Introduction to Elementary Particles Instructor’s Solution Manual

I II Introduction to Elementary Particles Instructor’s Solution Manual 29th August 2008 V Acknowledgments: I tha...

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I

II

Introduction to Elementary Particles Instructor’s Solution Manual

29th August 2008

V

Acknowledgments: I thank Robin Bjorkquist, who wrote and typeset many of the solutions in the first four chapters; Neelaksh Sadhoo, who typeset solutions from the first edition; and all those who sent me solutions or suggestions. I have tried to make every entry clear and accurate, but please: if you find an error, let me know ([email protected]). I will post errata on my web page http://academic.reed.edu/physics/faculty/griffiths.html

VI

Contents

1

Historical Introduction to the Elementary Particles

2

Elementary Particle Dynamics 9

3

Relativistic Kinematics 17

4

Symmetries 37

5

Bound States 57

6

The Feynman Calculus 79

7

Quantum Electrodynamics 97

8

Electrodynamics and Chromodynamics of Quarks

9

Weak Interactions 171

10

Gauge Theories 209

1

147

Contents

11

Neutrino Oscillations 233

12

What’s Next 237

A

The Dirac Delta Function 247

VII

1

1

Historical Introduction to the Elementary Particles Problem 1.1

For an undeflected charged particle, qE = qvB =⇒ v = With just a magnetic field, qvB = m

E . B

q v v2 E =⇒ = = 2 . R m BR B R

Problem 1.2

r0 = 10−15 m; h¯ = 6.58 × 10−22 MeV s; c = 3.00 × 108 m/s; h¯ h¯ c 1 so m = = 98.7 MeV/c2 . = 2r0 c 2r0 c2 Observed mπ = 138 MeV/c2 .

Off by a factor of 1.4.

Problem 1.3

r0 = 10−15 m; h¯ = 6.58 × 10−22 MeV s;

c = 3.00 × 108 m/s;

me = 0.511 MeV/c2 . h¯ h¯ h¯ c 1 so pmin = = = 98.7 MeV/c . 2 2r0 2r0 c q = p2min c2 + m2e c4 = 98.7 MeV .

∆x∆p ≥ Emin

The energy of an electron emitted in the beta decay of tritium is < 17 keV.

2

1 Historical Introduction to the Elementary Particles

Problem 1.4

1 [2 ( m N + m Ξ ) − m Σ ] . 3 = 938.9; mΞ = 1318.1; mΣ = 1190.5.

mΛ = mN

So mΛ =

1 [2 (2257.0) − 1190.5] = 1107.8 MeV/c2 . 3

Observed mΛ = 1115.7 MeV/c2 .

Off by 0.7%.

Problem 1.5

i 1 2 1h 2 2 mK + m2K − m2π = 4mK − m2π . 3 3 mK = 495.67; mπ = 138.04. i 1h m2η = 9.637 × 105 = 3.212 × 105 ⇒ mη = 566.8 MeV/c2 . 3 m2η =

Actually mη = 547.3 MeV/c2 .

Off by 3.5%.

Problem 1.6

M∆ − MΣ∗ = 1232 − 1385 = −153 MΣ∗ − MΞ∗ = 1385 − 1533 = −148. Average: −151.

∴ MΩ = MΞ∗ + 151 = 1533 + 151 = 1684 MeV/c2 . Actually MΩ = 1672 MeV/c2 .

Off by 0.7%.

Problem 1.7

(a) ∆− −→ n + π − or Σ− + K0 Σ∗+ −→ p + K¯0 ; Σ+ + π 0 ; Ξ∗− −→ Σ0 + K − ;

Σ− + K¯0 ;

Σ+ + η; Λ + K− ;

Σ0 + π + ; Ξ0 + π − ;

Λ + π+;

Ξ0 + K +

Ξ− + π 0 ;

Ξ− + η

3

(b) Kinematically allowed: ∆− −→ n + π − Σ∗+ −→ Σ+ + π 0 ;

Σ0 + π + ;

Ξ∗− −→ Ξ0 + π − ;

Ξ− + π 0

Λ + π+

Problem 1.8

(a) With a strangeness of −3, the Ω− would have to go to Ξ0 + K − or Ξ− + K¯0 to conserve S and Q. But the ΞK combination is too heavy (at least 1808 MeV/c2 , whereas the Ω− is predicted – see Problem 1.6 – to have a mass of only 1684). (b) About 0.5 cm ; t = d/v = (5 × 10−3 m)/(3 × 107 m/s) = 2 × 10−10 s. (Actually, t = 0.8 × 10−10 s.)

Problem 1.9

Σ+ − Σ− = 1189.4 − 1197.4 = −8.0 p − n + Ξ0 − Ξ− = 938.3 − 939.6 + 1314.8 − 1321.3 = −7.8

3% difference.

Problem 1.10

Roos lists a total of 30 meson types; in the first column is the particle name at the time, in the second column the quoted mass (in MeV/c2 ), and in the third its current status.

4


meson

mass

status

π K K3 χ2 κ3 K ∗∗ f K5∗ χ1 κ2 κ1 ψ5 K1∗ ϕ3 ω

138 496 1630 1340 1275 1260 1253 1150 1045 1040 1020 990 888 885 781

π K dead f 0 (1370)? f 1 (1285)? K1 (1270)? f 2 (1270)? dead a0 (980)? dead φ(1020) dead K ∗ (892) η 0 (985)? ω (782)

exotic

yes

yes

yes

meson

mass

ρ ρ2 ρ1 ψ4 K1∗∗ δ α ψ3 ζ η ϕ2 ψ2 ϕ1 ψ1 ω ABC

755 780 720 760 730 645 625 597 556 549 520 440 395 330 317

status ρ(770) f 0 (600)? f 0 (600)? dead dead dead dead dead dead η dead dead dead dead dead

exotic

yes

yes

yes yes

Problem 1.11

From the last column in Problem 1.10 I count 7 exotic species, all of them now dead. Of the surviving particles (of course) none is exotic.

Problem 1.12

¯ ¯ ¯ ud,¯ du, ¯ dd); 1 quark (u): one (uu); 2 quarks (u, d): four (uu, 3 quarks (u, d, s): nine; 4 quarks (u, d, s, c): sixteen; 5 quarks (u, d, s, c, b): twenty-five; 6 quarks (u, d, s, c, b, t): thirty-six. The general formula for n flavors is n2 .

Problem 1.13

1 quark (u) =⇒ 1 baryon (uuu); 2 quarks (u, d) =⇒ 4 baryons (uuu, uud, udd, ddd); 3 quarks (u, d, s) =⇒ 10 baryons (baryon decuplet).

5

For n quarks, we can have n ways

all three quarks the same : two the same, one different : all three different :

n(n − 1) ways

n(n − 1)(n − 2)/6 ways.

[For the third type of combination, divide by six to cover the equivalent permutations (uds = usd = dus = dsu = sud = sdu).] So the total is n + n(n − 1) + n(n − 1)(n − 2)/6

=n + n2 − n + n(n − 1)(n − 2)/6 n = [6n + (n − 1)(n − 2)] 6 n = 6n + n2 − 3n + 2 6 n 2 = n + 3n + 2 6 n(n + 1)(n + 2) = . 6 Thus for four quarks we have 20 baryon types, for five quarks, 35, and for six quarks, 56.

Problem 1.14

uuu uud udd ddd uus uds dds uss dss sss 10 have C = 0

uuc udc ddc usc dsc ssc 6 have C = 1

ucc dcc scc 3 have C = 2

ccc 1 has C = 3

6


Problem 1.15

uu¯ ud¯ du¯ dd¯ us¯ su¯ ds¯ sd¯

cu¯ cs¯ cd¯ 3 have C = 1

uc¯ dc¯ sc¯ 3 have C = −1

cc¯

ss¯ 10 have C = 0 ¯ (including cc)

Problem 1.16

qq¯

meson

mass

year

uu¯ ud¯ dd¯ us¯ ds¯ ss¯ cu¯ cd¯

π0

134.98 139.57 134.98 493.68 497.65 547.51 1864.5 1869.3

1950 1947 1950 1949 1947 1962 1976 1976

(*)

π+ π 0 (*) K+ K0 η (*) D0 D+

qq¯

meson

mass

year

cs¯ cc¯ ub¯ db¯ sb¯ cb¯ bb¯

Ds+

1968.2 2980.4 5279.0 5279.4 5367.5 6286 9460.3

1977 1980 1983 1983 1993 1998 1977

ηc (1S) B+ B0 Bs0 Bc+ Υ(1S)

All masses are in MeV/c2 ; (*) indicates that the particle is a combination of different quark states. qqq baryon mass year qqq baryon mass year uuu uud udd ddd uus uds dds uss dss sss

∆++ p n ∆− Σ+ Λ Σ− Ξ0 Ξ− Ω−

1232 938.27 939.57 1232 1189.37 1115.68 1197.45 1314.83 1321.31 1672.45

1951 1911 1932 1951 1963 1950 1965 1963 1963 1964

uuc udc ddc usc dsc ssc ucc dcc scc ccc

Σ++ c Λ+ c Σ0c Ξ+ c Ξ0c Ω0c Ξ++ cc Ξ+ cc Ω+ cc Ω++ ccc

2454.02 2286.46 2453.76 2467.9 2471.0 2697.5

1975 1975 1975 ? ? ?

3518.9

2002

7

qqq

baryon

mass

year

uub udb ddb usb dsb ssb ucb dcb scb ccb

Σ+ b Λ0b Σ− b Ξ0b Ξ− b Ω− b Ξ+ cb Ξ0cb Ω0cb Ω+ ccb

5807.8 5624 5815.2 5792 5792.9

2007 ? 2007 1995 2007

qqq

baryon

ubb dbb sbb cbb bbb

Ξ0bb Ξ− bb Ω− bb Ω0ccb Ω− bbb

mass

year

Blank spaces indicate that the particle has not yet been found (2008).

Problem 1.17

N:

3 × 336 − 62 = 946 ;

Σ:

2 × 336 + 540 − 62 = 1150 ;

actually 1193;

error: −3.6%.

Λ:

2 × 336 + 540 − 62 = 1150 ;

actually 1116;

error: 3.0%.

Ξ:

336 + 2 × 540 − 62 = 1354 ;

actually 1318;

error: 2.7%.

actually 939;

error: 0.7%.

Problem 1.18

Quarks and leptons:

ccc (q = −1) : e− ccn (q = − 32 ) : u¯ cnn (q = − 13 ) : d nnn (q = 0) : ν¯e

c¯c¯c¯ (q = 1) : e+ c¯c¯n¯ (q = 23 ) : u c¯n¯ n¯ (q = 13 ) : d¯ n¯ n¯ n¯ (q = 0) : νe

(The neutrinos could be switched, but this seems the most “natural” assignment.)

Mediators:

cccn¯ n¯ n¯ (q = −1) : nnnc¯c¯c¯ (q = +1) : cccc¯c¯c¯ (q = 0) : nnnn¯ n¯ n¯ (q = 0) :

W− W+ Z0 or vice versa γ

Gluons: We need matching triples of particles and antiparticles,

8


ccnc¯c¯n¯ (3 different orderings for each triple, so 9 possibilities); cnnc¯n¯ n¯ (3 different orderings for each triple, so 9 possibilities); leading to a total of 18 possibilities.

Problem 1.19

There are at least four distinct answers, depending on the particle in question: • Antiparticles (such as the positron) annihilate with the corresponding particle (the electron, in this case), and since there are lots of electrons in the lab, positrons don’t stick around long enough to have any role in ordinary chemical processes. But if you could work in a total vacuum you could make atoms and molecules of antimatter, and all of chemistry would proceed just the same as with ordinary matter. • Most elementary particles (such as muons, pions, and intermediate vector bosons) are intrinsically unstable; they disintegrate spontaneously in a tiny fraction of a second—not long enough to do any serious chemistry. You can make short-lived “exotic atoms”, with (say) muons in orbit around the nucleus instead of electrons. Some of these systems last long enough to do spectroscopy. • Neutrinos interact so feebly with matter that they have no impact on chemistry, even though we are in fact bathed in them all the time. • Quarks are the basic constituents of protons and neutrons, so in an indirect way they do play a fundamental role in chemistry. (And gluons play a fundamental role in holding the nucleus together.) But because of confinement, they do not occur as free particles, only in composite structures, so they don’t act as “individuals”.

9

2

Elementary Particle Dynamics Problem 2.1

Fg =

Gm2 ; r2

Fe =

e2 4πe0 r2

Fg 4πe0 Gm2 = 2.4 × 10−43 . = Fe e2

Problem 2.2

Problem 2.3

Eight are built on

; eight are built on

; one is special.

10

2 Elementary Particle Dynamics

Problem 2.4 e

e

e

e

g

g e

e

e

e

Energy and momentum are conserved at each vertex. Thus, for the virtual photon in the “horizontal” diagram E = 2me c2

and

p=0

=⇒

m = 2me

and

and in the “vertical” diagram, E=0

and

p=0

=⇒

m=0

and

v = 0.

v=0

11

Problem 2.5 d

u

W s X- s d

d

p-

u s L d

u

W

W

s X- s d

u

p-

u

d n d

(a) Ξ− −→ n + π − would be favored kinematically, but since two s quarks have to be converted, it requires an extra W − (hence two extra weak vertices), and this makes it much less likely. Λ + π − is favored. Experimentally, 99.887% go to Λ + π − ; the branching ratio for n + π − is less than 1.9 × 10−5 . (b) D0 −→ K − + π + : Neither vertex crosses generations: u W c D0 u

d

p+

s u K

D0 −→ π − + π + : One vertex crosses generations: u W c D0 u

d

d u

p+

p-

D0 −→ π − + K + : Both vertices cross generations:

12


u

+ s K

W c D0 u

d u

p-

Because weak vertices within a generation are favored, K − + π + is most likely, K + + π − least. Experimentally, the branching ratios are: 3.8% for K − + π + , 0.14% for π + + π − , and 0.014% for K + + π − . (c) The b quark prefers to go to c (Vcb = 0.042, whereas Vub = 0.004), so B should go to D.

Problem 2.6

e

W+

e

W

_

W+

e e

Z

W

Problem 2.7

(a) Impossible (charge conservation) (b) Possible, electromagnetic (c) Impossible (energy conservation) (d) Possible, weak (e) Possible, electromagnetic (f) Impossible (muon number conservation)

_

W+

e e

g

W

_

13

(g) Possible, strong (h) Possible, weak (i) Impossible (baryon number conservation) (j) Possible, strong (k) Impossible (baryon and lepton number conservation) (l) Possible, strong (m) Possible, strong (n) Impossible (charge conservation) (o) Possible, weak (p) Impossible (charge conservation) (q) Possible, electromagnetic (r) Possible, weak (s) Possible, weak (t) Possible, strong (u) Possible, electromagnetic (v) Possible, weak

Problem 2.8

(a) K + −→ µ+ + νµ + γ; weak and electromagnetic interactions are involved:

e W

m

ne e g

e nm

14


(b) Σ+ −→ p + γ; weak and electromagnetic interactions are involved: W s S+ u u

W u

g

d u p u

Problem 2.9

The lifetime tells us this is an OZI-suppressed strong interaction. Evidently the B meson must weigh more than half the Υ (just as the D weighs more than half the ψ). Thus the B meson should weigh more than 4730 MeV/c2 (and it does).

Problem 2.10

d pu

y' c c

u + d p c y c

Here’s a typical contributing diagram. Yes, it is a strong interaction. Yes, it is OZI-suppressed. We should expect a lifetime around 10−20 seconds.

Problem 2.11

(a) Particle X has charge +1 and strangeness 0; it was presumably a proton. (b) K − + p −→ K0 + K + + Ω− (strong); Ω− −→ Ξ0 + π − (weak); Ξ0 −→ Λ0 + π 0 (weak), π 0 −→ γ + γ (electromagnetic), and both photons undergo pair production γ −→ e+ + e− (electromagnetic); Λ0 −→ π − + p (weak).

15

Problem 2.12

A pion seems most likely—it requires only one weak vertex, with no generation crossing, and it’s light (hence kinematically favored): p

uud

u u d

p

d u

p+

W-

17

3

Relativistic Kinematics Problem 3.1

(

x 0 = γ( x − vt) =⇒ x 0 /γ = x − vt 2 t0 = γ t − cv2 x =⇒ vt0 /γ = vt − vc2 x

)

Adding these two equations: v2 1 0 x + vt0 = x 1 − 2 = x/γ2 ∴ x = γ ( x 0 + vt0 ). γ c    x 0 = γ ( x − vt) =⇒ v x0 = v x − v2 t  2 2 2 c γ c c  t0 = γ t − v2 x =⇒ t0 /γ = t − v2 x  c c Adding these two equations: 10 v 0 v2 t + 2 x = t 1 − 2 = t/γ2 γ c c

∴ t = γ t0 +

v 0 x c2

.

Also, y = y0 , and z = z0 . This confirms Eq. 3.3.

Problem 3.2

(a) From the Lorentz transformations (Eq. 3.1), t0 = γ t − h

∴ t A 0 − tB 0 = γ t A − tB −

v x c2

.

v (x A − xB ) . c2 i

If simultaneous in S (so that t A = t B ), then t A 0 = t B 0 + γ cv2 ( x B − x A ). (b) γ= q

1 1−

9 25

5 = ; 4

x B − x A = 4 km = 4 × 103 m.

18

3 Relativistic Kinematics

t A 0 = tB 0 +

3 × 103 m 5 3 c · · 4 × 103 m = t B 0 + = t B 0 + 10−5 s. 8 m/s 2 3 × 10 5 4 c A came on 10−5 seconds later.

So B went on first ;

Problem 3.3

(a) The dimension along the direction of motion undergoes length contraction, but the other two dimensions do not. Thus, volumes transform according to V = V 0 /γ. (b) The number of moleculues N is invariant. Density is ρ = N/V, so density transforms by ρ = N/V = γN/V 0 = γρ0 .

Problem 3.4

(a) d = vt = 0.998 × 3 × 108 m/s 2.2 × 10−6 s = 659 m . (b) γ= q

1 1 − (0.998)2

= 15.8.

d = v (γt) = γ(659) = (15.8)(659) = 10,400 m . (c) No.

Yes.

They only travel 10, 400(2.6 × 10−8 )/(2.2 × 10−6 ) = 123 m.

Problem 3.5

d = 600 m; τ = 2.2 × 10−6 s. The half-life t1/2 is related to the lifetime τ by t1/2 = (ln 2)τ. Thus, d = v(γt1/2 ) = p

v 1 − (v/c)2

(ln 2)τ =⇒ v = 0.80 c

No.

19

Problem 3.6

+ 13 c = 65 c = 10 12 c, but the (slower), so bullet does reach target.

(a) Velocity of bullet relative to ground = getaway car goes

3 4c

=

9 12 c

1 2c

(b) v=

1 c + 13 c 2 1 1 + 12 3

which is less than 34 c =

=

5 6c 7 6

21 28 c,

=

5 20 c= c, 7 28

so bullet doesn’t reach target.

Problem 3.7



γ γβ γβ γ M= 0 0 0 0  γ −γβ −γβ γ ΛM =   0 0 0 0

0 0 1 0 0 0 1 0

 0 0  0 1  γ 0 γβ 0  0  0 0 1

γβ γ 0 0

0 0 1 0

  2 γ (1 − β2 ) 0 0   0 γ2 (1 − β2 ) 0  = 0 0 0  0 0 1

0 0 1 0

 0 0 =1 0 1

Problem 3.8

From Eq. 3.8:

( x0 0 )2 − ( x1 0 )2 = γ2 ( x0 − βx1 )2 − γ2 ( x1 − βx0 )2 h 1 0 = γ2 ( x 0 )2 − 2βx x + β2 ( x 1 )2 i 1 0 −( x1 )2 + 2βx x − β2 ( x 0 )2 h i = γ2 ( x 0 )2 (1 − β2 ) − ( x 1 )2 (1 − β2 ) h i = γ2 (1 − β2 ) ( x 0 )2 − ( x 1 )2 = ( x 0 )2 − ( x 1 )2 . Since x2 0 = x2 and x3 0 = x3 , it follows that

( x 0 0 )2 − ( x 1 0 )2 − ( x 2 0 )2 − ( x 3 0 )2 = ( x 0 )2 − ( x 1 )2 − ( x 2 )2 − ( x 3 )2 . X

20


Problem 3.9

aµ = (3, −4, −1, −2) bµ = (5, 0, −3, −4) a2 = 42 + 12 + 22 = 21 b2 = 02 + 32 + 42 = 25 a · b = (4)(0) + (1)(3) + (2)(4) = 11 a2 = 32 − a2 = 9 − 21 = -12,

so aµ is spacelike.

b2 = 52 − b2 = 25 − 25 = 0,

so bµ is lightlike.

a · b = (3)(5) − a · b = 15 − 11 = 4.

Problem 3.10

(a) 10 (b) 6

(s00 , s11 , s22 , s33 , and s01 , s02 , s03 , s12 , s13 , s23 ) (the latter six – the first four are zero in this case).

(c) sνµ 0 = Λκν Λσ sκσ = Λκν Λσ sσκ µ

µ

= Λσ Λκν sσκ = sµν 0 , so sµν 0 is symmetric. µ

aνµ 0 = Λκν Λσ aκσ = Λκν Λσ (− aσκ ) µ

µ

= −Λσ Λκν aσκ = − aµν 0 , µ

so antisymmetry is also preserved by Lorentz transformations. (d) sνµ = gνκ gµλ sκλ = gνκ gµλ sλκ

= gµλ gνκ sλκ = sµν , so sµν is symmetric. aνµ = gνκ gµλ aκλ = gνκ gµλ (− aλκ )

= − gµλ gνκ aλκ = − aµν , so aµν is antisymmetric.

21

(e) sµν aµν = sνµ aνµ

(just by renaming the summation indices)

µν

= s (− aµν ) (by symmetry/antisymmetry) = −(sµν aµν ). But if x = − x, then x = 0. (f) Let sµν = 21 (tµν + tνµ ) and aµν = 12 (tµν − tνµ ). Clearly sµν = sνµ , aµν = − aνµ , and tµν = sµν + aµν .

Problem 3.11

γ= q

1 3 2 5

1−

1 = q

16 25

5 η = γ(c, v x , vy , vz ) = 4

5 = . 4

µ

3 c, c, 0, 0 5

=

c (5, 3, 0, 0) 4

Problem 3.12 µ

µ

µ

µ

p A + p B = pC + p D Multiply by Λνµ to transform to S0 : Λνµ ( p A + p B ) = Λνµ ( pC + p D ) µ

µ

µ

µ

ν0 pνA 0 + pνB 0 = pC + pνD 0

Problem 3.13

p2 = m2 c2 > 0 (Eq. 3.43), so it’s timelike (Eq. 3.25); if m = 0, then p2 = 0, so it’s lightlike . The 4-momentum of a virtual particle could be anything .

Problem 3.14

Say the potato weighs 41 kg at room temperature (20◦ C), and we heat it to the boiling point (100◦ C). If it were pure water (specific heat c = 1 Cal/kg◦ C=

22


4000 J/kg◦ C), the heat required would be 1 ◦ kg (80◦ C) = 8 × 104 J. Q = mc∆T = (4000 J/kg C) 4 The increase in mass is ∆m =

Q 8 × 104 J ∼ = = 10−12 kg . (Not a substantial addition!) c2 (3 × 108 m/s)2

Problem 3.15

(4-vectors) =⇒ pµ = pπ − pν =⇒ p2µ = p2π + p2ν − 2pπ · pν .

pπ = pµ + pν p2µ = m2µ c2 ,

p2π = m2π c2 ,

p2ν = 0;

Eπ Eν − pπ · pν , but pπ · pν = 0, c c Eπ = γmπ c2 and Eν = |pν |c. So

pπ · pν = But

so

m2µ c2 = m2π c2 − 2

Eπ Eν . c2

(m2π − m2µ )c . Also, |pπ | = γmπ v. 2γmπ h i 1 − (m2µ /m2π ) (m2π − m2µ )c |pν | v tan θ = (where β = ). = = 2 |pπ | 2γmπ γmπ v c 2βγ

(m2π − m2µ )c2 = 2γmπ |pν |c =⇒ |pν | =

So

Problem 3.16 µ

Before the collision, in the lab frame, pTOT = pTOT 2 =

EA + mB c c

2

− p2A =

EA c

+ m B c, p A , so

E2A + 2E A m B + m2B c2 − p2A c2

= 2E A m B + (m2B + m2A )c2 (I used E2A − p2A c2 = m2A c4 ). After the collision, in the CM frame, at threshold: pTOT 0 = ((m1 + m2 + . . . + mn )c, 0) = ( Mc, 0); µ

( pTOT 0 )2 = ( Mc)2 .

µ

But pTOT is conserved (same before as after, in either frame) and pTOT 2 is invariant (same value in both frames), so 2E A m B + (m2A + m2B )c2 = M2 c2 ,

or E A =

( M2 −m2A −m2B )c2 . 2m B

23

[Note that this generalizes the result in Example 3.4. There, m A = m B = m p ; M = 4m p ;

so

EA =

(16−1−1)m2p c2 2m p

= 7m p c2 .]

Problem 3.17

EA =

( M2 − m2A − m2B )c2 2m B

(a) M = 2m p + mπ0 = 2(938.27) + 134.98 = 2011.52 m A = m B = m p = 938.27 E=

(2011.52)2 − 2(938.27)2 = 1218 MeV 2(938.27)

(b) M = 2m p + mπ + + mπ − = 2(938.27) + 2(139.57) = 2155.68 m A = m B = m p = 938.27 E=

(2155.68)2 − 2(938.27)2 = 1538 MeV 2(938.27)

(c) M = 2m p + mn = 2(938.27) + 939.57 = 2816.11 m A = mπ − = 139.57; E=

m B = m p = 938.27

(2816.11)2 − (139.57)2 − (938.27)2 = 3747 MeV 2(938.27)

(d) M = mK0 + mΣ0 = 497.65 + 1192.6 = 1690.3 m A = mπ − = 139.57; E=

m B = m p = 938.27

(1690.3)2 − (139.57)2 − (938.27)2 = 1043 MeV 2(938.27)

24


(e) M = m p + mΣ+ + mK0 = 938.27 + 1189.4 + 497.65 = 2625.3 m A = m B = m p = 938.27 E=

(2625.3)2 − 2(938.27)2 = 2735 MeV 2(938.27)

Problem 3.18

For the second reaction (K − + p → Ω− + K0 + K + ), M = mΩ− + mK0 + mK+ = 1672.45 + 497.67 + 493.68 = 2663.80, m A = mK− = 493.68

and

m B = m p = 938.27,

so the threshold energy for the K − is EK− =

(2663.80)2 − (493.68)2 − (938.27)2 = 3182.3 MeV. 2(938.27)

In the first reaction, at threshold, the K − goes in the forward direction, and the other particles emerge as a group (there is no point in “wasting” energy in transverse motion, or in internal motion of the p, p, K + ). p

(before)

p

C K(after)

We have, in effect, an outgoing “particle” C, of mass mC = 2m p + mK = 2(938.27) + 493.68 = 2370.22 MeV/c2 . Conservation of energy/momentum says p1 + p2 = pC + pK , or p1 − pK = pC − p2 . Squaring: p21 + p2K − 2p1 · pK = p2C + p22 − 2pC · p2 , E EK EC 2 2 2 2 2 2 2 2 − p1 · pK = mC c + m p c − 2 mpc . m c + mK c − 2 p c c c Now EC = E + m p c2 − EK ,

p1 · pK = |p1 ||pK |,

|p1 | =

1 c

q

E2 − ( m p c2 )2 ,

25

and q q 1 1 1 a EK2 − (mK c2 )2 = (3182.3)2 − (439.68)2 = (3143.8) ≡ . c c c c q h i − EEK + a E2 − (m p c2 )2 = 12 (mC c2 )2 − (mK c2 )2 − (m p c2 )( E + m p c2 − EK ), q a E2 − (m p c2 )2 = E( EK − m p c2 ) + b,

|pK | =

where b≡

1 2

h

i (mC c2 )2 − (mK c2 )2 + (m p c2 )( EK − m p c2 ) = 4.7926 × 106 .

Squaring, h i 2 a2 E2 − (m p c2 )2 = E2 EK − m p c2 + b2 + 2bE EK − m p c2 , 2 2 E2 a2 − EK − m p c2 − 2bE EK − m p c2 − b2 + a2 m p c2 = 0. Numerically, EK − m p c2 = 2244.0, 2 = 4.84775 × 106 , a2 − EK − m p c2

2b( EK − m p c2 ) = 2.15091 × 1010 , 2 = 3.16697 × 1013 . b2 + a2 m p c2 So E2 4.84775 × 106 − E 2.15091 × 1010 − 3.16697 × 1013 = 0, or

E2 − E(4436.9) − 6.5329 × 106 = 0. q 1 1 2 6 4436.9 ± (4436.9) + 4(6.5329 × 10 ) = (4436.9 ± 6768.9). E= 2 2

But E has to be positive, so we need the upper sign: E = 5602.9 ⇒ T = 5602.9 − 938.3 = 4664.6 MeV .

26


Problem 3.19 µ

µ

µ

µ

µ

µ

(a) p A = p B + pC , or pC = p A − p B . Square both sides: p2C = p2A + p2B − 2p A · p B . Now,

p2A = m2A c2 ,

But

p2B = m2B c2 ,

p2C = m2C c2 ,

p A · pB =

E A EB − p A · pB . c c

p A = 0, and E A = m A c2 , so m2C c2 = m2A c2 + m2B c2 − 2m A EB , 2m A EB = −(m2C − m2A − m2B )c2 , EB =

(−m2C + m2A + m2B ) 2 c ; 2m A

EC =

so

or

(m2A − m2B + m2C ) 2 c 2m A

(b) EB2 − p2B c2 = m2B c4 =⇒ p2B =

(m2A + m2B − m2C )2 c2 4m2A m2B c2 EB2 − m2B c2 = − 2 c 4m2A 4m2A

q c m4A + m4B + m4C + 2m2A m2B − 2m2A m2C − 2m2B m2C − 4m2A m2B 2m A q c λ(m2A , m2B , m2C ) = 2m A

|p B | =

∴ | p B | = | pC | =

c q λ(m2A , m2B , m2C ) 2m A

(c) The decay is kinematically forbidden if m A < m B + mC (not enough energy to produce the final particles, in CM frame).

Problem 3.20

(a) Eµ = 109.8 MeV;

Eν = 29.8 MeV

(b) Eγ = 67.5 MeV (c) Eπ + = 248.1 MeV;

Eπ0 = 245.6 MeV

(d) E p = 943.6 MeV;

Eπ − = 172.1 MeV

(e) EΛ = 1135 MeV;

EK+ = 536.6 MeV

27

Problem 3.21

The velocity of the muon is 1

γ= s

1−

= q

v=

2 = q 2

m2π −mµ m2π +m2µ

m2π −m2µ m2π +m2µ

c (Example 3.3). So

(m2π + m2µ ) (m2π + m2µ )2 − (m2π − m2µ )2

(m2π + m2µ ) m4π + m4µ + 2m2π m2µ − m4π − m4µ + 2m2π m2µ

=

m2π + m2µ 2mπ mµ

.

Lab lifetime is γτ, so distance is vγτ, or ! m2π + m2µ m2π − m2µ m2π − m2µ c d= τ = cτ. 2mπ mµ 2mπ mµ m2π + m2µ d=

(139.6)2 − (105.7)2 (3 × 108 m/s)(2.20 × 10−6 s) = 186 m. 2(139.6)(105.7)

Problem 3.22

(a) The minimum is clearly Emin = m B c2 , with B at rest. For maximum EB we want B going one way, C, D, . . . moving as a unit in the opposite direction (there is no point in wasting energy in relative motion of the other particles). Thus the others act as a single particle, of mass M = mC + m D + · · · , and we can simply quote the result for two-body decays (Problem 3.19a): Emax =

m2A + m2B − M2 2 c . 2m A

In case that argument is not completely compelling, let’s look more closely at the case of just three outgoing particles. For maximum EB we certainly want C and D to come off opposite to B, and the only question is how best to apportion the momentum between them. Let’s say pC = − xpB ,

p D = ( x − 1)p B

(so that pC + pD = −pB ).

28


Here x ranges from 0 (C at rest) to 1 (D at rest). From conservation of energy, m A c2 = EB + EC + ED . But q q EC = m2C c4 + x2 p2B c2 , ED = m2D c4 + ( x − 1)2 p2B c2 , and p2B c2 = EB2 − m2B c4 . So q q m A c2 = EB + m2C c4 + x2 ( EB2 − m2B c4 ) + m2D c4 + ( x − 1)2 ( EB2 − m2B c4 ). Differentiate with respect to x: dEB 1 1 dEB 2 2 4 2 √ + 2x ( EB − m B c ) + x 2EB 0= dx 2 F dx 1 1 dEB 2 2 4 2 √ + 2( x − 1)( EB − m B c ) + ( x − 1) 2EB , 2 dx q q √ √ F ≡ m2C c4 + x2 ( EB2 − m2B c4 ); ≡ m2D c4 + ( x − 1)2 ( EB2 − m2B c4 ). For the maximum we want dEB /dx = 0, so 1 1 0 = √ 2x ( EB2 − m2B c4 ) + √ 2( x − 1)( EB2 − m2B c4 ), 2 F 2 x ( x − 1) √ + √ = 0, F

p √ x = (1 − x ) F

Therefore x2 m2D c4 + ( x − 1)2 ( EB2 − m2B c4 ) = (1 − x )2 m2C c4 + x2 ( EB2 − m2B c4 ) , x2 m2D c4 = (1 − x )2 m2C c4 =⇒ xm D = (1 − x )mC ; mC . x= (mC + m D )

x (mC + m D ) = mC ,

What is the corresponding EB ? p p √ √ m A c2 = EB + F + , but x = (1 − x ) F, so p √ 1−x p 1 = F= −1 F, and hence x x 2

mAc = E +

= E+

p 1 1p −1 F = E+ F x x

p

F+

r

1 2 4 m c + E2 − m2B c4 . x2 C

( E ≡ EB )

29

(m A c2 − E)2 = (mC + m D )2 c4 + E2 − m2B c4 =⇒ m2A c4 − 2Em A c2 + E2 = (mC + m D )2 c4 + E2 − m2B c4 2Em A = m2A + m2B − (mC + m D )2 c2 . 2 m A + m2B − (mC + m D )2 2 Emax = c . 2m A Are we sure this is a maximum? Maybe it’s a minimum, and the maximum occurs at the end of the interval (x = 0 or x = 1). We can test for this by calculating E at the end points: At x = 0, q m A c2 = E + mC c2 + m2D c4 + E2 − m2B c4 . q (m A − mC )c2 − E = (m2D − m2B )c4 + E2

(m A − mC )2 c4 − 2E(m A − mC )c2 + E2 = (m2D − m2B )c4 + E2 (m A − mC )2 + m2B − m2D 2 c 2( m A − m C ) (For x = 1, just interchange C and D.) Now E0 =

m2B (m2B − m2D ) 2 ( m C + m D )2 ( E − E ) = m + − − m + m − max 0 A A C mA mA (m A − mC ) c2 2 (m A − mC ) m B − (mC + m D )2 + m A mC (m A − mC ) − m A (m2B − m2D ) = . m A (m A − mC ) So

2m A (m A − mC ) ( Emax − E0 ) c2 = m A m2B − m2C − m2D − 2mC m D + m A mC − m2C − m2B + m2D

− mC (m2B − m2C − m2D − 2mC m D ) = mC m A (−2mC − 2m D + m A ) − m2B + m2C + m2D + 2mC m D . 2m A (m A − mC ) ( Emax − E0 ) m C c2

= (m2A + m2C + m2D − 2m A mC − 2m A m D + 2mC m D ) − m2B = (m A − mC − m D )2 − m2B . But m A > m B + mC + m D =⇒ (m A − mC − m D ) > m B , so the right side is positive, and therefore Emax > E0 . (By the same token, Emax > E1 (at x = 1), which differs only by C ↔ D.) Conclusion: Emax is a maximum, not a minimum.

30


(b) Emin = me c2 = 0.511 MeV ;

Emax =

(m2µ + m2e ) 2 c = 52.8MeV. 2mµ

Problem 3.23

(a) The CM moves at speed v relative to lab. Classically, u = v + v = 2v, but relativistically the velocities add by Eq. 3.5: u=

v+v 2v = 2 . 1 + vv 1 + vc2 c2

Solve for v :

u v2 = 2v =⇒ 2 v2 − 2v + u = 0 c2 c q ! r 2 +2 ± 4 − 4 uc2 c2 u2 =⇒ v = = 1± 1− 2 2 cu2 u c u+u

2

If we use the + sign, for small u we get v ≈ 2 cu , which is wrong (it should be u2 ), so we need the − sign: ! r c2 u2 v= 1− 1− 2 . u c (b) In the CM frame, then: 1

γ2 =

1−

1

=

v2 c2

1−

c2 u2

1−2

1−

1

=

2

2

1 − 2 uc 2 + 1 + 2 uc 2 Let γ0 ≡

q

r 1 2 1− u2

q

1−

u2 c2

u2 c2

+1−

u2 c2

u2 c2

= 2

q

1−

u2 c2

− 1−

u2 c2

.

(the γ in the lab frame). Then

c

r

u2 1 u2 1 = 0; = 1 − 0 2 . Therefore 2 2 γ c c (γ ) h i 1 − (γ10 )2 ( γ 02 − 1) (γ0 − 1)(γ0 + 1) 1 2 i = γ = h = = ( γ 0 + 1) 0 − 1) 0 − 1) 1 1 2 ( γ 2 ( γ 2 2 γ 0 − ( γ 0 )2 1−

γ=

q

1 0 2 (γ

+ 1)

31

(c) So

T = (γ − 1)mc2 =

h

√1 2

p

i γ0 + 1 − 1 mc2 .

2 1 0 T + 1 . Therefore Hence ( γ + 1) = 2 mc2 2 T 2T 2 2T 4T 0 γ = −1 + 2 + 1 = + + 2 + 1. 2 2 4 2 4 mc mc m c m c T 0 = (γ0 − 1)mc2 (kinetic energy in lab), so 2T 2 4T 2T 2 T 2 T0 = + mc = + 4T = 4T 1 + . X mc2 mc2 2mc2 m2 c4

Problem 3.24

Let p A be the 4-momentum of A before the collision, and let q A be the 4momentum after the collision, in the CM frame; p0A , q0A are the corresponding quantities in the Breit fame. Now p A · q A = p0A · q0A (since the 4-dimensional dot product is invariant): E2A E0A2 − p · q = − p0A · q0A . A A c2 c2 (Note: The incoming A and outgoing A have the same energy in CM – only the direction of A’s momentum changes. Likewise, the incoming and outgoing A have same energy (E0A ) in Breit frame, since their momenta are opposite.) Now p A · q A = p2A cos θ, where p2A =

1 2 1 E − m2A c4 , and p0A · q0A = −p0A2 = − 2 E0A2 − m2A c4 . c2 A c

So E 02 E2A 1 1 − 2 E2A − m2A c4 cos θ = 2A + 2 E0A2 − m2A c4 ; 2 c c c c E2A (1 − cos θ ) + m2A c4 cos θ = 2E0A2 − m2A c4 . 1 + cos θ θ θ 1 − cos θ E0A2 = E2A + m2A c4 = E2A sin2 + m2A c4 cos2 2 2 2 2 q or (dropping the subscripts) E0 = E2 sin2 2θ + m2 c4 cos2 2θ .

32


y

y'

p

q

A

p'

A

q/2 q

q

A

x q'

p

B

B

B

Velocity makes angle θ/2 with incident A. ing p A and q A – see diagram.) vE 0 q Ax = γ q Ax − =0 c c

q'

A

x' p'

B

(v points along the line bisect-

c2 |q A | cos 2θ c2 q A x c2 θ =⇒ v = = = cos E E E 2 s 2 θ m A c2 c 1− v = cos 2 E

q 1 2 2 4 E − mAc c

Problem 3.25

(a) 1 c2 1 = 2 c 1 = 2 c

s+t+u =

h

( p A + p B )2 + ( p A − p C )2 + ( p A − p D )2

i

h

p2A + 2p A · p B + p2B + p2A − 2p A · pC + p2C + p2A − 2p A · p D + p2D h i m2A c2 + m2B c2 + m2C c2 + m2D c2 + 2p A · ( p A + p B − pC − p D )

Conservation of energy and momentum requires p A + p B = pC + p D , so s + t + u = m2A + m2B + m2C + m2D .

X

(b) c2 s = p2A + 2p A · p B + p2B = p2A + 2

E A EB − p A · pB c c

+ p2B

i

33

In the CM frame, p A · pB = −p2A , so p2A =

E2 E2A − p2A =⇒ p2A = 2A − p2A = −p A · pB 2 c c

EB2 E2 − p2B = 2B − p2A 2 c c s q E2 EB =⇒ = p2B + p2A = p2B + 2A − p2A . c c

and

p2B =

So 

s

E c2 s = p2A + 2  A c

 2 2 E E p2B + 2A − p2A + 2A − p2A  + p2B c c

E2 E c2 s + p2A − p2B − 2 2A = 2 A c c

s p2B +

E2A − p2A c2

Squaring both sides,

(c2 s + p2A − p2B )2 − E2A =

4 4 E E 4 2 4 2 2 2 2 2 2 A A ( c s + p = ( p − p − p ) E ) E + 4 + 4 B A A A A c2 c2 B c4 c4

(c2 s + p2A − p2B )2 4s

ECM A =

(s + m2A − m2B )c2 √ . 2 s

(c) In the lab frame, pB = 0 and EB = m B c2 , so c2 s = p2A + 2 Elab A =

E A m B c2 + p2B = m2A c2 + 2m B E A + m B c2 c c

(s − m2A − m2B )c2 . 2m B

(d) In the CM frame, ( p A + p B

)2

=

EA E + B c c

2

=

2 √ 2 ETOT CM = =⇒ ETOT sc 2 c

Problem 3.26

1 1 s = 2 ( p A + p B )2 = 2 c c

"

E A + EB c

#

2

− (p A + p B )

2

34


In the CM frame, p A + pB = 0 and E A = EB =

p

p2 c2 + m2 c4 , so

4(p2 c2 + m2 c4 ) 4(p2 + m2 c2 ) . X = 2 c c2 # " 1 1 E A − EC 2 2 2 − ( p A − pC ) t = 2 ( p A − pC ) = 2 c c c

s=

1 c2

But E A = EC and (p A − pC )2 = p2A + p2C − 2p A · pC = 2p2 (1 − cos θ ), so t=

−2p2 (1 − cos θ ) . c2

X

u is the same as t, except that p A · pD = −p A · pD = −p2 cos θ: u=

−2p2 (1 + cos θ ) . c2

X

Problem 3.27 g

E

q

e

g

(before)

E', pg

f

e (after)

p

e

In this problem I’ll use pe for the three-momentum of the charged particle, and pγ for the three-momentum of the photon. Conservation of momentum: pe sin φ = pγ sin θ =⇒ sin φ =

E0 sin θ cpe s

E0 E = pγ cos θ + pe cos φ = cos θ + pe c c

1−

E0 sin θ pe c

or E − E0 cos θ =

q

p2e c2 − ( E0 sin θ )2 .

=⇒ p2e c2 = E2 − 2EE0 cos θ + E02 cos2 θ + E02 sin2 θ =⇒ p2e c2 = E2 − 2EE0 cos θ + E02

2

35

Conservation of energy: q p E + mc2 = E0 + m2 c4 + p2e c2 = E0 + m2 c4 + E2 − 2EE0 cos θ + E02 02 E − E0 + mc2 )2 = E2 + E + m2 c4 − 2EE0 + 2Emc2 − 2E0 mc2 02 = m2 c4 + E2 − 2EE0 cos θ + E hc hc 2mc2 ( E − E0 ) = 2EE0 (1 − cos θ ); E0 = 0 , E = λ λ 1 h2 c2 1 =⇒ m c2 hc (1 − cos θ ) − 0 = λ λ λλ0 0 λ −λ h h mc = 0 (1 − cos θ ) =⇒ λ0 = λ + (1 − cos θ ) 0 mc λλ λλ

37

4

Symmetries Problem 4.1

A A A A A A

movement of corners → A, B → B, C → C → A, B → C, C → B → B, B → A, C → C → B, B → C, C → A → C, B → B, C → A → C, B → A, C → B

symmetry operation I Ra Rc R− Rb R+

Problem 4.2

I I I R+ R+ R− R− Ra Ra Rb Rb Rc Rc

R+ R+ R− I Rc Ra Rb

R− R− I R+ Rb Rc Ra

Ra Ra Rb Rc I R+ R−

Rb Rb Rc Ra R− I R+

Rc Rc Ra Rb R+ R− I

The group is not Abelian; the multiplication table is not symmetrical across the main diagonal (for example, R+ R a = Rb , but R a R+ = Rc ).

38

4 Symmetries

Problem 4.3

(a) 

 100 D ( I ) = 0 1 0 ; 001



 0 0 1 D ( R − ) = 1 0 0 ; 0 1 0



 10 0 D ( R a ) = 0 0 1 ; 01 0



 0 0 1 D ( R b ) = 0 1 0 ; 1 0 0



 0 1 0 D ( R c ) = 1 0 0 . 0 0 1

(b) I, R+ and R− are represented by 1; R a , Rb , and Rc are represented by −1. This representation is not faithful.

Problem 4.4

a

d

c b

The square has 8 symmetry operations: doing nothing (I), clockwise rotation through 90◦ (R+ ), rotation through 180◦ (Rπ ), counterclockwise rotation through 90◦ (R− ), and flipping about the vertical axis a (R a ), the horizontal axis b (Rb ), or the diagonal axes c (Rc ) or d (Rd ).

I R+ Rπ R− Ra Rb Rc Rd

I I R+ Rπ R− Ra Rb Rc Rd

R+ R+ Rπ R− I Rd Rc Ra Rb

Rπ Rπ R− I R+ Rb Ra Rd Rc

R− R− I R+ Rπ Rc Rd Rb Ra

Ra Ra Rc Rb Rd I Rπ R+ R−

Rb Rb Rd Ra Rc Rπ I R− R+

Rc Rc Rb Rd Ra R− R+ I Rπ

Rd Rd Ra Rc Rb R+ R− Rπ I

The group is not Abelian (for example, R+ R a = Rc , but R a R+ = Rd ).

39

Problem 4.5

(a) Let A and B be two n × n unitary matrices. Then f )∗ ( AB) = B˜ ∗ ( A˜ ∗ A) B = B˜ ∗ B = 1, ( AB so AB is also unitary (the set of n × n unitary matrices is closed under multiplication). The usual matrix identity is unitary. All unitary matrices have inverses (A−1 = A˜ ∗ ), which are themselves unitary. Finally, matrix multiplication is associative, so U (n) a group. (b) Suppose that the unitary matrices A and B both have determinant 1. Then det( AB) = det( A) det( B) = 1. We already know that AB is unitary, so the set of n × n unitary matrices with determinant 1 is closed. The matrix identity has determinant 1, and det( A−1 ) = 1/ det( A) = 1, so the set contains the appropriate inverse and identity elements; SU (n) is a group. (c) O(n), the real subset of U (n) is closed by the same argument that U (n) is closed: f )( AB) = B˜ ( AA ˜ ) B = BB ˜ = 1. ( AB The identity, inverse, and associativity requirements are still met, so O(n) is a group. (d) The set of n × n orthogonal matrices is closed, as is the set of n × n matrices with determinant 1. Therefore, SO(n) is closed. If A ∈ SO(n), so is A−1 , and I ∈ SO(n); SO(n) is a group.

Problem 4.6 y'

y

A

ay f

q ax

a x = a cos φ;

ay = a sin φ

a'y

x'

a'x x

( a = length of A)

a0x = a cos(φ − θ ) = a(cos φ cos θ + sin φ sin θ ) = cos θa x + sin θay a0y = a sin(φ − θ ) = a(sin φ cos θ − cos φ sin θ ) = cos θay − sin θa x

40

4 Symmetries

∴

0 ax cos θ sin θ ax = . a0y − sin θ cos θ ay

˜ = RR

cos θ sin θ − sin θ cos θ

cos θ − sin θ sin θ cos θ

(cos2 θ + sin2 θ ) (cos θ sin θ − sin θ cos θ ) (sin θ cos θ − cos θ sin θ ) (sin2 θ + cos2 θ )

=

R=

cos θ sin θ − sin θ cos θ

.

=

1 0 0 1

so R is orthogonal. X det( R) = cos2 θ + sin2 θ = 1

Group is SO(2).

R ( θ1 ) R ( θ2 ) cos θ1 sin θ1 cos θ2 sin θ2 = − sin θ1 cos θ1 − sin θ2 cos θ2 (cos θ1 cos θ2 − sin θ1 sin θ2 ) (cos θ1 sin θ2 + sin θ1 cos θ2 ) = (− sin θ1 cos θ2 − cos θ1 sin θ2 ) (− sin θ1 sin θ2 + cos θ1 cos θ2 ) cos(θ1 + θ2 ) sin(θ1 + θ2 ) = = R ( θ1 + θ2 ) = R ( θ2 + θ1 ) − sin(θ1 + θ2 ) cos(θ1 + θ2 )

= R ( θ2 ) R ( θ1 ). Yes, it is Abelian.

Problem 4.7

˜ MM =

1 0 0 −1

1 0 0 −1

=

1 0 0 1

Yes, it is in O(2).

det( M ) = −1. No, it is not in SO(2). ax ax M = , so a0x = a x ; a0y = − ay . ay − ay No, this is not a possible rotation of the xy plane. (It’s a reflection in the x axis.)

Problem 4.8

For a spinning solid sphere, I = 25 mr2 and ω = v/r, so 1 2 2 v 5 h¯ h¯ = L = Iω = mr , v= . 2 5 r 4 mr

41

An electron has mass m = 9 × 10−31 kg and radius r < 10−18 m, giving v > 1014 m/s, much faster than the speed of light. Evidently this classical model cannot be taken literally.

Problem 4.9

(a) Before: 2 × 2 × 2 = 8 states. (↑↑↑, ↑↑↓, ↑↓↑, ↑↓↓, ↓↑↑, ↓↑↓, ↓↓↑, ↓↓↓) After: s = 23 is 4 states; s = 12 is 2 states, and occurs twice, so we get 4 + 2 + 2 = 8. X (b) 2 and 1 −→ 3, 2 or 1. 3 and 1/2 −→ 7/2 −→ 5/2 2 and 1/2 −→ 5/2 −→ 3/2 1 and 1/2 −→ 3/2 −→ 1/2

(8 states) (6 states) (6 states) (4 states) (4 states) (2 states) 8 + 6 + 6 + 4 + 4 + 2 = 30 states in all

We get total angular momenta 27 ,

5 2

(twice),

3 2

(twice), and 12 .

We had 5 × 3 × 2 = 30 states to begin with, and 30 at the end. X

Problem 4.10

Combining the spins of p and e, we can get s = 1 or s = 0, but not s = 12 . So this process would violate conservation of angular momentum. (If the p–e system carries orbital angular momentum we could achieve s > 1, but still only integer values: 12 remains inaccessible.) 1 2

would do , for combining s = 1 with s =

ing s = 0 with s = s = 1 with s =

3 2

1 2

yields 12 ). But

yields

5 3 1 2 , 2 , and 2 .

3 2

1 2

yields

3 2

and

1 2

(and combin-

would also be OK , since combining

If we allow for orbital angular momentum

in the final state, any half-integer spin would work. But likely possibility.

1 2

seems the most

42

4 Symmetries

Problem 4.11

The proton has spin 12 and the pion has spin 0, so the total spin in the final state is 21 . This must to give s∆ = 32 . Thusil h combine with orbital angular h i momentum could be 1

(s = 12 ) + (l = 1) ⇒

3 2

or

1 2

or 2

(s = 12 ) + (l = 2) ⇒

5 2

or

3 2

.

Problem 4.12 1 2

× 1 Clebsch–Gordan table, I read q q | 23 21 i = 13 |1 1i| 12 − 12 i + 23 |1 0i | 12 12 i . | {z }

From the

spin up

Probability of spin up is 2/3.

Problem 4.13

From the 2 × 2 table, I read q q q 18 |2 0i|2 0i = 35 |4 0i + 0|3 0i + − 27 |2 0i + 0|1 0i + 15 |0 0i. So you could get j = 4, probability

18 35 ;

j = 2, probability 72 ;

or j = 0, probability 15 .

(What this means, of course, is that the total angular momentum squared could come out 4(5)h¯ 2 = 20¯h2 ( j = 4) or 2(3)h¯ 2 = 6¯h2 ( j = 2) or 0( j = 0).) Do the probabilities add to 1? 18 2 1 18 + 10 + 7 35 + + = = = 1. X 35 7 5 35 35

Problem 4.14

From the 2 ×

3 2

table, I read q q 3 3 3 3 1 | 52 − 21 i = 27 | 2 1 i| − i + 70 2 2 35 |2 0i| 2 − 2 i q q 5 6 + − 14 |2 −1i| 23 12 i + − 35 |2 −2i| 23

3 2 i.

43

So you could get m1 = 1, probability 27/70; or 0, probability 3/35; or − 1, probability 5/14; or − 2, probability 6/35. (That is, Sz for the spin-2 particle could be h¯ , 0, −h¯ , or −2¯h.) Do the probabilities add to 1? 27 3 5 6 27 + 6 + 25 + 12 70 + + + = = = 1. X 70 35 14 35 70 70

Problem 4.15

√ ! h¯ h¯ 1/√2 = χ+ ; eigenvalue . 2 2 1/ 2 ! √ h¯ 0 1 h¯ h¯ 1/ √2 Sˆ x χ− = = − χ− ; eigenvalue − . 2 10 2 2 −1/ 2 h¯ Sˆ x χ+ = 2

01 10

Problem 4.16

2 2 1 1 | a|2 + |b|2 = √ (α + β) + √ (α − β) 2 2 1 = [(α∗ + β∗ ) (α + β) + (α∗ − β∗ ) (α − β)] 2 1 2 = | α | + α ∗ β + β ∗ α + | β |2 + | α |2 − α ∗ β − β ∗ α + | β |2 2 = |α|2 + | β|2 = 1. X

Problem 4.17

(a) h¯ 2

0 −i i 0

α α −i (h¯ /2) β λα =λ =⇒ = β β i (h¯ /2)α λβ

h¯ −i β = λα; 2

h¯ h¯ 2 h¯ 2 h¯ i α = λβ ⇒ −i β = λ λβ ⇒ β = λ2 β ⇒ λ = ± 2 2 i¯h 4 2

44

4 Symmetries

Plus sign: h¯ h¯ −i β = α ⇒ β = iα ⇒ χ+ = 2 2

α α = . β iα

Minus sign: h¯ h¯ −i β = − α ⇒ β = −iα ⇒ χ− = 2 2

α −β

=

α −iα

.

√ In both cases normalization (|α|2 + | β|2 = 1) gives α = 1/ 2. Conclusion: 1 1 Eigenspinors are χ± = √ . Eigenvalues are ±h¯ /2. 2 ±i (b) 1 1 1 α 1 = cχ+ + dχ− = c √ + d√ = β 2 i 2 −i

√1 (c + d ) 2 √i (c − d ) 2

!

√ 1 i 1 α = √ (c + d) β = √ (c − d) ⇒ 2(α − iβ) = 2c ⇒ c = √ (α − iβ) 2 2 2 √ 1 or 2(α + iβ) = 2d ⇒ d = √ (α + iβ) 2 You could get

± 12 h¯ , probability 12 |α ∓ iβ|2 .

Problem 4.18

(a) From Eq. 4.24, probability of 21 h¯ is 2 1 2 2 9 1 √ (α + β) = √1 √ √ + = . 2 2 10 5 5 Probability of − 12 h¯ is 2 1 √ (α − β) = 2

1 1 2 2 1 √ 2 √5 − √5 = 10 .

(b) From Problem 4.17, probability of 12 h¯ is 1 1 2i 2 1 1 1 √ √ √ − = (1 − 2i )(1 + 2i ) = (1 + 4) = . 2 10 10 2 5 5 Probability of − 12 h¯ is 1 1 2i 2 1 √ 2 √5 + √5 = 2 .

45

(c) Probability of 21 h¯ is |α|2 =

1 4 ; probability of − 12 h¯ is | β|2 = . 5 5

Problem 4.19

(a) σx2 =

0 1 1 0

0 −i i 0

=

0 −i i 0

0 −i i 0

10 01

X

10 X 01 1 0 1 0 10 X = σz2 = 0 −1 0 −1 01

σy2 =

01 10

=

(b) σx σy =

σz σy σz σx σx σz

0 −i i 0

=

0 1 1 0

i 0 0 −i

=i

1 0 0 −1

= iσz

−i 0 = −iσz 0 i 0 −i 1 0 0 i 0 1 = = =i = iσx i 0 0 −1 i 0 1 0 1 0 0 −i 0 −i = = = −iσx 0 −1 i 0 −i 0 1 0 01 01 0 −i = = =i = iσy 0 −1 10 −1 0 i 0 01 1 0 0 −1 = = = −iσy 10 0 −1 1 0

σy σx = σy σz

01 10

=

Problem 4.20

(a)

[σi , σj ] = σi σj − σj σi = δij + ieijk σk − δji − ie jik σk But δij = δji ,

e jik = −eijk .

So [σi , σj ] = 2ieijk σk .

46

4 Symmetries

(b)

{σi , σj } = σi σj + σj σi = δij + ieijk σk + δji + ie jik σk = 2δij (c)

(σ · a)(σ · b) = ∑ σi ai σj b j = ∑ ai b j (σi σj ) = ∑ ai b j (δij + ieijk σk ) i,j

i,j

i,j

= ∑ ai b j δij + i ∑ eijk ai b j σk = a · b + iσ · (a × b) i,j

i,j

Problem 4.21

(a) Noting that σz2 = 1, σz3 = σz , etc.,

(iθσz )2 (iθσz )3 eiθσz = 1 + iθσz + + +··· 2 3! θ4 θ3 θ5 θ2 + − · · · + iσz θ − + +··· = 1− 2 4! 3! 5! = cos θ + iσz sin θ In particular eiπσz /2 = cos π/2 + iσz sin π/2 = iσz . (b) Similarly (all we need is σy2 = 1) eiθσy = cos θ + iσy sin θ, so ◦

◦

U = cos 90 − i sin 90 σy = −iσy = −i 1 0 −1 1 0 U = = , 0 1 0 0 1

0 −i i 0

=

0 −1 1 0

which is spin down.

(c) θ·σ 1 θ·σ 2 1 θ·σ 3 + +··· + −i −i U (θ) = e−iθ·σ /2 = 1 + −i 2 2 2 3! 2 But (θ · σ )2 = θ · θ + iσ · (θ × σ ) = θ 2 . So (θ · σ ) 1 θ 2 1 θ2 U (θ) = 1 − i − + i (θ · σ ) 3 + · · · 2 2 2 3! 2 " # " # 2 4 1 θ 1 θ θ·σ θ 1 θ 3 + −··· −i − +··· = 1− 2 2 4! 2 θ 2 3! 2 θ θ ˆ = cos − i (θ · σ ) sin . 2 2

47

Problem 4.22

(a) U = cos

θ θ − i (θˆ · σ ) sin ; 2 2

U † = cos

θ θ + i (θˆ · σ † ) sin 2 2

But σ † = σ (the Pauli matrices are Hermitian), so U † = cos

θ θ + i (θˆ · σ ) sin . 2 2

θ θ θ θ ˆ ˆ cos + i (θ · σ ) sin UU = cos − i (θ · σ ) sin 2 2 2 2 θ θ θ = cos2 − i (θˆ · σ ) sin cos 2 2 2 θ θ θ ˆ + i (θ · σ ) cos sin + (θˆ · σ )(θˆ · σ ) sin2 2 2 2 θ θ = cos2 + sin2 = 1. X 2 2 †

[Note: I used Problem 4.20 (c) to show that

(θˆ · σ )(θˆ · σ ) = θˆ · θˆ + iσ · (θˆ × θˆ) = 1.] (b) θˆz (θˆx − i θˆy ) ˆθ · σ = θˆx 0 1 + θˆy 0 −i + θˆz 1 0 = 10 i 0 0 −1 (θˆx + i θˆy ) −θˆz θ 1 0 θˆz (θˆx − i θˆy ) − i sin 0 1 −θˆz 2 (θˆx + i θˆy ) (cos 2θ − i θˆz sin 2θ ) −i sin 2θ (θˆx − i θˆy ) = −i sin θ (θˆx + i θˆy ) (cos θ + i θˆz sin θ )

U (θ) = cos

θ 2

2

2

2

θ θ θ θ θ ˆ ˆ det U = cos − i θz sin cos + i θz sin + sin2 (θˆx + i θˆy )(θˆx − i θˆy ) 2 2 2 2 2 θ θ θ θ θ = cos2 + θˆz2 sin2 + sin2 (θˆx2 + θˆy2 ) = cos2 + sin2 (θˆx2 + θˆy2 + θˆz2 ) 2 2 2 2 2 θ θ = cos2 + sin2 = 1. X 2 2

48

4 Symmetries

Problem 4.23

(a) Call the three column vectors χ+ , χ0 , χ− . We have: 

Sˆz χ+ = h¯ χ+ ;

Sˆz χ0 = 0;

Sˆz χ− = −h¯ χ− ;

so

 1 0 0 Sˆz = h¯  0 0 0 . 0 0 −1

(b)

√ √ Sˆ+ |1 0i = h¯ 2 − 0|1 1i = 2¯h|1 1i; √ √ Sˆ+ |1 − 1i = h¯ 2 − 0|1 0i = 2¯h|1 0i.           0 0 1 0 1 √ √ ∴ Sˆ+ 0 = 0; Sˆ+ 1 = 2¯h 0 ; Sˆ+ 0 = 2¯h 1 . 1 0 0 0 0 √   0 2¯h √0 ∴ Sˆ+ = 0 0 2¯h. 0 0 0 √ √ Sˆ− |1 1i = h¯ 2 + 0|1 0i = 2¯h|1 0i; √ √ Sˆ− |1 0i = h¯ 2 + 0|1 − 1i = 2¯h|1 − 1i; Sˆ− |1 − 1i = 0.   0 0 √0 ∴ Sˆ− =  2¯h √0 0. 0 2¯h 0 Sˆ+ |1 1i = 0;

(c)    01 0 0 −i 0 h¯ 1 h¯ 1 Sˆ x = (Sˆ+ + Sˆ− ) = √ 1 0 1 ; Sˆy = (Sˆ+ − Sˆ− ) = √  i 0 −i  2 2i 2 01 0 2 0 i 0 

(d) Represent the four spin 32 states (ms = + 23 , + 12 , − 12 , − 32 ) by the column vectors         1 0 0 0 0 1 0 0  ,  ,  ,  . 0 0 1 0 0

0

0

1

49

3 1 = h¯ | 32 32 i; Sˆz | 32 12 i = h¯ | 23 12 i; 2 2 1 3 Sˆz | 23 − 12 i = − h¯ | 23 − 12 i; Sˆz | 32 − 32 i = − h¯ | 23 − 32 i 2 2   3 0 0 0 0 1 0 0 h ¯ . =⇒ Sˆz =  2  0 0 −1 0  0 0 0 −3 q √ 3 3 3 3 3 3¯h| 2 2 i; Sˆ+ | 32 32 i = 0; Sˆ+ | 32 12 i = h¯ 15 4 − 4|2 2i = q 1 3 1 3 1 Sˆ+ | 32 − 12 i = h¯ 15 4 + 4 | 2 2 i = 2¯h | 2 2 i; q √ 3 3 3 1 Sˆ+ | 32 − 32 i = h¯ 15 − | − i = 3¯h| 2 − 12 i; 4 4 2 2 Sˆz | 23

3 2i

 √ 0 3¯h 0 0 =⇒ Sˆ+ =  0 0 0 0 Sˆ− | 32

3 2i

= h¯

Sˆ− | 32

1 2i

= h¯

 0 0 2¯h √0  . 0 3¯h 0 0

√

q

15 4

− 34 | 32 12 i =

q

15 4

+ 14 | 32 − 12 i = 2¯h| 32 − 21 i;

Sˆ− | 32 − 12 i = h¯

q

15 4

√0  3¯h =⇒ Sˆ− =   0 0 

3¯h| 23

− 34 | 23 − 32 i = 0 0 0 0 2¯h √0 0 3¯h

√

1 2 i;

3¯h| 32 − 32 i;

 0 0 . 0 0

√ 0 3 0 √  1 h ¯ 3 0 2  Sˆ x = (Sˆ+ + Sˆ− ) =  2 2  0 2 √0 0 0 3 

 0  √0  ; 3 0

 √ 3i 0 0 0 − √  1 ˆ h¯  0   3i 0 −2i √ ˆ ˆ Sy = ( S+ − S− ) =  . 2i 2 0 2i √0 − 3i  0 0 3i 0 

Sˆ− | 32 − 32 i = 0

50

4 Symmetries

Problem 4.24

Ω − = |0 0i;

Σ + = |1 1i;

Ξ0 = | 12 21 i;

ρ + = |1 1i;

η = |0 0i;

Problem 4.25

(a) 1 2 +0 = X 3 3 1 1 1 1 Q=− + +0 = − X 2 2 3 3 1 1 1 Q = 0+ −1 = − X 2 3 3 1 1 Q= + 2 2

u: d: s:

(b) 1 1 1 2 − Q=− + − +0 = − u¯ = 2 2 3 3 1 1 1 1 d¯ = | 21 21 i; Q = + − +0 = X 2 2 3 3 1 1 1 s¯ = |0 0i; Q = 0 + − +1 = X 2 3 3

| 21

1 2 i;

X

Problem 4.26

(a) 1 Q = I3 + ( A + S + C + B + T ) 2 (b) I3 =

1 (U + D ) ; 2

A=

1 (U + C + T − D − S − B ) 3

K¯ 0 = | 12 21 i.

51

(c) 1 1 1 (U + C + T − D − S − B ) + S + C + B + T Q = (U + D ) + 2 2 3

=

1 [2(U + B + T ) + ( D + S + B)] 3

Problem 4.27 2 Itot = (I1 + I2 )2 = I12 + I22 + 2I1 · I2 .

Here I12 = I22 =

1 2

1 +1 2

3 = . 4

3 3 + + 2I1 · I2 , or I1 · I2 = − 34 . 4 4 3 3 = 1(1 + 1) = 2 in triplet, so 2 = + + 2I1 · I2 , or I1 · I2 = 14 . 4 4

2 Itot = 0 in singlet state, so 0 = 2 Itot

Problem 4.28

(a)

M a = M f = M3 ; 2 M3 + 3 1 M c = M d = M3 + 3

1 M ; 3 1 2 M ; 3 1 √ √ M g = Mh = Mi = M j = ( 2/3)M3 − ( 2/3)M1 .

Mb = Me =

(b) σa : σb : σc : σd : σe : σ f : σg : σh : σi : σj = 9|M3 |2 : |2M3 + M1 |2 : |M3 + 2M1 |2 : |M3 + 2M1 |2 : |2M3 + M1 |2 : 9|M3 |2 : 2|M3 − M1 |2 : 2|M3 − M1 |2 : 2|M3 − M1 |2 : 2|M3 − M1 |2 (c) σa : σb : σc : σd : σe : σ f : σg : σh : σi : σj = 9 : 4 : 1 : 1 : 4 : 9 : 2 : 2 : 2 : 2

52

4 Symmetries

Problem 4.29

Using the method in Section 4.3: q q ( π − + p : |1 − 1i| 21 12 i = 13 | 32 − 12 i − 23 | 21 − 12 i

|1 1i| 21 12 i = | 32 23 i q q  1 2 3 1 1 1 1 0 + Σ0 :  − i| 1 0 i = | − i + | − 12 i K |   2 2 2 q3 2 q3 2 K + + Σ− : | 21 12 i|1 − 1i = 13 | 23 − 12 i − 23 | 21 − 12 i    + K + Σ+ : | 21 12 i|1 1i = | 23 32 i √ √ 2 2 1 2 M c = M3 ; M a = M3 − M1 ; M b = M3 + M1 3 3 3 3 2 1 ∴ σa : σb : σc = |M3 − M1 |2 : |M3 + 2M1 |2 : |M3 |2 . 9 9 π+ + p :

If the I = 3/2 channel dominates, M3 M1 , so σa : σb : σc = 2 : 1 : 9. If the I = 1/2 channel dominates, M1 M3 , so σa : σb : σc = 2 : 4 : 9.

Problem 4.30

(

K− + p : K¯ 0 + p :

  Σ0 + π 0 :       Σ+ + π − :   Σ+ + π 0 :      Σ0 + π + :

q q | 12 − 12 i| 12 21 i = 12 |1 0i − 12 |0 0i | 12 12 i| 21 12 i = |1 1i q q |1 0i|1 0i = 23 |2 0i − 13 |0 0i q q q |1 1i|1 − 1i = 16 |2 0i + 12 |1 0i + 13 |0 0i q q |1 1i|1 0i = 12 |2 1i + 12 |1 1i q q |1 0i|1 1i = 12 |2 1i − 12 |1 1i

q q (a) M a = − 12 13 M0 ; Itot = 0 q q q q (b) Mb = 12 12 M1 + 12 13 M0 ; q (c) Mc = 12 M1 ; Itot = 1 q (d) Md = − 12 M1 ; Itot = 1

Itot = 0 or 1

53

σa : σb : σc : σd =

1 1 |M0 |2 : | 6 2

q

1 3 M0

+

q

2 1 2 M1 |

:

1 1 |M1 |2 : |M1 |2 . 2 2

If I = 0 dominates, σa : σb : σc : σd = 1 : 1 : 0 : 0 If I = 1 dominates, σa : σb : σc : σd = 0 : 1 : 2 : 2

Problem 4.31

If M1 M3 , Eq. 4.50 becomes σa : σc : σj = 0 : 2 : 1, so σtot (π + + p) = 0, σtot (π − + p) = 3, and σtot (π + + p)/σtot (π − + p) = 0. There is no sign of a resonance in σtot (π + + p) at 1525, 1688, or 2190, so these must be I = 1/2. (The ratios are not zero, of course—evidently there is a lot of resonant background.) There is a clear resonance in σtot (π + + p) at 1920; this must be I = 3/2. (The ratio is σtot (π + + p)/σtot (π − + p) = 43/36 = 1.2, which is not very close to 3—presumably this is again due to nonresonant background.) The nomenclature should be N (1525), N (1688), ∆(1920), N (2190). The Particle Physics Booklet lists N (1520, N (1680), ∆(1920), N (2190) (and there are others with roughly the same mass).

Problem 4.32

n

Σ∗ 0 :

|1 0i    ( a) Σ+ + π − :  

q q q |1 1i|1 − 1i = 16 |2 0i + 12 |1 0i + 13 |0 0i q q (b) Σ0 + π 0 : |1 0i|1 0i = 23 |2 0i − 13 |0 0i  q q q    (c) Σ− + π + : |1 − 1i|1 1i = 1 |2 0i − 1 |1 0i + 1 |0 0i 6 2 3 q q M a = 12 M1 , Mb = 0, Mc = − 12 M1 . σa : σb : σc = 1 : 0 : 1 I’d expect to see about 50 decays each to Σ+ + π − and Σ− + π + , but none to Σ0 + π 0 .

Problem 4.33

(a) Isospin must be zero. (b) The deuterons carry I = 0, so the isospin on the left is zero. The α has I = 0, and π has I = 1, so the isospin on the right is one. This process does not conserve isospin, and hence is not a possible strong interaction.

54

4 Symmetries

(c) There are five possible 4-nucleon states:

(nnnn), (nnnp) = 4 H, (nnpp) = 4 He, (nppp) = 4 Li, ( pppp) = 4 Be In principle they could form an I = 2 multiplet, but since 4 H and 4 Li do not exist, this is out. No: 4 Be and (nnnn) should not exist. (4 H, 4 He, and 4 Li could make an I = 1 multiplet, but, again, 4 H and 4 Li do not exist, so this too is out. Evidently four nucleons bind only in the I = 0 combination, making 4 He.)

Problem 4.34

(a) Suppose f is an eigenfunction of P: P f = λ f . Then P2 f = P ( P f ) = P ( λ f ) = λ ( P f ) = λ ( λ f ) = λ2 f . But P2 = 1, so λ2 = 1, and hence λ = ±1. X (b) Let f ± ( x, y, z) ≡

1 2

[ f ( x, y, z) ± f (− x, −y, −z)] . Then f = f + + f − . Note

that P f± =

1 2

[ f (− x, −y, −z) ± f ( x, y, z)] = ± f ± ,

so f ± is an eigenfunction of P, with eigenvalue ±1.

Problem 4.35

(a) No. P would transform a (left-handed) neutrino into a right-handed neutrino, which doesn’t even exist (in the massless limit). (b) The K has intrinsic parity −1, so the two-pion decay violates conservation of parity.

55

Problem 4.36

(a) Equation 4.60 says G = (−1) I C, so π : I = 1, C = 1 ⇒ G = −1 ρ : I = 1, C = −1 ⇒ G = 1 ω : I = 0, C = −1 ⇒ G = −1 η : I = 0, C = 1 ⇒ G = 1 η 0 : I = 0, C = 1 ⇒ G = 1 φ : I = 0, C = −1 ⇒ G = −1 f 2 : I = 0, C = 1 ⇒ G = 1 (b) [[?????]]

Problem 4.37

(a) Since the η and the π’s have spin zero, the final state would need to have orbital angular momentum zero: l = 0 (by conservation of angular momentum). But this means that the parity of the final state is ( Pπ )( Pπ )(−1)l = (−1)(−1)(−1)0 = 1. However, the parity of the η is −1, so this decay is forbidden by conservation of parity. (b) The G-parity of the final state is (−1)3 = −1 (Eq. 4.61). But the G-parity of the η is (−1)0 (1) = +1 (Eq. 4.60 and Table 4.6). So η −→ 3π violates conservation of G-parity, and hence is a forbidden strong interaction.

Problem 4.38

Same baryon number as antiparticle ⇒ meson; same charge as antiparticle ⇒ ¯ since these are already their own antiparneutral. We’re not interested in qq, itlces; we need q1 q¯2 , with Q1 = Q2 = Q: Q = −1/3 : ds¯ ↔ sd,¯ db¯ ↔ bd,¯ sb¯ ↔ bs¯. Q = 2/3 :

uc¯ ↔ cu¯

(only, since t forms no bound states).

¯ 0. So the candidate mesons are K0 ↔ K¯ 0 , B0 ↔ B¯ 0 , Bs0 ↔ B¯ s0 , D0 ↔ D

56

4 Symmetries

Neutron/antineutron oscillation would violate conservation of baryon number. The vector mesons decay by the strong interaction long before they would have a chance to interconvert.

Problem 4.39

First establish the convention for positive charge, as in Section 4.4.3: it is the charge carried by the lepton preferentially produced in the decay of the longlived neutral K meson. Then define right-handedness: it is the helicity of the charged lepton produced in the decay of a positively charged pion. Finally, define “up” as the direction away from the earth, and “front” as the side our eyes are on; cross “up” with “front”, using the right-hand rule, and the result is the side our hearts are on. (Identifying the particle types is easy—just make a list in order of increasing mass.)

Problem 4.40

Penguin:

W

t

b

d pu

-0

B

u p+ d

d

Tree:

d W B0

b d

u

u d

p-

p+

57

5

Bound States Problem 5.1

(a) md − m p − mn = 1875.6 − 938.27 − 939.57 = −2.2, so the binding energy is 2.2 MeV , which is only 0.12% of the total. No , it’s not relativistic. (b) mπ − md − mu = 140 − 340 − 336 = −536, so the binding energy is 536 MeV , which is 3.8 times the total. Yes , it is relativistic.

Problem 5.2

Here n = 1, l = 0, ml = 0, so " #1/2 2 31 e−r/a L10 (2r/a)Y00 (θ, φ) e−iE1 t/¯h . Ψ100 = a 2 But

L10 = 1,

and

Ψ100 =

√ Y00 = 1/ 4π 2 a3/2

so

1 1 e−r/a √ e−iE1 t/¯h = √ e−r/a e−iE1 t/¯h , 2 π πa3

where E1 = −me4 /2¯h2 . Now ∇2 f (r ) = (1/r2 )(d/dr )[r2 d f /dr ], so 1 1 −r/a 2 − iE1 t/¯h 1 d 2 ∇ Ψ100 = √ e r − e a r2 dr πa3 1 1 1 1 2 −r/a 1 2 − iE1 t/¯h − r/a e − = Ψ100 . = √ 2re − r e − a r2 a ar a2 πa3 The left side of the Schrödinger equation is h¯ 2 1 2 e2 lhs = − − Ψ100 − Ψ100 . 2 2m a ar r

58

5 Bound States

But a = h¯ 2 /me2 , so the second term cancels the third, leaving lhs = −

h¯ 2 me4 Ψ100 . Ψ = − 100 2ma2 2¯h2

Meanwhile, the right side is ∂Ψ100 iE1 me4 i¯h = i¯h − Ψ100 = E1 Ψ100 = − 2 Ψ100 . ∂t h¯ 2¯h

X

Normalization requires Z

|Ψ|2 r2 sin θ dr dθ dφ = 1.

Here

1 −2r/a e , and sin θ dθ dφ = 4π, so πa3 Z Z ∞ 1 4 a 3 2 2 −2r/a 2 |Ψ100 | r sin θ dr dθ dφ = = 1. X 4π e r dr = 3 2 3 2 πa a 0 Z

|Ψ100 |2 =

Problem 5.3

There are four states: " #1/2 2 3 1 Ψ200 = e−r/2a L11 (2r/2a) Y00 e−iE2 t/¯h 2a 4(2)3 r −r/2a −iE2 t/¯h 1 √ 2− e e a 4a 2πa #1/2 " 2 3 1 −r/2a 2r e = L30 (2r/2a) Y10 e−iE2 t/¯h 2a 2a 4(6)3

=

Ψ210

1 √ re−r/2a cos θe−iE2 t/¯h 4a2 2πa " #1/2 2 3 1 −r/2a 2r = e L30 (2r/2a) Y1±1 e−iE2 t/¯h 2a 2a 4(6)3

=

Ψ21±1

= ∓

8a2

1 √ πa

re−r/2a sin θe±iφ e−iE2 t/¯h

where E2 = −me4 /8¯h2 , and I used r 1 3 0 0 Y0 = √ , Y1 = cos θ, 4π 4π

Y1±1

r

=∓

3 sin θ e±iφ , 8π

59

and L11 ( x ) = −2x + 4, L30 = 6.

Problem 5.4

E=c

q

1/2 p2 . p2 + m2 c2 = mc2 1 + 2 2 m c

Binomial expansion: 1 1 (1 + e)1/2 = 1 + e − e2 + · · · ⇒ 2 8 2 2 2 (p ) p p2 p4 2 − . E ≈ mc2 1 + = mc + − 2m 8m3 c2 2m2 c2 8m4 c4 So T = E − mc2 ≈

p2 p4 − . 2m 8m3 c2

X

Problem 5.5

From Eq. 5.19, 4 3 4 3 1 4 2 1 = E2 − α mc − − E2 − α mc − 4(2)4 2 2 4(2)4 1 2 1 1 4 1 4 2 (0.511 × 106 ) = 4.53 × 10−5 eV . α mc = = 32 32 137

E3/2 − E1/2

4

2

1 −1 4

By comparison, E2 − E1 = E1

3 3 = − E1 = − (−13.6) = 10.2 eV 4 4

so the fine structure is smaller by a factor of nearly a million.

Problem 5.6

From Eq. 5.20, ∆ES = α5 mc2

1 13 5 2 α mc k(2, 0) ≈ 3 32 4(2)

(using k(2, 0) ≈ 13). From Eq. 5.21, 1 1 1 5 2 2 ∆EP = α5 mc2 k ( 2, 1 ) − ≈ − α mc π (3/2) 32 3π 4(2)3

60

5 Bound States

(I dropped k (2, 1) < 0.05 in comparison with 2/3π = 0.21). So the Lamb shift is 1 5 2 13.2 1 5 2 α mc = (0.511 × 106 ) ∆E = ∆ES − ∆EP ≈ 13 + 3π 32 32 137

= 4.37 × 10−6 eV . E = hν ⇒ ν =

E 4.37 × 10−6 = 1.057 × 109 Hz . = h 2π (6.58 × 10−16 )

(Not bad!)

Problem 5.7

In the absence of any splitting there are 16 completely degenerate states (four orbital states, and for each of these two electron spin states and two proton spin states). Fine structure splits these into two distinct energy levels, one for j = 1/2 and one for j = 3/2 (the 2S1/2 and 2P1/2 states remain degenerate, 2P3/2 is slightly higher). The Lamb shift lifts the 2S1/2 /2P1/2 degeneracy, so there are now three distinct energy levels. Hyperfine splitting further separates each of these into two, according to the value of f (0 or 1, if j = 1/2; 1 or 2, if j = 3/2). So there are 6 energy levels in all: n = 2, l = 0, j = 1/2, f = 0 (one state); n = 2, l = 0, j = 1/2, f = 1 (three states); n = 2, l = 1, j = 1/2, f = 0 (one state); n = 2, l = 1, j = 1/2, f = 1 (three states); n = 2, l = 1, j = 3/2, f = 1 (three states); n = 2, l = 1, j = 3/2, f = 2 (five states)—still 16 states in all. Using Eq. 5.23 with n = 2, j = 1/2, f = 0 (lower sign), and l = 0, 1: γp m −1 1 m ∆ES = =− α4 mc2 α4 mc2 γ p , mp 16 (1/2)(1/2) 4 mp γp −1 1 m m α4 mc2 =− α4 mc2 γ p . ∆EP = mp 16 (1/2)(3/2) 12 m p 1 m ∆E = ∆ES − ∆EP = − α4 mc2 γ p 6 mp 1 4 1 .511 (0.511 × 106 )(2.79) = 3.67 × 10−7 eV . = − 6 938 137 This is about a tenth the Lamb shift.

61

Problem 5.8

For n = 3 we have l = 0 (S), l = 1 ( P), and l = 2 ( D ). For l = 0 we can have either the singlet or the triplet spin configuration (since l = 0, j = s automatically). For l = 1 we can have the singlet (with j = 1 automatically) or the tripet (for which j can be 0, 1, or 2). For l = 2 we have the singlet (with j = 2 automatic) or the triplet (for which j can be 1, 2, or 3). So there are 10 levels in all. But it turns out that 3 P2 and 3 D2 are degenerate, so in fact there are just 9 distinct energy levels. The unperturbed energy of all these states is (Eq. 5.27) E3 = −α2 mc2

1 −13.6 = = −0.756 eV. 36 18

The fine/hyperfine correction is (Eq. 5.29) 11 (1 + e/2) (1 + e/2) 4 2 1 −5 Efs = α mc − = (2.68 × 10 ) 0.1146 − 54 96 (2l + 2) (2l + 1) (with e given by Eq. 5.30). In addition there is an annihilation correction, which applies only to the triplet S state (Eq. 5.31) Eann = α4 mc2

1 = 1.342 × 10−5 eV. 108

In units of µeV (10−6 eV), the corrections are: 1

S0 : l = 0, s = 0, j = 0, e = 0;

∆E = 26.8(0.1146 − 1) = -23.7

3

S1 : l = 0, s = 1, j = 1, e = − 43 ;

∆E = 26.8(0.1146 − 31 ) + 13.42 = 7.56

1

P1 : l = 1, s = 0, j = 1, e = 0;

∆E = 26.8(0.1146 − 13 ) = -5.86

3

7 P2 : l = 1, s = 1, j = 2, e = − 10 ; ∆E = 26.8(0.1146 −

3

P1 : l = 1, s = 1, j = 1, e = 12 ;

3

P0 : l = 1, s = 1, j = 0, e = 2;

1

D2 : l = 2, s = 0, j = 2, e = 0;

3

D3 : l = 2, s = 1, j = 3, e = − 10 21 ;

3

D2 : l = 2, s = 1, j = 2, e = 16 ;

3

D1 : l = 2, s = 1, j = 1, e = 56 ;

13 60 ) = -2.74 5 ∆E = 26.8(0.1146 − 12 ) = -8.10 ∆E = 26.8(0.1146 − 32 ) = -14.8 ∆E = 26.8(0.1146 − 51 ) = -2.29 16 ∆E = 26.8(0.1146 − 105 ) = -1.01 13 ∆E = 26.8(0.1146 − 60 ) = -2.74 ∆E = 26.8(0.1146 − 17 60 ) = -4.52

62

5 Bound States 10 3S

1

0

1

D

3

P2

1P 1

3

D3

2

3

D2

3

D1

3

P1

-10

3

P0

-20 1S 0

-30 1S state

3

1P state

3S state

P states

1

D state

3

D states

Problem 5.9

The φ is quasi-bound , because the OZI-allowed decay into a pair of K’s is kinematically permissible (mφ = 1019 > 2mK = 990).

Problem 5.10

The parameters we have to work with (in solving the Schrödinger equation) are m (units of kg = J · s2 /m2 ),

h¯ (units of J · s),

F0 (units of N = J/m).

From these we must construct an energy: γ

J = mαh¯ β F0 =

J · s2 m2

α

(J · s) β

J m

γ

= Jα+ β+γ s2α+ β m−2α−γ .

Evidently 2α + β = 0 ⇒ β = −2α; 2α + γ = 0 ⇒ γ = −2α; α + β + γ = −3α = 1, so α = −1/3, β = γ = 2/3, and hence E=

m−1/3h¯ 2/3 F02/3 a

where a is some numerical factor.

(h¯ F0 )2 = m

1/3 a,

63

Problem 5.11

For the ψ’s, M = 2mc + En = 2500 + En : F0

M1

M2

M3

M4

500 1000 1500 Expt

2800 3000 3200 3097

3200 3600 4000 3686

3500 4100 4500 4039

3700 4400 5100 4159

For the Υ’s, M = 2mb + En = 9000 + En : F0

M1

500 9300 1000 9500 1500 9700 Expt 9460 Level spacings:

M2

M3

M4

9700 10100 10500 10023

10000 10600 11000 10355

10200 10900 11600 10579

F0

M2 − M1

M3 − M2

M4 − M3

500 1000 1500 Expt (ψ) Expt (Υ)

400 600 800 589 563

300 500 600 353 332

200 400 500 120 224

Evidently F0 is between 500 and 1000; of the three choices, 500 MeV/fm is best (though 1000 looks better for the masses themselves). The results are not exact, of course, because (a) the potential (Eq. 5.35) is just an approximation, (b) the parameters used in creating Table 5.2 are very rough (in particular, the quark mass is about right for the c, but way off for the b, and the strong coupling is only taken to one significant digit). All things considered, the agreement is surprisingly good.

Problem 5.12

For the pseudoscalar mesons, S1 · S2 = − 34 h¯ 2 , so M = m1 + m2 +

4m2u

3 2 1 m2u ( 159 ) − h ¯ = m + m − 477 . 2 1 4 m1 m2 m1 m2 h¯ 2

64

5 Bound States

π : m1 = m2 = mu ⇒ M = 308 + 308 − 477 = 139 308 K : m1 = mu , m2 = ms ⇒ M = 308 + 483 − 477 = 487 483 uu¯ : same as π : 139, dd¯ : same as π : 139 308 2 ss¯ : m1 = m2 = ms ⇒ M = 483 + 483 − 477 = 772 483 1 1 2 η : M = (139) + (139) + (772) = 561 6 6 3 1 1 1 η 0 : M = (139) + (139) + (772) = 350 (?!) 3 3 3 For the vector mesons, S1 · S2 = 14 h¯ 2 , so M = m1 + m2 +

4m2u h¯

2

(159)

1 2 h¯ 4

1 m2 = m1 + m2 + 159 u . m1 m2 m1 m2

ρ : m1 = m2 = mu ⇒ M = 308 + 308 + 159 = 775 308 = 892 K ∗ : m1 = mu , m2 = ms ⇒ M = 308 + 483 + 159 483 2 308 φ : m1 = m2 = ms ⇒ M = 483 + 483 + 159 = 1031 483 ω : m1 = m2 = mu ⇒ M = 308 + 308 + 159 = 775

Problem 5.13

(a) For the pseudoscalar mesons (see Problem 5.12): M = m1 + m2 − 477

m2u . m1 m2

308 2 = 2471 ηc : m1 = m2 = mc ⇒ M = 1250 + 1250 − 477 1250 308 D0 : m1 = mc , m2 = mu ⇒ M = 1250 + 308 − 477 = 1440 1250 (308)2 = 1658 Ds+ : m1 = mc , m2 = ms ⇒ M = 1250 + 483 − 477 (1250)(483)

The observed masses are 2980, 1865, and 1968 , respectively.

65

For the vector mesons (Problem 5.12): M = m1 + m2 + 159

m2u . m1 m2

308 2 = 2510 ψ : m1 = m2 = mc ⇒ M = 1250 + 1250 + 159 1250 308 D ∗ 0 : m1 = mc , m2 = mu ⇒ M = 1250 + 308 + 159 = 2539 1250 (3082 Ds∗ + : m1 = mc , m2 = ms ⇒ M = 1250 + 483 + 159 = 1758 (1250)(483)

The observed masses are 3097, 2007, and 2112 , respectively. (b) For the pseudoscalar bottom mesons, we have 308 = 4775 4500 (308)2 sb¯ : m1 = ms , m2 = mb ⇒ M = 483 + 4500 − 477 = 4960 (438)(4500)

ub¯ : m1 = mu , m2 = mb ⇒ M = 308 + 4500 − 477

(308)2 = 5742 (1250)(4500) 308 2 = 8998 bb¯ : m1 = m2 = mb ⇒ M = 4500 + 4500 − 477 4500 cb¯ : m1 = mc , m2 = mb ⇒ M = 1250 + 4500 − 477

The observed masses are 5279, 5368, 6286, ??? , respectively. For the vector bottom mesons: ub¯ : m1 = mu , m2 = mb ⇒ M = 308 + 4500 + 159

308 4500

2

= 4809

(308)2 = 4990 (483)(4500) (308)2 : m1 = mc , m2 = mb ⇒ M = 1250 + 4500 + 159 = 5753 (1250)(4500) 308 2 : m1 = m2 = mb ⇒ M = 4500 + 4500159 = 9001 4500

sb¯ : m1 = ms , m2 = mb ⇒ M = 483 + 4500 + 159 cb¯ bb¯

The observed masses are ???, ???, ???, and 9460 , respectively. Evidently the effective masses of the heavy quarks are somewhat larger than their “bare“ masses as listed in the Particle Physics Booklet—hardly a surprise, since the same is true of the light quarks.

66

5 Bound States

Problem 5.14

The proton is composed of u, u, and d. To make it antisymmetric in 1 ↔ 2, with appropriate normalization, we want p =

√1 (ud − du )u 2

. The other five

corners are constructed in the same way: n=

√1 (ud − du )d 2

Ξ0 =

√1 (us − su )s 2

,

Σ+ =

√1 (us − su )u 2

,

,

Ξ− =

√1 (ds − sd )s 2

.

Σ− =

√1 (ds − sd )d 2

,

Actually, the overall phase of p is arbitrary, but having chosen it, the others are not entirely arbitrary. In terms of isospin, p = | 12 12 i; applying the isospin lowering operator (see Problem 4.23(a) for the angular momentum analog): q 1 1 I− p = I− | 2 2 i = 12 ( 32 ) − 12 (− 21 )| 12 − 12 i = | 21 − 12 i = n. For three particles, I = I1 + I2 + I3

⇒

I− = I1 − + I2 − + I3 −

(where the subscript indicates the particle acted upon). Meanwhile, for the quarks I− u = d and I− d = 0, so √1 [ I1 − + I2 − + I3 − ] (udu − duu ) 2 1 √ [( I− u )du + u ( I− d )u + ud ( I− u ) − ( I− d )uu − d ( I− u )u − du ( I− u )] 2 √1 [ ddu + 0 + udd − 0 − ddu − dud ] 2 √1 (udd − dud ) = n. X 2

I− p =

= = =

Evidently these phases are consistent. Now Σ+ = |1 1i, and (Eq. 4.77) q √ √ I− Σ+ = I− |1 1i = 1(2) − 1(0)|1 0i = 2|1 0i = 2Σ0 . So (noting that s is an isosinglet, so I− s = 0): I− Σ+ =

= = =

√1 2 √1 2

√1 2

√1 2

[ I1 − + I2 − + I3 − ] (usu − suu)

[( I− u)su + u( I− s)u + us( I− u) − ( I− s)uu − s( I− u)u − su( I− u)]

[dsu + 0 + usd − 0 − sdu − sud] √ (dsu + usd − sdu − sud) = 2Σ0 ⇒ Σ0 =

1 2

[(us − su)d + (ds − sd)u]

67

Finally, Λ is constructed from u, d, and s; the most general linear combination of the various permutations that is antisymmetric in 1 ↔ 2 is Λ = α(ud − du)s + β(us − su)d + γ(ds − sd)u. Orthogonalize with respect to Σ0 : 1 [(us − su)d + (ds − sd)u] · [α(ud − du)s + β(us − su)d + γ(ds − sd)u] 2 1 = [α(0) + β(2) + γ(2)] = 0 ⇒ γ = − β. 2 Orthogonalize with respect to ψ A : 1 √ [uds − usd + dsu − dus + sud − sdu] 6 · [α(ud − du)s + β(us − su)d + γ(ds − sd)u] 1 = √ [α(2) + β(−2) + γ(2)] = 0 ⇒ α = β − γ = 2β. 6 So Λ = β [2(ud − du)s + (us − su)d − (ds − sd)u] . Normalize: β2 [2(ud − du)s + (us − su)d − (ds − sd)u]

· [2(ud − du)s + (us − su)d − (ds − sd)u] 1 = β2 [4(2) + 2 + 2] = 12β2 = 1 ⇒ β = √ . 12 Λ=

√1 12

[2(ud − du)s + (us − su)d − (ds − sd)u] .

Problem 5.15

We want a color singlet made from q and q¯ (see Eq. 5.46). This is the color analog to the flavor singlet (Eq. 5.42): 1 √ (r¯r + bb¯ + g g¯ ) 3

68

5 Bound States

Problem 5.16

Equation 5.62 says √ 2 ψ= [ψ12 (s)ψ12 ( f ) + ψ23 (s)ψ23 ( f ) + ψ13 (s)ψ13 ( f )] , 3 so

hψ|ψi =

2h hψ12 (s)|ψ12 (s)ihψ12 ( f )|ψ12 ( f )i + hψ12 (s)|ψ23 (s)ihψ12 ( f )|ψ23 ( f )i 9 +hψ12 (s)|ψ13 (s)ihψ12 ( f )|ψ13 ( f )i + hψ23 (s)|ψ12 (s)ihψ23 ( f )|ψ12 ( f )i

+hψ23 (s)|ψ23 (s)ihψ23 ( f )|ψ23 ( f )i + hψ23 (s)|ψ13 (s)ihψ23 ( f )|ψ13 ( f )i +hψ13 (s)|ψ12 (s)ihψ13 ( f )|ψ12 ( f )i + hψ13 (s)|ψ23 (s)ihψ13 ( f )|ψ23 ( f )i i +hψ13 (s)|ψ13 (s)ihψ13 ( f )|ψ13 ( f )i . These states are normalized (hψ12 (s)|ψ12 (s)i = 1, etc.), but theyare not orthogonal. From Eqs. 5.51, 5.52, and 5.53, using the “typical“ state 21 21 :

hψ12 (s)|ψ23 (s)i = hψ12 (s)|ψ13 (s)i = hψ13 (s)|ψ23 (s)i =

1 2 1 2 1 2

(↑↓↑ − ↓↑↑) · (↑↑↓ − ↑↓↑) = − 12 = hψ23 (s)|ψ12 (s)i (↑↓↑ − ↓↑↑) · (↑↑↓ − ↓↑↑) = (↑↑↓ − ↓↑↑) · (↑↑↓ − ↑↓↑) =

1 2 1 2

= hψ13 (s)|ψ12 (s)i = hψ23 (s)|ψ13 (s)i.

From the figures on pp. 185-186, using the “typical“ state in the upper right corner:

hψ12 ( f )|ψ23 ( f )i = hψ12 ( f )|ψ13 ( f )i = hψ13 ( f )|ψ23 ( f )i =

hψ|ψi =

1 2 1 2 1 2

(udu − duu) · (uud − udu) = − 12 = hψ23 ( f )|ψ12 ( f )i (udu − duu) · (uud − duu) = (uud − duu) · (uud − udu) =

1 2 1 2

= hψ13 ( f )|ψ12 ( f )i = hψ23 ( f )|ψ13 ( f )i.

2h 1 1 1 1 1 1 + − 12 − 12 + 12 + − − + 1 + 2 2 2 2 2 9 i 6 2 9 2 1 1 1 + 21 2 + 2 2 + 1 = 9 3 + 4 = 9 2 = 1. X

69

Problem 5.17

From Eq. 5.62, using Eqs. 5.51-53 and the diagrams on pp. 185-186, I find: √ 2 1 1 + 1 1 √ (↑↓↑ − ↓↑↑) √ (usu − suu) |Σ : 2 2 i = 3 2 2 1 1 + (↑↑↓ − ↑↓↑)(uus − usu) + (↑↑↓ − ↓↑↑)(uus − suu) 2 2 1 = √ [usu(2 ↑↓↑ − ↓↑↑ − ↑↑↓) 3 2 + suu(2 ↓↑↑ − ↑↓↑ − ↑↑↓) + uus(2 ↑↑↓ − ↑↓↑ − ↓↑↑)]

=

|Λ :

1 h √ 2u(↑)s(↓)u(↑) − u(↓)s(↑)u(↑) − u(↑)s(↑)u(↓) 3 2 + 2s(↓)u(↑)u(↑) − s(↑)u(↓)u(↑) − s(↑)u(↑)u(↓) i + 2u(↑)u(↑)s(↓) − u(↑)u(↓)s(↑) − u(↓)u(↑)s(↑) .

− 12 i √ 2 1 1 √ (↑↓↓ − ↓↑↓) √ (2uds − 2dus + usd − sud − dsu + sdu) = 3 2 12 1 1 + √ (↓↑↓ − ↓↓↑) √ (2sud − 2sdu + dus − dsu − uds + usd) 2 12 1 1 + √ (↑↓↓ − ↓↓↑) √ (2usd − 2dsu + uds − sdu − dus + sud) 2 12 1 ↑↓↓ − ↓↓↑+ ↓↓↑) = √ [uds(2 ↑↓↓ −2 ↓↑↓ − ↓↑↓ + 6 3 ↑↓↓ + + dus(−2 ↑↓↓ +2 ↓↑↓ + ↓↑↓ − ↓↓↑− ↓↓↑) 1 2

+ ↓↓↑ +2 ↑↓↓ −2 ↓↓↑) + usd(↑↓↓ − ↓↑↓ ↓↑↓− ↓↑↓ +2 ↓↑↓ −2 ↓↓↑ + ↓↓↑) + sud(− ↑↓↓+ ↑↓↓− − ↓↓↑ −2 ↑↓↓ +2 ↓↓↑) + dsu(− ↑↓↓ + ↓↑↓ ↓↑↓+ ↓↓↑)] ↓↑↓ −2 ↓↑↓ +2 ↓↓↑ − + sdu( ↑↓↓− ↑↓↓+ 1 = √ {uds(↑↓↓ − ↓↑↓) + dus(↓↑↓ − ↑↓↓) + usd(↑↓↓ − ↓↓↑) 2 3 + sud(↓↑↓ − ↓↓↑) + dsu(↓↓↑ − ↑↓↓) + sdu(↓↓↑ − ↓↑↓)}

70

5 Bound States

=

1 h √ u(↑)d(↓)s(↓) − u(↓)d(↑)s(↓) + d(↓)u(↑)s(↓) − d(↑)u(↓)s(↓) 2 3 + u(↑)s(↓)d(↓) − u(↓)s(↓)d(↑) + s(↓)u(↑)d(↓) − s(↓)u(↓)d(↑) i + d(↓)s(↓)u(↑) − d(↑)s(↓)u(↓) + s(↓)d(↓)u(↑) − s(↓)d(↑)u(↓) .

Problem 5.18

For notational simplicity, use ψij = −ψji for the spin-1/2 functions (Eqs. 5.51, 5.52, and 5.53) and φij = −φji for the octet flavor states (pp. 185-186). The following structure is antisymmetric, as you can easily check: ψ = A [ψ12 (φ31 + φ32 ) + ψ23 (φ12 + φ13 ) + ψ31 (s) (φ23 + φ21 )]

Problem 5.19

(a) First construct the wave functions, using Eq. 5.62 (with 5.51, 5.52, and 5.53 for spin and the figures on pp. 185-186 for flavor): √ 2 1 | p : 12 12 i = (↑↓↑ − ↓↑↑)(udu − duu) 3 2 1 1 + (↑↑↓ − ↑↓↑)(uud − udu) + (↑↑↓ − ↓↑↑)(uud − duu) 2 2 1 = √ [udu(2 ↑↓↑ − ↓↑↑ − ↑↑↓) + perms] 3 2 1 = √ [2u(↑)d(↓)u(↑) − u(↓)d(↑)u(↑) − u(↑)d(↑)u(↓) + perms] . 3 2

√ 2 1 (↑↓↑ − ↓↑↑)(udd − dud) 3 2 1 1 + (↑↑↓ − ↑↓↑)(dud − ddu) + (↑↑↓ − ↓↑↑)(udd − ddu) 2 2 1 = √ [udd(−2 ↓↑↑ + ↑↓↑ + ↑↑↓) + perms] 3 2 1 = √ [−2u(↓)d(↑)d(↑) + u(↑)d(↓)d(↑) + u(↑)d(↑)d(↓) + perms] . 3 2

|n :

1 1 2 2i

=

71

√ 2 1 (↑↓↑ − ↓↑↑)(usu − suu) 3 2 1 1 + (↑↑↓ − ↑↓↑)(uus − usu) + (↑↑↓ − ↓↑↑)(uus − suu) 2 2 1 = √ [usu(2 ↑↓↑ − ↓↑↑ − ↑↑↓) + perms] 3 2 1 = √ [2u(↑)s(↓)u(↑) − u(↓)s(↑)u(↑) − u(↑)s(↑)u(↓) + perms] . 3 2

|Σ+ :

1 1 2 2i

=

√ 2 1 |Σ : = (↑↓↑ − ↓↑↑)(dsd − sdd) 3 2 1 1 + (↑↑↓ − ↑↓↑)(dds − dsd) + (↑↑↓ − ↓↑↑)(dds − sdd) 2 2 1 = √ [dsd(2 ↑↓↑ − ↓↑↑ − ↑↑↓) + perms] 3 2 1 = √ [2d(↑)s(↓)d(↑) − d(↓)s(↑)d(↑) − d(↑)s(↑)d(↓) + perms] . 3 2 −

1 1 2 2i

√ 2 1 (↑↓↑ − ↓↑↑)(uss − sus) |Ξ : = 3 2 1 1 + (↑↑↓ − ↑↓↑)(sus − ssu) + (↑↑↓ − ↓↑↑)(uss − ssu) 2 2 1 = √ [uss(−2 ↓↑↑ + ↑↓↑ + ↑↑↓) + perms] 3 2 1 = √ [−2u(↓)s(↑)s(↑) + u(↑)s(↓)s(↑) + u(↑)s(↑)s(↓) + perms] . 3 2 0

1 1 2 2i

√ 2 1 = (↑↓↑ − ↓↑↑)(dss − sds) |Ξ : 3 2 1 1 + (↑↑↓ − ↑↓↑)(sds − ssd) + (↑↑↓ − ↓↑↑)(dss − ssd) 2 2 1 = √ [dss(−2 ↓↑↑ + ↑↓↑ + ↑↑↓) + perms] 3 2 1 = √ [−2d(↓)s(↑)s(↑) + d(↑)s(↓)s(↑) + d(↑)s(↑)s(↓) + perms] . 3 2 −

1 1 2 2i

In each of these cases “perm“ stands for three permutations (two in addition to the one listed), in which the odd quark occupies the first, second, or third

72

5 Bound States

position; Σ0 and Λ are a little trickier, because they involve all three quarks, and there are six permutations:

| Σ0 :

1 1 2 2i

=

√ 1 2 √ (↑↓↑ − ↓↑↑)(usd − sud + dsu − sdu) 3 2 2

1 + √ (↑↑↓ − ↑↓↑)(dus − dsu + uds − usd) 2 2 1 + √ (↑↑↓ − ↓↑↑)(uds − sdu + dus − sud) 2 2 1 = [usd(2 ↑↓↑ − ↓↑↑ − ↑↑↓) + (6 perms)] 6 1 = [2u(↑)s(↓)d(↑) − u(↓)s(↑)d(↑) − u(↑)s(↑)d(↓) + (6 perms)] . 6

√ 2 1 √ (↑↓↑ − ↓↑↑)(2uds − 2dus + usd − sud − dsu + sdu) |Λ : = 3 2 6 1 + √ (↑↑↓ − ↑↓↑)(2sud − 2sdu + dus − dsu − uds + usd) 2 6 1 + √ (↑↑↓ − ↓↑↑)(2usd − 2dsu + uds − sdu − dus + sud) 2 6 1 = √ [3uds(↑↓↑ − ↓↑↑) + (6 perms)] 6 3 1 = √ [u(↑)d(↓)s(↑) − u(↓)d(↑)s(↑) + (6 perms)] . 2 3 1 1 2 2i

Now use Eq. 5.67 to determine the magnetic moment of the proton, as in Example 5.3:

73

2 √ u(↑)d(↓)u(↑) 3 2 h¯ h¯ h¯ h¯ 2 2 √ √ (2µu − µd )| ai ( µ S µ − µ + µ | ai = )| a i = u u ∑ i iz d 2 2 2 23 2 3 2 2 2 µ a = h a| ∑(µi Siz )| ai = (2µu − µd ); h¯ 9 −1 |bi ≡ √ u(↓)d(↑)u(↑) 3 2 h¯ 1 h¯ h¯ h¯ −1 √ |bi = − √ µd |bi ( µ S − µ + µ + µ )| b i = u u ∑ i iz d 2 2 2 23 2 3 2 2 1 µb = hb| ∑(µi Siz )|bi = µ ; h¯ 18 d −1 |ci ≡ √ u(↑)d(↑)u(↓) 3 2 −1 h¯ h¯ 1 h¯ h¯ √ )| c i = ( µ S + µ − µ |ci = − √ µd |ci µ u u ∑ i iz d 2 2 2 23 2 3 2 1 2 µ . µc = hc| ∑(µi Siz )|ci = h¯ 18 d

| ai ≡

The total is the sum (times three for the three permutations): 2 2 µ = 3(µ a + µb + µc ) = 3 (2µu − µd ) + µd = 13 (4µu − µd ) . 9 18 Comparing the wave functions, we see that no new calculation is required for n, Σ+ , Σ− , Ξ0 , and Ξ− (for some of them there is an overall minus sign in the wave function, but this squares out; sometimes the permutation listed is different, but this doesn’t affect the answer). n : (same as p, only with u ↔ d) :

µ=

1 3 (4µd

− µu ) .

Σ+ : (same as p, only with d → s) :

µ=

1 3 (4µu

− µs ) .

Σ− : (same as n, only with u → s) :

µ=

1 3 (4µd

− µs ) .

Ξ0 : (same as n, only with d → s) :

µ=

1 3 (4µs

− µu ) .

Ξ− : (same as Ξ0 , only with u → d) :

µ=

1 3 (4µs

− µd ) .

74

5 Bound States

As for Σ0 ,

| ai ≡

1 u(↑)s(↓)d(↑) 3

h¯ h¯ h¯ 1 h¯ − µs + µd | ai = (µu − µs + µd )| ai 2 2 2 23 2 1 µ a = h a| ∑(µi Siz )| ai = (µu − µs + µd ); h¯ 9 −1 |bi ≡ u(↓)s(↑)d(↑) 6 h¯ −1 h¯ h¯ h¯ 1 ( µ S + µ + µ )| b i = − µ |bi = − (−µu + µs + µd )|bi s u ∑ i iz d 6 2 2 2 26 2 1 µb = hb| ∑(µi Siz )|bi = (−µu + µs + µd ); h¯ 36 −1 u(↑)s(↑)d(↓) |ci ≡ 6 h¯ h¯ 1 h¯ h¯ −1 ∑(µi Siz )|ci = 6 µu 2 + µs 2 − µd 2 |ci = − 2 6 (µu + µs − µd )|ci 1 2 ( µ u + µ s − µ d ). µc = hc| ∑(µi Siz )|ci = h¯ 36

∑(µi Siz )|ai =

1 3

µu

The total is the sum (times six for the six permutations): µ = 6( µ a + µ b + µ c ) 1 1 1 = 6 (µu − µs + µd ) + (−µu + µs + µd ) + (µu + µs − µd ) 9 36 36

=

1 3 (2µu

+ 2µd − µs .

Finally, for Λ, 1 √ u(↑)d(↓)s(↑) 2 3 1 h¯ h¯ h¯ h¯ 1 √ √ (µu − µs + µs )| ai µ − µ + µ | ai = )| a i = ( µ S u s ∑ i iz d 2 2 2 22 3 2 3 2 1 µ a = h a| ∑(µi Siz )| ai = ( µ u − µ d + µ s ); h¯ 12 −1 |bi ≡ √ u(↓)d(↑)s(↑) 2 3 −1 h¯ h¯ h¯ h¯ 1 √ ( µ S )| b i = − µ + µ + µ |bi = − √ (−µu + µd + µs )|bi u s ∑ i iz d 2 2 2 22 3 2 3 2 1 µb = hb| ∑(µi Siz )|bi = (−µu + µd + µs ). h¯ 12

| ai ≡

75

The total is the sum (times six for the six permutations): µ = 6( µ a + µ b 1 1 (µu − µd + µs ) + (−µu + µd + µs ) =6 12 12 = µs . (b) In units of the nuclear magneton, e¯h/2m p c, µu =

2 mp , 3 mu

µd = −

1 mp , 3 md

µs = −

1 mp 3 ms

(Eq. 5.66), so, using mu = md = 336,

ms = 538,

, and m p = 938,

the magnetic moments should be mp 1 2 mp 1 mp 938 p: 4 − − = = = 2.79 3 3 mu 3 mu mu 336 1 mp 2 mp 1 2 mp 2 938 4 − =− = -1.86 n: − =− 3 3 mu 3 mu 3 mu 3 336 1 mp 1 938 Λ: − =− = -0.58 3 ms 3 538 2 mp 8 1 1 1 mp 938 + 4 + = 2.68 Σ : − − = 3 3 mu 3 ms 9 336 538 mp 1 2 mp 1 mp 1 mp 2 1 0 Σ : 2 + − − − = + 3 3 mu 3 mu 3 ms 9 mu ms 938 2 1 = + = 0.81 9 336 538 1 1 mp 938 4 1 1 mp − Σ : 4 − − − = − + = -1.05 3 3 mu 3 ms 9 336 538 1 1 mp 2 mp 938 4 2 Ξ0 : 4 − − = − − = -1.40 3 3 ms 3 mu 9 538 336 1 1 mp 1 mp 938 4 1 Ξ− : 4 − − − = − + = -0.46 3 3 ms 3 mu 9 538 336

76

5 Bound States

Problem 5.20

Following Problem 5.19, but with the wave function from Problem 5.18:

|p :

1 1 2 2i

=

A [(↑↓↑ − ↓↑↑)(duu − uud + udu − uud) 2 + (↑↑↓ − ↑↓↑)(udu − duu + uud − duu)

− (↑↑↓ − ↓↑↑)(uud − udu + duu − udu)] A = [duu(↑↓↑ − ↓↑↑ −2 ↑↑↓ +2 ↑↓↑ + ↓↑↑ − ↑↑↓) + perms] 2 3 = A [d(↑)u(↓)u(↑) − d(↑)u(↑)u(↓) + perms] . 2

|n :

1 1 2 2i

=

A [(↑↓↑ − ↓↑↑)(ddu − udd + ddu − dud) 2 + (↑↑↓ − ↑↓↑)(udd − dud + udd − ddu)

− (↑↑↓ − ↓↑↑)(dud − ddu + dud − udd)] A [udd(− ↑↓↑ + ↓↑↑ +2 ↑↑↓ −2 ↑↓↑ − ↓↑↑ + ↑↑↓) + perms] 2 3 = A [u(↑)d(↑)d(↓) − u(↑)d(↓)d(↑) + perms] . 2

=

For the proton, 3A d(↑)u(↓)u(↑) 2 3A h¯ h¯ 3A h¯ h¯ ( µ S )| a i = µ − µ + µ | ai = (µd )| ai u u ∑ i iz d 2 2 2 2 2 2 3A 2 2 µd ; µ a = h a| ∑(µi Siz )| ai = h¯ 2 3A |bi ≡ − d(↑)u(↑)u(↓) 2 3A h¯ h¯ h¯ h¯ 3A ∑(µi Siz )|bi = − 2 µd 2 + µu 2 − µu 2 |bi = − 2 2 µd |bi 3A 2 2 µb = hb| ∑(µi Siz )|bi = µd . h¯ 2

| ai ≡

The total is the sum (times three for the three permutations): 3A 2 µ p = 3( µ a + µ b ) = 6 µd 2 (negative, since µd is). Comparing the wave functions, the neutron calculation is identical (except for an overall minus sign, which squares away), only with

77

u ↔ d. So µn = 6

3A 2

2 µu .

Evidently µn µu = = µp µd

2 mp 3 mu

1 mp m / − = −2 d = −2 3 md mu

(assuming, in the final step, that mu ≈ md , in the spirit of Eq. 5.68). No: this is totally inconsistent with the experimental value (-0.68).

Problem 5.21

The flavor states are ρ+ = −ud,¯ ρ− = du¯ (Eq. 5.38), and the ms = 1 spin state is ↑↑ (Eq. 4.15), so

|ρ+ : 1 1i = −u(↑)d¯(↑),

|ρ− : 1 1i = d(↑)u¯ (↑).

Following the method of Section 5.6.2 (Eq. 5.67), and noting that in this case Sz = h¯ /2 for both quarks: µM =

2 h M 1|(µ1 S1z + µ2 S2z )| M 1i = h M 1|(µ1 + µ2 )| M 1i h¯

Thus µρ− = ( µd − µu ) = − µρ+

µρ+ = ( µu − µd ) ,

X

(tne magnetic moment of an antiquark is opposite to that of the quark). Now, to the extent that mu = md , Eq. 5.66 says µu = −2µd , so µρ+ = −2µd − µd = −3µd . Meanwhile, for the proton (Example 5.3) µp =

1 3

(4µu − µd ) =

1 3

(−8µd − µd ) = −3µd = µρ+ . X

Problem 5.22

In this case m1 = m2 = ms , m3 = mu , so S1 · S2 ( S1 + S2 ) · S u M = 2ms + mu + A0 + . ms mu m2s Now J = S1 + S2 + Su , so J 2 = (S1 + S2 )2 + 2(S1 + S2 ) · Su + Su2 ,

78

5 Bound States

but J 2 = Su2 = (3/4)h¯ 2 (since both the quark and the Ξ have spin-1/2), so

(S1 + S2 ) · Su = − 12 (S1 + S2 )2 . The flavor wave function is symmetric in 1 ↔ 2, so the spin function must also be symmetric—the two s quarks are in the triplet (spin 1) state:

(S1 + S2 )2 = 2¯h2 , (S1 + S2 ) · Su = −h¯ 2 , S1 · S2 = 14 h¯ 2 . " # h¯ 2 h¯ 2 4 h¯ 2 0 1 0 M = 2ms + mu + A − − = m + 2m + A . u s ms mu 4 mu ms 4m2s m2s This confirms Eq. 5.83. Numerically, using mu = 363, ms = 538, A0 = 50(2mu /¯h)2 : 4 1 − M = 2(538) + 363 + 50(363)2 = 1327 MeV/c2 . (538)2 (538)(363) (The observed value is 1315.)

Problem 5.23

For hydrogen (Problem 5.2) ψ100 = √

1 πa3

e−r/a

⇒

|ψ100 (0)|2 =

1 . πa3

For positronium the Bohr radius is doubled (Eq. 5.28): |ψ100 (0)|2 = 1/[π (2a)3 ]. Now a = h¯ 2 /me2 (Eq. 5.13) and α = e2 /¯hc (Eq. 5.11), so a = h¯ /αmc, and hence

|ψ100 (0)|2 =

1 αmc 3 . π 2¯h

79

6

The Feynman Calculus Problem 6.1

Number there at time t: N (t) = N0 e−Γt . Number still there at time t + dt: N (t + dt) = N0 e−Γ(t+dt) = N0 e−Γt e−Γdt ∼ = N0 e−Γt (1 − Γ dt). Number that decay between t and t + dt: N (t) − N (t + dt) ∼ = N0 e−Γt (1 − 1 + Γ dt) = N0 e−Γt Γ dt. Fraction that decay between t and t + dt: Γe−Γt dt. Probability that an individual selected at random from the initial sample will decay between t and t + dt: p(t)dt = Γe−Γt dt. Mean lifetime: τ=

Z ∞ 0

tp(t) dt = Γ

Z ∞ 0

te

−Γt

dt = Γ

1 Γ2

=

1 . Γ

[Note the distinction between the probability that a given particle will decay in the next instant dt (which is Γ dt) and the probability that an individual in the initial sample will decay between t and t + dt (which is Γe−Γt dt); the difference is that in the latter case there are fewer and fewer around to decay (hence the factor e−Γt ).]

Problem 6.2

N (t1/2 ) =

1 N0 = N0 e−Γt1/2 ⇒ eΓt1/2 = 2 ⇒ Γt1/2 = ln 2 ⇒ t1/2 = τ ln 2. 2

80

6 The Feynman Calculus

Problem 6.3

(a) N = 106 e−(2.2×10 (approximately).

−5 )/ (2.2×10−6 )

= 106 e−10 = 4.58 × 10−5 × 106 = 46

(b) e−1/(2.6×10

−8 )

7

= e−3.85×10 = 10log10 e

(−3.85×107 )

7

= 10−1.67×10 = 10−16,700,000

= 100.434

(−3.85×107 )

(pretty small!)

Problem 6.4

b

f

r

fm

fm

q

ˆ Use polar coordinates (r, φ): v = r˙ rˆ + r φ˙ φ. Energy is conserved: E=

1 2 1 mv + V (r ) = m(r˙ 2 + r2 φ˙ 2 ) + V (r ). 2 2

Angular momentum is conserved: ˙ ˆ or L = mr2 φ. L = r × mv = mr rˆ × (r˙ rˆ + r φ˙ φˆ ) = mr2 φ˙ (rˆ × φˆ ) = mr2 φ˙ z, √ In the distant past, E = 12 mv20 and L = bmv0 , so L = bm 2E/m. So conservation of angular momentum says r √ ˙φ = L = bm 2E/m = b 2E , m mr2 mr2 r2 and conservation of energy becomes E=

1 2 1 2 b2 2E 1 Eb2 mr˙ + mr 4 + V (r ) = mr˙ 2 + 2 + V (r ), or 2 2 2 r r m

81

r˙ 2 =

2 Eb2 E − 2 − V (r ) . m r

However we’re not interested in r as a function of t, but rather r as a function of φ: r dr dr dφ dr ˙ dr b 2E r˙ = = = φ= . dt dφ dt dφ dφ r2 m It pays to change variables, letting u ≡ 1/r, so that dr dr du 1 du = =− 2 . dφ du dφ u dφ r r du b 2E 2E du = − b . Thus r˙ = −r2 dφ r2 m m dφ 2 2 du 2E b2 V du 1 1 V 2 2E b = 1− 2 − =⇒ = 2 − 2 − 2 , or m dφ m E dφ r b r b E r r du 1 b2 V 1 V b du = 1− 2 − = 1 − b2 u2 − ; √ = dφ. dφ b E b E r 1 − b2 u2 − V/E Integrate in from distant past (φ = 0, r = ∞ =⇒ u = 0) to point of closest approach (φ = φm , r = rmin ⇒ u = umax , the point at which du/dφ = 0, which is to say, where the radical goes to zero): φm = b

Z um

√

0

du

then θ = π − 2φm

;

1 − b2 u2 − V/E

(see diagram). k = ku2 : r2

(a) So far, this is true for any potential. Now we put in V (r ) = φm = b

Z um 0

= q where u2m = φm = q

du q

1 − b2 u2 − Ek u2 Z um

1 1+

= √

k

b2 E

du p

0

u2m

− u2

Z um

b b2

+ k/E

0

du q

1 b2 +k/E

− u2

,

1 . b2 +k/E

1 1+

k b2 E

sin−1

u um 1 = q u m 0 1+

θ = π − 2φm = π − π q

1 1+

k

b2 E

k b2 E

sin−1 (1) =

= π 1− p

π 1 q 2 1+

1 1 + (k/b2 E)

! .

k b2 E

82


(b) According to Eq. 6.10,

b db dσ . In this case, = dΩ sin θ dθ

dθ 1 1 =π db 2 [1 + (k/b2 E)]3/2 b E dσ = dΩ sin θ kπ

k b + E 2

kπ k 2 k −3/2 − 3 =− b2 + E E E b

3/2 .

But we want it in terms of θ :

θ 1 k 1 = p =⇒ 1 + 2 = π b E (1 − πθ )2 1 + (k/b2 E)

1−

1 − 1 + 2 πθ − k 1 =⇒ 2 = − 1 = b E (1 − πθ )2 (1 − πθ )2

θ2 π2

=

θ (2π − θ ) ( π − θ )2

( π − θ )2 k ( π − θ )2 b2 E = ; b2 = k θ (2π − θ ) E θ (2π − θ ) 3 dσ b4 E k 3/2 1 k 2 ( π − θ )4 E π ∴ = 1+ 2 = dΩ sin θ kπ sin θ E2 θ 2 (2π − θ )2 kπ π − θ b E dσ π 2 k (π − θ ) = dΩ sin θ E θ 2 (2π − θ )2 (c) σ=

=

Z

dσ π2 k dΩ = dΩ E

2π 3 k E

Z π 0

Z

1 (π − θ ) (sin θ dθ dφ) 2 sin θ θ (2π − θ )2

(π − θ ) dθ = Infinity. − θ )2

θ 2 (2π

Problem 6.5

q

r2 + m22 c2 +

q

r2 + m23 c2 . Square: q q m21 c2 = r2 + m22 c2 + r2 + m23 c2 + 2 r2 + m22 c2 r2 + m23 c2 m1 c =

c2 2 (m1 − m22 − m23 ) − r2 = 2

q

r2 + m22 c2

q

r2 + m23 c2 .

Square again:

c4 2 (m1 − m22 − m23 )2 − r2 c2 (m21 − m22 − m23 ) + r4 = r4 + r2 c2 (m22 + m23 ) + m22 m23 c4 4 i c4 h 2 (m1 − m22 − m23 )2 − 4m22 m23 = r2 c2 (m22 + m23 + m21 − m22 − m23 ) = r2 m21 c2 4

83

i c2 h 4 m1 + m42 + m43 − 2m21 m22 − 2m21 m23 + 2m22 m23 − 4m22 m23 2 4m1 c q 4 r= m1 + m42 + m43 − 2m21 m22 − 2m21 m23 − 2m22 m23 X 2m1

r2 =

Problem 6.6

Plug M = αmπ c into Eq. 6.35, with S = 1/2: Γ=

|p| α2 c 2 |p|. ( αm c ) = π 16π¯h 16π¯hm2π c So Γ = α2 mπ c2 /32π¯h.

But Eγ = 21 mπ c2 ⇒ |pγ | = Eγ /c = 12 mπ c.

32π (6.58 × 10−22 )(137)2 1 32π¯h = = 2 = 9.2 × 10−18 s. Γ 135 α m π c2

τ=

The experimental value is 8.4 × 10−17 s, so this estimate is off by a factor of 10.

Problem 6.7

(a) In the CM frame, p2 = −p1 , so p1 = p1 · p2 =

E1 E2 , p1 , p2 = , −p1 . c c

E1 E2 E E − p1 · (−p1 ) = 1 2 2 + p21 c c c

2

2 2

( p1 · p2 ) − ( m1 m2 c ) = = But m21 c2 =

E1 E2 + p21 c2

2

− ( m1 m2 c2 )2

E12 E22 E E + 2 1 2 2 p21 + p41 − m21 m22 c4 4 c c

E12 E22 E22 2 2 2 2 − p , and m c = − p = − p21 . 2 2 1 c2 c2 c2

( p1 · p2 )2 − ( m1 m2 c2 )2 E2 E2 E E = 1 4 2 + 2 1 2 2 p21 + p41 − c c

E12 − p21 c2 |

!

E22 − p21 c2

{z

( E2 + E2 ) E2 E22 1 −p21 1 2 2 +p41 c c4

=

1 2 2 1 p ( E + E22 + 2E1 E2 ) = 2 p21 ( E1 + E2 )2 . c2 1 1 c

!

}

So:

84


∴

p

( p1 · p2 )2 − ( m1 m2 c2 )2 =

(b) In the lab frame, p1 =

1 |p |( E + E2 ). c 1 1

E1 , p , p2 = (m2 c, 0), so p1 · p2 = E1 m2 . c 1

∴ ( p1 · p2 )2 − (m1 m2 c2 )2 = E12 m22 − m21 m22 c4 = m22 ( E12 − m21 c4 ) = m22 (p21 c2 ). ∴

p

( p1 · p2 )2 − (m1 m2 c2 )2 = |p1 |m2 c.

Problem 6.8

Start with Eq. 6.47: dσ = dΩ

h¯ c 8π

2

S|M|2 |p f | . ( E1 + E2 )2 |pi |

This is for the CM frame, but here the target particle is so heavy that it barely moves, and hence the lab and the CM are effectively the same. Particle 1 bounces off particle 2; its energy is unchanged, and hence so too is the magnitude of its momentum: |p f | = |pi |, and ( E1 + E2 ) = ( E1 + m2 c2 ) ≈ m2 c2 , so 2 dσ |M|2 h¯ c = . dΩ 8π ( m b c2 )2 (S = 1, since 1 and 2 are different particles).

Problem 6.9

Using the result of Problem 6.7 (b) in Eq. 6.38: h¯ 2 S dσ = 4 | p1 | m 2 c

1 4π

2 Z

2

|M| δ

× δ3 (p1 + p2 − p3 − p4 )

E1 + E2 − |p3 | − |p4 | c

d 3 p3 d 3 p4 . |p3 | |p4 |

Here p2 = 0 and E2 = m2 c2 . Doing the p4 integral: h¯ 2 S dσ = 64π 2 |p1 |m2 c

Z

2

|M| δ

E1 + m2 c − |p3 | − |p1 − p3 | c

d 3 p3 . |p3 ||p1 − p3 |

85

Now d3 p3 = |p3 |2 d|p3 | dΩ, and (p1 − p3 )2 = |p1 |2 + |p3 |2 − 2|p1 ||p3 | cos θ (where θ is the scattering angle for particle 3). So ∞ |p3 |2 d|p3 | dσ h¯ 2 S 2 p |M| = dΩ 64π 2 |p1 |m2 c 0 |p3 | |p1 |2 + |p3 |2 − 2|p1 ||p3 | cos θ q E1 + m2 c − |p3 | − |p1 |2 + |p3 |2 − 2|p1 ||p3 | cos θ ×δ c p Let z ≡ |p3 | + |p1 |2 + |p3 |2 − 2|p1 ||p3 | cos θ; then

Z

dz 1 2|p3 | − 2|p1 | cos θ = 1+ p 2 d|p3 | 2 |p1 | + |p3 |2 − 2|p1 ||p3 | cos θ p |p1 |2 + |p3 |2 − 2|p1 ||p3 | cos θ + |p3 | − |p1 | cos θ p = |p1 |2 + |p3 |2 − 2|p1 ||p3 | cos θ z − |p1 | cos θ = p . |p1 |2 + |p3 |2 − 2|p1 ||p3 | cos θ p

d|p3 | dz = . Therefore: z − |p1 | cos θ |p1 |2 + |p3 |2 − 2|p1 ||p3 | cos θ

h¯ 2 S dσ = 2 dΩ 64π |p1 |m2 c

Z

|M|2

|p3 | dz δ (z − |p1 | cos θ )

E1 + m2 c − z c

[Here |p3 | is a function of z.]

=

h¯ 2 S |M|2 64π 2 |p1 |m2 c

|p3 | E1 c

+ m2 c − |p1 | cos θ E [Here |p3 | is the value when z = 1 + m2 c.] c 2 h¯ S|M|2 |p3 | = 8π m2 |p1 |( E1 + m2 c2 − |p1 |c cos θ )

86


Problem 6.10

(a) Insert the result from Problem 6.7(b) into Eq. 6.38 (and note that p2 = 0): h¯ 2 S 1 4|p1 |m2 c (4π )2

d3 p3 d3 p4 q p23 + m21 c2 p24 + m22 c2 2 Z q q E1 + E2 S h¯ 2 2 2 2 2 2 2 |M| δ − p3 + m1 c − p4 + m2 c = 8π |p1 |cm2 c

dσ =

Z

|M|2 δ4 ( p1 + p2 − p3 − p4 ) q

× δ3 (p1 − p3 − p4 ) q

d3 p3 d3 p4 q . p23 + m21 c2 p24 + m22 c2

Doing the p4 integral, and setting d3 p3 = |p3 |2 d|p3 | dΩ: 2 Z h¯ S dσ |M|2 = dΩ 8π |p1 |cm2 q q δ Ec1 + m2 c − p23 + m21 c2 − (p1 − p3 )2 + m22 c2 q q × |p3 |2 dp3 . 2 2 2 2 2 2 p3 + m1 c (p1 − p3 ) + m2 c Now

(p1 − p3 )2 = p21 + p23 − 2p1 · p3 = p21 + p23 − 2|p1 | |p3 | cos θ, where θ is the scattering angle. Let r ≡ |p3 |, for simplicity: 2 Z h¯ S dσ = |M|2 dΩ 8π |p1 |cm2 q q δ Ec1 + m2 c − r2 + m21 c2 − r2 − 2r |p1 | cos θ + p21 + m22 c2 q q r2 dr. × 2 2 2 2 2 2 2 r + m1 c r − 2r |p1 | cos θ + p1 + m2 c Let z ≡

q

r2 + m21 c2 +

q

r2 − 2r |p1 | cos θ + p21 + m22 c2 . Then

dz r r − |p1 | cos θ +q = q dr r2 + m21 c2 r2 − 2r |p1 | cos θ + p21 + m22 c2 q rz − |p1 | cos θ r2 + m21 c2 q = q . r2 + m21 c2 r2 − 2r |p1 | cos θ + p21 + m22 c2 dσ = dΩ

h¯ 8π

2

S |p1 |cm2

Z

+ m2 c − z q r2 dz |M|2 2 2 2 rz − |p1 | cos θ r + m1 c δ

E1 c

87

(here r is a function, implicitly, of z) 2 h¯ S r2 dσ q = |M|2 dΩ 8π |p1 |cm2 r ( E1 /c + m2 c) − |p1 | cos θ r2 + m21 c2 (now q r is the valueqof |p3 | dictated by conservation of energy and momentum, so r2 + m21 c2 = p23 + m21 c2 = E3 /c). dσ = dΩ

h¯ 8π

2

p23 S|M|2 . |p1 |m2 |( E1 + m2 c2 )|p3 | − E3 |p1 | cos θ |

Note that E3 and p3 are functions of θ (for a given incident energy E1 ). The final result can be written in a number of different (equivalent) ways; perhaps the tidiest is 2 dσ h¯ S|M|2 |p3 |3 . = dΩ 8π |p1 |m2 E1 E3 m2 − E4 m21 c2 (b) In this case |p1 | = E1 /c and |p3 | = E3 /c, so 2 E32 /c2 dσ h¯ S|M|2 c = 2 dΩ 8π E1 m2 |( E1 + m2 c ) E3 /c − ( E1 /c) E3 cos θ | 2 h¯ S|M|2 E3 = . 2 8π E1 m2 | E1 + m2 c − E1 cos θ | Now p1 + p2 = p3 + p4 ⇒ p1 − p3 = p4 − p2 p21 + p23 − 2p1 · p3 = p24 + p22 − 2p2 · p4 ⇒ p1 · p3 = p2 · p4 − m22 c2 E1 E3 E E − |p1 ||p3 | cos θ = 2 2 4 − p2 · p4 − m22 c2 = m2 E4 − m22 c2 c2 c E1 E3 (1 − cos θ ) = m2 c2 ( E4 − m2 c2 ) = m2 c2 ( E1 + E2 − E3 − m2 c2 )

= ( E1 − E3 )m2 c2 h i E3 E1 (1 − cos θ ) + m2 c2 = E1 m2 c2 ,

1 E3 = . E1 + m2 − E1 cos θ E1 m2 c2 2 2 h¯ S|M|2 E32 h¯ E3 dσ 2 = = S|M| . X dΩ 8π E1 m2 E1 m2 c2 8πm2 E1 c

Problem 6.11

(a) No

[See (b).]

c2

88


(b) Allowed if (and only if) n A , n B , and nC are either all even or all odd. Proof: (necessary) Take the allowed diagram and snip every internal 0 = N ’external’ lines, where N is line. We now have n0A = n0B = nC the number of vertices. When we now reconnect the internal lines, each join removes two ’external’ lines of one species. Thus when they’re all back together we have n A = N − 2I A , n B = N − 2IB , and nC = N − 2IC , where I A is the number of internal A lines, and so on. Clearly, they’re all even, or all odd, depending on the number of vertices. (sufficient) Given n A , n B , and nC , pick the largest of them (say, n A ) and draw that number of vertices, with A, B, C as ’external’ lines on each one. Now just connect up B lines in pairs (converting two ’external’ lines into one internal line, each time you do so), until you’re down to n B – as long as n A and n B are either both even or both odd, you will obviously be able to do so. Now do the same for nC . We have constructed a diagram, then, with n A external A lines, n B external B lines, and nC external C lines. (c) In view of (b) we’ll need either 3 B’s and one C or 3 C’s and one B: A −→ B + B + B + C

or A −→ B + C + C + C.

C A

C

A

B B

B

A

B

(a) Start with the four vertices: 2

3

1

4

C C

C

Problem 6.12

B

A

89

An external A will attach to each of these (with 4-momentum labeled by the number of the vertex). The B from vertex 1 could connect to vertex 2 (in which case the C line goes to 3 or 4), or to 3 (in which case the C line goes to 2 or 4), or to 4 (in which case the C line goes to 1 or 2). So there are six diagrams in all: A

A

A C

B

1

A A

A

A

A

B C 4

C 2

B 5

A

A

B C

C A

A

3

A

A

C

B

A

C

A

A

A

B C

A B

C

B

A

A

C B

B

B

C

A

A

C

B 6

A

(b) Let’s start with diagram 2:

A p

2

p1

A

(−ig)4

p3

C

q1 B

ZZZZ

A

q2 B C

q

4

q3 p4

A

i i i i (2π )4 q21 − m2B c2 q22 − m2C c2 q23 − m2B c2 q24 − m2C c2

× δ4 ( p1 + q4 − q1 )(2π )4 δ4 ( p2 + q1 − q2 )(2π )4 δ4 (q2 − q3 − p3 ) × (2π )4 δ4 (q3 − p4 − q4 ) = g4 ×

ZZZZ

d4 q1 d4 q2 d4 q3 d4 q4 (2π )4 (2π )4 (2π )4 (2π )4

δ4 ( p1 + q4 − q1 ) δ4 ( p2 + q1 − q2 ) δ4 ( q2 − q3 − p3 ) q21 q22 q23

δ4 ( q3 − p4 − q4 ) 4 d q1 d4 q2 d4 q3 d4 q4 q24

Do the q4 integral (⇒ q4 = q1 − p1 ) and the q3 integral (⇒ q3 = q2 − p3 ):

= g4

ZZ

δ4 ( p2 + q1 − q2 ) δ4 ( q2 − p3 − p4 − q1 + p1 ) 4 d q1 d4 q2 q21 q22 (q2 − p3 )2 (q1 − p1 )2

90


Do the q2 integral (⇒ q2 = p2 + q1 ), and drop the subscript on q1 :

= g4

δ4 ( p2 + q − p3 − p4 − q + p1 ) d4 q 2 2 2 2 + q ) ( p2 + q − p3 ) ( q − p1 )

Z

q2 ( p

Cancel (2π )4 δ4 ( p1 + p2 − p3 − p4 ), and multiply by i:

M=i

g 4 Z 1 d4 q. 2 2 2π q ( q + p2 ) ( q + p2 − p3 )2 ( q − p1 )2

The diagram 5 is the same, only with mC ↔ m B . Since, however, we assumed mC = m B = 0, the amplitude is identical to diagram 2. For diagram 1 (and hence also 3) we switch p3 ↔ p4 :

M=i

g 4 Z 1 d4 q; 2π q2 ( q + p2 )2 ( q + p2 − p4 )2 ( q − p1 )2

for diagrams 4 and 6 we switch p2 ↔ p3 :

M=i

g 4 Z 1 d4 q. 2 2 2π q ( q + p3 ) ( q + p3 − p2 )2 ( q − p1 )2

Thus the full amplitude is

M = 2i

g 4 Z 1 1 2π q2 ( q − p1 )2 ( q + p2 )2 ( q + p2 − p3 )2

1 1 + + ( q + p2 )2 ( q + p2 − p4 )2 ( q + p3 )2 ( q + p3 − p2 )2

d4 q.

(There are other ways of writing this, of course, which amount to redefining the integration variable q.)

Problem 6.13

With mC = 0, Eq. 6.55 becomes 1 1 2 + M=g ( p4 − p2 )2 ( p3 − p2 )2 With m B = 0,

( p4 − p2 )2 = p24 + p22 − 2p4 · p2 = m2B c2 + m2A c2 − 2p4 · p2 = m2A c2 − 2p4 · p2 ( p3 − p2 )2 = p23 + p22 − 2p3 · p2 = m2B c2 + m2A c2 − 2p3 · p2 = m2A c2 − 2p3 · p2 .

91

Now E1 = E2 and E3 = E4 , so conservation of energy ( E1 + E2 = E3 + E4 ) says E1 = E2 = E3 = E4 ≡ E. Thus: E E E E p1 = , p1 ; p2 = , −p1 ; p3 = , p3 ; p4 = , −p3 , c c c c so p2 · p4 = E2 /c2 − p1 · p3 ;

p2 · p3 = E2 /c2 + p1 · p3 .

1 1 M=g + 2 2 m2A c2 − 2E2 /c2 + 2p1 · p3 m A c − 2E2 /c2 − 2p1 · p3 # " 2 m2A c2 − 2E2 /c2 2 =g 2 m2A c2 − 2E2 /c2 − 4(p1 · p3 )2

!

2

Now p1 · p3 = |p1 ||p3 | cos θ (Fig. 6.10), and E2 − m2A c2 ; c2

E2 E2 − m2B c2 = 2 . 2 c c So (p1 · p3 )2 = E2 /c2 − m2A c2 ( E2 /c2 ) cos2 θ. ) ( m2A c2 − 2E2 /c2 2 M = 2g 2 m2A c2 − 2E2 /c2 − 4( E2 /c2 ) E2 /c2 − m2A c2 cos2 θ p21 =

p23 =

( E1 + E2 )2 = 4E2 ,

dσ = dΩ

g2h¯ c 8π

2

× h

=

h¯ g2 c3 8π

2

S|M|2 |p f | . Here S = 1/2, ( E1 + E2 )2 |pi | q |p f | = E/c, |pi | = E2 /c2 − m2A c2 so

dσ = dΩ

According to Eq. 6.47,

h¯ c 8π

1 E/c q 2E2 E2 /c2 − m2 c2 A m2A c2 − 2E2 /c2

m2A c2 − 2E2 /c2

2

2

i2 − 4( E2 /c2 ) E2 /c2 − m2A c2 cos2 θ m2A c4 − 2E2

1 2E

q

2

E2 − m2A c4

h

m2A c4 − 2E2

2

2

i2 − 4E2 E2 − m2A c4 cos2 θ

The total cross-section is σ=

Z

dσ dΩ = dΩ

Z

dσ sin θ dθ dφ = 2πA dΩ

Z π 0

sin θ dθ , ( a − b cos2 θ )2

where

92


A≡

h¯ g2 c3 8π

2

1 2E

q

E2 − m2A c4

2 a ≡ m2A c4 − 2E2 ; Z π 0

sin θ dθ = ( a − b cos2 θ )2

m2A c4 − 2E2

2

;

b ≡ 4E2 E2 − m2A c4 . Z 1

dx ( a − bx2 )2

(letting x ≡ cos θ ) " r # 1 b 1 1 −1 = + √ tanh . a a−b a ab −1

Here a − b = (m A c2 )4 − 4E(m A c2 )2 + 4E4 − 4E2 ( E2 − m2A c4 ) = (m A c2 )4 , so σ = (2π )

h¯ g2 c3 8π

2

1 2E

q

E2 − m2A c4

(

1 ( m A c2 )4 

q

) 2 − m2 c4 E 1 A  + tanh−1  q 2 c4 2 2E − m 2 2 A 2E2 − m A c4 2E E2 − m A c4 " q 2 2 3 2 2E E − m2A c4 1 h¯ g c 1 = 2π 8 ( m A c2 )4 E2 E2 − m2A c4 #  q 2E E2 − m2A c4 1  . + tanh−1  (2E2 − m2A c4 ) 2E2 − m2A c4 2E

Problem 6.14

Quoting Eq. 6.55, with mC = 0: 1 1 2 M=g + . ( p4 − p2 )2 ( p3 − p2 )2

93

B

(3) A

(2)

(1)

q

A

(4) After

Before

B

Here p1 =

E ,p , c

p2 = (mc, 0),

p3 =

E3 (1, pˆ 3 ), c

p4 =

E4 (1, pˆ 4 ). c

( p3 − p2 )2 = p23 + p22 − 2p2 · p3 = 0 + (mc)2 − 2mE3 , ( p4 − p2 )2 = p24 + p22 − 2p2 · p4 = 0 + (mc)2 − 2mE4 = ( p1 − p3 )2 = p23 + p21 − 2p1 · p3 EE3 E3 2 = 0 + (mc) − 2 − |p| cos θ c c2 So

E3 ( E − |p|c cos θ ) mc2 But the incident proton is nonrelativistic, so E ≈ mc2 |p|c, and hence E4 ≈ EE3 /mc2 ≈ E3 . Meanwhile, conservation of energy says E + mc2 = E3 + E4 ≈ 2E3 , so E3 ≈ E4 ≈ mc2 , and therefore E4 =

( p4 − p2 )2 ≈ mc2 − 2(mc)2 = −(mc)2 = ( p3 − p2 )2 g 2 1 1 M = g2 − − = − 2 . mc (mc)2 (mc)2 Using the result of Problem 6.9: 2 1 2 dσ h¯ E3 2 |M| |p3 | = where |p3 | = ≈ mc 2 dΩ 8π c m|p|[ E + mc − |p|c cos θ ) 2 g 4 h¯ g2 2 1 h¯ mc 1 ≈ 4 = . 2 8π mc 8π m|p|(2mc2 ) |p|(mc)5 But |p| ≈ mv, so, in the nonrelativistic limit, dσ = dΩ

h¯ g2 8π (mc)3

2 c v

σ = 4π

h¯ g2 8π (mc)3

2 c . v

94


Problem 6.15

p3

p2

(a)

B

q A

A

B

p2

q

C p

p1

B 4

A

p3

A

p

B

C

p1

4

Diagram 1: Z

=

(−ig)2

q2

i d4 q (2π )4 δ4 ( p1 − p4 − q)(2π )4 δ4 (q + p2 − p3 ) 2 2 (2π )4 − mC c

−ig2 (2π )4 δ4 ( p1 + p2 − p3 − p4 ), ( p1 − p4 )2 − m2C c2

M1 =

g2 . ( p1 − p4 )2 − m2C c2

Diagram 2: Z

=

(−ig)2

i d4 q 4 4 4 4 ( 2π ) δ ( p + p − q )( 2π ) δ ( q − p − p ) 2 3 1 4 2 (2π )4 q2 − m C c2

g2 −ig2 4 4 ( 2π ) δ ( p + p − p − p ) , M = . 2 3 2 1 4 2 ( p1 + p2 )2 − m C c2 ( p1 + p2 )2 − m2C c2 " # 1 1 2 M = M1 + M2 = g + ( p1 − p4 )2 − m2C c2 ( p1 + p2 )2 − m2C c2

(b)

A

p

p2

1

Before

p B

q

p4 B

A

3

After

E E E p1 = , p , p2 = , − p1 , p 3 = , p3 , c 1 c c 2E E2 ( p1 + p2 ) = , 0 , so ( p1 + p2 )2 = 4 2 c c

p4 =

E , −p3 c

95

p1 − p4 = (0, p1 + p3 ),

so

( p1 − p4 )2 = −(p1 + p3 )2 = −(p21 + p23 + 2p1 · p3 ) = −(2p21 + 2|p1 ||p1 | cos θ ) = −4p21 cos2 θ/2. ! g2 c2 1 M= − 2 . 4 E2 p1 cos2 θ/2 2

|M|2 |p4 | ; |p4 | = |p1 |, (2E)2 |p1 | !2 2 2 2 h¯ c 1 g c2 1 = − 2 8π (2E)2 4 E2 p1 cos2 θ/2

dσ = dΩ

But

p21 = E2 /c2 − m2A c2 ,

h¯ c 8π

so

! 1 c2 c2 c2 − − = E2 E2 ( E2 − m2 c4 ) cos2 θ/2 p21 cos2 θ/2 2 2 2 4 2 2 2 2 4 2 2 ( E − m c ) cos θ/2 − E 2 E cos θ/2 − 1 − m c cos θ/2 =c =c E2 ( E2 − m2 c4 ) cos2 θ/2 E2 ( E2 − m2 c4 ) cos2 θ/2 2 2 2 4 2 2 2 2 4 2 E sin θ/2 + m c cos θ/2 2 E tan θ/2 + m c = −c = −c E2 ( E2 − m2 c4 ) cos2 θ/2 E2 ( E2 − m2 c4 )   g2h¯ c3 E2 tan2

dσ =  64π dΩ

θ 2

2 + m2 c4 

E3 ( E2 − m2 c4 )

.



(c)

A

p

1

p2

p B

Before 2 h¯ |M|2 8πm B c  E  p1 = , p1 , p2 = (m B c, 0)   c E  p3 = , p3 , p4 = (m B c, 0)   c

dσ = dΩ

p4

A 3

q B

After

From these, we get:

96


( p1 − p4 ) =

( p1 − p4 )

2

E − m B c, p1 ; c

− m2C c2

E − mB c c

2

E − mB c c

2

E + mB c c

2

E + mB c c

2

= =

( p1 + p2 )2 − m2C c2 = =

⇒M=g

2

2 m2B c2

.

( p1 + p2 ) =

dσ = dΩ

E + m B c, p1 . c

− p21 − m2C c2

−

E2 − m2A c2 c2

− m2C c2 ∼ = m2B c2 .

− m2C c2 ∼ = m2B c2 .

− p21 − m2C c2

−

h¯ g2 4πm3B c3

E2 − m2A c2 c2

!2 .

(d)

σ=

Z

dσ dΩ = 4π dΩ

dσ dΩ

1 =⇒ σ = π

h¯ g2 2m3B c3

!2 .

97

7

Quantum Electrodynamics Problem 7.1

From Eq. 3.8, 0

x0 = γ( x0 − βx1 ) 0

x1 = γ( x1 − βx0 ) 0

x2 = x2 0

x3 = x3 . Lowering a spatial index costs a minus sign, so x00 = γ( x0 + βx1 )

− x10 = γ(− x1 − βx0 ) ⇒ x10 = γ( x1 + βx0 ) x20 = x2 x30 = x3 . These are the transformation rules for covariant 4-vectors. Now

(∂µ φ)0 =

∂φ ∂φ ∂x ν ∂x ν = = ( ∂ ν φ ), ∂x ν ∂x µ 0 ∂x µ 0 ∂x µ 0

but the inverse Lorentz transformations (Eq. 3.3) say 0

0

0

0

x0 = γ( x0 + βx1 ) x1 = γ( x1 + βx0 ) x2 = x2

0 0

x3 = x3 , from which it follows that ∂x0 ∂x

00

= γ,

∂x0 ∂x

10

= γβ,

∂x1 ∂x

00

= γβ,

∂x1 ∂x

10

= γ,

∂x2 ∂x

20

= 1,

∂x3 ∂x3

0

= 1,

98

7 Quantum Electrodynamics

and all the rest are zero. Accordingly,

( ∂0 φ ) 0 = ( ∂0 φ ) ( ∂1 φ ) 0 = ( ∂0 φ ) ( ∂2 φ ) 0 = ( ∂2 φ ) ( ∂3 φ ) 0 = ( ∂3 φ )

∂x0

∂x1

00

+ ( ∂1 φ )

0

+ ( ∂1 φ )

0

= ( ∂2 φ )

0

= ( ∂3 φ ),

∂x ∂x0 ∂x1 ∂x2

∂x2 ∂x3 ∂x3

0

= γ [(∂0 φ) + β(∂1 φ)]

0

= γ [(∂1 φ) + β(∂0 φ)]

∂x0 ∂x1 ∂x1

so ∂µ φ transforms as a covariant 4-vector (justifying the placement of the index).

Problem 7.2

1 0 1 0 1 0 =2 = 2g00 , 0 −1 0 −1 0 1 1 0 0 σi 0 σi 1 0 + 0 −1 0 −1 −σi 0 −σi 0 0 σi 0 −σi 0 0 + = = 2g0i , i 0 0 σ 0 −σi 0 0 σi 0 σj 0 σj 0 σi + −σi 0 −σ j 0 −σ j 0 −σi 0 i j j i −σ σ 0 −σ σ 0 −{σi , σ j } 0 + = . 0 −σi σ j 0 −σ j σi 0 −{σi , σ j }

{ γ0 , γ0 } = 2( γ0 )2 = 2 { γ0 , γ i } = = { γi , γ j } = =

But {σi , σ j } = 2δij (Problem 4.20), so

{γi , γ j } = −2δij

1 0 0 1

Conclusion: {γµ , γν } = 2gµν for all µ, ν. X

= 2gij .

99

Problem 7.3

From Eq. 7.46,  u (1) † u (1) = N ∗ 1 0

c( p x −ipy ) cpz ( E+mc2 ) ( E+mc2 )

  N  

1 0

  

cpz  ( E+mc2 )   c( p x +ipy ) 2 ( E+mc )

(

c2 ( p x − ipy )( p x + ipy ) c2 p2z = | N |2 1 + 0 + + ( E + mc2 )2 ( E + mc2 )2 o n | N |2 2 2 2 2 2 2 2 = ( E + mc ) + c p + c ( p + p ) z x y ( E + mc2 )2 n o | N |2 2 2 2 2 ) + c p ( E + mc . = ( E + mc2 )2

)

But c2 p2 = E2 − m2 c4 = ( E − mc2 )( E + mc2 ), so u (1) † u (1) =

2E| N |2 | N |2 2 2 { E + mc + E − mc } = . ( E + mc2 ) ( E + mc2 ) 2E| N |2

Likewise, u(2)† u(2) = (E+mc2 ) (same calculation with last two terms switched). From Eq. 7.43, this is equal to 2E/c, so r

2E 2E| N |2 = , c ( E + mc2 )

or | N | =

E + mc2 . c

Similarly, starting with Eq. 7.47,  cpz ( E+mc2 )    c( px +ipy )  2  ( E+mc )  0 

v (2) † v (2) = | N | 2


1

 

1 0

 

p2 c2 = |N| +1 ( E + mc2 )2 n o | N |2 2 2 2 2 = ( E + mc )( E − mc ) + ( E + mc ) ( E + mc2 )2 | N |2 2E| N |2 = E − mc2 + E + mc2 = 2 ( E + mc ) ( E + mc2 ) 2

(and the same for v(1) ). So the normalization constant is the same for all of them.

100


Problem 7.4

 u (1) † u (2) = | N | 2 1 0

(

= |N|

2


    



0 1

 

c( p x −ipy )   ( E+mc2 )  cpz − (E+mc2 )

c2 pz ( p x − ipy ) c2 pz ( p x − ipy ) 0+0+ − ( E + mc2 )2 ( E + mc2 )2

)

= 0. X

 cpz ( E+mc2 )    c( px +ipy )  2  ( E+mc )  1 

v(1)† v(2) = −| N |2

c( p x +ipy ) −cpz ( E+mc2 ) ( E+mc2 )

(

= −| N |

2

0

 

c2 pz ( p x + ipy ) c2 pz ( p x + ipy ) − +0+0 ( E + mc2 )2 ( E + mc2 )2

u (1) † v (1) = | N | 2 1 0



)

= 0. X

c( p x −ipy ) ( E+mc2 )   −cpz  c( p x −ipy )  cpz  (E+mc2 )  2 2   ( E+mc ) ( E+mc )



= | N |2

 

1 0



0 1



c( p x − ipy ) c( p x − ipy ) +0+0+ 2 ( E + mc ) ( E + mc2 )

= 2( p x − ipy ).

NO: u(1) and v(1) are not orthogonal.

Problem 7.5

In the nonrelativistic limit p = mv and E = mc2 , so cpz cmvz vz = = , 2c E + mc2 2mc2

c( p x ± ipy ) cm(v x ± ivy ) (v x ± ivy ) = = . 2 2 2c E + mc 2mc

So u A is of order 1, but u B is of order v/c. X

101

Problem 7.6

√ p In this case p x = py = 0, and cpz = c|p| = E2 − m2 c4 = ( E − mc2 )( E + mc2 ).      √ 1 1 E + mc2 √     0 0   √   E + mc2   . r (E−mc2 )  = √1 √ 0 √ 2 )( E + mc2 )  = u (1) = N  ( E − mc       2 E − mc c c 2     2 ( E+mc ) ( E+mc ) 0 0 0  0 1  E + mc2   . = √   c √ 0 2 − E − mc     0 0 √ √ √   2 2 − ( E−mc )( E+mc )  2  E + mc2    = √1 − E − mc  ( E+mc2 ) √ v (1) =     c c   0 √ 0 E + mc2 1 

u (2)

√

√

v (2)

 E − mc2  1   . √ 0 = −√   2 E + mc  c 0

From Eq. 7.51, 

1 h¯  0 Sz =  2 0 0

0 −1 0 0

0 0 1 0

 0 0  , 0  −1

so S z u (1) =

h¯ (1) h¯ h¯ h¯ u , S z u (2) = − u (2) , S z v (1) = − v (1) , S z v (2) = v (2) . X 2 2 2 2

Eigenvalues are

h¯ 2,

− h2¯ , − h2¯ , h2¯ , respectively .

Problem 7.7

Let u(±) ≡ au(1) + bu(2) . This automatically satisfies the Dirac equation (7.49); we need to pick the constants a and b such that u(±) is a (normalized) eigenstate of the helicity operator:

(pˆ · Σ) u(±) = ±u(±) .

102


Now p·Σ =

p·σ 0 , 0 p·σ

and p · σ =

pz ( p x − ipy ) , ( p x + ipy ) − pz

so

(p · Σ) u(±) =

=

=

=

             p·σ 0 cpz  + bN  c( px −ipy )  aN     (E+mc2 )  0 p·σ     (E+mc2 )      c ( p + ip )   − cp x y z   2 2 ( E+mc ) ( E+mc )      0 1     ( p · σ )   (p · σ )    1 0  + b   N a  c(p·σ ) p x − ipy    c (p· σ )  pz     E+mc2 E+mc2 p x + ipy − pz      pz p x − ipy         p x + ipy   − pz      = ±|p|u(±) N a  cp2  + b   0         E+mc2   cp2   0 2 E+mc      0   1            1 0    cp     z ±|p| N a  + b  c( px −ipy )  . 2      (E+mc )   (E+mc2 )    c( p x +ipy )   −cpz   2 2      







1 0

( E+mc )

0 1

( E+mc )

Equating the top elements of the last two equations (you can check for yourself that the other three components yield the same condition), ±|p| − pz apz + b( p x − ipy ) = ±|p| a, or b = a. p x − ipy So  u(±)

   = Na   

1



 ±|p|− pz  p x −ipy   ± c |p|  ( E+mc2 )  c|p|(|p|∓ pz ) ( E+mc2 )( p x −ipy )

 Na = p z ± |p|

 p z ± |p|  p x + ipy    ±c|p|( pz ±|p|)  = A = A  (E+mc  2)  

     

p z ± |p|

 (|p|2 − p2z )  p x −ipy  ±c|p|( pz ±|p|)   ( E+mc2 )  ±c|p|(|p|2 − p2z ) 2 ( E+mc )( p x −ipy )



±c|p|( p x +ipy ) ( E+mc2 )

u ± c |p| u ( E+mc2 )



! ,

103

where

u≡

p z ± |p| p x + ipy

and A ≡ Na/( pz ± |p|) is determined by normalization: ! u c2 p2 ± c | p | 2 2 † † (±) † (±) u = | A| u† E+mc2 u† = | A | u u + u u u ± c |p| u (e + mc2 )2 E+mc2 p z ± |p| E2 − m2 c4 = | A |2 1 + ( p ± | p |) ( p − ip ) z x y p x + ipy ( E + mc2 )2 2 E − mc p2z ± 2pz |p| + p2 + p2x + p2y = | A |2 1 + E + mc2 2E = | A |2 2|p| (|p| ± pz ) . ( E + mc2 ) Using the convention in Eq. 7.43, this is equal to 2E/c, so

| A |2 =

( E + mc2 ) . X 2|p|c (|p| ± pz )

Problem 7.8

(a) Equation 7.19 says γ0 p0 − γ · p − mc = 0. Multiply from the left by cγ0 , noting that (γ0 )2 = 1: H ≡ cp0 = cγ0 (γ · p + mc). (b)

[ H, Li ] =

∑ eijk [ H, r j pk ] = ∑ eijk cγ0 γl [ pl , r j ] pk = ∑ eijk cγ0 γl (−i¯hδ jl ) pk j,k

= −i¯hcγ

j,k,l

0

∑ eijk γ p

j k

j,k,l

0

= −i¯hcγ (γ × p)i ,

j,k

[ H, L] = −i¯hcγ0 (γ × p). (c) Noting that γ0 commutes with Σ, we have c¯h h¯ [cγ0 (γ · p + mc), Σi ] = p j [ γ0 γ j , Σ i ]; 2 2 ∑ j 0 σj 0 σj 1 0 γ0 γ j = = ; 0 −1 −σj 0 σj 0 0 σj σj 0 σj 0 0 σj 0 [σj , σi ] 0 j [ γ γ , Σi ] = − = σj 0 0 σj 0 σj σj 0 [σj , σi ] 0

[ H, Si ] =

104


[σj , σi ] = −2i ∑ eijk σk ,

so

k

[γ γ , Σi ] = −2i ∑ eijk 0 j

k

c¯h [ H, Si ] = 2

0 σk σk 0

∑ p (−2ieijk )γ j

= −2iγ0 ∑ eijk γk , k

0 k

γ = −i¯hcγ0 (p × γ)i ;

j,k

[ H, S] = i¯hcγ0 (γ × p). Thus

[ H, J] = [ H, L] + [ H, S] = −i¯hcγ0 (γ × p) + i¯hcγ0 (γ × p) = 0, and hence J is conserved (but not L or S). (d) h¯ 2 σ 2 0 h¯ 2 . S2 = Σ2 = 0 σ2 4 4 But σ 2 = σ12 + σ22 + σ32 = 1 + 1 + 1 = 3, so S2 =

3¯h2 4

(times the unit matrix, of course). But every spinor is an eigenstate of the unit matrix (with eigenvalue 1), so every spinor is an eigenstate of S2 , with eigenvalue 34 h¯ 2 . Evidently s = 1/2, so the Dirac equation describes particles of spin 1/2 (no surprise for us, but it may have been for Dirac).

Problem 7.9



iγ2 = i

0 σ2 −σ2 0



2 (1)∗

iγ u

0 0 0 0 0 −1 = N 0 −1 0 1 0 0 

iγ2 u(2)∗

   0 0 0 1 −i   0  = 0 0 −1 0   0 −1 0 0 0 1 0 0 0 0    c( p x −ipy )  1 1  mc2 )    (E+cp 0 z   −  0   cpz  = N  (E+mc2 )  = v(1) .      0  (E+mc2 )    0 c( p x −ipy ) 0 2 1 ( E+mc )     z  0 − (E+cpmc 2) 1    c( px +ipy )  1     0   c( px +ipy )  = N − (E+mc2 )  = v(2) .    0   (E+mc2 )    −1 cp z 0 − (E+mc2 ) 0

0 0 = i 0 −i

0 0 0 0 0 −1 = N 0 −1 0 1 0 0

0 0 i 0

0 i 0 0 

105

Problem 7.10

Formally, change the sign of m (in Eq. 7.20), which changes the sign of m in Eqs. 7.23, 7.24, 7.27, 7.28 (ψ A now represents antiparticles, and ψB particles), 7.30, 7.33, and 7.35. In Eq. 7.42 we now use the minus sign for (1) and (2), and the plus sign for (3) and (4), but the results are the same from here on. The point is that only the superficial notation is affected, not the physical content of the theory.

Problem 7.11

Following the hint, the Dirac equation in the primed frame becomes ν ∂x µ ∂ν (Sψ) − mc(Sψ) = 0. i¯hγ ∂x µ 0 Multiply on the left by S−1 : ∂x ν i¯h S−1 γµ S ∂ν ψ − mcψ = 0. ∂x µ 0 The Dirac equation in the original frame is i¯hγν ∂ν ψ − mcψ = 0, so solutions transform into solutions provided that ∂x ν S −1 γ µ S = γν . ∂x µ 0 Now, the inverse Lorentz transformations (Eq. 3.3) yield (Problem 7.1) ∂x0 ∂x0

0

= γ,

∂x0 ∂x1

0

= γβ,

∂x1 ∂x0

0

= γβ,

∂x1 ∂x1

0

= γ,

∂x2 ∂x2

0

= 1,

∂x3 ∂x3

0

= 1,

and all the rest are zero. Evidently S must satisfy the following four conditions: (1) γ0 = S−1 γ0 S γ + S−1 γ1 S γβ ⇒ Sγ0 = (γ0 S)γ + (γ1 S)γβ, (2) γ1 = S−1 γ0 S γβ + S−1 γ1 S γ ⇒ Sγ1 = (γ0 S)γβ + (γ1 S)γ, (3) γ2 = S−1 γ2 S ⇒ Sγ2 = γ2 S, (4) γ3 = S−1 γ3 S ⇒ Sγ3 = γ3 S.

106


It remains to check that S = a+ + a− γ0 γ1 satisfies these four equations. First note that γ0 and γ1 anticommute with γ2 and γ3 , and hence the product γ0 γ1 commutes with them. So (3) and (4) are satisfied. As for (1) and (2): Sγ0 = a+ γ0 + a− γ0 γ1 γ0 = a+ γ0 − a− γ0 γ0 γ1 = a+ γ0 − a− γ1 , Sγ1 = a+ γ1 + a− γ0 γ1 γ1 = a+ γ1 − a− γ0 , γ0 S = a + γ0 + a − γ0 γ0 γ1 = a + γ0 + a − γ1 , γ1 S = a + γ1 + a − γ1 γ0 γ1 = a + γ1 − a − γ1 γ1 γ0 = a + γ1 + a − γ0 . So equation (1) requires a+ γ0 − a− γ1 = ( a+ γ0 + a− γ1 )γ + ( a+ γ1 + a− γ0 )γβ

= γ( a+ + βa− )γ0 + γ( a− + βa+ )γ1 . Now γ= p

γ2 − 1 ⇒β= ⇒β = γ2 1 − β2 1

2

p

(γ − 1)(γ + 1) , γ

so γ( a+ + βa− ) = γ

"q

p 1 2 ( γ + 1) −

(γ − 1)(γ + 1) γ

q

# 1 2 ( γ − 1)

q

1 2 ( γ + 1)( γ − γ + 1) = a+ ; # " q p q ( γ − 1 )( γ + 1 ) 1 γ( a− + βa+ ) = γ − 12 (γ − 1) + 2 ( γ + 1) γ q = 12 (γ − 1)(−γ + γ + 1) = − a− ,

=

and hence (1) is satisfied. Likewise, equation (2) says a+ γ1 − a− γ0 = ( a+ γ0 + a− γ1 )γβ + ( a+ γ1 + a− γ0 )γ

= γ( a− + βa+ )γ0 + γ( a+ + βa− )γ1 , and this is confirmed by the previous two results.

Problem 7.12

If the parity transformation is to carry solutions (to the Dirac equation) into solutions, we again require

S −1 γ µ S

∂x ν = γν . ∂x µ 0

107

But this time x µ 0 = ( x0 , − x1 , − x2 , − x2 ), so ∂x0 ∂x

00

= 1,

∂x1 ∂x

∂x2

= −1,

10

∂x

20

= −1,

∂x3 ∂x3

0

= −1,

and all the rest are zero. So S−1 γ0 S ⇒ Sγ0 = γ0 S, (2) γ1 = − S−1 γ1 S ⇒ Sγ1 = −γ1 S, (3) γ2 = − S−1 γ2 S ⇒ Sγ2 = −γ2 S, (4) γ3 = − S−1 γ3 S ⇒ Sγ3 = −γ3 S.

(1) γ0 =

Evidently S is a 4 × 4 matrix that commutes with γ0 , and anticommutes with the other three. The obvious solution in S = γ0 (though you could multiply this by any number of modulus 1).

Problem 7.13

(a) From Eq. 7.53: 

S=

a+ a− σ1 a− σ1 a+

a+ 0 = 0 a−

0 a+ a− 0

0 a− a+ 0

 a− 0 . 0 a+

It’s real, and symmetric, so S† = S. †

2

S S=S =

a+ a− σ1 a− σ1 a+

a+ a− σ1 a− σ1 a+

=

a2+ + a2− σ12 2a+ a− σ1 . 2a+ a− σ1 a2+ + a2− σ12

But σ12 = 1, and a2+ + a2− = 21 (γ + 1) + 12 (γ − 1) = γ, while q q q 1 2a+ a− = −2 · ( γ + 1) ( γ − 1) = − γ2 − 1 = − γ 2 q v = −γ 1 − (1 − β2 ) = −γβ. (β ≡ ) c So S† S = γ

1 − βσ1 . − βσ1 1

X

s 1−

1 γ2

108


(b)

a+ a− σ1 S γ S= a− σ1 a+ 2 a+ a− σ1 a+ a− σ1 ( a+ − a2− ) 0 = = a− σ1 a+ − a− σ1 − a+ 0 ( a2− − a2+ ) † 0

But

( a2+

−

a2− )

=

a+ a− σ1 a− σ1 a+

1 0 0 −1

1 1 2 ( γ + 1) − 2 ( γ − 1)

= 1, so

S † γ0 S

=

1 0 0 −1

= γ0 .

X

Problem 7.14 †

¯ 5 ψ)0 = (ψ0 γ0 γ5 ψ0 ) = (Sψ)† γ0 γ5 (Sψ) = ψ† S† γ0 γ5 Sψ. But (ψγ 01 a+ a− σ1 a σ a a+ a− σ1 0 1 γ5 S = = − 1 + = = Sγ5 . 10 a− σ1 a+ a+ a− σ1 a− σ1 a+ 1 0 So

S† γ0 γ5 S = S† γ0 Sγ5 = γ0 γ5 (Problem 7.13b), and hence ¯ 5 ψ)0 = ψ† γ0 γ5 ψ = ψγ ¯ 5 ψ. X (ψγ

Problem 7.15

(γµ pµ − mc)u = 0 =⇒ u† (γµ† pµ − mc) = 0 =⇒ u† (γµ† γ0 pµ − γ0 mc) = 0. But γµ† γ0 = γ0 γµ (see below), so u† γ0 (γµ pµ − mc) = 0, or u¯ (γµ pµ − mc) = 0. Similarly, (γµ pµ + mc)v = 0 ⇒ v¯ (γµ pµ + mc) = 0 (same as above, with sign of m reversed). Proof that

γµ† γ0 = γ0 γµ :

† 1 0 1 0 = = γ0 , so it holds for µ = 0. 0 −1 0 −1 † 0 σi 0 −(σi )† . ( γi )† = = −σi 0 (σi )† 0

γ0† =

109

But (σi )† = σi :

†

0 1 (σ ) = = = σ1 ; 1 0 † 0 −i 0 −i 2 † = = σ2 ; (σ ) = i 0 i 0 † 1 0 1 0 3 † = = σ3 . (σ ) = 0 −1 0 −1 0 −σi So (γi )† = = −γi . But γi anticommutes with γ0 , so γi γ0 = −γ0 γi . σi 0 Therefore (γi )† γ0 = −γi γ0 = γ0 γi . So it holds for µ = i = 1, 2, 3 also. X 1 †

01 10

Problem 7.16

In the notation of Eq. 7.42 and 7.46, 1 0 uA † 0 2 † † ¯ = u γ u = N u A uB uu = N 2 u†A u A − u†B u B . 0 −1 uB In particular, for u(1) : 1 c2 pz ¯ = N2 1 0 uu − p ( p − ip ) z x y 0 p x + ipy ( E + mc2 )2 h i 2 2 N c 2 2 2 2 2 2 2 ( p + p + p ) = ( E + mc ) − c p = N2 1 − z x y ( E + mc2 )2 ( E + mc2 )2 h i 2 ( E + mc ) 1 2 2 2 4 2 2 = E + 2Emc + m c − c p c ( E + mc2 )2 1 1 2 2 4 = 2Emc + 2m c = 2mc2 ( E + mc2 ) = 2mc. X c( E + mc2 ) c( E + mc2 ) It’s essentially the same for u(2) ; for v(1) and v(2) we use Eq. 7.47: p x − ipy 0 c2 2 ¯ = N vv − 01 ( p x + ipy ) − pz − pz 1 ( E + mc2 )2 i h N2 c2 2 2 2 2 2 2 2 = N2 ( p + p + p ) − 1 = c p − ( E + mc ) z y ( E + mc2 )2 x ( E + mc2 )2 h i ( E + mc2 ) 1 = c2 p2 − E2 − 2Emc2 − m2 c4 2 2 c ( E + mc ) −1 −1 2 2 4 = = 2Emc + 2m c 2mc2 ( E + mc2 ) = −2mc. X c( E + mc2 ) c( E + mc2 )

110


Problem 7.17

¯ µ ψ. Then Let aµ ≡ ψγ 0 ¯ 0 ψ)0 = ψ0† γ0 γ0 ψ0 = (Sψ)† γ0 γ0 (Sψ) = ψ† S† γ0 γ0 Sψ. a0 = (ψγ

¯ 0 ψ)0 = ψ† (S† S)ψ. But γ0 γ0 = 1, so (ψγ 1 − βσ1 (Problem 7.13a) − βσ1 1 0 σ1 1 0 0 σ1 = γ − βγ = γ − βγ σ1 0 0 −1 −σ1 0

S† S = γ

= γ(γ0 γ0 ) − βγ(γ0 γ1 ) = γ0 (γγ0 − βγγ1 ). 0 a0 = ψ† γ0 (γγ0 − βγγ1 )ψ = ψ¯ (γγ0 − βγγ1 )ψ = γa0 − βγa1 .X

Similarly 0 ¯ 1 ψ)0 = ψ† S† γ0 γ1 Sψ. a1 = (ψγ

S † γ0 γ1 S = ( a + + a − γ0 γ1 ) γ0 γ1 ( a + + a − γ0 γ1 )

= a2+ γ0 γ1 + 2a+ a− (γ0 γ1 )2 + a2− (γ0 γ1 )3 . ( γ0 γ1 )2 =

1 0 0 −1

0 σ1 − σ1 0

2

=

0 σ1 σ1 0

2

=

( σ 1 )2 0 0 ( σ 1 )2

= 1.

S† γ0 γ1 S = ( a2+ + a2− )γ0 γ1 + 2a+ a− = γγ0 γ1 − γβγ0 γ0

= γ0 (γγ1 − βγγ0 ) (using results from Problem 7.13a). 0 a1 = ψ† γ0 (γγ1 − βγγ0 )ψ = ψ¯ (γγ1 − βγγ0 )ψ = γa1 − βγa0 . X

Finally, 0

¯ 2 ψ)0 = ψ† S† γ0 γ2 Sψ. a2 = (ψγ γ2 S = γ2 ( a+ + a− γ0 γ1 ) = ( a+ + a− γ0 γ1 )γ2 = Sγ2 (since γ2 anticommutes with γ0 and γ1 ). 0 ¯ 2 ψ = a2 . a2 = ψ† S† γ0 Sγ2 ψ = ψ† γ0 γ2 ψ = ψγ

X

111

(I used S† γ0 S = γ0 .) Same for a3 . Under parity, ψ0 = γ0 ψ (Eq. 7.61), so 0

¯ 0 ψ ) 0 = ψ † γ0 γ0 γ0 γ0 ψ = ψ † ( γ0 )2 ψ a0 = (ψγ ¯ 0 ψ = a0 = ψγ

(because (γ0 )2 = 1) (so a0 is invariant under P)

0

¯ i ψ ) † = ψ † γ0 γ0 γ i γ0 ψ = ψ † γ i γ0 ψ ai = (ψγ ¯ i ψ = − ai = ψ† (−γ0 γi )ψ = −ψγ

(so ”spatial” parts change sign).

Problem 7.18

Equation 7.61 =⇒ Pψ = γ0 ψ, so Pψ(1) = ψ(1) , Pψ(2) = ψ(2) , Pψ(3) = −ψ(3) , and Pψ(4) = −ψ(4) . Evidently ψ(1) and ψ(2) (the electron states) are eigenstates of P with eigenvalue +1 (positive intrinsic parity), while ψ(3) and ψ(4) (the positron states) have eigenvalue −1. If P = −γ0 , the parities are reversed, but it is still the case that particle and antiparticle have opposite parity.

Problem 7.19

(a) From Eqs. 7.15 and 7.69:

σµν

{γµ , γν } = 2gµν ⇒ γµ γν + γν γµ = 2gµν , i = (γµ γν − γν γµ ) ⇒ γµ γν − γν γµ = −2iσµν . 2

Adding, 2γµ γν = 2 ( gµν − iσµν ) ,

or

γµ γν = gµν − iσµν .

(b) 0 σ2 0 σ1 γ1 γ2 − γ2 γ1 = iγ1 γ2 = i − σ1 0 − σ2 0 3 1 2 σ 0 −σ σ 0 = = Σ3 . = i 0 σ3 0 − σ1 σ2

σ12 =

i 2

I used the anticommutation relation for the gamma matrices, and σ1 σ2 = iσ3 (Problem 4.19); note that because σi is not part of a 4-vector, we do not distinguish upper and lower indices.

112


Similarly, σ

σ

13

23

0 σ1 0 σ3 = γ γ − γ γ = iγ γ = i − σ1 0 − σ3 0 1 3 2 −σ σ 0 −σ 0 = i = = − Σ2 . 0 − σ1 σ3 0 − σ2 i 2

1 3

3 1

1 3

0 σ2 = γ γ − γ γ = iγ γ = i − σ2 0 2 3 1 −σ σ 0 σ 0 = i = = Σ1 . 0 − σ2 σ3 0 σ1 i 2

2 3

3 2

2 3

0 σ3 − σ3 0

Problem 7.20

(a) Start with Eq. 7.73: ∂µ F µν = (4π/c) J ν . For the case ν = 0: ∂Ey ∂ 00 ∂ ∂ ∂ ∂Ex ∂Ez F + 1 F10 + 2 F20 + 3 F30 = + + = ∇·E ∂x ∂y ∂z ∂x0 ∂x ∂x ∂x 4π 4π 0 J = cρ = 4πρ. So ∇ · E = 4πρ. X = c c

∂µ F µ0 =

For the case ν = 1: ∂(− By ) ∂ 01 ∂ ∂ ∂ ∂(− Ex ) ∂Bz F + 1 F11 + 2 F21 + 3 F31 = + + 0 ∂(ct) ∂y ∂z ∂x ∂x ∂x ∂x ∂By 1 ∂Ex ∂Bz 1 ∂E = − + − = − + (∇ × B) c ∂t ∂y ∂z c ∂t x 4π 1 4π = J = Jx . c c

∂µ F µ1 =

This is the x component of

∇×B−

1 ∂E 4π = J X c ∂t c

(the y component comes from ν = 2, and the z component from ν = 3). (b) Take the divergence of Eq. 7.73: ∂ν ∂µ F µν =

4π ∂ν J ν . c

But ∂ν ∂µ is symmetric in µ ↔ ν (by the equality of cross-derivatives), whereas F µν is antisymmetric, so the left side is zero (Problem 3.10e), and hence ∂µ J µ = 0. X

113

Problem 7.21

The continuity equation (Eq. 7.74) says 0 = ∂ µ J µ = ∂0 J 0 + ∂1 J 1 + ∂2 J 2 + ∂3 J 3 =

∂Jy ∂(cρ) ∂Jx ∂Jz + + + ∂(ct) ∂x ∂y ∂z

∂ρ = −∇ · J. ∂t

=⇒

Integrate over some volume V, with surface S: Z V

∂ρ d dτ = ∂t dt

Z

dQ ρ dτ = =− dt V

Z V

∇ · J dτ = −

Z S

J · dA

(where Q is the total charge in V, and I used the divergence theorem in the last step). This says that the rate of change of the charge in V is minus the flux of charge out through the surface—no charge simply disappears or is created from nothing. In particular, if we pick a volume such that J is zero at the surface, then Q is constant.

Problem 7.22

A00 = A0 + ∂0 λ = 0

⇒

∂0 λ = − A0

λ(r, t) = −c

Z t −∞

⇒

∂λ = −cA0 . ∂t

A0 (r, t0 ) dt0 .

(You don’t have to use −∞ as the lower limit—any constant would do.) 1 ∂2 λ 1 ∂2 t 2 − ∇ λ = − A0 (r, t0 ) dt0 + c∇2 c ∂t2 −∞ c2 ∂t2 Z t 1 ∂ = − ∇2 A0 (r, t0 ) dt0 . [ A0 (r, t)] + c c ∂t −∞

λ =

Z

Z t −∞

A0 (r, t0 ) dt0

2

But Aµ = 0 ⇒ ∇2 A0 (r, t0 ) = (1/c2 )(∂2 /∂t0 ) A0 (r, t0 ), so t 1 ∂A0 1 ∂ 0 λ = − + A0 (r, t ) c ∂t c ∂t0

−∞

1 ∂A0 1 ∂A0 1 ∂A0 (r, t0 ) = − + − c ∂t c ∂t c ∂t0

−∞

1 ∂A0 (r, t0 ) =− c ∂t0

= 0. X −∞

114


Problem 7.23

(a) µ

µ

λ = ∂ ∂µ λ = i¯hκa ∂ ∂µ e

−ip· x/¯h

i = i¯hκa − h¯

2

pµ pµ e−ip· x/¯h = 0 X

(Eq. 7.90). (b) Aµ 0 = Aµ + ∂µ λ = ae−ip· x/¯h eµ + ∂µ i¯hκae−ip· x/¯h i = ae−ip· x/¯h eµ + i¯hκa − pµ e−ip· x/¯h = ae−ip· x/¯h (eµ + κ pµ ) . h¯ Thus eµ 0 = eµ + κ pµ .

115

Problem 7.24

∑

u(s) u¯ (s)

s=1,2

  1      0   = | N |2  cpz  1 0 − cpz 2 − c( px −ip2y ) E+mc E+mc   E+mc2     c( px +ipy ) E+mc2



0 1

    



    +  c( px −ipy )  0 1 − c( px +ip2y ) cpz 2 E+mc E+mc   E+mc2     cpz − E+mc2   −c( p x −ipy ) −cpz  1 0  2 2 E + mc E + mc      0 0 0 0 E + mc2    −c2 pz ( p x −ipy )  = −c2 p2z  cpz 0   c  ( E+mc2 )2 ( E+mc2 )2   E+mc2   2 p ( p +ip )  − c2 ( p2x + p2y ) c ( p + ip ) − c x y z x y  0 E+mc2



0 0  1 0  +  c( px −ipy ) 0 E+mc2  −cp 0 E+mcz 2  E+mc2

( E+mc2 )2

0 −c( p x +ipy ) E+mc2 −c2 ( p2x + p2y ) ( E+mc2 )2 c2 pz ( p x +ipy ) ( E+mc2 )2

( E+mc2 )2

 0     cpz    2 E+mc  2 c pz ( p x −ipy )   ( E+mc2 )2     2 2  −c pz  ( E+mc2 )2

 − pz −( p x − ipy )   E+mc2 0 −( p x + ipy ) pz   c . 2 = cp   pz ( p x − ipy ) − E+mc2 0   2 ( p x + ipy ) − pz 0 − E+cpmc2 c

But p · σ =

pz p x − ipy p x + ipy − pz  E c

∑

s=1,2

 u(s) u¯ (s) =  

+ mc 0

0

E c

and

0 + mc

(p · σ )

E2 − m2 c4 c2 p2 = = E − mc2 , so 2 E + mc E + mc2



−(p · σ)

− Ec

+ mc 0 0 − Ec + mc

  

E 0 0 σ = γ −p· + mc = γµ pµ + mc. X −σ 0 c

116


∑

v(s) v¯ (s)

s=1,2

 c( p −ip )  x y   mc2   E+cp  z   = | N |2  − E+mc2  c( px +ip2y ) − cpz 2 0 −1 E+mc   E+mc  0    1   cpz    E+mc2   c( px +ipy )    E+mc2  c( p x −ipy ) cpz +  E+mc2 E+mc2 −1 0    1    0  2 2 2  c ( p x + py ) −c2 pz ( p x −ipy ) −c( p x −ipy )   0  ( E+mc2 )2 E+mc2   (2E+mc2 )2   −c pz ( p x +ipy ) c2 p2z cpz E + mc2    0 E+mc 2 = ( E+mc2 )2   (E+mc2 )2    c  0 0 0 0      c( p x +ipy )  −cpz 0 −1 E+mc2 E+mc2   2 2 2 c pz ( p x −ipy ) c pz z − E+cpmc 0   2 ( E+mc2 )2  2(E+mc2 )2    c pz ( px +ipy ) c2 ( p2x + p2y ) −c( px +ipy )     0 +  (E+mc2 )2 ( E+mc2 )2 E+mc2    c( p x −ipy ) cpz   −1 0   E+mc2 E+mc2   0 0 0 0  cp2  0 − pz −( p x − ipy ) 2 E + mc   cp2   0 −( p x + ipy ) pz 2   E + mc =  2 E + mc  pz  0 ( p x − ipy ) − c

− pz ! 0

p x + ipy 

E−mc2 c

  =  

=

0

− E+cmc 

0

−(p · σ)

E−mc2 c

2

− E+cmc 0 E+mc2 0 − c

(p · σ )

E 0 γ − p · γ − mc = γµ pµ − mc. c

2

  !  

X

Problem 7.25

e(1) = (1, 0, 0),

e(2) = (0, 1, 0),

∑

s=1,2

(s) (s) ∗ ei e j

=

p = (0, 0, p) ⇒ pˆ = (0, 0, 1).

(1) (1) ∗ ei e j

(2) (2) ∗

+ ei e j

is zero unless i = j; it is 1 for i = j = 1 and i = j = 2, but 0 for i = j = 3, so δij doesn’t quite do it, but if we subtract off pˆ i pˆ j , which is 1 for i = j = 3 and 0

117

otherwise, that will fix the problem:

∑

s=1,2

(s) (s) ∗

ei e j

= δij − pˆ i pˆ j . X

Problem 7.26

e 1

Before

m

e

m

2

3

4

After

From Problem 7.6,   a+ 0  u (1) =   a −  , u (2) = 0 p where a± ≡ ( Ee ± mc2 )/c;



     0 0 b+  b+   a+  0   , quadu(3) =   , u(4) =   , 0 0  b−  b− a− 0 q b± ≡ ( Eµ ± Mc2 )/c. [Note: The bottom

entry in u(2) is not −b− (nor is it − a− in u(3)) – unlike Problem 7.6. The point is that (2) and (3) are going in the −z direction, so pz = −|p| for them.] From Eq. 7.106, n o ge2 M=− [u¯ (3)γ0 u(1)][u¯ (4)γ0 u(2)] − [u¯ (3)γi u(1)][u¯ (4)γi u(2)] 2 ( p1 − p3 ) (where i is summed from 1 to 3). 

u¯ (3)γ0 u(1) = 0 a+

   a+ a+ 0 0 0   0     0 a− γ γ   a−  = 0 a+ 0 a−  a−  = 0. 0 0 

u¯ (3)γi u(1) = 0 a+ 0 a−

1 0 0 −1

i

0 σ −σi 0

 a+ 0    a−  0

 a−   0 a a  = 0 a+ σi − + 0 a− σi + = 0 a+ 0 − a−   a+  0 0 i −σ 0 σi σi σi 1 i 11 12 11 = 2a+ a− 0 1 = 2a+ a− 0 1 = 2a+ a− σ21 . i σi i 0 σ21 σ21 22 

σi

118


 0  b+    0 

u¯ (4)γi u(2) = b+ 0 b− 0

1 0 0 −1

i

0 σ −σi 0

b− 



0 i i 0  σ b−  0  = b+ 0 σ i + σ = b+ 0 − b− 0  b− 0  b− b+ 0  −σi b+ σi σi σi 0 i 11 12 12 = 2b+ b− 1 0 = 2b b = 2b+ b− σ12 . 1 0 + − i σi i 1 σ21 σ 22 22

M=

8ge2 ge2 (2a+ a− 2b+ b− )σ 21 · σ 12 = ( a+ a− )(b+ b− ), 2 ( p1 − p3 ) ( p1 − p3 )2

where I used σ 21 · σ 12 = (1)(1) + (i )(−i ) + (0)(0) = 2 in the last step. Now r r Ee2 − m2 c4 p2e c2 ( a+ a− ) = = = |pe |, (b+ b− ) = |pµ |, and |pe | = |pµ |. c2 c2 8ge2 p2e . ( p1 − p3 )2 Ee Ee , pe , p 3 = , −pe ; so ( p1 − p3 ) = (0, 2pe ), ( p1 − p3 )2 = 0 − 4p2e . p1 = c c So M =

∴M=

8ge2 p2e = −2ge2 . −4p2e

Problem 7.27

There are two diagrams:

p2

p4

p4

q

q p1

p2

p

3

p1

p

3

119

Diagram 1: Z

i (q + mc) ν ( ig γ ) u ( 1 ) eν∗ (3) eµ∗ (4) v¯ (2)(ige γµ ) 2 e q − m2 c2

d4 q × (2π )4 δ4 ( p1 − q − p3 )(2π )4 δ4 ( p2 + q − p4 ) (2π )4 ( p − p + mc) ∗ e (3) u (1) = −ige2 v¯ (2)e∗ (4) 1 32 ( p1 − p3 ) − m2 c2

× (2π )4 δ4 ( p2 + p1 − p3 − p4 ) | {z } erase

M1 =

ge2 [v¯ (2)e∗ (4)( p1 − p3 + mc)e∗ (3)u(1)] . ( p1 − p3 )2 − m2 c2

Diagram 2: The same as Diagram 1, only with 3↔4. So

M2 =

ge2 [v¯ (2)e∗ (3)( p1 − p4 + mc)e∗ (4)u(1)] . ( p1 − p4 )2 − m2 c2

Problem 7.28

i† h † [v¯ ( a)Γ2 v(b)]∗ = v( a)† γ0 Γ2 v(b) = v(b)† Γ2† γ0 v( a)

= v(b)† γ0 γ0 Γ2† γ0 v( a) = v¯ (b)Γ¯ 2 v( a)     ∑ [v¯(a)Γ1 v(b)] [v¯(a)Γ2 v(b)]∗ = ∑ v¯(a)Γ1  ∑ v(b)v¯(b) Γ¯ 2 v(a). spins s( a) s(b) Invoking the completeness relation (Eq. 7.99),

∑ v(b)v¯(b) = ( pb − mb c) ,

s(b)

and defining Q ≡ Γ1 ( pb − mb c)Γ¯ 2 ,

∑

[v¯ ( a)Γ1 v(b)] [v¯ ( a)Γ2 v(b)]∗ =

spins

∑ v¯(a)Qv(a) = ∑ ∑ v¯(a)i Qij v(a) j

s( a)

=

 

s( a) i,j

 

∑ Qij  ∑ v(a) j v¯(a)i  = ∑ Qij ( pa − ma c) ji = ∑ [Q( pa − ma c)]ii i,j

s( a)

i,j

i

= Tr [ Q( p a − m a c)] = Tr [Γ1 ( pb − mb c)Γ¯ 2 ( p a − m a c)] .

120


The others are identical, except that the completeness relation carries a plus sign for particle states. So

∑

[u¯ ( a)Γ1 v(b)] [u¯ ( a)Γ2 v(b)]∗ = Tr [Γ1 ( pb − mb c)Γ¯ 2 ( p a + m a c)] ,

∑

[v¯ ( a)Γ1 u(b)] [v¯ ( a)Γ2 u(b)]∗ = Tr [Γ1 ( pb + mb c)Γ¯ 2 ( p a − m a c)] .

spins

spins

Problem 7.29

(a) If ν = 0, note that γ0† = γ0 and γ0 γ0 = 1, so γ0 γ0† γ0 = γ0 γ0 γ0 = γ0 . X 0 σi 0 −σi† i i † If ν = i = 1, 2, 3, then γ = =⇒ (γ ) = . But −σi 0 σi† 0 σi† = σi , so (γi )† = −γi . Moreover, γi γ0 = −γ0 γi , so γ0 γi† γ0 = − γ0 γ i γ0 = γ0 γ0 γ i = γ i . X (b) Γ¯ ≡ γ0 Γ† γ0 = γ0 (γa γb · · · γc )† γ0 = γ0 (γc† · · · γb† γa† )γ0 . Insert γ0 γ0 = 1 between every pair: h i Γ¯ = γ0 γc† (γ0 γ0 ) · · · (γ0 γ0 )γb† (γ0 γ0 )γa† γ0

= (γ0 γc† γ0 ) · · · (γ0 γb† γ0 )(γ0 γa† γ0 ) = γc · · · γb γa . X

Problem 7.30

According to Eq. 7.112, M1 = A [u¯ (4)Γ1 u(1)], where A≡

ge2 ( p1 − p3 )2 − (mc)2

and

Γ1 ≡ e(2)( p1 − p3 + mc)e(3)∗ .

First sum over the outgoing electron spins (s4 ) and average over the incoming electron spins (s1 ). Casimir’s trick (Eq. 7.125) gives h|M1 |2 ie

=

A2 1 2 A ∑ [u¯ (4)Γ1 u(1)] [u¯ (4)Γ1 u(1)]∗ = Tr [Γ1 ( p1 + mc)Γ¯ 1 ( p4 + mc)] , 2 2 s1 ,s4

121

where Γ¯ 1 ≡ γ0 Γ1† γ0 = γ0 [e(2)( p1 − p3 + mc)e(3)∗ ]† γ0 i h = γ0 e(3)† ( p1† − p3† + mc)e(2)∗ † γ0 h ih ih i = γ0 e(3)† γ0 γ0 ( p1† − p3† + mc)γ0 γ0 e(2)∗ † γ0

= e(3)( p1 − p3 + mc)e(2)∗ . Similarly, by Eq. 7.113, M2 = B [u¯ (4)Γ2 u(1)], where B≡

ge2 ( p1 + p2 )2 − (mc)2

h|M2 |2 ie =

and

Γ2 ≡ e(3)∗ ( p1 + p2 + mc)e(2).

B2 Tr [Γ2 ( p1 + mc)Γ¯ 2 ( p4 + mc)] , Γ¯ 2 = e(2)∗ ( p1 + p2 + mc)e(3). 2

hM1 M2∗ ie =

AB 2

∑ [u¯ (4)Γ1 u(1)] [u¯ (4)Γ2 u(1)]∗

s1 ,s4

= hM2 M1∗ ie =

AB 2

AB Tr [Γ1 ( p1 + mc)Γ¯ 2 ( p4 + mc)] , 2

∑ [u¯ (4)Γ2 u(1)] [u¯ (4)Γ1 u(1)]∗

s1 ,s4

=

AB Tr [Γ2 ( p1 + mc)Γ¯ 1 ( p4 + mc)] . 2

Now sum and average over the photon spins:

h|M1 |2 i =

A2 4

∑ Tr

s2 ,s3

h

e(2)µ γµ ( p1 − p3 + mc)e(2)∗ν γν ( p1 + mc)e(3)κ γκ

i × ( p1 − p3 + mc)e(2)∗λ γλ ( p4 + mc) . Writing the completeness relation for photons (Eq. 7.105) as ( 0, if µ or ν is 0 (s) (s) ∑ eµ eν = Qµν , where Qµν ≡ δ − pˆ pˆ , otherwise s1 ,s2 ij i j

h|M1 |2 i =

h A2 Qµλ Qνκ Tr γµ ( p1 − p3 + mc)γν ( p1 + mc)γκ 4 i

× ( p1 − p3 + mc)γλ ( p4 + mc) .

h|M2 |2 i =

h B2 Qµλ Qνκ Tr γµ ( p1 + p2 + mc)γν ( p1 + mc)γκ 4 i

× ( p1 + p2 + mc)γλ ( p4 + mc) .

122


hM1 M2∗ i =

h AB Qµλ Qνκ Tr γµ ( p1 − p3 + mc)γν ( p1 + mc)γλ 4 i

× ( p1 + p2 + mc)γκ ( p4 + mc) .

hM2 M1∗ i =

h AB Qµλ Qνκ Tr γµ ( p1 + p2 + mc)γν ( p1 + mc)γλ 4 i

× ( p1 − p3 + mc)γκ ( p4 + mc) .

Conclusion: ( h 1 h|M| i = Qµλ Qνκ A2 Tr γµ ( p1 − p3 + mc)γν ( p1 + mc)γκ 4 i × ( p1 − p3 + mc)γλ ( p4 + mc) h i + B2 Tr γµ ( p1 + p2 + mc)γν ( p1 + mc)γκ ( p1 + p2 + mc)γλ ( p4 + mc) h i + ABTr γµ ( p1 − p3 + mc)γν ( p1 + mc)γλ ( p1 + p2 + mc)γκ ( p4 + mc) 2

h

+ ABTr γ ( p1 + p2 + mc)γ ( p1 + mc)γ ( p1 − p3 + mc)γ ( p4 + mc) µ

ν

λ

κ

i

)

where A, B, and Qµν are defined above.

Problem 7.31

(a) Tr A ≡ ∑ Aii , so

(1) Tr ( A + B) = ∑( A + B)ii = ∑ Aii + ∑ Bii = Tr A + Tr B.

(2) Tr (αA) = ∑(αA)ii = ∑ α( Aii ) = αTr A.

(3) Tr ( AB) = ∑( AB)ii = ∑ ∑( Aij Bji ) = ∑ ∑ Bji Aij i

j

i

j

= ∑( BA) jj = Tr ( BA). j

(b) gµν gµν = g00 g00 + g01 g01 + · · · + g33 g33 , but gµν = 0 unless µ = ν

= g00 g00 + g11 g11 + g22 g22 + g33 g33 = (1)(1) + (−1)(−1) + (−1)(−1) + (−1)(−1) = 4. X

123

(c) µ ν ν µ µ ν ν µ ab + ba = ( aµ γ )(bν γ ) + (bν γ )( aµ γ ) = aµ bν (γ γ + γ γ ) = aµ bν (2gµν ) = 2aµ bµ = 2a · b X

Problem 7.32

(a) γµ γν + γν γµ = 2gµν

=⇒ gµν γµ γν + gµν γν γµ = 2gµν gµν = 2 · 4 = 8 γµ γµ + γµ γµ = 2γµ γµ = 8 =⇒ γµ γµ = 4. X

γµ γν γµ = γµ (2gµν − γµ γν ) = 2γµ gµν − γµ γµ γν

= 2γν − 4γν = −2γν . X γµ γν γλ γµ = γµ γν (2gλµ − γµ γλ ) = 2 gλµ γµ γν − (γµ γν γµ ) γλ | {z } | {z } −2γν

γλ

= 2(γλ γν + γν γλ ) = 2(2gλν ) = 4gνλ . X γµ γ ν γ λ γ σ γ µ

= γµ γν γλ (2gσµ − γµ γσ ) = 2 gσµ γµ γν γλ − γµ γν γλ γµ γσ | {z } | {z } γσ

4gνλ

= 2γσ γν γλ − 4gνλ γσ = 2γσ 2gνλ − γλ γν − 4gνλ γσ = 4gνλ γσ − 2γσ γλ γν − 4gνλ γσ = −2γσ γλ γν . X (b) γµ aγµ = γµ ( aν γν )γµ = aν (γµ γν γµ ) = aν (−2γν ) = −2( aν γν ) = −2a X γµ abγµ = γµ ( aν γν )(bλ γλ )γµ = aν bλ (γµ γν γλ γµ ) = aν bλ (4gνλ ) = 4a · b X γµ abcγµ = γµ ( aν γν )(bλ γλ )(cσ γσ )γµ = aν bλ cσ (γµ γν γλ γσ γµ )

= aν bλ cσ (−2γσ γλ γν ) = −2(cσ γσ )(bλ γλ )( aν γν ) = −2cba X

124


Problem 7.33

(a) (10) The trace of the product of an odd number of gamma matrices is zero. Proof : Insert γ5 γ5 = 1 : Tr (γµ γν · · · γσ ) = Tr (γµ γν · · · γσ γ5 γ5 ). But γ5 anticommutes with every γµ , so, pulling one γ5 through to the left we pick up a factor of (−1)n , if there are n γ’s: Tr (γµ γν · · · γσ ) = (−1)n Tr (γ5 γµ γν · · · γσ γ5 )

= (−1)n Tr (γµ γν · · · γσ γ5 γ5 ) [by Tr ( AB) = Tr ( BA)]. So Tr (γµ γν · · · γσ ) = (−1)n Tr (γµ γν · · · γσ ). So if n is odd, the trace is zero. X (11) 

1 0 Tr (1) = Tr  0 0

0 1 0 0

0 0 1 0

 0 0  = 1 + 1 + 1 + 1 = 4. X 0 1

(12) 1 [Tr (γµ γν ) + Tr (γν γµ )] 2 (since they’re equal, by trace rule 3) 1 1 = Tr (γµ γν + γν γµ ) = Tr (2gµν ) = gµν Tr (1) = 4gµν . X 2 2

Tr (γµ γν ) =

(13) Tr (γµ γν γλ γσ ) h i = Tr (2gµν − γν γµ )γλ γσ

= 2gµν Tr (γλ γσ ) − Tr (γν γµ γλ γσ ) i h = 8gµν gλσ − Tr γν (2gµλ − γλ γµ )γσ = 8gµν gλσ − 2gµλ Tr (γν γσ ) + Tr (γν γλ γµ γσ ) h i = 8gµν gλσ − 8gµλ gνσ + Tr γν γλ (2gµσ − γσ γµ ) = 8gµν gλσ − 8gµλ gνσ + 2gµσ Tr (γν γλ ) − Tr (γν γλ γσ γµ ).

125

But the last term is equal to −Tr (γµ γν γλ γσ ) by trace theorem 3, so 2Tr (γµ γν γλ γσ ) = 8( gµν gλσ − gµλ gνσ + gµσ gµλ ), or Tr (γµ γν γλ γσ ) = 4( gµν gλσ − gµλ gνσ + gµσ gµλ ) X (b) Tr (ab) = aµ bν Tr (γµ γν ) = 4gµν aµ bν = 4a · b. X Tr (abc d) = aµ bν cλ dσ Tr (γµ γν γλ γσ )

= 4aµ bν cλ dσ ( gµν gλσ − gµλ gνσ + gµσ gνλ ) = 4 [( a · b)(c · d) − ( a · c)(b · d) + ( a · d)(b · c)] X

Problem 7.34

(a) Theorem 14 Tr(γ5 ) = Tr

0 1 1 0

= 0. X

Theorem 15 Tr(γ5 γµ γν ) is antisymmetric in µ ↔ ν. Proof: from the fundamental anticommutator, h i Tr(γ5 {γµ , γν }) = Tr γ5 (2gµν ) = 2gµν Tr(γ5 ) = 0

= Tr(γ5 γµ γν ) + Tr(γ5 γν γµ ) so Tr(γ5 γµ γν ) = −Tr(γ5 γν γµ ). But there is no general antisymmetric second-rank tensor (there’s gµν , but it’s symmetric), so Tr(γ5 γµ γν ) = 0. X [In case you don’t like that argument, note that if µ = ν, then γµ γν = ±1, and we are left with Tr(γ5 ), which is 0. If µ 6= ν, then (after appropriate anticommutations and again reducing the square of any gamma matrix to ±1) Tr(γ5 γµ γν ) = Tr(iγ0 γ1 γ2 γ3 γµ γν ) reduces to ±i times the trace of the product of two distinct gamma’s, which is zero by Theorem 12.] Theorem 16 By the same argument, Tr(γ5 γµ γν γλ γσ ) is completely antisymmetric under interchange of any two indices, but there does exist a general antisymmetric fourth-rank tensor, namely the Levi-Civita symbol, eµνλσ (Eq. 7.127). All we have to do is calculate one permutation of the indices, and the others are all determined by invoking antisymmetry: Tr(γ5 γ0 γ1 γ2 γ3 ) = Tr[γ5 (−iγ5 )] = −iTr(1) = −4i Since e0123 ≡ −1, it follows that Tr(γ5 γµ γν γλ γσ ) = 4ieµνλσ . X

126


(b) Theorem 15’ Tr(γ5 ab) = aµ bν Tr(γ5 γµ γν ) = 0. X Theorem 16’ Tr(γ5 abc d) = aµ bν cλ dσ Tr(γ5 γµ γν γλ γσ ) = 4iaµ bν cλ dσ eµνλσ . X

Problem 7.35

(a) Since eµνλσ is zero unless the four indices are all different, eµνλσ eµνλτ is zero unless σ = τ: eµνλσ eµνλτ = Aδτσ , and it remains only to determine the number A, by working out a particular case. Pick σ = τ = 3: eµνλ3 eµνλ3 = Aδ33 = A. Here µνλ must be some permutation of 012, and there are six such permutations (all equal), so eµνλ3 eµνλ3 = 6e0123 e0123 = −6. Evidently A = −6, and hence eµνλσ eµνλτ = −6δτσ . X (b) This time λ could be θ or τ (and σ the other), so eµνλσ eµνθτ = A δθλ δτσ − δτλ δθσ (it’s obviously antisymmetric in τ ↔ θ). To determine A, pick (for example) λ = θ = 2, σ = τ = 3: eµν23 eµν23 = A δ22 δ33 − δ32 δ23 = A = e0123 e0123 + e1023 e1023 = −2. So

eµνλσ eµνθτ = −2 δθλ δτσ − δτλ δθσ . X

(c) This time there are six terms: eµνλσ eµφθτ = A δφν δθλ δτσ − δφν δτλ δθσ + δτν δφλ δθσ − δτν δθλ δφσ + δθν δτλ δφσ − δθν δφλ δτσ . In particular, eµ123 eµ123 = A = e0123 e0123 = −1, so eµνλσ eµφθτ = − δφν δθλ δτσ − δφν δτλ δθσ + δτν δφλ δθσ − δτν δθλ δφσ + δθν δτλ δφσ − δθν δφλ δτσ .

127

(d) There are 24 terms: eµνλσ eωφθτ µ µ µ µ µ µ = − δω δφν δθλ δτσ − δω δφν δτλ δθσ + δω δτν δφλ δθσ − δω δτν δθλ δφσ + δω δθν δτλ δφσ − δω δθν δφλ δτσ µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

ν λ σ ν λ σ λ σ σ σ λ σ − δφ δω δθ δτ + δφ δω δτ δθ − δφ δτν δω δθ + δφ δτν δθλ δω − δφ δθν δτλ δω + δφ δθν δω δτ ν λ σ ν λ σ λ σ σ σ λ σ + δθ δω δφ δτ − δθ δω δτ δφ + δθ δτν δω δφ − δθ δτν δφλ δω + δθ δφν δτλ δω − δθ δφν δω δτ ν λ σ ν λ σ λ σ σ σ λ σ − δτ δω δφ δθ + δτ δω δθ δφ − δτ δθν δω δφ + δτ δθν δφλ δω − δτ δφν δθλ δω + δτ δφν δω δθ

(I got the overall factor by picking µνλσ = ωφθτ = 0123.)

Problem 7.36

(a) h i Tr γµ γν (1 − γ5 )γλ (1 + γ5 )γλ [but {γ5 , γλ } = 0] i h [but (1 + γ5 )2 = 2(1 + γ5 )] = Tr γµ γν γλ (1 + γ5 )(1 + γ5 )γλ i h i h = 2Tr γµ γν γλ (1 + γ5 )γλ = 2Tr γλ γµ γν γλ (1 + γ5 )

= 2(4gµν )Tr(1 + γ5 ) = 32gµν .

(I used 8, 11, and 14 on p. 253.) (b) Dropping traces of odd products of gamma matrices: Tr [( p + mc)(q + Mc)( p + mc)(q + Mc)] i h = Tr ( pq + mcq + Mc p + mMc2 )( pq + mcq + Mc p + mMc2 )

= Tr( pq pq) + 2mMc2 Tr( pq) + 2mMc2 Tr(q p) + m2 c2 Tr(qq) + M2 c2 Tr( p p) + m2 M2 c4 Tr(1)

= 4[( p · q)( p · q) − p2 q2 + ( p · q)( p · q)] + 4mMc2 [4( p · q)] + m2 c2 (4q2 ) + M2 c2 (4p2 ) + m2 M2 c4 (4) = 8( p · q)2 − 4m2 M2 c4 + 16mMc2 ( p · q) + 8m2 M2 c4 + 4m2 M2 c4 h i2 = 8[( p · q)2 + 2mMc2 ( p · q) + m2 M2 c4 ] = 8 ( p · q) + mMc2 .

128


Problem 7.37

|M|2 = |M1 |2 + |M2 |2 + M1∗ M2 + M1 M2∗ , h|M|2 i = h|M1 |2 i + h|M2 |2 i + hM1 M2∗ i + hM1 M2∗ i∗ . Here

M1 = −

ge2 [u¯ (3)γµ u(1)] u¯ (4)γµ u(2) ; 2 ( p1 − p3 )

M2 = +

ge2 [u¯ (4)γµ u(1)] u¯ (3)γµ u(2) . 2 ( p1 − p4 )

From Eq. 7.129, with m = M = 0:

h|M1 |2 i =

8ge4 [( p1 · p2 )( p3 · p4 ) + ( p1 · p4 )( p2 · p3 )] . ( p1 − p3 )4

And, switching 3 ↔ 4,

h|M2 |2 i =

hM1 M2∗ i

8ge4 [( p1 · p2 )( p3 · p4 ) + ( p1 · p3 )( p2 · p4 )] . ( p1 − p4 )4

1 − ge4 = 4 ( p1 − p3 )2 ( p1 − p4 )2 × ∑ [u¯ (3)γµ u(1)] u¯ (4)γµ u(2) [u¯ (4)γν u(1)]∗ [u¯ (3)γν u(2)]∗ | {z } | {z } spins [u¯ (1)γν u(4)]

[u¯ (2)γν u(3)]

{z

|

}

F

where

F = ∑ u¯ (3)γ

∑ u(1)u¯ (1)

µ

s3

! γ

ν

∑ u(4)u¯ (4)

! γµ

s4

s1

∑ u(2)u¯ (2)

( p1 +mc)

Q

= ∑ Qij

∑ u(3)u¯ (3) s3

i,j

|

! ji

{z

( p3 +mc) ji

= ∑( Q p3 )ii = Tr ( Q p3 )

}

= Tr (γ p1 γ p4 γµ p2 γν p3 ) µ

ν

i

γν u ( 3 )

s2

{z } | {z } | {z ( p4 +mc) ( p2 +mc) = ∑ u¯ (3) (γµ p1 γν p4 γµ p2 γν ) u(3) = ∑ ∑ u¯ (3)i Qij u(3) j | {z } s3 i,j s3 |

!

}

129

hM1 M2∗ i

− ge4 Tr (γµ p1 γν p4 γµ p2 γν p3 ). = 4( p1 − p3 )2 ( p1 − p4 )2

But γµ p1 γν p4 γµ = −2 p4 γν p1 (theorem 9 on p. 253), so Tr (γµ p1 γν p4 γµ p2 γν p3 ) = −2Tr ( p4 γν p1 p2 γν p3 ),

[and γν p1 p2 γν = 4p1 · p2 , (theorem 8) ] = −8( p1 · p2 )Tr ( p4 p3 ) = −8( p1 · p2 )(4)( p3 · p4 ) (theorem 12’). Thus

hM1 M2∗ i =

8ge4 ( p · p2 )( p3 · p4 ) = hM1 M2∗ i∗ ( p1 − p3 )2 ( p1 − p4 )2 1

(since it is clearly real). Now,

( p1 − p3 )2 = p21 + p23 − 2p1 · p3 = (mc)2 + (mc)2 − 2p1 · p3 = −2p1 · p3 ( p1 − p4 )2 = p21 + p24 − 2p1 · p4 = −2p1 · p4 . So

h|M|2 i =

2ge4 [( p1 · p2 )( p3 · p4 ) + ( p1 · p4 )( p2 · p3 )] ( p1 · p3 )2

+

2ge4 [( p1 · p2 )( p3 · p4 ) + ( p1 · p3 )( p2 · p4 )] ( p1 · p4 )2

+2

2ge4 ( p · p )( p · p ). ( p1 · p3 )( p1 · p4 ) 1 2 3 4

But

( p1 + p2 ) = ( p3 + p4 ) =⇒ p21 + p22 + 2p1 · p2 = p23 + p24 + 2p3 · p4 =⇒ p1 · p2 = p3 · p4

(since p2i = m2 c2 = 0).

Likewise p1 − p3 = p4 − p2

=⇒ p21 + p23 − 2p1 · p3 = p24 + p22 − 2p2 · p4 =⇒ p1 · p3 = p2 · p4 ; and p1 − p4 = p3 − p2 =⇒ p1 · p4 = p2 · p3 .

130


So 2ge4 h|M| i = ( p1 · p3 )2 ( p1 · p4 )2 2

(

( p1 · p4 )4 + ( p1 · p3 )4 )

+ ( p1 · p4 )2 ( p1 · p2 )2 + ( p1 · p3 )2 ( p1 · p2 )2 + 2( p1 · p3 )( p1 · p4 )( p1 · p2 )2 | {z }

where 



  2 2  = ( p1 · p2 )2  (| p1 · p4 ) + ( p1 · p3 ){z+ 2( p1 · p3 )( p1 · p4}) ( p1 · p4 + p1 · p3 )2

where, in turn,

( p1 · p4 + p1 · p3 )2 = [ p1 · ( p3 + p4 )]2 = [ p1 · ( p1 + p2 )]2 h i2 h i2 = p21 + ( p1 · p2 ) = mc2 + ( p1 · p2 ) = ( p1 · p2 )2 i h 2ge4 4 4 4 ( p · p ) + ( p · p ) + ( p · p ) . 2 3 1 1 1 4 ( p1 · p3 )2 ( p1 · p4 )2

h|M|2 i =

Problem 7.38

(a)

h|M|2 i =

8ge4 [( p1 · p2 )( p3 · p4 ) + ( p1 · p4 )( p2 · p3 )] ( p1 − p3 )4

( p1 − p3 )2 = p21 + p23 − 2p1 · p3 = −2p1 · p3 h|M|2 i =

e

2ge4 [( p1 · p2 )( p3 · p4 ) + ( p1 · p4 )( p2 · p3 )] ( p1 · p3 )2

p

p2

1

Before

p

q

p4

m m

e

3

After

131

E E , pi , p 2 = , − pi , c c 2 E E p4 = , −p f , = p2i = p2f , c c2

p3 =

p1 =

E ,p c f

,

E2 cos θ c2

pi · p f =

E2 E2 + p2i = 2 2 = p3 · p4 ; 2 c c E2 E2 E2 θ p1 · p4 = 2 + pi · p f = 2 (1 + cos θ ) = 2 2 cos2 = p2 · p3 ; 2 c c c E2 E2 E2 2 θ p1 · p3 = 2 − pi · p f = 2 (1 − cos θ ) = 2 2 sin . 2 c c c p1 · p2 =

2ge4 h|M| i = 4( E/c)4 sin4 θ/2 2

E4 θ E4 4 4 + 4 4 cos4 2 c c

2ge4

=

1 + cos4 θ/2 . sin4 θ/2

(b) Put result of part (a) into Eq. 6.47, with S = 1, |pi | = |p f |, E1 + E2 = 2E: dσ = dΩ

h¯ c 8π

2

ge4 2E2

1 + cos4 θ/2 sin4 θ/2

.

Problem 7.39

e

p

p2

1

p e

Before

e

3

q

p4 After

e

(a)

h|M|2 i =

h i 2ge4 ( p1 · p2 )4 + ( p1 · p3 )4 + ( p1 · p4 )4 . 2 2 ( p1 · p3 ) ( p1 · p4 )

E E p1 = , p , p2 = , − pi , c i c E E2 , −p f , p4 = = p2i = p2f , c c2

p3 =

E ,p c f

pi · p f =

,

E2 cos θ c2

132


E2 E2 + p2i = 2 2 ; 2 c c E2 E2 p1 · p3 = 2 − pi · p f = 2 (1 − cos θ ); c c E2 E2 p1 · p4 = 2 + pi · p f = 2 (1 + cos θ ). c c p1 · p2 =

2ge4 ( E/c)4 (1 − cos θ )2 ( E/c)4 (1 + cos θ )2 " 4 2 4 # 4 2 E E2 E (1 − cos θ ) + (1 + cos θ ) × 2 2 + c c2 c2 h i 2ge4 4 4 = 16 + ( 1 − cos θ ) + ( 1 + cos θ ) (1 − cos2 θ )2 2ge4 16 + 1 − 4 cos θ + 6 cos2 θ − 4 cos3 θ + cos4 θ = sin4 θ + 1 + 4 cos θ + 6 cos2 θ + 4 cos3 θ + cos4 θ

h|M|2 i =

2ge4 2 4 18 + 12 cos θ + 2 cos θ sin4 θ 2 4ge4 4ge4 9 + 6 cos2 θ + cos4 θ = 3 + cos2 θ = 4 4 sin θ sin θ 2 2 2ge2 4 2 2 = . (4 − sin θ ) = 2ge 1 − sin2 θ sin2 θ

=

(b) Put this into Eq. 6.47, with S = 12 , |pi | = |p f |, and E1 + E2 = 2E: dσ = dΩ

h¯ c 8π

2

2 2 1 1 h¯ cge2 4 4 2 = 2ge 1 − 1− . 2 8πE 2(2E)2 sin2 θ sin2 θ

This is different from the result for e + µ → e + µ (Problem 7.38). As explained in the footnote on page 247, the non-relativistic results for e + e → e + e and e + µ → e + µ should be the same, but for the extreme relativistic regime this is no longer the case.

Problem 7.40

Restoring the complex conjugation, Eq. 7.158 reads √ Msinglet = −2 2ige2 (e3∗ × e4∗ )z .

133

|Msinglet |2 = 8ge4 e3∗ × e4∗

z

e3 × e4

h|Msinglet |2 i = 8ge4 e3ij e3kl ∑ e3k e3 i∗ s3

=

8ge4 e3ij e3kl

z

= 8ge4 e3ij e3 i∗ e4 ∗j e3kl e3k e4l .

∑ s4

δki − pˆ 3 k pˆ 3 i δlj − pˆ 4 l pˆ 4 j

e4l e4 ∗j

h i = 8ge4 e3ij e3ij − e3ij e3il pˆ 4 l pˆ 4 j − e3ij e3kj pˆ 3 k pˆ 3 i + e3ij pˆ 3 i pˆ 4 j e3kl pˆ 3 k pˆ 4 l h = 8ge4 (e312 e312 + e321 e321 ) − δ33 δjl − δ3l δ3j pˆ 4 l pˆ 4 j i − δ33 δik − δ3k δ3i pˆ 3 k pˆ 3 i + ( pˆ 3 × pˆ 4 )z ( pˆ 3 × pˆ 4 )z i h ˆ 3 + pˆ 3 z pˆ 3 z + ( pˆ 3 × pˆ 4 )z ( pˆ 3 × pˆ 4 )z . ˆ 4 + pˆ 4 z pˆ 4 z − pˆ 3 · p = 8ge4 2 − pˆ 4 · p From Eq. 7.136: pˆ 4 z = −1,

pˆ 3 z = 1,

pˆ 3 × pˆ 4 = 0,

so h|Msinglet |2 i = 16ge4 , consistent with Eq. 7.163. X This method works because, having already put the electron/positron pair in the singlet configuration, any contribution from the photon triplet is automatically zero, and hence it doesn’t matter whether you include it or not.

Problem 7.41

Problem 7.42 p2

p4

q1 q

q

4

q

2

p1

p

3

3

134


Applying the Feynman rules, Z

−igµλ −igκν [LOOP] [u¯ (4) (ige γν ) u(2)] q22 q21

[u¯ (3) (ige γµ ) u(1)]

× (2π )4 δ4 ( p1 − p3 − q2 )(2π )4 δ4 (q2 − q3 − q4 )(2π )4 δ4 (q3 + q4 − q1 ) × (2π )4 δ4 (q1 + p2 − p4 )

d4 q1 d4 q2 d4 q3 d4 q4 (2π )4 (2π )4 (2π )4 (2π )4

where “LOOP“ stands for "

i (q + mc) i (q + mc) −Tr (ige γ ) 24 2 2 (ige γκ ) 23 2 2 q4 − m c q3 − m c

#

λ

The q2 integral, using δ4 ( p1 − p3 − q2 ), sends q2 → p1 − p3 ≡ q (for short). The q1 integral, using δ4 (q1 + p2 − p4 ), sends q1 → p4 − p2 , and the two remaining delta functions ⇒ q3 + q4 = p1 − p3 = p4 − p2 , so q1 is also q (of course). The q3 integral, using δ4 (q − q3 − q4 ), sends q3 → q − q4 , and we erase the final delta function (2π )2 δ4 ( p1 − p3 + p2 − p4 ). The is still an integral over q4 , which (for simplicity) we rename k. Multiplying by i: M = ( ) ige4 d4 k Tr γµ (k + mc)γν (q − k + mc) µ − 4 [u¯ (3)γ u(1)] [u¯ (4)γν u(2)] X q (2π )4 [k2 − m2 c2 ][(q − k)2 − m2 c2 ]

Problem 7.43

Problem 7.44

e

p

p2

1

p

Before p1 =

m

E ,p , c

q

p4

m

p3 =

q ≡ p1 − p3 = 0, p − p0 ,

E 0 ,p , c

|p | = |p0 |,

e

3

After p · p0 = p2 cos θ,

2 q2 = −(p − p0 )2 = − p2 + p0 − 2p · p0

−2p2 (1 − cos θ ) = −4p2 sin2 θ/2. X

135

Problem 7.45

The fractional correction to the fine structure constant is ∆α α (0) = f ( x ), α 3π

where

x=

4p2 sin2 θ/2 . m2 c2

For a head-on collision θ = 180◦ . At v = (0.1)c this is nonrelativistic, so |p| = mv: 4m2 v2 x= = 4(0.1)2 = 0.04 1, m2 c2 so f ( x ) ≈ x/5 = 0.008 (Eq. 7.185). ∆α 0.008 = = 6.2 × 10−6 . α 3π (137) At E = 57.8 GeV, by contrast, we have a highly relativistic electron, so |p| ≈ E/c, and 4E2 4(57.8 × 103 )2 x= = = 5.12 × 1010 1, (mc2 )2 (0.511)2 so (Eq. 7.185) f ( x ) ≈ ln x = ln 5.12 × 1010 = 24.7. ∆α 24.7 = = 0.0191, α 3π (137)

α ( q2 ) =

1 1 [1 + 0.0191] = 7.44 × 10−3 = 137 134

Problem 7.46

Problem 7.47

[See Sakurai, Advanced Quantum Mechanics, pp. 228-229.]

136


Problem 7.48

g

p1

p2

e

p

e

3

(a) (( 4 4 (((( ( (2π ) δ ( p − p − p3 ) [u¯ (3)(ige )v(2)] ( ( 2 1 (

M = − ge [u¯ (3)v(2)] h|M|2 i = ge2 Tr [( p2 − m2 c)( p3 + m3 c)] h i = ge2 Tr ( p2 p3 ) − (me c)2 Tr (1) h i h i = ge2 4p2 · p3 − 4(me c)2 = 4ge2 p2 · p3 − (me c)2 . Now p1 = p2 + p3 =⇒ p21 = p22 + p23 + 2p2 · p3

=⇒ (mγ c)2 = 2(me c)2 + 2p2 · p3 1 =⇒ p2 · p3 = (mγ c)2 − (me c)2 . 2 2 2 1 2 2 So h|M| i = 4ge (mγ c) − 2(me c) = 2ge2 c2 (m2γ − 4m2e ) . 2 From Eq. 6.35, with S = 1, we have (in the CM): Γ=

| p3 | 2g2 c2 (m2γ − 4m2e ). 8π¯hm2γ c e

Conservation of energy ⇒ mγ c2 = E2 + E3 = 2E3 , so E32 =

1 cq 2 1 2 4 mγ − 4m2e mγ c = m2e c4 + p23 c2 ⇒ p23 = (m2γ − 4m2e )c2 , |p3 | = 4 4 2 Γ=

2ge2 c2 (m2γ − 4m2e ) c q ge2 c2 2 − 4m2 = m (m2 − 4m2e )3/2 . γ e 2 8π¯hm2γ γ 8π¯hm2γ c

137

(b) τ=

8π¯hm2γ c4 2¯h(mγ c2 )2 1 = = 3/2 3/2 Γ ge2 [(mγ c2 )2 − 4(me c2 )2 ] α ( m γ c2 )2 − 4( m e c2 )2 |{z} | {z } 4πα

=

negligible

2(6.58 × 10−22 )

c2 )2

2¯h(mγ 2¯h = 6 × 10−22 seconds . = = 2 3 (1/137) (300) α(mγ c ) α ( m γ c2 )

Problem 7.49

p2

m

p4

m

q e

p1

p

3

e

(a) Z

[u¯ (3)(ige )u(1)]

q2

−i [u¯ (4)(ige )u(2)] − ( m γ c )2

×(2π )4 δ4 ( p1 − p3 − q)(2π )4 δ4 ( p2 − p4 + q) M=

d4 q (2π )4

− ge2 [u¯ (3)u(1)] [u¯ (4)u(2)] . ( p1 − p3 )2 − ( m γ c )2

(b)

h|M|2 i =

1 ge4 4 [( p1 − p3 )2 − (mγ c)2 ]2

× Tr [( p1 + m1 c)( p3 + m3 c)] Tr [( p2 + m2 c)( p4 + m4 c)] | {z }| {z } F

where F = Tr ( p1 p3 ) + (me c)2 Tr (1) = 4p1 · p3 + 4(me c)2

138


and = Tr ( p2 p4 ) + (mµ c)2 Tr (1) = 4( p2 · p4 ) + 4(mµ c)2 .

h|M|2 i =

1 (2ge )4 ( p1 · p3 ) + ( m e c )2 ( p2 · p4 ) + ( m µ c )2 . 4 [( p1 − p3 )2 − (mγ c)2 ]2

e

p

p2

1

p

q

p4

m

Before

m

e

3

After

(c)

p1 =

E ,p , c 1

p21 = p23 =

p2 = E2 , c2

E , −p1 , c

p3 =

E , p3 , c

p4 =

E , −p3 . c

p1 − p3 = (0, p1 − p3 )

( p1 − p3 )2 = −(p1 − p3 )2 = −p21 − p23 + 2p1 · p3 = −2

E2 E2 E2 + 2 2 cos θ = −2 2 (1 − cos θ ). 2 c c c

E2 E2 − p · p = (1 − cos θ ) = p2 · p4 . 3 1 c2 c2 2 2 E 1 (2ge )4 h|M|2 i = ( 1 − cos θ ) . 4 [−2( E/c)2 (1 − cos θ ) − (mγ c)2 ]2 c2 p1 · p3 =

From Eq. 6.47, dσ = dΩ

h¯ c 8π

2

h|M|2 i = 4E2

h¯ cge2 8πE

2

E2 (1 − cos θ ) 2E2 (1 − cos θ ) + (mγ c2 )2

(d) If E mγ c2 ,

2 dσ 1 h¯ cge2 E = (1 − cos θ )2 dΩ 4 4π (mγ c2 )2

2 .

139

dσ sin θ dθ dφ dΩ 2 Z π h¯ cge2 E 2π (1 − 2 cos θ + cos2 θ ) sin θ dθ = 4 4π (mγ c2 )2 0 | {z }

σ=

Z

#

where #=

π 2 cos3 θ 8 − cos θ + cos θ − = 2+0+ 3 = 3 3 0 2

1 σ= 3π

h¯ ge2 E 2m2γ c3

!3 .

(e) From Problem 6.8, dσ = dΩ

h¯ 8πmµ c

2

h|M|2 i.

From Example 7.7, θ ( m γ c )2 ; 2 θ ( p1 · p3 ) = m2e c2 + 2p2 sin2 ∼ = ( m e c )2 ; 2 ( p2 · p4 ) = ( m µ c )2

( p1 − p3 )2 = −4p2 sin2

ih i 4 h ∼ 1 (2ge ) 2(me c)2 2(mµ c)2 = h|M| i = 4 4 (mγ c) 2

dσ = dΩ

h¯ 8πmµ c

2

me mµ 4ge2 m2γ

!2

=

me mµ 4ge2 m2γ

h¯ ge2 me 2πcm2γ

!2

!2 .

Unlike the Rutherford formula, this one is independent of E and of θ. dσ 1 = σ = 4π dΩ π

h¯ ge2 me cm2γ

!2 .

Alternatively, setting mγ = 0, and treating the muon as heavy and recoilless,

140


e

E, p

E, p1

m

p1 =

q

m

Before

e

3

After

E E 0 , p , p3 = , p , p2 = p4 = (mµ c, 0); ( p1 − p3 ) = (0, p − p0 ), c c 2

( p1 − p3 )2 = −(p − p0 )2 = −(p2 + p0 − 2p · p0 ) = −2p2 (1 − cos θ ) = −4p2 sin2 (θ/2); p1 · p3 =

E2 E2 0 − p · p = − p2 cos θ = m2e c2 + p2 − p2 cos θ ≈ m2e c2 , c2 c2

(since |p| me c), and p2 · p4 = m2µ c2 . Inserting this into the result of part (b): 1 (2ge )4 2 2 2 2 2 2 2 2 h|M| i = m c + m c m c + m c = e e µ µ 4 (4p2 sin2 θ/2)2 2

dσ = dΩ

h¯ 8πmµ c

2

2

h|M| i =

h¯ ge2 me mµ c2

ge2 me mµ c2

!2

p2 sin2 θ/2

!2

8πmµ cm2e v2 sin2 θ/2 2 2 2 h¯ c4πe2 /¯hc e2 h¯ cge2 = = = , 8πme v2 sin2 θ/2 8πme v2 sin2 θ/2 2me v2 sin2 θ/2

which is precisely the Rutherford formula (Eq. 7.132).

141

Problem 7.50

p2

p2

p4

p4

q

q p

p1

p1

3

p

3

(a) Z

v¯ (2)(ige )

i (q + me c) q2 − ( m e c )2

ige u(1)

× (2π )4 δ4 ( p1 − p3 − q)(2π )4 δ4 ( p2 + q − p4 ) M (1) =

d4 q (2π )4

ge2 [v¯ (2)( p1 − p3 + me c)u(1)] . [( p1 − p3 )2 − (me c)2 ]

To get M(2) , switch 3 ↔ 4; the total, then, is

M=

ge2 v¯ (2)

p1 − p4 + me c p1 − p3 + me c + [( p1 − p3 )2 − (me c)2 ] [( p1 − p4 )2 − (me c)2 ]

u (1) .

(b) Set me = mγ = 0. Note that

( p1 − p3 )2 = p21 + p23 − 2p1 · p3 = m2e c2 + m2γ c2 − 2p1 · p3 = −2p1 · p3 , and ( p1 − p4 )2 = −2p1 · p4 . Also note that p1 u(1) = me cu(1) = 0.

M=

ge2 p3 + p4 v¯ (2) u (1). 2 p1 · p3 p1 · p4

142


2 p3 1 ge2 p4 p3 + p4 p p2 Tr + 4 2 p1 · p3 p1 · p4 1 p1 · p3 p1 · p4 g 4 1 1 e = Tr ( p3 p1 p3 p2 ) + Tr ( p4 p1 p4 p2 ) 2 2 ( p1 · p3 ) ( p1 · p4 )2 1 + [Tr ( p3 p1 p4 p2 ) + Tr ( p4 p1 p3 p2 )] ( p1 · p3 )( p1 · p4 ) " g 4 1 e = 4 p3 · p p3 · p1 p3 · p2 − 3 p1 · p2 + p3 · p2 p1 · p3 2 2 (p · p )

h|M|2 i =

1

3

1 p4 · p p · p p · p2 − 4 p1 · p2 + p4 · p2 p1 · p4 ( p1 · p4 )2 4 1 4 1 + p · p p · p − p3 · p4 p1 · p2 ( p1 · p3 )( p1 · p4 ) 3 1 4 2

+

+ p3 · p2 p1 · p4 + p4 · p1 p3 · p2 − p4 · p3 p1 · p2 + p4 · p2 p1 · p3

#

2 " p2 · p3 p2 · p4 p2 · p4 p2 · p3 = 2 +2 +2 +2 p1 · p3 p1 · p4 p1 · p4 p1 · p3 # ( p · p )( p · p ) −2 1 2 3 4 ( p1 · p3 )( p1 · p4 ) ( p1 · p2 )( p3 · p4 ) p2 · p4 4 p2 · p3 − + . = ge p1 · p3 p1 · p4 2( p1 · p3 )( p1 · p4 )

ge2 2

(c) In the CM frame, p1 · p3 = p2 · p4 =

E2 (1 − cos θ ); c2

p1 · p2 = p3 · p4 = 2

E2 . c2

p1 · p4 = p2 · p3 =

E2 (1 + cos θ ); c2

So h|M|2 i is

h i ge4 ( p2 · p3 )( p1 · p4 ) + ( p2 · p4 )( p1 · p3 ) − 12 ( p1 · p2 )( p3 · p4 ) ( p1 · p3 )( p1 · p4 ) 4 h i ge4 E 2 2 = ( 1 + cos θ ) + ( 1 − cos θ ) − 2 ( E/c)4 (1 − cos θ )(1 + cos θ ) c ge4 cos2 θ = 1 + 2 cos θ + cos2 θ + 1 − 2 cos θ + cos2 θ − 2 = 2ge4 . 2 (1 − cos θ ) sin2 θ

h|M|2 i = 2ge4 cot2 θ .

143

(d) Using Eq. 6.47, with S = 1/2, dσ = dΩ

h¯ c 8π

2 2 4 1 2ge cot2 θ h¯ cge2 cot θ . = 16πE 2 (2E)2

No , σ is not finite. The θ integral is Z π 0

cot2 θ sin θ dθ =

Z π cos2 θ 0

sin θ

dθ;

near θ = 0, cos θ ≈ 1 while sin θ ≈ θ, so the integral goes like 1 dθ ∼ ln θ → ∞. θ 0

Z 0

Problem 7.51 ∗

(a) Assume ψ = iγ2 ψ∗ ; we want to prove that ψ0 = iγ2 ψ0 . Multiply the first (on the left) by S, and note that S is real, and commutes with γ2 (Eq. 7.53): ∗

ψ0 = Sψ = iSγ2 ψ∗ = iγ2 Sψ∗ = iγ2 (Sψ)∗ = iγ2 ψ0 . X (b) ψ=

χA χB

0 σ2 = iγ ψ = i −σ2 0 2 ∗

χ∗A χ∗B

=

iσ2 χ∗B , −iσ2 χ∗A

so χ A = iσ2 χ∗B

and

χ B = −iσ2 χ∗A . X

[These two equations are redundant. If you take the complex conjugate of the second, and multiply by σ2 , and note that σ2 is imaginary and (σ2 )2 = 1, σ2 χ∗B = iσ2 σ2∗ χ A = −i (σ2 )2 χ A = −iχ A , so χ A = iσ2 χ∗B .] Recall that 0 0

a = a γ − a · γ = a

0

0 1 0 0 σ a −a · σ . −a· = 0 −1 −σ 0 a · σ − a0

For Majorana particles, the Dirac equation (i¯h∂ψ − mc ψ = 0) becomes, i¯h

∂0 − σ · ∇ σ · ∇ − ∂0

χ −iσ2 χ∗

− mc

χ −iσ2 χ∗

= 0,

144


or (for the “upper“ components) i¯h [∂0 χ + i (σ · ∇)σ2 χ∗ ] − mcχ = 0. X [For the “lower“ components we have i¯h [∂0 (iσ2 χ∗ ) + (σ · ∇)χ] + mc(iσ2 χ∗ ) = 0. Multiply by −i and then take the complex conjugate: i¯h [−σ2∗ ∂0 χ − i (σ ∗ · ∇)χ∗ ] + mcσ2∗ χ = 0. Use σ2∗ = −σ2 , and multiply by σ2 : h i i¯h (σ2 )2 ∂0 χ − iσ2 (σ ∗ · ∇)χ∗ − mc(σ2 )2 χ = 0. But (σ2 )2 = 1, so i¯h [∂0 χ − iσ2 (σ ∗ · ∇)χ∗ ] − mcχ = 0. Evidently the two equations are consistent provided that σ2 σi∗ = −σi σ2 . If i is 1 or 3, the Pauli matrix is real, and this is just the statement that σ1 and σ3 anticommutate with σ2 ; if i = 2 the matrices commute, but we pick up an extra minus sign from the fact that σ2 is imaginary. So it works.] (c) The most general linear combination of plane wave solutions to the Dirac equation is   c( p x −ipy ) ip· x/¯h cpz ip· x/¯h a1 e−ip· x/¯h + a3 E+ e − a e 4 E+mc2 mc2    a e−ip· x/¯h − a cpz eip· x/¯h − a c( px +ipy ) eip· x/¯h  3 E+mc2 4 E+mc2  2  ψ=   a1 cpz e−ip· x/¯h + a2 c( px −ipy ) e−ip· x/¯h − a4 eip· x/¯h  2  E+mc2  E+mc c( p x +ipy ) −ip· x/¯h cpz − ip · x/¯ h ip · x/¯ h − a2 E+mc2 e + a3 e a1 E+mc2 e For a Majorana particle,    α α   β   β ∗  =  ∗  =  0 −i α  − β  −i α∗ i 0 β∗ 

ψ=

χA −iσ2 χ∗A

So the fourth row is the complex conjugate of the first row (and the third is minus the conjugate of the second, but this leads to exactly the same constraint): a1

c( p x + ipy ) −ip· x/¯h cpz e − a2 e−ip· x/¯h + a3 eip· x/¯h E + mc2 E + mc2 ∗ c( p x − ipy ) ip· x/¯h cpz ip· x/¯h e − a e = a1 e−ip· x/¯h + a3 4 E + mc2 E + mc2 c( p x + ipy ) −ip· x/¯h cpz = a1∗ eip· x/¯h + a3∗ e − a4∗ e−ip· x/¯h , E + mc2 E + mc2

145

from which it follows that a3 = a1∗ and a4 = a2∗ . Picking a1 = 1, a2 = 0 for χ(1) , and a1 = 0, a2 = 1 for χ(2) , and adopting a convenient normalization: χ

(1)

χ (2)

( E + mc2 )e−ip· x/¯h + c( p x − ipy )eip· x/¯h = , −cpz eip· x/¯h −cpz eip· x/¯h = . ( E + mc2 )e−ip· x/¯h − c( p x + ipy )eip· x/¯h

The general solution (for a given E and p) is an arbitrary linear combination of these.

147

8

Electrodynamics and Chromodynamics of Quarks Problem 8.1

(a) Referring to the figure on page 277, Z

[v¯ (2)(ige γµ )u(1)]

−igµν [u¯ (3)(−ige Qγν )v(4)] q2

× (2π )4 δ4 ( p1 + p2 − q)(2π )4 δ4 (q − p3 − p4 ) = −i

d4 q (2π )4

Qge2 [v¯ (2)γµ u(1)] u¯ (3)γµ )v(4) . 2 ( p1 + p2 )

M=

Qge2 [v¯ (2)γµ u(1)] u¯ (3)γµ )v(4) . ( p1 + p2 )2

X

(b) Using Casimir’s trick, 2 1 Qge2 h|M| i = ∑ ( p1 + p2 )2 4 spins × [v¯ (2)γµ u(1)] u¯ (3)γµ v(4) [v¯ (4)γν u(3)] [u¯ (1)γν v(2)] 2 1 Qge2 = ∑ ( p1 + p2 )2 4 spins × Tr [γµ ( p1 + mc)γν ( p2 − mc)] Tr γµ ( p4 − Mc)γν ( p3 + Mc) . X 2

(c) Dropping traces of odd products of gamma matrices, Tr [γµ ( p1 + mc)γν ( p2 − mc)] = Tr (γµ p1 γν p2 ) − (mc)2 Tr (γµ γν ) = p1 κ p2 λ Tr γµ γκ γν γλ − (mc)2 Tr (γµ γν ) = p1 κ p2 λ 4 gµκ gνλ − gµν gκλ + gµλ gνκ − (mc)2 4gµν i h µ µ = 4 p1 p2ν − gµν ( p1 · p2 ) + p1ν p2 − gµν (mc)2 .

148

8 Electrodynamics and Chromodynamics of Quarks

Likewise, Tr γµ ( p4 − Mc)γν ( p3 + Mc) h io n = 4 p4 µ p3 ν + p4 ν p3 µ − gµν ( p3 · p4 ) + ( Mc)2 . So 2 n h io 1 Qge2 µ ν ν µ µν 2 h|M| i = 4 p p + p p − g ( p · p ) + ( mc ) 2 2 2 1 1 1 4 ( p1 + p2 )2 h io n × 4 p3 µ p4 ν + p4 µ p3 ν − gµν ( p3 · p4 ) + ( Mc)2 2 Qge2 =4 2( p1 · p3 )( p2 · p4 ) + 2( p1 · p4 )( p2 · p3 ) 2 ( p1 + p2 ) 2

−2 ( p1 · p2 )( p3 · p4 ) − 2( p1 · p2 )( Mc)2 − 2( p1 · p2 )( p3 · p4 ) −2 ( p3 · p4 )(mc)2 + 4( p1 · p2 )( p3 · p4 ) + 4( p1 · p2 )( Mc)2 + 4( p3 · p4 )(mc)2 + 4(mMc2 )2 2 Qge2 =8 ( p1 · p3 )( p2 · p4 ) + ( p1 · p4 )( p2 · p3 ) 2 ( p1 + p2 ) +( p1 · p2 )( Mc)2 + ( p3 · p4 )(mc)2 + 2(mMc2 )2 . X (d)

e

p

p

p2

1

e

p1 =

E ,p , c

( p1 + p2 )2 =

After

q p2 = 2E c

E , −p , c

2 ,

q

p4

Before

p1 · p2 =

p3 =

q

3

E 0 ,p , c

2 E + p2 , c

p4 =

E 0 , −p ; c

2 E 2 + p0 ; c 2 E p1 · p4 = p2 · p3 = + p · p0 . c p3 · p4 =

2 E − p · p0 , c 2 2 E E 2 2 02 p = − (mc) , p = − ( Mc)2 , p · p0 = |p||p0 | cos θ. c c 2 2 E E 2 p1 · p2 = 2 − (mc) , p3 · p4 = 2 − ( Mc)2 ; c c

p1 · p3 = p2 · p4 =

149

" # E 4 0 2 ( p1 · p3 )( p2 · p4 ) + ( p1 · p4 )( p2 · p3 ) = 2 + (p · p ) c ( " # " # ) E 4 E 2 E 2 2 2 2 = 2 + − (mc) − ( Mc) cos θ c c c

h|M|2 i ( " # " # ) 2 E 4 Qge2 E 2 E 2 2 2 2 2 =8 + − (mc) − ( Mc) cos θ c c c 4( E/c)2 " # " # ! E 2 E 2 2 2 2 2 2 2 + 2 − (mc) ( Mc) + 2 − ( Mc) (mc) + 2(mMc ) c c 2 ( 4 E Qge2 = + ( ME)2 + (mE)2 2 c ( E/c) # " # ) " 2 E E 2 − (mc)2 − ( Mc)2 cos2 θ + c c ( 2 2 2 mc Mc2 2 4 = Q ge 1 + + E E ) " 2 # 2 2 # " Mc2 mc 2 1− cos θ . X + 1− E E

Problem 8.2

From Eq. 6.47 and Problem 8.1: ( 2 2 2 2 dσ mc Mc2 h¯ c ( Qge2 )2 1+ + = dΩ 8π E E (2E)2 " # )√ 2 2 # " 2 Mc2 mc E2 − M 2 c4 1− 1− cos2 θ √ E E E2 − m2 c4 σ=

Z

dσ sin θ dθ dφ, dΩ

Z 2π 0

dφ = 2π,

Z π 0

sin θ dθ = 2,

Z π 0

cos2 θ sin θ dθ =

2 . 3

150


σ =

( " √ 2 2 2 # E2 − M 2 c4 mc Mc2 (2π ) √ 2 1+ + E E E2 − m2 c4 # ) " 2 2 # " 2 2 Mc2 mc + 1− 1− 3 E E 2 √ 2 2 2 h¯ cα E − M 2 c4 h mc2 Mc2 √ 3+3 +3 E E E E2 − m2 c4 2 2 2 i 2 2 mc Mc2 mc2 Mc2 +1− − + E E E E 2 √ 2 2 2 E − M 2 c4 h 1 mc2 h¯ cα 1 Mc2 √ 1+ + E 2 E 2 E E2 − m2 c4 2 2 2 i 1 mc Mc2 + 4 E E " 2 p 2 # " 2 # 1 Mc2 1 − ( Mc2 /E)2 1 mc2 h¯ cα p 1+ 1+ . X E 2 E 2 E 1 − (mc2 /E)2

h¯ cQ4πα 16πE

=

πQ2 12

=

πQ2 3

=

πQ2 3

2

Problem 8.3

There’s a second diagram in the elastic case, and this means that the kinematic factors do not cancel , as they do for the muons.

Problem 8.4

From Eq. 8.13, n h io qµ Lµν = 2 ( p1 · q) p3ν + ( p3 · q) p1ν + qν (mc)2 − ( p1 · p3 ) . q = p4 − p2 = p1 − p3 ⇒ p1 · q = p21 − p1 · p3 = (mc)2 − p1 · p3 = − p3 · q, so

h i qµ Lµν = 2 (mc)2 − p1 · p3 [ p3ν − p1ν + qν ] = 0.

In the “Breit“ frame (Problem 3.24), E E p1 = , p , p3 = , −p , c c

so

qµ = (0, 2p),

and if we choose the z axis to lie along p, then q has only a z component— that is, there exists an inertial frame in which qµ = (0, 0, 0, q). In this frame

151

qµ Lµν = qL3ν = 0, so L3ν = 0, and (since Lµν is symmetric) this means also that Lν3 = 0. Written out as a matrix, then,      · · · x · · · 0 · · · 0  · · · 0  · · · x   · · · 0 µν     Lµν =   · · · 0 , and hence L Kνµ =  · · · 0  · · · x  x x x x 0 0 0 0 0000 does not depend on the “x“ entries (since they are all multiplied by 0; we could just as well set all these entries to zero, in which case qµ K µν = 0. Moreover, since this is a covariant equation, it will then hold in any reference frame. (Note that K µν is a symmetric matrix.)

Problem 8.5

Using the first footnote on page 281, p · q = −q2 /2: i K4 2 ν K5 h K2 ν ν 2 ν ( p · q ) p + q q + ( p · q ) q + q p qµ K µν = −K1 qν + ( Mc)2 ( Mc)2 ( Mc)2 1 K5 2 K2 ν 2 2 2 ν = q −K1 ( Mc) + K4 q − q +q p − + K5 = 0. 2 2 ( Mc)2 K5 2 K2 p·q 2 2 2 −K1 ( Mc) + K4 q − q +q − + K5 = 2 2 ( Mc)2 q2 K4 2 K5 = K1 − q + q2 − K2 + 2K5 = 0. 2 2 ( Mc) 2( Mc)2 q2 2 2 q K4 = ( Mc) K1 − K2 + K5 2 + . 2( Mc)2

pν qµ K

µν

Use this to eliminate K4 in line 2: K5 2 q2 ν 2 2 q −K1 ( Mc) + ( Mc) K1 − K2 + K5 2 + − q 2 2( Mc)2 K + q2 pν − 2 + K5 = 0. 2

( Mc)2 qν (−K2 + 2K5 ) +

q2 p ν (−K2 + 2K5 ) = 0 ⇒ K2 = 2K5 . 2

Put this into the expression for q2 K4 : q2 K4 = ( Mc)2 K1 − 2K5 + K5 2 +

q2 2( Mc)2

= K1 ( Mc)2 + K5

X

q2 . 2

152


K4 = K1

( Mc)2 K5 + . X 2 q2

Problem 8.6

From Eq. 8.13, for a “Dirac” proton n h io µ µ K µν = 2 p2 p4ν + p2ν p4 + gµν ( Mc)2 − ( p2 · p4 ) . Use p4 = q + p2 , and drop the subscript on p2 : n h io K µν = 2 pµ (qν + pν ) + pν (qµ + pµ ) + gµν ( Mc)2 − p · (q + p) q2 2 µ ν ν µ µ ν µν 2 − ( Mc) = 2 p q + p q + 2p p + g ( Mc) + 2 (where I used p · q = −q2 /2 and p2 = ( Mc)2 ), or K µν = 2 ( pµ qν + pν qµ ) + 4pµ pν + gµν q2 . Comparing the generic expression (Eq. 8.18):

K

µν

qµ qν K2 1 µ 1 ν µ ν = K1 − g + 2 + p + q p + q 2 2 q ( Mc)2 K K2 K 2 2 µ ν ν µ µ ν µν µ ν K1 = (p q + p q ) +p p − g K1 + q q + , 2( Mc)2 ( Mc)2 q2 4( Mc)2

we read off

µν

K2 = 4( Mc)2

and

K1 = − q 2 .

Meanwhile, from Eq. 8.24, K1 = −q2 G2M K2 = (2Mc)

⇒

2 2 GE

GM = 1

− [q2 /(2Mc)2 ]

1 − [q2 /(2Mc)2 ]

⇒

GE = 1 .

153

Problem 8.7

Putting Eqs. 8.13 and 8.18 into Eq. 8.14: h io ge4 n µ ν ν µ µν 2 2 p p + p p + g ( mc ) − ( p · p ) 3 1 1 3 1 3 q4 qµ qν K2 1 1 + pµ + 2 qµ × K1 − gµν + 2 pν + 2 qν q ( Mc)2 ( 2ge4 p1 · qp3 · q 2 2 + (mc) − p1 · p3 = 4 K1 −2p1 · p3 − 4(mc) + 4p1 · p3 + 2 q2 q " K2 1 1 + 2 p · p + p · q p · p + p · q 3 3 1 1 2 2 ( Mc)2 #) h i 2 + p + 21 q · p + 12 q (mc) − p1 · p3

h|M|2 i =

2g4 = 4e q

(

( p · q)( p3 · q) K1 ( p1 · p3 ) − 3(mc)2 + 2 1 q2 " K2 2( p1 · p)( p3 · p) + ( p1 · p)( p3 · q) + ( p1 · q)( p3 · p) + ( Mc)2

+

1 2 ( p1

2

· q)( p3 · q) + p + ( p · q) +

1 2 4q

h

2

(mc) − ( p1 · p3 )

i

#)

Now q = p4 − p2 = p1 − p3

⇒

h i q2 = p21 + p23 − 2p1 · p3 = 2 (mc)2 − p1 · p3

p1 · q = p21 − p1 · p3 = (mc)2 − p1 · p3 = 12 q2 p3 · q = p1 · p3 − p23 = p1 · p3 − (mc)2 = − 12 q2 p2 = ( Mc)2 ,

p · q = − 12 q2

So the coefficient of K1 reduces to 2 2 q2 q − = 2( p1 · p3 ) − 4(mc)2 , ( p1 · p3 ) − 3(mc)2 + 2 2 2 q

154


and the coefficient of K2 /( Mc)2 becomes 2 2 2 q q 1 q2 q 2( p1 · p)( p3 · p) + ( p1 · p) − + ( p3 · p ) + − 2 2 2 2 2 2 2 2 q q q + ( Mc)2 − + 2 4 2 2 q q2 q2 2 = 2( p1 · p)( p3 · p) + −( p1 − p3 ) · p − + ( Mc) − 2 4 4

= 2( p1 · p)( p3 · p) +

q2 ( Mc)2 . 2

Thus 2g4 h|M| i = 4e q

(

2

h i K1 2( p1 · p3 ) − 4(mc)2 ) K2 q2 2 + 2( p1 · p)( p3 · p) + ( Mc) 2 ( Mc)2

X

Problem 8.8

From the text, p = ( Mc, 0), p1 =

E E0 (1, pˆ i ), p3 = (1, pˆ f ), p1 · p = ME, p3 · p = ME0 , c c

EE0 EE0 EE0 ˆ ˆ ( 1 − p · p ) = ( 1 − cos θ ) = 2 sin2 θ/2. i f c2 c2 c2 From Problem 8.7 (with m → 0) p1 · p3 =

h i EE0 q2 = 2 (mc)2 − ( p1 · p3 ) = −4 2 sin2 θ/2. c Putting this into Eq. 8.19:

h|M|2 i =

(

4ge4 4( EE0 /c2 ) sin2 θ/2

2

K1 2

+ K2

EE0 sin2 θ/2 c2

EE0 EE0 − sin2 θ/2 c2 c2 )

( ge4 c2 2K1 sin2 θ/2 + K2 cos2 θ/2 . = 4EE0 sin4 θ/2

)

X

155

Problem 8.9

In the lab frame (see p. 282), p1 =

E (1, pˆ i ), c

p3 =

E0 (1, pˆ f ), c

pˆ i · pˆ f = cos θ,

p2 = ( Mc, 0).

From energy-momentum conservation, p4 = p1 + p2 − p3 =

E − E0 Epˆ i − E0 pˆ f Mc + , c c

! .

But p24 = ( Mc)2 , so

2

( Mc) =

E − E0 Mc + c

2

−

Epˆ i − E0 pˆ f c

!2

( E − E0 )2 E2 + E0 2 − 2EE0 cos θ − c2 c2 0 2EE ( 1 − cos θ ) , = ( Mc)2 + 2M( E − E0 ) − c2

= ( Mc)2 + 2M( E − E0 ) +

or

EE0 EE0 M ( E − E0 ) = 2 (1 − cos θ ) = 2 2 sin2 θ/2. c c E 2E . X ME = E0 M + 2 sin2 θ/2 ; E0 = c 1 + (2E/Mc2 ) sin2 θ/2

Problem 8.10

For a “Dirac“ proton (Problem 8.6), K1 = −q2 and K2 = (2Mc)2 , and the Rosenbluth formula (Eq. 8.23) becomes dσ = dΩ

α¯h 4ME sin2 θ/2

2

i E0 h −2q2 sin2 θ/2 + (2Mc)2 cos2 θ/2 . E

If mc2 E Mc2 , then E0 ≈ E (Problem 8.9), and q2 is

( p1 − p3 )2 = p21 + p23 − 2p1 · p3 = 2m2 c2 − 2

EE0 E2 (1 − cos θ ) ≈ −4 2 sin2 θ/2, 2 c c

156


so dσ ≈ dΩ

α¯h 4ME sin2 θ/2

2

2 2

2

4M c cos θ/2 =

α¯hc 2E sin2 θ/2

2

But for a highly relativistic electron (mc2 E), E ≈ |p|c, so dσ ≈ dΩ

α¯h 2|p| sin2 θ/2

2

cos2 θ/2.

Compare this with the Mott formula (Eq. 7.131): dσ = dΩ

α¯h 2p2 sin2 θ/2

2 h

i (mc)2 + p2 cos2 θ/2 ,

which reduces (in the case mc |p|) to dσ ≈ dΩ

α¯h 2|p| sin2 θ/2

2

cos2 θ/2.

X

Problem 8.11

[The answer is given in the text.]

Problem 8.12

From Eq. 8.28:   1 r = 0 , 0

  0 b = 1 , 0

  0 g = 0 , 1

so, under the action of U, 

0 r → 0 1  0 b → 0 1  0 g → 0 1

    1 0 1 0 0 1 0 = 0 = g 0 0 0 1     1 0 0 1 0 1 1 = 0 = r 0 0 0 0     1 0 0 0 0 1 0 = 1 = b. 0 0 1 0

cos2 θ/2.

157

¯ Thus (Eq. 8.29) ¯ b¯ → r¯, g¯ → b. Likewise, r¯ → g, √ |3i → |30 i = ( g g¯ − r¯r )/ 2 = α|3i + β|8i √ √ = α(r¯r − bb¯ )/ 2 + β(r¯r + bb¯ − 2g g¯ )/ 6 β α β α β = g g¯ −2 √ + r¯r √ + √ + bb¯ − √ + √ . 6 6 6 2 2 Evidently

√ 3 1 β √ = −2 √ , β = − 2 6 2 α β α 1 1 1 = √ − √ , α=− −√ = √ + √ 2 6 2 2 2 2 2 √ β α 1 3 0 = −√ + √ = √ − √ = 0. X 6 2 2 2 2 6 Similarly

√ |8i → |80 i = ( g g¯ + r¯r − 2bb¯ )/ 6 = γ|3i + δ|8i √ √ = γ(r¯r − bb¯ )/ 2 + δ(r¯r + bb¯ − 2g g¯ )/ 6 δ γ δ γ δ ¯ = g g¯ −2 √ + r¯r √ + √ + bb − √ + √ . 6 6 6 2 2 Evidently δ 1 1 √ = −2 √ , δ = − 2 6 6 √ δ 1 1 γ γ 3 √ = √ +√ = √ − √ , γ= 2 6 6 2 2 2 6 √ 2 γ 3 2 δ 1 −√ = −√ + √ = − √ − √ = −√ . X 6 6 6 2 2 2 2 6

Problem 8.13

There are 64 matrix products here, and we need to calculate the trace of each one. I think it’s time for Mathematica. First define the 8 lambda matrices (Eq. 8.34), then construct a matrix whose ij element is Tr(λi λ j ):

158

8 Electrodynamics and Chromodynamics of Quarks m1 = 880, 1, 0 MZ /2), d, s, and b (and in the case of the quarks, three colors):

∑(|cV |2 + |c A |2 ) = 3(0.5000) + 3(0.2514) + 6(0.2867) + 9(0.3695) = 7.2999. f

f

f

The branching ratios (Γi /Γtotal ) are: e, µ, τ νe , νµ , ντ u, c d, s, b

0.2514/7.2999 = 0.0344, or 0.5000/7.2999 = 0.0685, or 3(0.2867)/7.2999 = 0.1178, 3(0.3695)/7.2999 = 0.1519,

: : : :

3% 7% or 12% or 15%

(c) From Eq. 9.91, gz2 =

ge2 4πα = = 0.5156, (0.2314)(0.7686) sin θw cos2 θw 2

so τ =

1 Γtotal

=

48π¯h gz2 MZ c2

1 48π (6.582 × 10−22 MeV s = (7.2999) (0.5156)(91188 MeV)(7.2999)

= 2.89 × 10−25 s . With a fourth generation, the lifetime would decrease. The precise factor depends on how many (and which) members of the 4th generation have masses less than MZ /2, but it should be approximately 3/4.

Problem 9.24

R =

=

σ(e+ e− → Z → hadrons) = σ (e+ e− → Z → µ+ µ− ) 9(0.2867 + 0.3695) = 23.5 . 0.2514

(The numbers are from Problem 9.23(b).)

∑q

q 2 cV

µ

cV

2

+

q 2 cA

2 µ + cA

196

9 Weak Interactions

Problem 9.25

First evaluate the coefficient in curly brackets:

0.5000 − 2(0.2314) + 4(0.2314)2 (0.2314)(0.7686)

2

= 1.998 ≈ 2.

Then σZ 1 1 (2E)4 x4 = , =2 σγ 16 [(2E)2 − ( MZ c2 )2 ]2 + (h¯ Γ Z MZ c2 )2 8 ( x 2 − 1)2 + a2 where (using gz2 = 0.5156, from Eq. 9.91) a≡

h¯ Γ Z 7.3gz2 = 0.025. = 48π M Z c2

200

150

100

50

0.0

0.2

0.4

0.6

0.8

1.0

1.2

Problem 9.26

(a) Given

( p − mc)u = 0;

u± ≡ 21 (1 ± γ5 )u

(upper sign for R, lower for L).

( p − mc)u± = =

) 1 p − mc)(1 ± γ5 )u = 12 ( p − mc u ± 12 ( p − mc)γ5 u 2 ( ± 12 γ5 (− p − mc)u = ∓mcγ5 u 6= 0

(unless m = 0). I used the fact that γ5 anticommutes with γµ , so pγ5 = −γ5 p. (b) P± =

1 2

1 1 ±1 0 1 1 0 ± = . 0 1 1 0 2 ±1 1

197

P± w = λw ⇒

1 2

1 ±1 ±1 1

wA wB

=λ

wA wB

⇒

w A ± wB wB ± w A

= 2λ

wA , wB

so w A ± w B = 2λw A , w B ± w A = 2λw B ; (2λ − 1)w A = ±w B , (2λ − 1)w B = ±w A .

(2λ − 1)2 w A = ±(2λ − 1)w B = w A ,

(4λ2 − 4λ + 1)w A = w A ,

and hence λ(λ − 1) = 0, If λ = 0, w B = ∓w A ; if λ = would be   1  0   w1 =   − 1  , w2 = 0

λ = 0, λ = 1 .

1, w B = ±w A . So a simple set of eigenspinors  0  1   ,  0  −1 

  0 1  w4 =  0 ,

  1 0  w3 =  1 ,

or

(λ = 0 for P+ , λ = 1 for P− );

(λ = 1 for P+ , λ = 0 for P− ).

1

0 (c)

o (1 ± γ5 )( p − mc) − ( p − mc)(1 ± γ5 ) n h io = 12 ( p − mc) − ( p − mc) ± γ5 ( p − mc) − ( p − mc)γ5

[ P± , ( p − mc)] =

1 2

n

= ± 12 γ5 ( p − mc + p + mc) = ±γ5 p 6= 0. But if a spinor is simultaneously an eigenstate of two noncommuting operators, A and B, which is to say, if Aw = λw and Bw = µw, then

[ A, B]w = ABw − BAw = µ( Aw) − λ( Bw) = µλw − λµw = 0. So w could be an eigenspinor of both P± and ( p − mc) only if γ5 pw = 0, or mγ5 w = 0, which is only possible if m = 0 (if γ5 w = 0, then P± w = (1/2)w = 0, and w = 0 is by definition not an eigenspinor).

Problem 9.27

u χL = (9.141); d0 L

jµ± = χ¯ L γµ τ ± χ L (9.136);

τ ± = 12 (τ 1 ± iτ 2 ) (9.137).

198

9 Weak Interactions

01 0 +i 10 i 1 01 0 −i = 10 i 2

τ+ = τ−

jµ+

= u¯ d¯0

1 2

L

γµ

01 00

0 1 ; 0 0 1 0 0 −i 0 0 = = . 0 1 0 2 2 0

−i 0

=

1 2

0 2 0 0

=

0 u d 0 ¯ = u¯ d L γµ = u¯ L γµ d0L 0 d L 0 L

¯ µ (1 − γ5 )2 d 0 = = u¯ 21 (1 + γ5 )γµ 12 (1 − γ5 )d0 = 14 uγ

1 ¯ 5 0 2 uγµ (1 − γ ) d

.

I used Table 9.2 for the chiral spinors, {γ5 , γµ } = 0, and (1 − γ5 )2 = 2(1 − γ5 ). jµ−

= u¯ d¯0

L

γµ

00 10

u 0 0 ¯ = u¯ d L γµ = d¯0L γµ u L 0 d L u L

= d¯0 12 (1 + γ5 )γµ 12 (1 − γ5 )u = 14 d¯0 γµ (1 − γ5 )2 u =

1 ¯0 5 2 d γµ ( 1 − γ ) u

.

Meanwhile, from Eq . 9.138, jµ3 =

=

3 1 ¯ 2 χ L γµ τ χ L 1 2

1 2

=

u¯ d¯0

γ L µ

u¯ L γµ u L − d¯0L γµ d0L =

1 4

1 0 0 −1

¯ µ uγ

u = d0 L

1 2

(1 − γ5 )u − d¯0 γ

u¯ d¯0 µ

γ L µ

(1 − γ5 ) d 0

u −d0

L

,

and, from Eqs. 9.144 and 9.131, 2

jµem =

∑ Qi (u¯ i γµ ui ) =

i =1

h

2 ¯ 1 ¯0 0 µ u − 3 d γµ d 3 uγ

i

.

Finally, from Eq. 9.140, ¯ µ u − 23 d¯0 γµ d0 − 12 uγ ¯ µ (1 − γ5 )u + 21 d¯0 γµ (1 − γ5 )d0 jµY = 2( jµem − jµ3 ) = 43 uγ h i ¯ µ 35 + γ5 u − d¯0 γµ 13 + γ5 d0 . = 12 uγ

Problem 9.28

According to Eq. 9.155, the quark and lepton couplings to the Z are of the form −igz jµ3 − sin2 θw jµem .

199

I’ll do the first generation (e, νe , u, d); the other generations are identical. (As explained in the first footnote to Section 9.6, it doesn’t matter whether you use d or d0 , so I’ll use d.) For the quarks (Problem 9.27 and Eq. 9.144): h i 1 ¯ ¯ µ (1 − γ5 )d ; jem = 2 (uγ ¯ µ (1 − γ5 )u − dγ jµ3 = 14 uγ µ 3 ¯ µ u ) − 3 ( dγµ d ). For the leptons (Eq. 9.138): jµ3 = 21 [ν¯L γµ νL − e¯L γµ e L ]. But ν¯L γµ νL = ν¯ 12 (1 + γ5 )γµ 12 (1 − γ5 )ν (and similarly for the electron term), while (1 + γ5 )γµ = γµ (1 − γ5 ), and (1 − γ5 )2 = 2(1 − γ5 ). So ¯ µ (1 − γ5 )ν − eγ ¯ µ (1 − γ5 ) e ], jµ3 = 41 [νγ

¯ µ e) jµem = −(eγ

while

(Eq. 9.131). The generic neutral weak vertex factor is (Eq. 9.90) igz γµ ( c V − c A γ 5 ) ; 2 picking out the relevant terms in −igz jµ3 − sin2 θw jµem , we find: Neutrino: i i h igz h ¯ µ (1 − γ5 ) ν = − ¯ µ 12 − 21 γ5 ν ⇒ cV = 12 , c A = νγ −igz 14 νγ 2

−

1 2

.

Electron: h i i igz h ¯ µ − 21 + 2 sin2 θw + 12 γ5 e ¯ µ (1 − γ5 )e + sin2 θw eγ ¯ µe = − eγ −igz − 14 eγ 2

⇒ cV = − 21 + 2 sin2 θw , c A = − 21 . Up Quark:

−igz

h

5 1 ¯ 2 4 uγµ (1 − γ ) u − 3

i i igz h ¯ µu = − ¯ µ 12 − 34 sin2 θw − 12 γ5 u sin2 θw uγ uγ 2

⇒ cV =

1 2

− 34 sin2 θw , c A =

1 2

.

Down Quark: i h i h ¯ µ (1 − γ5 )d + 1 sin2 θw dγ ¯ µ d = − igz dγ ¯ µ − 1 + 2 sin2 θw + 1 γ5 d −igz − 14 dγ 3 2 3 2 2

⇒ cV = − 12 + 32 sin2 θw , c A = − 21 .

200

9 Weak Interactions

Problem 9.29

The two basic hadronic modes, ¯ (1) τ → ντ + d + u,

¯ (2) τ → ντ + s + u,

have decay rates Γ1 = 3 cos2 θC Γe ,

Γ2 = 3 sin2 θC Γe

(3 for color), where Γe is the rate for τ → ντ + e + ν¯e (from Problem 9.5). So Γtot = Γe + Γµ + Γ1 + Γ2 = 2Γe + 3Γe (sin2 θC + cos2 θC ) = 5Γe . Evidently τ=

2τleptons 1 1 = = = (0.4)(8.16 × 10−13 ) = 3.26 × 10−13 s Γtot 5Γe 5

(the experimental value lifetime is 2.91 × 10−13 s), and the branching ratios are Γµ Γe 1 = = = 20% , Γtot Γtot 5

Γhadrons 3 = = 60% . Γtot 5

The experimental branching ratios are 17.8% (electrons), 17.4% (muons), and 61% for hadrons. [I got the latter figure from the Particle Physics Booklet, by adding the dominant hadronic modes: π − π 0 ντ (25.5%), π − ντ (10.9%), π − π − π + ντ (9.3%), π − π 0 π 0 ντ (9.3%), π − π − π + π 0 ντ (4.6%), π − π 0 π 0 π 0 ντ (1.0%). This doesn’t count the many modes with branching ratios below 1%; by my estimate they add to about 4%, raising the total hadronic ratio to about 65%.]

201

Problem 9.30

(a) Treat all secondaries as massless. The principal decay modes are c → s + e+ + νe :

Γ = |Vcs |2 Γ0

c → s + µ+ + νµ :

Γ = |Vcs |2 Γ0

c → d + e+ + νe :

Γ = |Vcd |2 Γ0

c → d + µ+ + νµ : c → s + u + d¯ :

Γ = |Vcd |2 Γ0 Γ = 3|Vcs |2 |Vud |2 Γ0

c → s + u + s¯ : c → d + u + d¯ :

Γ = 3|Vcs |2 |Vus |2 Γ0

c → d + u + s¯ :

Γ = 3|Vcd |2 |Vus |2 Γ0

Γ = 3|Vcd |2 |Vud |2 Γ0

where Γ0 ≡

mc mµ

5

Γµ ,

and Γµ is the decay rate of the muon. In terms of the Cabibbo angle(stick with CKM matrix elements if you prefer), Γtot = Γ0 (cos2 θC + cos2 θC + sin2 θC + sin2 θC + 3 cos2 θC cos2 θC

+ 3 cos2 θC sin2 θC + 3 sin2 θC cos2 θC + 3 sin2 θC sin2 θC ) i h = Γ0 cos2 θC + sin2 θC 2 + 3 cos2 θC + sin2 θC = 5Γ0 . So the lifetime is 1 mµ 5 1 105.66 5 τ= (2.197 × 10−6 ) = 1.9 × 10−12 s . τµ = 5 mc 5 1250 (b) In this (pretty gross) approximation, the lifetime of the D should be about 1.9 × 10−12 s . The observed lifetime of the D + is 1.04 × 10−12 s, and that of D0 is 0.41 × 10−12 s, so we’re not too far off. The branching ratios for semileptonic (e or µ) and hadronic modes are the same as for the τ (Problem 9.29): 20% (each) and 60%, respectively. According to the Particle Physics Booklet the dominant electronic decays of the D + are K¯ 0 e+ νe (8.6%), K¯ ∗0 e+ νe (5.6%), K − π + e+ νe (4.5%), for a total of 18.7%, and the dominant muonic decays are K¯ 0 µ+ νµ (9.5%), K¯ ∗0 µ+ νµ (5.5%), K − π + µ+ νµ (4.0%), for a total of 19.0%. The rest are hadronic, so the agreement is very good. The D0 decays are much more complicated, and the agreement is not so hot: a total of 6.2% for the electronic modes and 5.6% for muonic modes.

202

9 Weak Interactions

(c) The principal decay modes of the b quark are b → c + e + ν¯e :

Γ = |Vcb |2 Γ0

b → c + µ + ν¯µ :

Γ = |Vcb |2 Γ0

b → c + τ + ν¯τ :

Γ = |Vcb |2 Γ0

b → u + e + ν¯e :

Γ = |Vub |2 Γ0

b → u + µ + ν¯µ :

Γ = |Vub |2 Γ0

b → u + τ + ν¯τ :

Γ = |Vub |2 Γ0

b → c + d + u¯ :

Γ = 3|Vcb |2 |Vud |2 Γ0

b → c + s + u¯ :

Γ = 3|Vcb |2 |Vus |2 Γ0

b → c + d + c¯ :

Γ = 3|Vcb |2 |Vcd |2 Γ0

b → c + s + c¯ :

Γ = 3|Vcb |2 |Vcs |2 Γ0

b → u + d + u¯ :

Γ = 3|Vub |2 |Vud |2 Γ0

b → u + s + u¯ :

Γ = 3|Vub |2 |Vus |2 Γ0

b → u + d + c¯ :

Γ = 3|Vub |2 |Vcd |2 Γ0

b → u + s + c¯ :

Γ = 3|Vub |2 |Vcs |2 Γ0

where Γ0 ≡

mb mµ

5

Γµ .

So h i Γtot = Γ0 3|Vcb |2 + 3|Vub |2 + 3(|Vcb |2 + |Vub |2 ) |Vud |2 + |Vus |2 + |Vcd |2 + |Vcs |2 h i = 3Γ0 |Vcb |2 + |Vub |2 1 + cos2 θC + sin2 θC + sin2 θC + cos2 θC h i = 9 (|Vcb |2 + |Vub |2 Γ0 = 9 (0.0422)2 + (0.0040)2 Γ0 = 0.0162 Γ0 . So the lifetime of the b quark (and hence of the B meson) is τ=

1 0.0162

mµ mb

5 τµ =

1 0.0162

105.66 4300

5

(2.197 × 10−6 ) = 1.2 × 10−12 s .

The observed lifetimes are: 1.64 × 10−12 s (for the B− ), 1.53 × 10−12 s (for the B0 ). The agreement is very good for both of them. Branching ratios: Γ0 |Vcb |2 + |Vub |2 Γ` Γhad 2 1 = = = 11% ; = = 67% . Γtot 9 Γtot 3 9Γ0 (|Vcb |2 + |Vub |2 )

203

The experimental leptonic branching ratios are 10.2% (for the B+ ) and 10.5% (for the B0 )—not bad! (d) The b quark cannot decay without crossing generations, whereas the c quark can go to an s, in the same generation. The smallness of the CKM matrix elements Vcb and Vub almost exactly compensates for the larger mass of the b quark, so the final lifetimes come out practically the same.

Problem 9.31

p1 t

p3 W p

b

2

M = i u¯ (2)

−igw µ √ γ (1 − γ5 ) u(1)eµ∗ (3) 2 2

2 gw [u¯ (2)γµ (1 − γ5 )u(1)][u¯ (1)γν (1 − γ5 )u(2)]eµ∗ (3)eν (3) 8 p3 µ p3 ν g2 ν 5 p + m c ) γ ( 1 − γ )( p + m c )] − g + h|M|2 i = w Tr[γµ (1 − γ5 )( µν 2 2 1 1 16 ( MW c)2

|M|2 =

The trace reduces to ν 5 2 µ 5 ν 5 Tr[γµ (1 − γ5 ) p p 2 ] + m1 m2 c Tr[ γ (1 − γ ) γ (1 − γ )] 1 γ (1 − γ )

ν 5 2 2 µ ν 5 5 = Tr[γµ p p 2 ] + m1 m2 c Tr[ γ γ (1 + γ )(1 − γ ) ] 1 γ (1 − γ ) {z } | 1−(γ5 )2 =0

ν 5 µ ν µ ν 5 = 2Tr[γµ p p p p p p 2 ] = 2[Tr( γ 2 ) + Tr( γ 2 γ )] 1 γ (1 − γ ) 1γ 1γ

= 2p1κ p2λ [Tr(γµ γκ γν γλ ) + Tr(γµ γκ γν γλ γ5 )] = 8p1κ p2λ gµκ gνλ − gµν gκλ + gµλ gνκ + ieµκνλ .

Because this is contracted with a symmetric tensor, the antisymmetric (final) term does not contribute. i p3 µ p3 ν g2 h µ µ h|M|2 i = w p1 p2ν − gµν ( p1 · p2 ) + p1ν p2 − gµν + 2 ( MW c)2 " # 2 2( p1 · p3 )( p2 · p3 ) − ( p1 · p2 ) p23 gw 2( p1 · p2 ) + = 2 ( MW c)2 g2 ( p · p )( p · p ) = w ( p1 · p2 ) + 2 1 3 22 3 . 2 ( MW c)

204

9 Weak Interactions

In the rest frame of the t, p1 = (mt c, 0),

p2 =

Eb ,p , c

p3 =

EW , −p , c

So p1 · p2 = mt Eb ,

p1 · p3 = mt EW ,

2 4 2 c ⇒ p2 = − p2 c2 = MW EW

p2 · p3 =

Eb EW + p2 ; c2

2 2 EW EW Eb EW 2 2 2 2 c ⇒ p · p = c ; − M + − MW 2 3 W c2 c2 c2

2 2 mt c2 = Eb + EW ⇒ Eb = mt c2 − EW ⇒ p2 · p3 = mt EW − MW c ;

p1 · p2 = mt (mt c2 − EW ) = m2t c2 − mt EW . 2 2 gw 2 2 2 h|M| i = mt EW (mt EW − MW c ) mt − mt EW + 2 ( MW c)2 g2 (mt EW )2 = w m2t c2 − mt EW − 2mt EW + 2 2 ( MW c)2 " # 2 mt EW g2 m t mt c2 − 3EW + 2 = w . 2 ( MW c)2 2

Using the result of Problem 3.19 (and setting mb → 0): EW =

2 − m2 2 m2t + MW m2 + MW b 2 c ≈ t c2 . 2mt 2mt

" # 2 ) c2 2 + M 4 ) c4 2m 3 (m2t + MW gw 2mt (m4t + 2m2t MW t 2 W mt c − h|M| i = + 2 2 mt ( MW c)2 4m2t ! 4 c4 2 m 3 3 1 1 gw 2 2 2 2 t m2t c2 − m2t c2 − MW c + + m2t c2 + MW c = 2 c2 2 2 2 2 MW 2 g c 2 2 c2 gw w 2 2 4 2 2 4 (m2t − MW mt MW − 2MW + mt = )(m2t + 2MW ). = 2 2MW 4MW 2

Quoting Eq. 6.35,

|p| Γ= 8π¯hm2t c

gw c 2MW

2

2 2 )(m2t + 2MW ). (m2t − MW

But (Problem 3.19b, with mb → 0) q 4 − 2m2 M2 2 )c m4t + MW (m2t − MW t W |p| ≈ c= , 2mt 2mt

205

so " #2 " # 2 ( m c2 )3 MW 2 gw MW 2 t 1+2 Γ = 1− mt mt 64π¯h( MW c2 )2 " #2 " # (0.653)2 (174000)3 80.4 2 80.4 2 = 1+2 1− 174 174 64π (6.582 × 10−22 )(80403)2

= 2.32 × 1024 /s. τ = 1/Γ = 4.3 × 10−25 s .

Problem 9.32

(a)

p2

p4

nm

m q

e

Z

u¯ (3)

W ne

p1

p

3

i −igw −igw 5 √ (1 − γ5 ) u (1) √ ¯ u ( 4 ) ( 1 − γ ) u ( 2 ) ( MW c)2 2 2 2 2

×(2π )4 δ4 ( p1 − p3 − q)(2π )4 δ4 ( p2 − p4 + q)

d4 q . (2π )4

Using the first delta function, the integral replaces q by p1 − p3 ; erasing (2π )4 δ( p1 + p2 − p3 − p4 ) and multiplying by i leaves 2 gw u¯ (3)(1 − γ5 )u(1) u¯ (4)(1 − γ5 )u(2) . 2 8( MW c)

M=

(b) 2

|M| =

2 gw 8( MW c)2

2 h

ih i∗ u¯ (3)(1 − γ5 )u(1) u¯ (3)(1 − γ5 )u(1) h ih i∗ × u¯ (4)(1 − γ5 )u(2) u¯ (4)(1 − γ5 )u(2) .

206

9 Weak Interactions

h

u¯ (3)(1 − γ5 )u(1)

i∗

h i† = u (3) † γ0 (1 − γ5 ) u (1) = u (1) † (1 − γ5 ) † ( γ0 ) † u (3)

= u¯ (1)γ0 (1 − γ5 )γ0 u(3) = u¯ (1)(1 + γ5 )γ0 γ0 u(3) = u¯ (1)(1 + γ5 )u(3).

|M|2 =

2 gw 8( MW c)2

2 h

ih i u¯ (3)(1 − γ5 )u(1) u¯ (1)(1 + γ5 )u(3) h ih i × u¯ (4)(1 − γ5 )u(2) u¯ (2)(1 + γ5 )u(4) .

2 h i 2 1 gw 5 5 h|M| i = Tr ( 1 − γ )( p + m c )( 1 + γ )( p + m c ) 3 3 1 1 2 8( MW c)2 h i 5 ×Tr (1 − γ5 )( p + m c )( 1 + γ )( p + m c ) . 2 2 4 4 2

But m2 = m3 = 0, so the m1 c in the first term (and m4 c in the last) are traces of an odd number of gamma matrices, and hence are zero.

h|M|2 i =

2 h i h i 2 gw 1 5 5 5 Tr (1 − γ5 ) p p p p 3 Tr (1 − γ ) 2 (1 + γ ) 1 (1 + γ ) 4 2 2 8( MW c)

But p(1 + γ5 ) = (1 − γ5 ) p, and (1 − γ5 )2 = 2(1 − γ5 ), so 2 gw h|M| i = 2 8( MW c)2 2

2

h i h i 5 Tr (1 − γ5 ) p p p p 3 Tr (1 − γ ) 2 1 4 .

Now, Tr(γ5 ab) = 0, and Tr(ab) = 4a · b, so 2 2 1 gw h|M| i = ( p1 · p3 )( p2 · p4 ) . 2 ( MW c)2 2

(c) Equation 6.47 says dσ = dΩ

h¯ c 8π

2

h|M|2 i |p f | . ( E1 + E2 )2 |pi |

Treating all four particles as massless, E E E p1 = , pi , p 2 = , − pi , p 3 = , pf , c c c

| pi | = | p f | =

E . c

p4 =

E , −p f c

;

207

So 2 2 2 2 E E E E − pi · p f = − cos θ = 2 sin2 (θ/2), c c c c 2 2 2 2 E E E E − pi · p f = − cos θ = 2 sin2 (θ/2) p2 · p4 = c c c c

p1 · p3 =

(where θ is the angle between the incoming electron and the outgoing neutrino). dσ = dΩ

=

h¯ c 8π

1 2

2

1 1 4E2 2

gw MW c

4

gw MW c

h¯ 8πc

#2 4 " 2 E 2 2 sin (θ/2) c

2

E2 sin4 (θ/2) .

(d) 2 Z dσ 1 gw 4 h¯ 1 σ = E2 dΩ = (1 − cos θ )2 sin θ dθ dφ dΩ 2 MW c 8πc 4 2 Z 1 gw 4 h¯ 2π π = E2 (1 − cos θ )2 sin θ dθ 2 MW c 8πc 4 0 {z } | π 8 1 3 (1−cos θ ) = Z

3

=

2 3π

gw MW c

4

h¯ 8c

2

E2 =

2 3π

0

2E h¯ gw 2 c3 8MW

3

!2 .

(e) The Standard Model (Eqs. 9.13 and 9.14, with mµ → 0) says dσ 1 = dΩ 2

2E h¯ cgw 4π ( MW c)2

2 ,

1 σ= 8π

2E h¯ gw 2 MW c3

!2 .

Both the Griffiths theory and the Standard Model say dσ/dΩ and σ are proportional to E2 , so the energy dependence offers no test. The obvious difference is that the Standard Model says dσ/dΩ is independent of the scattering angle, whereas my theory says it goes like sin4 (θ/2)—it’s zero in the forward direction (θ = 0), and rises to a maximum at θ = 180◦ (back-scattering). So the thing to do is count the number of muons coming off at various directions ; if it’s constant, the Standard Model wins, but I say more of them will come out along the direction of the incident electron.

208

9 Weak Interactions

Problem 9.33

From the 2006 Particle Physics Booklet, Section 11.2:

|Vud | = 0.97377,

|Vub | = 0.00431,

|Vcb | = 0.230,

|Vcb | = 0.0416,

|Vtd | = 0.0074,

|Vub | = 0.9991,

so ∗ Vud Vub (0.97377)(0.00431) |z1 | = ∗ = (0.230)(0.0416) = 0.44; Vcd Vcb Vtd Vtb∗ (0.0074)(0.9991) | z2 | = ∗ = (0.230)(0.0416) = 0.77. Vcd Vcb The phases are discussed in the Particle Physics Booklet, Section 11.3: γ ≡ arg(−z1 ) = 63◦ ;

β ≡ arg(−1/z2 ) =

1 2

sin−1 (0.687) = 22◦ .

If we write z2 = |z2 |eiθ , then −1/z2 = (1/|z2 |)ei(π −θ ) , so (π − θ ) = β, and hence θ = π − β = 180◦ − 22◦ = 158◦ . In the complex plane, then, the three numbers 1, z1 , and z2 look like this: (-0.71, 0.29) (1,0) (0.20, -0.39)

The unitarity triangle itself takes the form

z1

z2 1

Here the coordinates of the tail of z1 are 0.44 cos 63◦ = 0.20, 0.44 sin 63◦ = 0.39; the coordinates of the head of z2 are 1 − 0.77 cos 22◦ = 0.29, 0.77 sin 22◦ = 0.29. Evidently the triangle does not (quite) close, with the current experimental values.

209

10

Gauge Theories Problem 10.1

(a) If r is the distance from the apex, then z = r cos α,

so

vr =

z˙ . cos α

The tangential velocity is ˙ vφ = r sin α φ˙ = z tan α φ, so T=

1 m m(v2r + v2φ ) = z2 sin2 α φ˙ 2 + z˙ 2 . U = mgz . 2 2 2 cos α

(b) m z2 sin2 α φ˙ 2 + z˙ 2 − mgz . 2 cos2 α ∂L m d ∂L m ∂L m ˙ ¨ = z ⇒ = z; = z sin2 α φ˙ 2 − mg. ∂z˙ dt ∂z˙ ∂z cos2 α cos2 α cos2 α L = T−U =

Euler-Lagrange for z says m m z¨ = z sin2 α φ˙ 2 − mg, or z¨ = z sin2 αφ˙ 2 − g cos2 α . 2 cos α cos2 α Similarly, m 2 d ∂L = z sin2 αφ˙ ⇒ 2 ˙ dt cos α ∂φ

∂L ∂φ˙

= m tan2 α 2zz˙ φ˙ + z2 φ¨ ;

Euler-Lagrange for φ says 2z˙ φ˙ m tan2 α 2zz˙ φ˙ + z2 φ¨ = 0, or φ¨ = − . z

∂L = 0. ∂φ

210

10 Gauge Theories

(c) Since ∂L/∂φ = 0, the Euler-Lagrange equation for φ says ∂L/∂φ˙ is a ˙ Physically, it is the constant—call it `: ` ≡ mz2 tan2 α φ. z component of angular momentum . (d) φ˙ =

`2 ` 2 ¨ ⇒ z = z sin α − g cos2 α, m tan2 α z2 m2 tan4 α z4

or cos2

z¨ =

α

`2 cot2 α −g m2 z3

.

[Conservation of energy says E = T+U =

m 2 2 ˙ 2 + z˙ 2 + mgz. z sin α φ 2 cos2 α

˙ Using ` to eliminate φ˙ now yields an equation for z.]

Problem 10.2

It is safest to work out a specific component—say µ = 0, ν = 1. First find all the terms in L that involve ∂0 A1 (or ∂0 A1 —which, of course, is −∂0 A1 ): i 1 h 0 1 ∂ A − ∂1 A0 ( ∂0 A1 − ∂1 A0 ) + ∂1 A0 − ∂0 A1 ( ∂1 A0 − ∂0 A1 ) + · · · 16π −1 0 1 ∂ A ( ∂0 A1 ) − ∂0 A1 ( ∂1 A0 ) − ∂1 A0 ( ∂0 A1 ) + ∂1 A0 ( ∂1 A0 ) = 16π 1 0 1 0 0 1 0 1 + ∂ A ( ∂1 A0 ) − ∂ A ( ∂0 A1 ) − ∂ A ( ∂1 A0 ) + ∂ A ( ∂0 A1 ) + · · ·

−

=

i −1 h −2 ( ∂0 A1 )2 + 4 ( ∂0 A1 ) ( ∂1 A0 ) + · · · 16π

Evidently

−1 −1 ∂L = [−4(∂0 A1 ) + 4(∂1 A0 )] = (−∂0 A1 + ∂1 A0 ) ∂ ( ∂0 A1 ) 16π 4π −1 0 1 = ∂ A + ∂1 A0 . 4π The other components are the same (except for a few signs when µ and ν are both spatial).

211

Problem 10.3

First note that ∂ν ∂µ (∂µ Aν − ∂ν Aµ ) = ∂µ ∂µ (∂ν Aν ) − ∂ν ∂ν (∂µ Aµ ) = 0. Now apply ∂ν to Eq. 10.19: ∂ν ∂µ (∂µ Aν − ∂ν Aµ ) +

mc 2 h¯

∂ν Aν = 0 ⇒ ∂ν Aν = 0. X

Hence the middle term in Eq. 10.19 is actually zero, leaving ∂µ ∂µ ( Aν ) +

mc 2 h¯

Aν = 0. X

Problem 10.4

i¯hc µ ∂L i¯hc ∂L = − γ ψ ⇒ ∂µ = − γµ ∂µ ψ; ¯ ¯ ∂(∂µ ψ) 2 ∂(∂µ ψ) 2 ∂L i¯hc µ = γ (∂µ ψ) − (mc2 )ψ. ∂ψ¯ 2 The Euler-Lagrange equation for ψ¯ says:

−

i¯hc µ i¯hc µ γ ∂µ ψ = γ (∂µ ψ) − (mc2 )ψ, 2 2

or

iγµ (∂µ ψ) −

mc h¯

ψ=0

(the Dirac equation). Likewise ∂L i¯hc ¯ µ i¯hc ∂L = ψγ ⇒ ∂µ = (∂µ ψ)γµ ; ∂(∂µ ψ) 2 ∂(∂µ ψ) 2 i¯hc ∂L ¯ = − (∂µ ψ¯ )γµ − (mc2 )ψ. ∂ψ 2 The Euler-Lagrange equation for ψ says: i¯hc i¯hc ¯ (∂µ ψ¯ )γµ = − (∂µ ψ¯ )γµ − (mc2 )ψ, 2 2

or

(the adjoint Dirac equation).

Problem 10.5

∂L 1 = ∂µ φ; ∗ ∂(∂µ φ ) 2

∂L 1 mc 2 = − φ. ∂(φ∗ ) 2 h¯

i (∂µ ψ¯ )γµ +

mc h¯

ψ¯ = 0

212

10 Gauge Theories

So the Euler–Lagrange equation leads to ∂µ

1 µ ∂ φ 2

=−

1 mc 2 φ, 2 h¯

or

∂µ ∂µ φ +

mc 2 h¯

φ=0 .

Similarly, 1 ∂L = ∂µ φ∗ ; ∂(∂µ φ) 2

mc 2 ∂L 1 mc 2 ∗ =− φ , so ∂µ ∂µ φ∗ + φ∗ = 0 . ∂φ 2 h¯ h¯

Thus φ and φ∗ both satisfy the Klein–Gordon equation. Obviously, the second equation is the complex conjugate of the first.

Problem 10.6

∂L = 0; ∂(∂µ ψ¯ )

∂L = i¯hcγµ ∂µ ψ − mc2 ψ − qγµ ψAµ . ∂ψ¯

So 0 = i¯hcγµ ∂µ ψ − mc2 ψ − qγµ ψAµ , or i¯hγµ (∂µ ψ) − mcψ =

q Aµ γµ ψ . c

Similarly, ∂L ¯ µ; = i¯hcψγ ∂(∂µ ψ)

∂L ¯ µ Aµ ; = −mc2 ψ¯ − qψγ ∂ψ

¯ µ Aµ , or i¯h(∂µ ψ¯ )γµ + mcψ¯ = − q ψγ ¯ µ Aµ . i¯hc(∂µ ψ¯ )γµ = −mc2 ψ¯ − qψγ c

Problem 10.7

¯ µ ψ] = cq (∂µ ψ¯ )γµ ψ + ψγ ¯ µ (∂µ ψ) . ∂µ J µ = ∂µ [cqψγ With electromagnetic coupling the Dirac equation says (Problem 10.6) q i h mc q i h mc ¯ µ Aµ . ψ+ Aµ γµ ψ ; (∂µ ψ¯ )γµ = i ψ¯ + ψγ γµ ( ∂µ ψ ) = −i h¯ h¯ c h¯ h¯ c Therefore, ∂µ J µ = cqi

h mc h¯

¯ + ψψ

i q ¯ µ ψ) Aµ − mc ψψ ¯ − q (ψγ ¯ µ ψ) Aµ = 0. X (ψγ h¯ c h¯ h¯ c

213

Problem 10.8

We simply replace the ”ordinary” derivatives by ”covariant” derivatives (Eq. 10.38), and add on a free Lagrangian for Aµ : 1 µν 1 1 mc 2 ∗ φ φ− L = (Dµ φ)∗ (D µ φ) − F Fµν . 2| 16π {z } 2 h¯ F

Here ∗ q q Aµ φ ∂µ φ + i Aµ φ h¯ c h¯ c q 2 iq ∗ µ ∗ µ = (∂µ φ) (∂ φ) + (∂µ φ) A φ − Aµ φ∗ (∂µ φ) + Aµ Aµ φ∗ φ. h¯ c h¯ c

F=

∂µ φ + i

So

L =

1 1 µν 1 mc 2 ∗ (∂µ φ)∗ (∂µ φ) − φ φ − F Fµν 2 2 h ¯ 16π {z } | | {z } free φ Lagrangian

free Aµ Lagrangian

1 q 2 ∗ µ iq φ φA Aµ . + [φ(∂µ φ∗ ) − φ∗ (∂µ φ)] Aµ + 2 h¯ c | 2¯hc {z } interaction term

To get the current, we apply the Euler-Lagrange equation (exploiting Eq. 10.17): q 2 δL 1 δL iq = − F µν ; (φ∂ν φ∗ − φ∗ ∂ν φ) + | φ |2 A ν . = δ(∂µ Aν ) 4π δAν 2¯hc h¯ c So

−

q 2 1 iq | φ |2 A ν , ∂µ F µν = (φ∂ν φ∗ − φ∗ ∂ν φ) + 4π 2¯hc h¯ c

or ∂µ F µν =

4π iq ∗ ν q2 (φ ∂ φ − φ∂ν φ∗ ) − 2 |φ|2 Aν . c 2¯h h¯ c

Evidently (Eq. 10.24) Jν =

iq ∗ ν q2 (φ ∂ φ − φ∂ν φ∗ ) − 2 |φ|2 Aν . 2¯h h¯ c

(Curiously, Aν itself appears in the last term.) And the divergence of J µ is i iq h ∗ µ ∗ µ ∂µ J µ = (∂ (∂ )( ∂µ φ∗ ) − φ(∂µ ∂µ φ∗ ) µ φ)( ∂ φ ) + φ ( ∂µ ∂ φ ) − µ φ 2¯h 2 q − 2 (∂µ φ∗ )φAµ + φ∗ (∂µ φ) Aµ + φ∗ φ(∂µ Aµ ) . h¯ c

214

10 Gauge Theories

Meanwhile, working out the Euler–Lagrange equation for φ: 1 µ iq ∂L φAµ ; = ( ∂ φ ) + ∂(∂µ φ∗ ) 2 2¯hc ∂L 1 mc 2 iq 1 q 2 = − φ − φAµ Aµ . (∂µ φ) Aµ + ∗ ∂φ 2 h¯ 2¯hc 2 h¯ c

iq h¯ c

(∂µ φ) Aµ + φ∂µ Aµ mc 2 q 2 iq φ− φAµ Aµ . =− (∂µ φ) Aµ + h¯ h¯ c h¯ c

⇒ ∂µ ∂ φ + µ

Similarly, ∂L 1 µ ∗ iq = (∂ φ) − φ∗ Aµ ; ∂(∂µ φ) 2 2¯hc ∂L iq 1 mc 2 ∗ 1 q 2 ∗ µ φ + φ A Aµ = − (∂µ φ∗ ) Aµ + ∂φ 2 h¯ 2¯hc 2 h¯ c

⇒ ∂µ ∂µ φ∗ −

iq h¯ c

(∂µ φ∗ ) Aµ + φ∗ ∂µ Aµ mc 2 q 2 iq ∗ =− φ + (∂µ φ∗ ) Aµ + φ∗ Aµ Aµ . h¯ h¯ c h¯ c

Combining these results, i φ∗ (∂µ ∂µ φ) − φ(∂µ ∂µ φ∗ ) mc 2 q q = φ∗ (∂µ φ) Aµ + φ ∂µ Aµ − i φ∗ φ + φ∗ (∂µ φ) Aµ h¯ c h¯ h¯ c iq2 ∗µ q mc 2 ∗ µ ∗ µ + 2 φ φA Aµ + φ (∂µ φ) A + φ ∂µ A + i φ∗ φ 2 h¯ c h¯ h¯ c q iq2 ∗µ µ ∗ φ φA Aµ + φ(∂ φ ) Aµ − 2 h¯ c h¯ c2 2q ∗ = φ (∂µ φ) Aµ + φ(∂µ φ)∗ Aµ + φ∗ φ ∂µ Aµ . h¯ c Thus ∂µ J µ =

q 2q ∗ φ (∂µ φ) Aµ + φ(∂µ φ)∗ Aµ + φ∗ φ ∂µ Aµ 2¯h h¯ c 2 q − 2 (∂µ φ∗ )φAµ + φ∗ (∂µ φ) Aµ + φ∗ φ∂µ Aµ = 0. X h¯ c

215

Problem 10.9

(a) δL =

∂L ∂L δφi + δ(∂µ φi ) ∂φi ∂(∂µ φi )

(summation on i implied). But the Euler-Lagrange equation says ∂L ∂L = ∂µ , and δ(∂µ φ) ≡ ∂µ (φi + δφi ) − ∂µ φi = ∂µ (δφi ), ∂φi ∂(∂µ φi ) so δL = ∂µ

∂L ∂(∂µ φi )

δφi +

∂L ∂µ (δφi ) = ∂µ ∂(∂µ φi )

∂L δφi . X ∂(∂µ φi )

(b) For an infinitesimal phase transformation, Eq. 10.26 says ψ → ei(δθ ) ψ = (1 + iδθ )ψ,

so

δψ = iψ(δθ );

δψ¯ = −i ψ¯ (δθ ).

Meanwhile (from Example 10.2), ∂L ¯ µ, = i¯hcψγ ∂(∂µ ψ)

∂L = 0, ∂(∂µ ψ¯ )

so Jµ =

∂L δψ ¯ µ )(iψ) = −h¯ c(ψγ ¯ µ ψ) , = (i¯hcψγ ∂(∂µ ψ) δθ

which (apart from a constant factor −q/¯h) agrees with Eq. 10.36. (c) This time δφ = iφ δθ, δφ∗ = −iφ∗ δθ, and (Problem 10.8) ∂L 1 iq ∗ µ = ∂µ φ∗ − φ A , ∂(∂µ φ) 2 2¯hc

∂L 1 iq = ∂µ φ + φAµ . ∂(∂µ φ∗ ) 2 2¯hc

So ∂L δφ ∂L δφ∗ + ∂(∂µ φ) δθ ∂(∂µ φ∗ ) δθ 1 µ ∗ iq ∗ µ 1 µ iq µ = ∂ φ − φ A (iφ) + ∂ φ+ φA (−iφ∗ ) 2 2¯hc 2 2¯hc

Jµ =

=

i q [ φ ∂ µ φ ∗ − φ ∗ ∂ µ φ ] + | φ |2 A µ , 2 h¯ c

which (apart from a factor −q/¯h) agrees with the result in Problem 10.8.

Problem 10.10

H=

a b ; c d

H† =

a∗ c∗ . b∗ d∗

216

10 Gauge Theories

Evidently H is Hermitian (H † = H) ⇔ a and d are real, and b = c∗ , four (real) numbers in all, and θ, a1 , a2 , a3 is one way of packaging them: 1 0 0 1 0 −i 1 0 θ1+τ·a = θ + a1 + a2 + a3 0 1 1 0 i 0 0 1 (θ + a3 ) ( a1 − ia2 ) = , ( a1 + ia2 ) (θ − a3 ) which is precisely the form for the most general Hermitian 2 × 2 matrix, with a = θ + a3 , b = a1 − ia2 = c∗ , and d = θ − a3 (which, of course, can be solved to get θ and a in terms of a, b, c, and d).

Problem 10.11

Equations 10.56 and 10.57 ⇒ q q ∂µ + i τ · A0µ ψ0 = S ∂µ + i τ · Aµ ψ. h¯ c h¯ c Now, ψ0 = Sψ, so ∂µ ψ0 = (∂µ S)ψ + S(∂µ ψ), and Eq. [1] becomes q q (∂µ S)ψ + S( ∂µ ψ )+i τ · A0µ Sψ = S( ∂µ ψ ) + i S τ · Aµ ψ, h¯ c h¯ c or

q q τ · A0µ S = i S τ · Aµ . h¯ c h¯ c Multiply by S−1 on the right: q q ( ∂ µ S ) S −1 + i τ · A0µ = i S τ · Aµ S−1 , h¯ c h¯ c or h¯ c τ · A0µ = S(τ · Aµ )S−1 + i ∂ µ S S −1 . X q

(∂µ S) + i

Problem 10.12

(a)

F

µν

→ = = =

2q 2q ν ν µ µ µ ∂ A +∂ λ+ (λ × A ) − ∂ A + ∂ λ + (λ × A ) h¯ c h¯ c 2q µ ∂ µ Aν − ∂ ν Aµ + ∂ µ ∂ ν λ − ∂ ν ∂ µ λ + [∂ (λ × Aν ) − ∂ν (λ × Aµ )] h¯ c 2q µ Fµν + [∂ λ × Aν + λ × (∂µ Aν ) − ∂ν λ × Aµ − λ × (∂ν Aµ )] h¯ c 2q Fµν + [λ × Fµν + Aµ × ∂ν λ − Aν × ∂µ λ] . h¯ c µ

ν

ν

[1]

217

(b) 2q µν µ ν ν µ → F + [λ × F + A × ∂ λ − A × ∂ λ] h¯ c 2q · Fµν + λ × Fµν + Aµ × ∂ν λ − Aν × ∂µ λ h¯ c 4q µν = Fµν Fµν + F · λ × Fµν + Fµν · Aµ × ∂ν λ − Aν × ∂µ λ h¯ c

µν

F Fµν

µν

(I dropped terms in λ2 , since we are only concerned with infinitesimal transformations). But F · (λ × F) = 0, since the cross-product of any vector is perpendicular to the vector itself (I dropped the indices, since they are irrelevant to this argument, which holds for each µν), and Fµν · Aµ × ∂ν λ − Aν × ∂µ λ = −2Fµν · Aν × ∂µ λ = 2 (Aν × Fµν ) · ∂µ λ In the first step I used F µν (tµν − tνµ ) = −2F µν tνµ , which holds because F µν is antisymmetric in µ ↔ ν, and in the second step I used the vector identity A · (B × C) = C · (A × B) = −(B × A) · C. So 8q 1 L → L− (Aν × Fµν ) · ∂µ λ. 16π h¯ c

Problem 10.13

We know from Problem 10.12 that

( ∂ µ Aν − ∂ ν Aµ ) → ( ∂ µ Aν − ∂ ν Aµ ) 2q + [ λ × ( ∂ µ Aν − ∂ ν Aµ ) + Aµ × ∂ ν λ − Aν × ∂ µ λ ] h¯ c So Fµν → (∂µ Aν − ∂ν Aµ ) 2q + [ λ × ( ∂ µ Aν − ∂ ν Aµ ) + Aµ × ∂ ν λ − Aν × ∂ µ λ ] h¯ c 2q 2q 2q − Aµ + ∂ µ λ + ( λ × Aµ ) × Aν + ∂ ν λ + ( λ × Aν ) h¯ c h¯ c h¯ c ( 2q 2q = Fµν + (Aµ × Aν ) + λ × Fµν + (Aµ × Aν ) + (Aµ × ∂ ν λ ) h¯ c h¯ c

− (Aν × ∂ µ λ ) − (Aµ × Aν ) − (Aµ × ∂ ν λ ) − ( ∂ µ λ × Aν ) ) 2q µ ν µ ν − [A × ( λ × A ) + ( λ × A ) × A ] h¯ c

218

10 Gauge Theories

(I dropped terms of second order in λ). Three pairs of terms cancel, leaving 2q Fµν → Fµν + (λ × Fµν ) h¯ c 2 2q + [ λ × (Aµ × Aν ) − Aµ × ( λ × Aν ) + ( λ × Aµ ) × Aν ] h¯ c The term in square brackets is zero, by virtue of the vector identity A × (B × C) + B × (C × A) + C × (A × B) = 0. So

F

µν

→F

µν

+

2q h¯ c

(λ × Fµν ) . X

Problem 10.14

(a) From the definition (Eq. 10.65) Fµν = ∂µ Aν − ∂ν Aµ −

2q (Aµ × Aν ) h¯ c

we have τ · Fµν = ∂µ (τ · Aν ) − ∂ν (τ · Aµ ) −

2q τ · (Aµ × Aν ) . h¯ c

Now, exploiting the result of Problem 4.20: µ

µ

(τ · Aµ ) (τ · Aν ) − (τ · Aν ) (τ · Aµ ) = Ai Aνj (τi τj − τj τi ) = 2iAi Aνj eijk τk = 2iτ · (Aµ × Aν ) , so iq [(τ · Aµ ) (τ · Aν ) − (τ · Aν ) (τ · Aµ )] . h¯ c Invoking the transformation rule (Eq. 10.58): τ · Fµν = ∂µ (τ · Aν ) − ∂ν (τ · Aµ ) +

h¯ c τ · A0µ = S τ · Aµ S−1 + i (∂µ S)S−1 , q we have τ·F

µν 0

h¯ c µ h¯ c ν −1 ν µ −1 −1 = ∂ S (τ · A ) S + i (∂ S)S − ∂ S (τ · A ) S + i (∂ S)S q q ( h¯ c µ h¯ c ν iq µ −1 −1 ν −1 −1 S (τ · A ) S + i (∂ S)S S (τ · A ) S + i (∂ S)S + h¯ c q q ) h¯ c ν h¯ c µ ν −1 −1 µ −1 −1 − S (τ · A ) S + i (∂ S)S S (τ · A ) S + i (∂ S)S . q q

µ

ν

−1

219

Expanding this out, τ · Fµν 0 = (∂µ S) (τ · Aν ) S−1 + S [τ · (∂µ Aν )] S−1 + S (τ · Aν ) ∂µ S−1 i h¯ c h + i ∂ µ ( ∂ ν S ) S −1 q ν − (∂ S) (τ · Aµ ) S−1 − S [τ · (∂ν Aµ )] S−1 − S (τ · Aµ ) ∂ν S−1 i h¯ c h − i ∂ ν ( ∂ µ S ) S −1 q ( iq + S ( τ · Aµ ) S −1 S ( τ · Aν ) S −1 h¯ c i i¯hc h S ( τ · Aµ ) S −1 ( ∂ ν S ) S −1 + ( ∂ µ S ) S −1 S ( τ · Aν ) S −1 + q 2 h i h¯ c − ( ∂ µ S ) S −1 ( ∂ ν S ) S −1 q

− S ( τ · Aν ) S −1 S ( τ · Aµ ) S −1 i i¯hc h − S ( τ · Aν ) S −1 ( ∂ µ S ) S −1 + ( ∂ ν S ) S −1 S ( τ · Aµ ) S −1 q ) 2 h i h¯ c + ( ∂ ν S ) S −1 ( ∂ µ S ) S −1 q The idea now is to pull out an S on the left and an S−1 on the right, by exploiting the hint: ∂µ S = −S ∂µ S−1 S, ∂ µ S −1 = − S −1 ( ∂ µ S ) S −1 . Then τ · Fµν 0 = iq S τ · ( ∂ µ Aν ) − τ · ( ∂ ν Aµ ) + [(τ · Aµ ) (τ · Aν ) − (τ · Aν ) (τ · Aµ )] S−1 h¯ c " + S − ∂ µ S −1 S ( τ · Aν ) − ( τ · Aν ) S −1 ( ∂ µ S ) + ∂ ν S −1 S ( τ · Aµ ) + ( τ · Aµ ) S −1 ( ∂ ν S ) + ∂ µ S −1 S ( τ · Aν ) − ( τ · Aµ ) S −1 ( ∂ ν S ) # + ( τ · Aν ) S −1 ( ∂ µ S ) − ∂ ν S −1 S ( τ · Aµ ) S −1 i h¯ c h µ ν ( ∂ ∂ S ) S −1 + ( ∂ ν S ) ∂ µ S −1 − ( ∂ ν ∂ µ S ) S −1 − ( ∂ µ S ) ∂ ν S −1 q i h¯ c h − i S − ∂µ S−1 SS−1 (∂ν S) + ∂ν S−1 SS−1 (∂µ S) S−1 , q

+i

220

10 Gauge Theories

or h¯ c h τ · Fµν 0 = S (τ · Fµν ) S−1 + i S ∂ν S−1 SS−1 (∂µ S) q i µ −1 −1 ν − ∂ S SS (∂ S) + ∂µ S−1 (∂ν S) − ∂ν S−1 (∂µ S) S−1

= S (τ · Fµν ) S−1 . X (b) Tr

h i i τ · Fµν 0 τ · Fµν 0 = Tr Sτ · Fµν S−1 Sτ · Fµν S−1 h i = Tr S−1 (Sτ · Fµν ) τ · Fµν = Tr (τ · Fµν ) τ · Fµν . X h

(c) µν µν Tr (τ · Fµν ) τ · Fµν = Fi Fµν j Tr(τi τj ) = Fi Fµν j Tr δij + ieijk τk µν

= Fi Fµν j δij Tr (1) = 2Fµν · Fµν . (I used the fact that the Pauli matrices are traceless, and the trace of the 2 × 2 unit matrix is 2.) Conclusion: the Lagrangian in Eq. 10.63,

L=−

1 µν F · Fµν , 16π

is invariant under finite gauge transformations. [But since a finite transformation can be built up from infinitesimal transformations, it is really unnecessary to check anything beyond the infinitesimal case, which is typically much easier.] X

Problem 10.15

According to Eq. 10.65, F

µν

≡ ∂ A −∂ A − µ

ν

ν

µ

2q h¯ c

(Aµ × Aν ) .

Putting this into the Lagrangian (Eq.10.69): ( 1 µν 1 µ 1 L=− F · Fµν − J · Aµ = − ( ∂ µ Aν − ∂ ν Aµ ) · ∂ µ Aν − ∂ ν Aµ 16π c 16π ) 2 2q 2q µ ν µ ν −2 ∂ µ Aν − ∂ ν Aµ · (A × A ) + (A × A ) · Aµ × Aν h¯ c h¯ c 1 − Jµ · Aµ . c

221

Referring to Eq. 10.17 (if this notation—with a 3-vector in the denominator— disturbs you, write it out in component form): 1 1 2q 1 ∂L =− 2 2 (Aµ × Aν ) = − Fµν . ( ∂ µ Aν − ∂ ν Aµ ) + ∂ ( ∂ µ Aν ) 4π 16π h¯ c 4π 2 i ∂L 2q 1 h 2q 2 = − 2Aµ × (∂µ Aν − ∂ν Aµ ) − 4Aµ × (Aµ × Aν ) ∂Aν 16π h¯ c h¯ c 1 1 2q 1 − Jν = − Aµ × Fµν − Jν . c 4π h¯ c c To isolate Aν for the derivatives, I used the vector identity A · (B × C) = B · (C × A). The Euler-Lagrange equation is ∂µ F

µν

=

2q h¯ c

Aµ × Fµν +

4π ν J . c

Before we cast this in the more familiar nonrelativistic form it is best to express the 3-vectors in index form (we will need the bold face for spatial 3vectors): 2q 4π ν µν µν ∂µ Fi = eijk A j µ Fk + J , h¯ c c i where i runs from 1 to 3, and summation over j and k is implied. With ν = 0 we have (in the notation of Eqs. 7.71, 7.72, and 7.79):

∇ · Ei = −

2q h¯ c

eijk A j · Ek + 4πρi .

For ν = 1, 2, and 3: 1 ∂Ei =− ∇ × Bi − c ∂t

2q h¯ c

4π eijk Vj Ek + A j × Bk + J . c i

These are the inhomogeneous “Maxwell“ equations; the homogeneous equations come from the definition of Fµν : 2q µν µ µ ν ν µ Fi = ∂ Ai − ∂ Ai − eijk A j Aνk . h¯ c For µ = 0 we get 1 ∂Ai Ei = −∇Vi − + c ∂t

2q h¯ c

eijk Vj Ak .

222

10 Gauge Theories

For µ = 1, ν = 2: Fi12 = ∂1 A2i − ∂2 A1i −

− Biz

2q h¯ c

eijk A1j A2k 2q = −∇ x Aiy + ∇y Ai x − eijk A j x Aky . h¯ c

Now eijk A j x Aky = eikj Ak x A j y = −eijk A j y Ak x , so eijk A j x Aky = 21 eijk [ A j x Aky − A j y Ak x ] = 21 eijk A j × Ak

z

.

Evidently Bi = ∇ × Ai + Hence

∇ · Bi =

1 2

2q h¯ c

eijk A j × Ak .

q eijk ∇ · A j × Ak . h¯ c

Meanwhile,

∇ × Ei = −

1 ∂ (∇ × Ai ) + c ∂t

But

∇ × Ai = Bi −

1 2

2q h¯ c

eijk ∇ × (Vj Ak ).

2q h¯ c

1 ∂ A j × Ak + ∇ × (Vj Ak ) 2c ∂t

eijk A j × Ak ,

so

∇ × Ei = −

1 ∂Bi + c ∂t

2q h¯ c

eijk

.

Problem 10.16

This is the 3 × 3 analog to Problem 10.10. 

 a b c H = d e f  ; g h j

 a∗ d∗ g∗ H † = b∗ e∗ h∗  . c∗ f ∗ j∗ 

Evidently H is Hermitian (H † = H) ⇔ a, e, and j are real, and b = d∗ , c = g∗ , f = h∗ , nine (real) numbers in all, and θ, a1 , a2 , . . . , a8 is one way of packaging

223

them:        0 1 0 0 −i 0 1 0 0 100 θ 1 + λ · a = θ 0 1 0 + a1 1 0 0 + a2  i 0 0 + a3 0 −1 0 0 0 0 0 0 0 0 0 0 001         00 1 0 0 −i 0 0 0 0 0 0 + a4 0 0 0 + a5 0 0 0  + a6 0 0 1 + a7 0 0 − i  10 0 i 0 0 0 1 0 0 i 0   1 0 0 1  + a8 √ 0 1 0  3 0 0 −2 √   ( a1 − ia2 )√ ( a4 − ia5 ) ( θ + a3 + a8 / 3) =  ( a1 + ia2 ) (θ − a3 + a8 / 3) a6 − ia7√ ) . ( a4 + ia5 ) ( a6 + ia7 ) (θ − 2a8 / 3) 

This is precisely the Hermitian 3 × 3 matrix, with √ form for the most general ∗ a =√(θ + a3 + a8 / 3), b = a1 − ia2 = d , c =√a4 − ia5 = g∗ , e = (θ − a3 + a8 / 3), f = a6 − ia7 = h∗ , and j = (θ − 2a8 / 3) (which, of course, can be solved to get θ and a in terms of a, b, c, d, e, f ,g, h, and j).

Problem 10.17

(a) First suppose that A is diagonal: d1 0   A= D =0 .  .. 0 

Then



j

d  1 0   j D =0 . . . 0

0 j d2 0 .. . 0

0 d2 0 .. . 0 0 0 j d3 .. . 0

0 0 d3 .. . 0

 ··· 0 ··· 0   ··· 0  . ..  .

· · · dn

 ··· 0  ··· 0   ··· 0  , ..   . j · · · dn

224

10 Gauge Theories

and hence 1 1 e D = 1 + D + D2 + D3 + · · · 2 3!   (1 + d1 + 21 d21 · · · ) 0 ··· 0   0 (1 + d2 + 12 d22 + · · · ) · · · 0   =   .. .. ..   . . . 1 2 0 0 · · · (1 + d n + 2 d n + · · · )  d1  e 0 ··· 0  0 e d2 · · · 0    =  . . ..  .  .. .. .  d 0 0 ··· e n Thus

det e D = ed1 ed2 ed3 · · · edn = ed1 +d2 +d3 +···+dn = eTr( D) . X

Now suppose that A is diagonalizable: d1 0   S−1 AS = D =  0 .  .. 0 

0 d2 0 .. . 0

0 0 d3 .. . 0

 ··· 0 ··· 0   ··· 0  . ..  .

· · · dn

Since Tr( AB) =Tr( BA), it follows that Tr( D ) = Tr(S−1 AS) = Tr(SS−1 A) = Tr( A), and (inserting a factor of SS−1 = 1 between adjacent A’s) S−1 A2 S = (S−1 AS)(S−1 AS) = D2 , S−1 A3 S = (S−1 AS)(S−1 AS)(S−1 AS) = D3 , etc. So S

1 2 1 3 1+ A+ A + A +··· S e S = S 2 3! 1 1 = 1 + D + D2 + D3 + · · · = e D . 2 3!

−1 A

−1

F

But det( AB) = det( A) det( B), so h i h i det S−1 e A S = [det(S−1 )] det e A [det(S)] = [det(S−1 S)] det e A = det e A .

225

Conclusion: det e A = det S−1 e A S = det e D = eTr( D) = eTr( A) . X Finally, the general case. Because S−1 AS = J, it follows from the same argument as before (F) that S −1 e A S = e J ,

and hence

det(e A ) = det(e J ).

Now J has only zeroes above the main diagonal: d1 x   J =x .  .. x 

0 d2 x .. . x

0 0 d3 .. . x

 ··· 0 ··· 0   ··· 0  , ..  .

· · · dn

(where the x’s stand for possibly nonzero elements), and this feature persists for powers of J:   j d1 0 0 · · · 0    x d2j 0 · · · 0    j   J j =  x x d3 · · · 0  , . . . ..   . . . . . . . j x x x · · · dn so



e d1 x   x eJ =   .  .  . x

0 e d2 x .. . x

0 0 e d3 .. . x

 ··· 0 ··· 0    ··· 0 . ..   .  · · · edn

Evaluating each determinant by minors along the top row: e d2 x  det e J = ed1  .  .. x 

0 ··· e d3 · · · .. . x ···

  d 0 e 3 ···  0 d1 d2  .. ..  = e e  . .  x ··· d e n

 0 ..  .  d e n

= ed1 ed2 ed3 · · · edn = ed1 +d2 +d3 +···+dn = eTr( J ) . But Tr( J ) = Tr(S−1 AS) = Tr( A), so det e A = det e J = eTr( J ) = eTr( A) . X

226

10 Gauge Theories

(b) From (a) we know that det eiλ·a = eTr(iλ·a) . But (Problem 10.16) Tr(iλ · a) = i

a8 a3 + √ 3

a a + − a3 + √8 + −2 √8 = 0. 3 3

So eTr(iλ·a) = e0 = 1,

and hence

det eiλ·a = 1. X

Problem 10.18

The first part is identical to Problem 10.11, with τ → λ. Equations 10.80 and 10.81 ⇒ q q ∂µ + i λ · A0µ ψ0 = S ∂µ + i λ · Aµ ψ. [1] h¯ c h¯ c Now, ψ0 = Sψ, so ∂µ ψ0 = (∂µ S)ψ + S(∂µ ψ), and Eq. [1] becomes q q λ · A0µ Sψ = S( ∂µ ψ ) + i S λ · Aµ ψ, (∂µ S)ψ + S( ∂µ ψ )+i h¯ c h¯ c or

q q λ · A0µ S = i S λ · Aµ . h¯ c h¯ c − 1 Multiply by S on the right: q q λ · A0µ = i S λ · Aµ S−1 , ( ∂ µ S ) S −1 + i h¯ c h¯ c or h¯ c λ · A0µ = S(λ · Aµ )S−1 + i ∂ µ S S −1 . X (Eq. 10.82) q

(∂µ S) + i

The second part is the same as Eqs. 10.58-10.61, only with τ → λ and λ → φ. The infinitesimal transformations (up to first order in φ) are: iq S = e−iqλ·φ/¯hc ∼ = 1 − λ · φ, h¯ c

iq S −1 ∼ = 1 + λ · φ, h¯ c

iq ∂µ S ∼ = − λ · ∂µ φ h¯ c

In this approximation Eq. 10.82 becomes iq iq λ · A0µ ∼ = 1 − λ · φ ( λ · Aµ ) 1 + λ · φ h¯ c h¯ c h¯ c iq iq +i − λ · ∂µ φ 1+ λ·φ q h¯ c h¯ c iq ∼ (λ · Aµ )(λ · φ) − (λ · φ)(λ · Aµ ) + λ · (∂µ φ) = λ · Aµ + h¯ c

227

The term in square brackets is a commutator; using Eq. 8.35: h i (λ · Aµ ), (λ · φ) = Aαµ φ β λα , λ β = Aαµ φ β 2i f αβγ λγ = −2iλ · φ × Aµ (summation on α, β, and γ from 1 to 8 implied; in the last step I adopted the generalized dot- and cross-product notation introduced in the text—see Eq. 10.84). Conclusion: 2q 0 ∼ λ · φ × Aµ ] λ · Aµ = λ · Aµ + λ · ( ∂ µ φ ) + h¯ c or 2q 0 ∼ Aµ = Aµ + ( ∂ µ φ ) + φ × Aµ . X (Eq. 10.83) h¯ c

Problem 10.19

In the notation suggested on page 370: h i Tµν = − p2 + (mc)2 gµν + pµ pν pµ pν −1 ( T −1 )µν = 2 g − µν p − (mc)2 (mc)2 Tµλ ( T

−1 λν

)

= = = =

(Eq. 10.92) (Eq. 10.95)

−1 pλ pν λν − p + (mc) gµλ + pµ pλ 2 g − p − (mc)2 (mc)2 λ ν λ ν pµ pλ p p p p − 2 gλν − gµλ gλν − 2 2 (mc) p − (mc) (mc)2 pµ pν pµ pν p µ p ν p2 δµν − − 2 + 2 2 2 (mc) p − (mc) (mc) [ p2 − (mc)2 ] h i pµ pν 2 2 2 2 − p + ( mc ) − ( mc ) + p = δµν . X δµν + (mc)2 [ p2 − (mc)2 ] h

2

2

i

Problem 10.20

There are three free Klein-Gordon Lagrangians (one for each particle type), and one interaction term, of the form iL = −igφ A φB φC (since the vertex factor is −ig): 1 1 m A c 2 2 1 1 m B c 2 2 L = (∂µ φ A )(∂µ φ A ) − φA + (∂µ φB )(∂µ φB ) − φB 2 2 h¯ 2 2 h¯ 2 1 1 mC c + (∂µ φC )(∂µ φC ) − φC2 − gφ A φB φC . 2 2 h¯

228

10 Gauge Theories

[Actually, for Lint to have the correct dimensions (energy/volume—p. 357), with g carrying the units of momentum (p. 213), when the scalar √ fields have the dimensions given on p. 357, we must have Lint = −(1/¯h h¯ c) gφ A φB φC , and now the naive construction of the vertex factor does not get the c’s and h¯ ’s right. See remark before Eq. 10.101.]

Problem 10.21

This describes a theory in which there are three kinds of particles: • a spin- 12 particle of mass m1 , • its antiparticle (likewise mass m1 ), for both of which the propagator is

q + m1 c i 2 , q − ( m1 c )2

and

• a spin-0 particle of mass m2 , for which the propagator is The primitive vertex is

p2

i . − ( m2 c )2

Vertex factor: iαY .

Problem 10.22

Equation 10.119 says φ1 = η + (µ/λ), φ2 = ξ, so ∂µ φ1 = ∂µ η,

∂µ φ2 = ∂µ ξ,

φ12 + φ22 = η 2 + 2η

µ2 µ + 2 + ξ2, λ λ

229

Substitute these into the Lagrangian (Eq. 10.115): 1 1 µ µ2 1 (∂µ η )∂µ η ) + (∂µ ξ )∂µ ξ ) + µ2 η 2 + 2η + 2 + ξ 2 2 2 2 λ λ 2 1 µ µ2 − λ2 η 2 + 2η + 2 + ξ 2 4 λ λ 1 1 1 µ3 1 µ4 + µ2 ξ 2 = ( ∂ µ η ) ∂ µ η ) + ( ∂ µ ξ ) ∂ µ ξ ) + µ2 η 2 + η + 2 2 λ 2 λ2 2 µ2 µ4 µ2 1 µ − λ2 η 4 + 4η 2 2 + 4 + ξ 4 + 4η 3 + 2η 2 2 + 2η 2 ξ 2 4 λ λ λ λ 3 2 µ µ µ + 4η 3 + 4ηξ 2 + 2ξ 2 2 λ λ λ 1 1 1 µ µ 2 2 1 2 2 1 (∂µ η )∂ η ) + (∂µ ξ )∂ ξ ) + η µ = −1− +ξ µ − 2 2 2 2 2 µ3 λ2 4 µ4 1 1 − + η (1 − 1) − η + ξ 4 + 2η 2 ξ 2 + 2 2 4 λ 4 λ 4 µ 1 1 − λµ(η 3 + ηξ 2 ) + 2 − 2 4 λ 1 λ2 4 1 (∂µ η )(∂µ η ) − µ2 η 2 + (∂µ ξ )(∂µ ξ ) − η + ξ 4 + 2η 2 ξ 2 = 2 2 4

L =

− λµ(η 3 + ηξ 2 ) +

µ4 . X 4λ2

Problem 10.23

ψ1 + ψ2 =

√

1 2 η ⇒ η = √ (ψ1 + ψ2 ); 2

ψ1 − ψ2 =

√

1 2 ξ ⇒ ξ = √ (ψ1 − ψ2 ) 2

So

(∂µ η )(∂µ η ) + (∂µ ξ )(∂µ ξ ) 1 = (∂µ ψ1 + ∂µ ψ2 )(∂µ ψ1 + ∂µ ψ2 ) + (∂µ ψ1 − ∂µ ψ2 )(∂µ ψ1 − ∂µ ψ2 ) 2 = (∂µ ψ1 )(∂µ ψ1 ) + (∂µ ψ2 )(∂µ ψ2 ); (η 2 + ξ 2 ) =

i 1h (ψ1 + ψ2 )2 + (ψ1 − ψ2 )2 = ψ12 + ψ22 . 2

230

10 Gauge Theories

Putting this into Eq. 10.120: µ2 1 (∂µ ψ1 )(∂µ ψ1 ) + (∂µ ψ2 )(∂µ ψ2 ) − (ψ1 + ψ2 )2 2 2 4 µλ λ µ4 − √ (ψ1 + ψ2 )(ψ12 + ψ22 ) − (ψ12 + ψ22 )2 + 2 16 4λ 2 2 2 µ2 2 1 µ 2 1 µ µ (∂µ ψ1 )(∂ ψ1 ) − ψ1 + (∂µ ψ2 )(∂ ψ2 ) − ψ2 − µ2 ψ1 ψ2 = 2 2 2 2 4 λ µλ µ4 − √ ψ13 + ψ1 ψ22 + ψ2 ψ12 + ψ23 − ψ14 + ψ24 + 2ψ12 ψ22 + 2 . 16 4λ 2 2

L =

Problem 10.24

L =

µ µ µ 1 1 1 2 ( ∂µ φ1 )( ∂ φ1 ) + 2 ( ∂µ φ2 )( ∂ φ2 ) + 2 ( ∂µ φ3 )( ∂ φ3 ) + 21 µ2 (φ12 + φ22 + φ32 ) − 41 λ2 (φ12 + φ22 + φ32 )2

This is invariant under rotations in φ1 , φ2 , φ3 space. The “potential energy” function is U = − 12 µ2 (φ12 + φ22 + φ32 ) + 14 λ2 (φ12 + φ22 + φ32 )2 and the minima lie on a sphere of radius µ/λ: φ12min + φ22min + φ32min = µ2 /λ2 To apply the Feynman calculus, we expand about a particular ground state (“the vacuum”)—we may as well pick φ1min = µ/λ,

φ2min = 0,

φ3min = 0.

As before, we introduce new fields, η, ξ, and ζ, which are the fluctuations about this vacuum state: η ≡ φ1 − µ/λ,

ξ ≡ φ2 ,

ζ ≡ φ3

Rewriting the Lagrangian in terms of these new field variables, we find : 1 1 1 L = (∂µ η )(∂µ η ) − µ2 η 2 + (∂µ ξ )(∂µ ξ ) + (∂µ ζ )(∂µ ζ ) 2 2 2 h i 2 λ 3 2 2 4 4 4 2 2 2 2 2 2 − µλ(η + ηξ + ηζ ) + η +ξ +ζ +2 η ξ +η ζ +ξ ζ 4

+

µ4 . 4λ2

231

The first term is a free Klein–Gordon Lagrangian (Eq. 10.11) for the field η, √ which evidently carries a mass mη = 2 µ}/c ; the second and third terms are free Lagrangians for the fields ξ and ζ, which are massless: mξ = mζ = 0 ; and the fourth term defines various couplings. Evidently in this theory there are 2 Goldstone bosons .

Problem 10.25

The last term in Eq. 10.129, −(1/16π ) F µν Fµν , is not altered by the transformation in Eq. 10.130. The previous two terms are the same as those in Eq. 10.115, and we have already shown (Problem 10.22) that they become

− µ2 η 2 −

µ4 λ2 4 η + ξ 4 + 2η 2 ξ 2 − λµ(η 3 + ηξ 2 ) + 2 . 4 4λ

It remains only to work out the first term: iq iq ∂µ − Aµ φ∗ ∂µ + Aµ φ h¯ c h¯ c iq q q iq = ∂µ φ1 − i∂µ φ2 − Aµ φ1 − Aµ φ2 ∂µ φ1 + i∂µ φ2 + Aµ φ1 − Aµ φ2 h¯ c h¯ c h¯ c h¯ c q iq ( ( µ µ (∂( ( i (∂( φ2 ) + = (∂µ φ1 )(∂µ φ1 ) + ( (∂ (∂µ φ1 ) Aµ φ2 µφ µ φ1 ) A φ1 − 1 )( h¯ c h¯ c q iq ( µ( µ (∂( ( i (∂( φ1 ) + (∂µ φ2 )(∂µ φ2 ) + (∂µ φ2 ) Aµ φ1 + (∂ −( µφ µ φ2 ) A φ2 2 )( h¯ c h¯ c q 2 q 2 iq µ q µ µ φ1 φ1 Aµ Aµ + i φ (∂φ1 ) Aµ φ1 + (∂ φ2 ) Aµ φ1 + − 1 φ2 Aµ A h¯ c h¯ c h¯ c h¯ c q 2 q 2 q µ iq µ µ − (∂ φ1 ) Aµ φ2 − (∂φ2 ) Aµ φ2 − i φ φ2 φ2 Aµ Aµ 1 φ2 Aµ A + h¯ c h¯ c h¯ c h¯ c 2q = (∂µ φ1 )(∂µ φ1 ) + (∂µ φ2 )(∂µ φ2 ) + −(∂µ φ1 )φ2 + (∂µ φ2 )φ1 Aµ h¯ c q 2 (φ12 + φ22 ) Aµ Aµ + h¯ c 2q h µ i µ = (∂µ η )(∂µ η ) + (∂µ ξ )(∂µ ξ ) + −(∂µ η )ξ + (∂µ ξ ) η + A h¯ c λ q 2 µ µ2 + η 2 + 2η + 2 + ξ 2 Aµ Aµ h¯ c λ λ

232

10 Gauge Theories

Putting it all together, 1 q 1 (∂µ η )(∂µ η ) + (∂µ ξ )(∂µ ξ ) + −(∂µ η )ξ + (∂µ ξ )η Aµ 2 2 h¯ c q 2 µ µ q 1 µ2 2 µ 2 η + 2η + 2 + ξ Aµ Aµ − µ2 η 2 + (∂µ ξ ) A + λ h¯ c 2 h¯ c λ λ 2 λ µ4 1 µν − η 4 + ξ 4 + 2η 2 ξ 2 − λµ(η 3 + ηξ 2 ) + 2 − F Fµν 16π 4 4λ 1 1 = (∂µ η )(∂µ η ) − µ2 η 2 + (∂µ ξ )(∂µ ξ ) 2 2 1 µν 1 q 2 µ2 µ Aµ A + − F Fµν + 16π 2 h¯ c λ2 q 2 µ q + −(∂µ η )ξ + (∂µ ξ )η Aµ + η Aµ Aµ h¯ c h¯ c λ 1 q 2 2 λ2 4 + η + ξ 2 Aµ Aµ − η + ξ 4 + 2η 2 ξ 2 − λµ(η 3 + ηξ 2 ) 2 h¯ c 4 4 µ q µ + (∂µ ξ ) Aµ + 2 . X λ h¯ c 4λ

L =

Problem 10.26

x

A

h h h

h

h

h

x

A

h

A

x

A

A

h

A

x

A

h

x

h

h

x

x

h

h

h

x

x

x x

233

11

Neutrino Oscillations Problem 11.1

To calculate the energy liberated in the creation of a sphere of radius R and uniform density ρ, imagine building it up by layers. When it is at radius r, a mass dm falls in (from infinity), giving up potential energy m dm G 4 G dE = G = ρ πr3 (ρ4πr2 dr ) = (4πρ)2 r4 dr, r r 3 3 so the total energy of the sphere is E =

=

GR5 GR5 (4πρ)2 = 15 15

3M R3

2

=

3 GM2 5 R

3 (6.67 × 10−11 )(1.99 × 1030 )2 J = 2.28 × 1041 J. 5 6.96 × 108

The solar luminosity is 3.85 × 1026 W (all numbers from the Particle Physics Booklet), so the lifetime is 2.28 × 1041 s = 5.91 × 1014 s = 1.87 × 107 yrs . 3.85 × 1026

Problem 11.2

(a) Follow the derivation of Eq. 11.7, only this time assume the rest energy dominates: |p|2 1 |p|2 |p|2 2 2 2 2 4 = mc + E = m c 1 + 2 2 ⇒ E ≈ mc 1 + 2 m2 c2 2m m c 1 m2 − m1 1 2 1 2 2 E2 − E1 = (m2 − m1 )c + |p| − = ( m2 − m1 ) c − T 2 m2 m1 m

≈ ( m2 − m1 ) c2

234

11 Neutrino Oscillations

(where m is the average of the two masses, and T is the kinetic energy). From Eq. 11.6, the period of the oscillations is τ=

2π 6.58 × 10−22 MeV s 2π¯h 2π¯h = = = 1.18 × 10−9 s . E2 − E1 ( m2 − m1 ) c2 3.49 × 10−12 MeV

(b) The lifetime of KS0 is much shorter: 8.94 × 10−11 s; the lifetime of K0L is much longer: 5.17 × 10−8 s. So a beam that starts out pure K0 (say) will become pure K0L long before it has a chance to oscillate to K¯ 0 .

Problem 11.3

(a) e

ne

e

e

ne

e Z

W ne

ne

(b and c) e

e

e

nm

nt

Z

e Z

nm (b)

(c)

nt

Problem 11.4

(a) Q x R

R+h

( R + h)2 = R2 + x2 − 2Rx cos(π − Θ) ⇒ R2 + 2Rh + h2 = R2 + x2 + 2Rx cos Θ;

235

x2 + (2R cos Θ) x − (2Rh + h2 ) = 0 ⇒ x =

√

R2 cos Θ2 + 2Rh + h2 − R cos Θ .

(We need the positive root, since x > 0.) (b)

Problem 11.5

(a) E = γmc2 = √

mc2 1 − v2 /c2

⇒ 1−

v 2 c

=

mc2 E

2

q ⇒ v = c 1 − (mc2 /E)2 .

Ultrarelativistic ⇒ E mc2 , so (keeping the first term in the binomial expansion) " " 2 # 2 # 1 mc2 1 1 1 mc2 v ≈ c 1− , . ≈ 1+ 2 E v c 2 E (b) The time it takes a neutrino to reach earth is t = L/v, so the difference in arrival times is " 2 2 # 1 L 1 mc2 1 mc2 1 τ = t2 − t1 = L ≈ 1+ −1− − v2 v1 c 2 E2 2 E1 ! L 1 L 1 ( E − E2 )( E1 + E2 ) . = − 2 = (mc2 )2 1 (mc2 )2 2c 2c E22 E1 E12 E22 s

2

2cτ L( E1 − E2 )( E1 + E2 ) s 2c(10) = (20)(30) MeV = 52 eV . 5 (1.7 × 10 )c(365 · 24 · 60 · 60)(10)(50)

mc = E1 E2

[Statistical analysis lowers this number a bit.]

Problem 11.6

(a)

236

11 Neutrino Oscillations

Wm

(b) ∆E ∆t ≥

nm

ne

e

h¯ h¯ 6.58 × 10−22 ⇒ ∆t = = = 4.09 × 10−27 s . 2 2MW c2 2 × 80.4 × 103

d = (4.09 × 10−27 )(3 × 108 ) = 1.23 × 10−18 m . In the time available, the neutrino travels nowhere near far enough for oscillations to occur. So this decay is still effectively forbidden.

237

12

What’s Next Problem 12.1

(a) Equation 10.132 gives the mass of the gauge particle under the Higgs mechanism: r √ qµ √ v gw h¯ c gw v √ = mA = 2 π ⇒ MW = 2 π 2 h¯ c . 2 4π λc c 2 2c2 Thus

√ 2MW c2 h¯ c v = . X gw

¯ (b) From Problem 10.21, the Yukawa coupling Lint = −αY ψψφ yields a naive vertex factor iαY . When the scalar field φ shifts, according to Eq. 10.130, φ = η + µ/λ = η + v, the Yukawa coupling generates a mass term for the quark or lepton field ψ: ¯ ¯ −αY ψψv = −mc2 ψψ

⇒

αY =

mc2 v

plus a residual Yukawa coupling whose vertex factor is evidently imc2 /v . [As I mentioned before Eq. 10.101, this procedure doesn’t get the constants √ right; the correct vertex factor is −imc2 /(v h¯ c).] (c) The relevant terms in Eq. 10.136 are µ q 2 µν 3 i Lint = i g η Aµ Aν − λµη . λ h¯ c The first term gives the hWW and hZZ couplings, with the vertex factor (see remark before Eq. 10.101) i

( Mm c2 )2 µν 4π µ q 2 µν g g = i h¯ c λ h¯ c v(h¯ c)3

238

12 What’s Next

(where Mm stands for the mass of√the W or the Z, as the case may be). [The correct vertex factor is i (2M2 c3 /v h¯ c) gµν .] The second term gives the hhh coupling, with vertex factor−iλµ = −iµ2 /v. But (comparing Eqs. 10.11 and 10.136) the mass of the Higgs is given by µ2 = 1 2 2 ( m h c/¯h ) , so the hhh vertex factor is

−i

1 m h c 2 . 2v h¯

[Actually, −3im2h c3 /v.]

Problem 12.2

p1

p2

h p3

f

f

√ (a) The vertex factor (Problem 12.1) is imc2 /v h¯ c, where m is the mass of the quark or lepton, so imc2 mc2 √ M = i v¯ (3) u(2) = − √ [v¯ (3)u(2)] . v h¯ c v h¯ c

2

h|M| i =

=

mc2 √ v h¯ c

2

mc2 √ v h¯ c

2 h

Tr [( p2 + mc)( p3 − mc)] 2 h i i mc2 √ Tr( p2 p3 − (mc)2 Tr(1) = 4 p2 · p3 − (mc)2 . v h¯ c

In the CM, 2 2 E E E E p2 = , p , p3 = , −p , p2 = − (mc)2 , p2 · p3 = + p2 , c c c c so 2

h|M| i = 4

mc2 √ v h¯ c

# 2 " 2 E 2 2 − (mc) . c

Conservation of energy says E = (mh c2 /2), so 2 2 mc2 m h c 2 mc3 √ √ h|M|2 i = 8 − (mc)2 = 2 m2h − 4m2 . 2 v h¯ c v h¯ c

239

The decay rate is (Eq. 6.35 with |p| =

Γ=

c 2

p

( E/c)2 − (mc)2 = (c/2)

q

m2h − 4m2 ):

q

2 3/2 m2h − 4m2 mc3 2 1 mc3 2 2 2 2 √ √ m − 4m = m − 4m 2 h h 8π¯hm2h c 8π¯hm2h v h¯ c v h¯ c

Or, using Eq. 12.1 to express v in terms of MW : 2 3/2 3/2 2 m2 c2 gw mc3 gw 2 2 2 2 m − 4m = m − 4m . h h 2 2MW c2 32π¯hm2h MW √ Finally, writing gw = 4παw : Γ=

1 8π¯hm2h

αw Γ= m c2 8¯h h

m MW

2 "

1−

2m mh

2 #3/2 .

(b) 3/2 1 − (2mb /mh )2 1 − (2mc /mh )2 3/2 2 4.3 1 − (8.6/120)2 = = (12.8)(0.9953)3/2 = 13 . 1.2 1 − (2.4/120)2

Γ(bb¯ ) = Γ(cc¯)

mb mc

2

3/2 1 − (2mb /mh )2 1 − (2mτ /mh )2 3/2 2 4.3 1 − (8.6/120)2 =3 = 3(5.84)(0.9958)3/2 = 17 . 1.78 1 − (3.55/120)2

Γ(bb¯ ) =3 Γ(τ + τ − )

mb mτ

2

Problem 12.3

p

p1 h

2

p3

W,Z W,Z

The vertex factor is (Problem 12.1): M 2 c3 2i √ gµν , v h¯ c

240

12 What’s Next

where M is the mass of the W or the Z, as the case may be. The amplitude is i 2M2 c3 µν ∗ 2M2 c3 h ∗ ∗ eµ (2) e ∗ µ (3) . M = ieµ (2) i √ g eν (3) = − √ v h¯ c v h¯ c 2h i 2M2 c3 √ |M|2 = eµ∗ (2)e∗ µ (3) [eν (2)eν (3)] . v h¯ c Summing over the outgoing spins, using Eq. 9.158: " # 2 µ p2 µ p2 ν p3 p3ν 2M2 c3 2 µν √ h|M| i = −g + − gµν + ( Mc)2 ( Mc)2 v h¯ c 2 p3 · p3 ( p2 · p3 )2 2M2 c3 p2 · p2 √ − + = 4− ( Mc)2 ( Mc)2 ( Mc)4 v h¯ c 2 h 2 i 2M2 c3 ( p2 · p3 )2 2c 4 2 √ √ 2 ( Mc ) 2+ = + ( p · p ) = . 2 3 ( Mc)4 v h¯ c v h¯ c In the CM frame E p2 = ,p , c

2 1 E E − ( Mc)2 , E = mh c2 , −p , p2 = c c 2 2 2 E c2 E + p2 = 2 − ( Mc)2 = (m2h − 2M2 ), p2 · p3 = c c 2

p3 =

i c4 4 mh − 4m2h M2 + 4M4 2( Mc)4 + ( p2 · p3 )2 = 2( Mc)4 + 4 c4 4 mh − 4m2h M2 + 12M4 , = 4 3 2 c √ h|M|2 i = m4h − 4m2h M2 + 12M4 . v h¯ c q p Put this into Eq. 6.35, with |p| = (mh c/2)2 − ( Mc)2 = (c/2) m2h − 4M2 , h

q

m2h − 4M2 c3 2 4 2 2 4 √ m − 4m M + 12M Γ = h h 8π¯hm2h c v h¯ c s " # m3h c5 M 2 M 4 2M 2 = S 1−4 + 12 1− . mh mh mh 16π¯h2 v2 √ Or, using Eq. 12.1 with gw = 4παw : S(c/2)

α w m h c2 Γ= S 16¯h

mh MW

2 "

1−4

M mh

2

+ 12

M mh

4 #

s

1−

2M mh

2 .

241

The statistical factor S is 1 for h → W + + W − , and 1/2 for h → Z + Z (because in this case there are two identical particles in the final state). (b) r 2 MW 4 MW 2 W + 12 m 1 − 2M 1−4 m mh h h Γ(WW ) r =2 2 4 2 . Γ( ZZ ) MZ 2MZ Z 1−4 M + 12 1 − m m m h

h

h

With MW = 80, MZ = 91, and mh = 200, Γ(WW ) = 2.8 . Γ( ZZ )

Problem 12.4

The Super-K tank holds 50,000 tons, or

(5 × 104 tons)(2000 lbs/ton)((4.45 N/lb)

1 9.8 m/s2

= 4.5 × 107 kg.

Roughly half of the mass is neutrons, but let’s imagine it is entirely protons; then the number of protons in the tank is 4.5 × 107 kg = 2.7 × 1034 protons. 1.67 × 10−27 kg Suppose the experiment is to run for one year, and in that time we hope to see 5 proton decays. The decay rate is 1/τ, so 5=

1 2.7 × 1034 (2.7 × 1034 )(1 yr) ⇒ τ = yr = 5 × 1033 yr . τ 5

The current limit is 1033 years, so we are very close to the practical limit on such experiments.

Problem 12.5

• Decay of the muon (Eq. 9.34):

• Decay of the neutron (Eq. 9.60):

4 ( m c2 )5 1 gw µ . 12(8π )3 h¯ ( MW c2 )4 4 ( m c2 )5 m − m 5 1 gw n p e Γ∼ . me 4π 3 24 h¯ ( MW c2 )4

Γ∼

242

12 What’s Next

• Semileptonic decays (Ex. 9.3):

Γ∼

4 ( ∆mc2 )5 gw 1 . 30π 3 24 h¯ ( MW c2 )4

The pion decay formula (Eq. 9.76) is not so helpful here, since we don’t know f π , but apart from numerical factors (including gw ), the generic formula seems to be ( Mc2 )4 τ ∼ h¯ , (mc2 )5 where m is the mass of the decaying particle and M is the mass of the mediator responsible. For proton decay, mc2 ≈ 1 GeV, Mc2 ≈ 1016 GeV, so τ ∼ 10−24

(1016 )4 = 1040 s, (1)5

or, since there are (365)(24)(60)(60) = 3 × 107 seconds in a year, τ ∼ 1032 yr . Obviously, tweaking the numerical factors could change this by several orders of magnitude, but essential point remains: this is an extremely long lifetime.

Problem 12.6

|M |2 − |M˜|2 = |M1 + M2 |2 − |M˜1 + M˜2 |2 = |M1 |2 + |M2 |2 + M1 M2∗ + M1∗ M2 − |M˜1 |2 − |M˜2 |2 − M˜1 M˜2∗ − M˜1∗ M˜2 = |M1 |eiφ1 eiθ1 |M2 |e−iφ2 e−iθ2 + |M1 |e−iφ1 e−iθ1 |M2 |eiφ2 eiθ2 −|M1 |eiφ1 e−iθ1 |M2 |e−iφ2 eiθ2 − |M1 |e−iφ1 eiθ1 |M2 |eiφ2 e−iθ2 n h i h io = |M1 ||M2 | ei(θ1 −θ2 ) ei(φ1 −φ2 ) − e−i(φ1 −φ2 ) + e−i(θ1 −θ2 ) e−i(φ1 −φ2 ) − ei(φ1 −φ2 ) h i = 2i |M1 ||M2 | sin(φ1 − φ2 ) ei(θ1 −θ2 ) − e−i(θ1 −θ2 ) = −4|M1 ||M2 | sin(φ1 − φ2 ) sin(θ1 − θ2 ). qed

243

Problem 12.7

y a'

a'y

a

ay q f

a'x

ax

x

(a) a x = a cos φ,

ay = a sin φ

a0x = a cos(θ + φ) = a [cos θ cos φ − sin θ sin φ] = cos θ a x − sin θ ay ,

X

a0y

X

= a sin(θ + φ) = a [sin θ cos φ + cos θ sin φ] = sin θ a x + cos θ ay ,

(b) a0 · b0 = a0x bx0 + a0y by0 = (cos θ a x − sin θ ay )(cos θ bx − sin θ by )

+ (cos θ ay + sin θ a x )(cos θ by + sin θ bx ) 2

= a x bx (cos θ + sin2 θ ) + a x by (− cos θ sin θ + sin θ cos θ ) + ay bx (− cos θ sin θ + sin θ cos θ ) + ay by (cos2 θ + sin2 θ ) = a x b x + a y by . X (c) sin(dθ ) ≈ dθ,

cos(dθ ) ≈ 1,

a0x = a x − (dθ ) ay , a0y = ay + (dθ ) a x .

(d) a0x bx0 + a0y by0 = a x − (dθ ) ay bx − (dθ )by + ay + (dθ ) a x by + (dθ )bx i h = a x bx 1 + (dθ )2 + a x by [−dθ + dθ ] + ay bx [dθ − dθ ] i h i h + ay by 1 + (dθ )2 = ( a x bx + ay by ) 1 + (dθ )2 = ( a x bx + ay by ) X (to first order in dθ).

244

12 What’s Next

Problem 12.8

(a) ¯ δφ∗ = (δφ)∗ = (2e† γ0 ψ)† = 2ψ† (γ0 )† e = 2ψ† γ0 e = 2ψe †

(recall that γ0 = γ0 ). Similarly δψ¯ = δ(ψ† γ0 ) = (δψ)† γ0 = (i/¯hc)(γµ e∂µ φ)† γ0 ¯ µ ∂µ φ∗ = (i/¯hc)e† γµ † γ0 ∂µ φ∗ = (i/¯hc)(e† γ0 )γ0 γµ † γ0 ∂µ φ∗ = (i/¯hc)eγ (recall that γ0 γ0 = 1 and γ0 γµ † γ0 = γµ —see Problem 7.15). (b) 1 µ ¯ )(∂µ φ) + (∂µ φ∗ )∂µ (eψ ¯ ) ∂ (δφ∗ )(∂µ φ) + (∂µ φ∗ )∂µ (δφ) = ∂µ (ψe 2 = (∂µ φ)(∂µ ψ¯ )e + e¯ (∂µ φ∗ )(∂µ ψ).

δL1 =

(c) ¯ µ ∂µ (δψ) = −(eγ ¯ µ ∂µ (γν e∂ν φ) ¯ ν ∂ν φ∗ )γµ ∂µ ψ + ψγ δL2 = i¯hc (δψ¯ )γµ (∂µ ψ) + ψγ ¯ µ γν (∂µ ∂ν φ)e. = −e¯ (∂ν φ∗ )γν γµ (∂µ ψ) + ψγ Now, γµ γν ∂µ ∂ν can be written more symmetrically as (1/2)(γµ γν + γν γµ )∂µ ∂ν = gµν ∂µ ∂ν = ∂2 , the d’Alembertian. So δL2 = −e¯ (∂ν φ∗ )γν γµ (∂µ ψ) + ∂µ [ψ¯ (∂µ φ)e] − (∂µ φ)(∂µ ψ¯ )e. The third term on the right is −δL1 + e¯ (∂µ φ∗ )(∂µ ψ), so δL2 = −δL1 + ∂µ [ψ¯ (∂µ φ)e] + e¯ (∂µ φ∗ )(∂µ ψ) − (∂ν φ∗ )γν γµ (∂µ ψ) . The term in curly brackets can be written as

(∂ν φ∗ ) [ gµν − γν γµ ] (∂µ ψ) = −iσµν (∂ν φ∗ )(∂µ ψ) (I used the anticommutation relation for the Dirac matrices, Eq. 7.15, and the definition of σµν , Eq. 7.69). But ∂µ {σµν [φ∗ (∂ν ψ) − (∂ν φ∗ )ψ]} = σµν (∂µ φ∗ )(∂ν ψ) + φ∗ (∂µ ∂ν ψ) − (∂µ ∂ν φ∗ )ψ − (∂ν φ∗ )(∂µ ψ)

= −2σµν (∂ν φ∗ )(∂µ ψ) (since σµν is antisymmetric, the contraction with ∂µ ∂ν is zero, and σµν tµν = −σµν tνµ for any tensor tµν ). Putting this together, we have i ¯ µ {σµν [φ∗ (∂ν ψ) − (∂ν φ∗ )ψ]} δL2 = −δL1 + ∂µ [ψ¯ (∂µ φ)e] + e∂ 2 = −δL1 + ∂µ Qµ .

245

(d)

mc 2 1 ¯ + φ∗ eψ ¯ ). (ψeφ δL3 = − [(δφ∗ )φ + φ∗ (δφ)] = − 2 h¯ mc ¯ µ (∂µ φ)e . ¯ µ (∂µ φ∗ )ψ − ψγ δL4 = −mc2 [(δψ¯ )ψ + ψ¯ (δψ)] = −i eγ h¯ ¯ (e) Using Eq. 10.15 (and its adjoint) to replace the derivatives of ψ (and ψ), mc ¯ µ e(∂µ φ) ¯ µ (∂µ φ∗ )ψ − eγ ¯ µ φ∗ (∂µ ψ) + (∂µ ψ¯ )γµ eφ + ψγ ∂µ Rµ = i −eγ h¯ mc 2 ¯ ] ¯ µ e(∂µ φ) − mc [eφ ¯ ∗ ψ + ψeφ ¯ µ (∂µ φ∗ )ψ + ψγ = i −eγ h¯ h¯ = δL3 + δL4 .

Problem 12.9

(a) c = 3.00 × 108 m/s; h¯ = 1.055 × 10−34 kg m2 /s; G = 6.67 × 10−11 m3 /(kg s2 ). The dimensions of c ph¯ q Gr are m p kg m2 q m3 r = m p+2q+3r s− p−q−2r kgq−r , s s kg s2 so for l P we want q = r;

− p − q − 2r = 0 ⇒ p = −3r;

p + 2q + 3r = 1 ⇒ q = 1/2.

Evidently l P = c−3/2h¯ 1/2 G1/2 =

q

h¯ G c3

= 1.6 × 10−35 m .

For t P we want q = r;

p + 2q + 3r = 0 ⇒ p = −5r;

− p − q − 2r = 1 ⇒ r = 1/2.

Evidently t P = c−5/2h¯ 1/2 G1/2 =

q

h¯ G c5

= 5.38 × 10−44 s .

For m P we want q − r = 1 ⇒ q = r + 1;

p + 2q + 3r = 0 ⇒ p = −5r − 2;

− p − q − 2r = 0 ⇒ 2r + 1 = 0 ⇒ r = −1/2.

246

12 What’s Next

Evidently m P = c1/2h¯ 1/2 G −1/2 = EP = m P c2 = 1.96 × 109 J =

q

c¯h G

= 2.18 × 10−8 kg .

1.96 × 109 eV = 1.23 × 1019 GeV . 1.60 × 10−19

(b) Coulomb (F = q1 q2 /r2 ) → Newton (F = Gm1 m2 /r2 ) ⇒ e2 → Gm2 , so r  q  (6.67×10−11 )(9.11×10−31 )2 = 4.18 × 10−23 (i) −34 8 Gm2 αG = = q (1.055×10 )(3.00×10 ) h¯ c   G c¯h = 1 (ii) h¯ c G

Problem 12.10

GMm v2 ⇒ v= = m r r2

q

GM r

.

Problem 12.11

Escape velocity: 1 2 Mm 2G 4 8π mv − G = 0 ⇒ v2 = πR3 ρ = GR2 ρ = H 2 R2 . 2 R R 3 3 So ρ=

3H 2 . 8πG

The Particle Physics Booklet gives H = 7.5 × 10−11 /yr = 2.4 × 10−18 /s, so ρ=

3(2.4 × 10−18 )2 = 1.0 × 10−26 kg/m3 . 8π (6.67 × 10−11 )

(For comparison, the density of water is 1000 kg/m3 .)

247

A

The Dirac Delta Function Problem A.1

(a) 2(12 ) + 7(1) + 3 = 12 . (b)

(π is outside the domain of integration.)

Zero

Problem A.2

g( x ) =

p

x2 + 1 − x − 1;

g0 ( x ) =

x 1 1 √ 2x − 1 = √ −1 2 x2 + 1 x2 + 1

p

x2 + 1 = x + 1 ⇒ x2 + 1 = x2 + 2x + 1 ⇒ x = 0; √ g0 (0) = −1; | g0 (0)| = 1, so δ( x2 + 1 − x − 1) = δ( x ) .

Zero of g :

Problem A.3

g( x ) = sin x =⇒ g0 ( x ) = cos x g( x ) = 0 =⇒ sin x = 0 =⇒ x = nπ

(n = 0, ±1, ±2, . . .)

g0 (nπ ) = cos nπ = (−1)n =⇒ | g0 (nπ )| = 1. ∞

δ(sin x ) =

∑

n=−∞

δ( x − nπ ) .

248

A The Dirac Delta Function

!3 Π

!2 Π

!Π

Π

0

2Π

3Π

Problem A.4

g( x ) ≡ x2 − 2x + y = 0 ⇒ x± = 1 ±

p

1 − y. p g ( x± ) = 2( x± − 1) = ± 1 − y;

0

0

g ( x ) = 2x − 2;

p | g0 ( x± )| = 2 1 − y.

1 δ( g( x )) = p [δ( x − x+ ) + δ( x − x− )] 2 1−y p p 0 < x± < 2 ⇔ 0 < 1 ± 1 − y < 2, or −1 < ± 1 − y < 1. But if a is in the range −1 p < a < 1, so too is − a, so x± are both in the domain of the integral provided 1 − y < 1, which is to say 0 < y < 1. Conclusion:

f (y) ≡

Z 2 0

( δ( g( x )) dx =

1/

p

1 − y,

0,

0

Introduction to Elementary Particles Instructor’s Solution Manual

Solution manual for Introduction to Elementary Particles

Introduction to elementary particles