ e−x /a cos dx = − dx 1 ∓ cos L −∞ L L L −∞ −∞

But

=−

√ V0 a √ 2 2 V0 √ π a ∓ a πe−(2πna/L) = − π 1 ∓ e−(2πna/L) , L L

same as (b).

(d) Af (x) = f (−x) (the parity operator). The eigenstates are even functions (with eigenvalue +1) and odd functions (with eigenvalue −1). The linear combinations we found in (c) are precisely the odd and even linear combinations of ψn and ψ−n .

Problem 6.8 Ground state is nondegenerate; Eqs. 6.9 and 6.31 ⇒ 3 2 3 a V0 E = a

a

sin2

1

2

= 8V0 sin

π 4

2

sin

0

π 2

π π π a a 3a x sin2 y sin2 z δ(x − )δ(y − )δ(z − ) dx dy dz a a a 4 2 4

2

sin

3π 4

1 1 = 8V0 (1) = 2V0 . 2 2

First excited states (Eq. 6.34): 2π a a 3a 2 π 2 π 2 Waa = 8V0 sin x sin y sin z δ(x − )δ(y − )δ(z − ) dx dy dz a a a 4 2 4 1 = 8V0 (1)(1) = 4V0 . 2 π π 2π a a 3a 2 sin = 8V0 x sin y sin2 z δ(x − )δ(y − )δ(z − ) dx dy dz a a a 4 2 4 1 1 = 8V0 (0) = 0. 2 2

Wbb

2

2

Wcc = 8V0

π π 2π a a 3a x sin2 y sin2 z δ(x − )δ(y − )δ(z − ) dx dy dz a a a 4 2 4

sin 1 = 8V0 (1)(1) = 4V0 . 2

2

Wab = 8V0 sin

Wac = 8V0 sin

π 4

π 4

sin

sin

π 2

π 2

sin(π) sin

2

sin

π 2

3π 2

sin

3π 2

sin

3π 4

sin

3π 4

= 0.

= 8V0

1 √ 2

(1)(1)(−1)

1 √ 2

= −4V0 .



Wbc = 8V0 sin 

3π sin sin(π) sin sin = 0. 2 2 4  (1 − λ) 0 −1 −1 0  = 4V0 D; det(D − λ) = 0 −λ 0 = −λ(1 − λ)2 + λ = 0 1 −1 0 (1 − λ)

π 4

1 0 W = 4V0  0 0 −1 0

161

π

λ = 0,

π

or (1 − λ)2 = 1 ⇒ 1 − λ = ±1 ⇒ λ = 0

⇒

or λ = 2.

So the first-order corrections to the energies are 0, 0, 8V0 .

Problem 6.9

  1 (a) χ1 =  0 , eigenvalue V0 ; 0

    0 0 χ2 =  1 , eigenvalue V0 ; χ3 =  0 , eigenvalue 2V0 . 0 1 [V0 (1 − 9) − λ] 0 0 0 [V0 − λ] 9V0 = 0; (b) Characteristic equation: det(H − λ) = 0 9V0 [2V0 − λ] [V0 (1 − 9) − λ][(V0 − λ)(2V0 − λ) − (9V0 )2 ] = 0 ⇒ λ1 = V0 (1 − 9). (V0 − λ)(2V0 − λ) − (9V0 )2 = 0 ⇒ λ2 − 3V0 λ + (2V02 − 92 V02 ) = 0 ⇒

V 9V02 − 4(2V02 − 92 V02 ) V0 0 = 3 ± 1 + 492 ≈ 3 ± (1 + 292 ) . 2 2 2 V0 V0 λ2 = 3 − 1 + 492 ≈ V0 (1 − 92 ); λ3 = 3 + 1 + 492 ≈ V0 (2 + 92 ). 2 2 λ=

(c)

3V0 ±



 −1 0 0 H = 9V0  0 0 1  ; 0 10

= 9V0

E31 = χ3 |H |χ3 = 9V0

  −1 0 0 0 0 0 1  0 0 10 0 1 0 1



  0 0 0 1  1  = 0 (no first-order correction). 0

|χm |H |χ3 |2 E32 = ; 0 E30 − Em m=1,2 χ2 |H |χ3 = 9V0 E30 − E20 = 2V0 − V0 = V0 .

χ1 |H |χ3 = 9V0

    −1 0 0 0

0 1 0 0  0 0 1   0  = 9V0 1 0 0  1  = 0, 0 1 0 1 0



  0 0 1 0  0  = 9V0 . 1

So E32 = (9V0 )2 /V0 = 92 V0 .

E3 = E30 + E31 + E32 = 2V0 + 0 + 92 V0 = V0 (2 + 92 )

Through second-order, then, (same as we got for λ3 in (b)).


162


(d) Waa = χ1 |H |χ1 = 9V0

Wbb = χ2 |H |χ2 = 9V0

Wab = χ1 |H |χ2 = 9V0

    −1 0 0 1

−1 1 0 0  0 0 1   0  = 9V0 1 0 0  0  = −9V0 . 0 1 0 0 0

    0 −1 0 0

0 0 1 0  0 0 1   1  = 9V0 0 1 0  0  = 0. 0 1 0 0 1

    −1 0 0 0

0 1 0 0  0 0 1   1  = 9V0 1 0 0  0  = 0. 0 1 0 0 1







Plug the expressions for Waa , Wbb , and Wab into Eq. 6.27:

1 1 1 E± −9V0 + 0 ± 92 V02 + 0 = (−9V0 ± 9V0 ) = {0, −9V0 }. = 2 2 To first-order, then, E1 = V0 − 9V0 , we got in (b).

E2 = V0 , and these are consistent (to first order in 9) with what

Problem 6.10 Given a set of orthonornal states {ψj0 } that are degenerate eigenfunctions of the unperturbed Hamiltonian: Hψj0 = E 0 ψj0 , construct the general linear combination, ψ0 =

ψj0 |ψl0 = δjl ,

n

αj ψj0 .

j=1

It too is an eigenfunction of the unperturbed Hamiltonian, with the same eigenvalue: H 0 ψ0 =

n

αj H 0 ψj0 = E 0

j=1

n

αj ψj0 = E 0 ψ 0 .

j=1

We want to solve the Schr¨ odinger equation Hψ = Eψ for the perturbed Hamiltonian H = H 0 + λH . Expand the eigenvalues and eigenfunctions as power series in λ: E = E 0 + λE 1 + λ2 E 2 + . . . ,

ψ = ψ 0 + λψ 1 + λ2 ψ 2 + . . . .

Plug these into the Schr¨ odinger equation and collect like powers: (H 0 + λH )(ψ 0 + λψ 1 + λ2 ψ 2 + . . . ) = (E 0 + λE 1 + λ2 E 2 + . . . )(ψ 0 + λψ 1 + λ2 ψ 2 + . . . )

⇒

H 0 ψ 0 + λ(H 0 ψ 1 + H ψ 0 ) + . . . = E 0 ψ 0 + λ(E 0 ψ 1 + E 1 ψ 0 ) + . . . c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


163

The zeroth-order terms cancel; to first order H 0 ψ1 + H ψ0 = E 0 ψ1 + E 1 ψ0 . Take the inner product with ψj0 : ψj0 |H 0 ψ 1 + ψj0 |H ψ 0 = E 0 ψj0 |ψ 1 + E 1 ψj0 |ψ 0 . But

ψj0 |H 0 ψ 1 = H 0 ψj0 |ψ 1 = E 0 ψj0 |ψ 1 , so the first terms cancel, leaving ψj0 |H ψ 0 = E 1 ψj0 |ψ 0 .

Now use

ψ0 =

n

αl ψl0 ,

and exploit the orthonormality of {ψl0 }:

l=1 n

αl ψj0 |H |ψl0

=E

l=1

1

n

αl ψj0 |ψl0 = E 1 αj ,

l=1

or, defining Wjl ≡ ψj0 |H |ψl0 ,

n

Wjl αl = E 1 αl .

l=1

This (the generalization of Eq. 6.22 for the case of n-fold degeneracy) is the eigenvalue equation for the matrix W (whose jlth element, in the {ψj0 } basis, is Wjl ); E 1 is the eigenvalue, and the eigenvector (in the {ψj0 } basis) is χj = αj . Conclusion: The first-order corrections to the energy are the eigenvalues of W. QED

Problem 6.11

m (a) From Eq. 4.70: En = − 22

e2 4π90

2

1 1 = − mc2 n2 2

1 e2 c 4π90

2

1 α2 mc2 = − . n2 2n2

(b) I have found a wonderful solution—unfortunately, there isn’t enough room on this page for the proof.

Problem 6.12 Equation 4.191 ⇒ V = 2En , e2 − 4π90

2 2 e 1 e2 1 m for hydrogen. V = − . ; En = − 2 4π90 r 2 4π90 n2

0 1 2 2 1 1 e m = −2 2 r 2 4π90 n2

⇒

0 1 1 1 1 me2 = = 2 2 r 4π90 n an2

So

(Eq. 4.72). QED


164


Problem 6.13 In Problem 4.43 we found (for n = 3, l = 2, m = 1) that s = 0 : 1 =

s = −1 :

0

0

1 r2 1 r3

1 = 1

4! 6!

3! = 6!

(s + 6)! 6!

3a 2

s .

(of course).

0 1 −1 1 1 1 2 5! 3a = · = = r 6! 2 6 3a 9a

s = −2 :

s = −3 :

6! (1) = 1 6!

rs =

3a 2 3a 2

−2 = −3

Eq. 6.55 says

2 1 4 = · 6 · 5 9a2 135a2

1 1 = 32 a 9a

Eq. 6.56 says

1 1 8 = = · 3 6 · 5 · 4 27a 405a3

.

1 2 = (5/2) · 27 · a2 135a2

.

1 1 Eq. 6.64 says = 3 2(5/2)3 · 27 · a 405a3

.

For s = −7 (or smaller) the integral does not converge: 1/r7 = ∞ in this state; this is reflected in the fact that (−1)! = ∞.

Problem 6.14 Equation 6.53 ⇒ Er1 = −

Er1

1 =− 2mc2

1 n+ 2

But Problem 2.12

1 Er1 = − 2mc2

From Eq. 2.69:

x4 =

2

1 ω −2 n+ 2 2

2

x4 =

2

1 1 E = (n + )ω, V = mω 2 x2 2 2

⇒

1 2 4 4 1 2 2 ω mω x + m ω x . 2 4

1 x2 = (n + ) , so 2 mω

⇒

1 n+ 2

1 2 E − 2EV + V 2 . Here 2mc2

1 2 ω 2 − n + 2

2

1 mω 4 2 ω 2 + m2 ω 4 x4 = − 2 x4 . 4 8c

2 2 a + a+ a− + a− a+ + a2− a2+ + a+ a− + a− a+ + a2− , 4m2 ω 2 +

2 n| a2+ a2− + a+ a− a+ a− + a+ a− a− a+ + a− a+ a+ a− + a− a+ a− a+ + a2− a2+ |n. 2 2 4m ω

(Note that only terms with equal numbers of raising and lowering operators will survive). Using Eq. 2.66, x4 =

2 n| a2+ n(n − 1) |n − 2 + a+ a− (n |n) + a+ a− (n + 1) |n 2 2 4m ω



165

+a− a+ (n |n) + a− a+ (n + 1) |n + a2− (n + 1)(n + 2) |n + 2

=

2 n| n(n − 1) n(n − 1) |n + n (n |n) + (n + 1) (n |n) 4m2 ω 2

+n (n + 1) |n + (n + 1) (n + 1) |n + (n + 1)(n + 2) (n + 1)(n + 2) |n

=

2 n(n − 1) + n2 + (n + 1)n + n(n + 1) + (n + 1)2 + (n + 1)(n + 2) 2 2 4m ω

=

Er1

2mω

2 (n − n + n + n + n + n + n + n + 2n + 1 + n + 3n + 2) = 2

2

2

2

2

mω 4 2 3 =− 2 · · 3(3n2 + 2n + 1) = − 2 2 8c 4m ω 32

2

2 ω 2 mc2

2mω

2 (6n2 + 6n + 3).

(2n2 + 2n + 1).

Problem 6.15 Quoting the Laplacian in spherical coordinates (Eq. 4.13), we have, for states with no dependence on θ or φ: d 2 d p2 = −2 ∇2 = − 2 r2 . r dr dr Question: Is it Hermitian? Using integration by parts (twice), and test functions f (r) and g(r): ∞ ∞ 1 d d 2 2 2 dg 2 2 2 dg f |p g = − f 2 f r 4πr dr = −4π r dr r dr dr dr dr 0 0 ∞ df dg dg ∞ = −4π2 r2 f − r2 dr dr 0 dr dr 0 ∞ ∞ ∞ d 2 2 dg 2 df 2 df = −4π r f − r g + r g dr dr 0 dr 0 dr dr 0 ∞ dg df = −4π2 r2 f + p2 f |g. − r2 g dr dr 0 The boundary term at infinity vanishes for functions f (r) and g(r) that go to zero exponentially; the boundary term at zero is killed by the factor r2 , as long as the functions (and their derivatives) are finite. So f |p2 g = p2 f |g, and hence p2 is Hermitian. Now we apply the same argument to 4 d p = 2 r dr 4

d 1 d 2 d r r , dr r2 dr dr 2


166


integrating by parts four times: ∞ d 1 d 4 4 2 d 2 dg f |p g = 4π f r r dr dr dr r2 dr dr 0 ∞ ∞ 1 d 1 d 4 2 d 2 dg 2 df d 2 dg = 4π r f r r dr r − dr r2 dr dr dr dr r2 dr dr 0 0 ∞ ∞ 1 d 1 d df d d 4 2 d 2 dg 2 dg 2 df 2 dg = 4π r f r − r r r dr + dr r2 dr dr dr dr dr r2 dr dr dr dr 0 0 ∞ d 1 d df d d dg 2 dg 2 dg 2 df = 4π4 r2 f r − r + r 2 dr r dr dr dr dr dr dr dr dr 0 ∞ dg d 1 d df r2 dr − r2 2 dr dr r dr dr 0

∞ d 1 d df d d dg 1 d 2 dg 2 dg 2 df 2 d 2 df = 4π4 r2 f g r − r + r − r r 2 2 dr r dr dr dr dr dr dr dr dr dr r dr dr 0 ∞ d d d df 1 + r2 r2 g dr dr dr r2 dr dr 0 ∞ 1 d 1 d 4 2 d 2 dg 2 d 2 df = 4π r f r −r g r dr r2 dr dr dr r2 dr dr 0 ∞ df dg df d d dg + p4 f |g − 4π4 r2 − r2 dr dr dr dr dr dr 0 This time there are four boundary terms to worry about. Infinity is no problem; the trouble comes at r = 0. If the functions f and g went to zero at the origin (as they do for states with l > 0) we’d be OK, but states with l = 0 go like exp(−r/na). So let’s test the boundary terms using f (r) = e−r/na ,

g(r) = e−r/ma .

In this case dg 1 2 −r/ma =− r e dr ma

dg 1 2 r − 2mar e−r/ma r2 = 2 dr (ma)

dg 1 2 1 r − 2mar e−r/ma . r2 = − e−r/na 2 dr na (ma) r2

d dr df d dr dr

This goes to zero as r → 0, so the second pair of boundary terms vanishes—but not the first pair: 1 d 2ma −r/ma 1 2 dg 1 − r = e r2 dr dr (ma)2 r d 1 d 1 2 dg 2(ma)2 + 2mar − r2 e−r/ma r = 2 3 2 dr r dr dr (ma) r dg d 1 d 1 r2 f 2(ma)2 + 2mar − r2 e−r/ma e−r/na r2 = 2 3 dr r dr dr (ma) This does not vanish as r → 0; rather, it goes to 2/ma. For these particular states, then, 1 8π4 1 4 f |p g = − + p4 f |g, a m n c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


167

or, tacking on the normalization factor, ψn00 = √

1 e−r/na , π (na)3/2

ψn00 |p4 ψm00 =

84 (n − m) + p4 ψn00 |ψm00 , a4 (nm)5/2

and hence p4 is not Hermitian, for such states.

Problem 6.16 (a) [L · S, Lx ] = [Lx Sx + Ly Sy + Lz Sz , Lx ] = Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ] = Sx (0) + Sy (−iLz ) + Sz (iLy ) = i(Ly Sz − Lz Sy ) = i(L × S)x . Same goes for the other two components, so [L · S, L] = i(L × S). (b) [L · S, S] is identical, only with L ↔ S: [L · S, S] = i(S × L). (c) [L · S, J] = [L · S, L] + [L · S, S] = i(L × S + S × L) = 0. (d) L2 commutes with all components of L (and S) , so

L · S, L2 = 0.

L · S, S 2 = 0.

(e) Likewise,

(f ) L · S, J 2 = L · S, L2 + L · S, S 2 + 2 [L · S, L · S] = 0 + 0 + 0 =⇒ L · S, J 2 = 0.

Problem 6.17 With the plus sign, j = l + 1/2 (l = j − 1/2) :

1 Equation 6.65 ⇒ Eso =

=

1 Efs

=

=

Er1

+

(En )2 2mc2

1 Eso

(En )2 = 2mc2

3+

Eq. 6.57 ⇒

Er1

(En )2 =− 2mc2

4n −3 . j

(En )2 n j(j + 1) − (j − 12 )(j + 12 ) − 34 mc2 (j − 12 )j(j + 12 ) n (En )2 n(j 2 + j − j 2 + 14 − 34 ) (En )2 = . 1 1 2 2 mc mc j(j + 12 ) (j − 2 )j(j + 2 )

4n 2n − +3+ j j(j + 12 )

1 (En )2 2n 4n 1 − 2 j + = 3 − . 2 2mc2 j(j + 12 ) j + 12


168

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY With the minus sign, j = l − 1/2 (l = j + 1/2) :

Equation 6.65 ⇒

=

1 Efs =

1 Eso

Eq. 6.57 ⇒

Er1

(En )2 =− 2mc2

4n −3 . j+1

(En )2 n j(j + 1) − (j + 12 )(j + 32 ) − 34 = mc2 (j + 12 )(j + 1)(j + 32 )

(En )2 n(j 2 + j − j 2 − 2j − 34 − 34 ) −n (En )2 = . 1 3 2 mc mc2 (j + 1)(j + 12 ) (j + 2 )(j + 1)(j + 2 )

2n 4n 2n (En )2 (En )2 1 − 3 + − 3 − = 1 + 2 j + 2mc2 j+1 2mc2 2 (j + 1)(j + 12 ) (j + 1)(j + 12 )

(En )2 = 2mc2

4n 3− j + 12

Problem 6.18

.

For both signs, then,

1 Efs

(En )2 = 2mc2

4n 3− j + 12

. QED

1 1 36 2πc 5 ; E1 = −13.6 eV; − = − E1 ⇒ λ = − 9 4 36 5 E1 36 (2π)(1.97 × 10−11 × 106 eV · cm) c = 1.97 × 10−11 MeV·cm; λ = = 6.55 × 10−5 cm= 655 nm. 5 (13.6 eV) 4n 3.00 × 108 m/s c (En )2 1 14 3− : = 4.58 × 10 Hz. ν= = Equation 6.66 ⇒ Ef s = λ 6.55 × 10−7 m 2mc2 j + 12 E30 − E20 = hν =

2πc = E1 λ

For n = 2: l = 0 or l = 1, so j = 1/2 or 3/2. Thus n = 2 splits into two levels :

j = 1/2 :

E21

(E2 )2 = 2mc2

j = 3/2 : E21 =

(E2 )2 2mc2

8 3− 1

3−

8 2

5 (E2 )2 5 =− =− 2 mc2 2

=−

2 (E1 )2 1 (13.6 eV)2 5 = −5.66 × 10−5 eV. = − 4 mc2 32 (.511 × 106 eV)

1 (E2 )2 1 = − (3.62 × 10−4 eV) = −1.13 × 10−5 eV. 2 mc2 32

For n = 3: l = 0, 1 or 2, so j = 1/2, 3/2 or 5/2. Thus n = 3 splits into three levels :

j = 1/2 :

E31

(E3 )2 = 2mc2

j = 3/2 :

E31

(E3 )2 = 2mc2

j = 5/2 : E31 =

(E3 )2 2mc2

12 3− 1 12 3− 2

3−

12 3

9 (E3 )2 =− = −9 mc2 2

1 92

(E1 )2 1 = − (3.62 × 10−4 eV) = −2.01 × 10−5 eV. mc2 18

=−

1 3 (E3 )2 = − (3.62 × 10−4 eV) = −0.67 × 10−5 eV. 2 mc2 54

=−

1 (E3 )2 1 (3.62 × 10−4 eV) = −0.22 × 10−5 eV. =− 2 mc2 162



169

0

E3 j= 5/2 j= 3/2 j=1/2 0

E2 1 2 3 4

5

6

j= 3/2 j= 1/2

There are six transitions here; their energies are (E30 + E31 ) − (E20 + E21 ) = (E30 − E20 ) + ∆E, where ∆E ≡ E31 − E21 . Let β ≡ (E1 )2 /mc2 = 3.62 × 10−4 eV. Then: 1 3 1 7 1 ( → ) : ∆E = − − β=− β = −8.80 × 10−6 eV. − 2 2 18 32 288

(

3 3 → ): 2 2

5 3 ( → ): 2 2

(

1 1 → ): 2 2

∆E =

1 1 11 − − − β= β = 4.61 × 10−6 eV. 54 32 864

1 1 65 ∆E = − + β= β = 9.08 × 10−6 eV. 162 32 2592 ∆E =

5 32

−

1 18

β=

29 β = 36.45 × 10−6 eV. 288

3 1 ( → ): 2 2

1 5 119 ∆E = − + β= β = 49.86 × 10−6 eV. 54 32 864

5 1 ( → ): 2 2

1 5 389 ∆E = − + β= β = 54.33 × 10−6 eV. 162 32 2592

Conclusion: There are six lines; one of them ( 12 → 32 ) has a frequency less than the unperturbed line, the other five have (slightly) higher frequencies. In order they are: 32 → 32 ; 52 → 32 ; 12 → 12 ; 32 → 12 ; 52 → 12 . The frequency spacings are:

ν2 − ν1 ν3 − ν3 ν4 − ν3 ν5 − ν4 ν6 − ν5

= = = = =

(∆E2 − ∆E1 )/2π (∆E3 − ∆E2 )/2π (∆E4 − ∆E3 )/2π (∆E5 − ∆E4 )/2π (∆E6 − ∆E5 )/2π

= = = = =

3.23 × 109 1.08 × 109 6.60 × 109 3.23 × 109 1.08 × 109

Hz Hz Hz Hz Hz


170


Problem 6.19

1 j+ 2

n − (j +

=



2 −

1 2)

+

n 1−

=

1 j+ 2

j+

1 2 2

α2

α

α α2 2n(j+ 12 )

1−

≈

− α2

α j+

2

≈

1 2

1 j+ 2

α

n − j + 12 + j + 12 −

1 1− 2

α2 2(j+ 12 )

=

α j+

2 1 2

1 α2 = (j + ) − . 2 2(j + 12 )

α α2 2(j+ 12 )

n−

α α2 ≈ . 1+ n 2n(j + 12 )

2 −1/2 −1/2 α α2 α2    1 + ≈ 1 + 1 +  

2 n2 n(j + 12 ) n − (j + 12 ) + j + 12 − α2 

≈1−

Enj

1 α2 2 n2

1+

α2 n(j + 12 )

+

3 α4 α2 α4 = 1 − + 8 n4 2n2 2n4

−n 3 1 + 4 j+2

.

α2 −n α2 n α4 3 α2 mc2 1+ 2 ≈ mc 1 − 2 + 4 + −1 =− 1 2 2n 2n 4 2n n j+2 j+ 2

13.6 eV α2 n =− 1+ 2 2 n n j+

1 2

3 − 4

1 2

3 − 4

,

confirming Eq. 6.67.

Problem 6.20 Equation 6.59 ⇒ B = B= =

e 1 L. 4π90 mc2 r3

Say L = , r = a; then

1 e 4π90 mc2 a3 (1.60 × 10−19 C)(1.05 × 10−34 J · s)

2 3 = 12 T. 4π 8.9 × 10−12 C /N · m2 (9.1 × 10−31 kg) (3 × 108 m/s) (0.53 × 10−10 m)

2

So a “strong” Zeeman field is Bext 10 T, and a “weak” one is Bext 10 T. Incidentally, the earth’s field (10−4 T) is definitely weak.



171

Problem 6.21 For n = 2, l = 0 (j = 1/2) or l = 1 (j = 1/2 or 3/2). The eight states are: |1 = |2 0

1 1 2 2

|2 = |2 0

1 2

|3 = |2 1

1 1 2 2

|4 = |2 1

1 2

 

− 12

(1/2)(3/2) + (3/4) 3/2 =1+ = 2. gJ = 1 +  2(1/2)(3/2) 3/2  

− 12

−1/2 (1/2)(3/2) − (1)(2) + (3/4) =1+ = 2/3. gJ = 1 +  2(1/2)(3/2) 3/2

In these four cases, Enj = − |5 = |2 1

3 3 2 2

|6 = |2 1

3 1 2 2

|7 = |2 1

3 2

|8 = |2 1

3 2

13.6 eV α2 2 3 5 1+ − = −3.4 eV 1 + α2 . 4 4 1 4 16

         

(3/2)(5/2) − (1)(2) + (3/4) 5/2 gJ = 1 + =1+ = 4/3.  2(3/2)(5/2) 15/2 − 12        3  − 2

α2 2 3 1 2 In these four cases, Enj = −3.4 eV 1 + − = −3.4 eV 1 + α . 4 2 4 16

5 2 E1 = −3.4 eV 1 + 16 α + µB Bext .

The energies are:

E2 = −3.4 eV 1 +

5 2 16 α

E3 = −3.4 eV 1 +

5 2 16 α

E4 = −3.4 eV 1 +

5 2 16 α

E5 = −3.4 eV 1 +

1 2 16 α

E6 = −3.4 eV 1 +

1 2 16 α

E7 = −3.4 eV 1 +

1 2 16 α

E8 = −3.4 eV 1 +

1 2 16 α

− µB Bext . + 13 µB Bext . − 13 µB Bext . + 2µB Bext . + 23 µB Bext . − 23 µB Bext . − 2µB Bext .


172

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY E

µBBext 5 (slope 2) 6 (slope 2/3)

-3.4 (1+α2/16) eV 7 (slope -2/3) 8 (slope -2) 1 (slope 1) 3 (slope 1/3)

-3.4 (1+5α2/16) eV

4 (slope -1/3) 2 (slope -1)

Problem 6.22 1 Efs = nlml ms |(Hr + Hso )|nlml ms = −

4n 2 ml ms En2 e2 . − 3 + 2 2 2 2mc l + 1/2 8π90 m c l(l + 1/2)(l + 1)n3 a3

    

2En2 E1 α2 2E1 − = = − (13.6 eV). (Problem 6.11.) mc2 mc2 n4 n4 2 2 2 Now 2 2 2 2 2 3 e e2 e (me ) e m   = α2 (13.6 eV).   8π9 m2 c2 a3 = 2 · 4π9 m2 c2 (4π9 2 )3 = 22 4π9 4π90 c 0 0 0 0 1 Efs

1 13.6 eV 2 3 m l ms 13.6 eV 2 3 l(l + 1) − ml ms = α − α + + = − . QED n3 (l + 1/2) 4n l(l + 1/2)(l + 1) n3 4n l(l + 1/2)(l + 1)

Problem 6.23 The Bohr energy is the same for all of them: E2 = −13.6 eV/22 = −3.4 eV. The Zeeman contribution is the 1 second term in Eq. 6.79: µB Bext (ml +2ms ). The fine structure is given by Eq. 6.82: Efs = (13.6 eV/8)α2 {· · · } = 2 (1.7 eV)α {· · · }. In the table below I record the 8 states, the value of (ml + 2ms ), the value of {· · · } ≡ l(l + 1) − ml ms 3 − , and (in the last column) the total energy, −3.4 eV [1−(α2 /2){· · · }]+(ml +2ms )µB Bext . 8 l(l + 1/2)(l + 1) State = |nlml ms |1 = |2 0 0 12 |2 = |2 0 0 − 12 |3 = |2 1 1 12 |4 = |2 1 − 1 − 12 |5 = |2 1 0 12 |6 = |2 1 0 − 12 |7 = |2 1 1 − 12 |8 = |2 1 − 1 12

(ml + 2ms ) 1 −1 2 −2 1 −1 0 0

{· · · } −5/8 −5/8 −1/8 −1/8 −7/24 −7/24 −11/24 −11/24

Total Energy -3.4 eV [1 + (5/16)α2 ] + µB Bext -3.4 eV [1 + (5/16)α2 ] − µB Bext -3.4 eV [1 + (1/16)α2 ] + 2µB Bext -3.4 eV [1 + (1/16)α2 ] − 2µB Bext -3.4 eV [1 + (7/48)α2 ] + µB Bext -3.4 eV [1 + (7/48)α2 ] − µB Bext -3.4 eV [1 + (11/48)α2 ] -3.4 eV [1 + (11/48)α2 ]



173

Ignoring fine structure there are five distinct levels—corresponding to the possible values of (ml + 2ms ): 2

(d = 1); 1

(d = 2); 0

(d = 2); −1

(d = 2); −2

(d = 1).

Problem 6.24 e e Bext · L + 2S = Bext 2ms = 2ms µB Bext (same as the Zeeman term in Eq. 6.79, 2m 2m α2 3 13.6 eV with ml = 0). Equation 6.67 ⇒ Enj = − (since j = 1/2). So the total energy is 1 + n − n2 n2 4

Equation 6.72 ⇒ Ez1 =

α2 3 13.6 eV 1+ 2 n− E=− + 2ms µB Bext . n2 n 4 13.6 eV 2 3 3 13.6 eV 2 = − 1 , which is the same as Fine structure is the α term: = − α n− α n4 4 n3 4n Eq. 6.82, with the term in square brackets set equal to 1. QED 1 Efs

2

Problem 6.25 1 Equation 6.66 ⇒ Efs =

1 Efs

E1 =− 32

α2 2

E22 2mc2

3−

8 3− j + 1/2

8 j + 1/2

=

E12 32mc2

3−

8 j + 1/2

;

E1 α2 (Problem 6.11), so =− 2 mc 2

8 13.6 eV 2 8 = α 3− =γ 3− . 64 j + 1/2 j + 1/2

8 For j = 1/2 (ψ1 , ψ2 , ψ6 , ψ8 ), Hfs1 = γ(3 − 8) = −5γ. For j = 3/2 (ψ3 , ψ4 , ψ5 , ψ7 ), Hfs1 = γ(3 − ) = −γ. 2 This confirms all the γ terms in −W (p. 281). Meanwhile, Hz = (e/2m)Bext (Lz + 2Sz ) (Eq. 6.71); ψ1 , ψ2 , ψ3 , ψ4 are eigenstates of Lz and Sz ; for these there are only diagonal elements: Hz =

e Bext (ml + 2ms ) = (ml + 2ms )β; 2m

Hz 11 = β;

This confirms the upper left corner of −W. Finally:  Hz 55 (Lz + 2Sz )|ψ5 = + 23 |1 0| 12 12     Hz 66   1 1 1  (Lz + 2Sz )|ψ6 = − 3 |1 0| 2 2 Hz 77 so Hz 88  (Lz + 2Sz )|ψ7 = − 23 |1 0| 12 − 12    Hz 56  1 1 1   (Lz + 2Sz )|ψ8 = − 3 |1 0| 2 − 2 Hz 78

= = = = = =

Hz 22 = −β;

Hz 33 = 2β;

Hz 44 = −2β.

(2/3)β, (1/3)β, −(2/3)β, −(1/3)β, √ Hz 65 = −(√2/3)β, Hz 87 = −( 2/3)β,

which confirms the remaining elements.


174


Problem 6.26 There are eighteen n = 3 states (in general, 2n2 ). WEAK FIELD Equation 6.67 ⇒ E3j

α2 3 3 α2 1 1 13.6 eV 1+ − = −1.51 eV 1 + − . =− 9 9 j + 1/2 4 3 j + 1/2 4

Equation 6.76 ⇒ Ez1 = gJ mj µB Bext . State |3 l j mj

gJ (Eq. 6.75)

l = 0, j = 1/2

|3 0

1 1 2 2

l = 0, j = 1/2

|3 0

1 2

l = 1, j = 1/2

|3 1

1 1 2 2

l = 1, j = 1/2

|3 1

1 2

l = 1, j = 3/2

|3 1

3 3 2 2

l = 1, j = 3/2

|3 1

3 1 2 2

l = 1, j = 3/2

|3 1

3 2

−

1 2

l = 1, j = 3/2

|3 1

3 2

− 32

l = 2, j = 3/2

|3 2

3 3 2 2

l = 2, j = 3/2

|3 2

3 1 2 2

l = 2, j = 3/2

|3 2

3 2

−

1 2

l = 2, j = 3/2

|3 2

3 2

− 32

l = 2, j = 5/2

|3 2

5 5 2 2

l = 2, j = 5/2

|3 2

l = 2, j = 5/2

1 3

1 j+1/2

−

2

1/4

2

1/4

2/3

1/4

2/3

1/4

4/3

1/12

4/3

1/12

4/3

1/12

4/3

1/12

4/5

1/12

4/5

1/12

4/5

1/12

4/5

1/12

6/5

1/36

5 3 2 2

6/5

1/36

|3 2

5 1 2 2

6/5

1/36

l = 2, j = 5/2

|3 2

5 2

− 12

6/5

1/36

l = 2, j = 5/2

|3 2

5 2

−

3 2

6/5

1/36

l = 2, j = 5/2

|3 2

5 2

− 52

6/5

1/36

− 12 − 12

1 4

Total Energy 2 −1.51 eV 1 + α4 + µB Bext 2 −1.51 eV 1 + α4 − µB Bext 2 −1.51 eV 1 + α4 + 13 µB Bext 2 −1.51 eV 1 + α4 − 13 µB Bext 2 −1.51 eV 1 + α12 + 2µB Bext 2 −1.51 eV 1 + α12 + 23 µB Bext 2 −1.51 eV 1 + α12 − 23 µB Bext 2 −1.51 eV 1 + α12 − 2µB Bext 2 −1.51 eV 1 + α12 + 65 µB Bext 2 −1.51 eV 1 + α12 + 25 µB Bext 2 −1.51 eV 1 + α12 − 25 µB Bext 2 −1.51 eV 1 + α12 − 65 µB Bext 2 −1.51 eV 1 + α36 + 3µB Bext 2 −1.51 eV 1 + α36 + 95 µB Bext 2 −1.51 eV 1 + α36 + 35 µB Bext 2 −1.51 eV 1 + α36 − 35 µB Bext 2 −1.51 eV 1 + α36 − 95 µB Bext 2 −1.51 eV 1 + α36 − 3µB Bext

STRONG FIELD Equation 6.79 ⇒ −1.51 eV + (ml + 2ms )µB Bext ; 13.6 eV 2 Equation 6.82 ⇒ α 27

1 l(l + 1) − ml ms α2 l(l + 1) − ml ms 1 − = −1.51 eV − . 4 l(l + 1/2)(l + 1) 3 l(l + 1/2)(l + 1) 4



Etot

1 = −1.51 eV(1 + α A) + (ml + 2ms )µB Bext , where A ≡ 3

175

2

l(l + 1) − ml ms 1 − l(l + 1/2)(l + 1) 4

.

These terms are given in the table below: State |n l ml ms l=0 l=0 l=1 l=1

|3 0 0

(ml + 2ms )

A

1

1/4

1 2

|3 0 0 − 12

−1

|3 1 1 12 |3 1 − 1 − 12

1/4

2

1/12

−2

l=1

|3 1 0

l=1

|3 1 0 − 12

l=1

|3 1 − 1

1 2

0

7/36

l=1

|3 1 1 − 12

0

7/36

l=2

|3 2 2 12

3

1/36

l=2

1 2

1/12

|3 2 − 2 − 12

l=2

|3 2 1

l=2

|3 2 − 1 − 12

l=2 l=2

|3 2 0

1

5/36

−1

5/36

−3

1 2

2

7/180

−2

7/180

1 2

|3 2 0 − 12

1/36

1

1/20

−1

1/20

l=2

|3 2 − 1 12

0

11/180

l=2

|3 2 1 − 12

0

11/180

l=2

|3 2 − 2 12

−1

13/180

l=2

1 2

|3 2 2 −

1

13/180

Total Energy −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 + −1.51 eV 1 +

α2 4 2

α 4

2

α 12 2

+ µB Bext − µB Bext + 2µB Bext

− 2µB Bext + µB Bext

α 12

5α2 36 5α2 36

− µB Bext

7α2 36 7α2 36

α2 36 2

α 36

+ 3µB Bext

− 3µB Bext + 2µB Bext

7α2 180 7α2 180 α2 20 2

α 20

2

11α 180

− 2µB Bext

+ µB Bext − µB Bext

11α2 180 13α2 180 13α2 180

− µB Bext + µB Bext

INTERMEDIATE FIELD As in the book, I’ll use the basis |n l j mj (same as for weak field); then the fine structure matrix elements are diagonal: Eq. 6.66 ⇒ 1 Efs

γ≡

E32 = 2mc2

12 3− j + 1/2

E12 = 54mc2

4 1− j + 1/2

E1 α 2 =− 108

4 1− j + 1/2

4 = 3γ 1 − j + 1/2

,

13.6 eV 2 1 1 1 α . For j = 1/2, Efs = −9γ; for j = 3/2, Efs = −3γ; for j = 5/2, Efs = −γ. 324

The Zeeman Hamiltonian is Eq. 6.71: Hz = 1 (Lz + 2Sz )µB Bext . The first eight states (l = 0 and l = 1) are the same as before (p. 281), so the β terms in W are unchanged; recording just the non-zero blocks of −W: ' (9γ − β), (9γ + β), (3γ − 2β), (3γ + 2β),

√

2 (3γ√− 23 β) 3 β 2 (9γ − 13 β) 3 β

( ' ,

√

2 (3γ√+ 23 β) 3 β 2 (9γ + 13 β) 3 β

( .


176


The other 10 states (l = 2) must first be decomposed into eigenstates of Lz and Sz : | 52

5 2

= |2 2| 12

1 2

=⇒ (γ − 3β)

| 52 − 52 = |2 − 2| 12 − 12 =⇒ (γ + 3β)

| 52

3 2

=

| 32

3 2

=

| 52

1 2

=

| 32

1 2

=

1 |2 5

2| 12 − 12 +

4 5 |2

2| 12 − 12 −

2 |2 5

1| 12 − 12 +

| 52 − 12 = | 32 − 12 =

| 52 − 32 = | 32 − 32 =

3 5 |2

1| 12 − 12 −

3 |2 5 2 5 |2

4 |2 5 1 5 |2



1| 12

1  2

1 5 |2

1| 12

1  2

3 |2 5

0| 12

1  2

0| 12

1  2

4 |2 5

0| 12 − 12 + 0| 12 − 12 −

2 5 |2

2 |2 5

− 1| 12 − 12 + − 1| 12 − 12 −

3 5 |2



' =⇒

1  2

− 1| 12

1  2

1 |2 5 4 5 |2

√

6 (γ √ − 35 β) 5 β 6 (3γ − 25 β) 5 β



− 1| 12

2 (γ − 95 β) 5β 2 β (3γ − 65 β) 5

=⇒

' =⇒



− 2| 12

1  2

− 2| 12

1  2

√

(

6 (γ √ + 35 β) 5 β 6 (3γ + 25 β) 5 β

=⇒

(

2 (γ + 95 β) 5β 2 β (3γ + 65 β) 5

[Sample Calculation: For the last two, letting Q ≡ 1 (Lz + 2Sz ), we have Q| 52 − 32 = −2 45 |2 − 1| 12 − 12 − 15 |2 − 2| 12 12 ; Q| 32 − 32 = −2 15 |2 − 1| 12 − 12 + 45 |2 − 2| 12 12 . 52 − 32 |Q| 52 − 32 = (−2) 45 − 15 = − 95 ; 32 − 32 |Q| 32 − 32 = (−2) 15 − 45 = − 65 ; 52 − 32 |Q| 32 − 32 = −2 45 15 + 15 45 = − 45 + 25 = − 25 = 32 − 32 |Q| 52 − 32 .] So the 18 × 18 matrix −W splits into six 1 × 1 blocks and six 2 × 2 blocks. We need the eigenvalues of the 2 × 2 blocks. This means solving 3 characteristic equations (the other 3 are obtained trivially by changing the sign of β):

2 3γ − β − λ 3

9 γ− β−λ 5

3 γ− β−λ 5

1 2 9γ − β − λ − β 2 = 0 =⇒ λ2 + λ(β − 12γ) + γ(27γ − 7β) = 0. 3 9

6 4 2 33 2 2 2 = 0. 3γ − β − λ − β = 0 =⇒ λ + λ(3β − 4γ) + γ 3γ − γβ + 2β 5 25 5

11 2 6 2 2 3γ − β − λ − β = 0 =⇒ λ + λ(β − 4γ) + γ 3γ − β = 0. 5 25 5

The solutions are: c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


λ = −β/2 + 6γ ± (β/2)2 + βγ + 9γ 2 λ = −3β/2 + 2γ ± (β/2)2 + 35 βγ + γ 2 λ = −β/2 + 2γ ± (β/2)2 + 15 βγ + γ 2

⇒

91 92 93 94 95

= = = = =

177

E3 − 9γ + β E3 − 3γ + 2β E3 − γ + 3β E3 − 6γ + β/2 + 9γ 2 + βγ + β 2 /4 E3 − 6γ + β/2 − 9γ 2 + βγ + β 2 /4

96 = E3 − 2γ + 3β/2 +

γ 2 + 35 βγ + β 2 /4

97 = E3 − 2γ + 3β/2 − γ 2 + 35 βγ + β 2 /4 98 = E3 − 2γ + β/2 + γ 2 + 15 βγ + β 2 /4 99 = E3 − 2γ + β/2 − γ 2 + 15 βγ + β 2 /4

(The other 9 9’s are the same, but with β → −β.) Here γ =

13.6 eV 2 324 α ,

and β = µB Bext .

In the weak-field limit (β γ): 2 94 ≈ E3 − 6γ + β/2 + 3γ 1 + β/9γ ≈ E3 − 6γ + β/2 + 3γ(1 + β/18γ) = E3 − 3γ + β. 3 1 95 ≈ E3 − 6γ + β/2 − 3γ(1 + β/18γ) = E3 − 9γ + β. 3 9 96 ≈ E3 − 2γ + 3β/2 + γ(1 + 3β/10γ) = E3 − γ + β. 5 6 97 ≈ E3 − 2γ + 3β/2 − γ(1 + 3β/10γ) = E3 − 3γ + β. 5 3 98 ≈ E3 − 2γ + β/2 + γ(1 + β/10γ) = E3 − γ + β. 5 2 99 ≈ E3 − 2γ + β/2 − γ(1 + β/10γ) = E3 − 3γ + β. 5 1.51 eV 2 2 Noting that γ = −(E3 /36)α = 36 α , we see that the weak field energies are recovered as in the first table. In the strong-field limit (β γ): 94 ≈ E3 − 6γ + β/2 + β/2 1 + 4γ/β ≈ E3 − 6γ + β/2 + β/2(1 + 2γ/β) = E3 − 5γ + β. 95 ≈ E3 − 6γ + β/2 − β/2(1 + 2γ/β) = E3 − 7γ. 7 96 ≈ E3 − 2γ + 3β/2 + β/2(1 + 6γ/5β) = E3 − γ + 2β. 5 97 ≈ E3 − 2γ + 3β/2 − β/2(1 + 6γ/5β) = E3 −

13 γ + β. 5

9 98 ≈ E3 − 2γ + β/2 + β/2(1 + 2γ/5β) = E3 − γ + β. 5 11 γ. 5 Again, these reproduce the strong-field results in the second table. In the figure below each line is labeled by the level number and (in parentheses) the starting and ending slope; for each line there is a corresponding one starting from the same point but sloping down. 99 ≈ E3 − 2γ + β/2 − β/2(1 + 2γ/5β) = E3 −


178

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 3(3)

E

6(9/5−> 2) 8(3/5−>1) 2(2)

Ε3

7(6/5−>1) 4(2/5−>1)

Ε 3 −γ

9(2/5−>0)

Ε 3 −3γ

1(1) 5(1/3−>0)

Ε 3 −9γ

Problem 6.27 I ≡ (a · rˆ)(b · rˆ) sin θ dθ dφ = (ax sin θ cos φ + ay sin θ sin φ + az cos θ)(bx sin θ cos φ + by sin θ sin φ + bz cos θ) sin θ dθ dφ.

2π

But 0

2π

sin φ dφ =

2π

cos φ dφ = 0

sin φ cos φ dφ = 0, so only three terms survive : 0

(ax bx sin2 θ cos2 φ + ay by sin2 θ sin2 φ + az bz cos2 θ) sin θ dθ dφ.

I=

2

But I= 0

cos φ dφ = π, 0

π

2π 2

sin φ dφ = 0

so

2π

2π

dφ = 2π, so 0

π(ax bx + ay by ) sin2 θ + 2πaz bz cos2 θ sin θ dθ.

π

sin3 θ dθ =

But 0

4 , 3

π

cos2 θ sin θ dθ = 0

2 , 3

2 4 4π 4π I = π(ax bx + ay by ) + 2πaz bz = (ax bx + ay by + az bz ) = (a · b). QED 3 3 3 3

[Alternatively, noting that I has to be a scalar bilinear in a and b, we know immediately that I = A(a·b), where ˆ then I = A = cos2 θ sin θ dθ dφ = A is some constant (same for all a and b). To determine A, pick a = b = k; 4π/3.] √ For states with l = 0, the wave function is independent of θ and φ (Y00 = 1/ 4π), so 0 1 ∞ 3(Sp · rˆ)(Se · rˆ) − Sp · Se 1 2 2 = |ψ(r)| r dr [3(Sp · rˆ)(Se · rˆ)] sin θ dθ dφ. r3 r3 0 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


179

The first angular integral is 3(4π/3)(Sp · Se ) = 4π(Sp · Se ), while the second is −(Sp · Se ) sin θ dθ dφ = −4π(Sp · Se ), so the two cancel, and the result is zero. QED [Actually, there is a little sleight-of-hand here, since for l = 0, ψ → constant as r → 0, and hence the radial integral diverges logarithmically at the origin. Technically, the first term in Eq. 6.86 is the field outside an infinitesimal sphere; the delta-function gives the field inside. For this reason it is correct to do the angular integral first (getting zero) and not worry about the radial integral.]

Problem 6.28

g ; we want reduced mass in a, but not in mp me (which come from From Eq. 6.89 we see that ∆E ∝ mp me a3 Eq. 6.85); the notation in Eq. 6.93 obscures this point.

(a) g and mp are unchanged; me → mµ = 207me , and a → aµ .

From Eq. 4.72, a ∝ 1/m, so

a 207 mµ (reduced) m µ mp 1 207 207 = = 186. = = · = = (9.11×10−31 ) aµ me mµ + mp me 1 + 207(me /mp ) 1.11 1 + 207 1.67×10−27 ) ∆E = (5.88 × 10−6 eV) (1/207) (186)3 = 0.183 eV. (b) g : 5.59 → 2;

a mp (reduced) m2e 1 1 = = · = . ap me me + me me 2 3 1 2 1.67 × 10−27 eV) = 4.82 ×10−4 eV. 5.59 9.11 × 10−31 2

m p → me ;

∆E = (5.88 × 10−6

a mm (reduced) m e mµ 1 207 . = = · = am me m e + mµ m e 208 3 2 1.67 × 10−27 207 −6 ∆E = (5.88 × 10 ) = 1.84×10−5 eV. 5.59 (207)(9.11 × 10−31 ) 208

(c) g : 5.59 → 2;

m p → mµ ;

Problem 6.29 Use perturbation theory: 1 1 e2 H = − − , 4π90 b r

for

0 < r < b. ∆E = ψ|H |ψ,

with ψ ≡ √

1 πa3

e−r/a .

b b b e2 1 1 1 −2r/a 2 1 e2 2 −2r/a −2r/a 4π r dr = − r e dr − re dr − e 4π90 πa3 b r π90 a3 b 0 0 0 b 2 2 e2 1 2r 2r a a 2 −2r/a a −2r/a −2r/a =− − − +a e e − r e −1 − −1 π90 a3 b 2 2 a 2 a 0

a2 −2b/a a3 a2 a 2 −2b/a a3 −2b/a 2b 2b e2 −1 − e −1 + − − b e − − =− + e π90 a3 2b 4b a 4 a 4b 4 2 2 3 2 2 e ab a a ab a a a =− e−2b/a − − − + + + −1 π90 a3 2 2 4b 2 4 4 b

e2 a a2 a e2 a a −2b/a a2 −2b/a =− + 1 + − 1 = 1 − + 1 + e e − . π90 a3 4 b 4 b 4π90 a b b

∆E = −


180


Let +2b/a = 9 (very small). Then the term in square brackets is: 2 2 92 93 1− + 1+ 1−9+ − + ··· 9 9 2 6 2 3 2✄ 92 2✄ 9 9 92 =1 + ✁9 − − + ( )93 + · · · = + ( )93 + ( )94 · · · ✁− ✄ +1 ✁ + ✄ − ✁9 − 2 ✁+ 2 6 3 6 ✄9 ✄9 e2 1 4b2 To leading order, then, ∆E = . 4π90 a 6a2 E = E1 = − ∆E e2 = E 4π90

me4 ; 2(4π90 )2 2

−

a=

4π90 2 ; me2

so

Ea = −

e2 . 2(4π90 )

2 2(4π90 ) 2b2 4 b = − . e2 3a2 3 a

Putting in a = 5 × 10−11 m: ∆E 16 4 10−15 = − × 10−10 ≈ −5 × 10−10 . =− −11 E 3 5 × 10 3 By contrast,

fine structure: ∆E/E ≈ α2 = (1/137)2 = 5 × 10−5 , hyperfine structure: ∆E/E ≈ (me /mp )α2 = (1/1800)(1/137)2 = 3 × 10−8 .

So the correction for the finite size of the nucleus is much smaller (about 1% of hyperfine).

Problem 6.30 (a) In terms of the one-dimensional harmonic oscillator states {ψn (x)}, the unperturbed ground state is |0 = ψ0 (x)ψ0 (y)ψ0 (z). E01 = 0|H |0 = ψ0 (x)ψ0 (y)ψ0 (z)|λx2 yz|ψ0 (x)ψ0 (y)ψ0 (z) = λx2 0 y0 z0 . But

y0 = z0 = 0.

So there is no change, in first order.

(b) The (triply degenerate) first excited states are   |1 = ψ0 (x)ψ0 (y)ψ1 (z) |2 = ψ0 (x)ψ1 (y)ψ0 (z)  |3 = ψ1 (x)ψ0 (y)ψ0 (z) In this basis the perturbation matrix is

Wij = i|H |j,

i = 1, 2, 3.

1|H |1 = ψ0 (x)ψ0 (y)ψ1 (z)|λx yz|ψ0 (x)ψ0 (y)ψ1 (z) = λx2 0 y0 z1 = 0, 2

2|H |2 = ψ0 (x)ψ1 (y)ψ0 (z)|λx2 yz|ψ0 (x)ψ1 (y)ψ0 (z) = λx2 0 y1 z0 = 0, 3|H |3 = ψ1 (x)ψ0 (y)ψ0 (z)|λx2 yz|ψ1 (x)ψ0 (y)ψ0 (z) = λx2 1 y0 z0 = 0, 1|H |2 = ψ0 (x)ψ0 (y)ψ1 (z)|λx2 yz|ψ0 (x)ψ1 (y)ψ0 (z) = λx2 0 0|y|11|z|0 2 =λ [using Problems 2.11 and 3.33]. |0|x|1|2 = λ 2mω 2mω 1|H |3 = ψ0 (x)ψ0 (y)ψ1 (z)|λx2 yz|ψ1 (x)ψ0 (y)ψ0 (z) = λ0|x2 |1y0 1|z|0 = 0, 2|H |3 = ψ0 (x)ψ1 (y)ψ0 (z)|λx2 yz|ψ1 (x)ψ0 (y)ψ0 (z) = λ0|x2 |11|y|00 z0 = 0. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


181



Eigenvalues of W :

 2 0 a0 W = a 0 0 , where a ≡ λ . 2mω 0 0 0 −E a 0 2 a −E 0 = −E 3 + Ea2 = 0 ⇒ E = {0, ±a} = 0, ±λ . 2mω 0 0 −E

Problem 6.31 (a) The first term is the nucleus/nucleus interaction, the second is the interaction between the nucleus of atom 2 and the electron in atom 1, the third is between nucleus 1 and electron 2, and the last term is the interaction between the electrons.

x x 2 1 1 x −1 1 = 1− 1+ + = + ... , R−x R R R R R so 2 x x 2 x x 2 x1 − x2 x1 − x2 1 1 2 2 − 1− + 1+ 1− 1+ + + + R R R R R R 1 e2 e2 x1 x2 2x1 x2 ≈ =− . − 2 4π90 R R 2π90 R3

1 e2 H ∼ = 4π90 R

(b) Expanding Eq. 6.99:

1

1 2 e2 2 x − x2− p+ + p2− + k x2+ + x2− − 2m 2 4π90 R3 +

1

1 2 e2 = (2x1 x2 ) = H 0 + H p1 + p22 + k x21 + x22 − 2m 2 4π90 R3

H =

(c)

ω± =

k m

e2 1∓ 2π90 R3 k

1/2

(Eqs. 6.96 and 6.98).

2 1 1 e2 e2 ∼ + ... . − = ω0 1 ∓ 2 2π90 R3 mω02 8 2π90 R3 mω02

2 1 1 1 e2 e2 ∆V ∼ − + = ω0 1 − 2 2 2π90 R3 mω02 8 2π90 R3 mω02 2 1 e2 e2 1 1+ − − ω0 2 2π90 R3 mω02 8 2π90 R3 mω02 2 2 2 1 e 1 1 1 e2 − = = − . 2ω0 4 2π90 R3 mω02 8 m2 ω03 2π90 R6 (d) In first order: E01 = 0|H |0 = −

e2 e2 ψ (x )ψ (x )|x x |ψ (x )ψ (x ) = − x0 x0 = 0. 0 1 0 2 1 2 0 1 0 2 2π90 R3 2π90 R3


182

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY In second order: ∞ |ψn |H |ψ0 |2 . Here |ψ0 = |0|0, |ψn = |n1 |n2 , so E0 − En n=1 2 ∞ ∞ e2 |n1 |x1 |0|2 |n2 |x2 |0|2 = [use Problem 3.33] 3 2π90 R E0,0 − En1 ,n2 n1 =1 n2 =1 2 e2 |1|x|0|2 |1|x|0|2 = [zero unless n1 = n2 = 1] 2π90 R3 ( 12 ω0 + 12 ω0 ) − ( 32 ω0 + 32 ω0 ) 2 2 2 2 e2 1 e 1 = − =− 2 3 . 3 2π90 R 2ω0 2mω0 8m ω0 2π90 R6

E02 =

Problem 6.32 (a) Let the unperturbed Hamiltonian be H(λ0 ), for some fixed value λ0 . Now tweak λ to λ0 + dλ. The perturbing Hamiltonian is H = H(λ0 + dλ) − H(λ0 ) = (∂H/∂λ) dλ (derivative evaluated at λ0 ). The change in energy is given by Eq. 6.9: dEn = En1 = ψn0 |H |ψn0 = ψn |

∂H |ψn dλ (all evaluated at λ0 ); ∂λ

so

∂En ∂H = ψn | |ψn . ∂λ ∂λ

[Note: Even though we used perturbation theory, the result is exact, since all we needed (to calculate the derivative) was the infinitesimal change in En .] (b) En = (n + 12 )ω;

H=−

2 d 2 1 + mω 2 x2 . 2 2m dx 2

(i) ∂En 1 = (n + ); ∂ω 2 V =

1 mω 2 x2 , 2

so

∂H = mωx2 ; ∂ω

1 so F-H ⇒ (n + ) = n|mωx2 |n. But 2

1 1 1 V = n| mω 2 x2 |n = ω(n + ); 2 2 2

(ii) ∂En 1 = (n + )ω; ∂ 2

2 ∂H d2 = =− ∂ m dx2

1 2 so F-H ⇒ (n + )ω = n|T |n, 2

or

−

2 d 2 2m dx2

∂H 1 1 2 d 2 + ω 2 x2 = − = ∂m 2m2 dx2 2 m

1 1 So F-H ⇒ 0 = − T + V , or m m Problems 2.12 and 3.31.

=

2 T;

T = 12 (n + 12 )ω.

(iii) ∂En = 0; ∂m

V = 12 (n + 12 )ω.

T = V .

2 d 2 − 2m dx2

1 + m

1 mω 2 x2 2

=−

1 1 T + V. m m

These results are consistent with what we found in



183

Problem 6.33 (a) ∂En 4 4me3 = En ; =− 2 2 2 ∂e 32π 90 (jmax + l + 1)2 e 4 e En = − e 2π90

0 1 0 1 2 2 m 1 1 8π90 1 e 8π90 E1 8π90 e2 m 1 , or = − 2 En = − 2 2 = − 2 − 2 = . 2 r r e e n e 2 4π90 n 4π90 2 n2

4π90 2 = a (by Eq. 4.72), so me2

But

∂H 2e 1 =− . So the F-H theorem says: ∂e 4π90 r

0 1 1 1 = 2 . r n a

(Agrees with Eq. 6.55.)

(b) ∂H 2 (2l + 1); = ∂l 2mr2

∂En 2En 2me4 =− = ; 2 2 2 ∂l 32π 90 (jmax + l + 1)3 n 0

2En 2 (2l + 1) − = n 2m

But −

4mE1 2 = 2, 2 a

1 r2

1

0 ,

or 0

so

1 r2

1 r2

1 =−

1 =

n3 (l

so F-H says

4mE1 4mEn =− 3 . 2 n(2l + 1) n (2l + 1)2

1 . + 12 )a2

(Agrees with Eq. 6.56.)

Problem 6.34 Equation 4.53 ⇒ u =

But

l(l + 1) 2mEn 2m − − 2 2 2 r

me2 1 = (Eq. 4.72), 2 4π90 a u =

and

−

e2 4π90

1 u. r

2m m 2mEn = 2 2 22

e2 4π90

2

1 1 = 2 2 . So 2 n a n

l(l + 1) 2 1 u. − + r2 ar n2 a2

s

∴

(ur u ) dr =

=−

ur

s

l(l + 1) 2 2 1 1 − + 2 2 u dr = l(l + 1)rs−2 − rs−1 + 2 2 rs 2 r ar n a a n a

d (urs )u dr = − dr

(u rs u ) dr − s

(urs−1 u ) dr.


184

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY Lemma 1:

Lemma 2:

d (urs )u dr = − dr

s

(ur u ) dr = −sr

2

2

(urs u ) dr = −

s−1

s+1

(u r

,

or

u ) dr = −

(u rs+1 u ) dr = −(s + 1)

⇒

urs−1 u dr

s (urs u ) dr = − rs−1 . 2

d s+1 u (r u ) dr = −(s + 1) dr

(u rs u) dr − s

(u rs u ) dr,

or:

s

(u r u ) dr −

(u rs u ) dr = −

2 s+1

(u rs+1 u ) dr.

(u rs+1 u ) dr.

Lemma 3: Use in Lemma 2, and exploit Lemma 1:

l(l + 1) 1 2 2 + (urs+1 u ) dr − s+1 r2 ar n2 a2

2 2 1 =− l(l + 1) (urs−1 u ) dr − (urs u ) dr + 2 2 (urs+1 u ) dr s+1 a n a s+1 s 2 2 1 s − 1 s−2 s =− l(l + 1) − r − − rs−1 + 2 2 − r s+1 2 a 2 n a 2 s−1 2 1 s = l(l + 1) rs−2 − rs−1 + 2 2 rs . s+1 a s+1 n a

(u rs u ) dr = −

Plug Lemmas 1 and 3 into : 2 1 l(l + 1)rs−2 − rs−1 + 2 2 rs a n a s−1 2 1 s(s − 1) s−2 s s−2 = −l(l + 1) r + rs−1 − 2 2 rs + r . s+1 a s+1 n a 2

2 s 2 s−1 s(s − 1) s s−1 1+ rs−2 = 0. r − r + l(l + 1) 1 + − n2 a2 a s+1 s+1 2 2s s+1

2s+1 s+1

2(s + 1) s 2 (s2 − 1) s−1 2 r − + 2s l + l − (2s + 1)r rs−2 = 0, or, finally, n2 a2 a 4 (s + 1) s sa2 2 r − a(2s + 1)rs−1 + (4l + 4l + 1 −s2 )rs−2 = 0. QED 2 n 4 (2l+1)2



185

Problem 6.35 (a) 1 1 − a n2

0 1 0 1 1 1 1 +0=0⇒ = 2 . r r n a

2 1 r − 3a1 + (2l + 1)2 − 1 a2 2 n 4 r =

0 1 1 1 2 a = 0 ⇒ 2 r = 3a − l(l + 1)a2 2 = 2 3n2 − l(l + 1) . r n n a n

a 2 3n − l(l + 1) . 2

2 a2 3 2 1 3 2 a 2 2 r − 5ar + − 4 a = 0 ⇒ r = 5a − l(l + 1) − (2l + 1) 3n (2l + 1)2 − 4 2 2 n 2 n 2 2 a2 3 2 a2 r = 15n2 − 5l(l + 1) − 4l(l + 1) − 1 + 4 = 15n2 − 9l(l + 1) + 3 2 n 2 2 2 2 2 3a n a = r2 = 5n2 − 3l(l + 1) + 1 ; 5n2 − 3l(l + 1) + 1 . 2 2 4 3 3 (2l + 1)2 − 9 a2 r = 0 =⇒ r − 7ar2 + 2 n 4 3 4 3 a n2 a2 2 r = 7a 5n − 3l(l + 1) + 1 − [4l(l + 1) − 8] a2 3n2 − l(l + 1) 2 n 2 4 2 2 . a3 4 2 2 = 35n − 21l(l + 1)n + 7n − [3l(l + 1) − 6] 3n − l(l + 1) 23 a = 35n4 − 21l(l + 1)n2 + 7n2 − 9l(l + 1)n2 + 3l2 (l + 1)2 + 18n2 − 6l(l + 1) 2 a3 = 35n4 + 25n2 − 30l(l + 1)n2 + 3l2 (l + 1)2 − 6l(l + 1) . 2 r3 =

n2 a3 35n4 + 25n2 − 30l(l + 1)n2 + 3l2 (l + 1)2 − 6l(l + 1) . 8

(b) 0 0+a

1 r2

1

1 − (2l + 1)2 − 1 a2 4

0

1 r3

1

0 =0⇒

1 r2

1

0 = al(l + 1)

1 1 . r3

(c) 0 al(l + 1)

1 r3

1

1 = ⇒ (l + 1/2)n3 a2

0

1 r3

1 =

1 . l(l + 1/2)(l + 1)n3 a3

Agrees with Eq. 6.64.


186


Problem 6.36 (a) |1 0 0 = √

1 πa3

−r/a

e

(Eq. 4.80),

But the θ integral is zero: 0

π

Es1

1 = 1 0 0|H |1 0 0 = eEext 3 πa

e−2r/a (r cos θ)r2 sin θ dr dθ dφ.

π sin2 θ cos θ sin θ dθ = = 0. So Es1 = 0. QED 2 0

 1 1 r −r/2a   √ |1 = ψ = 1 − e 2 0 0   2a  2πa 2a   1 1   re−r/2a sin θeiφ  |2 = ψ2 1 1 = − √ 2 8a πa (b) From Problem 4.11: 1 1   |3 = ψ2 1 0 = √ re−r/2a cos θ  2  4a  2πa   1 1   re−r/2a sin θe−iφ  |4 = ψ2 1−1 = √ πa 8a2  π   1|Hs |1 = {. . . } cos θ sin θ dθ = 0    0 π    2  2|Hs |2 = {. . . } sin θ cos θ sin θ dθ = 0     0 π    2  3|Hs |3 = {. . . } cos θ cos θ sin θ dθ = 0     0 π    2 4|Hs |4 = {. . . } sin θ cos θ sin θ dθ = 0     All matrix elements of Hs are zero  0 2π   except 1|Hs |3 and 3|Hs |1 1|Hs |2 = {. . . } eiφ dφ = 0 (which are complex conjugates,  0 2π    so only one needs to be evaluated). −iφ   1|Hs |4 = {. . . } e dφ = 0    0  2π     2|Hs |3 = {. . . } e−iφ dφ = 0    0  2π    −2iφ  2|Hs |4 = {. . . } e dφ = 0    0  2π    −iφ   3|Hs |4 = {. . . } e dφ = 0 0

r −r/2a −r/2a e 1− re cos θ(r cos θ)r2 sin θ dr dθ dφ 2a

∞ eEext r −r/a 4 2 = 1 − (2π) cos θ sin θ dθ r dr e 2πa8a3 2a 0 0 ∞ ∞ eEext 2 1 eEext 1 4 −r/a 5 −r/a 5 6 = 4!a 5!a r e dr − r e dr = − 8a4 3 2a 0 12a4 2a 0 eEext 5 5 = = eaEext (−3) = −3aeEext . 24a 1 − 12a4 2

1|Hs |3 = eEext √

1 1 1 1 √ 2 2πa 2a π2πa 4a



0 0 W = −3aeEext  1 0

0 0 0 0

1 0 0 0

 0 0 . 0 0



187

We need the eigenvalues of this matrix. The characteristic equation is: −λ 0 1 0 −λ 0 0 0 −λ 0 0 −λ 0 0 3 2 2 2 1 0 −λ 0 = −λ 0 −λ 0 + 1 0 0 = −λ(−λ) + (−λ ) = λ (λ − 1) = 0. 0 0 −λ 0 0 −λ 0 0 0 −λ The eigenvalues are 0, 0, 1, and −1, so the perturbed energies are E2 , E2 , E2 + 3aeEext , E2 − 3aeEext .

Three levels.

    0 0 1 0    (c) The eigenvectors with eigenvalue 0 are |2 =   0  and |4 =  0 ; the eigenvectors with eigenvalues ±1 0 1   1 1  0   . So the “good” states are ψ2 1 1 , ψ2 1−1 , √1 (ψ2 0 0 + ψ2 1 0 ), √1 (ψ2 0 0 − ψ2 1 0 ). are |± ≡ √   ±1 2 2 2 0 pe 4 = −e

1 1 πa 64a4

2π

But

cos φ dφ = 0

r2 e−r/a sin2 θ r sin θ cos φî + r sin θ sin φˆj + r cos θkˆ r2 sin θ dr dθ dφ.

2π

sin φ dφ = 0, 0

0

π

4 π sin θ = 0, sin3 θ cos θ dθ = 4

pe 4 = 0.

so

0

Likewise

pe 2 = 0.

1 pe ± = − e (ψ1 ± ψ3 )2 (r)r2 sin θ dr dθ dφ 2 2 1 1 1 r r ˆ 2 sin θ dr dθ dφ =− e 1− ± cos θ e−r/a r(sin θ cos φ î + sin θ sin φ ˆj + cos θ k)r 2 2 2πa 4a 2a 2a 2 e kˆ 1 r r =− 2π 1 − r3 e−r/a cos θ sin θ dr dθ. ± cos θ 2 2πa 4a2 2a 2a π π But 0 cos θ sin θ dθ = 0 cos3 θ sin θ dθ = 0, so only the cross-term survives: pe ±

1 r e ˆ 1− r cos θ r3 e−r/a cos θ sin θ dr dθ = − 3k ± 8a a 2a

π ∞ e e 2 r 4 −r/a 1 2 5 6 ˆ ˆ 1− =∓ cos θ sin θ dθ dr = ∓ k k r e 4!a − 5!a 4 8a 2a 8a4 3 2a 0 0 1 5 5 ˆ = ∓ekˆ 24a 1 − = ±3aek. 12a4 2

Problem 6.37 (a) The nine states are: c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

188

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY  l=0:   l = 1 :           l=2:             

|3 0 0 |3 1 1 |3 1 0 |3 1 − 1 |3 2 2 |3 2 1 |3 2 0 |3 2 − 1 |3 2 − 2

= R30 Y00 = R31 Y11 = R31 Y10 = R31 Y1−1 = R32 Y22 = R32 Y21 = R32 Y20 = R32 Y2−1 = R32 Y2−2

Hs contains no φ dependence, so the φ integral will be: n l m|Hs |n l m = {· · · }

2π

e−imφ eim φ dφ,

which is zero unless

m = m.

0

π 2 For diagonal elements: n l m|Hs |n l m = {· · · } 0 [Plm (cos θ)] cos θ sin θ dθ. But (p. 137 in the text) m Pl is a polynomial (even or odd) in cos θ, multiplied (if m is odd) by sin θ. Since sin2 θ = 1 − cos2 θ, 2 [Plm (cos θ)] is a polynomial in even powers of cos θ. So the θ integral is of the form

π 2j+1

(cos θ) 0

π (cos θ)2j+2 sin θ dθ = − = 0. All diagonal elements are zero. (2j + 2) 0

There remain just 4 elements to calculate: m = m = 0 : 3 0 0|Hs |3 1 0, 3 0 0|Hs |3 2 0, 3 1 0|Hs |3 2 0; m = m = ±1 : 3 1 ± 1|Hs |3 2 ± 1. 3 0 0|Hs |3 1 0 = eEext

R30 R31 r3 dr

8 2 1 1 1 √ 3/2 R30 R31 r dr = √ 3/2 27 a 27 6 a a 3

Let

Y00 Y10 cos θ sin θ dθ dφ.

2r 2r2 1− + 3a 27a2

From Table 4.7 : r −r/3a 3 e−r/3a 1 − r dr. re 6a

x ≡ 2r/3a:

5 ∞ 3a x2 24 x 4 −x √ 1−x+ 1− x e dx 6 4 35 2a4 2 0 ∞ a 5 1 a 5 5 5 1 = √ 1 − x + x2 − x3 x4 e−x dx = √ 4! − 5! + 6! − 7! 4 12 24 4 12 24 2 2 0 2 2 √ = −9 2a.

R30 R31 r3 dr =

Y00 Y10

√ √ π 3 32 3 3 cos θ cos θ sin θ dθ dφ = 2π = . cos θ sin θ dθ = 4π 2 3 3 0 '√ ( √ √ 3 3 0 0|Hs |3 1 0 = eEext (−9 2a) = −3 6aeEext . 3 3 0 0|Hs |3 2 0 = eEext R30 R31 r3 dr Y00 Y20 cos θ sin θ dθ dφ.

1 cos θ sin θ dθ dφ = √ 4π

3 4π

√


CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY Y00 Y20

1 cos θ sin θ dθ dφ = √ 4π

5 16π

3 1 0|Hs |3 2 0 = eEext

(3 cos2 θ − 1) cos θ sin θ dθ dφ = 0.

189

3 0 0 |Hs |3 2 0 = 0.

R31 R32 r3 dr

Y10 Y20 cos θ sin θ dθ dφ.

r −r/3a 2 −r/3a 3 8 1 1 4 1 1 √ 3/2 √ R31 R32 r dr = 1 − r e r dr re 6a 27 6 a a 81 30 a3/2 a2 √ 7 ∞ 24 3a x 6 −x a 9 5 1 = √ 1− x e dx = √ a. 6! − 7! = − 4 4 2 38 5a6 2 24 5 0 3

Y10 Y20

3 5 sin θ cos θ dθ dφ = cos θ(3 cos2 θ − 1) cos θ sin θ dθ dφ 4π 16π √ √

π π 15 15 3 1 2 4 2 = 2π − cos5 θ + cos3 θ = √ . (3 cos θ − cos θ) sin θ dθ = 8π 4 5 3 15 0 0 '

√ ( √ 9 5 2 √ = eEext − = −3 3aeEext . a 2 15 ±1 ∗ ±1 3 1 ± 1|Hs |3 2 ± 1 = eEext R31 R32 r3 dr Y1 Y2 cos θ sin θ dθ dφ. 3 1 0|Hs |3 2 0

±1 ∗

Y1

(' ( 3 15 ∓ sin θe∓iφ sin θ cos θe±iφ cos θ sin θ dθ dφ Y2±1 cos θ sin θ dθ dφ = ∓ 8π 8π √ π π 3 5 3√ cos3 θ cos5 θ 2 2 = 2π + cos θ(1 − cos θ) sin θ dθ = 5 − 8π 4 3 5 0 0 1 = √ . 5 ' √ ( 5 9 1 √ 3 1 ± 1|Hs |3 2 ± 1 = eEext −9 = − aeEext . a 2 2 5 '

Thus the matrix representing Hs is (all empty boxes are zero; all numbers multiplied by −aeEext ): (b) The perturbing matrix (below) breaks into a 3×3 block, two 2 × 2 blocks, and two 1 × 1 blocks, so we can work out the eigenvalues in each block separately. √ √   −λ 2 0 20 √ √ √0 3 × 3 : 3 3  2 0 1  ; 2 −λ 1 = −λ3 + λ + 2λ = −λ(λ2 − 3) = 0; 0 1 −λ 0 1 0 √ λ = 0, ± 3

2×2:

9 2

01 10

⇒ ;

E11 = 0, E21 = 9aeEext , E31 = −9aeEext . −λ 1 2 1 −λ = λ − 1 = 0 ⇒ λ = ±1.


190

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY 300 310 320 311 321 31-1 32-1 322 32-2 300

3√6

310 3√6 320

3√3 3√3 9/2

311 321

9/2 9/2

31-1 32-1

9/2

322 32-2

9 9 aeEext , E51 = − aeEext . From the other 2×2 we get E61 = E41 , E71 = E51 , and from the 1×1’s we 2 2 0 (degeneracy 3) (9/2)aeEext (degeneracy 2) −(9/2)aeEext (degeneracy 2) get E81 = E91 = 0. Thus the perturbations to the energy (E3 ) are: 9aeEext (degeneracy 1) (degeneracy 1) −9aeEext

E41 =

Problem 6.38 µ0 gd e2 1 Sd · Se ; Eq. 6.91 ⇒ Sd · Se = (S 2 − Se2 − Sd2 ). 3πmd me a3 2

Electron has spin 12 , so Se2 = 12 32 2 = 34 2 ; deuteron has spin 1, so Sd2 = 1(2)2 = 22 .

1 3 2 2 2 Total spin could be 32 [in which case S 2 = 32 52 2 = 15 4 ] or 2 [in which case S = 4 ]. Thus   1 15 2 3 2

 2 4 − 4 − 22 = 12 2  3 µ0 gd e2 2 Sd · Se = ; the difference is 2 , so ∆E = .

1 3 2 3 2 2 2πmd me a3 2 2 − − 2 = − 2 4 4 1 Equation 6.89 ⇒ Ehf =

1 1 2gd e2 2 2gd 4 3 g d mp ⇒ µ = , so ∆E = = = ∆Ehydrogen (Eq. 6.98). 0 2 2 2 3 2 2 4 c 90 c 4π90 md me c a md me c a 2 gpmd c 2 g p md 4 5.59 ch Now, λ = = λh , and since md = 2mp , λd = , so λd = (21 cm) = 92 cm. ν ∆E 3 gd mp 3 1.71 But

µ0 90 =

Problem 6.39 (a) The potential energy of the electron (charge −e) at (x, y, z) due to q’s at x = ±d alone is: eq 1 1 V =− . Expanding (with d x, y, z) : + 4π90 (x + d)2 + y 2 + z 2 (x − d)2 + y 2 + z 2 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


≈

1 (x ± d)2 + y 2 + z 2 1 d

1∓

= (x2 ± 2dx + d2 + y 2 + z 2 )−1/2 = (d2 ± 2dx + r2 )−1/2 =

3 4x2 x r2 − 2+ d 2d 8 d2

=

191 1 d

1±

2x r2 + 2 d d

−1/2

x 1 1 1 ∓ + 2 (3x2 − r2 ) . d d 2d

eq x 2eq x 1 1 eq 2 2 2 2 V =− (3x2 − r2 ) 1 − + 2 (3x − r ) + 1 + + 2 (3x − r ) = − − 4π90 d d 2d d 2d 4π90 d 4π90 d3 = 2βd2 + 3βx2 − βr2 ,

where

e q . 4π90 d3

β≡−

Thus with all six charges in place H = 2(β1 d21 + β2 d22 + β3 d23 ) + 3(β1 x2 + β2 y 2 + β3 z 2 ) − r2 (β1 + β2 + β3 ). QED 1 (b) 1 0 0|H |1 0 0 = e−2r/a H r2 sin θ dr dθ dφ πa3 3 (β1 + β2 + β3 ) = V0 + 3 e−2r/a (β1 x2 + β2 y 2 + β3 z 2 )r2 sin θdr dθ dφ − r2 e−2r/a r2 sin θ dr dθ dφ. πa πa3 I1 ≡

r e

I2 ≡

r sin θ dr dθ dφ = 4π 0

−2r/a

e

2 −2r/a 2

2

2

2

∞

a r4 e−2r/a dr = 4π4!( )5 = 3πa5 . 2

2

(β1 x + β2 y + β3 z )r sin θ dr dθ dφ

r4 e−2r/a (β1 sin2 θ cos2 φ + β2 sin2 θ sin2 φ + β3 cos2 θ) sin θ dr dθ dφ. 2π 2π 2π But cos2 φ dφ = sin2 φ dφ = π, dφ = 2π. So 0 0 0 ∞ π = r4 e−2r/a dr π(β1 + β2 ) sin2 θ + 2πβ3 cos2 θ sin θ dθ. 0 0 π π 4 2 3 But sin θ dθ = , cos2 θ sin θ dθ = . So 3 3 0

0 a 5 4π 4π = 4! (β1 + β2 ) + β3 = πa5 (β1 + β2 + β3 ). 2 3 3

=

3 (β1 + β2 + β3 ) πa5 (β1 + β2 + β3 ) − 3πa5 = V0 . πa3 πa3   |2 0 0 = R20 Y00       |2 1 1 = R21 Y11 (functional forms in Problem 4.11). (c) The four states are −1 |2 1 − 1 = R21 Y1       |2 1 0 = R21 Y10

Diagonal elements: n l m|H |n l m = V0 + 3 β1 x2 + β2 y 2 + β3 z 2 − (β1 + β2 + β3 )r2 . 1 0 0|H |1 0 0 = V0 +


192

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY For |2 0 0, x2 = y 2 = z 2 = 13 r2 (Y00 does not depend on φ, θ; this state has spherical symmetry), so 2 0 0|H |2 0 0 = V0 .

(I could have used the same argument in (b).)

n2 a2 2 From Problem 6.35(a), r2 = 5n − 3l(l + 1) + 1 , so for n = 2, l = 1 : r2 = 30a2 . Moreover, 2 2π 2π since x2 = {. . . } cos2 φ dφ = {. . . } sin2 φ dφ = y 2 , and x2 + y 2 + z 2 = r2 , it follows 0

0

1 2 1 (r − z 2 ) = 15a2 − z 2 . So all we need to calculate is z 2 . 2 2 1 1 r2 e−r/a cos2 θ(r2 cos2 θ)r2 sin θ dr dθ dφ 2 1 0|z 2 |2 1 0 = 2πa 16a4

that x2 = y 2 =

=

1 16a5

∞

r6 e−r/a dr

0

π

cos4 θ sin θ dθ = 0

2 1 6!a7 = 18a2 ; 5 16a 5

x2 = y 2 = 15a2 − 9a2 = 6a2 .

2 1 0|H |2 1 0 = V0 + 3(6a2 β1 + 6a2 β2 + 18a2 β3 ) − 30a2 (β1 + β2 + β3 ) = V0 − 12a2 (β1 + β2 + β3 ) + 36a2 β3 . 1 1 2 1 ± 1|z |2 1 ± 1 = πa 64a4

2

1 = 32a5

∞

6 −r/a

r e

π

1 (1 − cos θ) cos θ sin θ dθ = 6!a7 32a5 2

dr

0

r2 e−r/a sin2 θ(r2 cos2 θ)r2 sin θ dr dθ dφ

0

2

2 2 − 3 5

= 6a2 ;

x2 = y 2 = 15a2 − 3a2 = 12a2 . 2 1 ± 1|H |2 1 ± 1 = V0 + 3(12a2 β1 + 12a2 β2 + 6a2 β3 ) − 30a2 (β1 + β2 + β3 ) = V0 + 6a2 (β1 + β2 + β3 ) − 18a2 β3 . Off-diagonal elements:

We need 2 0 0|H |2 1 0, 2 0 0|H |2 1 ± 1, 2 1 0|H |2 1 ± 1, and 2 1 − 1|H |2 1 1.

Now n l m|V0 |n l m = 0, by orthogonality, and n l m|r2 |n l m = 0, by orthogonality of Ylm , so all we need are the matrix elements of x2 and y 2 (|z 2 | = −|x2 | − |y 2 ). For 2 0 0|x2 |2 1 ± 1 and 2π 2π 2π 2 1 0|x2 |2 1 ± 1 the φ integral is 0 cos2 φe±iφ dφ = 0 cos3 φ dφ ± i 0 cos2 φ sin φ dφ = 0, and the same goes for y 2 . So 2 0 0|H |2 1 ± 1 = 2 1 0|H |2 1 ± 1 = 0. π π For 2 0 0|x2 |2 1 0 and 2 0 0|y 2 |2 1 0 the θ integral is 0 cos θ(sin2 θ) sin θ dθ = sin4 θ/40 = 0, so 2 0 0|H |2 1 0 = 0.

Finally:

1 1 2 1 − 1|x |2 1 1 = − r2 e−r/a sin2 θe2iφ (r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ πa 64a4 ∞ π 2π 1 5 6 −r/a =− r e dr sin θdθ e2iφ cos2 φ dφ 64πa5 0 0 0 2

6!a7

=−

16/15

π/2

16 π 1 6!a7 = −6a2 . 64πa5 15 2


CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY For y 2 , the φ integral is

2π 0

193

e2iφ sin2 φ dφ = −π/2, so 2 1 − 1|y 2 |2 1 1 = 6a2 , and 2 1 − 1|z 2 |2 1 1 = 0.

2 1 − 1|H |2 1 1 = 3 β1 (−6a2 ) + β2 (6a2 ) = −18a2 (β1 − β2 ).

The perturbation matrix is: 2 1 0 2 1 1 2 1 -1 0 0 0 V0 − 12a2 (β1 + β2 ) + 24a2 β3 0 0 0 V0 + 6a2 (β1 + β2 ) − 12a2 β3 −18a2 (β1 − β2 ) 0 −18a2 (β1 − β2 ) V0 + 6a2 (β1 + β2 ) − 12a2 β3 A B The 2 × 2 block has the form ; its characteristic equation is (A − λ)2 − B 2 = 0, so A − λ = ±B, B A or 2 2 2 2

2 0 0 V0 0 0 0

0 0 1 0 1 1 1 -1

λ = A ∓ B = V0 + 6a2 (β1 + β2 ) − 12a2 β3 ± 18a2 (β1 − β2 ) =

91 9 The first-order corrections to the energy (E2 ) are therefore: 2 93 94

V0 + 24a2 β1 − 12a2 β2 − 12a2 β3 , V0 − 12a2 β1 + 24a2 β2 − 12a2 β3 .

= V0 = V0 − 12a2 (β1 + β2 − 2β3 ) = V0 − 12a2 (−2β1 + β2 + β3 ) = V0 − 12a2 (β1 − 2β2 + β3 )

(i) If β1 = β2 = β3 , then 91 = 92 = 93 = 94 = V0 : one level (still 4-fold degenerate). (ii) If β1 = β2 = β3 , then 91 = V0 , 92 = V0 − 24a2 (β1 − β3 ), 93 = 94 = V0 + 12a2 (β1 − β3 ): three levels (one remains doubly degenerate). (iii) If all three β’s are different, there are four levels (no remaining degeneracy).

Problem 6.40 (a) (i) Equation 6.10:

0

H = E00 = H = ψ00 =

(H 0 − E00 )ψ01 = −(H − E01 )ψ00 .

2 2 2 e2 1 2 2 ∇ − =− ∇ + , since − 2m 4π90 r 2m ar 2 − . 2ma2 eEext r cos θ; E01 = 0 (Problem6.36(a)). 1 √ e−r/a ; ψ01 = f (r)e−r/a cos θ. πa3

a=

4π90 2 . me2


194

CHAPTER 6. TIME-INDEPENDENT PERTURBATION THEORY Equation 4.13 ⇒

f e−r/a d cos θ d d 2 d −r/a + r f e sin θ (cos θ) r2 dr dr r2 sin θ dθ dθ

−r/a cos θ d fe 1 d = r2 f − f e−r/a + 2 − sin2 θ 2 r dr a r sin θ dθ

cos θ 2 cos θ −r/a 1 2 1 −r/a 2 −r/a = 2r f f − − + r − + f e fe f e f r2 a a a2 r2

2 1 1 1 1 = cos θe−r/a f − f + 2 f + 2 f − f − 2f 2 . a a a r r

∇2 ψ01 =

Plug this into Eq. 6.10:

2 1 1 1 2 1 1 11 1 −r/a − f − f + 2f + 2 f − f cos θe − 2f 2 + 2f − f 2 = −eEext r cos θ √ e−r/a , 2m a a a r r ar a πa3

2 f − f a

1

1 + 2f − 2f 2 = r r

f (r) = A + Br + Cr2 ,

Now let

so

2meEext √ 2 πa3

f = B + 2Cr

r=

4γ r, a

and f = 2C.

where

γ≡

meEext √ . 22 πa

Then

2 2 2 4γ 2C − (B + 2Cr) + (B + 2Cr) − 2 (A + Br + Cr2 ) = r. a r r a Collecting like powers of r: r−2 : r−1 : r0 : r1 :

A = 0. 2B − 2B = 0 (automatic). 2C − 2B/a + 4C − 2C = 0 ⇒ B = 2aC. −4C/a = 4γ/a ⇒ C = −γ.

Evidently the function suggested does satisfy Eq. 6.10, with the coefficients A = 0, B = −2aγ, C = −γ; the second-order correction to the wave function is ψ01 = −γr(r + 2a)e−r/a cos θ. (ii) Equation 6.11 says, in this case: E02

meEext √ eEext e−r/a (r cos θ)r(r + 2a)e−r/a cos θ r2 sin θ dr dθ dφ = = −√ πa3 22 πa ∞ π m(eEext )2 4 −2r/a =− 2π r (r + 2a)e dr cos2 θ sin θ dθ 2πa2 2 0 0 2 a 5 cos3 θ π eEext a 6 = −m + 2a 4! 5! − a 2 2 3 1

ψ00 |H |ψ01

= −m

eEext a

2

27 6 a 8

2 = −m 3

2

3eEext a 2

0

2 .



195

(b) (i) This is the same as (a) [note that E01 = 0, as before, since ψ00 is spherically symmetric, so cos θ = 0] except for the r-dependence of H . So Eq. ⇒ 1 1 1 2mep 2β mep 1 √ √ f + 2f = − 2 , where β ≡ . − − 2f 2 = − r a r r 4π90 2 πa3 r2 4π90 2 πa3 The solution this time it obvious: f (r) = β (constant). [For the general solution we would add the general solution to the homogeneous equation (right side set equal to zero), but this would simply reproduce the unperturbed ground state, ψ00 , which we exclude—see p. 253.] So ψ01 = βe−r/a cos θ. (ii) The electric dipole moment of the electron is

pe = −er cos θ = −eψ00 +ψ01 |r cos θ|ψ00 +ψ01 = −e ψ00 |r cos θ|ψ00 + 2ψ00 |r cos θ|ψ01 + ψ01 |r cos θ|ψ01 . But the first term is zero, and the third is higher order, so 1 pe = −2e √ β e−r/a (r cos θ)e−r/a cos θ r2 sin θ dr dθ dφ πa3 ∞ π mep a 4 2 me2 p 3 −2r/a 2 = −2e 2π 3! r e dr cos θ sin θ dθ = − 2 3 2 3 4π90 πa 90 πa 2 3 0 0 4 me2 p 3a 2 me2 pa =− = −p. =− 90 2 πa3 8 3 4π90 2 Evidently the dipole moment associated with the perturbation of the electron cloud cancels the dipole moment of the nucleus, and the total dipole moment of the atom is zero. (iii) The first-order correction is zero (as noted in (i)). The second-order correction is E02

mep cos θ −r/a √ e e−r/a cos θ r2 sin θ dr dθ dφ = =√ r2 πa3 4π90 2 πa3 ∞ π a 2 (ep)2 (ep)2 −2r/a 2 = −m 2π e dr cos θ sin θ dθ = −2m (4π90 )2 2 πa3 (4π90 )2 2 a3 2 3 0 0 4 2 2 p 4 4 p me = = E1 . − 3 2(4π90 )2 2 e2 a2 3 ea ψ00 |H |ψ01

1

ep − 4π90


196

CHAPTER 7. THE VARIATIONAL PRINCIPLE

Chapter 7

The Variational Principle Problem 7.1 (a)

∞

V = 2αA

2

−2bx2

xe

2

dx = 2αA

0

2 b α H = +√ . 2m 2πb

Hmin

2 = 2m

∞ 1 −2bx2 αA2 α 2b α − e = 2b = 2b π = √2bπ . 4b 0

2 1 α α m ∂H = − √ b−3/2 = 0 =⇒ b3/2 = √ ; b= ∂b 2m 2 2π 2π 2

mα √ 2π2

2/3

α +√ 2π

'√

2π2 mα

(1/3

α2/3 2/3 = 1/3 m (2π)1/3

mα √ 2π2

2/3 .

1/3 1 3 α 2 2 . +1 = 2 2 2πm

(b)

∞

V = 2αA

2

4 −2bx2

x e 0

2 b 3α H = + . 2m 16b2

Hmin

2 = 2m

Problem 7.2

2

Normalize: 1 = 2|A|

0

∞

3αm 42

3 dx = 2αA 8(2b)2

2

π 3α = 2b 16b2

π 2b

2b 3α . = π 16b2

2 3α ∂H 3αm = − = 0 =⇒ b3 = ; b= ∂b 2m 8b3 42 1/3

3α + 16

42 3αm

2/3

α1/3 4/3 1/3 −1/3 = 3 4 m2/3

π 1 π dx = 2|A|2 3 = 3 |A|2 . A = (x2 + b2 )2 4b 2b

3αm 42

1/3

1 1 + 2 4

.

3 = 4

3α4 4m2

1/3 .

2b3 . π



197

∞ 1 1 2 d2 2 Kinetic Energy: T = − |A| dx. 2 + b2 ) dx2 2m (x2 + b2 ) −∞ (x d2 1 −2 4x 2(3x2 − b2 ) d −2x But = 2 + 2x 2 = , so = 2 2 2 2 2 2 2 2 2 3 dx (x + b ) dx (x + b ) (x + b ) (x + b ) (x2 + b2 )3

∞ ∞ 2 2b3 ∞ (3x2 − b2 ) 1 1 42 b3 2 T = − dx = − dx − 4b dx 3 2m π 0 (x2 + b2 )4 πm (x2 + b2 )3 (x2 + b2 )4 0 0

42 b3 2 3π 2 5π =− = 3 − 4b . πm 16b5 32b7 4mb2 ∞ 3 1 x2 1 2 2 2 2b π Potential Energy: V = mω |A| 2 dx = mω = mω 2 b2 . 2 + b2 )2 2 (x π 4b 2 0 H =

2 1 + mω 2 b2 . 4mb2 2

Hmin

2 = 4m

√

Problem 7.3

∂H 2 1 2 2 4 + mω b = 0 =⇒ b = =⇒ b2 = √ =− . ∂b 2mb3 2m2 ω 2 2 mω

2mω 1 1 + mω 2 √ = ω 2 2 mω

'√

2 1 + √ 4 2 2

(

√ =

2 1 ω = 0.707 ω > ω. 2 2

  A(x + a/2), (−a/2 < x < 0), ψ(x) = A(a/2 − x), (0 < x < a/2),   0, (otherwise).

2 3 a/2 12 2 2 a 3 a3 2 21 a 1 = |A| 2 − x dx = −2|A| −x |A| ; A = = |A| = (as before). 2 3 2 3 3 12 a3 0 0  A, (−a/2 < x < 0), dψ  a d2 ψ a = Aδ x + = −A, (0 < x < a/2), − 2Aδ(x) + Aδ x − .  dx dx2 2 2  0, (otherwise).

a/2

2

a

2 a a 2 2 2 a ψ Aδ x + − 2Aδ(x) + Aδ x − dx = 2Aψ(0) = A 2m 2 2 2m m 2 2 2 a 12 =6 (as before). = 2m a3 ma2 a 2 α α 2 V = −α |ψ|2 δ(x) dx = −α|ψ(0)|2 = −αA2 = −3 . H = T + V = 6 −3 . 2 a ma2 a T = −

Hmin

∂ α 2 2 +3 2 =0⇒a=4 H = −12 . 3 ∂a ma a mα mα mα2 3 3 2 mα 2 3mα2 = =6 − 3α − − = m 42 42 2 8 4 82

> −

mα2 . 22


198


Problem 7.4 (a) Follow the proof in §7.1: ψ = ∞

∞

cn ψn , where ψ1 is the ground state. Since ψ1 |ψ = 0, we have:

n=1

cn ψ1 |ψ = c1 = 0; the coefficient of the ground state is zero. So

n=1 ∞

H =

En |cn |2 ≥ Efe

n=2

(b)

1 = |A|

2

∞

|cn |2 = Efe , since En ≥ Efe for all n except 1.

n=2

∞

2 −2bx2

x e −∞

T = −

2 |A|2 2m

2

∞

−∞

1 dx = |A| 2 8b

xe−bx

2

2

π =⇒ |A|2 = 4b 2b

2b . π

d2 −bx2 xe dx dx2

2 2 2 2 d −bx2 − 2bx2 e−bx = −2bxe−bx − 4bxe−bx + 4b2 x3 e−bx e dx ∞

−2bx2 2b π π 2 22 b 2b 1 2 2 4 2 3 −6bx + 4b x e T = − dx = − 4b 2 2 −6b + 4b 2 2m π m π 8b 2b 32b 2b 0 42 b 3 3 32 b =− − + = . m 4 8 2m ∞ 2b π 1 3 3mω 2 1 2 2 2 −2bx2 2 2 V = mω |A| 2 = . x e x dx = mω 4b 2 2 2 π 32b 2b 8b −∞ d2 xe−bx dx2

H = Hmin

=

32 b 3mω + ; 2m 8b

∂H 32 3mω 2 m2 ω 2 mω 2 = − . = 0 =⇒ b = =⇒ b = 2 2 ∂b 2m 8b 4 2 3 32 mω 3mω 2 2 3 3 = + = ω + = ω. 2m 2 8 mω 4 4 2

This is exact, since the trial wave function is in the form of the true first excited state.

Problem 7.5 0 (a) Use the unperturbed ground state (ψgs ) as the trial wave function. The variational principle says 0 0 0 0 0 0 0 0 0 0 0 ψgs |H|ψgs ≥ Egs . But H = H 0 + H , so ψgs |H|ψgs = ψgs |H 0 |ψgs + ψgs |H |ψgs . But ψgs |H 0 |ψgs = 0 0 0 Egs (the unperturbed ground state energy), and ψgs |H |ψgs is precisely the first order correction to the 0 1 ground state energy (Eq. 6.9), so Egs + Egs ≥ Egs . QED 2 2 (b) The second order correction (Egs ) is Egs =

|ψ 0 |H |ψgs |2 m . But the numerator is clearly positive, 0 − E0 Egs m

m=gs

0 0 2 and the denominator is always negative (since Egs < Em for all m), so Egs is negative.



199

Problem 7.6 He+ is a hydrogenic ion (see Problem 4.16); its ground state energy is (2)2 (−13.6 eV), or −54.4 eV. It takes 79.0 − 54.4 = 24.6 eV to remove one electron.

Problem 7.7 I’ll do the general case of a nucleus with Z0 protons. Ignoring electron-electron repulsion altogether gives ψ0 =

Z03 −Z0 (r1 +r2 )/a e , πa3

(generalizing Eq. 7.17)

and the energy is 2Z02 E1 . Vee goes like 1/a (Eqs. 7.20 and 7.25), so the generalization of Eq. 7.25 is Vee = − 54 Z0 E1 , and the generalization of Eq. 7.26 is H = (2Z02 − 54 Z0 )E1 . If we include shielding, the only change is that (Z − 2) in Eqs. 7.28, 7.29, and 7.32 is replaced by (Z − Z0 ). Thus Eq. 7.32 generalizes to

5 5 2 2 H = 2Z − 4Z(Z − Z0 ) − Z E1 = −2Z + 4ZZ0 − Z E1 . 4 4

∂H 5 5 = −4Z + 4Z0 − E1 = 0 =⇒ Z = Z0 − . ∂Z 4 16

Hmin

2 5 5 5 5 = −2 Z0 − + 4 Z0 − Z0 − Z0 − E1 16 16 4 16 5 25 5 5 25 2 2 = −2Z0 + Z0 − + 4Z0 − Z0 − Z0 + E1 4 128 4 4 64 5 25 (16Z0 − 5)2 E1 = E1 , = 2Z02 − Z0 + 4 128 128

generalizing Eq. 7.34. The first term is the naive estimate ignoring electron-electron repulsion altogether; the second term is Vee in the unscreened state, and the third term is the effect of screening. 5 11 = = 0.688. The effective nuclear charge is less than 1, as expected. 16 16 112 121 = E1 = E1 = −12.9 eV. 128 128

Z0 = 1 (H− ): Z = 1 − Hmin

27 5 272 729 = = 1.69 (as before); Hmin = E1 = E1 = −77.5 eV. 16 16 128 128 43 1849 5 432 Z0 = 3 (Li+ ): Z = 3 − = = 2.69 (somewhat less than 3); Hmin = E1 = E1 = −196 eV. 16 16 128 128

Z0 = 2 (He): Z = 2 −

Problem 7.8 1 1 1 1 e−2r2 /a d3 r D = aψ0 (r1 ) ψ0 (r1 ) = aψ0 (r2 ) ψ0 (r2 ) = a 3 r2 r1 πa r1 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

200

CHAPTER 7. THE VARIATIONAL PRINCIPLE 1 = πa3

2 −a

√

e

1 [. . . ] = rR

r 2 +R2 −2rR cos θ

r+R

e−2y/a y dy = −

|r−R|

1 2 2π r sin θ dr dθ dφ = r πa3

∞

π

r

2 −a

e

0

√

r 2 +R2 −2rR cos θ

sin θ dθ dr.

0

a a a −2(r+R)/a r+R+ e − e−2|r−R|/a |r − R| + 2rR 2 2

2 a −2R/a ∞ −2r/a a D= 2 − r+R+ e e dr a 2R 2 0 −2R/a

R

−e

2R/a

e 0

=−

1 aR

e−2R/a a 2

+e−2R/a

2

a 2 2

a R−r+ dr − e2R/a 2

∞

−2r/a

e R

a r−R+ dr 2

R ∞ 2 a a −2r/a ∞ 2r 2r 2R/a a −2r/a − − e e − 1 − e2R/a −R + − e − 1 a 2 2 2 a R 0 R

a2 aR a2 aR a2 a2 aR a2 a2 2R a2 aR a2 a2 2R a2 + + + + + + − − + − + − − − 4 2 4 2 4 4 2 4 4 a 4 2 4 4 a 4

=−

1 aR

=−

1 −2R/a 2 a a −2R/a a + aR + −a2 =⇒ D = e − 1+ e aR R R

e−2R/a

a a 2r/a R a a + R+ − e−2R/a R + e 2 2 2 2 0

e2r/a

(confirms Eq. 7.47).

1 1 1 X = aψ0 (r1 ) ψ0 (r2 ) = a 3 e−r1 /a e−r2 /a d3 r r1 πa r1

=

1 πa2

√

e−r/a e−

[. . . ] = −

X=

r 2 +R2 −2rR cos θ/a

2π 1 2 r sin θ dr dθ dφ = r πa2

∞

re−r/a

0

π

√

e−

r 2 +R2 −2rR cos θ/a

sin θ dθ dr.

0

a −(r+R)/a (r + R + a) − e−|r−R|/a (|r − R| + a) e rR

2 a −R/a ∞ −2r/a − e e (r + R + a)dr a2 R 0 −R/a

R

−e

(R − r + a)dr − e

R/a

−R/a

e

a 2 2

−2r/a

e

0

2 =− aR

∞

(r − R + a)dr

R

+ (R + a)

a 2

−R/a

−e

R2 (R + a)R − 2


CHAPTER 7. THE VARIATIONAL PRINCIPLE −e

R/a

=−

201

∞ ∞ 2 a 2r −2r/a R/a a −2r/a − − 1 (−R + a) − e e −e 2 2 a R R

2 2 a R2 aR a2 aR a2 a2 2R a2 e−R/a + + − R2 − aR + + − − − aR 4 2 2 2 2 2 4 a 4

2 −R/a aR R2 R −R/a =− e − =⇒ X = e − 1+ aR 2 2 a

(confirms Eq. 7.48).

Problem 7.9 There are two changes: (1) the 2 in Eq. 7.38 changes sign . . . which amounts to changing the sign of I in Eq. 7.43; (2) the last term in Eq. 7.44 changes sign . . . which amounts to reversing the sign of X. Thus Eq. 7.49 becomes

D−X H = 1 + 2 E1 , and hence Eq. 7.51 becomes 1−I Etot D−X 2 1/x − (1 + 1/x) e−2x − (1 + x)e−x 2a −1−2 = −1 + − 2 = −E1 R 1−I x 1 − (1 + x + x2 /3)e−x

2 1 − (1 + x + x2 /3)e−x − 1 + (x + 1)e−2x + (x + x2 )e−x = −1 + x 1 − (1 + x + x2 /3)e−x

2 (1 + x)e−2x + 23 x2 − 1 e−x = −1 + . x 1 − (1 + x + x2 /3)e−x

F (x) =

The graph (with plus sign for comparison) has no minimum, and remains above −1, indicating that the energy is greater than for the proton and atom dissociated. Hence, no evidence of bonding here.

F(x) -0.5 -0.6

(+)

(−)

-0.7 -0.8 -0.9 2

4

6

8

x

-1.1


202


Problem 7.10 According to Mathematica, the minimum occurs at x = 2.493, and at this point F = 0.1257. E1 1 −(0.1257)E1 2 mω = V = − 2 F , so ω = . a a m mp m p 1 = mp . mp + m p 2 3 × 108 m/s (0.1257)(13.6 eV) ω= = 3.42 × 1014 /s. (0.529 × 10−10 m) (938 × 106 eV)/2

Here m is the reduced mass of the proton: m =

1 1 ω = (6.58 × 10−16 eV · s)(3.42 × 1014 /s) = 0.113 eV (ground state vibrational energy). 2 2 Mathematica says that at the minimum F = −1.1297, so the binding energy is (0.1297)(13.6 eV) = 1.76 eV. Since this is substantially greater than the vibrational energy, it stays bound. The highest vibrational energy is 1 1.76 given by (n + 12 )ω = 1.76 eV, so n = − = 7.29. I estimate eight bound vibrational states (including 0.226 2 n = 0).

Problem 7.11 (a) 1= T = V = = H =

|ψ| dx = |A| 2

2

a/2

cos

2

πx

dx = |A|

⇒

A=

2 . a

a 2 2 d ψ π π 2 2 − ψ 2 dx = ψ 2 dx = . 2m dx 2m a 2ma2 a/2 1 1 mω 2 a 3 π/2 2 2 2 2 22 2 2 πx x ψ dx = mω x cos y cos2 y dy mω dx = 2 2 a −a/2 a a π −π/2 2

π/2 mω 2 a2 y 3 y 1 y cos 2y mω 2 a2 π 2 + − sin 2y + −1 . = π3 6 4 8 4 4π 2 6 −π/2 π 2 2 mω 2 a2 π 2 mω 2 a π 2 ∂H π 2 2 + + − 1 ; = − − 1 =0 ⇒ 2ma2 4π 2 6 ∂a ma3 2π 2 6 2

2

−a/2 2

a=π

Hmin

2a

mω

2 π 2 /6 − 1

1/4 .

2 π 2 2 mω π 2 /6 − 1 mω 2 π 2 2 = + −1 π 2mπ 2 2 4π 2 6 mω π 2 /6 − 1 1 π2 1 1 = ω − 2 = ω(1.136) > ω. 2 3 2 2

[We do not need to worry about the kink at ±a/2. It is true that d2 ψ/dx2 has delta functions there, but since ψ(±a/2) = 0 no “extra” contribution to T comes from these points.] c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


203

(b) Because this trial function is odd, it is orthogonal to the ground state, so by Problem 7.4 H will give an upper bound to the first excited state. a πx 1 1= |ψ|2 dx = |B|2 sin2 dx = |B|2 a ⇒ B = √ . a a −a 2 2 2 2 2 2 d ψ π π ψ 2 dx = T = − ψ 2 dx = . 2m dx 2m a 2ma2 a πx 1 1 1 mω 2 a 3 π 2 2 V = mω 2 x2 ψ 2 dx = mω 2 x2 sin2 y sin y dy dx = 2 2 a −a a 2a π −π 2

π mω 2 a2 y 3 y 1 y cos 2y mω 2 a2 2π 2 = − − sin 2y − −1 . = 4π 2 2π 3 6 4 8 4 3 −π π 2 2 mω 2 a2 2π 2 mω 2 a 2π 2 ∂H π 2 2 H = + + −1 ; =− −1 =0 ⇒ 2ma2 4π 2 3 ∂a ma3 2π 2 3 a=π

Hmin

mω

2 2π 2 /3 − 1

1/4 .

2 π 2 2 mω 2π 2 /3 − 1 mω 2 2π 2 2 + −1 π = 2mπ 2 2 4π 2 3 mω 2π 2 /3 − 1 1 4π 2 1 3 = ω − 2 = ω(3.341) > ω. 2 3 2 2

Problem 7.12 We will need the following integral repeatedly:

2l−k−1

∞ Γ k+1 Γ xk 1 2 2 dx = 2l−k−1 . l 2 2 2b Γ(l) (x + b ) 0 (a) 1 4n−1

b4n−1 Γ(2n) 2 Γ 2

. |ψ| dx = 2|A| ⇒A= 2n Γ(2n) Γ 12 Γ 4n−1 (x2 + b2 ) −∞ 0 2 ∞ 2 1 d d2 ψ 2 2 ∞ −2nx − dx ψ dx = − A 2 2 n 2 2 n+1 2m −∞ dx2 2m −∞ (x + b ) dx (x + b ) n2 2 ∞ 1 1 2(n + 1)x2 dx A n+1 − n+2 2 2 n m (x2 + b2 ) (x2 + b2 ) −∞ (x + b ) ∞ ∞ 2n2 2 1 x2 A 2n+1 dx − 2(n + 1) 2n+2 dx m (x2 + b2 ) (x2 + b2 ) 0 0

2n2 b4n−1 Γ(2n) 2(n + 1) Γ 32 Γ 4n+1 2 n(4n − 1) 1 Γ 12 Γ 4n−1 2 2 1 4n−1

− = . m Γ 2 Γ 2b4n−1 Γ(2n + 1) 2b4n−1 Γ(2n + 2) 4mb2 (2n + 1) 2

1= T = = = =

∞

2

2

∞

1

|A|2 Γ dx = 4n−1 b


204

CHAPTER 7. THE VARIATIONAL PRINCIPLE ∞ x2 1 2 2 ψ x dx = mω 2A 2n dx 2 (x2 + b2 ) −∞ 0 3 4n−3

b4n−1 Γ(2n) mω 2 b2 1 Γ 2 Γ 2 = mω 2 1 4n−1 4n−3 = . 2b Γ(2n) 2(4n − 3) Γ 2 Γ 2

1 V = mω 2 2

∞

2 2

2 n(4n − 1) mω 2 b2 ∂H 2 n(4n − 1) mω 2 b + ; = − + =0 ⇒ 4mb2 (2n + 1) (4n − 3) ∂b 2mb3 (2n + 1) (4n − 3)

1/4 n(4n − 1)(4n − 3) b = . mω 2(2n + 1) mω 2 2(2n + 1) n(4n − 1)(4n − 3) 2 n(4n − 1) mω Hmin = + 4m (2n + 1) n(4n − 1)(4n − 3) 2(4n − 3) mω 2(2n + 1) 1 2n(4n − 1) 8n2 − 2n 1 1 = ω = ω > ω. 2 2 (2n + 1)(4n − 3) 2 8n − 2n − 3 2 H =

(b)

b4n−3 Γ(2n) |B|2 Γ 32 Γ 4n−3 2

. 2|B| ⇒ B = dx = 2n b4n−3 Γ(2n) Γ 32 Γ 4n−3 (x2 + b2 ) 0 2 2 2 ∞ x d 2nx2 1 − dx − B n+1 2 2 n 2 2 n 2m (x2 + b2 ) −∞ (x + b ) dx (x + b ) 2 B 2 ∞ 4nx 4n(n + 1)x3 −2nx x − − dx n+1 + n+2 2m −∞ (x2 + b2 )n (x2 + b2 )n+1 (x2 + b2 ) (x2 + b2 ) ∞ ∞ 4n2 B 2 x2 x4 3 2n+1 dx − 2(n + 1) 2n+2 dx 2m (x2 + b2 ) (x2 + b2 ) 0 0

5 4n−1 Γ 2n2 b4n−3 Γ(2n) 3 Γ 32 Γ 4n−1 2(n + 1) 32 n(4n − 3) 2 2 Γ 2 3 4n−3

− = . m Γ 2 Γ 2b4n−1 Γ(2n + 1) 2b4n−1 Γ(2n + 2) 4mb2 (2n + 1) 2

1= T = = = =

2

∞

1 V = mω 2 2B 2 2 H = b = Hmin =

=

x2

∞

x4 2n

b4n−3 Γ(2n) 1 2 Γ dx = mω 2 3 4n−3 4n−5 2 2b Γ 2 Γ 2

5 4n−5

3 mω 2 b2 2 Γ 2 = . Γ(2n) 2 (4n − 5)

(x2 + b2 ) 2 3 n(4n − 3) 3 mω 2 b2 ∂H 32 n(4n − 3) 3mω 2 b + ; = − + =0 ⇒ 4mb2 (2n + 1) 2 (4n − 5) ∂b 2mb3 (2n + 1) (4n − 5)

1/4 n(4n − 3)(4n − 5) . mω 2(2n + 1) 3 mω 2 32 n(4n − 3) mω 2(2n + 1) n(4n − 3)(4n − 5) + 4m (2n + 1) n(4n − 3)(4n − 5) 2 (4n − 5) mω 2(2n + 1) 3 2n(4n − 3) 8n2 − 6n 3 3 = ω ω > ω. 2 2 (2n + 1)(4n − 5) 2 8n − 6n − 5 2 0



205

(c) As n → ∞, ψ becomes more and more “gaussian”. In the figures I have plotted the trial wave functions for n = 2, n = 3, and n = 4, as well as the exact states (heavy line). Even for n = 2 the fit is pretty good, so it is hard to see the improvement, but the successive curves do move perceptably toward the correct result.

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 1

2

Analytically, for large n, b ≈

x2 + b2

n

mω

1

4

3

n · 4n · 4n 2 · 2n

1/4 =

2

4

3

2n , so mω

n n x2 mωx2 = b2n 1 + 2 ≈ b2n 1 + b 2n

→

2

b2n emωx

/2

.

Meanwhile, using Stirling’s approximation (Eq. 5.84), in the form Γ(z + 1) ≈ z z e−z : 2n−1 b4n−1 Γ(2n) b4n−1 1 b4n−1 (2n − 1)2n−1 e−(2n−1) 2n − 1

√ A = 1 ≈ √ √ 2n − 3/2.

3 1 ≈ 2n−3/2 3 π 2n − π e 2n − 2 Γ 2 Γ 2n − 2 e−(2n−3/2) 2 1 1 − 2n 1 3 3 1 1 But ≈ 1 − 1 + ≈1+ − =1+ ; 3 2n 4n 4n 2n 4n 1 − 4n n 2 2n−1 2 √ 2n − 1 1 1 1/4 so → e ≈ 1 + = e. 4n 1 + 1/4n 2n − 32 1/4 b4n−1 √ √ 2n 4n−1 2n = √ e 2n = ⇒ A≈ b2n−1/2 . So b π π πe 1/4 1/4 mω 1/4 −mωx2 /2 mω 1/4 −mωx2 /2 2n 2n 2n−1/2 1 −mωx2 /2 ψ ≈ b e = e = e , π b2n π 2n π 2

which is precisely the ground state of the harmonic oscillator (Eq. 2.59). So it’s no accident that we get the exact energies, in the limit n → ∞.

Problem 7.13 1 = |A|

2

−2br 2 2

e

2

r sin θ dr dθ dφ = 4π|A|

∞

2 −2br 2

r e 0

dr = |A|

2

π 3/2 2b

⇒ A=

2b π

3/4 .


206

CHAPTER 7. THE VARIATIONAL PRINCIPLE e2 V = − |A|2 4π 4π90 2 T = − |A|2 2m

∞

−2br 2

e 0

1 2 e2 r dr = − r 4π90

2b π

3/2

1 e2 4π = − 2 4b 4π90

2b . π

e−br (∇2 e−br ) r2 sin θ dr dθ dφ −2b

2 2 2 2 1 d d 1 d But (∇2 e−br ) = 2 r2 e−br = 2 −2br3 e−br = 2 3r2 − 2br4 e−br . r dr dr r dr r 3/2 3/2 ∞ 2 2 1 − π π 2b 2b 3 2 4 −2br 2 3 = πb4 − 2b (4π)(−2b) (3r − 2br )e dr = 2 2m π m π 8b 2b 32b 2b 0 2 2b 3 3 32 b = 4πb − = . m π 8b 16b 2m 2

32 b e2 H = − 2 2m 4π90 Hmin

2b ; π

2

∂H 32 e2 = − ∂b 2m 4π90

2 1 √ =0 π b

⇒

√

e2 b= 4π90

2 2m . π 32

2 2 2 4 2 4m2 2 2 2m m e2 e2 8 e2 e − − 2 = 4π90 π 94 4π90 π 4π90 π 32 4π90 2 3π 3π 2 2 m e 8 8 =− 2 = E1 = −11.5 eV. 2 4π90 3π 3π

32 = 2m

Problem 7.14 Let ψ = √

1 πb3

e−r/b (same as hydrogen, but with a → b adjustable). From Eq. 4.191, we have T = −E1 =

2 2 for hydrogen, so in this case T = . 2 2ma 2mb2 ∞ −µr e2 4π 1 1 e2 4 ∞ −(µ+2/b)r e2 4 e2 −2r/b e 2 V = − e dr = − e r dr = − = − . r 3 3 3 2 2 4π90 πb 0 r 4π90 b 0 4π90 b (µ + 2/b) 4π90 b(1 + µb 2 ) H =

2 1 e2 − . 2 2 2mb 4π90 b(1 + µb 2 )

∂H 2 1 2 e2 µ e2 (1 + 3µb/2) =− 3 + = − + + =0 ∂b mb 4π90 b2 (1 + µb/2)2 b(1 + µb/2)3 mb3 4π90 b2 (1 + µb/2)3 2 m

4π90 e2

=b

(1 + 3µb/2) , (1 + µb/2)3

or

b

⇒

(1 + 3µb/2) = a. (1 + µb/2)3

This determines b, but unfortunately it’s a cubic equation. So we use the fact that µ is small to obtain a suitable approximate solution. If µ = 0, then b = a (of course), so µa 1 =⇒ µb 1 too. We’ll expand in powers of µb: 2

3µb 9 3µb µb 3 6 2 2 2 a≈b 1+ ≈ b 1 − (µb) + (µb) = b 1 − (µb) . 1− +6 2 2 2 4 4 4 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 7. THE VARIATIONAL PRINCIPLE Since the 34 (µb)2 term is already a b≈ ≈ a 1+ 1 − 34 (µb)2

207

a second-order correction, we can replace b by a:

3 (µa)2 . 4

1 2 e2 − 2 2 3 3 2 2 2 4π9 0 a 1 + (µa) 2ma 1 + 4 (µa) 1 + 12 (µa) 4

µa 2 3 e2 1 µa 2 3 2 2 1 − 2 (µa) − 1−2 ≈ 1 − (µa) +3 2ma2 4 4π90 a 4 2 2

3 3 3 3 = −E1 1 − (µa)2 + 2E1 1 − µa + (µa)2 − (µa)2 = E1 1 − 2(µa) + (µa)2 . 2 4 4 2

Hmin =

Problem 7.15 (a) H= λ=

1 2

Ea h h Eb

Ea + Eb ±

; det(H − λ) = (Ea − λ)(Eb − λ) − h2 = 0 =⇒ λ2 − λ(Ea + Eb ) + Ea Eb − h2 = 0.

Ea2 + 2Ea Eb + Eb2 − 4Ea Eb + 4h2

⇒ E± =

1 2

Ea + Eb ±

(Ea − Eb )2 + 4h2 .

(b) Zeroth order: Ea0 = Ea , Eb0 = Eb . First order: Ea1 = ψa |H |ψa = 0, Eb1 = ψb |H |ψb = 0. Second order: |ψb |H |ψa |2 h2 |ψa |H |ψb |2 h2 Ea2 = =− ; Eb2 = = ; Ea − Eb Eb − Ea Eb − Ea Eb − Ea E − ≈ Ea −

h2 ; (Eb − Ea )

E+ ≈ Eb +

h2 . (Eb − Ea )

(c) H = cos φ ψa + sin φ ψb |(H 0 + H )| cos φ ψa + sin φ ψb = cos2 φ ψa |H 0 |ψa + sin2 φ ψb |H 0 |ψb + sin φ cos φ ψb |H |ψa + sin φ cos φ ψa |H |ψb = Ea cos2 φ + Eb sin2 φ + 2h sin φ cos φ. ∂H = −Ea 2 cos φ sin φ + Eb 2 sin φ cos φ + 2h(cos2 φ − sin2 φ) = (Eb − Ea ) sin 2φ + 2h cos 2φ = 0. ∂φ

tan 2φ = −

or

2h = −9 where Eb − Ea

sin2 2φ(1 + 92 ) = 92 ;

cos 2φ = √

∓1 1 + 92

9≡

sin 2φ = √

2h . Eb − Ea

±9 ; 1 + 92

(sign dictated by tan 2φ =

sin 2φ

1 − sin2 2φ

= −9;

sin2 2φ = 92 (1 − sin2 2φ);

cos2 2φ = 1 − sin2 2φ = 1 −

92 1 = ; 1 + 92 1 + 92

sin 2φ = −9). cos 2φ


208

CHAPTER 7. THE VARIATIONAL PRINCIPLE 1 1 1 1 1 1 2 √ √ ; sin φ = (1 − cos 2φ) = . cos φ = (1 + cos 2φ) = 1∓ 1± 2 2 2 2 1 + 92 1 + 92

1 1 1 9 1 1 (Eb − Ea + 2h9) √ Hmin = Ea 1 ∓ √ + Eb 1 ± √ ± h√ = E a + Eb ± 2 2 2 1 + 92 1 + 92 1 + 92 1 + 92 2

(Eb − Ea ) + 2h (Eb2h (Eb − Ea + 2h9) (E − Ea )2 + 4h2 −Ea ) √ b = = = (Eb − Ea )2 + 4h2 , 2 2 2 2 1+9 (Eb − Ea ) + 4h 1 + (Eb4h −Ea )2

But

So

1 Ea + Eb ± (Eb − Ea )2 + 4h2 we want the minus sign (+ is maximum) 2 = 12 Ea + Eb − (Eb − Ea )2 + 4h2 .

Hmin =

(d) If h is small, the exact result (a) can be expanded: E± = 1 =⇒ E± ≈ 2

so

Ea + Eb ± (Eb − Ea ) 1 +

E+ ≈ E b +

h2 , (Eb − Ea )

2h2 (Eb − Ea )2

E− ≈ E a −

1 2

(Ea + Eb ) ± (Eb − Ea ) 1 +

4h2 (Eb −Ea )2

.

1 2h2 = Ea + Eb ± (Eb − Ea ) ± , 2 (Eb − Ea )

h2 , (Eb − Ea )

confirming the perturbation theory results in (b). The variational principle (c) gets the ground state (E− ) exactly right—not too surprising since the trial wave function Eq. 7.56 is almost the most general state (there could be a relative phase factor eiθ ).

Problem 7.16 For the electron, γ= −e/m, so E± = ±eBz /2m (Eq. 4.161). For consistency with Problem 7.15, Eb > Ea , eBz eBz 1 0 so χb = χ+ = , Ea = E − = − . , χa = χ− = , Eb = E + = 0 1 2m 2m (a)

eBx 01 χa |H |χa = m 2

eBx 10 χb |H |χb = 2m

eBx 01 χa |H |χb = 2m

01 10

1 eBx 0 0 1 = = 0; 1 0 2m

01 10

1 = 0; 0

01 10

0 eBx eBx 1 0 1 = = . So 0 1 2m 2m

eBx 1 0 χb |H |χa = 2m

0 1 1 0

eBx 0 = ; 1 2m

h=

eBx , 2m

and the conditions of Problem 7.15 are met. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


209

(b) From Problem 7.15(b), Egs

h2 e eBz (eBx /2m)2 ≈ Ea − =− − = − (Eb − Ea ) 2m (eBz /m) 2m

(c) From Problem 7.15(c), Egs =

Egs

1 =− 2

eBz m

1 2

Ea + Eb −

2 +4

eBx 2m

Bx2 . Bz + 2Bz

(Eb − Ea )2 + 4h2 (it’s actually the exact ground state).

2 = −

e 2 Bz + Bx2 2m

(which was obvious from the start, since the square root is simply the magnitude of the total field).

Problem 7.17 (a) 1 1 1 r1 = √ (u + v); r2 = √ (u − v); r12 + r22 = (u2 + 2u · v + v 2 + u2 − 2u · v + v 2 ) = u2 + v 2 . 2 2 2 (∇21 + ∇22 )f (r1 , r2 ) =

∂2f ∂2f ∂2f ∂2f ∂2f ∂2f + 2 + 2 + + 2 + 2 2 2 ∂x1 ∂y1 ∂z1 ∂x2 ∂y2 ∂z2

∂f ∂f ∂ux ∂f ∂vx 1 = + =√ ∂x1 ∂ux ∂x1 ∂vx ∂x1 2 ∂2f 1 ∂ =√ 2 ∂x1 2 ∂x1 =

1 2

So

∂f ∂f + ∂ux ∂vx

∂f ∂f − ∂ux ∂vx

=

∂2f ∂2f + 2 2 ∂ux ∂vx

;

.

∂f ∂f ∂ux ∂f ∂vx 1 = + =√ ∂x2 ∂ux ∂x2 ∂vx ∂x2 2

∂f ∂f − ∂ux ∂vx

∂ 2 f ∂vx ∂ 2 f ∂ux ∂ 2 f ∂vx ∂ 2 f ∂ux + + + ∂u2x ∂x1 ∂ux ∂vx ∂x1 ∂vx ∂ux ∂x1 ∂vx2 ∂x1

;

1 =√ 2

∂2f ∂2f ∂2f − 2 + ∂u2x ∂ux ∂vx ∂vx2

∂2f ∂2f + 2 ∂x1 ∂x22

∂f ∂f + ∂ux ∂vx

1 =√ 2

∂2f ∂2f ∂2f +2 + 2 2 ∂ux ∂ux ∂vx ∂vx

∂2f 1 ∂ =√ ∂x22 2 ∂x2 1 = 2

∂ 2 f ∂vx ∂ 2 f ∂ux ∂ 2 f ∂vx ∂ 2 f ∂ux + − − ∂u2x ∂x2 ∂ux ∂vx ∂x2 ∂vx ∂ux ∂x2 ∂vx2 ∂x2

. , and likewise for y and z: ∇21 + ∇22 = ∇2u + ∇2v .

2 1 λ H =− (∇2u + ∇2v ) + mω 2 (u2 + v 2 ) − mω 2 2v 2 2 4

2m2 2 2 1 1 1 2 2 2 ∇u + mω u + − ∇v + mω 2 v 2 − λmω 2 v 2 . QED = − 2m 2 2m 2 2 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

.

210


√ √

(b) The energy is 32 ω (for the u part) and 32 ω 1 − λ (for the v part): Egs = 32 ω 1 + 1 − λ . (c) The ground state for a one-dimensional oscillator is ψ0 (x) =

mω 1/4 π

e−mωx

2

/2

(Eq. 2.59).

So, for a 3-D oscillator, the ground state is ψ0 (r) = ψ(r1 , r2 ) =

H =

mω 3/2 π

mω 3/4 π

e−mωr

2

/2

, and for two particles

e− 2 (r1 +r2 ) . (This is the analog to Eq. 7.17.) mω

2

2

3 3 ω + ω + Vee = 3ω + Vee (the analog to Eq. 7.19). 2 2

mω 3 λ Vee = − mω 2 4 π

e−

2 2 mω (r1 +r2 )

(r1 − r2 )2 d3 r1 d3 r2

(the analog to Eq. 7.20).

r12 −2r1 ·r2 +r22

The r1 · r2 term integrates to zero, by symmetry, and the r22 term is the same as the r12 term, so mω 3 2 2 mω λ Vee = − mω 2 2 e− (r1 +r2 ) r12 d3 r1 d3 r2 4 π ∞ ∞ mω 3 2 λ 2 2 −mωr22 / 2 = − mω (4π) e r2 dr2 e−mωr1 / r14 dr1 2 π 0 0 2 4 5 π π 3 3 8m ω 1 = − λω. = −λ π3 4 mω mω 8 mω mω 4 3 λ H = 3ω − λω = 3ω 1 − . 4 4 The variational principle says this must exceed the exact ground-state energy (b); let’s check it: √ √ λ λ λ √ λ2 3 3ω 1 − > ω 1 + 1 − λ ⇔ 2 − > 1 + 1 − λ ⇔ 1 − > 1 − λ ⇔ 1 − λ + > 1 − λ. 4 2 2 2 4

It checks. In fact, expanding the exact answer in powers of λ, Egs ≈ 32 ω(1 + 1 − 12 λ) = 3ω 1 − λ4 , we recover the variational result.

Problem 7.18

1==

|ψ| d r1 d r2 = |A| 2 3

3

2

ψ12

3

d r1

ψ22

3

d r2 + 2

3

ψ1 ψ2 d r1

ψ1 ψ2 d r2 +

3

ψ22

3

d r1

ψ12

3

d r2

= |A|2 (1 + 2S 2 + 1), where

S≡

3

ψ1 (r)ψ2 (r) d r =

(Z1 Z2 )3 πa3

−(Z1 +Z2 )r/a

e

2a3 y 3 4 y 3 = 4πr dr = 3 . a 2 (Z1 + Z2 )3 x 2



A2 =

1

6

211

.

2 1 + (y/x)

2 e2 H=− (∇21 + ∇22 ) − 2m 4π90

1 1 + r1 r2

+

e2 1 , 4π90 |r1 − r2 |

2 Z1 e2 Z2 2 2 Hψ = A − ψ1 (r1 )ψ2 (r2 ) (∇ + ∇2 ) − + 2m 1 4π90 r1 r2 2 Z1 e2 Z2 + − ψ2 (r1 )ψ1 (r2 ) + (∇21 + ∇22 ) − 2m 4π90 r1 r2

+A

where Vee ≡

e2 4π90

Z1 − 1 Z2 − 1 Z2 − 1 Z1 − 1 ψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 )ψ1 (r2 ) + Vee ψ, + + r1 r2 r1 r2

1 e2 . 4π90 |r1 − r2 |

The term in first curly brackets is (Z12 + Z22 )E1 ψ1 (r1 )ψ2 (r2 ) + (Z22 + Z12 )ψ2 (r1 )ψ1 (r2 ), so Hψ = (Z12 + Z22 )E1 ψ

Z1 − 1 Z2 − 1 e2 Z2 − 1 Z1 − 1 ψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 )ψ1 (r2 ) + Vee ψ +A + + 4π90 r1 r2 r1 r2 H =

(Z12

+

Z22 )E1

+ Vee + A

2

e2 4π90

Z1 − 1 Z2 − 1 Z2 − 1 Z1 − 1 × ψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 )ψ1 (r2 ) + + . ψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 )ψ1 (r2 ) r1 r2 r1 r2

1 1 = (Z1 − 1)ψ1 (r1 ) ψ1 (r1 ) + (Z2 − 1)ψ2 (r2 ) ψ2 (r2 ) r1 r2 1 + (Z2 − 1)ψ1 (r1 ) ψ2 (r1 )ψ2 (r2 )|ψ1 (r2 ) r1 1 1 + (Z1 − 1)ψ1 (r1 )|ψ2 (r1 )ψ2 (r2 ) ψ1 (r2 ) + (Z1 − 1)ψ2 (r1 ) ψ1 (r1 )ψ1 (r2 )|ψ2 (r2 ) r2 r1 1 1 + (Z2 − 1)ψ2 (r1 )|ψ1 (r1 )ψ1 (r2 ) ψ2 (r2 ) + (Z2 − 1)ψ2 (r1 ) ψ2 (r1 ) r2 r1 1 + (Z1 − 1)ψ1 (r2 ) ψ1 (r2 ) r2 0 1 0 1 1 1 1 1 = 2(Z1 − 1) + 2(Z1 − 1) + 2(Z1 − 1)ψ1 |ψ2 ψ1 ψ2 + 2(Z2 − 1)ψ1 |ψ2 ψ1 ψ2 . r 1 r 2 r r


212

CHAPTER 7. THE VARIATIONAL PRINCIPLE 0 1 1 1 Z1 = ψ1 (r) ψ1 (r) = ; r 1 r a

But

+A

2

e2 4π90

0 1 Z2 1 = , r 2 a

so

H = (Z12 + Z22 )E1

1 1 1 2 (Z1 − 1)Z1 + (Z2 − 1)Z2 + (Z1 + Z2 − 2)ψ1 |ψ2 ψ1 ψ2 + Vee . a a r 3

ψ1 |ψ2 = S = (y/x) , so

And

2 1 (Z1 Z2 )3 a y3 y3 −(Z1 +Z2 )r/a ψ1 ψ2 = 4π e r dr = = . r πa3 2a3 Z1 + Z2 2ax2 2 y 3 y 3 e 2 2 1 H = (x2 − y 2 )E1 + A2 + Vee Z1 + Z22 − (Z1 + Z2 ) + (x − 2) 2 4π90 a x 2x2

y6 1 1 1 = (x2 − y 2 )E1 + 4E1 A2 x2 − y 2 − x + (x − 2) 5 + Vee . 2 2 2 x e2 1 ψ ψ 4π90 |r1 − r2 | 1 e ψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 )ψ1 (r2 ) A2 ψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 ) + ψ1 (r2 ) = 4π90 |r1 − r2 |

e 1 1 ψ1 (r1 )ψ2 (r2 ) + 2ψ1 (r1 )ψ2 (r2 ) ψ2 (r1 )ψ1 (r2 ) = A2 2ψ1 (r1 )ψ2 (r2 ) |r1 − r2 | 4π90 |r1 − r2 | e A2 (B + C), where =2 4π90

Vee =

B ≡ ψ1 (r1 )ψ2 (r2 )

B=

Z13 Z23 (πa3 )2

1 ψ1 (r1 )ψ2 (r2 ); |r1 − r2 |

e−2Z1 r1 /a e−2Z2 r2 /a

e−2Z2 r2 /a

+

− 2r1 r2 cos θ2

r22

d3 r2

Z2 r1 πa3 −2Z2 r1 /a 1− 1+ = 3 e Z2 r1 a Z 3 Z 3 (πa3 ) B = 1 3 22 4π (πa ) Z23 4Z 3 = 31 a

0

∞

∞

−2Z1 r1 /a

e 0

1 ψ2 (r1 )ψ1 (r2 ). |r1 − r2 |

1 d3 r1 d3 r2 . As on pp 300-301, the r2 integral is |r1 − r2 |

1 r12

C ≡ ψ1 (r1 )ψ2 (r2 )

(Eq. 7.24, but with a →

2 a). Z2

1 Z2 r1 −2Z2 r1 /a 1− 1+ r12 dr1 e r1 a

Z2 2 −2(Z1 +Z2 )r1 /a −2Z1 r1 /a −2(Z1 +Z2 )r1 /a r1 e dr1 r e − r1 e − a 1


CHAPTER 7. THE VARIATIONAL PRINCIPLE 4Z 3 = 31 a =

C=

=

a 2Z1

2

−

a 2(Z1 + Z2 )

2

Z2 2 − a

213

a 2(Z1 + Z2 )

3 =

Z13 a

1 1 Z2 − − Z12 (Z1 + Z2 )2 (Z1 + Z2 )3

y2 Z1 Z2 Z1 Z2 y2 = . 1+ 1 + a(Z1 + Z2 ) (Z1 + Z2 )2 4ax 4x2

Z13 Z23 (πa3 )2

(Z1 Z2 )3 (πa3 )2

e−Z1 r1 /a e−Z2 r2 /a e−Z2 r1 /a e−Z1 r2 /a

e−(Z1 +Z2 )(r1 +r2 )/a

1 d3 r1 d3 r2 |r1 − r2 |

1 d3 r1 d3 r2 . |r1 − r2 |

4 The integral is the same as in Eq. 7.20, only with a → Z1 +Z a. Comparing Eqs. 7.20 and 7.25, we see that the 2 integral itself was 2 5 πa3 5 2 5 (Z1 Z2 )3 5π 2 20 (Z1 Z2 )3 5 y6 45 a5 = = = . π a . So C = 4a 8 256 (πa3 )2 256 (Z1 + Z2 )5 a (Z1 + Z2 )5 16a x5

Vee = 2

e 4π90

A

2

y2 4ax

y2 1+ 2 4x

5 y6 y2 5y 4 y2 2 + = 2A (−2E1 ) 1+ 2 + 4 . 16a x5 4x 4x 4x

y6 y2 y2 2 1 2 2 1 2 1 5y 4 2 x − y − x + (x − 2) 5 − 1+ 2 + 4 x − y − 2 [1 + (y/x)6 ] 2 2 x [1 + (y/x)6 ] 4x 4x 4x

H = E1

2

E1 = 6 (x + y 6 ) E1 = 6 (x + y 6 )

y4 1 2 6 1 2 1 y6 y6 y2 5y 6 6 6 2 (x − y )(x + y ) − 2x x − y − x + + − 5+ + 2 2 2 x4 x 4x 16x3 16x5 2

E1 = (x6 + y 6 )

1 1 1 1 5 x + x y − x6 y 2 − y 8 − 2x8 + x6 y 2 + 2x7 − x2 y 6 + 2xy 6 − x5 y 2 − x3 y 4 − xy 6 2 2 2 8 8 8

2 6

1 6 2 1 5 2 1 3 4 11 6 1 8 −x + 2x + x y − x y − x y + xy − y . 2 2 8 8 2 8

7

Mathematica finds the minimum of H at x = 1.32245, y = 1.08505, corresponding to Z1 = 1.0392, Z2 = 0.2832. At this point, Hmin = 1.0266E1 = −13.962 eV, which is less than −13.6 eV—but not by much!

Problem 7.19 The calculation is the same as before, but with me → mµ (reduced), where mµ (reduced) = mµ 1+ =1+ 2mp

m µ md mµ 2mp mµ = = . From Problem 6.28, mµ = 207me , so mµ + md mµ + 2mp 1 + mµ /2mp

207 2

(9.11 × 10−31 ) 207 me = 1.056; mµ (reduced) = = 196 me . −27 (1.67 × 10 ) 1.056

This shrinks the whole molecule down by a factor of almost 200, bringing the deuterons much closer together, as desired. The equilibrium separation for the electron case was 2.493 a (Problem 7.10), so for muons, R = 2.493 (0.529 × 10−10 m) = 6.73 × 10−13 m. 196 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

214


Problem 7.20 (a) −

2 ∂ 2 ψ ∂ 2 ψ = Eψ. + 2m ∂x2 ∂y 2

Y

d2 X d2 Y 2mE + X = − 2 XY ; dx2 dy 2

d2 X = −kx2 X; dx2

Let ψ(x, y) = X(x)Y (y).

d2 Y = −ky2 Y, dy 2

1 d2 X 1 d2 Y 2mE + =− 2 . X dx2 Y dy 2

with kx2 + ky2 =

2mE . The general solution to the y equation is 2

Y (y) = A cos ky y + B sin ky y; the boundary conditions Y (±a) = 0 yield ky =

nπ π with minimum . 2a 2a

[Note that ky2 has to be positive, or you cannot meet the boundary conditions at all.] So 2 π2 2 E ≥ kx + 2 . For a traveling wave kx2 has to be positive. Conclusion: Any solution with E < 2m 4a π 2 2 will be a bound state. 8ma2 (b)

y a II

I a

x

Integrate over regions I and II (in the figure), and multiply by 8.

∞

a

y a

2

y x , v ≡ , dx = a du, dy = a dv. a a x=a y=0 1 ∞ ∞ 1 e−2αu (1 − v)3 = A2 a2 (1 − v)2 e−2αu du dv = A2 a2 × 3 2α 1 1 0 0

III = A2

=

1−

e−2αx/a dx dy.

Let u ≡

A2 a2 A2 a2 −2α . (−1) − e−2α = e 6α 6α



II =

1 2 A 2

a

a

1− x=0

1 = A2 a2 2

y=0

1

1

xy a2

2

e−2α dx dy

2 −2α

(1 − uv) e 0

0

215

1 du dv = A2 a2 e−2α 2

1

(v 2 − 3v + 3)dv, 0

2 ∂2 2 ∂2 ψ| 2 + 2 |ψ = −8 (JI + JII ). [Ignore roof-lines for the moment.] 2m ∂x ∂y 2m

0 2

2 ✼ y ∂ y −αx/a ∂ 1− 1 − dx dy e−αx/a e = A2 + a ∂x2 ∂y 2 a x=a y=0 2 2✄ 2 2 2 2 ∞ a y α 1 α α A a −2α 2 −2αx/a =A 1− e dx dy = III = = A2 αe−2α . e a a a 6 α 6 a ✚ x=a y=0 ✁

JII

1 (1 − uv)3 dv −3v 0

2 2 e2α A a −2α 11 2 2 −2α 9α = 1 ⇒ A2 = 2 8 + A a e e . 6α 36 2a (6 + 11α)

Normalizing:

H = −

3

0

1 1 (1 − v) − 1 1 = − A2 a2 e−2α dv = A2 a2 e−2α 2 3 0 v 6 3 1 2 1 v 11 2 2 −2α v = A2 a2 e−2α . − 3 + 3v = A a e 6 3 2 36 0 1

1

∞

a

xy −α ∂ 2 xy −α ∂2 e 1 − e dx dy = 0. + a2 ∂x2 ∂y 2 a2 0 0 xy ∂ y ∂2 1 − = − 2 = 0, and likewise for ∂ 2 /∂y 2 .] [Note that ∂x2 a2 ∂x a

JI =

1 2 A 2

Hso

far

a

a

1−

2 2 α −2α e = − A2 . 3 m

Now the roof-lines; label them as follows: I. Right arm: at y = 0 : KI . II. Central square: at x = 0 and at y = 0 : KII . III. Boundaries: at x = ±a and at y = ±a : KIII . 2 ∞ a |y| −αx/a ∂ |y| −αx/a 2 ∂2 2 KI = 4 − 1− 1− + 2 dx dy. A e e 2m a ∂x2 ∂y a x=a y=−a

|y| = y θ(y) − θ(−y) ,

∂ |y| 1 ✟✟ + ✟ ✟✟ , 1− = − θ(y) − θ(−y) + ✟ yδ(y) yδ(y) ∂y a a 2 ∂ 1 2 |y| = − δ(y) − δ(−y)(−1) = − δ(y). 1− 2 ∂y a a a c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

216

CHAPTER 7. THE VARIATIONAL PRINCIPLE 22 2 KI = − A m

∞

−2αx/a

e x=a ♣

a

dx y=−a

|y| 2 1− − δ(y) dy a a ♠

∞ a −2α e−2αx/a e−2α 2 = e ♣= =− ; ♠=− , (−2α/a) a (−2α/a) 2α a 22 A2 a 22 −2α 2 ✁ e−2α − 2 ; =− KI = A . e a m ✁ mα 2α ✁

KII

2 x|y| −α ∂ x|y| −α ∂2 1− 2 e 1 − 2 e dx dy = 4A + 2 a ∂x2 ∂y a x=0 y=−a

a a 2 2 2 −2α x|y| 2x =− 1− 2 − 2 δ(y) dx dy A e m a a x=0 y=−a a 22 A2 −2α 2 22 −2α 2 =− e e − ✁2✁ xdx; KII = A . m m ✁a 0 2

2 − 2m

a

a

a2 2

KIII

2 a a+5 2 ∂ ∂2 =8 − ψ + 2 ψ dx dy. 2m ∂x2 ∂y y=0 x=a−5

In this region (x, y both positive)

ψ=A

    2 −α   1 − xy/a e (x < a)   ψ=A ,   −αx/a   1 − y/a e (x > a)  

or

y x −α[θ(a−x)+ x θ(x−a)] a 1− . θ(x − a) + θ(a − x) e a a

✟ −α θ(a−x)+ x θ(x−a) ∂ψ y 1 x ✟ ✘ ] ✘ a ✟− x) e [ ✘− a) + θ(a − x) − δ(a =A − ✘ δ(x ∂x a a a✟ ✟

y α x αx ✘ ✘ ✘ −α[θ(a−x)+ x θ(x−a)] ✘ ✘ a ✘ − a) ✘− x) − θ(x − a) − ✘✘δ(x + 1− α✘ δ(a θ(x − a) + θ(a − x) e a a a a [Note: f (x) = xδ(x) should be zero—but perhaps we should check that this is still safe when we’re planning to take it’s derivative: df /dx = δ(x) + x dδ/dx :

df dδ dδ g dx = g δ(x) + x dx = g(0) + gx dx dx dx dx ✿0 ✘ d dg ✘ ✘ = g(0) + ✘ gxδ(x)| (gx)δ(x)dx = g(0) − g+x δ(x)dx ✘✘ x=0 − dx dx = g(0) − g(0) − (xg )|x=0 = 0. This confirms that f (x) can be taken to be zero even when differentiated.] So δ(x − a) −

x 1 δ(a − x) = (a − x)δ(a − x) = 0. Hence the cancellations above, leaving a a



217

x ∂ψ y = A − 2 θ(a − x)e−α[θ(a−x)+ a θ(x−a)] ∂x a

α y x −α[θ(a−x)+ x θ(x−a)] a − θ(x − a) 1 − θ(x − a) + θ(a − x) e a a a

x α y y x = Ae−α[θ(a−x)+ a θ(x−a)] − 2 θ(a − x) − θ(x − a) 1 − θ(x − a) + θ(a − x) a a a a x A y y = − e−α[θ(a−x)+ a θ(x−a)] θ(a − x) + αθ(x − a) 1 − . a a a ∂2ψ y y A −α[θ(a−x)+ xa θ(x−a)] − δ(a − x) + αδ(x − a) 1 − =− e ∂x2 a a a

✟ ✘ + 1 θ(x − a) + x δ(x✟−✟a) ✘−✘x) − α −✘ δ(a ✟ a a ✟ integral 0

A −α αy y = − e δ(x − a) α − − . a a a A αy y − dx dy ψ(x, y) − e−α δ(x − a) α − a a a y=0 x=a−5 42 A −α a αy y αy y αy 2 42 A2 −2α a y2 = α− α−2 e − dy = e − + 2 + 2 dy ψ(a, y) ma a a ma a a a a y=0 0

KIII = −

42 m

a

a+5

A(1−y/a)e−α

42 A2 −2α α a2✄ α a3 1 a3 1 a2 = αa − ✁ 2 e + 2 + 2 − ma a 2 a 3 a 3 a 2 ✁ ✁

a 42 A2 −2α αa a ✁ ✁ ✁ ✟−✟ ✟− + αa αa = e + ✟ m 2 3 3 α 1 1 3 − 6 = 6 (2α−1)

=

42 A2 (2α − 1)e−2α ; 6m

KIII =

22 A2 (2α − 1)e−2α . 3m

2 2 α −2α 22 −2α 2 22 −2α 2 22 2 e e e A (2α − 1)e−2α H = − A2 + A + A + 3 m mα m 3m

2 2 2 2A2 e−2α 2 3 A2 e−2α 2 − α + + 2 + (2α − 1) = − α + + 3 + 2α − 1 = m 3 α 3 3m α ✘ ✘ 2 −2α 2 2 −2α 2 −2α ✟

9 ✚ ✘ 2A e 2 2 3 2A e e2α α 2e ✟ = α+2+ = α2 + 2α + 3 = ✁ α + 2α + 3 2 3m α 3mα 3 m✚ α ✁2 a (6 + 11α) =

32 (α2 + 2α + 3) . ma2 (6 + 11α)

32 (6 + 11α)(2α + 2) − (α2 + 2α + 3)(11) dH =0 = dα ma2 (6 + 11α)2

⇒ (6 + 11α)(2α + 2) = 11(α2 + 2α + 3).

12α + 12 + 22α2 + 22α = 11α2 + 22α + 33 ⇒ 11α2 + 12α − 21 = 0. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

218


√ (12)2 + 4 · 11 · 21 −6 ± 36 + 231 α= = 22 11 −6 ± 16.34 10.34 = = [α has to be positive] = 0.940012239. 11 11 −12 ± 2

Hmin

2 6 2 32 2(α + 1) . = (α + 1) = 1.058 = 2 2 ma 11 11 ma ma2

But Ethreshold =

π 2 2 2 = 1.2337 , 8 ma2 ma2

so E0 is definitely less than Ethreshold .


CHAPTER 8. THE WKB APPROXIMATION

219

Chapter 8

The WKB Approximation Problem 8.1

a

p(x) dx = nπ,

with n = 1, 2, 3, . . . and

p(x) =

2m[E − V (x)] (Eq. 8.16).

0

Here

a

p(x) dx =

√

0

2mE

a 2

+

2m(E − V0 )

4 ⇒ E + E − V0 + 2 E(E − V0 ) = 2m

Square again:

nπ a

a 2

=

√

2 = 4En0 ;

2m

a √ 2

E+

E − V0 = nπ

2 E(E − V0 ) = (4En0 − 2E + V0 ).

2

4E(E − V0 ) = 4E 2 − 4EV0 = 16En0 + 4E 2 + V02 − 16EEn0 + 8En0 V0 − 4EV0

2

⇒ 16EEn0 = 16En0 + 8En0 V0 + V02 ⇒ En = En0 +

V2 V0 + 0 0. 2 16En

V0 Perturbation theory gave En = En0 + ; the extra term goes to zero for very small V0 (or, since En0 ∼ n2 ), for 2 large n.

Problem 8.2 (a) dψ i = f eif / ; dx

d2 ψ i = 2 dx

if /

f e

i + (f )2 eif /

i 1 2 if / = . f − 2 (f ) e

d2 ψ p2 i 1 2 if / p2 f e = − ψ =⇒ − (f ) = − 2 eif / =⇒ if − (f )2 + p2 = 0. QED 2 2 2 dx c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

220


(b) f = f0 + f1 + 2 f2 + · · · =⇒ (f )2 = (f0 + f1 + 2 f2 + · · · )2 = (f0 )2 + 2f0 f1 + 2 [2f0 f2 + (f1 )2 ] + · · · f = f0 + f1 + 2 f2 + · · · .

i(f0 + f1 + 2 f2 ) − (f0 )2 − 2f0 f1 − 2 [2f0 f2 + (f1 )2 ] + p2 + · · · = 0.

0 : (f0 )2 = p2 ;

1 : if0 = 2f0 f1 ; 2 : if1 = 2f0 f2 + (f1 )2 ; . . . df0 df1 i ±p i i f0 i d (c) = = ±p =⇒ f0 = ± p(x)dx + constant ; = = ln p =⇒ f1 = ln p + const. dx dx 2 f0 2 ±p 2 dx 2 if i i i ψ = exp = exp ± p(x) dx + ln p + K = exp ± p dx p−1/2 eiK/ 2 C i = √ exp ± p dx . QED p

Problem 8.3 1 γ=

1 |p(x)| dx =

2a

2m(V0 − E) dx =

0

2a 2m(V0 − E).

√ T ≈ e−4a 2m(V0 −E)/ .

From Problem 2.33, the exact answer is 1

T = 1+

V02 4E(V0 −E)

sinh2 γ

.

Now, the WKB approximation assumes the tunneling probability is small (p. 322)—which is to say that γ is large. In this case, sinh γ = 12 (eγ − e−γ ) ≈ 12 eγ , and sinh2 γ ≈ 14 e2γ , and the exact result reduces to 16E(V0 − E) 1 e−2γ . ≈ T ≈ 2 V02 V 2γ 0 1 + 16E(V0 −E) e The coefficient in { } is of order 1; the dominant dependence on E is in the exponential factor. In this sense T ≈ e−2γ (the WKB result).

Problem 8.4 I take the masses from Thornton and Rex, Modern Physics, Appendix 8. They are all atomic masses, but the electron masses subtract out in the calculation of E. All masses are in atomic units (u): 1 u = 931 MeV/c2 . The mass of He4 is 4.002602 u, and that of the α-particle is 3727 MeV/c2 . U238 : Z = 92, A = 238, m = 238.050784 u → Th234 : m = 234.043593 u. r1 = (1.07 × 10−15 m)(238)1/3 = 6.63 × 10−15 m. E = (238.050784 − 234.043593 − 4.002602)(931) MeV = 4.27 MeV. V =

2E = m

(2)(4.27) × 3 × 108 m/s = 1.44 × 107 m/s. 3727



221

90 γ = 1.980 √ − 1.485 90(6.63) = 86.19 − 36.28 = 49.9. 4.27

τ=

(2)(6.63 × 10−15 ) 98.8 7.46 × 1021 21 e s = 7.46 × 10 s = yr = 2.4 × 1014 yrs. 1.44 × 107 3.15 × 107

Po212 : Z = 84, A = 212, m = 211.988842 u → Pb208 : m = 207.976627 u. r1 = (1.07 × 10−15 m)(212)1/3 = 6.38 × 10−15 m. E = (211.988842 − 207.976627 − 4.002602)(931) MeV = 8.95 MeV. V =

2E = m

(2)(8.95) × 3 × 108 m/s = 2.08 × 107 m/s. 3727

82 γ = 1.980 √ − 1.485 82(6.38) = 54.37 − 33.97 = 20.4. 8.95

τ=

(2)(6.38 × 10−15 ) 40.8 e s = 3.2 × 10−4 s. 2.08 × 107

These results are way off—but note the extraordinary sensitivity to nuclear masses: a tiny change in E produces enormous changes in τ . √ Much more impressive results are obtained when you plot the logarithm of lifetimes against 1/ E, as in Figure 8.6. Thanks to David Rubin for pointing this out. Some experimental values are listed below (all energies in MeV): A E τ 238 4.198 4.468 × 109 yr A E τ 236 4.494 2.342 × 107 yr 224 7.488 0.79 s 5 234 4.775 2.455 × 10 yr Uranium (Z = 92): Protactinium (Z = 91): 222 8.540 2.9 ms 232 5.320 68.9 yr 220 9.650 0.78 µs 230 5.888 20.8 day 218 9.614 0.12 ms 228 6.680 9.1 min 226 7.570 0.35 s A 232 Thorium (Z = 90): 230 228 226

E τ 4.012 1.405 × 1010 yr 4.687 7.538 × 104 yr 5.423 1.912 yr 6.337 30.57 min

A 226 224 Radium (Z = 88): 222 220 218

E 4.784 5.685 6.559 7.455 8.389

τ 1600 yr 3.66 day 38 s 18 ms 25.6 µs


222


Problem 8.5 (a) V (x) = mgx. (b) 2 d 2 ψ d2 ψ 2m2 g − + mgxψ = Eψ =⇒ = 2m dx2 dx2 2

E x− mg

E . Let y ≡ x − , and α ≡ mg

2m2 g 2

1/3 .

d2 ψ E d2 ψ 3 = α yψ. Let z ≡ αy = α(x − = zψ. This is the Airy equation (Eq. 8.36), and ), so dy 2 mg dz 2 the general solution is ψ = aAi(z) + bBi(z). However, Bi(z) blows up for large z, so b = 0 (to make ψ E normalizable). Hence ψ(x) = aAi α(x − mg ) . Then

(c) Since V (x) = ∞ for x < 0, we require ψ(0) = 0; hence Ai [α(−E/mg)] = 0. Now, the zeros of Ai are an (n = 1, 2, 3, . . . ). Abramowitz and Stegun list a1 = −2.338, a2 = −4.088, a3 = −5.521, a4 = −6.787, 2 1/3 αEn mg etc. Here − an , or En = −( 12 mg 2 2 )1/3 an . In this case = an , or En = − an = −mg mg α 2m2 g 1 1 2 2 2 2 −34 J·s)2 = 5.34 × 10−68 J3 ; ( 12 mg 2 2 )1/3 = 3.77 × 10−23 J. 2 mg = 2 (0.1 kg)(9.8 m/s ) (1.055 × 10 E1 = 8.81 × 10−23 J,

E2 = 1.54 × 10−22 J,

E3 = 2.08 × 10−22 J,

E4 = 2.56 × 10−22 J.

(d) 2T = x

dV (Eq. 3.97); dX

here

But T + V = H = En , For the electron,

1 mg 2 2 2

so

1/3

dV = mg, dx 3 V = En , 2

so

x

dV = mgx = V , dx

or V =

1 (9.11 × 10−31 )(9.8)2 (1.055 × 10−34 )2 = 2 x =

T =

2 En . But V = mgx, 3

E1 = 1.84 × 10−32 J = 1.15 × 10−13 eV.

so

1/3

1 V . 2 so x =

2En . 3mg

= 7.87 × 10−33 J.

2(1.84 × 10−32 ) = 1.37 × 10−3 = 1.37 mm. 3(9.11 × 10−31 )(9.8)

Problem 8.6 (a)

x2

Eq. 8.47 =⇒ 0

x2

p(x) dx = 0

√

1 p(x) dx = (n − )π, where p(x) = 2m(E − mgx) and E = mgx2 =⇒ x2 = E/mg. 4

x2 √ 2 E − mgx dx = 2m − (E − mgx)3/2 3mg 0 0 2 2 1 2 2 1 =− (E − mgx2 )3/2 − E 3/2 = E 3/2 . 3 mg 3 mg

2m

x2



223

V(x) E mgx

x2

1 1 √ (2E)3/2 = (n − )π, 4 3 mg

or

En =

9

8π

2

x

mg 2 2 (n − 14 )2

1/3

.

(b)

9 2 2 2 π mg 8

1/3

9 2 = π (0.1)(9.8)2 (1.055 × 10−34 )2 8

1/3

= 1.0588 × 10−22 J.

2/3 3 = 8.74 × 10−23 J, 4 2/3 7 −22 E2 = (1.0588 × 10 ) = 1.54 × 10−22 J, 4 2/3 11 −22 E3 = (1.0588 × 10 ) = 2.08 × 10−22 J, 4 2/3 15 E4 = (1.0588 × 10−22 ) = 2.56 × 10−22 J. 4 E1 = (1.0588 × 10−22 )

These are in very close agreement with the exact results (Problem 8.5(c)). In fact, they agree precisely (to 3 significant digits), except for E1 (for which the exact result was 8.81 × 10−23 J). (c) From Problem 8.5(d), x =

n=

2En , 3mg

so

1=

2 (1.0588 × 10−22 ) 3 (0.1)(9.8)

n−

1 4

2/3

, or

n−

1 4

2/3 = 1.388 × 1022 .

1 + (1.388 × 1022 )3/2 = 1.64 × 1033 . 4

Problem 8.7

x2

p(x) dx = x1

1 n− 2

1 n− 2

x2

π = mω −x2

π;

p(x) =

1 2m E − mω 2 x2 ; 2

2E − x2 dx = 2mω mω 2

0

x2

x22

−

x2

1 x2 = −x1 = ω

2E . m

x2 −1 2 2 2 dx = mω x x2 − x + x2 sin (x/x2 ) 0


224


En = n − 12 ω (n = 1, 2, 3, . . . )

π π πE 2E = mωx22 = mω . 2 2 mω 2 ω

= mωx22 sin−1 (1) =

Since the WKB numbering starts with n = 1, whereas for oscillator states we traditionally start with n = 0, letting n → n + 1 converts this to the usual formula En = (n + 12 )ω. In this case the WKB approximation yields the exact results.

Problem 8.8 (a) 1 mω 2 x22 = En = 2

1 n+ 2

ω

(counting n = 0, 1, 2, . . . );

x2 =

(2n + 1) . mω

(b) Vlin (x) =

1 1 mω 2 x22 + (mω 2 x2 )(x − x2 ) =⇒ Vlin (x2 + d) = mω 2 x22 + mω 2 x2 d. 2 2

V (x2 + d) − Vlin (x2 + d) = V (x2 )

1 2 2 mω (x2

x2 + 2x2 d + d2 − x22 − 2x2 d = 2 = x22

+ d)2 − 12 mω 2 x22 − mω 2 x2 d 1 2 2 2 mω x2

d x2

2 = 0.01.

d = 0.1 x2 .

(c)

2m α= mω 2 x2 2

1/3

(Eq. 8.34),

2m2 ω 2 (2n + 1)2 2 ≥ (50)3 ; 2 m2 ω 2 nmin = 125.

so

0.1 x2

or (2n + 1)2 ≥

2m2 ω 2 x2 2

1/3

2m2 ω 2 4 ≥ 5 =⇒ x2 2

(50)3 = 62500; 2

2n + 1 ≥ 250;

1/3

n≥

≥ 50.

249 = 124.5. 2

However, as we saw in Problems 8.6 and 8.7, WKB may be valid at much smaller n.

Problem 8.9 Shift origin to the turning point.  0 1 1   (x < 0) De− x |p(x )| dx  |p(x)| ix ψWKB = x i 1   Be 0 p(x ) dx + Ce− 0 p(x ) dx (x > 0)  |p(x)| c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


225

overlap 1

overlap 2

ψP ψ

ψ

WKB

WKB

E

Nonclassical

Classical

x

0 patching region

Linearized potential in the patching region:

d2 ψp 2mV (0) = xψp = −α3 xψp , whereα ≡ dx2 2

V (x) ≈ E + V (0)x. Note : V (0) is negative.

2m|V (0)| 2

1/3 .

ψp (x) = aAi(−αx) + bBi(−αx). (Note change of sign, as compared with Eq. 8.37). p(x) =

2m[E − E − V (0)x] =

−2mV (0)x =

2m|V (0)|x =

√

√ α3 2 x = α3/2 x.

Overlap region 1 (x < 0):

0

|p(x )| dx = α

0

3/2

x

√

−x

dx = α

3/2

x

ψWKB ≈

0 2 2 3/2 2 3/2 3/2 − (−x ) = (−αx)3/2 . = 3 α (−x) 3 3 x

3/2 2 1 De− 3 (−αx) . For large positive argument (−αx 1) : 1/2 α3/2 (−x)1/4

3/2 3/2 2 2 1 1 ψp ≈ a √ e− 3 (−αx) + b √ e 3 (−αx) . Comparing ⇒ a = 2D 2 π(−αx)1/4 π(−αx)1/4

π ; α

b = 0.

Overlap region 2 (x > 0):

x

|p(x )| dx = α3/2

0

ψWKB ≈

x

√

x dx = α3/2

0

1 1/2 α3/4 x1/4

2

Bei 3 (αx)

3/2

x 2 3/2 2 3/2 (x ) = 3 (αx) . 3 0

+ Ce−i 3 (αx) 2

3/2

. For large negative argument (−αx −1) :

1 iπ/4 i 2 (αx)3/2 1 2 π a 3/2 −iπ/4 −i 23 (αx)3/2 3 √ ψp (x) ≈ a √ (remember : b = 0). sin + e − e e (αx) = e 3 4 π(αx)1/4 π(αx)1/4 2i a α iπ/4 a α −iπ/4 Comparing the two: B = , C=− . e e 2i π 2i π c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

226


Inserting the expression for a from overlap region 1 : B = −ieiπ/4 D; C = ie−iπ/4 D. For x > 0, then, x

x i π 2D −iD i x p(x ) dx +i π π 1 4 − e− 0 p(x ) dx −i 4 ψWKB = = p(x ) dx + e 0 sin . 0 4 p(x) p(x) Finally, switching the origin back to x1 :   x1 1 D − |p(x )| dx   x   e , (x < x ) ; 1   |p(x)|

ψWKB (x) = x 2D 1 π    sin p(x )dx + , (x > x1 ).    x1 4 p(x)

QED

Problem 8.10 At x1 , we have an upward-sloping turning point. Follow the method in the book. Shifting origin to x1 :  1 i 0 p(x ) dx − i x0 p(x ) dx  x  (x < 0) + B Ae  p(x) ψWKB (x) = x 1 1 1 x |p(x )| dx   Ce 0 + D− 0 |p(x )| dx (x > 0)  p(x)

In overlap region 2, Eq. 8.39 becomes ψWKB ≈

1

1/2 α3/4 x1/4

whereas Eq. 8.40 is unchanged. Comparing them =⇒ a = 2D

In overlap region 1, Eq. 8.43 becomes ψWKB ≈

2

Ce 3 (αx)

π , α

3/2

+ De− 3 (αx) 2

b=C

3/2

,

π . α

1 i 23 (−αx)3/2 −i 23 (−αx)3/2 Ae , + Be 1/2 α3/4 (−x)1/4

and Eq. 8.44 (with b = 0) generalizes to

a 2 π 2 π b 3/2 3/2 √ ψp (x) ≈ √ sin + cos + (−αx) + (−αx) 3 4 3 4 π(−αx)1/4 π(−αx)1/4 1 i 23 (−αx)3/2 iπ/4 −i 23 (−αx)3/2 −iπ/4 = √ . Comparing them =⇒ e + (ia + b)e e (−ia + b)e 2 π(−αx)1/4 ) A= A=

α π

−ia + b 2

C − iD eiπ/4 ; 2

iπ/4

e

;

B=

B=

α π

ia + b 2

e−iπ/4 . Putting in the expressions above for a and b :

C + iD e−iπ/4 . 2



227

These are the connection formulas relating A, B, C, and D, at x1 . At x2 , we have a downward-sloping turning point, and follow the method of Problem 8.9. First rewrite the middle expression in Eq. 8.52: 1 x2 x x 1 1 1 |p(x )| dx + |p(x )| dx − 1 x2 |p(x )| dx − |p(x )| dx x2 x2 ψWKB = . + De x1 Ce x1 |p(x)| x Let γ ≡ x12 |p(x)| dx, as before (Eq. 8.22), and let C ≡ De−γ , D ≡ Ceγ . Then (shifting the origin to x2 ):  0 0 1 1 1 |p(x )| dx − |p(x )| dx  x x  , (x < 0); C e + D e  |p(x)| ψWKB = x i 1   (x > 0). F e 0 p(x ) dx ,  p(x) 1/3 √ 2m|V (0)| In the patching region ψp (x) = aAi(−αx) + bBi(−αx), where α ≡ ; p(x) = α3/2 x. 2 0 2 In overlap region 1 (x < 0): |p(x )| dx = (−αx)3/2 , so 3 x   1 π  23 (−αx)3/2 − 23 (−αx)3/2   e + D e C  a = 2 D 1/2 α3/4 (−x)1/4 α Comparing =⇒ 3/2 3/2 2 2 a b π    ψp ≈ √ e− 3 (−αx) + √ e 3 (−αx)  b = C 2 π(−αx)1/4 π(−αx)1/4 α

ψWKB ≈

x

In overlap region 2 (x > 0):

p(x ) dx =

0

ψp ≈ √

3/2 2 2 1 (αx)3/2 =⇒ ψWKB ≈ 1/2 3/4 1/4 F ei 3 (αx) . 3 α x

a 2 π 2 π b 3/2 3/2 √ sin + cos + (αx) + (αx) 3 4 3 4 π(αx)1/4 π(αx)1/4

3/2 3/2 2 2 1 iπ −i π 4 ei 3 (αx) 4 e−i 3 (αx) (−ia + b)e . Comparing =⇒ (ia + b) = 0; = √ + (ia + b)e 2 π(αx)1/4 F =

α π

C =

−ia + b 2

iπ/4

e

α b = e−iπ/4 F, π

=b

1 D = 2

α iπ/4 . e π

b=

π −iπ/4 F; e α

π −iπ/4 a=i F. e α

α i a = e−iπ/4 F. D = eγ e−iπ/4 F ; π 2

C=

i −γ −iπ/4 e e F. 2

These are the connection formulas at x2 . Putting them into the equation for A: −γ C i −γ −iπ/4 e A= F − ieγ e−iπ/4 F eiπ/4 = i − iD eiπ/4 = e e − eγ F. 2 4 4 2 F e−2γ 1 T = = = 2 . −γ A (eγ − e 4 )2 1 − (e−γ/2 )2 If γ 1, the denominator is essentially 1, and we recover T = e−2γ

(Eq. 8.22).


228


Problem 8.11

x2 x2 √ 1 α ν π = 2 Equation 8.51 ⇒ n − 2m(E − αx ) dx = 2 2mE 1 − xν dx; 2 E 0 0 1/ν 1/ν 1 1 −1 α ν zE E z ≡ x , so x = ; dx = z ν dz. Then E α α ν

n−

1 2

√ π = 2 2mE √ = 2 2mE

E

1 1 ν+2

E α E α

1/ν 1/ν

1 ν

1

√ √ 1 z ν −1 1 − z dz = 2 2mE

0

E α

1/ν

E = αxν2 . Let

1 Γ(1/ν)Γ(3/2) ν Γ( ν1 + 32 )

√ 1/ν + 1) 12 π √ Γ( ν1 + 1) E = . 2πmE 1 3 α Γ( ν + 2 ) Γ( ν1 + 32 )

Γ( ν1

(n − 1 )π 1/ν Γ( ν1 + 32 ) = √ 2 ; α Γ( ν1 + 1) 2πm

( 2ν ) π Γ( ν1 + 32 ) ν+2 1 α. En = n− 2 2mα Γ( ν1 + 1)

π Γ(2) 2α 1 1 For ν = 2: En = n− α = (n − ) . 2 2mα Γ(3/2) 2 m

For a harmonic oscillator, with α = 12 mω 2 , En = n − 12 ω (n = 1, 2, 3, . . . ).

Problem 8.12 2 a2 V (x) = − sech2 (ax). Eq. 8.51 =⇒ m √ = 2 2a

x2

sech2 (ax) +

0

E=−

2 a2 sech2 (ax2 ) m

1 dx = a

−1 √ √ z 1−z

Limits :

1 z

1 n− 2

x2

π = 2 0

2 a2 2 2m E + sech (ax) dx m

mE dx. 2 a2

defines x2 . Let b ≡ −

mE , 2 a2

1 1 1 √ dz = − √ dz. 2a z 1 − z 2 z

Then

x = 0 =⇒ z = sech2 (0) = 1 x = x2 =⇒ z = sech2 (ax2 ) = −

z ≡ sech2 (ax),

mE =b 2 a2

1 n− 2

.

n−

1 2

so that x =

√ 1 sech−1 z, and hence a

z2 √ √ z−b 1 √ π = 2 2a − dz. 2a z1 z 1 − z

π=

√ 2 b

1

1 z

z−b dz. 1−z

1 (z − b) 1 b z−b = = − . 1−z z (1 − z)(z − b) (1 − z)(z − b) z (1 − z)(z − b)



1 n− 2

√

1

229

1

1

1

dz − b dz 2 (1 − z)(z − b) b z −b + (1 + b)z − z

1 √ 1−z √ −1 (1 + b)z − 2b −1 = 2 −2 tan − b sin z−b z(1 − b) b √ √ √ √ π √ π √ π −1 −1 −1 = 2 −2 tan (0) + 2 tan (∞) − b sin (1) + b sin−1 (−1) = 2 0 + 2 − b − b 2 2 2 √ √ √ √ (n − 12 ) 1 1 √ = 2π(1 − b); b=1− √ = 1 − b; n− . 2 2 2

π =

2

b

√ √ Since the left side is positive, the right side must also be: (n − 12 ) < 2, n < 12 + 2 = 0.5 + 1.414 = 1.914. So the only possible n is 1; there is only one bound state (which is correct—see Problem 2.51). For n = 1,

√

9 1 1 1 b=1− √ + = − √ ; 8 2 8 2

1 b=1− √ ; 2 2

2 a2 E1 = − m

9 1 −√ 8 2

= −0.418

2 a2 . m

2 a2 . Not bad. m

The exact answer (Problem 2.51(c)) is −0.5

Problem 8.13 n−

1 4

r0

2m [E − V0 ln(r/a)] dr;

E = V0 ln(r0 /a) defines r0 . r0 r0 √ = 2m V0 ln(r0 /a) − V0 ln(r/a) dr = 2mV0 ln(r0 /r) dr.

π =

0

0

Let x ≡ ln(r0 /r),

1 n− 4

1 n− 4

r0 =

π =

0

or r = r0 e−x =⇒ dr = −r0 e−x dx.

so ex = r0 /r,

x2

2mV0 (−r0 )

√

−e

xe

dx. Limits :

x1

π =

∞

2mV0 r0

√

−x

xe

dx =

r = 0 =⇒ x1 = ∞ r = r0 =⇒ x2 = 0

2mV0 r0 Γ(3/2) =

0

2π 1 n− mV0 4

⇒

3 En+1 − En = V0 ln n + 4

.

√ 2mV0 r0

π . 2

1 2π 2π 1 n− . + V0 ln En = V0 ln = V0 ln n − a mV0 4 4 a mV0

1 − V0 ln n − 4

= V0 ln

n + 3/4 n − 1/4

, which is indeed independent of m (and a).


230


Problem 8.14

1 n− 2

π = r1

where A ≡ −

r2

2 l(l + 1) e2 1 − 2m E + 4π90 r 2m r2

e2 1 4π90 E

1 n− 2

π =

√

B≡−

and

r2

−2mE

√

r1

2 l(l + 1) 2m E

dr =

√

−2mE

1 n− 2

π =

√

r2

−2mE

r1

−1 +

r1

B A − 2 dr, r r

are positive constants, since E is negative.

−r2 + Ar − B dr. r − r2 + Ar − B = (r − r1 )(r2 − r).

Let r1 and r2 be the roots of the polynomial in the numerator:

r2

√ (r − r1 )(r2 − r) √ 2 π √ dr = −2mE ( r2 − r1 ) . r 2

√ √ 1 2 n− = −2mE (r2 + r1 − 2 r1 r2 ) . But − r2 + Ar − B = −r2 + (r1 + r2 )r − r1 r2 2 =⇒ r1 + r2 = A; r1 r2 = B. Therefore

1 2 n− 2

=

√

√ √ −2mE A − 2 B = −2mE

e2 = 4π90 e2 4π90

−

−

'

e2 1 − −2 4π90 E

2 l(l + 1) − 2m E

(

2m − 2 l(l + 1). E

2m 1 = 2 n − + l(l + 1) ; E 2

−

E = 2m

42

(e2 /4π90 )2 2 . n − 12 + l(l + 1)

−(m/22 )(e2 /4π90 )2 −13.6 eV E= 2 = 2 . 1 1 n − 2 + l(l + 1) n − 2 + l(l + 1)

Problem 8.15 (a)

D

(i) ψWKB (x) =

|p(x)|

(ii) ψWKB (x) =

1 p(x)

1 −

e

i

Be

x x2

|p(x )| dx

x2 x

p(x ) dx

(x > x2 ); + Ce− i

x2 x

p(x ) dx

(x1 < x < x2 );


CHAPTER 8. THE WKB APPROXIMATION (iii) ψWKB (x) =

1 |p(x)|

Equation 8.46 =⇒ (ii) ψWKB

1

Fe

x1 x

231

|p(x )| dx

+ Ge− 1

x1 x

|p(x )| dx

x2

1 2D π sin = p(x )dx + x 4 p(x)

(0 < x < x1 ).

(x1 < x < x2 ).

To effect the join at x1 , first rewrite (ii): x2 x

2D 1 x π π 1 1 2D (ii) ψWKB = p(x ) dx − p(x ) dx + p(x ) dx − θ − sin sin = − , x1 x1 4 x1 4 p(x) p(x) where θ is defined in Eq. 8.58. Now shift the origin to x1 :   10 0 1 1   x |p(x )| dx + Ge− x |p(x )| dx   (x < 0) F e   |p(x)|

ψWKB = . x 2D π 1    p(x ) dx − θ − (x > 0)  sin  −  0 4 p(x) Following Problem 8.9: ψp (x) = aAi(−αx) + bBi(−αx), with α ≡

0

Overlap region 1 (x < 0):

|p(x )| dx =

x

2m|V (0)| 2

1/3

√ ; p(x) = α3/2 x.

2 (−αx)3/2 . 3

 2 3/2 1 − 23 (−αx)3/2  3 (−αx) F e + Ge  π 1/2 α3/4 (−x)1/4 =⇒ a = 2G ; 3/2 2 a b  − 23 (−αx)3/2 (−αx) α 3  √ ≈ √ e + e 2 π(−αx)1/4 π(−αx)1/4

ψWKB ≈ ψp

x

Overlap region 2 (x > 0): 0

=⇒ ψWKB

ψp ≈ √

π . α

2 (αx)3/2 . 3

2D 2 π 3/2 , ≈ − 1/2 3/4 1/4 sin (αx) − θ − 3 4 α x

a 2 π 2 π b 3/2 3/2 √ sin + cos + (αx) + (αx) . 3 4 3 4 π(αx)1/4 π(αx)1/4

Equating the two expressions:

=√

p(x ) dx =

b=F

1 πα1/4

−2D 1 i 2 (αx)3/2 −iθ −iπ/4 −i 23 (αx)3/2 iθ iπ/4 3 e e − e e e e 1/2 α3/4 2i

b 2 3/2 3/2 3/2 2 2 a i 2 (αx)3/2 iπ/4 e − e−i 3 (αx) e−iπ/4 + e 3 ei 3 (αx) eiπ/4 + e−i 3 (αx) e−iπ/4 2i 2

 π −iθ −iπ/4    −2D e e = (a + ib)eiπ/4 , α =⇒ π iθ iπ/4    2D = (−a + ib)e−iπ/4 , e e α

 π −iθ    ie (a + ib) = 2D α π iθ   (a − ib) = −2D ie  α

or or


232

CHAPTER 8. THE WKB APPROXIMATION   π π   −iθ iθ   2a = 2D  − e ) ⇒ a = 2D i(e sin θ,  α α =⇒ π π     2ib = 2D  i(e−iθ + eiθ ) ⇒ b = 2D cos θ.  α α Combining these with the results from overlap region 1 =⇒ 2G

π = 2D α

π sin θ, α

or

G = D sin θ; D

Putting these into (iii) : ψWKB (x) =

|p(x)|

F

1

2 cos θe

π = 2D α

x1 x

|p(x )| dx

π cos θ, α

+ sin θe− 1

or

x1 x

F = 2D cos θ.

|p(x )| dx

(0 < x < x1 ).

(b) 1

Odd(−) case: (iii) =⇒ ψ(0) = 0 ⇒ 2 cos θe 1

x1

|p(x )| dx =

0

x 1 0

|p(x )| dx

+ sin θe− 1

x 1 0

|p(x )| dx

= 0.

1 φ, with φ defined by Eq. 8.60. So sin θe−φ/2 = −2 cos θeφ/2 , or tan θ = −2eφ . 2

d|p(x)| 1 D φ/2 −φ/2 Even(+) case: (iii) =⇒ ψ (0) = 0 ⇒ − 2 cos θe + sin θe 2 (|p(x)|)3/2 dx 0

x x 1 1 1 1 1 |p(x )| dx 1 |p(x )| dx − 0 0 + − |p(0)| + sin θe 2 cos θe |p(0)| = 0. |p(x)| D

Now

√ 1 d|p(x)| d 1 dV dV d|p(x)| = , and 2m[V (x) − E] = 2m √ = 0, so = 0. dx dx 2 V − E dx dx 0 dx 0

2 cos θeφ/2 = sin θe−φ/2 , or tan θ = 2eφ . Combining the two results: tan θ = ±2eφ . QED (c) tan θ = tan

sin n + 12 π + 9 1 (−1)n cos 9 cos 9 1

= n+ =− ≈− . π+9 = 1 2 (−1)n+1 sin 9 sin 9 9 cos n + 2 π + 9

1 1 −φ 1 1 φ So − ≈ ±2e , or 9 ≈ ∓ e , or θ − n + π ≈ ∓ e−φ , 9 2 2 2

so θ ≈

1 n+ 2

1 π ∓ e−φ . QED 2

[Note: Since θ (Eq. 8.58) is positive, n must be a non-negative integer: n = 0, 1, 2, . . . . This is like harmonic oscillator (conventional) numbering, since it starts with n = 0.] c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


233 V(x)

-a

x1

a

x2

x

(d)

1 2m E − mω 2 (x − a)2 dx. Let z = x − a (shifts the origin to a). 2 x1

2 z2 1 1 2 2 = 2m E − mω z dz, where E = mω 2 z22 . 0 2 2

z2 z2 2 mω mω 2 −1 π mω 2 −1 2 2 2 2 2 = mω z z2 − z + z2 sin (z/z2 ) = z2 sin (1) = z , z2 − z dz = 2 2 0 0

1 θ =

=

x2

π mω 2E πE = . 2 mω 2 ω

πE Putting this into Eq. 8.61 yields ≈ ω

1 n+ 2

1 π ∓ e−φ , 2

or

En±

≈

1 n+ 2

ω ∓

ω −φ e . QED 2π

(e) + − 1 Ψ(x, t) = √ ψn+ e−iEn t/ + ψn− e−iEn t/ =⇒ 2 |Ψ(x, t)|2 =

− + − + 1 +2 |ψn | + |ψn− |2 + ψn+ ψn− ei(En −En )t/ + e−i(En −En )t/ . 2

En− − En+ ω 1 ω ≈ 2 e−φ = e−φ , so 2π π 1 + 2 ω −φ |Ψ(x, t)|2 = ψn (x) + ψn− (x)2 + ψn+ (x)ψn− (x) cos e t . 2 π

(Note that the wave functions (i), (ii), (iii) are real ). But

It oscillates back and forth, with period

τ=

2π 2π 2 φ e . QED = (ω/π) e−φ ω

(f ) 1 φ=2

0

x1

x1 √ mω 2 1 2 2m mω 2 (x − a)2 − E dx = 2mE (x − a)2 − 1 dx. 2 2E 0


234

CHAPTER 8. THE WKB APPROXIMATION Let z ≡

m ω(a − x), 2E

dx = −

so

 

 m  ωa ≡ z0 x = 0 =⇒ z = Limits: . 2E   x = x1 =⇒ radicand = 0 =⇒ z = 1

2E 1 dz. m ω

z0 2√ 2E 1 z0 2 4E z0 2 4E 1 2 φ= 2mE z − 1 dz = z − 1 dz = z z − 1 − ln(z + z 2 − 1) m ω 1 ω 1 ω 2 1 2E z0 z02 − 1 − ln z0 + z02 − 1 , = ω where z0 = aω

m . 2E

V (0) =

1 2 2 2 mω a ,

so

m V (0) E ⇒ ω 2 a2 E ⇒ aω 2

m 1, 2E

or z0 1.

In that case φ≈

2E 2 2E 2 2E 2 2 m mωa2 z0 − ln(2z0 ) ≈ z0 = a ω = . ω ω ω 2E

This, together with Eq. 8.64, gives us the period of oscillation in a double well.

Problem 8.16 (a) En ≈

n2 π 2 2 . 2m(2a)2

With n = 1,

E1 =

π 2 2 . 8ma2

(b) V(x) -a

a

x

V0 E1 V(x)

x0

a

-a

E

x

tunneling E1

(c) 1 x0 V 0 − E1 . |p(x)| dx. αx0 = V0 − E1 ⇒ x0 = a α p(x) = 2m [E − V (x)]; V (x) = −αx, E = E1 − V0 . √ √ √ √ = 2m(E1 − V0 + αx) = 2mα x − x0 ; |p(x)| = 2mα x0 − x. √ √

x0 x0 √ 1√ 2 2mα 2 2mα 3/2 3/2 γ = − (x0 − x) 2mα x0 − x dx = = 3 (x0 − a) . 3 a a γ =



235

Now x0 − a = (V0 − E1 − aα)/α, and αa 2 /ma2 ≈ E1 V0 , so we can drop E1 and αa. Then √ 3/2 8mV03 2 2mα V0 γ≈ . = 3 α 3α Equation 8.28 ⇒ τ =

4a 2γ e , v

where τ=

1 π 2 2 π 2 2 ⇒ v2 = , mv 2 ≈ 2 2 8ma 4m2 a2

or v =

π . 2ma

So

4a 8ma2 2γ 2ma e2γ = e . π π

(d)

2

(8) 9.1 × 10−31 10−10 e2γ = 2 × 10−19 e2γ ; −34 π (1.05 × 10 ) 3 (8) (9.1 × 10−31 ) (20 × 1.6 × 10−19 ) γ = = 4.4 × 104 ; (3) (1.6 × 10−19 ) (7 × 106 ) (1.05 × 10−34 )

τ = 2 × 10−19 × 1038,000 s = 1038,000 yr. τ =

8.8×104 4 e2γ = e8.8×10 = 10log e = 1038,000 .

Seconds, years . . . it hardly matters; nor is the factor out front significant. This is a huge number—the age of the universe is about 1010 years. In any event, this is clearly not something to worry about.

Problem 8.17 Equation 8.22 ⇒ the tunneling probability: T = e−2γ , where 1 x0 γ = 2m(V − E) dx. Here V (x) = mgx, E = 0, x0 = R2 + (h/2)2 − h/2 (half the diagonal). 0 √ x x0 2m √ m 2 3/2 0 2m 3/2 = mg x1/2 dx = 2g x = 2g x0 . 3 3 0 0 √ I estimate: h = 10 cm, R = 3 cm, m = 300 gm; let g = 9.8 m/s2 . Then x0 = 9 + 25 − 5 = 0.83 cm, and (2)(0.3) γ= (2)(9.8) (0.0083)3/2 = 6.4 × 1030 . −34 (3)(1.05 × 10 ) Frequency of “attempts”: say f = v/2R. We want the product of the number of attempts (f t) and the probability of toppling at each attempt (T ), to be 1: v −2γ 2R 2γ =1 ⇒ t= e e . 2R v Estimating the thermal velocity: 12 mv 2 = 12 kB T (I’m done with the tunneling probability; from now on T is the temperature, 300 K) ⇒ v = kB T /m.

13×1030 30 30 m 2γ 0.3 = (5 × 108 ) × 105.6×10 s t = 2R e = 2(0.03) e12.8×10 = 5 × 108 10log e −23 kB T (1.4 × 10 )(300) t

= 16 × 105.6×10

30

yr.

Don’t hold your breath.


236

CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY

Chapter 9

Time-Dependent Perturbation Theory Problem 9.1 ψnlm = Rnl Ylm . From Tables 4.3 and 4.7: ψ100 = √

ψ210 = √

1 πa3

e−r/a ;

1 32πa3

ψ200 = √

r −r/2a cos θ; e a

1 8πa3

1−

r −r/2a ; e 2a

ψ21±1 = ∓ √

1 64πa3

r r/2a sin θ e±iφ . e a

But r cos θ = z and r sin θe±iφ = r sin θ(cos φ ± i sin φ) = r sin θ cos φ ± ir sin θ sin φ = x ± iy. So |ψ|2 is an even function of z in all cases, and hence z|ψ|2 dx dy dz = 0, so Hii = 0. Moreover, ψ100 is even in z, and so are ψ200 , ψ211 , and ψ21−1 , so Hij = 0 for all except H100,210 = −eE √

eE =− √ 4 2πa4 or

∞

1 πa3

√

1

1 3 32πa a

r4 e−3r/2a dr

0

eE e−r/a e−r/2a z 2 d3 r = − √ 4 2πa4

π

cos2 θ sin θ dθ 0

0

2π

e−3r/2a r2 cos2 θ r2 sin θ dr dθ dφ

eE dφ = − √ 4! 4 2πa4

2a 3

5

2 2π = − 3

28 √

35 2

eEa,

−0.7449 eEa.

Problem 9.2 i −iω0 t c˙a = − Hab e cb ;

i iω0 t c˙b = − Hba e ca . Differentiating with respect to t :

i iω0 t i iωo t i −iω0 t i iω0 t iω0 t c¨b = − Hba iω0 e − Hab e ca + e c˙a = iω0 − Hba e ca − Hba e cb , or c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY c¨b = iω0 c˙b −

237

1 1 2 |H |2 cb . Let α2 ≡ 2 |Hab | . Then c¨b − iω0 c˙b + α2 cb = 0. 2 ab

This is a linear differential equation with constant coefficients, so it can be solved by a function of the form cb = eλt :

i 1 λ2 − iω0 λ + α2 = 0 =⇒ λ = iω0 ± −ω02 − 4α2 = (ω0 ± ω) , where ω ≡ ω02 + 4α2 . 2 2 The general solution is therefore

cb (t) = Aei(ω0 +ω)/2 + Bei(ω0 −ω)/2 = eiω0 t/2 Aeiωt/2 + Be−iωt/2 , or

cb (t) = eiω0 t/2 [C cos (ωt/2) + D sin (ωt/2)] . But cb (0) = 0,

so C = 0,

and hence

cb (t) = Deiω0 t/2 sin (ωt/2) . Then

ω0 i iω0 t iω0 iω0 t/2 ω iω0 t/2 ω c˙b = D e sin (ωt/2) + e cos (ωt/2) = Deiω0 t/2 cos (ωt/2) + i sin (ωt/2) = − Hba e ca . 2 2 2 ω ca =

i ω −iω0 t/2 ω0 D cos (ωt/2) + i e sin (ωt/2) . But ca (0) = 1, 2 Hba ω

ω0 ca (t) = e−iω0 t/2 cos (ωt/2) + i sin (ωt/2) , ω 2Hba iω0 t/2 cb (t) = e sin (ωt/2) , iω

where ω ≡

ω02 + 4

so

i ω 2 D = 1. Conclusion: Hba

|2 |Hab 2 .

2 | ω02 4|Hab 2 sin (ωt/2) + sin2 (ωt/2) 2 2 2 ω ω 1 |H |2 = cos2 (ωt/2) + 2 ω02 + 4 ab sin2 (ωt/2) = cos2 (ωt/2) + sin2 (ωt/2) = 1. ω 2

|ca |2 + |cb |2 = cos2 (ωt/2) +

Problem 9.3 This is a tricky problem, and I thank Prof. Onuttom Narayan for showing me the correct solution. The safest approach is to represent the delta function as a sequence of rectangles: (1/29), −9 < t < 9, δ5 (t) = 0, otherwise. Then Eq. 9.13 ⇒

 t < −9 :         t > 9 :

ca (t) = 1, cb (t) = 0, ca (t) = a, cb (t) = b,

  iα −iω0 t  e cb ,  c˙a = − 25     −9 < t < 9 :   ∗   iω0 t ca . c˙b = − iα 25 e c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

238


In the interval −9 < t < 9,

d2 cb iα |α|2 −iα −iω0 t iα∗ i29 dcb dcb iα∗ iω0 t iω0 t iω e iω − c − = − = iω = − e c + e c cb . 0 a b 0 b 0 dt2 29 29 29 α∗ dt 29 dt (29)2 Thus cb satisfies a homogeneous linear differential equation with constant coefficients: d2 cb |α|2 dcb + − iω cb = 0. 0 dt2 dt (29)2 Try a solution of the form cb (t) = eλt : iω0 ± |α|2 λ − iω0 λ + =0⇒λ= 2 (29)

2

or λ= The general solution is

iω0 iω ± , where ω ≡ 2 2

−ω02 − |α|2 /(9)2 , 2

ω02 + |α|2 /(9)2 .

cb (t) = eiω0 t/2 Aeiωt/2 + Be−iωt/2 .

But cb (−9) = 0 ⇒ Ae−iω5/2 + Beiω5/2 = 0 ⇒ B = −Ae−iω5 , so

cb (t) = Aeiω0 t/2 eiωt/2 − e−iω(5+t/2) .

Meanwhile

iω 2i9 −iω0 t 2i9 −iω0 t/2 iω0 iωt/2 −iω(5+t/2) iωt/2 −iω(5+t/2) + e c ˙ = e A − e + e e e b α∗ α∗ 2 2 9 −iω0 t/2 = − ∗e A (ω + ω0 )eiωt/2 + (ω − ω0 )e−iω(5+t/2) . α

ca (t) =

But

Thus

9 i(ω0 −ω)5/2 29ω i(ω0 −ω)5/2 α∗ i(ω−ω0 )5/2 e A [(ω + ω ) + (ω − ω )] = − e A, so A = − . e 0 0 α∗ α∗ 29ω 1 −iω0 (t+5)/2 ca (t) = (ω + ω0 )eiω(t+5)/2 + (ω − ω0 )e−iω(t+5)/2 e 2ω

ω(t + 9) ω0 ω(t + 9) −iω0 (t+5)/2 =e cos + i sin ; 2 ω 2

iα∗ iω0 (t−5)/2 iω(t+5)/2 ω(t + 9) iα∗ iω0 (t−5)/2 −iω(t+5)/2 −e sin cb (t) = − e =− e e . 29ω 9ω 2 ca (−9) = 1 = −

ω0 a = ca (9) = e−iω0 5 cos(ω9) + i sin(ω9) , ω

b = cb (9) = −

iα∗ sin(ω9). 9ω

This is for the rectangular pulse; it remains to take the limit 9 → 0: ω → |α|/9, a → cos

|α|

ω0 9 +i sin |α|

|α|

→ cos

|α|

,

so

iα∗ b→− sin |α|

|α|

,



239

and we conclude that for the delta function ca (t) =

1, t < 0, cos(|α|/), t > 0;

 t < 0,  0, cb (t) = α∗  −i sin(|α|/), t > 0. α Obviously, |ca (t)|2 + |cb (t)|2 = 1 in both time periods. Finally, Pa→b = |b|2 = sin2 (|α|/).

Problem 9.4 (a)  i −iω0 t   Eq. 9.10 =⇒ c˙a = − ca Haa + cb Hab e  (these are exact, and replace Eq. 9.13). i iω0 t  Eq. 9.11 =⇒ c˙b = − cb Hbb + ca Hba e Initial conditions: Zeroth order:

ca (0) = 1,

ca (t) = 1,

cb (0) = 0.

cb (t) = 0.

  i t   c (t) = 1 − H (t ) dt  c˙a = − i Haa a =⇒ 0 aa First order: t i  c˙ = − i H eiω0 t =⇒   cb (t) = − Hba (t )eiω0 t dt  b ba 0

t

2 i t 1 i t 1+ |ca | = 1 − H (t ) dt H (t ) dt = 1 + H (t ) dt = 1 (to first order in H ). 0 aa 0 aa 0 aa

i t iω0 t i t −iω0 t |cb |2 = − Hba (t )e dt Hab (t )e dt = 0 (to first order in H ). 0 0 2

So |ca |2 + |cb |2 = 1 (to first order). (b) i d˙a = e

t 0

Haa (t ) dt

t i i i −iω0 t Haa ca + e 0 Haa (t ) dt c˙a . But c˙a = − ca Haa + cb Hab e

Two terms cancel, leaving

t i i i t −iω0 t d˙a = − e 0 Haa (t ) dt cb Hab e . But cb = e− 0 Hbb (t ) dt db . i it i −iω0 t −iω0 t = − e 0 [Haa (t )−Hbb (t )]dt Hab e db , or d˙a = − eiφ Hab e db .


240

CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY Similarly, i d˙b = e

t 0

Hbb (t ) dt

t i i i iω0 t Hbb cb + e 0 Hbb (t ) dt c˙b . But c˙b = − cb Hbb + ca Hba e .

t i i it iω0 t = − e 0 Hbb (t ) dt ca Hba e . But ca = e− 0 Haa (t ) dt da . i i it iω0 t iω0 t e da = − e−iφ Hba e da . QED = − e 0 [Hbb (t )−Haa (t )]dt Hba

(c) Initial conditions: Zeroth order:

ca (0) = 1 =⇒ da (0) = 1;

da (t) = 1,

cb (0) = 0 =⇒ db (0) = 0.

db (t) = 0.

i d˙a = 0 =⇒ da (t) = 1 =⇒ ca (t) = e−

First order:

i i iω0 t d˙b = − e−iφ Hba e =⇒ db = − i it cb (t) = − e− 0 Hbb (t )dt

t

t

t 0

Haa (t ) dt

.

e−iφ(t ) Hba (t )eiω0 t dt =⇒

0

e−iφ(t ) Hba (t )eiω0 t dt .

0

These don’t look much like the results in (a), but remember, we’re only working to first order in H , t so ca (t) ≈ 1 − i 0 Haa (t ) dt (to this order), while for cb , the factor Hba in the integral means it is already first order and hence both the exponential factor in front and e−iφ should be replaced by 1. Then t cb (t) ≈ − i 0 Hba (t )eiω0 t dt , and we recover the results in (a).

Problem 9.5 c(0) a (t) = a,

Zeroth order:

First order:

 i −iω0 t ib t −iω0 t  (1)  b =⇒ ca (t) = a − H (t )e dt .  c˙a = − Hab e 0 ab t i iω0 t ia  (1)   c˙b = − Hba e a =⇒ cb (t) = b − Hba (t )eiω0 t dt . 0

Second order:

c(2) a (t)

(0)

cb (t) = b.

ib =a−

i −iω0 t ia t iω0 t c˙a = − Hab e H (t )e dt =⇒ b− 0 ba 0

t

Hab (t )e−iω0 t dt

a − 2

0

t

Hab (t )e−iω0 t

t

Hba (t )eiω0 t dt

dt .

0

To get cb , just switch a ↔ b (which entails also changing the sign of ω0 ): t t ia t iω0 t b (2) iω0 t −iω0 t cb (t) = b − H (t )e dt − 2 H (t )e Hab (t )e dt dt . 0 ba 0 ba 0



241

Problem 9.6 i (2) (1) For H independent of t, Eq. 9.17 =⇒ cb (t) = cb (t) = − Hba

t

eiω0 t dt =⇒

0

t

H = − ba eiω0 t − 1 . Meanwhile Eq. 9.18 =⇒ ω0 0 t t t 1 1 (2) 2 −iω0 t iω0 t 2 1 ca (t) = 1 − 2 |Hab | e e dt dt = 1 − 2 |Hab | 1 − e−iω0 t dt iω0 0 0 0 ' ( t

1 −iω0 t i e−iω0 t i 2 2 t + = 1+ |H | |H | t + −1 . e = 1+ ω0 2 ab iω0 ω0 2 ab iω0

(2) cb (t)

i eiω0 t = − Hba iω0

0

For comparison with the exact answers (Problem 9.2), note first that cb (t) is already first order (because of the Hba in front), whereas ω differs from ω0 only in second order, so it suffices to replace ω → ω0 in the exact formula to get the second-order result: cb (t) ≈

1 iω0 t/2 2Hba 2Hba H eiω0 t/2 sin (ω0 t/2) = eiω0 t/2 − e−iω0 t/2 = − ba eiω0 t − 1 , e iω0 iω0 2i ω0

in agreement with the result above. Checking ca is more difficult. Note that 2 2 2 |2 4|Hab |Hab | | | |Hab ω0 |Hab ω = ω0 1 + 1 + 2 = ω ≈ ω + 2 ; . ≈ 1 − 2 0 0 2 2 2 ω 0 2 ω 0 2 ω 0 2 ω ω 0 2 Taylor expansion:  2 2 ω0 t |Hab | t | t |Hab   + ≈ cos (ω t/2) − sin (ω0 t/2)  cos(x + 9) = cos x − 9 sin x =⇒ cos (ωt/2) = cos 0 2 2 2 ω ω 0 0 2 2 ω0 t |Hab | t |H | t   ≈ sin (ω0 t/2) + ab 2 cos (ω0 t/2) +  sin(x + 9) = sin x + 9 cos x =⇒ sin (ωt/2) = sin 2 ω 0 2 ω0 ω0 t ω0 t ω0 t ω0 t |H |2 t |H |2 |H |2 t ca (t) ≈ e−iω0 t/2 cos sin − ab 2 sin + i 1 − 2 2ab 2 + ab 2 cos 2 ω0 2 ω0 2 ω0 2 2 ω0 t ω0 t ω0 t ω0 t |H | ω0 t 2i = e−iω0 t/2 cos sin + i sin − ab2 t sin − i cos + 2 2 ω0 2 2 ω0 2

2 |H | 2i 1 = e−iω0 t/2 eiω0 t/2 − ab2 −iteiω0 t/2 + eiω0 t/2 − e−iω0 t/2 ω0 ω 2i

|H |2 1 1 −iω0 t i 2 = 1 − ab2 −it + t + |H | − 1 , as above. 1 − e−iω0 t = 1 + e ω0 ω0 ω0 2 ab iω0

Problem 9.7 (a) c˙a = −

i Vab eiωt e−iω0 t cb ; 2

c˙b = −

i Vba e−iωt eiω0 t ca . 2


242

CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY Differentiate the latter, and substitute in the former: Vba i(ω0 − ω)ei(ω0 −ω)t ca + ei(ω0 −ω)t c˙a 2

Vba i(ω0 −ω)t Vab −i(ω0 −ω)t Vba i(ω0 −ω)t |Vab |2 = i(ω0 − ω) −i e e e −i ca − i cb = i(ω0 − ω)c˙b − cb . 2 2 2 (2)2

c¨b = −i

d2 cb dcb |Vab |2 |Vab |2 + i(ω − ω0 ) cb = 0. Solution is of the form cb = eλt : λ2 + i(ω − ω0 )λ + = 0. + 2 2 dt dt 4 42

1 (ω − ω0 ) |Vab |2 2 λ= = i − , with ωr defined in Eq. 9.30. −i(ω − ω0 ) ± −(ω − ω0 ) − ± ω r 2 2 2

i −

General solution: cb (t) = Ae

(ω−ω0 ) +ωr 2

t

i −

+ Be

(ω−ω0 ) +ωr 2

t

= e−i(ω−ω0 )t/2 Aeiωr t + Be−iωr t ,

or, more conveniently: cb (t) = e−i(ω−ω0 )t/2 [C cos(ωr t) + D sin(ωr t)] . But cb (0) = 0, so C = 0 :

ω0 − ω i(ω0 −ω)t/2 i(ω0 −ω)t/2 cb (t) = De e sin(ωr t). c˙b = D i sin(ωr t) + ωr e cos(ωr t) ; 2

2 i(ω−ω0 )t 2 i(ω−ω0 )t/2 ω0 − ω ca (t) = i e c˙b = i e D i sin(ωr t) + ωr cos(ωr t) . But ca (0) = 1 : Vba Vba 2 2 −iVba 1=i Dωr , or D = . Vba 2ωr i(ω0 −ω)t/2

cb (t) = −

i Vba ei(ω0 −ω)t/2 sin(ωr t), 2ωr

ω0 − ω sin(ωr t) . ca (t) = ei(ω−ω0 )t/2 cos(ωr t) + i 2ωr

(b) Pa→b (t) = |cb (t)| = 2

and the denominator,

|Vab | 2ωr

2 sin2 (ωr t).

The largest this gets (when sin2 = 1) is

|Vab |2 /2 , 4ωr2

4ωr2 = (ω − ω0 )2 + |Vab |2 /2 , exceeds the numerator, so P ≤ 1 (and 1 only if ω = ω0 ).

2 2 ω0 − ω |Vab | sin2 (ωr t) + sin2 (ωr t) 2ωr 2ωr (ω − ω0 )2 + (|Vab |/)2 = cos2 (ωr t) + sin2 (ωr t) = cos2 (ωr t) + sin2 (ωr t) = 1. 4ωr2

|ca |2 + |cb |2 = cos2 (ωr t) +

(c) If

|Vab | (ω − ω0 ) , 2

2

Eq. 9.28.

2

then ωr

1 ≈ |ω − ω0 |, 2

and Pa→b

0 |Vab |2 sin2 ω−ω 2 t ≈ , 2 (ω − ω0 )2

confirming

(d) ωr t = π =⇒ t = π/ωr .



243

Problem 9.8 Spontaneous emission rate (Eq. 9.56): A =

R=

π |℘|2 ρ(ω), 390 2

with

ρ(ω) =

ω 3 |℘|2 . Thermally stimulated emission rate (Eq. 9.47): 3π90 c3

ω3 π 2 c3 (eω/kB T − 1)

(Eq. 9.52).

So the ratio is

A ω 3 |℘|2 390 2 π 2 c3 eω/kB T − 1 = · · = eω/kB T − 1. R 3π90 c3 π|℘|2 ω 3 The ratio is a monotonically increasing function of ω, and is 1 when eω/kb t = 2,

or

ω = ln 2, kB T ν=

ω=

kB T ln 2,

or

ν=

ω kB T = ln 2. For T = 300 K, 2π h

(1.38 × 10−23 J/K)(300 K) ln 2 = 4.35 × 1012 Hz. (6.63 × 10−34 J · s)

For higher frequencies, (including light, at 1014 Hz), spontaneous emission dominates.

Problem 9.9 (a) Simply remove the factor eω/kB T − 1 in the denominator of Eq. 5.113:

ρ0 (ω) =

ω 3 . π 2 c3

(b) Plug this into Eq. 9.47: Rb→a =

ω 3 ω 3 |℘|2 π |℘|2 2 3 = , 2 390 π c 3π90 c3

reproducing Eq. 9.56.

Problem 9.10 N (t) = e−t/τ N (0) (Eqs. 9.58 and 9.59). so t/τ = ln 2, or t1/2 = τ ln 2.

After one half-life, N (t) = 12 N (0), so

1 2

= e−t/τ , or

2 = et/τ ,

Problem 9.11 28 √ a. As 35 2 for x and y, we noted that |1 0 0, |2 0 0, and |2 1 0 are even (in x, y), whereas |2 1 ± 1 is odd. So the only

In Problem 9.1 we calculated the matrix elements of z; all of them are zero except 1 0 0|z|2 1 0 =


244


non-zero matrix elements are 1 0 0|x|2 1 ± 1 and 1 0 0|y|2 1 ± 1. Using the wave functions in Problem 9.1: 1 ∓1 1 √ 1 0 0|x|2 1 ± 1 = √ e−r/a re−r/2a sin θ e±iφ (r sin θ cos φ)r2 sin θ dr dθ dφ 3 3 a πa 8 πa ∞ π 2π 1 3 4 −3r/2a =∓ r e dr sin θ dθ (cos φ ± i sin φ) cos φ dφ 8πa4 0 0 0 5 ∓1 2a 4 27 = 4! a. (π) = ∓ 8πa4 3 3 35 5 2π ∓1 2a 4 1 0 0|y|2 1 ± 1 = 4! (cos φ ± i sin φ) sin φ dφ 8πa4 3 3 0 5 ∓1 2a 4 27 = 4! (±iπ) = −i 5 a. 4 8πa 3 3 3 1 0 0|r|2 0 0 = 0;

√ 27 2 ˆ 1 0 0|r|2 1 0 = a k; 35

℘2 = 0 (for |2 0 0 → |1 0 0),

Meanwhile,

E2 − E 1 1 ω= =

and |℘|2 = (qa)2

E1 − E1 4

=−

1 0 0|r|2 1 ± 1 =

27 ˆ ˆj , and hence a ∓ i − i 35

215 (for |2 1 0 → 1 0 0 and |2 1 ± 1 → |1 0 0). 310 3E1 , 4

so for the three l = 1 states:

2 33 E13 (ea)2 215 1 c 29 E13 e2 a2 210 E1 = − = 6 3 10 3 8 4 3 8 2 2 3 3π90 c 3 π 90 c 3 mc a 2 10 8 13.6 (3.00 × 10 m/s) 1 2 = 8 = 6.27 × 108 /s; τ = = 1.60 × 10−9 s 3 0.511 × 106 (0.529 × 10−10 m) A

A=−

for the three l = 1 states (all have the same lifetime); τ = ∞

for the l = 0 state.

Problem 9.12 [L2 , z] = [L2x , z] + [L2y , z] + [L2z , z] = Lx [Lx , z] + [Lx , z]Lx + Ly [Ly , z] + [Ly , z]Ly + Lz [Lz , z] + [Lz , z]Lz   [Lx , z] = [ypz − zpy , z] = [ypz , z] − [zpy , z] = y[pz , z] = −iy, [Ly , z] = [zpx − xpz , z] = [zpx , z] − [xpz , z] = −x[pz , z] = ix, But  [Lz , z] = [xpy − ypx , z] = [xpy , z] − [ypx , z] = 0. So:

[L2 , z] = Lx (−iy) + (−iy)Lx + Ly (ix) + (ix)Ly = i(−Lx y − yLx + Ly x + xLy ).

But

Lx y = Lx y − yLx + yLx = [Lx , y] + yLx = iz + yLx , Ly x = Ly x − xLy + xLy = [Ly , x] + xLy = −iz + xLy .



245

So: [L2 , z] = i(2xLy − iz − 2yLx − iz) =⇒ [L2 , z] = 2i(xLy − yLx − iz). 2 2 . L , [L , z] = 2i [L2 , xLy ] − [L2 , yLx ] − i[L2 , z] . = 2i [L2 , x]Ly + x[L2 , Ly ] − [L2 , y]Lx − y[L2 , Lx ] − i(L2 z − zL2 ) . But

[L2 , Ly ] = [L2 , Lx ] = 0 (Eq. 4.102), so

. 2 2 L , [L , z] = 2i (yLz − zLy − ix) Ly − 2i (zLx − xLz − iy) Lx − i L2 z − zL2 , or 2 2 2 L , [L , z] = −2 2yLz Ly = −2

−2zL2y

−

−2ixLy + 2xLz Lx + 2iyLx − L z + zL

2zL2x

2

−2z(L2x +L2y +L2z )+2zL2z

2yLz Ly − 2ixLy + 2xLz Lx + 2iyLx + 2zL2z − 2zL2 − L2 z + zL2

2

2 2 2 = −2 zL + L z − 4 (yLz − ix) Ly + (xLz + iy) Lx + zLz Lz 2

Lz y

2

Lz x

= 22 zL2 + L2 z − 42 (Lz yLy + Lz xLx + Lz zLz ) = 22 (zL2 + L2 z). QED Lz (r·L)=0

Problem 9.13 |n 0 0 =

Rn0 (r)Y00 (θ, φ)

1 = √ Rn0 (r), 4π

so

1 n 0 0|r|n 0 0 = 4π

ˆ dx dy dz. Rn 0 (r)Rn0 (r)(x î + y ˆj + z k)

But the integrand is odd in x, y, or z, so the integral is zero.

Problem 9.14 (a)    |2 1 0  |2 1 1 |3 0 0 → → |1 0 0.   |2 1−1

(|3 0 0 → |2 0 0 and |3 0 0 → |1 0 0 violate ∆l = ±1 rule.)

(b) From Eq. 9.72:

ˆ 2 1 0|r|3 0 0 = 2 1 0|z|3 0 0 k.

From Eq. 9.69:

2 1 ± 1|r|3 0 0 = 2 1 ± 1|x|3 0 0 î + 2 1 ± 1|y|3 0 0 ˆj.

From Eq. 9.70:

± 2 1 ± 1|x|3 0 0 = i2 1 ± 1|y|3 0 0.

Thus

|2 1 0|r|3 0 0|2 = |2 1 0|z|3 0 0|2

and |2 1 ± 1|r|3 0 0|2 = 2|2 1 ± 1|x|3 0 0|2 ,


246

CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY so there are really just two matrix elements to calculate. ψ300 = R30 Y00 . From Table 4.3:

ψ21m = R21 Y1m ,

Y10 Y00

cos θ sin θ dθ dφ =

∗ Y1±1

3 4π

1 4π

π 2

cos θ sin θ dθ 0

0

Y00

sin θ cos φ dθ dφ = ∓

1 = ∓ 4π

2

3 8π

1 4π

2π

√ √ π 3 3 2 cos3 θ 1 dφ = (2π) = − =√ . 4π 3 2 3 3 0

π

2π

sin3 θ dθ 0

cos φe∓iφ dφ

0

2π

2π 3 4 1 1 2 cos φ dφ ∓ i cos φ sin φ dφ = ∓ √ (π ∓ 0) = ∓ √ . 2 3 π 6 6 0 0

From Table 4.7:

∞ ∞ 2r 2 r −r/2a 1 2 r 2 −r/3a 3 3 √ K ≡ 1− e R21 R30 r dr = √ r dr e + 3 a 27 a 24a3/2 27a3/2 0 a 0 6 7 5 ∞ 1 2 6 6 2 a 2 2 6 = √ 1 − u + u2 u4 e−5u/6 du = √ 4! a4 − 5! + 6! 3 3 27 5 3 5 27 5 9 2a 9 2 0 a 4! 65 2 a 4! 65 27 3 4 √ 2 = √ 5 − 6 · 5 + 63 = √ = 6 2 a. 6 6 3 27 5 9 2 5 9 2 5 So: 1 R21 (Y1±1 )∗ (r sin θ cos φ)R30 Y00 r2 sin θ dr dθ dφ = K ∓ √ . 6 1 2 1 0|z|3 0 0 = R21 Y10 (r cos θ)R30 Y00 r2 sin θ dr dθ dφ = K √ . 3

2 1 ± 1|x|3 0 0 =

|2 1 0|r|3 0 0|2 = |2 1 0|z|3 0 0|2 = K 2 /3; |2 1 ± 1|r|3 0 0 |2 = 2|2 1 ± 1|x|3 0 0 |2 = K 2 /3. Evidently the three transition rates are equal, and hence 1/3 go by each route. (c) For each mode, decay rate is

ω 3 e2 |r|2 A= ; 3π90 c3

here

E3 − E2 1 ω = =

E1 E1 − 9 4

=−

5 E1 , 36

so the total

2 9 3 2 5 E1 c 2 E1 e2 1 2 7 34 √ R=3 − 2a = 6 36 3π90 c3 3 56 5 mc2 a 9 2 2 13.6 3 × 108 1 =6 /s = 6.32 × 106 /s. τ = = 1.58 × 10−7 s. 6 −10 5 0.511 × 10 0.529 × 10 R



247

Problem 9.15 (a) Ψ(t) =

cn (t)e−iEn t/ ψn .

cn e−iEn t/ En ψn +

HΨ = i

∂Ψ ; ∂t

H = H0 + H (t);

cn e−iEn t/ H ψn = i

H0 ψn = En ψn . So

i c˙n e−iEn t/ ψn + i − cn En e−iEn t/ ψn .

The first and last terms cancel, so cn e−iEn t/ H ψn = i c˙n e−iEn t/ ψn . Take the inner product with ψm :

cn e−iEn t/ ψm |H |ψn = i

c˙n e−iEn t/ ψm |ψn . ψm |ψn = δmn , and define

Assume orthonormality of the unperturbed states,

cn e−iEn t/ Hmn = i c˙m e−iEm t/ ,

(b) Zeroth order: i c˙N = − HN N,

c˙m = −

or

Hmn ≡ ψm |H |ψn .

i cn Hmn ei(Em −En )t/ . n

cm (t) = 0 for m = N . Then in first order: i t cN (t) = 1 − H (t ) dt , whereas for m = N : 0 NN

cN (t) = 1, or

i c˙m = − HmN ei(Em −EN )t/ , or

cm (t) = −

i

t

HmN (t )ei(Em −EN )t / dt .

0

(c) t i(EM −EN )t/

i(EM −EN )t / e e −1 i i(EM −EN )t / e dt = − HM N = −HM N i(EM − EN )/ EM − EN 0 0 HM EM − E N N =− ei(EM −EN )t/2 2i sin t . (EM − EN ) 2

i cM (t) = − HM N

PN →M = |cM |2 =

t

2 4|HM N| sin2 (EM − EN )2

EM − EN t . 2

(d) 1 t iωt i cM (t) = − VM N e + e−iωt ei(EM −EN )t / dt 2 0 t ei(ω+EM −EN )t / ei(−ω+EM −EN )t / iVM N + =− . 2 i(ω + EM − EN )/ i(−ω + EM − EN )/ 0


248

CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY If EM > EN , the second term dominates, and transitions occur only for ω ≈ (EM − EN )/: iVM N EM − EN − ω 1 cM (t) ≈ − ei(EM −EN −ω)t/2 2i sin t , so 2 (i/)(EM − EN − ω) 2 PN →M = |cM |2 =

(EM

|VM N |2 sin2 − EN − ω)2

EM − EN − ω t . 2

If EM < EN the first term dominates, and transitions occur only for ω ≈ (EN − EM )/: iVM N EM − EN + ω 1 i(EM −EN +ω)t/2 cM (t) ≈ − 2i sin e t , and hence 2 (i/)(EM − EN + ω) 2

PN →M =

(EM

|VM N |2 sin2 − EN + ω)2

EM − EN + ω t . 2

Combining the two results, we conclude that transitions occur to states with energy EM ≈ EN ± ω, and PN →M =

EM − EN ± ω t . 2

Vba = −℘E0 (Eq. 9.34). The rest is as before (Section 9.2.3), leading to Eq. 9.47:

(e) For light, RN →M =

(EM

|VM N |2 sin2 − EN ± ω)2

π |℘|2 ρ(ω), with ω = ±(EM − EN )/ 390 2

(+ sign ⇒ absorption, − sign ⇒ stimulated emission).

Problem 9.16 For example (c): cN (t) = 1 −

|cN |2 = 1 +

i H t; NN

cm (t) = −2i

1 |H |2 t2 , 2 N N

|cm |2 = 1 +

m

HmN ei(Em −EN )t/2 sin (Em − EN )

|cm |2 = 4

|2 |HmN sin2 (Em − EN )2

Em − EN t 2

(m = N ).

Em − EN t , so 2

Em − EN |HmN |2 t2 2 2 |H | + 4 sin t . 2 N N (Em − EN )2 2 m=N

This is plainly greater than 1! But remember: The c’s are accurate only to first order in H ; to this order the |H |2 terms do not belong. Only if terms of first order appeared in the sum would there be a genuine problem with normalization. For example (d): cN

i = 1 − VN N

0

t

t sin(ωt ) i i cos(ωt ) dt = 1 − VN N =⇒ cN (t) = 1 − VN N sin(ωt). ω 0 ω


CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY VmN cm (t) = − 2

|cN |2 = 1 +

249

ei(Em −EN +ω)t/ − 1 ei(Em −EN −ω)t/ − 1 + (Em − EN + ω) (Em − EN − ω)

|VN N |2 sin2 (ωt); (ω)2

(m = N ).

So

and in the rotating wave approximation

Em − EN ± ω |VmN |2 2 |cm | = t (m = N ). sin (Em − EN ± ω)2 2 / Again, ostensibly |cm |2 > 1, but the “extra” terms are of second order in H , and hence do not belong (to first order). / You would do better to use 1 − m=N |cm |2 . Schematically: cm = a1 H + a2 H 2 + · · · , so |cm |2 = 2

a21 H 2 + 2a1 a2 H 3 + · · · , whereas cN = 1 + b1 H + b2 H 2 + · · · , so |cN |2 = 1 + 2b1 H + (2b2 + b21 )H 2 + · · · . Thus knowing cm to first order (i.e., knowing a1 ) gets you |cm |2 to second order, but knowing cN to first order (i.e., b1 ) does not get you |cN |2 to second order (you’d also need / b22). It is precisely this b2 term that would cancel the “extra” (second-order) terms in the calculations of |cm | above.

Problem 9.17 (a) Equation 9.82 ⇒ c˙m = −

i c˙m = − cm V0 (t);

− i

cm (t) = cm (0)e

i cn Hmn ei(Em −En )t/ . Here Hmn = ψm |V0 (t)|ψn = δmn V0 (t). n

dcm i i = − V0 (t) dt ⇒ ln cm = − cm

t 0

V0 (t ) dt

1 . Let Φ(t) ≡ −

t

V0 (t ) dt + constant.

V0 (t ) dt ;

cm (t) = eiΦ cm (0). Hence

0

|cm (t)|2 = |cm (0)|2 , and there are no transitions.

Φ(T ) = −

1

T

V0 (t) dt. 0

(b) Eq. 9.84 ⇒ cN (t) ≈ 1 − i Eq. 9.85 ⇒ cm (t) = −

i

t

t

   

V0 (t ) dt = 1 + iΦ.

cN (t) = 1 + iΦ(t), cm (t) = 0 (m = N ).   δmN V0 (t )ei(Em −EN )t / dt = 0 (m = N ). 

0

0

The exact answer is cN (t) = eiΦ(t) , cm (t) = 0, and they are consistent, since eiΦ ≈ 1 + iΦ, to first order.


250


Problem 9.18 Use result of Problem 9.15(c). H12

Here

En =

n2 π 2 2 , 2ma2

E2 − E1 =

so

3π 2 2 . 2ma2

π 2π sin x V0 sin x dx a a 0

a/2

sin 3π 2V0 sin πa x π 1 3π 4V0 V0 a x = − sin − sin = . = a 2(π/a) 2(3π/a) π 2 3 2 3π

2 = a

a/2

0

Eq. 9.86 =⇒ P1→2 = 4

4V0 3π

2ma2 3π 2 2

2

2

sin

[Actually, in this case H11 and H22 are nonzero: a/2 π V0 2 H11 x dx = , = ψ1 |H |ψ1 = V0 sin2 a a 2 0

3π 2 t 4ma2

H22

=

16ma2 V0 sin 9π 3 2

2 = ψ2 |H |ψ2 = V0 a

2

3π 2 T 4ma2

.

a/2 2

sin 0

V0 2π x dx = . a 2

However, this does not affect the answer, for according to Problem 9.4, c1 (t) picks up an innocuous phase factor, while c2 (t) is not affected at all, in first order (formally, this is because Hbb is multiplied by cb , in Eq. 9.11, and in zeroth order cb (t) = 0).]

Problem 9.19 Spontaneous absorption would involve taking energy (a photon) from the ground state of the electromagnetic field. But you can’t do that, because the gound state already has the lowest allowed energy.

Problem 9.20 (a) H = −γB · S = −γ (Bx Sx + By Sy + Bz Sz ) ; γ 0 1 0 −i 1 0 H = −γ (Bx σx + By σy + Bz σz ) = − Bx + By + Bz 1 0 i 0 0 −1 2 2 γ γ Bx − iBy Brf (cos ωt + i sin ωt) Bz B0 =− =− −Bz −B0 2 Bx + iBy 2 Brf (cos ωt − i sin ωt) γ Brf eiωt B0 = − . −iωt −B0 2 Brf e (b) iχ˙ = Hχ ⇒

γ γ a˙ a Brf eiωt Brf eiωt b B0 B0 a i ˙ = − =− b b 2 Brf e−iωt −B0 2 Brf e−iωt a −B0 b 

  a˙ = i γ B0 a + Brf eiωt b = i Ωeiωt b + ω0 a , 2 2

γ i −iωt   b˙ = −i B0 b − Brf e−iωt a = a − ω0 b . Ωe 2 2

⇒



251

(c) You can decouple the equations by differentiating with respect to t, but it is simpler just to check the quoted results. First of all, they clearly satisfy the initial conditions: a(0) = a0 and b(0) = b0 . Differentiating a: iω ω i ω a˙ = a + −a0 sin(ω t/2) + [a0 (ω0 − ω) + b0 Ω] cos(ω t/2) eiωt/2 2 2 ω 2 i iωt/2 ω ωa0 cos(ω t/2) + i [a0 (ω0 − ω) + b0 Ω] sin(ω t/2) = e 2 ω + iω a0 sin(ω t/2) + [a0 (ω0 − ω) + b0 Ω] cos(ω t/2) Equation 9.90 says this should be equal to

i iωt i iωt/2 Ω Ωe b + ω0 a = e Ωb0 cos(ω t/2) + i [b0 (ω − ω0 ) + a0 Ω] sin(ω t/2) 2 2 ω ω 0 + ω0 a0 cos(ω t/2) + i [a0 (ω0 − ω) + b0 Ω] sin(ω t/2) . ω By inspection the cos(ω t/2) terms in the two expressions are equal; it remains to check that ω Ω ω0 [a0 (ω0 − ω) + b0 Ω] + iω a0 = i [b0 (ω − ω0 ) + a0 Ω] + i [a0 (ω0 − ω) + b0 Ω] , ω ω ω which is to say i

a0 ω(ω0 − ω) + b0 ωΩ + a0 (ω )2 = b0 Ω(ω − ω0 ) + a0 Ω2 + a0 ω0 (ω0 − ω) + b0 ω0 Ω, or

a0 ωω0 − ω 2 + (ω )2 − Ω2 − ω02 + ω0 ω = b0 [Ωω − ω0 Ω + ω0 Ω − ωΩ] = 0.

Substituting Eq. 9.91 for ω , the coefficient of a0 on the left becomes 2ωω0 − ω 2 + (ω − ω0 )2 + Ω2 − Ω2 − ω02 = 0. The check of b(t) is identical, with a ↔ b, ω0 → −ω0 , and ω → −ω. (d) Ω b(t) = i sin(ω t/2)e−iωt/2 ; ω

P (t) = |b(t)| = 2

Ω ω

2

sin2 (ω t/2).

(e) P(ω) 1

1/2

∆ω

ω0

ω

The maximum (Pmax = 1) occurs (obviously) at ω = ω0 . P =

1 2

⇒ (ω − ω0 )2 = Ω2 ⇒ ω = ω0 ± Ω,

so

∆ω = ω+ − ω− = 2Ω.


252


(f ) B0 = 10, 000 gauss = 1 T; Brf = 0.01 gauss = 1 × 10−6 T. ω0 = γB0 . Comparing Eqs. 4.156 and gp e , where gp = 5.59. So 6.85, γ = 2mp ω0 (5.59)(1.6 × 10−19 ) gp e B0 = = (1) = 4.26 × 107 Hz. 2π 4πmp 4π(1.67 × 10−27 ) 2Brf ∆ω Ω γ ∆ν = = (4.26 × 107 )(2 × 10−6 ) = 85.2 Hz. = = 2Brf = νres 2π π 2π B0

νres =

Problem 9.21 (a) H = −qE · r = −q(E0 · r)(k · r) sin(ωt). Write E0 = E0 n ˆ, k =

H = −q

ωˆ k. Then c

E0 ω qE0 ω =− (ˆ n · r)(kˆ · r) sin(ωt). Hba b|(ˆ n · r)(kˆ · r)|a sin(ωt). c c

This is the analog to Eq. 9.33: Hba = −qE0 b|ˆ n · r|a cos ωt. The rest of the analysis is identical to the dipole case (except that it is sin(ωt) instead of cos(ωt), but this amounts to resetting the clock, and clearly has no effect on the transition rate). We can skip therefore to Eq. 9.56, except for the factor of 1/3, which came from the averaging in Eq. 9.46:

A=

2 5 ω3 q2 ω2 ˆ · r)|a|2 = q ω |b|(ˆ |b|(ˆ n · r)( k n · r)(kˆ · r)|a|2 . π90 c3 c2 π90 c5

(b) Let the oscillator lie along the x direction, so (ˆ n · r) = n ˆ x x and kˆ · r = kˆx x. For a transition from n to n , we have 2 q 2 ω 5 ˆ A= k n |(a2+ +a+ a− +a− a+ +a2− )|n, n ˆ |n |x2 |n|2 . From Example 2.5, n |x2 |n = x x π90 c5 2m¯ ω where ω ¯ is the frequency of the oscillator, not to be confused with ω, the frequency of the electromagnetic wave. Now, for spontaneous emission the final state must be lower in energy, so n < n, and hence the only surviving term is a2− . Using Eq. 2.66: n |x2 |n =

n | n(n − 1)|n − 2 = n(n − 1) δn ,n−2 . 2m¯ ω 2m¯ ω

Evidently transitions only go from |n to |n − 2, and hence ω=

En − En−2 1 ω − (n − 2 + 12 )¯ ω = 2¯ ω. = (n + 12 )¯

n |x2 |n =

n(n − 1) δn ,n−2 ; mω

Rn→n−2 =

2 q2 ω5 ˆ 2 ( k n ˆ ) n(n − 1). x x π90 c5 m2 ω 2

It remains to calculate the average of (kˆx n ˆ x )2 . It’s easiest to reorient the oscillator along a direction rˆ, making angle θ with the z axis, and let the radiation be incident from the z direction (so kˆx → kˆr = cos θ). c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 9. TIME-DEPENDENT PERTURBATION THEORY Averaging over the two polarizations (î and ˆj): ˆ n2r = 1 2

1 2

î2r + ˆjr2 =

253 1 2

sin2 θ cos2 φ + sin2 θ sin2 φ =

sin2 θ. Now average overall directions: π 1 1 1 kˆr2 n ˆ 2r = (1 − cos2 θ) cos2 θ sin θ dθ sin2 θ cos2 θ sin θ dθ dφ = 2π 4π 2 8π 0

π 1 1 2 2 cos3 θ cos5 θ 1 = = − + − = . 4 3 5 4 3 5 15 0

R=

1 q 2 ω 3 n(n − 1). 15 π90 m2 c5

Comparing Eq. 9.63:

2 ω R(forbidden) = (n − 1) 2 . R(allowed) 5 mc

For a nonrelativistic system, ω mc2 ; hence the term “forbidden”. (c) If √ both the initial state and the final state have l = 0, the wave function is independent of angle (Y00 = 1/ 4π), and the angular part of the integral is: 4π ˆ ˆ (Eq. 6.95). a|(ˆ n · r)(k · r)|b = · · · (ˆ n · r)(kˆ · r) sin θ dθ dφ = · · · (ˆ n · k) 3 But n ˆ · kˆ = 0, since electromagnetic waves are transverse. for forbidden transitions.

So R = 0 in this case, both for allowed and

Problem 9.22 [This is done in Fermi’s Notes on Quantum Mechanics (Chicago, 1995), Section 24, but I am looking for a more accessible treatment.]


254

CHAPTER 10. THE ADIABATIC APPROXIMATION

Chapter 10

The Adiabatic Approximation Problem 10.1 (a) Let

(mvx − 2

2Eni at)/2w

= φ(x, t). Φn =

nπ 2 sin x eiφ , w w

so

nπ nπ nπ ∂φ √ ∂Φn 2 2 1 1 nπx iφ iφ v sin = 2 − x e + − 2 v cos x e + sin x i eiφ ∂t 2 w3/2 w w w w w w ∂t

nπ v nπxv ∂φ = − cot − x +i Φn . 2w w2 w ∂t

i

∂φ Ei a 1 2Eni a v v 2 i = − − 2 mvx − 2En at = − n − φ. ∂t 2 w w w w

nπ ∂Φn v nπxv Eni a v cot = −i + x + i + i φ Φn . ∂t 2w w2 w w w

2 ∂ 2 Φn HΦn = − . 2m ∂x2 ∂φ mvx = . ∂x w

∂Φn = ∂x

nπ nπ ∂φ 2 nπ 2 iφ iφ i cos x e + sin x e . w w w w w ∂x

nπ nπ ∂Φn mvx = cot x +i Φn . ∂x w w w

imb nπ nπ ∂ 2 Φn mvx 2 nπ 2 2 nπ x + Φn + cot x +i = − csc Φn . 2 ∂x w w w w w w So the Schr¨ odinger equation (i∂Φn /∂t = HΦn ) is satisfied ⇔

nπ v nπxv Eni a v −i cot + x +i +i φ 2w w2 w w w 2 =− 2m

imv nπ nπ nπ 2 mvx 2 2 nπ − x + + cot x +i csc w w w w w w


CHAPTER 10. THE ADIABATIC APPROXIMATION Cotangent terms: − i

nπxv w2

?

=−

255

2 nπ mvx nπvx 2 i = −i 2 . 2m w w w

Remaining trig terms on right:

−

nπ 2 w

csc2

nπ nπ 2 nπ nπ 2 1 − cos2 (nπx/w) nπ 2 cot2 . x + x =− = − w w w w w sin2 (nπx/w)

This leaves:

v Ei a v mvx2 − 2Eni at nπ 2 imv m2 v 2 x2 ? i +i n +i − − 2 2 = + 2w w w 2w 2m w w w 2 2 iv✁✁ Eni a mv 2✚ x✚ x✚ vEni at ? n2 π 2 iv mv 2✚ − − ✚ + + ✁✁ − ✚ =− w 2mw ✁2 ✁2 ✚2w ✚2w

−

Eni a E i a2 ? n2 π 2 n2 π 2 2 a2 n2 π 2 (w − vt) = − n = − ⇔− =− = r.h.s. 2 w w 2mw 2ma w 2mw

So Φn does satisfy the Schr¨ odinger equation, and since Φn (x, t) = (· · · ) sin (nπx/w), it fits the boundary conditions: Φn (0, t) = Φn (w, t) = 0. (b) Equation 10.4 =⇒ Ψ(x, 0) = Multiply by

2 sin a

cn Φn (x, 0) =

2 n π x e−imvx /2a a

cn

nπ 2 2 sin x eimvx /2a . a a

and integrate:

a nπ 2 π 2 2 nπ nπ Ψ(x, 0) sin cn sin x e−imvx /2a dx = x sin x dx = c n . a 0 a a 0 a a δnn

So, in general:

cn =

2 a

2 cn = π

a

a nπ 2 2 cn = e−imvx /2a sin x Ψ(x, 0)dx. In this particular case, a 0 a

e−imvx

0

π

2

/2a

sin

nπ a

sin

e−iαz sin(nz) sin(z) dz. 2

π x dx. Let a

π x ≡ z; a

dx =

a dz; π

mva 2 mvx2 mvz 2 a2 = = z . 2 2a 2a π 2π 2

QED

0


256


(c) e−iE1 t/ ⇒ ω =

w(Te ) = 2a ⇒ a + vTe = 2a ⇒ vTe = a ⇒ Te a/v;

Ti =

8π

2π 4 ma2 2ma2 = . 2 2 π π

mav 2π 2

= 8πα 1,

Ψ(x, t) =

so

Ti =

4ma2 . π

Adiabatic ⇒ Te Ti ⇒

α 1. Then cn =

2 π

E1 2π ⇒ Ti = = 2π , or ω E1 a 4ma2 4 mav ⇒ 1, or v π π

π

sin(nz) sin(z)dz = δn1 .

Therefore

0

πx 2 i 2 sin ei(mvx −2E1 at)/2w , w w

which (apart from a phase factor) is the ground state of the instantaneous well, of width w, as required 2 mva by the adiabatic theorem. (Actually, the first term in the exponent, which is at most mva 2a = 2 1 and could be dropped, in the adiabatic regime.) (d) t t 1 π2 π 2 2 1 1 dt = − − 2 2m 2m v a + vt 0 (a + vt ) 0 π2 1 1 π 2 vt π 2 t =− − =− =− . 2mv a w 2mv aw 2maw πx i mvx2 2 So (dropping the term, as explained in (c)) Ψ(x, t) = sin e−iE1 at/w 2w w w i 2 2 2 E at 2 π at π t πx iθ (since − 1 = − =− = θ): sin e . Ψ(x, t) = w 2ma2 w 2maw w w θ(t) = −

1

can be written

This is exactly what one would naively expect: For a fixed well (of width a) we’d have Ψ(x, t) = Ψ1 (x)e−iE1 t/ ; for the (adiabatically) expanding well, simply replace a by the (time-dependent) width w, and integrate to get the accumulated phase factor, noting that E1 is now a function of t.

Problem 10.2 To show:

i

∂χ = Hχ, ∂t

where χ is given by Eq. 10.31 and H is given by Eq. 10.25.

∂χ = ∂t  λt

λt

(ω1 −ω) λ − sin cos − i cos α2 e−iωt/2 − 2 2 λ 2   

λt

(ω1 +ω) λ sin α2 eiωt/2 + cos λt 2 − sin 2 − i λ 2

iω 2 iω 2

cos

λt

cos

λt

2

2

− −

i(ω1 −ω) λ i(ω1 +ω) λ

sin

λt

sin

λt

2

2

cos

α

sin

α

2

2

e−iωt/2 eiωt/2

   



257

Hχ =   (ω1 −ω) i(ω1 +ω) λt α −iωt/2 λt λt α iωt/2 −iωt ) − i sin( ) cos e + e sin α cos( ) − sin( ) sin e cos α cos( λt 2 λ 2 2 2 λ 2 2  ω1      2 (ω −ω) i(ω +ω) λt α −iωt/2 λt λt α iωt/2 1 1 eiωt cos α cos( λt ) − i sin( ) cos e − cos α cos( ) − sin( ) sin e 2 λ 2 2 2 λ 2 2 (1) Upper elements: λ (ω1 − ω) λt α iω i(ω1 − ω) λt α λt λt ✚ ✚ i − i cos cos − − sin cos − sin cos ✚ ✚ ✁ 2 λ 2 2 λ 2 2 2✁ ✚ 2 ✚ 2 ✁ λt α (ω1 − ω) λt α λt (ω1 + ω) λt ? ✁ ω1 cos = sin −i sin cos α cos✚✚ + cos −i sin sin α , 2 λ 2 2 λ 2 2 2 ✚ 2 ✁

where = 2 sin The sine terms:

sin

α α cos✚✚ 2✚ 2

λt 2

iω(ω1 − ω) ω1 (ω1 − ω) iω1 (ω1 + ω) ? 2 α = 0. −iλ − + cos α + 2 sin λ λ λ 2

i λt 2 ω✚ + (ω12 − ωω1 ) cos α + (ω12 + ωω1 )(1 − cos α) sin −ω✟2 − ω12 + 2ωω1 cos α − ωω1 + ✚ ✟ λ 2 i λt ✘✘ ✘✘ = − sin ωω✘ ω12✘ cos α − ωω1 cos α + ω12 + ✘ ωω✘ ω12✘ cos α − ωω1 cos α = 0. −ω✟12 + 2ωω1 cos α − ✘ 1 +✘ 1 −✘ ✟ λ 2 The cosine terms: λt α λt cos ω) + ✚ ω − ω1 cos α − ω1 2 sin2 (ω1 − ✚ = −ω1 cos [−1 + cos α + (1 − cos α)] = 0. 2 2 2 (2) Lower elements: α✟ iω α✟ λ (ω1 + ω) λt i(ω1 + ω) λt λt λt ✟ ✟ i −i cos sin + − sin sin − sin cos ✁ 2 λ 2 ✟✟ 2 2 λ 2 ✟✟ 2 2 2✁ ✁ α (ω1 − ω) λt α λt i(ω1 + ω) λt α λt ? ✁ ω1 ✚ ✚ = −i sin 2 sin✚ cos2 − cos − sin cos α sin✚ . cos 2 λ 2 2 2 λ 2 2 ✚ 2 ✚ 2 ✁ The sine terms:

α iω (ω + ω) λt iω(ω1 + ω) iω1 (ω1 − ω) ? 1 1 sin −iλ + + 2 cos2 − cos α = 0. 2 λ λ 2 λ i λt 2 2 − ω12 + 2ωω1 cos α + ωω1 + ✚ ω✚ + (ω12 − ωω1 )(1 + cos α) − (ω12 + ωω1 ) cos α sin −✚ ω✚ λ 2 i λt 2 ✘✘ ✘✘ = ωω✘ ωω✘ ω12✘ cos α − ωω1 cos α − ✘ ω12✘ cos α − ωω1 cos α = 0. sin −ω12 + 2ωω1 cos α + ✘ 1 + ω1 − ✘ 1 +✘ λ 2 The cosine terms: λt λt 2 α cos (ω1 + ✚ + ω1 cos α = cos [ω1 − ω1 (1 + cos α) + ω1 cos α] = 0. ω) − ✚ ω − ω1 2 cos 2 2 2 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

258

CHAPTER 10. THE ADIABATIC APPROXIMATION As for Eq. 10.33: λt ω (ω1 − ω cos α) λt λt cos α sin α2 cos −i sin e−iωt/2 iωt 2 α + i sin α sin e−iωt/2 e sin 2 −eiωt cos α2 2 λ 2 λ 2

=

α , β

with

λt iω1 λt α iω α α λt α= cos − sin cos + cos α cos + sin α sin e−iωt/2 sin 2 λ 2 2 λ 2 2 2 cos(α− α 2 )=cos

i(ω1 − ω) λt α λt − sin cos e−iωt/2 = cos 2 λ 2 2

α 2

(confirming the top entry).

λt λt iω1 λt α iω α α β= cos sin − sin sin + cos α sin − sin α cos eiωt/2 2 λ 2 2 λ 2 2 2 λt i(ω1 + ω) λt α = cos − sin sin eiωt/2 2 λ 2 2

sin( α 2 −α)=− sin

α 2

(confirming the bottom entry).

λt λt λt (ω1 − ω cos α)2 ω2 2 2 2 sin sin α sin + + 2 λ2 2 λ2 2 λt λt 1 = cos2 + 2 ω12 − 2ωω1 cos α + ω 2 cos2 α + ω 2 sin2 α sin2 2 λ 2

|c+ |2 + |c− |2 = cos2

= cos2

λt 2

+ sin2

λt 2

ω 2 +ω12 −2ωω1 cos α=λ2

= 1.

Problem 10.3 (a) ψn (x) =

nπ 2 sin x . In this case R = w. w w

nπ 2 nπ nπ √ ∂ψn 1 1 sin = 2 − x + − x ; x cos ∂R 2 w3/2 w w w2 w c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


259

1 w ∂ψn ∂ψn ψn ψn = dx ∂R ∂R 0 w nπ nπ nπ 1 2nπ w =− 2 sin2 x sin x dx − 3 x cos x dx w 0 w w w w 0

0

1 2nπ 2 sin( w x) 1 w nπ w 2nπ =− 2 − 3 x dx x sin w 2 w 0 w w 1 w 2 2nπ nπ wx 2nπ =− sin − x − cos x 2w w3 2nπ w 2nπ w 0

w2 1 nπ 1 1 − =− − cos(2nπ) = − + = 0. 2w w3 2nπ 2w 2w

So Eq. 10.42 =⇒ γn (t) = 0.

(If the eigenfunctions are real, the geometric phase vanishes.)

(b) Equation 10.39 =⇒ θn (t) =

θn = −

n2 π 2 2mv

w2

w1

1

t

0

n2 π 2 2 n2 π 2 dt = − 2 2mw 2m

1 n2 π 2 dw = w2 2mv

1 dt dw; w2 dw

w2 n2 π 2 1 1 1 . = − w w1 2mv w2 w1

(c) Zero.

Problem 10.4 √ ψ=

mα −mα|x|/2 e . Here R = α,

so

√ √ ∂ψ m 1 1 mα m|x| −mα|x|/2 −mα|x|/2 √ e + . e = − 2 ∂R 2 α ∂ψ ψ = ∂R

√

√ 2 mα 1 m m mα m m2 α −2mα|x|/2 − |x| e = − 4 |x| e−2mα|x|/ . 3 2 2 α 2

0 2 2 2

1 ∞ ∂ψ m2 α ∞ −2mαx/2 m m 2m2 α −2mαx/2 ψ e dx − 4 xe dx = 2 =2 − 4 ∂R 22 0 2mα 2mα 0 1 1 − = 0. So Eq. 10.42 =⇒ γ(t) = 0. = 2α 2α mα2 E=− 2 , 2

1 so θ(t) = −

0

T

mα2 − 2 2

dt =

m 23

α2

α1

α2

dt m dα = 3 dα 2 c

α2

α1

α2 dα =

m 3 α2 − α13 . 2 6 c


260


Problem 10.5 According to Eq. 10.44 the geometric phase is

Rf

γn (t) = i

ψn |∇R ψn · dR.

Ri

Now

ψn |ψn = 1,

so

∇R ψn |ψn = ∇R ψn |ψn + ψn |∇R ψn = ψn |∇R ψn ∗ + ψn |∇R ψn = 0, and hence ψn |∇R ψn is pure imaginary. If ψn is real, then, ψn |∇R ψn must in fact be zero. Suppose we introduce a phase factor to make the (originally real) wave function complex: ψn = eiφn (R) ψn ,

where ψn is real.

Then ∇R ψn = eiφn ∇R ψn + i(∇R φn )eiφn ψn . So

ψn |∇R ψn = e−iφn eiφn ψn |∇R ψn + ie−iφn (∇R φn )eiφn ψn |ψn . But ψn |ψn = 1, ψn |∇R ψn = 0 γn (t)

Rf

=i

(as we just found), so ψn |∇R ψn = i∇R φn ,

and

and Eq. 10.44 =⇒

i∇R (φn ) · dR = − [φn (Rf ) − φn (Ri )] , so Eq. 10.38 gives:

Ri

Ψ n (x, t) = ψn (x, t)e− i

t 0

En (t )dt −i[φn (Rf )−φn (Ri )]

e

.

The wave function picks up a (trivial) phase factor, whose only function is precisely to kill the phase factor we put in “by hand”: t i Ψ n (x, t) = ψn (x, t)e− 0 En (t )dt eiφn (Ri ) = Ψn (x, t)eiφn (Ri ) . In particular, for a closed loop φn (Rf ) = φn (Ri ), so γn (T ) = 0.

Problem 10.6 H=

e B · S. Here m

B = B0 sin θ cos φ î + sin θ sin φ ˆj + cos θ kˆ ; take spin matrices from Problem 4.31.

   √  2 0 0 010 0 −i 0 eB0  √ sin θ cos φ 1 0 1 + sin θ sin φ  i 0 −i + cos θ  0 0 √ 0  H= m 2 010 0 i 0 0 0 − 2  √ −iφ 2 cos θ e sin θ 0 eB0  iφ  e sin θ 0 e−iφ =√ √ sin θ . 2m iφ 0 e sin θ − 2 cos θ 



We need the “spin up” eigenvector:

Hχ+ =

eB0 χ+ . m

√ √      2 cos θa + e−iφ sin θb = √2a. 2 cos θ e−iφ sin θ 0 a a  (i) √  eiφ sin θ   b  = 2  b  =⇒ (ii) eiφ sin θa + e−iφ sin θc = 2b. 0 e−iφ √ sin θ √ √  iφ c c 0 e sin θ − 2 cos θ (iii) eiφ sin θb − 2 cos θc = 2c. √


CHAPTER 10. THE ADIABATIC APPROXIMATION (i) ⇒ b =

√

iφ

2e

1 − cos θ sin θ

a=

Thus c = e2iφ tan2 (θ/2) a; 2 |a| 1 + tan2 (θ/2) = |a|2 2

a = e−iφ cos2 (θ/2);

Pick

√

iφ

2e tan (θ/2) a;

(ii) is redundant. 1 cos(θ/2) then

 −iφ cos2 (θ/2) √ e χ+ =  2 sin (θ/2) cos (θ/2) . eiφ sin2 (θ/2)

261 (iii) ⇒ b =

Normalize:

√

−iφ

2e

1 + cos θ sin θ

c=

√

2e−iφ cot (θ/2) c.

|a|2 + 2 tan2 (θ/2)|a|2 + tan4 (θ/2)|a|2 = 1 ⇒

4

b=

= 1 ⇒ |a|2 = cos4 (θ/2) . √

2 sin(θ/2) cos(θ/2)

and

c = eiφ sin2 (θ/2),

and



This is the spin-1 analog to Eq. 10.57.

1 ∂χ+ ˆ 1 ∂χ+ ˆ ∂χ+ rˆ + θ+ φ ∂r r ∂θ r sin θ ∂φ     −e−iφ cos (θ/2) sin (θ/2) −ie−iφ cos2 (θ/2) 1 √ 2 1 ˆ   φ. 0 = 2 cos (θ/2) − sin2 (θ/2) /2 θˆ + r r sin θ 2 iφ iφ ie sin (θ/2) e sin (θ/2) cos (θ/2)

∇χ+ =

1− cos2 (θ/2) [cos (θ/2) sin (θ/2)] + sin (θ/2) cos (θ/2) cos2 (θ/2) − sin2 (θ/2) r . + sin2 (θ/2) [sin (θ/2) cos (θ/2)] θˆ . 1 - 2 + cos (θ/2) −i cos2 (θ/2) + sin2 (θ/2) i sin2 (θ/2) φˆ r sin θ i 4 = sin (θ/2) − cos4 (θ/2) φˆ r sin θ i 2 = sin (θ/2) + cos2 (θ/2) sin2 (θ/2) − cos2 (θ/2) φˆ r sin θ i i ˆ = (1)(− cos θ) φˆ = − cot θ φ. r sin θ r 1 ∂ i −i ∂ i sin θ i ∇ × χ+ |∇χ+ = sin θ − cot θ rˆ = 2 (cos θ) rˆ = 2 rˆ = 2 rˆ. r sin θ ∂θ r r sin θ ∂θ r sin θ r χ+ |∇χ+ =

Equation 10.51 =⇒ γ+ (T ) = i

i 2 r dΩ = −Ω. r2

Problem 10.7 (a) Giving H a test function f to act upon: 1 Hf = ∇ − qA · ∇f − qAf + qϕf 2m i i

1 q q = − 2 ∇ · (∇f ) − ∇ · (Af ) − A · (∇f ) + q 2 A · Af + qϕf. 2m i i (∇·A)f +A·(∇f )


262

CHAPTER 10. THE ADIABATIC APPROXIMATION But ∇ · A = 0 and ϕ = 0 (see comments after Eq. 10.66), so Hf =

1 2 2 − ∇ f + 2iqA · ∇f + q 2 A2 f , 2m

i∇

(b) Apply

But

− qA ·

∇ − qA i

or

H=

1 2 2 − ∇ + q 2 A2 + 2iqA · ∇ . QED 2m

to both sides of Eq. 10.78:

2 Ψ=

q ig ∇ − qA · e ∇Ψ = −2 ∇ · (eig ∇Ψ ) − eig A · ∇Ψ . i i i

∇ · (eig ∇Ψ ) = ieig (∇g) · (∇Ψ ) + eig ∇ · (∇Ψ )

∇g =

and

q A,

so the right side is

q −i2 eig A · ∇Ψ − 2 eig ∇2 Ψ + iqeig A · ∇Ψ = −2 eig ∇2 Ψ . QED

Problem 10.8 (a) Schr¨ odinger equation: 2 d 2 ψ − = Eψ, 2m dx2

or

d2 ψ = −k 2 ψ dx2

(k ≡

√

2mE/)

0 < x < 12 a + 9, 1 2 a + 9 < x < a.

Boundary conditions: ψ(0) = ψ( 12 a + 9) = ψ(a) = 0. Solution: (1) 0 < x < 12 a + 9 :

ψ(x) = A sin kx + B cos kx. But ψ(0) = 0 ⇒ B = 0,

k( 12 a + 9) = nπ ψ( 12 a + 9) = 0 ⇒ or else A = 0. (2) 12 a + 9 < x < a : ψ( 12 a

and

(n = 1, 2, 3, . . . ) ⇒ En = n2 π 2 2 /2m(a/2 + 9)2 ,

ψ(x) = F sin k(a − x) + G cos k(a − x). But ψ(a) = 0 ⇒ G = 0,

k( 12 a − 9) = n π + 9) = 0 ⇒ or else F = 0.

and

(n = 1, 2, 3, . . . ) ⇒ En = (n )2 π 2 2 /2m(a/2 − 9)2 ,

   either E1 = The ground state energy is

  or else

π 2 2 (n = 1), with F = 0, 2m( 12 a + 9)2 π 2 2 E1 = (n = 1), with A = 0. 2m( 12 a − 9)2

Both are allowed energies, but E1 is (slightly) lower (assuming 9 is positive), so the ground state is   ψ(x) =

0,

2 1 2 a+5

sin

πx 1 2 a+5

, 0 ≤ x ≤ 12 a + 9; 1 2a

+ 9 ≤ x ≤ a.



263

ψ(x)

a x

a_ a_ + ε 2 2

(b) 2 d 2 ψ − +f (t)δ(x− 12 a−9)ψ = Eψ 2m dx2

⇒

ψ(x) =

A sin kx, 0 ≤ x < 12 a + 9, F sin k(a − x), 12 a + 9 < x ≤ a,

√

where

k≡

2mE .

Continuity in ψ at x = 12 a + 9 : 1

A sin k 2 a + 9 = F sin k a − 12 a − 9 = F sin k 12 a − 9

sin k 12 a + 9

. ⇒F =A sin k 12 a − 9

Discontinuity in ψ at x = 12 a + 9 (Eq. 2.125): −F k cos k(a−x)−Ak cos kx =

2mf A sin kx ⇒ F cos k 12 a − 9 +A cos k 12 a + 9 = − 2

2mf 2 k

A sin k

1

2a

sin k 12 a + 9 2T 1

cos k 12 a − 9 + A cos k 12 a + 9 = − A sin k 12 a + 9 . z sin k 2 a − 9

2T sin k 12 a + 9 cos k 12 a − 9 + cos k 12 a + 9 sin k 12 a − 9 = − sin k 12 a + 9 sin k 12 a − 9 . z 1

2T 1 1 sin k 2 a + 9 + 2 a − 9 = − cos k 12 a + 9 − 12 a + 9 − cos k 12 a + 9 + 12 a − 9 . z 2 A

sin ka = −

T (cos 2k9 − cos ka) z

(c) sin z =

T (cos z − 1) z

⇒

⇒

z sin z = T [cos z − cos(zδ)].

z cos z − 1 = = − tan(z/2) T sin z

⇒

z tan(z/2) = − . T

Plot tan(z/2) and −z/T on the same graph, and look for intersections:

tan(z/2) π

2π

3π

z

-z/T


+9 .

264

CHAPTER 10. THE ADIABATIC APPROXIMATION As t : 0 → ∞, T : 0 → ∞, and the straight line rotates counterclockwise from 6 o’clock to 3 o’clock, 2 π 2 so the smallest z goes from π to 2π, and the ground state energy goes from ka = π ⇒ E(0) = 2ma2 2 π 2 (appropriate to a well of width a) to ka = 2π ⇒ E(∞) = (appropriate for a well of width a/2. 2m(a/2)2 T z

(d) Mathematica yields the following table: (e) Pr =

Ir 1 , = Ir + Il 1 + (Il /Ir )

0 3.14159

1 3.67303

5 4.76031

20 5.72036

100 6.13523

1000 6.21452

where

a/2+5 1 1 Il = x− sin(2kx) A sin kx dx = A 2 4k 0 0 a a a 1 1 29 1 29 2 2 +9 − sin 2k +9 = A 1+ − sin ka + ka =A 2 2 4k 2 4 a ka a

a 1 = A2 1 + δ − sin(z + zδ) . 4 z

a/2+5

2

2

2

a

F 2 sin2 k(a − x) dx. Let

Ir = a/2+5

0

= −F 2

u ≡ a − x, du = −dx.

a/2−5

sin2 ku du = F 2

sin2 ku du = 0

a/2−5

1 a 2 F 1 − δ − sin(z − zδ) . 4 z

Il A2 A2 [1 + δ − (1/z) sin(z + zδ)] sin2 k(a/2 − 9) sin2 [z(1 − δ)/2] . But (from (b)) = 2 = = . Ir F [1 − δ − (1/z) sin(z − zδ)] F2 sin2 k(a/2 + 9) sin2 [z(1 + δ)/2]

I+ 1 1 = , where I± ≡ 1 ± δ − sin z(1 ± δ) sin2 [z(1 ∓ δ)/2]. Pr = . I− z 1 + (I+ /I− ) Using δ = 0.01 and the z’s from (d), Mathematica gives T Pr

0 0.490001

1 0.486822

5 0.471116

20 0.401313

100 0.146529

1000 0.00248443

As t : 0 → ∞ (so T : 0 → ∞), the probability of being in the right half drops from almost 1/2 to zero—the particle gets sucked out of the slightly smaller side, as it heads for the ground state in (a). (f )

T=0

T=1

T=5

T=20

T=100

T=1000



Problem 10.9 (a) Check the answer given:

t

xc = ω

f (t ) sin [ω(t − t )] dt =⇒ xc (0) = 0.

0

265

t

x˙ c = ωf (t) sin [ω(t − t)] + ω 2

f (t ) cos [ω(t − t )] dt = ω 2

0

x ¨c = ω f (t) cos [ω(t − t)] − ω

t

f (t ) cos [ω(t − t )] dt ⇒ x˙ c (0) = 0.

0

2

3

t

f (t ) sin [ω(t − t )] dt = ω 2 f (t) − ω 2 xc .

0

Now the classical equation of motion is m(d2 x/dt2 ) = −mω 2 x + mω 2 f . For the proposed solution, m(d2 xc /dt2 ) = mω 2 f − mω 2 xc , so it does satisfy the equation of motion, with the appropriate boundary conditions. (b) Let

z ≡ x − xc

(so ψn (x − xc ) = ψn (z), and z depends on t as well as x).

∂Ψ i dψn 1 m xc mω 2 = (−x˙ c )ei{} + ψn ei{} − (n + )ω + m¨ xc (x − ) − x˙ 2c + f xc ∂t dz 2 2 2 2

1 x˙ 2 mω 2 [ ] = −(n + )ω + 2x(f − xc ) + x2c − c2 . 2 2 ω

∂Ψ dψn i{} 1 x˙ 2c mω 2 2 . = −x˙ c e + iΨ −(n + )ω + 2x(f − xc ) + xc − 2 ∂t dz 2 2 ω ∂Ψ i dψn i{} = e + ψn ei{} (mx˙ c ); ∂x dz

∂2Ψ d2 ψn i{} dψn i{} i = e +2 e (mx˙ c ) − 2 2 ∂x dz dz

mx˙ c

2 ψn ei{} .

2 ∂ 2 Ψ 1 + mω 2 x2 Ψ − mω 2 f xΨ 2m ∂x2 2 2 2 d2 ψn i{} 2 dψn i{} imx˙ c 1 2 mx˙ c =− e − Ψ + mω 2 x2 Ψ − mω 2 f xΨ. 2 e + 2 2m dz 2m dz 2m 2

HΨ = −

But −

2 d 2 ψ n 1 1 + mω 2 z 2 ψn = (n + )ωψn , 2m dz 2 2 2

so

✟ 1 ✟✟ 1 dΨ✟ m 1 n HΨ = (n✟ +✟ )ωΨ − mω 2 z 2 Ψ − ix ˙ c ✟ ei{} + x˙ 2c Ψ + mω 2 x2 Ψ − mω 2 f xΨ ✟ 2 2 2 ✟ dz ✟ 2

✟ dψ ∂Ψ 1 ✟ mω 2 1 2 ✟ ✟ ? n i{} 2 ✟ ✟ = i − Ψ −(n + + x − x ˙ ) = −i x ˙ e )ω + (2xf − 2xx c c c ∂t 2 ω2 c ✟✟ dz ✟✟ 2 2 1 m 1 ✘ ? mω ✘2✘ − mω 2 z 2 + x˙ 2c + mω 2 x2 − ✘ mω fx = − 2 2 2 2

?

z 2 − x2 = −2xxc + x2c ;

' ✟ − 2xxc + 2xf ✟

x2c

1 − 2 x˙ 2c ω

(

?

z 2 = (x2 − 2xxc + x2c ) = (x − xc )2 .


266


(c) Eq. 10.90 ⇒ H = − H=

−

1 2 ∂ 2 1 + mω 2 x2 − 2xf + f 2 − mω 2 f 2 . Shift origin: 2 2m ∂x 2 2

u ≡ x − f.

2 ∂ 2 1 1 2 2 2 2 − . + u f mω mω 2m ∂u2 2 2

The first term is a simple harmonic oscillator in the variable u; the second is a constant (with respect to position). So the eigenfunctions are ψn (u), and the eigenvalues are harmonic oscillator ones, (n + 1 less the constant: En = (n + 12 )ω − 12 mω 2 f 2 . 2 )ω, t 1 d d (d) Note that sin [ω(t − t )] = cos [ω(t − t )], so x (t) = f (t ) cos [ω(t − t )] dt , or c ω dt dt 0 t t t df df xc (t) = f (t ) cos [ω(t − t )] − cos [ω(t − t )] dt = f (t) − cos [ω(t − t )] dt dt dt 0 0 0 (since f (0) = 0). Now, for an adiabatic process we want df /dt very small; specifically:

df ωf (t) dt

(0 < t ≤ t). Then the integral is negligible compared to f (t), and we have xc (t) ≈ f (t). (Physically, this says that if you pull on the spring very gently, no fancy oscillations will occur; the mass just moves along as though attached to a string of fixed length.) (e) Put xc ≈ f

into Eq. 10.92, using Eq. 10.93: i

Ψ(x, t) = ψn (x, t)e

2 −(n+ 12 )ωt+mf˙(x−f /2)+ mω 2

t 0

f 2 (t )dt

.

The dynamic phase (Eq. 10.39) is 1 θn (t) = −

0

t

1 mω 2 En (t ) dt = −(n + )ωt + 2 2

t

f 2 (t ) dt ,

Ψ(x, t) = ψn (x, t)eiθn (t) eiγn (t) ,

so

0

˙ confirming Eq. 10.94, with the geometric phase given (ostensibly) by γn (t) = m f (x − f /2). But the eigenfunctions here are real, and hence(Problem 10.5) the geometric phase should be zero. The point is that ˙ (in the adiabatic approximation) f˙ is extremely small (see above), and hence in this limit m f (x−f /2) ≈ 0 (at least, in the only region of x where ψn (x, t) is nonzero).

Problem 10.10 (a) c˙m = −

j

∂ψn iγn i(θn −θm ) δjn eiγn ψm |ψ˙j ei(θj −θm ) = −ψm | e e ∂t cm (t) = cm (0) −

t

ψm | 0

⇒

∂ψn iγn i(θn −θm ) e e dt . ∂t



267

(b) From Problem 10.9: ψn (x, t) = ψn (x − f ) = ψn (u),

where

u ≡ x − f,

and ψn (u) is the nth state of the ordinary harmonic oscillator;

But

pˆ =

∂ , i ∂u

m|p|n = i

so

ψm |

i ∂ψn = − f˙m|p|n, ∂t

∂ψn ∂u ∂ψn ∂ψn = = −f˙ . ∂t ∂u ∂t ∂u

where (from Problem 3.33):

√ mω √ mδn,m−1 − nδm,n−1 . Thus: 2

√ mω √ ∂ψn ˙ ψm | =f mδn,m−1 − nδm,n−1 . ∂t 2 Evidently transitions occur only to the immediately adjacent states, n ± 1, and (1) m = n + 1 : cn+1 = −

t mω √ f˙ n + 1 eiγn ei(θn −θn+1 ) dt . 2 0

But γn = 0, because the eigenfunctions are real (Problem 10.5), and (Eq. 10.39)

1 1 1 1 θn = − (n + )ωt =⇒ θn − θn+1 = −(n + ) + (n + 1 + ) ωt = ωt. 2 2 2

So

cn+1

t mω √ =− n+1 f˙eiωt dt . 2 0

(2) m = n − 1: t cn−1 = − 0

mω √ ˙ −f n eiγn ei(θn −θn−1 ) dt ; 2

1 1 θn − θn−1 = −(n + ) + (n − 1 + ) ωt = −ωt. 2 2

cn−1 =

mω √ n 2

t

f˙e−iωt dt .

0


268

CHAPTER 11. SCATTERING

Chapter 11

Scattering Problem 11.1 (a)

q1 b

θ φ

r q2

Conservation of energy:

E=

1 m(r˙ + r2 φ˙ 2 ) + V (r), 2

˙ So Conservation of angular momentum: J = mr2 φ. r˙ 2 +

du dφ

V (r) =

2

1 − 2 u

2m = 2 (E − V ) − u2 ; J

du J 2 J du u =− . Then: dφ m m dφ

du = dφ

q1 q2 1 . 4π9 r

J φ˙ = . mr2

J2 2 = (E − V ). We want r as a function of φ (not t). 2 2 m r m

dr dr du dφ r˙ = = = dt du dφ dt

where

2m (E − V ) − u2 ; J2

Also, let u ≡ 1/r.

J du − m dφ

dφ =

2 +

Then

J2 2 2 u = (E − V ), or m2 m

du

du = , 2m I(u) 2 J 2 (E − V ) − u

where


CHAPTER 11. SCATTERING I(u) ≡

269

2m (E − V ) − u2 . Now, the particle q1 starts out at r = ∞ (u = 0), φ = 0, and the point J2

of closest approach is rmin (umax ), Φ :

umax

Φ= 0

du √ . It now swings through an equal angle Φ I

on the way out, so

Φ + Φ + θ = π,

or

θ = π − 2Φ.

θ =π−2

umax

0

du . I(u)

So far this is general ; now we put in the specific potential: 2mE 2m q1 q2 − 2 u − u2 = (u2 − u)(u − u1 ), where u1 and u2 are the two roots. J2 J 4π90 (Since du/dφ = I(u), umax is one of the roots; setting u2 > u1 , umax = u2 .) I(u) =

θ =π−2

u2

du

= π + 2 sin−1

u −2u + u1 + u2 2 u2 − u1 0

(u2 − u)(u − u1 ) 0 u1 + u2 −1 −1 = π + 2 sin (−1) − sin u2 − u1 π u u1 + u2 + u 1 2 −1 −1 = π + 2 − − sin = −2 sin . 2 u2 − u1 u2 − u1

Now J = mvb, E = 12 mv 2 , where v is the incoming velocity, so J 2 = m2 b2 (2E/m) = 2mb2 E, and hence 2m/J 2 = 1/b2 E. So 1 1 1 q1 q2 q1 q2 A 1 I(u) = 2 − 2 , so − I(u) = u2 + 2 u − 2 . u − u2 . Let A ≡ b b E 4π90 4π90 E b b

To get the roots:

Thus

  2 2 A A 1 1 4 2b A A . u2 + 2 u − 2 = 0 =⇒ u = + 2 = 2 −1 ± 1 + − 2± b b 2 b b4 b 2b A

  2 A  2b  u2 = 2 −1 + 1 + , 2b A  −1

θ = 2 sin



 1 2

,

or

1 + (2b/A)

2b A

2 =

  2 A  2b  u1 = 2 −1 − 1 + ; 2b A

cos2 (θ/2) 1 − sin2 (θ/2) = ; sin2 (θ/2) sin2 (θ/2)

1 2

1 + (2b/A)

θ = sin ; 2

2b = cot(θ/2), A

or

1+

b=

2b A

u1 + u2 = u2 − u1

2 =

−1 2

1 + (2b/A)

1 ; sin (θ/2) 2

q1 q2 cot(θ/2). 8π90 E


.

270


(b) b D(θ) = sin θ = (c)

db 1 . Here db = q1 q2 − . dθ dθ 8π90 E 2 sin2 (θ/2)

1 q1 q2 cos(θ/2) q1 q2 1 = 2 sin(θ/2) cos(θ/2) 8π90 E sin(θ/2) 8π90 E 2 sin2 (θ/2)

σ=

D(θ) sin θ dθ dφ = 2π

q1 q2 8π90 E

2

π

0

q1 q2 16π90 E sin2 (θ/2)

2 .

sin θ dθ. sin4 (θ/2)

This integral does not converge, for near θ = 5 0 (and again near π) we have sin θ ≈ θ, sin(θ/2) ≈ θ/2, so 5 the integral goes like 16 0 θ−3 dθ = − 8θ−2 0 → ∞.

Problem 11.2 r θ x

Two dimensions:

eikr ψ(r, θ) ≈ A eikx + f (θ) √ . r

One dimension:

ψ(x) ≈ A eikx + f (x/|x|)e−ikx .

Problem 11.3 Multiply Eq. 11.32 by Pl (cos θ) sin θ dθ and integrate from 0 to π, exploiting the orthogonality of the Legendre polynomials (Eq. 4.34)—which, with the change of variables x ≡ cos θ, says π 2 Pl (cos θ)Pl (cos θ) sin θ dθ = δll . 2l + 1 0 The delta function collapses the sum, and we get (1) 2il jl (ka) + ikal hl (ka) = 0, and hence (dropping the primes) al = −

jl (ka) (1)

ikhl (ka)

. QED



271

Problem 11.4 Keeping only the l = 0 terms, Eq. 11.29 says that in the exterior region:

eikr eikr sin(kr) sin(kr) (1) ψ ≈ A j0 (kr) + ika0 h0 (kr) P0 (cos θ) = A + ika0 −i =A + a0 kr kr kr r

(r > a).

In the internal region Eq. 11.18 (with nl eliminated because it blows up at the origin) yields ψ(r) ≈ bj0 (kr) = b

sin(kr) kr

(r < a).

The boundary conditions hold independently for each l, as you can check by keeping the summation over l and exploiting the orthogonality of the Legendre polynomials:

sinka eika sin ka (1) ψ continuous at r = a: A + a0 =b . ka a ka (2) ψ discontinuous at r = a: Integrating the radial equation across the delta function gives

2 2 2 2 l(l + 1) d u − u dr ⇒ − dr + αδ(r − a) + ∆u + αu(a) = 0, or 2 2 2m dr 2m r 2m Now

u = rR,

so u = R+rR ;

∆u = ∆R+a∆R = a∆R =

∆u =

2mα u(a). 2

2mα 2mα β aR(a), or ∆ψ = 2 ψ(a) = ψ(a). 2 a

✭✭ A b ✟✟✟ β sin(ka) A b ✭ika ✭a✭ sin ka = b k cos(ka) + a0 ik 2 eika − 2 ✭sin(ka) k cos(ka) + ✟ . − ✭✭+ 0 ke ✭ ka ka ka ka2 a ka ✟

β A cos(ka) + ia0 keika = b cos(ka) + sin(ka) . ka

The indicated terms cancel (by (1)), leaving

Using (1) to eliminate b:

β A cos(ka) + ia0 keika = cot(ka) + sin(ka) + a0 keika A. ka

✘ + ia keika = cos(ka) ✘ + β sin(ka) + a k cot(ka)eika + β a0 eika . ✘ ✘✘ cos(ka) 0 0 ✘ ✘✘ ka a ika

ia0 ke

β β cos(ka) 1 1 + i cot(ka) + i = sin(ka). But ka 1, so sin(ka) ≈ ka, and cot(ka) = ≈ . ka ka sin(ka) ka

i ia0 k(1 + ika) 1 + (1 + β) = β; ka

a0 = −

aβ . 1+β

i i ia0 k 1 + (1 + β) + ika − 1 − β ≈ ia0 k (1 + β) = β. ka ka

Equation 11.25 ⇒ f (θ) ≈ a0 = −

Equation 11.27

⇒ σ = 4πD = 4π

aβ 1+β

aβ . 1+β

Equation 11.14 ⇒ D = |f |2 =

aβ 1+β

2

2 .


.

272


Problem 11.5 (a) In the region to the left

ψ(x) = Aeikx + B −ikx

(x ≤ −a).

In the region −a < x < 0, the Schr¨ odinger equation gives − where k =

2m(E + V0 )/.

h2 d 2 ψ − V0 ψ = Eψ 2m dx2

⇒

d2 ψ = −k ψ dx2

The general solution is ψ = C sin(k x) + D cos(k x)

But ψ(0) = 0 implies D = 0, so ψ(x) = C sin(k x)

(−a ≤ x ≤ 0).

The continuity of ψ(x) and ψ (x) at x = −a says Ae−ika + Beika = −C sin(k a), Divide and solve for B:

ikAe−ika − ikB ika = k C cos(k a).

ikAe−ika − ikBeika = −k cot(k a), Ae−ika + Beika

ikAe−ika − ikBeika = −Ae−ika k cot(k a) − Beika k cot(k a), Beika [−ik + k cot(k a)] = Ae−ika [−ik − k cot(k a)] .

−2ika k − ik cot(k a) . B = Ae k + ik cot(k a) (b)

|B|2 = |A|2

k − ik cot(k a) k + ik cot(k a) · = |A|2 . k + ik cot(k a) k − ik cot(k a)

(c) From part (a) the wave function for x < −a is ikx

ψ(x) = Ae

−2ika

+ Ae

k − ik cot(k a) −ikx e . k + ik cot(k a)

But by definition of the phase shift (Eq. 11.40) ψ(x) = A eikx − ei(2δ−kx) .

k − ik cot(k a) = −e2iδ . k + ik cot(k a) √ This is exact. For a very deep well, E V0 , k = 2mE/ 2m(E + V0 )/ = k , so

−2ika −ik cot(k a) e δ = −ka. = −e2iδ ; e−2ika = e2iδ ; ik cot(k a)

so

e−2ika



273

Problem 11.6 1 iδl e sin δl , k

From Eq. 11.46,

al =

and Eq. 11.33,

But (Eq. 11.19)

hl (x) = jl (x) + inl (x),

(1)

eiδl sin δl = i

al = i

jl (ka) (1) khl (ka)

, it follows that eiδl sin δl = i

jl (ka) (1)

hl (ka)

.

so

1 1 − i(n/j) jl (ka) (n/j) + i =i =i = , jl (x) + inl (x) 1 + i(n/j) 1 + (n/j)2 1 + (n/j)2

(writing (n/j) as shorthand for nl (ka)/jl (ka)). Equating the real and imaginary parts: cos δl sin δl =

(n/j) ; 1 + (n/j)2

sin2 δl =

1 . 1 + (n/j)2

Dividing the second by the first, I conclude that 1 tan δl = , (n/j)

or

−1

δl = tan

jl (ka) . nl (ka)

Problem 11.7 r > a : u(r) = A sin(kr + δ); r < a : u(r) = B sin kr + D cos kr = B sin kr,

because u(0) = 0 =⇒ D = 0.

Continuity at r = a =⇒ B sin(ka) = A sin(ka + δ) =⇒ B = A

sin(ka + δ) sin(ka + δ) . So u(r) = A sin kr. sin(ka) sin(ka)

From Problem 11.4, du sin(ka + δ) β β ∆ k cos(ka) = A sin(ka + δ). = u(a) ⇒ Ak cos(ka + δ) − A dr r=a a sin(ka) a cos(ka + δ) −

sin(ka + δ) β cos(ka) = sin(ka + δ), sin(ka) ka

sin(ka) cos(ka + δ) − sin(ka + δ) cos(ka) = sin(ka − ka − δ) =

− sin δ = β

β sin(ka) [sin(ka) cos δ + cos(ka) sin δ] , ka

sin2 (ka) [cos δ + cot(ka) sin δ] ; ka

cot δ = − cot(ka) −

β sin(ka + δ) sin(ka), ka

ka ; β sin2 (ka)

−1 = β

sin2 (ka) [cot δ + cot(ka)] . ka

cot δ = − cot(ka) +

ka . β sin2 (ka)


274


Problem 11.8 eikr 1 =⇒ ∇G = − 4πr 4π

G=−

1 1 ikr ∇e + eikr ∇ r r

=⇒

1 1 1 1 2 ikr ikr ikr 2 ∇ G = ∇ · (∇G) = − 2 ∇ · (∇e ) + ∇ (e ) + e ∇ . 4π r r r 2

But ∇

1 1 = − 2 rˆ; r r

∇(eikr ) = ikeikr rˆ;

(see reference in footnote 12) =⇒ ∇ e

2 ikr

1 ∇ = −4πδ 3 (r). So r 2

But

∇2 eikr = ik∇ · (eikr rˆ) = ik

1 d 2 ikr (r e ) r2 dr

ik = 2 (2reikr + ikr2 eikr ) = ikeikr r

2 + ik ; r

ikr 1 ikr 2 1 1 ikr 3 2 − 2 rˆ · ike rˆ + ike + ik − 4πe δ (r) . ∇ G=− 4π r r r 2

eikr δ 3 (r) = δ 3 (r), so

2ik 2ik k 2 eikr 1 ikr ∇ G = δ (r) − − 2 + 2 − e = δ 3 (r) + k 2 = δ 3 (r) − k 2 G. 4π r r r 4πr 2

3

(∇2 + k 2 )G = δ 3 (r). QED

Therefore

Problem 11.9 ψ=√

1 πa3

−r/a

e

;

e2 2 1 V =− =− 4π90 r ma r

√ (Eq. 4.72);

k=i

−2mE i = . a

In this case there is no “incoming” wave, and ψ0 (r) = 0. Our problem is to show that ik|r−r0 | m e − V (r0 )ψ(r0 ) d3 r0 = ψ(r). 2 2π |r − r0 | We proceed to evaluate the left side (call it I): −|r−r0 |/a m e 2 1 1 −r0 /a 3 √ I= − e d r0 − 3 2π2 ma |r − r | r 0 0 πa √ − r2 +r2 −2rr0 cos θ/a −r0 /a 0 e e 1 1 √ = r02 sin θ dr0 dθ dφ. 2 2 2πa πa3 r + r0 − 2rr0 cos θ r0 (I have set the z0 axis along the—fixed—direction r, for convenience.) Doing the φ integral (2π): √ ∞ π − r 2 +r02 −2rr0 cos θ/a 1 e −r0 /a I= √ sin θ dθ dr0 . The θ integral is r0 e r2 + r02 − 2rr0 cos θ a πa3 0 0 c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


π

0

275

√2 2 e− r +r0 −2rr0 cos θ/a a −(r+r0 )/a a −√r2 +r02 −2rr0 cos θ/a π −|r−r0 |/a e . e = − − e sin θ dθ = − rr0 rr0 0 r2 + r02 − 2rr0 cos θ

1 I=− √ r πa3 1 =− √ r πa3 1 =− √ r πa3 1 =− √ r πa3

e−r0 /a e−(r0 +r)/a − e−|r0 −r|/a dr0 0 ∞ r e−r/a e−2r0 /a dr0 − e−r/a dr − er/a ∞

0

0

e−2r0 /a dr0

r

a ∞ e − e−2r0 /a (r) − e −e 2 2 r a 1 a r/a −2r/a −r/a −r/a −r/a =√ − re − e e e = ψ(r). QED e 2 2 πa3 −r/a

a

∞

−r/a

r/a

Problem 11.10 For the potential in Eq. 11.81, Eq. 11.88 =⇒ 2m f (θ) = − 2 V0 κ

0

a

2mV0 r sin(κr) dr = − 2 κ

1 (κa)3 ; 3!

a 1 r 2mV0 sin(κr) − cos(κr) = − 2 3 [sin(κa) − κa cos(κa)] , κ2 κ κ 0

For low-energy scattering (ka 1):

where (Eq. 11.89) κ = 2k sin(θ/2). sin(κa) ≈ κa −

1 cos(κa) = 1 − (κa)2 ; 2

so

2mV0 1 1 2 mV0 a3 3 3 f (θ) ≈ − 2 3 κa − (κa) − κa + (κa) = − , κ 6 2 3 2

in agreement with Eq. 11.82.

Problem 11.11 sin(κr) =

=

So

1 2i

1 iκr e − e−iκr , 2i

so

∞

e−µr sin(κr) dr =

0

1 2i

∞

e−(µ−iκ)r − e−(µ+iκ)r dr

0

∞

κ 1 e−(µ−iκ)r e−(µ+iκ)r 1 1 1 µ + iκ − µ + iκ = 2 = . − − = −(µ − iκ) −(µ + iκ) 0 2i µ − iκ µ + iκ 2i µ2 + κ2 µ + κ2

f (θ) = −

2mβ 2mβ κ =− 2 2 . QED 2 κ µ2 + κ2 (µ + κ2 )


276


Problem 11.12 Equation 11.91 =⇒ D(θ) = |f (θ)|2 =

σ=

D(θ) sin θ dθ dφ = 2π

2k sin(θ/2) ≡ x, µ

Let

σ = 2π

2

2mβ 2

so

2mβ 2

2

2mβ 2

1 µ4

2 sin(θ/2) =

2

1 , (µ2 + κ2 )2

π

0

where Eq. 11.89 ⇒ κ = 2k sin(θ/2).

1

2 2 sin(θ/2) cos(θ/2) dθ. 2 1 + (2k/µ) sin2 (θ/2)

µ x, k

and

cos(θ/2) dθ =

µ dx. Then k

1 µ 2 x1 x θ = 0 =⇒ x = x0 = 0, dx. The limits are 2 2 θ = π =⇒ x = x1 = 2k/µ. µ4 k x0 (1 + x )

So

2k/µ 2 2

2mβ 1 1 1 1 1 2mβ σ = 2π − 1− =π 2 (µk)2 2 (1 + x2 ) 0 2 (µk)2 1 + (2k/µ)2 2

2 2mβ 1 1 1 4(k/µ)2 4mβ 2mE =π = π . But k 2 = , so 2 (µk)2 1 + 4k 2 /µ2 2 µ2 µ2 + 4k 2 2 σ=π

4mβ µ

2 (µk)2

1 . + 8mE

Problem 11.13 (a) V (r) = αδ(r − a). Eq. 11.80 =⇒ f = −

2mα f = − 2 a2 ;

D = |f | = 2

2mα 2 a 2

m 2π2

V (r) d3 r = −

2

m α4π 2π2

;

σ = 4πD = π

4mα 2 a 2

∞

δ(r − a)r2 dr.

0

2 .

(b) Eq. 11.88 =⇒ f = −

2m α 2 κ

0

∞

rδ(r − a) sin(κr) dr = −

2mα a sin(κa) 2 κ

(κ = 2k sin(θ/2)).

(c) Note first that (b) reduces to (a) in the low-energy regime (ka 1 =⇒ κa 1). Since Problem 11.4 was also for low energy, what we must confirm is that Problem 11.4 reproduces (a) in the regime for which the Born approximation holds. Inspection shows that the answer to Problem 11.4 does reduce to f = −2mαa2 /2 when β 1, which is to say when f /a 1. This is the appropriate condition, since (Eq. 11.12) f /a is a measure of the relative size of the scattered wave, in the interaction region.



277

Problem 11.14 F=

1 q1 q2 rˆ; 4π90 r2

F⊥ =

1 q1 q2 cos φ; 4π90 r2

b cos φ = , r

so

1 q1 q2 b ; 4π90 r3

F⊥ =

dt =

dx . v

F φ

q1 r

∞

−∞

F⊥ dt =

1 q1 q2 b 4π90 v

dx =2 2 (x + b2 )3/2

∞

0

∞

−∞

dx . But (x2 + b2 )3/2

∞ dx 2x = 2, = √ 2 2 3/2 2 2 2 b2 (x + b ) b x + b 0

I⊥ 1 q1 q2 q1 q2 1 tan θ = = . = 1 2 mv 4π90 b( 2 mv ) 4π90 bE 1 q1 q2 b= = 4π90 E tan θ

b x

q2

I⊥ =

q1 q2 8π90 E

r

−1

θ = tan

so

I⊥ =

1 2q1 q2 . 4π90 bv

q1 q2 . 4π90 bE

(2 cot θ).

The exact answer is the same, only with cot(θ/2) in place of 2 cot θ. So I must show that cot(θ/2) ≈ 2 cot θ, for small θ (that’s the regime in which the impulse approximation should work). Well: cot(θ/2) =

cos(θ/2) 1 2 ≈ = , sin(θ/2) θ/2 θ

for small θ,

while 2 cot θ = 2

cos θ 1 ≈ 2 . So it works. sin θ θ

Problem 11.15 First let’s set up the general formalism. From Eq. 11.101:

ψ(r) = ψ0 (r) + g(r − r0 )V (r0 )ψ0 (r0 ) d3 r0 + g(r − r0 )V (r0 ) g(r0 − r1 )V (r1 )ψ0 (r1 ) d3 r1 d3 r0 + · · ·

Put in ψ0 (r) = Aeikz , mA 2π2 m 2 + A 2π2

ψ(r) = Aeikz −

g(r) = −

m eikr : 2π2 r

eik|r−r0 | V (r0 )eikz0 d3 r0 |r − r0 |

ik|r0 −r1 | eik|r−r0 | e V (r0 ) V (r1 )eikz1 d3 r1 d3 r0 . |r − r0 | |r0 − r1 |


278


In the scattering region r r0 , Eq. 11.73 =⇒

eikr −ik·r0 eik|r−r0 | ≈ e , |r − r0 | r

with

k ≡ kˆ r,

so

m eikr ψ(r) = A eikz − e−ik·r0 V (r0 )eikz0 d3 r0 2π2 r

ik|r0 −r1 | m 2 eikr e −ik·r0 ikz1 3 3 e V (r1 )e V (r0 ) d r1 d r0 2π2 r |r0 − r1 |

f (θ, φ) = −

m 2π2

ei(k −k)·r V (r) d3 r +

ik|r−r0 | m 2 e −ik·r ikz0 3 d3 r. e V (r) )e d r V (r 0 0 2π2 |r − r0 |

I simplified the subscripts, since there is no longer any possible ambiguity. For low-energy scattering we drop the exponentials (see p. 414): m f (θ, φ) ≈ − 2π2

m 2 1 3 V (r) d r + V (r) V (r0 ) d r0 d3 r. 2π2 |r − r0 | 3

Now apply this to the potential in Eq. 11.81: a 1 1 V (r0 ) d3 r0 = V0 r02 sin θ0 dr0 dθ0 dφ0 . |r − r0 | |r − r | 0 0 Orient the z0 axis along r, so |r − r0 | = r2 + r02 − 2rr0 cos θ0 .

a π 1 1 3 2 V (r0 ) d r0 = V0 2π r0 sin θ0 dθ0 dr0 . But |r − r0 | r2 + r02 − 2rr0 cos θ0 0 0 0

π

1

1 sin θ0 dθ0 = 2 2 rr0 r + r0 − 2rr0 cos θ0

r2

+

r02

π 1 2/r, r0 < r; − 2rr0 cos θ0 = [(r0 + r) − |r0 − r|] = 2/r0 , r0 > r. rr 0 0

Here r < a (from the “outer” integral), so r

3

a 1 1 1r 1 2 1 2 3 2 2 2 V (r0 ) d r0 = 4πV0 + (a − r ) = 2πV0 a − r . r dr0 + r0 dr0 = 4πV0 |r − r0 | r 0 0 r 3 2 3 r

V (r)

a 3 1 1 2 2 1 a5 32 2 2 5 3 3 2 2 2 2a a − r r dr = 8π V0 a V (r0 ) d r0 d r = V0 (2πV0 )4π − π V0 a . = |r − r0 | 3 3 3 5 15 0

f (θ) = −

m 4 3 m 2 32 2 2 5 V πa + π V0 a = − 0 2π2 3 2π2 15

2mV0 a3 32

4 mV0 a2 1− . 5 2

Problem 11.16

d2 2 + k G(x) = δ(x) dx2

1 (analog to Eq. 11.52). G(x) = √ 2π

eisx g(s) ds

(analog to Eq. 11.54).



279

d2 1 1 1 2 2 2 isx +k G= √ (−s + k )g(s)e ds = δ(x) = . eisx ds =⇒ g(s) = √ dx2 2π 2π 2π(k 2 − s2 )

G(x) =

1 2π

G(x) = −

1 2π

1 G(x) = + 2π

∞

−∞

eisx ds. Skirt the poles as in Fig. 11.10. − s2

k2

?

?

eisx s+k eisx s−k

In either case, then, G(x) = − ψ(x) = G(x − x0 )

1 1 ds = − 2πi s−k 2π 1 1 ds = 2πi s+k 2π i ik|x| . e 2k

For x > 0, close above:

eikx eisx = −i . For x < 0, close below: s + k s=k 2k

eisx e−ikx . = −i s − k s=−k 2k

(Analog to Eq. 11.65.)

2m i 2m V (x0 )ψ(x0 ) dx0 = − 2 2k 2

eik|x−x0 | V (x0 )ψ(x0 ) dx0 ,

plus any solution ψ0 (x) to the homogeneous Schr¨ odinger equation: 2 d 2 ψ0 (x) = 0. So: + k dx2 im ψ(x) = ψ0 (x) − 2 k

∞

−∞

eik|x−x0 | V (x0 )ψ(x0 ) dx0 .

Problem 11.17 For the Born approximation let ψ0 (x) = Aeikx , and ψ(x) ≈ Aeikx .

im ∞ ik|x−x0 | ψ(x) ≈ A eikx − 2 e V (x0 )eikx0 dx0 k −∞

im x ik(x−x0 ) im ∞ ik(x0 −x) = A eikx − 2 e V (x0 )eikx0 dx0 − 2 e V (x0 )eikx0 dx0 . k −∞ k x

x ∞ im im ψ(x) = A eikx − 2 eikx V (x0 ) dx0 − 2 e−ikx e2ikx0 V (x0 ) dx0 . k k −∞ x Now assume V (x) is localized; for large positive x, the third term is zero, and

im ∞ ikx ψ(x) = Ae V (x0 ) dx0 . This is the transmitted wave. 1− 2 k −∞ For large negative x the middle term is zero:

im −ikx ∞ 2ikx0 ikx ψ(x) = A e − 2 e e V (x0 )dx0 . k −∞ c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

280


Evidently the first term is the incident wave and the second the reflected wave: 2 m 2 ∞ 2ikx . R= e V (x) dx 2 k −∞ If you try in the same spirit to calculate the transmission coefficient, you get 2

2 m 2 ∞ im ∞ T = 1 − 2 V (x)dx = 1 + 2 V (x)dx , k −∞ k −∞ which is nonsense (greater than 1). The first Born approximation gets R right, but all you can say to this order is T ≈ 1 (you would do better using T = 1 − R).

Problem 11.18 Delta function:

V (x) = −αδ(x).

∞

−∞

e2ikx V (x) dx = −α,

so

R=

mα 2 2 k

,

or, in terms of energy (k 2 = 2mE/2 ): R=

m2 α2 mα2 = ; 2mE2 22 E

mα2 . 22 E

T =1−R= 1−

The exact answer (Eq. 2.141) is

1 mα2 22 E

≈1−

mα2 , 22 E

so they agree provided E

mα . 22

1+ −V0 (−a < x < a) Finite square well: V (x) = . 0 (otherwise) a ∞ a e2ikx V0 V0 e2ika − e−2ika 2ikx 2ikx = − sin(2ka). e V (x) dx = −V0 e dx = −V0 =−k 2ik 2i k −∞ −a −a

2 m 2 V 0 So R = 2 sin(2ka) . k k

V0 T =1− sin 2E

2a √ 2mE

2 .

If E V0 , the exact answer (Eq. 2.169) becomes 2 2 V0 V0 2a √ 2a √ −1 T ≈1+ 2mE =⇒ T ≈ 1 − 2mE , sin sin 2E 2E so they agree provided E V0 .

Problem 11.19 The Legendre polynomials satisfy Pl (1) = 1 (see footnote 30, p. 124), so Eq. 11.47 ⇒ f (0) =

∞ 1 (2l + 1)eiδl sin δl . Therefore k

Im[f (0)] =

l=0

∞ 1 (2l + 1) sin2 δl , k l=0

and hence (Eq. 11.48): σ=

4π Im[f (0)]. QED k



281

Problem 11.20 Using Eq. 11.88 and integration by parts: 2 2 2m ∞ 2mA ∞ d 1 f (θ) = − 2 rAe−µr sin(κr) dr = − 2 − e−µr sin(κr) dr κ 0 κ 0 dr 2µ ∞ ∞ 2mA −µr 2 −µr 2 d − = sin(κr) e e [sin(κr)] dr 2µ2 κ dr 0 0 √ ∞ mA π −κ2 /4µ mA −µr 2 = e cos(κr) dr = − 2 0−κ √ e µ2 κ µ 2 µ 0 √ 2 mA π = − 2 3/2 e−κ /4µ , where κ = 2k sin(θ/2) (Eq. 11.89). 2 µ From Eq. 11.14, then, dσ πm2 A2 −κ2 /2µ e , = dΩ 44 µ3 and hence σ = = = =

2 2 dσ πm2 A2 e−4k sin (θ/2)/2µ sin θ dθ dφ dΩ = dΩ 44 µ3 π 2 m2 A2 π −2k2 sin2 (θ/2)/µ e sin θ dθ; write sin θ = 2 sin(θ/2) cos(θ/2) 24 µ3 0 π 2 m2 A2 1 −2k2 x2 /µ 2π 2 m2 A2 1 −2k2 x2 /µ e 2x 2 dx = xe dx 24 µ3 0 4 µ3 0 1 2π 2 m2 A2 µ −2k2 x2 /µ π 2 m2 A2 −2k2 /µ − e e = − − 1 4 µ3 4k 2 24 µ2 k 2

and let x ≡ sin(θ/2)

0

π 2 m2 A 2 −2k2 /µ = 1 − e . 24 µ2 k 2


282

CHAPTER 12. AFTERWORD

Chapter 12

Afterword Problem 12.1 Suppose, on the contrary, that α|φa (1)|φb (2) + β|φb (1)|φa (2) = |ψr (1)|ψs (2), for some one-particle states |ψr and |ψs . Because |φa and |φb constitute a complete set of one-particle states (this is a two-level system), any other one-particle state can be expressed as a linear combination of them. In particular, |ψr = A|φa + B|φb , and |ψs = C|φa + D|φb , for some complex numbers A, B, C, and D. Thus α|φa (1)|φb (2) + β|φb (1)|φa (2) = A|φa (1) + B|φb (1) C|φa (2) + D|φb (2) = AC|φa (1)|φa (2) + AD|φa (1)|φb (2) + BC|φb (1)|φa (2) + BD|φb (1)|φb (2). (i) (ii) (iii) (iv)

Take Take Take Take

the the the the

inner inner inner inner

product product product product

with with with with

φa (1)|φb (2)|: φa (1)|φa (2)|: φb (1)|φa (2)|: φb (1)|φb (2)|:

α = AD. 0 = AC. β = BC. 0 = BD.

(ii) ⇒ either A = 0 or C = 0. But if A = 0, then (i) ⇒ α = 0, which is excluded by assumption, whereas if C = 0, then (iii) ⇒ β = 0, which is likewise excluded. Conclusion: It is impossible to express this state as a product of one-particle states. QED


APPENDIX. LINEAR ALGEBRA

283

Appendix A

Linear Algebra Problem A.1 (a) Yes; two-dimensional. (b) No; the sum of two such vectors has az = 2, and is not in the subset. Also, the null vector (0,0,0) is not in the subset. (c) Yes; one-dimensional.

Problem A.2 (a) Yes;

1, x, x2 , . . . , xN −1 is a convenient basis. Dimension: N.

(b) Yes; 1, x2 , x4 , . . . . (c) No.

Dimension N/2 (if N is even) or (N + 1)/2 (if N is odd).

The sum of two such “vectors” is not in the space.

(d) Yes; (x − 1), (x − 1)2 , (x − 1)3 , . . . , (x − 1)N −1 . (e) No.

Dimension: N − 1.

The sum of two such “vectors” would have value 2 at x = 0.

Problem A.3 Suppose |α = a1 |e1 + a2 |e2 + · · · an |en and |α = b1 |e1 + b2 |e2 + · · · + bn |en . Subtract: 0 = (a1 − b1 )|e1 + (a2 − b2 )|e2 + · · · + (an − bn )|en . Suppose aj = bj for some j; then we can divide by (aj − bj ) to get: |ej = −

(a1 − b1 ) (a2 − b2 ) (an − bn ) |e1 − |e2 − · · · − 0|ej − · · · − |en , (aj − bj ) (aj − bj ) (aj − bj )

so |ej is linearly dependent on the others, and hence {|ej } is not a basis. If {|ej } is a basis, therefore, the components must all be equal (a1 = b1 , a2 = b2 , . . . , an = bn ). QED


284


Problem A.4 (i) e1 |e1 = |1 + i|2 + 1 + |i|2 = (1 + i)(1 − i) + 1 + (i)(−i) = 1 + 1 + 1 + 1 = 4. "e1 " = 2. |e 1 =

1 1 iˆ (1 + i) î + ˆj + k. 2 2 2

e 1 |e2

1 1 = (1 − i)(i) + (3) + 2 2

(ii)

−i 2

1=

1 (i + 1 + 3 − i) = 2. 2

ˆ |e 2 ≡ |e2 − e 1 |e2 |e 1 = (i − 1 − i)î + (3 − 1)ˆj + (1 − i)kˆ = (−1)î + (2)ˆj + (1 − i)k. e 2 |e 2 = 1 + 4 + 2 = 7.

1 ˆ |e 2 = √ [−î + 2ˆj + (1 − i)k]. 7

(iii) e 1 |e3 =

1 28 = 14; 2

√ 2 e 2 |e3 = √ 28 = 8 7. 7

|e 3 = |e3 − e 1 |e3 |e 1 − e 2 |e3 |e 2 = |e3 − 7|e1 − 8|e 2 ˆ = (0 − 7 − 7i + 8)î + (28 − 7 − 16)ˆj + (0 − 7i − 8 + 8i)kˆ = (1 − 7i)î + 5ˆj + (−8 + i)k. "e 3 "2 = 1 + 49 + 25 + 64 + 1 = 140.

1 ˆ |e 3 = √ [(1 − 7i)î + 5ˆj + (−8 + i)k]. 2 35

Problem A.5 From Eq. A.21:

α|β α|β γ|γ = γ| |β − |α = γ|β − γ|α. From Eq. A.19: α|α α|α

α|β α|β |α|β|2 γ|β = β|γ = β| |β − |α = β|β − β|α = β|β − , which is real. α|α α|α α|α ∗

α|β α|β γ|α∗ = α|γ = α| |β − |α = α|β − α|α = 0. γ|α = 0. So (Eq. A.20) : α|α α|α γ|γ = β|β −

|α|β|2 ≥ 0, and hence |α|β|2 ≤ α|αβ|β. QED α|α



285

Problem A.6 α|β = (1 − i)(4 − i) + (1)(0) + (−i)(2 − 2i) = 4 − 5i − 1 − 2i − 2 = 1 − 7i; α|α = 1 + 1 + 1 + 1 = 4;

β|β = 16 + 1 + 4 + 4 = 25;

cos θ =

β|α = 1 + 7i;

1 + 49 1 =√ ; 4 · 25 2

θ = 45◦ .

Problem A.7 Let

|γ ≡ |α + |β;

γ|γ = γ|α + γ|β.

γ|α∗ = α|γ = α|α + α|β =⇒ γ|α = α|α + β|α. γ|β∗ = β|γ = β|α + β|β =⇒ γ|β = α|β + β|β. "(|α + |β)"2 = γ|γ = α|α + β|β + α|β + β|α. But α|β + β|α = 2Re(α|β) ≤ 2|α|β| ≤ 2

α|αβ|β (by Schwarz inequality), so

"(|α + |β)" ≤ "α" + "β" + 2"α""β" = ("α" + "β") , and hence "(|α + |β)" ≤ "α" + "β". QED 2

2

2

2

Problem A.8 

(a)

 1 1 0 2 1 3 . 3i (3 − 2i) 4

   −3 (1 + 3i) 3i (−2 + 0 − 1) (0 + 1 + 3i) (i + 0 + 2i)  (4 + 0 + 3i) (0 + 0 + 9) (−2i + 0 + 6) = (4 + 3i) 9 (6 − 2i) . 6i (6 − 2i) 6 (4i + 0 + 2i) (0 − 2i + 6) (2 + 0 + 4) 

(b)



(c)

   (−2 + 0 + 2) (2 + 0 − 2) (2i + 0 − 2i) 0 0 0 0 3 . BA =  (0 + 2 + 0) (0 + 0 + 0) (0 + 3 + 0)  =  2 (−i + 6 + 4i) (i + 0 − 4i) (−1 + 9 + 4) (6 + 3i) −3i 12   −3 (1 + 3i) 3i 9 (3 − 2i) . [A, B] = AB − BA =  (2 + 3i) (−6 + 3i) (6 + i) −6 

(d)

 −1 2 2i  1 0 −2i . i 3 2 

(e)

 −1 1 −i  2 0 3 . −2i 2i 2


286

APPENDIX. LINEAR ALGEBRA 

(f )

 −1 2 −2i  1 0 2i  . −i 3 2

(g) 4 + 0 + 0 − 1 − 0 − 0 = 3. (h) 

B−1

1˜ = C; 3

|13 02 |  C = − 03 −i 2 0 −i 1 0

BB−1

   0 0| − | 0i 13 | 2 0 −i i 2 2| −i 20    i 2 2−i − | i 3 | = −3i 3 −6 . 2 0 i 0 2 − 0 0 |0 1|

   (4 + 0 − 1) (−6i + 0 + 6i) (2i + 0 − 2i) 3 0 1 1 (0 + 0 + 0) (0 + 3 + 0) (0 + 0 + 0)  = 0 3 = 3 3 (2i + 0 − 2i) (3 + 9 − 12) (−1 + 0 + 4) 0 0

det A = 0 + 6i + 4 − 0 − 6i − 4 = 0.

No;

B−1

  2 −3i i 1 0 3 0 . = 3 −i −6 2

   0 1 0 0 0 = 0 1 0 . 3 0 0 1

A does not have an inverse.

Problem A.9 (a) 

   −i + 2i + 2i 3i  2i + 0 + 6  = 6 + 2i . −2 + 4 + 4 6 (b)

 2 −i −2i 2 1 − i = −2i − 2i(1 − i) + 0 = −2 − 4i. 0

    2 0 −i 2 4

i 2i 2 0 1 0  1 − i = i 2i 2 1 − i = 4i + 2i(1 − i) + 2(3 − i) = 8 + 4i. i 3 2 0 3−i



(c)



(d)    i 2i (−1 + i)

2i 2 (1 + i) 0 = 4i (−2 + 2i) 2 4 (2 + 2i)

 0 0 . 0



287

Problem A.10 (a) S =

1 ˜ (T + T); 2

A=

1 ˜ (T − T). 2

(b) R =

1 (T + T∗ ); 2

M=

1 (T − T∗ ). 2

(c) H =

1 (T + T† ); 2

K=

1 (T − T† ). 2

Problem A.11 @ ki = (ST)ik = (ST)

n

Sij Tjk =

j=1

n

˜ S) ˜ ki ⇒ ST @ =T ˜ S. ˜ QED T˜kj S˜ji = (T

j=1

˜∗ = T† S† . QED @ ∗ = (T ˜ S) ˜ ∗=T ˜ ∗S (ST)† = (ST) (T−1 S−1 )(ST) = T−1 (S−1 S)T = T−1 T = I ⇒ (ST)−1 = T−1 S−1 . QED U† = U−1 , W† = W−1 ⇒ (WU)† = U† W† = U−1 W−1 = (WU)−1 ⇒ WU is unitary. H = H† , J = J† ⇒ (HJ)† = J† H† = JH; the product is hermitian ⇔ this is HJ, i.e. ⇔ [H, J] = 0 (they commute). (U + W)† = U† + W† = U−1 + W−1 = (U + W)−1 . ?

No; the sum of two unitary matrices is not unitary.

(H + J)† = H† + J† = H + J. Yes; the sum of two hermitian matrices is hermitian.

Problem A.12 U† U = I

(U† U)ik = δik =⇒

=⇒

n

† Uij Ujk =

j=1 (j)

Construct the set of n vectors a

i

≡ Uij (a

(j)

a(i)† a(k) =

n

∗ Uji Ujk = δik .

j=1

is the j-th column of U; its i-th component is Uij ). Then

n

∗

a(i) j a(k) j =

j=1

n

∗ Uji Ujk = δik ,

j=1

so these vectors are orthonormal. Similarly, UU† = I

=⇒

(UU† )ik = δik

=⇒

n

† Uij Ujk =

j=1

n

∗ Ukj Uij = δki .

j=1

This time let the vectors b(j) be the rows of U: b(j) i ≡ Uji . Then (k)† (i)

b

b

=

n j=1

∗ b(k) j b(i) j

=

n

∗ Ukj Uij = δki ,

j=1

so the rows are also orthonormal.


288


Problem A.13 ˜ ∗ ) = (det H) ˜ ∗ = (det H)∗ ⇒ det H is real. H† = H (hermitian) ⇒ det H = det(H† ) = det(H ˜ ∗ = | det U|2 = det I = 1, so det U = 1. U† = U−1 (unitary) ⇒ det(UU† ) = (det U)(det U† ) = (det U)(det U) ˜ = S−1 (orthogonal) ⇒ det(SS) ˜ = (det S)2 = 1, so det S = ±1. ˜ = (det S)(det S) S

Problem A.14 (a) 

î = cos θ î + sin θ ˆj;

ˆj = − sin θ î + cos θ ˆj;

 cos θ − sin θ 0 ˆ Ta =  sin θ cos θ 0 . kˆ = k. 0 0 1

y

y' θ

x' θ x z, z'

(b) 

î = ˆj;

ˆ ˆj = k;

kˆ = î.

 0 0 1 Tb = 1 0 0 . 0 1 0

y, x'

x, z'

z, y'

(c) 

î = î;

ˆj = ˆj;

ˆ kˆ = −k.

 1 0 0 Tc = 0 1 0  . 0 0 −1


APPENDIX. LINEAR ALGEBRA (d)

289



cos θ ˜ a Ta = − sin θ T 0     010 001 1 ˜ b Tb = 0 0 1 1 0 0 = 0 T 100 010 0

   sin θ 0 cos θ − sin θ 0 1 0 cos θ 0  sin θ cos θ 0 = 0 1 0 1 0 0 1 0 0    0 0 1 0 0 1 ˜ c Tc = 0 1 0  0 1 0 . T 0 1 0 0 −1 0

det Ta = cos2 θ + sin2 θ = 1.

det Tb = 1.

 0 0 . 1    0 0 1 0 0 1 0  = 0 1 0 . 0 −1 0 0 1

det Tc = -1.

Problem A.15 y

y'

θ

x, x' θ z

î = î;

ˆ ˆj = cos θ ˆj + sin θ k;

z'

  1 0 0 Tx (θ) = 0 cos θ − sin θ . 0 sin θ cos θ

kˆ = cos θ kˆ − sin θ ˆj.

y, y'

x' θ x z

θ z'



ˆ î = cos θ î − sin θ k;

ˆj = ˆj;

kˆ = cos θ kˆ + sin θ î.



î = ˆj;

ˆj = −î;

ˆ kˆ = k.

 0 −1 0 S = 1 0 0 . 0 0 1

 cos θ 0 sin θ 1 0 . Ty (θ) =  0 − sin θ 0 cos θ 

S−1

 0 1 0 = −1 0 0 . 0 0 1


290


STx S−1

STy S−1



 −1 0 1 0 0 0 0 cos θ 0 1 0 sin θ  −1 0 0 0 0 − cos θ 0 1 − sin θ



   −1 0 cos θ 0 sin θ 0 1 0 0 0  0 1 0  −1 0 0 0 1 − sin θ 0 cos θ 0 0 1     −1 0 0 cos θ sin θ 1 0 0 0 0 −1 0 0  = 0 cos θ sin θ  = Tx (−θ). 0 1 0 − sin θ cos θ 0 − sin θ cos θ

0 = 1 0  0 = 1 0 0 = 1 0  0 = 1 0

  0 0 1 0 − sin θ −1 0 0 cos θ 0 0 1    1 0 cos θ 0 sin θ 0 − sin θ =  0 1 0  = Ty (θ). 0 cos θ − sin θ 0 cos θ

Is this what we would expect? Yes, for rotation about the x axis now means rotation about the y axis, and rotation about the y axis has become rotation about the −x axis—which is to say, rotation in the opposite direction about the +x axis.

Problem A.16 From Eq. A.64 we have Af Bf = SAe S−1 SBe S−1 = S(Ae Be )S−1 = SCe S−1 = Cf . Suppose S† = S−1 and He = He† (S unitary, He hermitian). Then Hf † = (SHe S−1 )† = (S−1 )† He† S† = SHe S−1 = Hf , so Hf is hermitian. In an orthonormal basis, α|β = a† b (Eq. A.50). So if {|fi } is orthonormal, α|β = af † bf . But bf = Sbe (Eq. A.63), and also af † = ae† S† . So α|β = ae† S† Sbe . This is equal to ae† be (and hence {|ei } is also orthonormal), for all vectors |α and |β ⇔ S† S = I, i.e. S is unitary.

Problem A.17 Tr(T1 T2 ) =

n

(T1 T2 )ii =

i=1

Is

T1 T2 T3 = T2 T1 T3 =

(T1 )ij (T2 )ji =

i=1 j=1

Tr(T1 T2 T3 ) = Tr(T2 T1 T3 )?

n n

00 10

10 00

00 10

01 00

10 00

=

=

(T2 )ji (T1 )ij =

j=1 i=1

No. 0 T1 = 0

01 00

n n

Counterexample: 1 0 0 , T2 = , 0 1 0 0 1 0 0 0 0 1 0

0 0 1 0 0 0 0 0

=

=

n

(T2 T1 )jj = Tr(T2 T1 ).

j=1

T3 =

1 0 . 0 0

1 0 0 0

=⇒ Tr(T1 T2 T3 ) = 1.

0 0 0 0

=⇒ Tr(T2 T1 T3 ) = 0.



291

Problem A.18 Eigenvalues: (cos θ − λ) − sin θ = (cos θ − λ)2 + sin2 θ = cos2 θ − 2λ cos θ + λ2 + sin2 θ = 0, or λ2 − 2λ cos θ + 1 = 0. sin θ (cos θ − λ)

λ=

2 cos θ ±

√

4 cos2 θ − 4 = cos θ ± − sin2 θ = cos θ ± i sin θ = e±iθ . 2

So there are two eigenvalues, both of them complex. Only if sin θ = 0 does this matrix possess real eigenvalues, i.e., only if θ = 0 or π. Eigenvectors: cos θ − sin θ α ±iθ α =e =⇒ cos θ α − sin θ β = (cos θ ± i sin θ)α ⇒ β = ∓iα. Normalizing: sin θ cos θ β β 1 =√ 2

(1)

a

1 ; −i

(2)

a

1 =√ 2

1 . i

Diagonalization: 1 (1) (S−1 )11 = a1 = √ ; 2 −1

S

1 =√ 2

1 1 ; −i i

1 1 2 1 1 1 = 2 1

STS−1 =

1 (2) (S−1 )12 = a1 = √ ; 2 1 inverting: S = √ 2

−i (1) (S−1 )21 = a2 = √ ; 2

i (2) (S−1 )22 = a2 = √ . 2

1 i . 1 −i

1 1 i i cos θ − sin θ 1 1 (cos θ + i sin θ) (cos θ − i sin θ) = −i sin θ cos θ −i i (sin θ − i cos θ) (sin θ + i cos θ) 2 1 −i iθ iθ iθ −iθ 1 i e 2e e e 0 0 = = . −i 0 2e−iθ 0 e−iθ −ieiθ ie−iθ 2

Problem A.19 (1 − λ) 1 = (1 − λ)2 = 0 =⇒ λ = 1 0 (1 − λ)

α α = =⇒ α + β = α =⇒ β = 0; β β

11 01

(only one eigenvalue).

a=

1 0

(only one eigenvector—up to an arbitrary constant factor). Since the eigenvectors do not span the space, this 1 0 matrix cannot be diagonalized. [If it could be diagonalized, the diagonal form would have to be , since 0 1 the only eigenvalue is 1. But in that case I = SMS−1 . Multiplying from the left by S−1 and on the right by S : S−1 IS = S−1 SMS−1 S = M. But S−1 IS = S−1 S = I. So M = I, which is false.]


292


Problem A.20 Expand the determinant (Eq. A.72) by minors, using the first column: (T22 − λ) . . . ... n .. . . + . . det(T − λ1) = (T11 − λ) Tj1 cofactor(Tj1 ). j=2 . . . (Tnn − λ) But the cofactor of Tj1 (for j > 1) is missing two of the original diagonal elements: (T11 − λ) (from the first column), and (Tjj − λ) (from the j-th row). So its highest power of λ will be (n − 2). Thus terms in λn and λn−1 come exclusively from the first term above. Indeed, the same argument applied now to the cofactor of (T11 − λ) – and repeated as we expand that determinant – shows that only the product of the diagonal elements contributes to λn and λn−1 : (T11 − λ)(T22 − λ) · · · (Tnn − λ) = (−λ)n + (−λ)n−1 (T11 + T22 + · · · + Tnn ) + · · · Evidently then, Cn = (−1)n , and Cn−1 = (−1)n−1 Tr(T). To get C0 – the term with no factors of λ – we simply set λ = 0. Thus C0 = det(T). For a 3 × 3 matrix: (T11 − λ) T12 T13 T21 (T22 − λ) T23 T31 T32 (T33 − λ) = (T11 − λ)(T22 − λ)(T33 − λ) + T12 T23 T31 + T13 T21 T32 − T31 T13 (T22 − λ) − T32 T23 (T11 − λ) − T12 T21 (T33 − λ) = −λ3 + λ2 (T11 + T22 + T33 ) − λ(T11 T22 + T11 T33 + T22 T33 ) + λ(T13 T31 + T23 T32 + T12 T21 ) + T11 T22 T33 + T12 T23 T31 + T13 T21 T32 − T31 T13 T22 − T32 T23 T11 − T12 T21 T33 = −λ3 + λ2 Tr(T) + λC1 + det(T),

with

C1 = (T13 T31 + T23 T32 + T12 T21 ) − (T11 T22 + T11 T33 + T22 T33 ).

Problem A.21 The characteristic equation is an n-th order polynomial, which can be factored in terms of its n (complex) roots: (λ1 − λ)(λ2 − λ) · · · (λn − λ) = (−λ)n + (−λ)n−1 (λ1 + λ2 + · · · + λn ) + · · · + (λ1 λ2 · · · λn ) = 0. Comparing Eq. A.84, it follows that Tr(T) = λ1 + λ2 + · · · λn and det(T) = λ1 λ2 · · · λn .

QED

Problem A.22 (a) [Tf1 , Tf2 ] = Tf1 Tf2 −Tf2 Tf1 = STe1 S−1 STe2 S−1 −STe2 S−1 STe1 S−1 = STe1 Te2 S−1 −STe2 Te1 S−1 = S[Te1 , Te2 ]S−1 = 0. c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


293

(b) Suppose SAS−1 = D and SBS−1 = E, where D and E are diagonal:    d1 0 · · · 0 e1 0 · · ·  0 d2 · · · 0   0 e2 · · ·    D=. , E=. . . ..  ..  ..  .. . . . 0 0 · · · dn 0 0 ···

 0 0  ..  . . en

Then [A, B] = AB − BA = (S−1 DS)(S−1 ES) − (S−1 ES)(S−1 DS) = S−1 DES − S−1 EDS = S−1 [D, E]S. But diagonal matrices always commute:



 d1 e1 0 · · · 0  0 d2 e2 · · · 0    DE =  . .  = ED, ..  .. . ..  0 0 · · · dn en

so

[A, B] = 0. QED

Problem A.23 (a) M† =

1 1 ; 1 −i

MM† =

2 (1 − i) , (1 + i) 2

M† M =

2 (1 + i) ; (1 − i) 2

[M, M† ] =

0 −2i = 0. 2i 0

No.

(b) Find the eigenvalues: (1 − λ) 1 = (1 − λ)(i − λ) − 1 = i − λ(1 + i) + λ2 − 1 = 0; 1 (i − λ) √ (1 + i) ± (1 + i)2 − 4(i − 1) (1 + i) ± 4 − 2i λ= = . 2 2 Since there are two distinct eigenvalues, there must be two linearly independent eigenvectors, and that’s enough to span the space. So this matrix is diagonalizable, even though it is not normal.

Problem A.24 Let |γ = |α + c|β, for some complex number c. Then γ|Tˆγ = α|Tˆα + cα|Tˆβ + c∗ β|Tˆα + |c|2 β|Tˆβ, and Tˆγ|γ = Tˆα|α + c∗ Tˆβ|α + cTˆα|β + |c|2 Tˆβ|β. Suppose Tˆγ|γ = γ|Tˆγ for all vectors. For instance, Tˆα|α = α|Tˆα and Tˆβ|β = β|Tˆβ), so cα|Tˆβ + c∗ β|Tˆα = cTˆα|β + c∗ Tˆβ|α, and this holds for any complex number c. In particular, for c = 1: α|Tˆβ+β|Tˆα = Tˆα|β+Tˆβ|α, while for c = i: α|Tˆβ−β|Tˆα = Tˆα|β−Tˆβ|α. (I canceled the i’s). Adding: α|Tˆβ = Tˆα|β. QED


294


Problem A.25 (a)

˜∗ = T† = T (b)

(c)

1 1−i = T. 1+i 0

√ (1 − λ) (1 − i) = −(1 − λ)λ − 1 − 1 = 0; λ2 − λ − 2 = 0; λ = 1 ± 1 + 8 = 1 ± 3 . (1 + i) (0 − λ) 2 2

1 (1 − i) α α =2 =⇒ α + (1 − i)β = 2α =⇒ α = (1 − i)β. (1 + i) 0 β β

1 |α|2 + |β|2 = 1 =⇒ 2|β|2 + |β|2 = 1 =⇒ β = √ . 3

λ1 = 2, λ2 = −1.

1 a(1) = √ 3

1 (1 − i) α α =− =⇒ α + (1 − i)β = −α; (1 + i) 0 β β

1 3 2|β|2 + |β|2 = 1 =⇒ |β|2 = 1; β = 4 2 (1)† (2)

a

a

2 . 3

1 a(2) = √ 6

1−i . 1

1 α = − (1 − i)β. 2

i−1 . 2

(i − 1) 1 1 = √ (1 + i) 1 = √ (i − 1 − 1 − i + 2) = 0. 2 3 2 3 2

(d) 1 (1) Eq. A.81 =⇒ (S−1 )11 = a1 = √ (1 − i); 3 1 (1) (S−1 )21 = a2 = √ ; 3 1 =√ 3

−1

S

−1

STS

√ (1 − i) (i −√ 1)/ 2 ; 1 2

1 (2) (S−1 )12 = a1 = √ (i − 1); 6

2 (2) (S−1 )22 = a2 = √ . 6 1 ) =√ 3

−1 †

S = (S

(1 + i)√ √1 . (−i − 1)/ 2 2

√ 1 (1 − i) (1 − i) (i −√ 1)/ 2 (1 + i) 0 1 2 √ 1 1 6 0 (1 + i)√ √1 2(1 − i) (1 − √ i)/ 2 2 0 = = . = 0 −1 2 − 2 3 −(1 + i)/ 2 2 3 0 −3

1 = 3

(1 + i)√ √1 −(1 + i)/ 2 2

(e) Tr(T) = 1;

det(T) = 0 − (1 + i)(1 − i) = −2.

Tr(STS−1 ) = 2 − 1 = 1.

det(STS−1 ) = −2.



295

Problem A.26 (a) det(T) = 8 − 1 − 1 − 2 − 2 − 2 = 0. (b)

Tr(T) = 2 + 2 + 2 = 6.

(2 − λ) i 1 −i (2 − λ) i = (2 − λ)3 − 1 − 1 − (2 − λ) − (2 − λ) − (2 − λ) = 8 − 12λ + 6λ2 − λ3 − 8 + 3λ = 0. 1 −i (2 − λ) −λ3 + 6λ2 − 9λ = −λ(λ2 − 6λ + 9) = −λ(λ − 3)2 = 0.

λ1 = 0, λ2 = λ3 = 3. 

λ1 + λ2 + λ3 = 6 = Tr(T).

λ1 λ2 λ3 = 0 = det(T).

 0 0 0 Diagonal form: 0 3 0 . 0 0 3

(c) 

  2 i 1 α −i 2 i  β  = 0 =⇒ 2α + iβ + γ = 0 . −iα + 2β + iγ = 0 =⇒ α + 2iβ − γ = 0 1 −i 2 γ Add the two equations: 3α + 3iβ = 0 =⇒ β = iα;

2α − α + γ = 0 =⇒ γ = −α.



(1)

a

 α 1 =  iα  . Normalizing: |α|2 + |α|2 + |α|2 = 1 =⇒ α = √ . 3 −α

(1)

a

  1 1   i . =√ 3 −1

     2 i 1 α α =⇒ −α + iβ + γ = 0,  2α + iβ + γ = 3α −i 2 i  β  = 3 β  =⇒ −iα + 2β + iγ = 3β =⇒ α − iβ − γ = 0,  1 −i 2 γ γ α − iβ + 2γ = 3γ =⇒ α − iβ − γ = 0. 

The three equations are redundant – there is only one condition here: α − iβ − γ = 0. We could pick γ = 0, β = −iα, or β = 0, γ = α. Then     α α (2) (3) a0 = −iα ; a0 =  0  . 0 α (2)

But these are not orthogonal, so we use the Gram-Schmidt procedure (Problem A.4); first normalize a0 : a(2)

  1 1   −i . =√ 2 0

† (3)

a(2) a0

        1 1/2 1 1

† (3) α α α (3) = √ 1 i 0 0 = √ . So a0 − (a(2) a0 ) a(2) = α 0 − −i = α  i/2  . 2 2 2 0 1 1 1


296

APPENDIX. LINEAR ALGEBRA |α|2

Normalize:

1 1 3 + + 1 = |α|2 = 1 4 4 2

=⇒

α=

2 . 3

a(3)

  1 1   i . =√ 6 2

Check orthogonality: †

a(1) a(2)

(1) † (3)

a

a

  1

1 1 = √ 1 −i −1 −i = √ (1 − 1 + 0) = 0. 6 6 0  

1 1 1 = √ 1 −i −1  i  = √ (1 + 1 − 2) = 0. 3 2 3 2 2

(d) S−1 is the matrix whose columns are the eigenvectors of T (Eq. A.81): S−1

√ √ 2 √3 1 √ =√ 2 i − 3i 6 −√2 0

STS−1

√ √  √ 2 −√ 2 i − 2 √ 1 S = (S−1 )† = √  3 3i 0 . 6 1 −i 2

 1 i ; 2

√ √ √   √ √ 2 i 1 2 − 2i − 2 √2 √3 1 √ √ = 2i − 3i 3 3i 0  −i 2 i   √ 6 1 −i 2 1 −i 2 − 2 0   √ 0 3 √3 3      0 −3 3 i 3i  0 0 6

     1 0 0 0 0 0 0 1 i  = 0 18 0  = 0 3 0 . 6 0 0 18 0 0 3 2

Problem A.27 ˆ α|U ˆ β = U ˆ †U ˆ α|β = α|β. (a) U ˆ |α = λ|α =⇒ U ˆ α|U ˆ α = |λ|2 α|α. But from (a) this is also α|α. So |λ| = 1. (b) U ˆ |α = λ|α, U ˆ |β = µ|β =⇒ |β = µU ˆ −1 |β, so U ˆ † |β = 1 |β = µ∗ |β (from (b)). (c) U µ † ˆ ˆ β|U α = λβ|α = U β|α = µβ|α, or (λ − µ)β|α = 0. So if λ = µ, then β|α = 0. QED

Problem A.28 (a) (i) 

 0 0 4 M2 = 0 0 0 ; 0 0 0



 0 0 0 M3 = 0 0 0 , 0 0 0

so



297

       1 1 5 100 013 0 0 4 1 eM = 0 1 0 + 0 0 4 + 0 0 0 = 0 1 4 . 2 0 0 1 001 000 0 0 0 

(ii) 2

M =

−θ2 0 0 −θ2

= −θ2 I;

M3 = −θ3 M;

M4 = θ4 I;

etc.

1 2 θ3 0 1 θ4 0 1 e =I+θ − θ I− + I + ··· −1 0 2 3! −1 0 4! 2 4 3 5 θ θ θ θ 0 1 = 1− + − ··· I + θ − + − ··· −1 0 2 4! 3! 5! 10 0 1 cos θ sin θ = cos θ + sin θ = . 01 −1 0 − sin θ cos θ M

(b)   SMS−1 = D = 

.. 0

M −1

Se S



0

d1

  for some S.

. dn

1 2 1 3 = S I + M + M + M + · · · S−1 . Insert SS−1 = I : 2 3!

1 1 SeM S−1 = I + SMS−1 + SMS−1 SMS−1 + SMS−1 SMS−1 SMS−1 + · · · 2 3! 1 2 1 3 D = I + D + D + D + · · · = e . Evidently 2 3! det(eD ) = det(SeM S−1 ) = det(S) det(eM ) det(S−1 ) = det(eM ). But 

d21

 D2 = 

0 ..

0

. d2n

  ,



d31

 D3 = 

0 ..

. d3n

  ,

  Dk = 

dk1

0 ..

.

  ,

so

dkn

0 0    d   e 1 d21 d31 0 0 0 0 d1  1  ..  1  ..     .. eD = I +  . . . +  +   + ··· =  . . . . 2 3! 2 3 dn 0 dn 0 dn 0 dn 0 e 





det(eD ) = ed1 ed2 · · · edn = e(d1 +d2 +···dn ) = eTr D = eTr M (Eq. A.68), so det(eM ) = eTr M . QED (c) Matrices that commute obey the same algebraic rules as ordinary numbers, so the standard proofs of ex+y = ex ey will do the job. Here are two: c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

298

APPENDIX. LINEAR ALGEBRA m

n

(i) Combinatorial Method: Use the binomial theorem (valid if multiplication is commutative): M+N

e

∞ ∞ n ∞ n 1 1 n 1 n Mm Nn−m = = (M + N) = Mm Nn−m . m n! n! m!(n − m)! n=0 n=0 m=0 n=0 m=0

Instead of summing vertically first, for fixed n (m : 0 → n), sum horizontally first, for fixed m (n : m → ∞, or k ≡ n − m : 0 → ∞)—see diagram (each dot represents a term in the double sum). eM+N =

∞ ∞ 1 m 1 k M N = eM eN . QED m! k! m=0 k=0

(ii) Analytic Method: Let S(λ) ≡ eλM eλN ;

dS = MeλM eλN + eλM NeλN = (M + N)eλM eλN = (M + N)S. dλ

(The second equality, in which we pull N through eλM , would not hold if M and N did not commute.) Solving the differential equation: S(λ) = Ae(M+N)λ , for some constant A. But S(0) = I, so A = 1, and hence eλM eλN = eλ(M+N) , and (setting λ = 1) we conclude that eM eN = e(M+N) . [This method generalizes most easily when M and N do not commute—leading to the famous Baker-CampbellHausdorf lemma.] 0 1 0 0 As a counterexample when [M, N] = 0, let M = , N= . Then M2 = N2 = 0, so 0 0 −1 0 11 1 0 1 1 1 0 0 1 M N M N e =I+M= , e =I+N= ; e e = = . 01 −1 1 0 1 −1 1 −1 1 But (M + N) =

0 1 cos(1) sin(1) , so (from a(ii)): eM+N = . −1 0 − sin(1) cos(1)

The two are clearly not equal. (d) eiH =

∞ ∞ ∞ 1 n n 1 1 i H =⇒ (eiH )† = (−i)n (H† )n = (−i)n Hn = e−iH (for H hermitian). n! n! n! n=0 n=0 n=0

(eiH )† (eiH ) = e−iH eiH = ei(H−H) = I, using (c).

So eiH is unitary.


299


N = New M = 1/e problem number (modified for 2/e) X = 2/e problem number (unchanged from 1/e) Chapter 1 2/e 1/e

Chapter 2 2/e 1/e

1 2N 3 4 5 6 7 8 9 10 11 12 13 14 15 16N 17N 18N

1 2 3 4 5 6 7N 8N 9N 10 11 12 13 14N 15 16 17 18 19N 20 21N 22 23 24 25N 26 27 28 29 30 31 32 33 34 35 36 37 38 39N 40N 41N 42 43 44

1 6 7 8 11 12 13 14 2 3 4 5 9M 10

1 2 3 5 6M 7

13M 14 37 17M 15 16 18 19M 20 22 23 24 25 26 27 28 29 30 31 32 33 41M 4M 36 3.48

38 40 39

Chapter 2 (cont.) 2/e 1/e 45 46 47 48N 49 50 51 52 53 54N 55N 56N

42 43 44 45 47 48M 34M, 35M 49

300


N = New M = 1/e problem number (modified for 2/e) (M) = 1/e problem number (distant model for 2/e) X = 2/e problem number (unchanged from 1/e) Chapter 3 2/e 1/e

Chapter 4 2/e 1/e

1N 2N 3N 4N 5N 6N 7N 8N 9N 10N 11 12 13 14 15N 16 17 18 19 20 21 22N 23N 24 25 26N 27N 28 29N 30N 31 32 33 34 35N 36N 37N 38N 39 40N

1 2 3 4 5 6 7 8 9 10 11 12 13 14N 15N 16 17 18 19 20 21N 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41N 42

(33M) (21M) (12M)

38 51 41M 39 42 43 44 45 46 57M

57M 25M

52M

53 56 50 49M

55

1 2 3 4 5 6 7M 8 9M 10 11 12 13

17 16 19 20 21 22 23 25 26 27 28 29 30 31M 32 33 34 35 36 37 38 39 40 41 42

Chapter 4 (cont.) 2/e 1/e 43 44N 45 46 47N 48N 49N 50 51 52 53N 54 55 56 57 58N 59 60 61

43 14 15

44 45M 46 47 48 49 50 51 52M 53

301



Chapter 6 2/e 1/e

Chapter 7 2/e 1/e

1 2 3N 4 5 6 7 8 9 10 11 12 13 14 15N 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32N 33 34 35 36 37

1 2 3 4 5 6 7 8 9 10N 11 12 13 14 15N 16 17 18 19 20 21 22 23 24 25 26 27 28 29N 30N 31N 32 33 34 35 36 37 38 39 40N

1 2 3 4 5 6 7 8 9 10 11N 12N 13 14 15 16 17 18 19 20N

1 2 3 4 5 6 7 8 9 10 11M 11M 12 13 14 15M 16M 17M 18 19M 20 21M 22 23 24 25 26 27M 28 29 30 31 32 33

1M 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

27 28 29 30 31 32 33 34

1 2M 3M 4 5 6 7 8 9 10

11 12 13 14 15 16 17

302



Chapter 9 2/e 1/e

Chapter 10 2/e 1/e

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16N 17N

1 2 3 4 5 6 7 8 9N 10 11 12 13 14 15 16 17 18 19 20 21 22N

1 2 3 4 5 6 7 8N 9 10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 2 3M 4 5 6 7 8 9 10 11 12 13 14 15 16 17 21 19M 20

1 3M 4 5 6 8 9 10 11M

303



Chapter 12 2/e 1/e

Appendix 2/e 1/e

1 2 3 4 5N 6N 7N 8 9 10 11 12 13 14 15 16 17 18 19N 20N

1N

1 2 3 4 5 6 7 8 9 10 11 12 13N 14N 15 16 17 18 19 20 21 22 23N 24 25 26 27 28

1 2 3 4

5 6 7 8 9 10 11 12 13 14 15

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.9 3.10. 3.11 3.12 3.16

3.13 3.14 3.15 3.17 3.18 3.19 3.20. 3.40M 3.21M 3.22 3.23 3.24 3.47

Introduction to Quantum Mechanics

Introduction to quantum mechanics

Introduction to quantum mechanics