Quantum Field Theory

Quantum Field Theory1 R. Clarkson Dr. D. G. C. McKeon2 January 13, 2003 1 Notes 2 email: taken by R. Clarkson for Dr...

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Quantum Field Theory1 R. Clarkson

Dr. D. G. C. McKeon2 January 13, 2003

1 Notes 2 email:

taken by R. Clarkson for Dr. McKeon’s Field Theory (Parts I and II) Class. [email protected]

2

Contents 1 Constraint Formalism 1.1 Principle of Least action: . . . . . . . 1.2 Hamilton’s Equations . . . . . . . . . 1.3 Poisson Brackets . . . . . . . . . . . 1.4 Dirac’s Theory of Constraints . . . . 1.5 Quantizing a system with constraints

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2 Grassmann Variables 2.1 Integration . . . . . . . . . . . . . . . . . . . 2.2 Poisson Bracket . . . . . . . . . . . . . . . . 2.3 Quantization of the spinning particle. . . . . 2.4 General Solution to the free Dirac Equation 2.5 Charge Conjugation . . . . . . . . . . . . . . 2.6 Majorana Spinors . . . . . . . . . . . . . . . 2.7 Time Reversal . . . . . . . . . . . . . . . . .

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7 7 8 9 11 24

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27 28 30 32 50 56 57 58

3 Bargmann-Wigner Equations

61

4 Gauge Symmetry and massless spin one particles 4.1 Canonical Hamiltonian Density . . . . . . . . . . . . . . . . . . . . . . . . .

71 74

5 (2nd ) Quantization, Spin and Statistics 5.1 Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Feynman Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Quantizing the Dirac Field . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83 83 88 92

6 Interacting Fields 97 6.1 Gauge Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 6.2 Heisenberg Picture of Q.M. . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 6.3 Wick’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 7 Electron-Positron Scattering

105 3

4

CONTENTS

8 Loop Diagrams 8.1 Feynman Rules in Momentum Space 8.2 Combinatoric Factors . . . . . . . . . 8.3 Cross Sections From Matrix elements 8.4 Higher order corrections . . . . . . . 8.5 Renormalization . . . . . . . . . . . . 8.6 Regularization . . . . . . . . . . . . . 8.7 Noether’s Theorem . . . . . . . . . . 9 Path Integral Quantization 9.1 Heisenberg-Dirac . . . . . . . . . . . 9.2 Wave Functions . . . . . . . . . . . . 9.3 Free Particle . . . . . . . . . . . . . . 9.4 Feynman Rules . . . . . . . . . . . . 9.5 Path Integrals for Fermion Fields . . 9.6 Integration over Grassmann Variables 9.7 Gauge Invariance . . . . . . . . . . . 10 Quantizing Gauge Theories 10.1 Quantum Mechanical Path Integral 10.2 Gauge Theory Quantization . . . . 10.3 Feynman Rules . . . . . . . . . . . 10.4 Radiative Corrections . . . . . . . . 10.5 Divergences at Higher orders . . . . 10.5.1 Weinberg’s Theorem . . . . 10.6 Renormalization Group . . . . . . . 10.6.1 Euler’s Theorem . . . . . . 10.6.2 Explicit Calculations . . . .

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11 Spontaneous Symmetry Breaking 11.1 O(2) Goldstone model: . . . . . . . . . . . 11.2 Coleman Weinberg Mechanism . . . . . . . 11.3 One loop Effective Potential in λφ4 model 11.4 Dimensional Regularization . . . . . . . . 11.5 Spontaneous Symmetry Breaking in Gauge 12 Ward-Takhashi-Slavnov-Taylor Identities 12.1 Dimensional Regularization with Spinors . 12.1.1 Spinor Self-Energy . . . . . . . . . 12.2 Yang-Mills Theory . . . . . . . . . . . . . 12.3 BRST Identities . . . . . . . . . . . . . . . 12.4 Background Field Quantization . . . . . .

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109 109 113 115 119 122 123 131

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141 141 142 147 157 160 160 164

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169 169 172 173 174 185 185 190 195 198

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207 212 215 216 220 225

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229 234 235 238 240 242

CONTENTS

5

13 Anomalies 249 13.1 Anomalies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 14 Instantons 257 14.1 Quantum Mechanical Example . . . . . . . . . . . . . . . . . . . . . . . . . . 257 14.2 Classical Solutions to SU (2) YM field equations in Euclidean Space . . . . . 261

6

CONTENTS

Chapter 1 Constraint Formalism 1.1

Principle of Least action:

Configuration space: (q1 , q2 , . . . , qn )

with q (i) (ti ) & q (f ) (tf ) fixed, the classical path is the one that minimizes S = |{z}

action

Z

tf

L(qi (t), q˙i (t))dt (L = Lagrangian)

(1.1.1)

ti

i.e. if qi (t) = qiclassical (t) + ε δq(t). As S = S(ε) has a minimum at ε = 0, dS(0) = 0 dε {δq(ti ) = 0 } {δq(tf ) = 0} 7

(1.1.2)

8

CHAPTER 1. CONSTRAINT FORMALISM dS(0) = 0 dε Z =

µ

¶ ∂L ∂L dt δqi + δ q˙i ∂qi ∂ q˙i ti Int. by parts ¶ Z tf µ ∂L d ∂L = dt − δqi ∂qi dt ∂ q˙i ti tf

and, as δqi is arbitrary, µ ¶ ∂L d ∂L = ∂qi dt ∂ q˙i

1.2

(1.1.3)

Hamilton’s Equations

(Legendre transforms) ∂L ∂ q˙i H = pi q˙i − L(qi , q˙i ) pi =

(1.2.1) (1.2.2)

Note that H does not explicitly depend on q˙i . i.e. ∂H ∂ q˙i

= pi −

∂L ∂ q˙i

= 0 Thus H = H(pi , qi ). Now: dH =

∂H ∂H dqi + dpi ∂qi ∂pi |{z} |{z} A

(1.2.3)

B

∂L ∂L = pi dq˙i +q˙i dpi − dqi − dq˙i | {z } ∂qi ∂ q˙i | {z } cancels cancels

but if we insert (1.2.1) → (1.1.3), we get

∴ dH =

∂L ∂qi

=

d (p ) dt i

q˙i dpi − p˙ i dqi |{z} |{z} A

From (1.2.3) and (1.2.4), we have:

= p˙ i ,

∂H ∂pi ∂H = − ∂qi

(1.2.4)

B

q˙i =

(1.2.5)

p˙i

(1.2.6)

1.3. POISSON BRACKETS

1.3

9

Poisson Brackets A = A(qi , pi ) B = B(qi , pi ) {A, B}P B =

So, we have

(1.3.1) (1.3.2)

X µ ∂A ∂B i

∂A ∂B − ∂qi ∂pi ∂pi ∂qi

¶

(1.3.3)

q˙i = {qi , H} p˙i = {pi , H}

(1.3.4) (1.3.5)

d A(qi (t), pi (t)) = {A, H} dt

(1.3.6)

We can generalize this:

−→ Suppose in defining

pi =

∂L ∂ q˙i

(1.2.1)

(1.3.7)

we cannot solve for q˙i in terms of pi . i.e. in H = pi q˙i − L(qi , q˙i ) ex: With S.H.O. m 2 k 2 q˙ − q 2 2 ∂L p = = mq˙ ∂ q˙

L =

→∴ H = pq˙ − L ³ p ´ · m ³ p ´2 kq 2 ¸ − = p − m 2 m 2 2 2 p kq = − 2m 2

i.e. can solve for q˙ in terms of p here. Suppose we used the cartesian coord’s to define system m m L = a˙ 2 + b˙ 2 + λ1 (a˙ − b) + λ2 (b˙ + a) 2 2 i.e. (a(t), b(t)) = (cos(θ(t)), sin(θ(t))) ˙ ˙ ˙ ∴ (a(t), ˙ b(t)) = (− sin(θ(t)) θ(t), cos(θ(t)) θ(t))

10

CHAPTER 1. CONSTRAINT FORMALISM

a2 + b 2 = 1

ex.

scale θ˙ = const. = 1

∴ a˙ = −b / b˙ = a

Dynamical Variables: a, b, λ1 , λ2 ½ ¾ d ∂L ∂L a˙ − b i = 1 = =0= ˙ (λ1 , λ2 ) → Lagrangian Multipliers ∂λi dt ∂ λ˙ i b+a i=2 The trouble comes when we try to pass to Hamiltonian;

pλ1 pλ2

∂L pi = ∂ q˙i ¾ =0 → Cannot solve for λ˙ i in terms of pλi , because these are 2 constraints =0

i.e. if pi = cannot be solved, then

∂L ∂ q˙i

(1.3.8)

∂2L ∂pi = cannot be inverted. ∂ q˙j ∂ q˙i ∂ q˙j

In terms of Lagrange’s equations: d ∂L ∂L ∂2L ∂2L (qi , q˙i ) = = q¨j + q˙j dt ∂ q˙i ∂qi ∂ q˙i ∂ q˙j ∂ q˙i ∂qj ∂2L ∂2L ∂L So q¨j = − q˙j + ∂ q˙i ∂ q˙j ∂ q˙i ∂qj ∂qi } | {z

(1.3.9)

∗

∗ → qi , q˙i specified at t = to

1 1 ∴ qi (t0 + δt) = qi (t0 ) + q˙i (t0 ) δt + qï (t0 ) (δt)2 + | {z } | {z } 2! 3! given

given

2

µ

¶ d3 qi (t0 ) (δt)3 + . . . dt3

Can solve for qï (t0 ) if we can invert ∂ ∂q˙i ∂Lq˙j . Thus, if a constraint occurs, qï (t0 ) cannot be determined from the initial conditions using

1.4. DIRAC’S THEORY OF CONSTRAINTS

11

Lagrange’s equations. i.e. from our example, ¨ i (t0 ) λi (t0 + δt) = λi (t0 ) + . . . + λ |{z} ∗

∗ → cannot be determined.

1.4

Dirac’s Theory of Constraints

implies a constraint χi (qi , pi ) = 0 (i.e. from our example pλ1 = pλ2 = 0), then we If pi = ∂∂L q˙i can define H0 = pi q˙i − L(qi , q˙i ) Constraints hold here (1.4.1) (So, for our example, (scaling m = 1), 1 2 ˙2 (a˙ + b ) + λ1 (a˙ − b) + λ2 (b˙ + a) 2 = 0 = p λ2 = a˙ + λ1 = b˙ + λ2

L = pλ1 pa pb

=0

∴ H0

z }| { = pa a˙ + pb b˙ + pλi λ˙ i −L

¤ 1£ (pa − λ1 )2 + (pb − λ2 )2 − 2 λ1 [pa − λ1 − b] − λ2 [pb − λ2 + a])

= pa (pa − λ1 ) + pb (pb − λ2 ) − The constraints must hold for all t, thus

d χi (q, p) = 0 dt = {χi , H}P B ≈ 0 zero if χi = 0 (“weakly” equal to zero)

(1.4.2)

where H = H0 + ci χi . Sept. 15/99 So, we have: H0 = pi q˙i − L(q, q) ˙ H = H0 + ci χi (qi , pi ) This consistency condition could lead d χi = [χi , H0 + ci χi ] ≈ 0 → dt to some additional constraints The constraints coming from the definition pi =

∂L ∂ q˙i

are called primary constraints.

(1.4.3) (1.4.4) (1.4.5)

12


Additional constraints are called secondary. We could in principle also have tertiary constraints, etc.. (In practice, tertiary constraints don’t arise). Suppose we have constraints χi . They can be divided into First class and Second class constraints. first class constraints → label φi second class constraints → label θi

(1.4.6) (1.4.7)

For a first class constraint, φi , [φi , χj ] ≈ 0 = αkij xk (for all j). θi is second class if it is not first class. We know that H = H 0 + c i χi = H 0 + a i φ i + b i θi d χi = [χi , H] ≈ 0 dt d Thus φi = [φi , H0 + aj φj + bj θj ] dt = [φi , H0 ] + aj [φi , φj ] + φj [φi , aj ] + bj [φi , θj ] + θj [φi , bj ] ≈ 0 ≈ [φi , H0 ] (true for any ai , bj ) ≈0

≈0

z }| { z }| { d θi = [θi , H0 ] + aj [θi , φj ] + φj [θi , aj ] +bj [θi , θj ] + θj [θi , bj ] |{z} dt ≈ 0 ≈ [θi , H0 ] + bj [θi , θj ]

≈0

this fixes bj . Note we have not fixed ai . Hence for each first class constraint there is an arbitrariness in H0 . To eliminate this arbitrariness we impose extra conditions on the system. (These extra conditions are called gauge conditions). We call these gauge conditions γi (one for each first class constraint φi ). Full set of constraints: {φi , θi , γi } = {Θi } H = H 0 + a i φ i + b i θi + c i γ i

(1.4.8)

Provided {φi , γj } 6≈ 0, then the condition d Θi ≈ 0 dt

(1.4.9)


13

fixes ai , bi , ci (All arbitrariness is eliminated). Dirac Brackets (designed to replace Poisson Brackets so as to eliminate all constraints from the theory). Note: [θi , θj ] 6≈ 0 (Could be weakly zero for particular i, j but not in general (overall)). dij = [θi , θj ] = −[θj , θi ](Antisymmetric matrix) ∴ det(dij ) 6= 0 Thus i, j must be even. Hence there are always an even number of 2nd class constraints. Now we define the Dirac Bracket. X [A, B]∗ = [A, B] − [A, θi ]d−1 (1.4.10) ij [θj , B] i,j

Properties of the Dirac Bracket 1. [θi , B]∗ = [θi , B] −

X k,l

[θi , θk ] d−1 [θ , B] | {z } kl l d | ik{z } δil

= [θi , B] − [θi , B] = 0

(1.4.11)

2. We know that 0 = [[A, B], C] + [[B, C], A] + [[C, A], B]

(1.4.12)

0 = [[A, B]∗ , C]∗ + [[B, C]∗ , A]∗ + [[C, A]∗ , B]∗

(1.4.13)

We can show that

3. If A is some 1st class quantity, i.e. if [A, χi ] ≈ 0 for any constraint χi , then ≈0

[A, B]

∗

X z }| { = [A, B] − [A, θi ] d−1 ij [θj , B] ij

= [A, B]

(1.4.14)

Note that if H = H 0 + a i φ i + b i θi

(1.4.15)

14


then H itself is first class. Hence,

dC = [C, H] dt

(1.4.16)

Thus by (3) above, dC ≈ [C, H]∗ . dt But [θi , C]∗ = 0 by (1). Thus, in H, we can set θi = 0 before computing [C, H]∗ . i.e.

(1.4.17)

=0

z}|{ we can use [C, H] and take H to be just H = H0 + ai φi + bi θi If we want to find (Provided we exchange P.B. for Dirac B.). Thus if we use the Dirac Bracket, we need not determine bi . If we include the gauge condition, we can treat Θi = {φi , θi , γi } (1.4.18) dC dt

∗

as a large set of 2nd class constraints, and if

Dij = {Θi , Θj }

then

[A, B]∗ = [A, B] −

X

−1 [A, Θi ]Dij [Θj , B]

(1.4.19) (1.4.20)

ij

Sept. 17/99 So, so far: 1 2 ˙2 (a˙ + b − a2 − b2 ) + λ1 (a˙ − b) + λ2 (b˙ + a) L = 2 ∂L = a˙ + λ1 pa = ∂ a˙ ∂L ˙ pb = = b + λ2 ∂ b˙ pλ1 = pλ2 = 0 ⇒ Constraint H0 = pi q˙i − L 0 0 ¸ z }| { z }| { · 1 2 2 2 2 = pa a˙ + pb b˙ + pλ1 λ˙ 1 + pλ2 λ˙ 2 − (a˙ + b˙ − a − b ) + λ1 (a˙ − b) + λ2 (b˙ + a) 2 → can’t make any sense of this (can’t express λ˙ i in terms of pλi ) unless we impose constraints. " 1 = pa (pa − λ1 ) + pb (pb − λ2 ) − ((pa − λ1 )2 + (pb − λ2 )2 − a2 − b2 ) + 2 # λ1 (pa − λ1 − b) + λ2 (pb − λ2 + a)

=

(pa − λ1 )2 (pb − λ2 )2 1 2 + + (a + b2 ) + λ1 b − λ2 a 2 2 2

1.4. DIRAC’S THEORY OF CONSTRAINTS dpλ1 dt

θ1 θ2 θ3 θ4

15

= [pλ1 , H] ≈ 0

¾ pa − λ 1 − b = 0 Secondary constraints pb − λ 2 + a = 0 (tertiary constraints don’t arise)

= = = =

p λ1 p λ2 p a − λ1 − b p b − λ2 + a → These are all Second class - i.e. [θ1 , θ3 ] = 1 = [θ2 , θ4 ] (No first class constraints → no gauge condition). H = H0 + ci θi → ci fixed by the condition θ˙i = [θi , H] ≈ 0 - or could move to Dirac Brackets, and let θi Need:  0  0 dij =   −1 0

= 0. 0 1 0 0 0 0 −1 −2

 0 1   = [θi , θj ] 2  0

→ [X, Y ]∗ = [X, Y ] − [X, θi ]d−1 ij [θj , Y ]

Can eliminate constraints sequentially instead of all at once (easier). Eliminate θ3 &θ4 initially. (1)

θ1

(1) θ2 (1) d12

(1)

∴ dij

= p a − λ1 − b

= p b − λ2 + a (1)

(1)

= [θ1 , θ2 ] = 2· ¸ ¸ · 0 2 0 − 12 (1) −1 = → (dij ) = 1 0 −2 0 2 −1/2

∴ [X, Y ]∗ = [X, Y ] −

(1) [X, θ1 ]

Now we need to eliminate

1/2

z}|{ z}|{ (1) (1) (1) d−1 [θ , Y ] − [X, θ ] d−1 2 2 12 21 [θ1 , Y ]

θ1

(2)

= p λ1

(2) θ2

= p λ2

16

CHAPTER 1. CONSTRAINT FORMALISM (2)

dij →

(2) d12

(2)

= = =

(2)

dij Finally,

(2)

= [θi , θj ]∗ (note *)

=

¶ 1 [pb − λ2 + a, pλ2 ] 0 − [pλ1 , pa − λ1 − b] − 2 µ ¶ 1 [pλ1 , λ1 ] − [pλ2 , λ2 ] 2 1 (2) − = −d21 ·2 ¸ ¸ · 0 − 12 0 2 (2) −1 −→ (dij ) = 1 −2 0 0 2 µ

(2)

We can finally see that

(2)

(2)

[X, Y ]∗∗ = [X, Y ]∗ − [X, θi ]∗ (dij )−1 [θj , Y ]∗ [a, pa ]∗∗ = 1 [b, pb ]∗∗ = 1

and all other fundamental Dirac Brackets are zero. i.e. [a, pb ] = 0 as pa − λ 1 − b = 0 pb − λ 2 + a = 0 and b2 (−a)2 a2 + b2 + + + (pa − b)b − (pb + a)a H0 = 2 2 2 = pa b − pb a So, H = pa b − pb a da = [a, H]∗∗ dt = [a, pa b − pb a]∗∗ = b db = [b, H]∗∗ dt = −a If we have gauge conditions & first class constraints φi : 0

z}|{ H = H 0 + a i φ i + b i θi

1.4. DIRAC’S THEORY OF CONSTRAINTS 1st stage: → get rid of 2nd class constraints. Do this by [ ] → [ ]∗ At this stage, H = H 0 + a i φi As the ai ’s are not fixed, dA = [A, H]∗ dt ≈ [A, H0 ]∗ + ai [A, φi ]∗ ai → Arbitrariness

γ = 0 must intersect qi (t) at one & only one point.

→ “Gribov Ambiguity” (to be avoided).

17

18


Sept. 21/99 Relativistic Free Particle

S ∝ arc length from xµ (τi ) to xµ (τf ) ¶ Z τf s µ µ ¶ µ dxµ dx = −m dτ dτ dτ τi

The m = const. of proportionality → gµν = (+, −, −, −). S = −m pµ =

Z

τf

√ dτ x˙ 2

(1.4.21)

τi

∂L mx˙ µ = −√ µ ∂ x˙ x˙ 2

(1.4.22)

Constraints m2 x˙ µ x˙ µ x˙ 2 0 = p2 − m2

pµ pµ =

(1.4.23)

H0 = pµ x˙ µ − L √ mx˙ µ x˙ µ = − √ − (−m x˙ 2 ) x˙ 2 = 0 ! H = H 0 + U i χi = γ(p2 − m2 ) (Pure Constraint!) x˙ µ =

∂H ∂pµ

dxµ = κ(2pµ ) dτ

(1.4.24)


19

This arbitrariness in x˙ µ is a reflection of the fat that in S, τ is a freely chosen parameter, i.e. S

= −m

Z

τ → τ (τ 0 )

∴

dxµ dτ

S −→ Now let κ =

1 dτ 0 . 2 dτ

r

dxµ dxµ dτ dτ dτ dτ = 0 dτ 0 dτ

dτ

dxµ dτ 0 dτ 0 dτ r Z dxµ dxµ = −m dτ 0 dτ 0 dτ 0 =

(1.4.25)

(1.4.26) (1.4.27)

Thus, (insert κ into (1.4.24)) dτ 0 µ dxµ = p and equate this with (1.4.26) dτ dτ dxµ = pµ dτ 0

(1.4.28) (1.4.29)

Gauge fixing in this case corresponds to a choice of the parameter τ . -The formalism of Dirac actually breaks down for gauge choices γ which are dependent on “time” (which in this case means on τ ). (Note, we can think of this reparameterization invariance τ → τ (τ 0 ) as being a form of diffeomorphism invariance in 0 + 1 dimensions, i.e. xµ (τ ) is a scalar field moving in 0 + 1 dimensions, and has a so-called “tangent space” which is four dimensional. Thus, this is a simpler version of G.R. where we have scalars φ0 (xµ ) moving in 3 + 1 dimensions with the diffeomorphism invariance xµ → xµ (x0µ ). Techniques in G.R. & in the single particle case often overlap. eq. of motion: √ L = −m x˙ 2 d ∂L ∂L 0 = − dτ ∂ x˙ µ ∂xµ Thus, d dτ

µ

¶ −mx˙ µ √ = 0 x˙ 2 m¨ xµ √ = 0 x˙ 2

Now identify τ with the arc length along the particle’s trajectory:

20


ds2 = dxµ dxµ

If dτ 2 = ds2 , then dxµ dxµ = 1 dτ dτ x˙ 2 = 1

(1.4.30) (1.4.31)

(τ is called the “proper time” in this instance). i.e. if dx = 0, ds2 = dt2 = dτ 2 . In this case, the equation of motion becomes m¨ xµ = 0 The corresponding action is Z m τf √ dτ x˙ 2 ← absence of S= means this is not invariant under τ → τ (τ 0 ). 2 τi d ∂L ∂L − = m¨ xµ = 0 dt ∂ x˙ µ ∂xµ H = pµ x˙ µ − L where pµ =

∂L ∂ x˙ µ

= mx˙ µ H = pµ H =

µ

pµ m

pµ pµ 2m

¶

−

m ³ pµ ´2 2 m


21

Other gauge choice: τ = x4 = t(breaks Lorentz invariance) Work directly from the action: S = −m If we’ve chosen τ = t, then S = −m eq. of motion:

Z

Z

dτ

dt

√

r

dt dt d~r d~r − dτ dτ dτ dτ

1 − ~v 2 ;

~v =

d~r dt

d ∂L ∂L − = 0 dt ∂~r˙ ∂~r µ ¶ m~v d √ = 0 ∴ dt 1 − ~v 2 m~v ∂L (momentum) =√ p~ = 1 − ~v 2 ∂~r˙ H = p~ · ~r˙ − L but p~2 =

m2~v 2 1−~v 2

= p~ · ~v − L

∴ p~2 (1 − ~v 2 ) = m2~v 2 p~2 − p~2~v 2 = m2~v 2 p~2 = ~v 2 (~p2 + m2 ) p~2 2 ~v = m2 + p~2 m2 + 6 p~2 − 6 p~2 p~2 2 = 1 − ~v = 1 − 2 m + p~2 m2 + p~2 s √ m2 ∴ 1 − ~v 2 = p~2 + m2 Thus, p~ ~v = p m2 + p~2 p~ m H = p~ · p − (−m) p m2 + p~2 m2 + p~2 p m mc2 p~2 + m2 ⇒ E = numerical value of H = √ = =q 2 1 − v2 1 − vc2

22


Note: E 2 = p~2 + m2 (E 2 − p~2 ) − m2 = 0 ∴ pµ = (~p, E) pµ pµ − m 2 = 0 Limit 0 Z m→ √ S = −m dτ x˙ 2 → 0 ??

i.e.

d dt

|~v | → 1 as m → 0.

µ

m~v √ 1 − ~v 2

¶

→ 0 ???

Circumvent by introducing a Lagrange multiplier e. ¶ Z µ 2 x˙ 1 2 + m e dτ S = − 2 e e = e(τ ) , xµ = xµ (τ ) m2 → 0 is well defined in S. As

d ∂L ∂L − =0 dτ |{z} ∂ e˙ ∂e =0

So,

x˙ 2 − 2 + m2 = 0 → e = e Thus, Z

"

2

r

mx˙ √ + m2 x˙ 2 Z √ = −m dτ x˙ 2

S = −

1 2

dτ

x˙ 2 m2 r

x˙ 2 m2

#

Can see that e field can be eliminated → really just a Lagrange multiplier that insures S 6= 0 when m = 0. Note: The action µ 2 ¶ Z x˙ 1 2 +m e (1.4.32) S=− dτ 2 e


23

is invariant under τ → τ + f (τ ) δxµ = x˙ µ f (τ ) δe = f˙e + ef ˙

(1.4.33) (1.4.34) (1.4.35)

Sept. 22/99 So, for the above system: ¶ µ 1 x˙ 2 2 +m e L = − 2 e eqn’s of motion: µ ¶ d ∂L ∂L x˙ 2 2 from 0 = − 2 +m − =0 e dτ ∂ e˙ ∂e µ ¶ µ ¶ d x˙ µ d ∂L ∂L 0 = from − =0 dτ e dτ ∂ x˙ µ ∂xµ x˙ µ ∂L =− (no constraint → can solve for x˙ µ in terms of pµ ). pµ = µ ∂ x˙ e ∂L pe = =0 (primary constraint) ∂ e˙ =0 z}|{ µ H0 = pµ x˙ + pe e˙ −L µ ¶¸ · 1 x˙ 2 2 µ +m e ; x˙ 2 = xµ xµ = −p2 e2 = pµ (−p e) − − 2 e 1 = − e(p2 − m2 ) 2 ½ · ¸¾ 1 2 2 p˙e = 0 = [pe , H0 ] = pe , e(p − m ) 2 1 0 = − (p2 − m2 ) (Secondary Constraint) 2 Both pe = 0 (gauge condition e = 1) and p2 − m2 = 0 (already discussed) are first calss. Note: Remember that S=

Z

d4 x

√

¡ ¢¡ ¢ g gµν ∂ µ φA (x) ∂ ν φA (x)

• action for a scalar field φA (x) in 3+1 Dim.

(1.4.36)

24


Vierbein (“deals with 4-d”) gµν = eaµ eaν q √ g = det(gµν ) = [det(eµν )]−1 = e−1

In 0 + 1 dimensions

S =

Z

1 dτ e

µ

d A φ dτ

¶µ

d A φ dτ

¶

Z (x˙ µ )2 (φ˙ A )2 −→ dτ = dτ e e e → “Einbein” (assoc. with 1 dim) Z

1.5

Quantizing a system with constraints 1 ˆ ˆ [A, B]commutator (c) [A, B]P B → i~ ex. [q, p]P B = 1 [ˆ q , pˆ]c = i~

(1.5.1) (1.5.2) (1.5.3)

If there are constraints ξi (q, p) then, ξi (ˆ q , pˆ) |ψ >phys. = 0

(1.5.4)

¶ µ 1 x˙ 2 2 ex. For L = − +m e 2 e χ1 = pe χ2 = p2 − m2 (2 constraints) Quantization conditions will be [xµ , pν ] = i~δµν

(p2 − m2 ) |ψ >phys

(1.5.5)

δµν → (+, +, +, +) ¶ µ ∂L → Contrav. = der. of covar. re: pµ = ∂ x˙ µ = 0 (Klein-Gordon eq.)

pµ = −i~

∂ ∂xµ

(1.5.6) (1.5.7)

1.5. QUANTIZING A SYSTEM WITH CONSTRAINTS If Φ(x) =< x|ψ >phys then "µ

∂ −i~ µ ∂x

¶2

#

− m2 Φ(x) = 0

Classical Motivation for Spin Brint, deVecchia & How, Nuclear P. 118, pg. 76 (1977)

25

26


Chapter 2 Grassmann Variables (Casa/booni, N.C. 33A) θ1 , θ2 → Grassmann Variables (call θ if only 1 present), where; θ1 θ2 = −θ2 θ1 θ12 = −θ12 = 0 ex. (these can only be . . . ) f (x1 , θ) = a(x) + b(x)θ 1 f (x, θ1 , θ2 ) = a(x) + bi (x)θi + c(x)²ij θi θj 2 (→ ²ij = −²ji )

(2.0.1) (2.0.2) (2.0.3) (2.0.4)

Calculus

dκ = 0(κ = const.) (2.0.5) dθ =0 =1 z }| ¶{ z}|{ µ (-ve sign in 2nd term because we’re moving dθ2 d dθ1 (θ1 θ2 ) = θ2 + θ 1 − d dθ1 dθ2 dθ1 through to θ2 ) dθ1 dθ = 1 dθ

= θ2

(2.0.6)

Similarly ¶ µ d dθ1 dθ2 (θ1 θ2 ) = θ2 + θ 1 − dθ2 dθ2 dθ2 = −θ1

(2.0.7) 27

28

CHAPTER 2. GRASSMANN VARIABLES

2.1

Integration

i.e.

Z

Z

Z

d dθ

dθ ↔

dθ c = 0

i.e. Integration & differentiation are identical

(2.1.1)

(int. of a constant c )

(2.1.2)

dθ θ = 1

(2.1.3)

For example; Z

µ ¶ d d dθ1 dθ2 (θ1 θ2 ) = θ1 θ2 dθ1 dθ2 d = (−θ1 ) dθ1 = −1Z = −

= −

dθ2 dθ1 θ1 θ2

Z

dθ1 dθ2 θ2 θ1

Another example: Z

dθ1 dθ2 f (x, θ1 , θ2 ) =

Z





1 dθ1 dθ2 |{z} a + bi θi + ²ij c θi θj  |{z} 2 (i)

(ii)

(i) → 0 (const.) (ii) → 0 (θi → int. over θj gives 0) µ ¶ Z 1 1 = c dθ1 dθ2 θ1 θ2 − θ2 θ1 2 2 = −c(x) Delta function:

→

Z

Z

dθ δ(θ)f (x, θ) = f (x, 0)

dθ δ(θ) [a(x) + b(x)θ] = a(x) =⇒ δ(θ) = θ

(2.1.4)

2.1. INTEGRATION

29

The following is an example to demonstrate the properties of different kinds of statistical models. Suppose there are three students: (T) Tom (D) Dick (H) Harry How many ways can two prizes be awarded to T, D, H? 1. Suppose there are two medals, (distinguishable awards) a Newton medal (N) and a Shakespear medal (S). T N S N

D H S N S

T D S N S

H N S N

T NS

D

H

NS NS

So, there are 9 different ways to award 2 distinguishable medals. 2. Two silver dollars (2 medals, indistinguishable) T $ $

D H $ $ $ $

T $$

D

H

$$ $$

So, there are 6 ways to award 2 indistinguishable medals. 3. Two positions (P) on football team (2 medals, indistinguishable). But! → now makes no sense for one person to receive 2 “medals” (one player can’t have 2 positions). T D P P P P

H P P

There are 3 ways to award to indistinguishable yet distinct “medals”. © ↔ particles Now call the prizes students ↔ states 1. = Maxwell-Bolzmann statistics 2. = Bose-Einstein 3. = Fermi-Dirac

30


Sept. 24/99 Returning to our discussion of Grassmann variables, the Lagrangian is now ³ ´ L = L xµ (τ ), θa (τ ), x˙ µ (τ ), θ˙a (τ ) where 0 = θa (τ )θb (τ 0 ) + θb (τ 0 )θa (τ ) ∂L ∂L and Πa = pµ = ∂ x˙ µ ∂ θ˙a

(2.1.5) (2.1.6) (2.1.7)

H = H(xµ , θa , pµ , Πa ) = q˙µ pµ + θ˙a Πa −L | {z }

(2.1.8)

∗

δ

Z

τf τ0

* - order important  Z τf  µ ∂L δL ∂L ˙a ∂L dτ L = dτ  δx ∂xµ + δθ δθa + δ x˙ ∂ x˙ µ + δ θ ∂ θ˙a τ0 | {z } | {z } int. by parts

int. by parts

   

d ∂L ∂L d ∂L ∂L → = a = , → µ µ ˙ ∂x dτ ∂ θ˙a ∂θ Z dτ ∂ x³ ´ = δ dτ q˙µ pµ + θ˙a Πa − H(q, p, θ, Π) µ ¶ Z ∂H ∂H ∂H ∂H = dτ δ q˙ p + q˙ δp + δ θ˙ Π + θ˙ δΠ − δq − δp − δθ − δΠ ∂q ∂P ∂θ ∂Π int. by parts: µ ¶ Z ∂H ∂H ∂H ∂H ˙ ˙ = dτ −p˙ δq + q˙ δp − Π δθ + θ δΠ − δq − δp − δθ − δΠ ∂q ∂p ∂θ ∂Π

And so, for this to be zero, we must have: ∂H , ∂q µ ∂H = + µ , ∂p

p˙µ = − q˙µ

2.2

Poisson Bracket

A(q, p, θ, Π), B(q, p, θ, Π) Note: • Anything Grassmann is odd • Anything else is even For the order of two quantities:

˙ a = − ∂H Π ∂θa ∂H θ˙a = − a ∂Π

(2.1.9) (2.1.10)

2.2. POISSON BRACKET

31

• Even/Even (doesn’t matter) • Even/Odd (doesn’t matter) • Odd/Odd → Switch the two, you pick up a -ve sign.

[A, B]P B =

                

∂A ∂B ∂q ∂p ∂A ∂B ∂q ∂p

− −

∂A ∂p ∂A ∂p

∂B ∂q ∂B ∂q

+ +

∂A ∂B ∂θ ∂Π ∂A ∂B ∂θ ∂Π

− +

∂B ∂A ∂θ ∂Π ∂A ∂B ∂Π ∂θ

1. A,B “even”

∂A ∂B ∂q ∂p

+

∂B ∂A ∂q ∂p

−

∂A ∂B ∂θ ∂Π

−

∂B ∂A ∂θ ∂Π

2. A,B “Odd”

∂A ∂B ∂q ∂p

−

∂B ∂A ∂q ∂p

+

∂A ∂B ∂θ ∂Π

+

∂B ∂A ∂θ ∂Π

3. A “Even”, B “Odd”

With this, we find that (Not trivial!): 0 = [[A, B] , C] + [[B, C] , A] + [[C, A] , B]

(2.2.1)

Look at the “spinning particle” (Not superparticle). Simplest action involving Grassmann Variables. (Brink et. al. NP B118) L = L (φµ (τ ), ψ µ (τ )) (φµ ≡ xµ (τ ), ψ → Grassmann) " # µ ¶¶ µ 2 1 1 φ˙ µ φ˙ µ x˙ µ 2 ˙ − iψ (τ ) ψµ (τ ) ; cf L = − = − +m e 2 e 2 2

(2.2.2)

Note: 1. ψ˙ µ ψ˙ µ = 0 (why we can’t have two ψ˙ in L). 2. i → needed so that L = L+ ˙ i.e. L = (−iψ ψ) (Re: (AB)+ = B + A+ ) ˙ = −iψ ψ˙ = L L+ = (+i)(ψ˙ + ψ + ) = iψψ 3. The following τ δφµ δe δψ µ is an invariance of S =

R

dτ L

→ = = =

τ + f (τ ) ˙ φf f˙e + ef ˙ µ ψ˙ f

(reparametrization invariance)

(2.2.3) (2.2.4) (2.2.5) (2.2.6)

32

CHAPTER 2. GRASSMANN VARIABLES Quantizing this leads to negative norm states in the Hilbert space associated with ψ 0 (τ ). < ψ 0 |ψ 0 >

< 0 (Don’t want.) 1 ~2 (ex L = − ∂µ Aλ ∂ µ Aλ → Aλ Aλ = (A0 )2 − A 2

0

(A = -ve norm state)

¶

We eliminate the unwanted negative norm state by building in an extra symmetry. Ã ! i 1 φ˙ 2 µ µ − iψµ ψ˙ − χφ˙ ψµ ; χ(τ ) is Grassmann L= (2.2.7) 2 e e 1. Add in δχ = f˙χ + χf ˙ 2. We have another invariance (α = α(τ ) is Grassmann) δα φµ = iαψ µ Ã ! ˙µ i φ δα ψ µ = α − χψ µ e 2e δα e = iαχ δα χ = 2α˙ d (...) He didn’t recall δα L = dτ

(2.2.8) (2.2.9) (2.2.10) (2.2.11) (2.2.12)

• φ - Scalar in 0+1 dimensions. • ψ µ - Spinor in 0+1 dimensions. • e - “einbein” in 0+1 dimensions. • χ - “gravitino” in 0+1 dimensions. L - analogous to SUGRA L in 3+1 dimensions.

2.3

Quantization of the spinning particle. Πµ

µ ¶ ∂L ∂ i ν ˙ = = − ψν ψ 2 ∂ ψ˙ µ ∂ ψ˙ µ i ψµ → Second Class constraint = 2 i θµ = Π µ − ψ µ = 0 2 ∴→ [θµ , θν ]P B 6= 0

(2.3.1) (2.3.2) (2.3.3) (2.3.4)

2.3. QUANTIZATION OF THE SPINNING PARTICLE.

33

Sept. 28/99 We could proceed with the Dirac formalism. We could also cheat and use the equations of motion. Recall: Ã ! i 1 φ˙ µ φ˙ µ − iψµ ψ˙ µ − χφ˙ µ ψ µ L = (2.3.5) 2 e e δα φµ = iαψ µ Ã ! ˙µ φ i − χψ µ δα ψ µ = α e 2e δα e = iαχ δα χ = 2α˙

(2.3.6) (2.3.7) (2.3.8) (2.3.9)

Eq. of motion for e: d dτ

µ

∂L ∂ e˙

¶

∂L ∂e 0 = φ˙ 2 − iχφ˙ µ ψ µ =

(2.3.10) (2.3.11)

For χ : φ˙ µ ψµ = 0 → eliminates -ve norm states If φ0 = v, vψν = 0

(2.3.12)

We could have gotten these two equations from the Hamiltonian formalism because these are the constraint equations that follow from the primary constraints ∂L =0 ∂ e˙ ∂L = =0 ∂ χ˙

pe =

(2.3.13)

pχ

(2.3.14)

These are both first class constraints → gauge conditions. ¾ ¾ pχ = 0 χ=0 ← “proper time” guage conditions. pe = 0 e=1

(2.3.15)

The equations of motion for: φµ : d dτ

µ

∂L ∂ φ˙ µ

¶

∂L ∂φµ Ã ! d 2φ˙ µ iχψ µ 0 = − dτ e e =

(2.3.16) (2.3.17)

34

CHAPTER 2. GRASSMANN VARIABLES ψµ: χφ˙ µ µ ˙ 0 = 2ψ − e It is possible to show that the Dirac Brackets are

(2.3.18)

[φµ , pν ]∗ = gµν

(2.3.19)

[ψµ , ψν ]∗ = −igµν

Ã

Remember that pψµ =

i ∂L = ψµ 2 ∂ ψ˙ µ

!

(2.3.20)

Quantization: 1. [ , ]∗ → i~1 [ , ](anti)−commutator Thus: (letting ~ = 1) h i φˆµ , pˆν = igµν → φˆµ pˆν − pˆν φˆµ = igµν − h i ψˆµ , ψˆν = gµν → ψˆµ ψˆν + ψˆν ψˆµ = gµν +

(2.3.21) (2.3.22)

2. χi |ψ > = 0 for any constraint χi . In the gauge χ = 0, e = 1

pµ =

i ∂L = φ˙ µ − χψ µ ˙ ∂ φµ |2e{z }

(2.3.23)

=0

Thus: φ˙ 2 = 0 ⇒ pˆ2 |Ψ > = 0 ˆ > =0 φ˙ · ψ = 0 ⇒ pˆ · ψ|Ψ If we let γµ ψˆµ = √ 2

(2.3.24) (2.3.25) (2.3.26)

and as ψ˙ µ = 0 in the Heisenberg picture, then γ µ is a constant and satisfies the algebra, {γ µ , γ ν } = 2g µν ({ , } → anticommutator).

(2.3.27)

If the coordinate representation Ψ(φ) =< φ|Ψ >

(2.3.28)

then [φˆµ , pˆν ] = igµν → pν = −i −iγ µ

∂ Ψ(φ) = 0 ∂φµ

(2.3.29)

∂ ˆ > = 0 becomes and pˆ · ψ|Ψ ∂φν (2.3.30)


35

This is the massless Dirac equation. For the massive Dirac equation, Ã ! Z ˙ ·ψ 1 φ˙ 2 iχ φ S= dτ − iψ · ψ˙ − + em2 + iψ5 ψ˙ 5 + imχψ5 2 e e

(2.3.31)

Following the same sort of argument as above, we find, · µ ¶ ¸ ∂ → γµ −i − m Ψ(φ) = 0 ∂φµ Dirac’s Approach Schrödinger Equation: (note first order time derivative) i~

∂ψ(x, t) = H(x, p)ψ(x, t) ∂t → p~ = −i~∇

(2.3.32) (2.3.33)

Klein-Gordon equation (K.G.): ¢ ¡ µ pµ p − m 2 ψ ∂2 (c, ~ = 1) = ∇2 − ∂t2 |{z}

0 = pµ pµ

(2.3.34) (2.3.35)

2nd order

From Dirac equation

∂ψ = (~ α · p~ + mβ) ψ (2.3.36) ∂t For ψ to also satisfy the K.G. equation, then µ ¶2 ∂ i ψ = (~ α · p~ + mβ)2 ψ (2.3.37) ∂t · ¸ 1 ∂2 2 2 (αi αj + αj αi )pi pj + 2m(αi β + βαi )pi + m β ψ (2.3.38) − 2ψ = ∂t 2 Thus {αi , αj } = 2δij (2.3.39) {αi , β} = 0 (2.3.40) β2 = 1 (2.3.41) ¸ · ¸ · 0 σi I 0 , αi = (2.3.42) β = 0 −I σi 0 ¸ ¸ · ¸ · ¸ · · 1 0 0 −i 0 1 1 0 (2.3.43) , σ3 = , σ2 = , σ1 = where I = 0 −1 i 0 1 0 0 1 i

36

CHAPTER 2. GRASSMANN VARIABLES Eddington showed that all representations of αi , β in 4-d are unitarily equivalent. Inclusion of the electromagnetic field: ~ p~ → p~ − eA H → H + eΦ

)

~ Φ → electromagnetic potentials) (A,

(2.3.44)

Thus, i

i ∂ψ h ~ + βm + eΦ ψ = α ~ · (~p − eA) ∂t

(2.3.45)

Aside: 1. This can be generalized n dimensions.o n ton/2 2 dimension of (~ α, β) is 2(n−1)/2 (n=even) (n=odd) 2. In even dimensions all (~ α, β) are unitarily equivalent, but in odd # dimensions there are 2 sets (~ α, β), (+~ α, −β) (not unitarily equivalent to each other → all sets are unitarily equivalent to one or the other, not both). Sept. 29/99 Backtrack [q, p] = i~ → [xi , pj ] = i~δij pj = −i~

∂ ∂xj

∂ψ = Hψ (H → E) ∂t ∂ ∴ E = i~ ∂t

i~

But also, xµ = (~x, t) (c = 1) pµ = (~p, E) (consistent with (2.3.46),(2.3.48) provided pµ = i ∂x∂ µ ).

(2.3.46) (2.3.47) (2.3.48)


37

i.e. (Metric (−, −, −, +)) pµ = gµν pν = (−~p, E) ← (~p = −i∇ , E = i

∂ ) ∂t

= i∂µ ∂ = i µ ∂x µ

¶ ∂ = i ∇, ∂t h i ∂ψ ~ + βm + eΦ ψ i = α ~ · (~p − eA) ∂t ¸ ¸ · · I 0 0 ~σ , β= →α ~ = 0 −I ~σ 0 ∴ ψ = 4 − components   ψ1  ψ2   =   ψ3  ψ4 We had i~

(2.3.49)

∂ψ p~2 = ψ ∂t 2m

Pauli equation ∂ i~ ∂t

·

ψ1 ψ2

¸

# ~ · ψ1 ¸ ~ 2 ~σ · B (~p − eA) + eΦ + = ψ2 2m 2m "

(2.3.50)

Let ψ be: 

 φ˜1 · ¸ · ¸  φ˜2  φ φ˜   e−imt = ψ= =  χ | {z } χ˜ χ˜1 ∗ | {z } χ˜2 Slowly

(2.3.51)

varying

* - Principle time dependence for “slowly moving” particles. (~ = 1). i.e. i

∂ψ ∂t

= Hψ = Eψ

(2.3.52)

→ ψ ∼ e−iEt = e−imt in rest frame.

(2.3.53)

38


Thus, ½·

∂ i ∂t

·

φ χ i ∂φ ∂t i ∂χ ∂t

¾

½·

¸ · ¸ · ¸¾ · ¸ ~ 0 ~σ · Π m 0 eΦ 0 φ = + e + e−imt ~ 0 −m 0 eΦ χ ~σ · Π 0 ¸ ¸ · ~ + mφ + eΦψ ~σ · Πχ + mφ (2.3.54) = ~ − mχ + eΦχ + mχ ~σ · Πφ

¸

−imt

(Assume ∂χ , eΦχ ≈ 0 → (χ is “small”) ) ∂t Thus from 2nd equation, χ= Hence:

~ ~σ · Πφ 2m

(2.3.55)

" # ~ 2 (~σ · Π) ∂φ = + eΦ φ i ∂t 2m

(2.3.56)

Hence, ~ = (~σ · Π) 2

µ·

(p − eA)3 {(p − eA)1 − i(p − eA)2 } {(p − eA)1 + i(p − eA)2 } (p − eA)3

¸¶2

(2.3.57)

But: σi σj = δij + i²ijk σk (σ )ab (σ i )cd = 2δad δbc − δab δcd ~ 2 = (σi σj )Πi Πj ∴ (~σ · Π) = (δij + i²ijk σk )(p − eA)i (p − eA)j ~ 2 − i(~p − eA) ~ × (~p − eA) ~ · ~σ = (~p − eA) i

(2.3.58) (2.3.59)

∂Aj (~p = −i∇) ∂xi ~ 2 + e(∇ × A) ~ · ~σ = (~p − eA) ~ 2 + eB ~ · ~σ = (~p − eA) → [pi , Aj ] = −i

“Conserved Current” ∂ψ = (~ α · p~ + βm)ψ = (−i∇ · α ~ + βm)ψ ∂t and ∂ψ + (−i) = ψ + (~ α+ · p~+ + β + m) ; (~ α+ = α ~ , β + = β) ∂t = ψ + (~ α · p~+ + βm) = (+i∇ψ + · α ~ + ψ + βm) i

(2.3.60)

(2.3.61)

2.3. QUANTIZATION OF THE SPINNING PARTICLE. ψ + × (2.3.60)

→ iψ +

(2.3.61) ×ψ → −i Subtract the two

µ

39

∂ψ = ψ + (−i∇ · α ~ ψ) + ψ + βmψ ∂t

∂ψ + ∂t

¶

(2.3.62)

¡ ¢ ψ = i(∇ · ψ + ) · α ~ ψ + ψ + βmψ

(2.3.63)

∂ ¡ + ¢ ∂ ψ ψ = −i∇ · (ψ + α ~ ψ) ⇒ (ψ + ψ) + ∇ · (ψ + α ~ ψ) = 0 ∂t ∂t ∂ → i.e. 0 = ρ + ∇ · ~j ∂t where j µ = (ψ + α ~ ψ, ψ + ψ) = (~j, ρ) → ∂µ j µ = 0 i

(2.3.64)

ψ + ψ → probability density ψ+α ~ ψ → probability flux.

(2.3.65) (2.3.66)

Free particle solution in the frame of reference where p~ψ = 0 (“rest frame”): i.e. − i∇ψ = 0 ∴ ψ = ψ(t) i

∂ ∂t

·

φ(t) χ(t)

¸

= βm

= ∴

·

·

φ χ



¸

m 0 0 −m

 →β= 

¸·

φ(t) = e−imt φ0 χ(t) = e+imt χ0 · −imt ¸ e φ0 ψ(t) = +imt e χ0

φ χ



1 1 −1

¸

−1

  

(2.3.67) (2.3.68) (2.3.69) (2.3.70)

→ e+imt = “negative energy” (associated with −

K.G. equation (−~p2 + p20 − m2 )ψ = 0

p

p~2 + m2 )

(2.3.71) p

p0 ψ = ± p~2 + m2 ψ ∂ψ i = (~ α · p~ + βm)ψ ∂t

(2.3.72) (2.3.73)

40


Oct. 1/99 

 e−imt ψ10  e−imt ψ20  ∂ψ  α · p~ + βm)ψ, perform a “Boost”. ψ=  e+imt ψ30  → to solve i ∂t = (~ e+imt ψ40

(2.3.74)

Rewrite the Dirac equation in a covariant form.

∂ψ = (−i~ α · ∇ + βm)ψ ∂t ¸ · ∂ α) − m ψ 0 = i (βψ) + i∇ · (β~ ∂t Remember that µ ¶ ∂ i∂µ = i∇, i ∂t Then we have (i∂µ γ µ − m) ψ = 0 where γ 0 = β = +γ0 γ i = βαi = −γi As {αi , αj } = 2δ ij {β, αi } = 0 β2 = 1  −  − {γ µ , γ ν } = 2g µν ; g µν =   − i

(2.3.75) (2.3.76)

(2.3.77) (2.3.78) (2.3.79) (2.3.80) (2.3.81)  +

  

0 σi −σ i 0

¸

Different Representations are:

Standard: Chiral Representation:

·

1 0 γ = 0 −1 · ¸ 0 1 0 γ = 1 0 0

¸

i

γ = i

γ =

·

·

0 σi −σ i 0

¸

(2.3.82)

(2.3.83) (2.3.84)

Often, aµ γ µ is written 6 a. Lorentz transformation: → Either a Boost or a rotation. Boost in x-direction: x − vt x0 = √ , 1 − v2 y0 = y z0 = z

−vx + t t0 = √ 1 − v2

(2.3.85) (2.3.86) (2.3.87)


41

Rotation (example): x0 = x cos(θ) − y sin(θ) y 0 = x sin(θ) + y cos(θ) We know g µν x0µ x0ν = gµν xµ xν = t 2 − x2 − y 2 − z 2 Linear:

(2.3.88)

x0µ = aµν xν

(2.3.89)

gµν aµλ aνσ = gλσ

(2.3.90)

→ The Lorentz transformation is an O(3,1) transformation. Under the Lorentz transformation we set ψ 0 (x0 ) = S(a)ψ(x) (S(a) = 4 × 4 matrix.) With µ ¶ µ ∂ 0 = iγ − m ψ 0 (x0 ) ∂x0µ ¶ µ µ ∂ − m ψ(x) 0 = iγ ∂x

(2.3.91) (2.3.92)

(i.e. should have both the same masses and the same γ µ matrices in each equation → not a ¡ ¢ γ 0µ , as in iγ 0µ ∂x∂0µ − m ψ 0 (x0 ) = 0 ) Note first: ∂x0λ ∂ ∂ = ∂xµ ∂xµ ∂x0λ ∂ = aλµ 0λ ∂x Hence, (substitute into (2.3.92)), · µ ¶ ¸ µ λ ∂ 0 = iγ aµ 0λ − m S −1 (a)ψ 0 (x0 ) ∂x → multiply on left by S(a). 0 =

·

¸ ¡ ¢ λ ∂ µ −1 i S(a)γ S (a) aµ 0λ − m ψ 0 (x0 ) ∂x

(2.3.93)

42

CHAPTER 2. GRASSMANN VARIABLES But change of variables should leave us with an equation in the same form. ∴ γ µ = S(a)γ ρ S −1 (a)aµρ

(2.3.94)

From this we determine S(a) Suppose the Lorentz transformation is infinitesimal: aµν = δνµ + ∆wνµ But gµν aµλ aνσ = gλσ ∴ gλσ = gµν (δλµ + ∆wλµ )(δσν + ∆wσν )

(2.3.95)

(2.3.96)

To order ∆w, we see

or

0 = gµν (∆wλµ δσν + δλµ ∆wσν ) ∆wλσ = −∆wσλ

(2.3.97)

Now take S(a) = S(I + ∆w) i = I − σµν ∆wµν 4 → Need σµν . We also have i S −1 (a) = I + σµν ∆wµν 4 Hence ¶ µ ¶ µ ¡ µ ¢ i i ρ µ αβ γδ γ I + σγδ ∆w δρ + ∆wρµ γ = I − σαβ ∆w 4 4

(2.3.98)

(2.3.99) (2.3.100)

→ must hold to order ∆w. Consequently, 0 = γ ρ ∆wρµ +

Solution:

i µ (γ σαβ − σαβ γ µ ) ∆wαβ 4

Hence ¢ i µ −1 ¡ µ γβ δα − γα δβµ (γ σαβ − σαβ γ µ ) = 4 2 σµν =

i [γµ , γν ] 2

(2.3.101)


43

For a finite Lorentz transformation, let 1 µν w N ¶N µ i wµν ∴ S(a) = lim 1 − σµν N →∞ 4 N i = exp{− σµν wµν } 4 and if x0µ = aµν xν ¾ ½ i µν 0 0 ψ(x) then ψ (x ) = exp − σµν w 4 ∆wµν =

(2.3.102)

(2.3.103) (2.3.104) (2.3.105)

If ∆wµν = ∆w(I)µν → ∆w = scalar, (I)µν = 4 × 4 matrix characterizing the Lorentz transformation. then for a rotation about the x-axis, x0 = x − ∆w y y 0 = ∆w x + y

(2.3.106) (2.3.107)

(→ infinitesimal version of “reg.” rotation.) In this case, 

0 −1 0  +1 0 0 ∆wµν = ∆w   0 0 0 0 0 0 (where rows/columns = [x, y, z, t]) ½ ¾ i µν Thus S(a) = exp − σµν w 4 = e−iσµν w/2

 0 0   0  0

44


But, i (γ2 γ1 − γ1 γ2 ) 2 = iγ2 γ1 · ¸· ¸ 0 σ2 0 σ1 = i −σ2 0 −σ1 0 · ¸ σ3 0 = (−1) 0 σ3

σ21 =

Thus

S(a) = exp

½

i w 2

·

σ3 0 0 σ3

¸¾

(2.3.108)

If w = 2π (1 revolution) ½

S(w = 2π) = exp iπ

·

σ3 0 0 σ3

¸¾

;

σ32 = 1

We know, (~a · ~σ )2 + ... 2! But → (~a · ~σ )2 = ~a2 [σi σj = δij + ²ijk σk ] µ ¶ ¶ µ |~a|3 ~a · ~σ ~a2 (~a2 )2 |~a| + + + ... + + ... 1+ 2! 4! |~a| 3! ~σ · ~a cosh |~a| + sinh |~a| |~a| ½ · ¸¾ σ3 0 exp iπ 0 σ3 cosh(iπ) + sinh(iπ) σ3 | {z } | {z }

e~a·~σ = 1 + (~a · ~σ ) + = = and so

S(w = 2π) = =

=−1

=0

= −1

Thus if w = 2π, ψ 0 (x0 ) = −ψ(x), and if w = 4π, ψ 0 (x0 ) = +ψ(x) ∴ All physical quantities need an even # of ψ’s. → in 3600 rotation, ψψψ ⇒ −ψψψ ψψψψ ⇒ +ψψψψ

(2.3.109)


45

Oct. 5/99 So, we have: Dirac Equation: 0 = (iγ µ ∂µ − m) ψ

(2.3.110)

{γ µ , γ ν } = 2g µν ; g µν = (+ − −−) ½ ¾ i µν ψ → exp − ω σµν ψ 4 µ ½ ¾¶+ i µν + + exp − σ ωµν ψ → ψ 4 i [γµ , γν ] with σµν = 2

(2.3.111) (2.3.112) (2.3.113) (2.3.114)

Properties: γ

0

=

·

0 1 1 0

¸

0 +

= (γ )

i

γ =

i [γ0 , γi ] 2 = iγ0 γi = −σi0

·

0 σi −σi 0

¸

= −(γ i )+

σ0i =

i.e. (σ0i )+ = = = =

(iγ0 γi )+ (−i)(γi+ γ0+ ) (−i)(−γi )(γ0 ) −σ0i

σij = σij+ Thus + γ0 σµν γ0 = σµν i.e. µ = 0, ν = i + γ0 σ0i γ0 = γ0 (−σ0i )γ0 = (γ0 )2 σ0i = σ0i µ = i, ν = j + γ0 σij γ0 = γ0 (+σij )γ0 = γ02 σij = σij

Thus if

ψ¯ = ψ + γ0

(2.3.115)

46


then under a Lorentz transformation, ¾ ¶+ µ ½ i µν ψ γ0 ψ¯ → exp − σµν ω 4 µ ½ ¾¶+ i + µν = ψ exp − ωµν σ γ0 4 µ ½ ¾¶ i + µν+ exp + ωµν σ = ψ γ0 ; 4 ¾ ½ i µν+ + γ0 ωµν σ = ψ γ0 γ0 exp | {z } 4 {z } ψ¯ | exp{ 4i ωµν σ µν } ½ ¾ i µν ¯ ¯ ωµν σ ∴ ψ → ψ exp 4 Hence ¾ ¾ ½ ψ → Uψ i µν → U = exp − ωµν σ ¯ −1 4 ψ¯ → ψU Thus, (for ex.)

γ02 = 1

¯ → ψU ¯ −1 U ψ = ψψ ¯ ψψ

(i.e. Lorentz invariant). Lorentz transformations involving parity (discrete transformations).

Re.

(x, y, z, t) → (−x, −y, −z, t) ψ(x) → ψ 0 (x0 ) = S(a)ψ(x) S(a)aµν γ ν S −1 (a) = γ µ

i

x0µ = aµν xν  −1  −1 aµν =   −1 −1

(2.3.116) (2.3.117) 

+1

i

Sγ S = −γ 0 −1 Sγ S = +γ 0

  

(2.3.118) (2.3.119) (2.3.120)

Thus S = eiφ γ 0 = P (Parity operator) i.e. 0 0 ψ (x ) = eiφ γ 0 ψ(x) (for parity operation)

(2.3.121) (2.3.122)


47

Complete set of γ matrices (4 × 4 matrices). I γ µ γ5 γ µ γ5 σ µν (1) (4) (1) (4) (6)

→ γ5 = iγ 0 γ 1 γ 2 γ 3 i µ ν σ µν = [γ , γ ] 2 © µ 5ª γ ,γ = 0 (anticommute!)

∴

I = γ5 = = = γ 0 γ5 = =

·

¸ · ¸ · 1 0 0 1 0 i , γ = , γ = 0 1 1 0 ¸· ¸· ¸· · 0 σ2 0 σ1 0 1 i −σ 2 0 −σ 1 0 1 0 · ¸ · ¸ −σ 1 0 −iσ 1 0 i 0 σ1 0 −iσ 1 ¸ · −1 0 0 1 · ¸· ¸ 0 1 −1 0 1 0 0 1 ¸ · 0 1 −1 0 i

γ γ5

σ

0i

σ ij

= (16 total) (total linearly independent) (2.3.123)

·

0 σi = −σ i 0 · ¸ 0 σi = σi 0

¸·

−1 0 0 1

(2.3.124) (2.3.125) (2.3.126) 0 σi −σ i 0

0 σ3 −σ 3 0

¸

¸

(2.3.128)

(2.3.129)

¸

·

¸· ¸ · ¸ 0 1 0 σi −σ i 0 = iγ γ = i =i 1 0 −σ i 0 0 σi · k ¸ · ¸· ¸ i σ 0 0 σi 0 σj ijk = −i² = −σ j 0 0 σk 2 −σ i 0 0 i

(2.3.127)

Note, for example, γ α γ β γ ρ = g αβ γ ρ − g αρ γ β + g βρ γ α − i²καβρ γκ γ5 (²0123 = +1)

(2.3.130)

(2.3.131) (2.3.132)

48


¯ Transformation properties of bilinears in ψ, ψ: ¯ ψψ

¯ 5ψ ψγ

→ scalar ¾ ψ → Sψ ¯ ¯ i.e. ¯ ¯ −1 ∴ ψψ → ψψ ψ → ψS ¯ −1 γ5 Sψ ψS

→

© ª • If S = exp − 4i ωµν σ µν ; [γ5 , σ µν ] = 0

i i i.e. γ5 (γ µ γ ν − γ ν γ µ ) − (γ µ γ ν − γ ν γ µ ) γ5 = 0 2 2

Thus, [γ5 , S] = 0 ¯ 5 ψ −→ ψγ ¯ 5 S −1 Sψ ψγ |{z} S

=

¯ 5ψ ψγ

• If S = P = eiφ γ 0 (i.e. if S = parity operator)

¯ 5 ψ −→ ψγ ¯ 0 e−iφ γ5 eiφ γ 0 ψ = −ψγ ¯ 5 γ0 γ0 ψ ψγ |{z} P

¯ 5ψ = −ψγ

(Picks up -ve → “Pseudo Scalar” ). So also, one can show that ¯ νψ ¯ µ ψ −→ aµ ψγ ψγ |{z} ν S

¯ 0ψ ¯ 0 ψ −→ ψγ ψγ |{z} P

¯ i ψ −→ −ψγ ¯ iψ ψγ |{z} P

(The above are vectors)

and also ¯ ν γ5 ψ ¯ µ γ5 ψ −→ aµ ψγ ψγ |{z} ν S

¯ 0 γ5 ψ ¯ 0 γ5 ψ −→ −ψγ ψγ |{z} P

¯ i γ5 ψ −→ +ψγ ¯ i γ5 ψ ψγ |{z} P

(The above are axial vectors)


49

and ¯ µν ψ −→ aµ aν ψσ ¯ λκ ψ ψσ |{z} λ κ S

¯ 0i ψ −→ −ψσ ¯ 0i ψ ψσ |{z} P

¯ ij ψ ¯ ij ψ −→ +ψσ ψσ |{z} P

(The above are Tensors)

²0123 = +1 ( Right Handed System) ²µνλσ |{z} −→ aµα aνβ aλγ aσδ ²αβγδ S

²

µνλσ

−→ −²µνλσ (²0123 = −1 → Left handed system) |{z} P

Thus as

σ µν γ5 = i²µνλσ σλσ

we have

¯ µν γ5 ψ is a pseudotensor ψσ

(in tensor ex., above, S eq. would hold, but signs on P ’s change). Oct. 6/99 Useful Identities γµ γα γ µ = [γµ γα + γα γµ −γα γµ ]γ µ | {z } 2gµα

= 2gµα γ µ − γα γµ γ µ ;

γα γβ γγ

1 1 γµ γ µ = (γµ γ µ + γ µ γµ ) = (2gµµ ) 2 2

1 = 2γα − γα (2(4)) 2 = −2γα = gαβ γγ − gαγ γβ + gβγ γα − i²καβγ γ κ γ5 = gµα γ µ − gµµ γα + gαµ γµ − i ²µκµα γ κ γ5 |{z}

(2.3.133)

=0

= γα − 4γα + γα = −2γα

(2.3.134)

So also, γµ γα γβ γ µ γµ γα γβ γδ γ µ Tr[γµ γβ ] Tr[γµ γβ γγ γδ ]

= = = =

4gαβ −2γδ γβ γα 4gµβ 4(gαβ gγδ − gαγ gβδ + gαδ gβδ )

(2.3.135) (2.3.136) (2.3.137) (2.3.138)

50


Tr[γµ1 . . . γµ2n+1 ] = = = ∴ Tr[γµ1 . . . γµ2n+1 ] =

Tr[γµ1 . . . γµ2n+1 γ5 γ5 ] ; γ5 γ5 = 1 Tr[γ5 γµ1 . . . γµ2n+1 γ5 ] (Using property of traces) (−1)2n+1 Tr[γ5 γµ1 . . . γµ2n+1 γ5 ] (Moving γ5 through all (2n+1) matrices) 0 (2.3.139)

Tr[γµ γβ γ5 ] = 0

(2.3.140)

Tr[γα γβ γγ γδ γ5 ] = 4i²αβγδ

(2.3.141)

γ5 = γ 5 = iγ 0 γ 1 γ 2 γ 3 = −iγ0 γ1 γ2 γ3

2.4

General Solution to the free Dirac Equation (iγ µ ∂µ − m)ψ = 0

(2.4.1)

If 0

γ =

·

+1 0 0 −1

¡ ¢ then in the rest frame i.e. ∂x∂ i ψ = 0 

 ψ = α1   |

Using Boosts: ½

e−imt 0 0 0 {z ψ10



¸



i

,

γ =

0

 −imt   +α2  e  0  0 | {z } ψ20



·

0 σi −σ i 0



¸

= −γi

  0   0    +α3  imt  +α4    e   0 | {z } | } ψ30

(2.4.2)

 0 0   0 

(2.4.3)

eimt {z } ψ40

¾ i µν S = exp − σµν ω → Can express in closed form. 4 ½ ¾ i −v 1 ψi = exp − ωσ0 ψi0 → cosh(ω) = √ , sinh(ω) = √ 2 1 − v2 1 − v2

(2.4.4) (2.4.5)

2.4. GENERAL SOLUTION TO THE FREE DIRAC EQUATION

51

Recall, if

then

But σµν = 2i [γµ , γν ], So,

x − vt = cosh(˜ ω x) − sinh(˜ ω t) x0 = √ 1 − v2 −vx + t t0 = √ = − sinh(˜ ω x) + cosh(˜ ω t) 1 − v2 1 cosh(˜ ω) = √ 1 − v2 v sinh(˜ ω) = √ 1 − v2

σ01 = iγ0 γ1 · ¸· ¸ 1 0 0 −σ1 = 0 −1 σ1 0 ¸ · 0 −σ1 (2.4.6) = i −σ1 0 ¾ ½ i ∴ S = exp − ωγ0 γ1 2 ¸¾ · ½ ω 0 σ1 (2.4.7) = exp − 2 σ1 0 ¸ ¸2 · · 1 0 0 σ1 2 = Recall, σ1 = 1 → 0 1 σ1 0 · · · ¸2 ¸3 ¸ ω 0 σ1 1 ³ ω ´3 0 σ1 1 ³ ω ´2 0 σ1 ∴ S = 1− − + ··· + σ1 0 σ1 0 2 σ1 0 2! 2 3! 2 ¡ ¢ ¡ ¢ ¸ · − sinh ¡ω2 ¢σ1 cosh¡ ω2¢ (Closed form) (2.4.8) = − sinh ω2 σ1 cosh ω2

But tanh(ω) = −v, and recall the trig identities,

sinh(x + y) = sinh(x) cosh(y) + cosh(x) sinh(y) ; ∴ sinh(2θ) = 2 sinh(θ) cosh(θ) cosh(2θ) = cosh2 (θ) + sinh2 (θ) then tanh

³ω ´ 2

tanh(ω) q 1 + 1 − tanh2 (ω) −v √ = 1 + 1 − v2 =

x=y=θ

52


But p =

√mv 1−v 2

E=

√m , 1−v 2

³ω ´

−p 2 E+m r ³ω ´ E+m cosh = 2 2m

tanh

∴

Hence,

=

½ ¾ i S = exp − ωσ01 (x-direction) 2  1 0 0 r p E+m 0 1 E+m  = p 1 2m  0 E+m p 0 0 E+m

p E+m

0 0 1

For a boost in an arbitrary direction:

(p± = p1 ± ip2 ) S =

r

   

(2.4.9)

(2.4.10) 

E+m  2m 

1 0

0 1

p3 E+m p+ E+m

p− E+m p3 E+m

p3 E+m p+ E+m

1 0

p− E+m p3 E+m

0 1

   

(2.4.11)

Call the columns of the above matrix W1 (p), W2 (p), W3 (p), W4 (p). General solutions to the Dirac equation are: ψ1 = W1 (p)e−ip·x ψ3 = W3 (p)eip·x Properties of Wi (p)

ψ2 = W2 (p)e−ip·x ψ4 = W4 (p)eip·x (Where imt → ip · x)

1. +1, r = 1, 2 −1, r = 3, 4 ¯ ( W = W + γ0 )

(6 p − ²r m)Wr (p) = 0 ; ²r = ¯ (p)(6 p − ²r m) = 0 ; W

½

¯ r (p)W r0 (p) = δrr0 ²r 2. W 3. (W r (p))+ · W r (p) = 0

E rr 0 δ m

→ 4th component of a vector.

¯ r γ µ W r0 W 0 = (W r )+ γ 0 γ µ W r → µ = 0 → γ 0γ 0 = 1 0 = (W r )+ W r pµ (~p, E) and = m m


53

4. Completeness: 4 X

¯ βr (p) = δαβ = ²r Wαr (p)W

r=1

Oct. 8/99 So we have the following;

W10

W30

 1  0  −imt  W20 =   0 e 0   0  0   imt W40 =   0 e 1

 1  0  −imt  , =   0 e 0   0  0  imt  , =   1 e 0





(2.4.12)

(2.4.13)

Now, boost to a frame where they have momentum p. For example,   1 r  0  E + m −ip·x  W1 (p) =  e pz  E+m  2m 0 Σ03 W1,3 = 0 Σ03 W2,4

=

px +ipy E+m 0 W1,3 0 −W2,4



 =  

Where Σ03

(2.4.15) (2.4.16)



1

−1

(2.4.14)

1 −1

 = 

·

σz σz

¸

(2.4.17)

To obtain an eigenstate of ~σ · Sˆ with eigenvalue ±1, we consider i

ˆ

0,s 0 e− 2 φS·~σ W1,2 = W1,2 (also for 3, 4) where cos φ = Sˆ · n ˆz (0,s)

Upon boosting Wi

(2.4.19)

, we obtain, (uz )

U (p, uz ) = W1 U (p, −uz ) =

V (p, −uz ) = V (p, uz ) =

where now as

(2.4.18)

(p)

(2.4.20)

(u ) W2 z (p) (u ) W3 z (p) (u ) W4 z (p)

(2.4.21) (2.4.22) (2.4.23)

54

CHAPTER 2. GRASSMANN VARIABLES 1. we have, (6 p − εr m)Wr(s) (p) = 0 so (6 p − m)U (p, uz ) = 0 and (6 p + m)V (p, uz ) = 0

(2.4.24) (2.4.25) (2.4.26)

2. and, as (0,s)

(0,s)

ˆ 1,3 ~σ · SW = W1,3 (0,s) (0,3) ˆ 2,4 ~σ · SW = −W2,4

(2.4.27) (2.4.28)

⇒ in the rest frame, (pr = (0, 0, 0, m)), ˆ ~σ · Su(0, 0, 0, m; s) = u(p, s) ˆ ~σ · Sv(0, 0, 0, m; s) = −v(p, s) ˆ For example ~σ · Su(0, 0, 0, m; −s) = −u(p, −s)

(2.4.29) (2.4.30)

In general, any spinor which is a solution to the Dirac equation can be described as a linear superposition of these four eigenstates characterized by 1. pµ (p2 = m2 ) “Mass shell condition” 2. Sign of p0 (i.e. in the rest frame,

p0 m

= ±1)

3. Eigenvalue of ~σ · Sˆ in the rest frame. Thus, ψ(~x, t) = =

Z

Z

d4 k ψ(km ) d4 k δ(k 2 − m2 )

(2.4.31) X s=±

{b(k, s)u(k, s) + d∗ (k, s)v(k, s)}

(2.4.32)

Note if, (iγ · ∂ − m)ψ = 0 then (iγ · ∂ + m)(iγ · ∂ − m)ψ = 0 (−γ · ∂γ · ∂ − m2 )ψ = 0 ³ ´ 1 µ ν µ ν ν µ µν 2 Note: γ · ∂ γ · ∂ = γ ∂µ γ ∂ν = (γ γ + γ γ )∂µ ∂ν = g ∂µ ∂ν = ∂ 2 ∴ (∂ 2 + m2 )ψ = 0 (K.G. Eqn.) i.e., as pµ = i∂µ , (p2 − m2 )ψ(p) = 0

(2.4.33) (2.4.34) (2.4.35)

(2.4.36) (2.4.37)


55

Aside: Z

Recall

∞

dx δ(cx)f (x) ;

Z−∞ ∞

=

−∞

−∞

2

³y ´ dy δyf |c| c

f (0) δ(x) ⇒ δ(cx) = |c| |c|

= So, note that; Z ∞

y = cx

2

dx δ(x − a )f (x) = =

Z

∞

dx δ ((x − a)(x + a)) f (x) (let a > 0)

−∞ Z a+ε

dx δ ((x − a)(2a)) f (x)

a−ε

+

Z

−a+ε

dx δ ((x + a)(−2a)) f (x)

f (a) f (−a) + |2a| | − 2a| f (a) + f (−a) = |2a| δ(x + a) + δ(x − a) thus δ(x2 − a2 ) = |2a| Z

4

2

(2.4.40)

Thus, ψ(~x, t) =

d4 k δ(k 2 − m2 )

Z

X£

3

dk

(2.4.42)

∞

b(k, s)u(k, s)θ(k0 )e−ik·x + d∗ (k, s)v(k, s)θ(−k0 )e+ik·x

s=±

Integrate over k0 and insert normalizing factor 2 • +ve energy solution: +

(2.4.41)

dk0 δ(k02 − ~k 2 − m2 )ψ(~k, k0 ) (2.4.43) −∞ p Ã Z Z 2 δ(k0 + ~k 2 + m2 ) p = d3 k dk0 2 ~k 2 + m2 p ! δ(k02 − ~k 2 + m2 ) p ψ(k) (2.4.44) + 2 ~k 2 + m2

2

d k δ(k − m )ψ(k) =

Z

Z

(2.4.39)

−a−ε

=

Hence for

(2.4.38)

ψ (~x, t) =

Z

d3 k (2π)3/2

r

q

m (2π)3

p , (E = ~k 2 + m2 )

mX b(k, s)e−ik·x u(k, s)θ(k0 ) E s=±

¤

(2.4.45)

(2.4.46)

56

CHAPTER 2. GRASSMANN VARIABLES • -ve energy solution: −

ψ (~x, t) =

Z

d3 k (2π)3/2

r

m X ∗ d (k, s)eik·x v(k, s)θ(−k0 ) E s=±

(2.4.47)

Oct 12/99

2.5

Charge Conjugation

Add in an external potential (recall 6 ∂ = γ µ ∂µ ) 0 = (i 6 ∂ − e 6 A − m) ψ(~x, t) (pµ → pµ − eAµ h i ∂ ~ →i ψ= α ~ · (~p − eA) + βm + eΦ ψ ∂t

(2.5.1)

0 = (i 6 ∂ + e 6 A − m)ψc

(2.5.2)

We can show that there is a wave function ψc (~x, t) satisfying,

Taking the complex conjugate of (2.5.1),

If we now set

0 = (γµ∗ (−i∂ µ − eAµ ) − m)ψ ∗ £ ¤ = Cγ0 γµ∗ (i∂ µ + eAµ ) + m (Cγ0 )−1 (Cγ0 )ψ ∗ (cγ0 )(γµ∗ )(Cγ0 )−1 = −γµ and ψc = Cγ0 ψ ∗

(2.5.3) (2.5.4)

then we recover (2.5.2). Now, as γ

0

=

·

0 1 1 0

¸

i

γ =

·

0 iσ i −iσ i 0

¸

(2.5.5)

we find that as γ 0 γ µ∗ γ 0 = γ µT (Cγ 0 )−1 = γ 0 C then (2.5.3) becomes CγµT C −1 = −γµ

(2.5.6)

A solution to this equation is C = iγ 2 γ 0 = C ∗ = −C −1 = −C + = −C T

(2.5.7) (2.5.8)

2.6. MAJORANA SPINORS

57

With this, (γµ (i∂ µ + eAµ ) − m)ψc = 0

(2.5.9)

ψ¯ = ψ + γ 0 = (ψ ∗ )T γ 0 then ψc = C ψ¯∗

(2.5.13) (2.5.14)

Note that ψ and ψc transform in the same way under a Lorentz transformation. If, −i ψ 0 (x0 ) = S(a)ψ(x) ; S(a) = exp( ω µν σµν ) (2.5.10) 4 then ψc0 (x0 ) = S(a)ψc (x) (2.5.11) (2.5.12) Note that as,

If we now set ψ =

Z

ª d3 k X © −ip·x ∗ +ip·x b(p, s)u(p, s)e + d (p, s)v(p, s)e (2π)3/2 s=±

then vc (p, s) = C u¯T (p, s) uc (p, s) = C v¯T (p, s)

(2.5.15)

(2.5.16) (2.5.17)

Consequently, charge conjugation takes one from a positive energy solution associated with a charge e to a negative energy solution associated with a charge −e. (You can map one solution to another by (2.5.16), (2.5.17).

2.6

Majorana Spinors

A Majorana spinor satisfies an extra condition that ψ = ψc

(Lorentz invariant condition)

(2.6.1)

As ψc = C ψ¯T Z d3 k X ∗ T ip·x (b C u¯ e + dC v¯T e−ip·x ) = (2π)3/2 s=±

(2.6.3)

b(p, s) = d(p, s)

(2.6.4)

(2.6.2)

then we see that if ψ satisfies the Majorana condition, then

Note that in Euclidean space, ψc = Cψ ∗T , and if ψ = ψc , then ψ = 0 (only way). i.e. it’s not always possible to have Majorana condition.

58


2.7

Time Reversal

Recall that the Parity operation takes (x, y, z, t) → (−x, −y, −z, t) ψ 0 (x0 ) = P ψ(x) → P = eiφ γ 0 With time, the situation is slightly different. We must preserve, h i ∂ψ ~ i = Hψ = α ~ · (−i∇ − eA) + βm + eΦ ψ ∂t and this must be equivalent to (~r0 = ~r, t0 = −t)

∂ψ 0 (t0 ) = H 0 ψ 0 (t0 ) ∂t0 h i 0 0 0 ~ with H = α ~ · (−i∇ − eA ) + βm + eΦ Z recall: Aµ (x) = dx0 DR (x − x0 )γ˙ µ (x0 ) i

¤Aµ = −4πjµ

(2.7.1) (2.7.2)

(2.7.3)

(2.7.4) (2.7.5) (2.7.6) (2.7.7)

~ 0 (t0 ) = −A(t). ~ Here, Φ0 (t0 ) = Φ(t) , A In going from t → −t, we have the same physical process, but using a camera that’s running backwards (sequence of events in reverse order). A Wigner time transformation takes us from ψ(t) to ψ 0 (t0 ). ψ 0 (t0 ) = T ψ(t) where T OT −1 = T O∗ T −1

(2.7.8) (2.7.9)

with 0 = any operator, T is anti-unitary, and T is an ordinary matrix. i.e. T iT −1 = T (−i)T −1 = −i T ψ = T ψ∗

(2.7.10) (2.7.11)

Starting with (2.7.3), h i ∂ψ ~ + βm + eΦ T −1 T ψ = T α ~ · (−i∇ − eA) | {z } ∂t h i ∗ ∂T ψ ~ + T β ∗ T −1 m + eΦ (T ψ ∗ ) = (T α ~ ∗ T −1 )(i∇ − eA) −i ∂t but t = −t0 h i ∂(T ψ ∗ ) ~ 0 ) + T β ∗ T −1 m + eΦ0 (T ψ∗) +i = (T α ~ ∗ T −1 )(i∇ + eA ∂t This is the same as (2.7.4) provided: Ti

(2.7.12) (2.7.13)

(2.7.14)

2.7. TIME REVERSAL

59

1. ψ 0 (t0 ) = T ψ ∗ (t) 2. T α ~ ∗ T −1 = −~ α 3. T β ∗ T −1 = +β The solution is: T = −iα1 α3 = +iγ 1 γ 3

(2.7.15) (2.7.16)

The Dirac equation is invariant under an SO(3, 1) transformation ψ 0 (x0 ) = S(a)ψ(x) (as well as P, C, T →Parity, Charge Conj., time). Under the rotation subgroup SO(3), these are spinors with s = 1/2. Oct. 13/99

60


Chapter 3 Bargmann-Wigner Equations The Bargman-Wigner fields are fields of higher spin (s = 1, 3/2, 2, 5/2, . . .) that are built up out of the spin (1/2) fields. Spin 1/2 : (iγµ ∂ µ − m)αβ ψβ = 0

(3.0.1)

These higher spin fields are associated with wave functions: ψα1 α2 ...α2s (x) → spin s , totally symmetric in these 2s indices i.e. ψα → spin 1/2 ψαβ = ψβα → spin 1 etc. The wave functions satisfy 2s equations: £ ¤ 0 = i(γµ )α1 α01 ∂ µ − mδα1 α01 ψα01 α2 ...α2s ¤ £ 0 = i(γµ )α2 α02 ∂ µ − mδα2 α02 ψα1 α02 ...α2s .. . ¤ £ 0 = i(γµ )α2s α02s ∂ µ − mδα2s α02s ψα1 α2 ...α02s In the frame where

∂ ψ ∂xi

·

(3.0.3) (3.0.4) (3.0.5)

= 0 (rest frame p~ = 0), we get ¸ · ∂ 0 0 = iγαi α0i 0 − mδαi α0i ψα1 ...α0i ...α2s ∂x

¸ 1 0 , the solutions generalize the spin with γ = 0 −1      1 0 0  0  −imt  1  −imt  0   W1 =  , W2 =  , W3 =   0 e  0 e  1 0 0 0 0

When there are 2s indices +ve energy solutions

61

(3.0.2)

(3.0.6)

1/2 solution, 

 +imt e , 



 0  0  +imt  W4 =  (3.0.7)  0 e 1

62

CHAPTER 3. BARGMANN-WIGNER EQUATIONS (1)

(1) δ1α1 δ1α2 . . . δ1α2s = ψα1 ...α2s (2) δ2α1 δ1α2 δ1α3 . . . δ1α2s + δ1α1 δ2α2 δ1α3 . . . δ1α2s + ... + δ1α1 δ1α2 δ1α3 . . . δ2α2s = ψα(2)1 ...α2s (3) δ2α1 δ2α2 δ1α3 . . . δ1α2s + . (..)

¡ 2s ¢ 2

symmetrized terms.

(2s+1)

(2s+1) δ2α1 δ2α2 δ2α3 . . . δ2α2s = ψα1 ...α2s Similarly, there are ¡ 1 (2s¢ + 1) negative energy solutions. The operator 2 σz αβ operating on ψβ is generalized to

2 (Σz )α1 β1 ,α2 β2 ,...,α2s β2s = (σz )α1 β1 δα2 β2 δα3 β3 . . . δα2s β2s + δα1 β1 (σz )α2 β2 δα3 β3 . . . δα2s β2s .. . + δα1 β1 δα2 β2 δα3 β3 . . . (σz )α2s β2s

(3.0.8)

This acts on ψβ1 ...β2s Σz ψ (1) = sψ (1) ¶ µ 1 (2) ψ (2) Σz ψ = s− 2

(3.0.9) (3.0.10)

Σz ψ (3) = (s − 1) ψ (3) .. . (2s+2) Σz ψ = (−s) ψ (2s+2)

(3.0.11) (3.0.12)

Thus ψ (i) are the (2s + 1) eigenvectors of (Σz ). These are the spin eigenstates. Now, let’s examine the spin-one equations (s = 1). Consider ψαβ (x) = ψβα (x). We have, ¢ ¡ µ (3.0.13) 0 = iγαβ ∂µ − mδαβ ψβγ ¡ µ ¢ 0 = iγγδ ∂µ − mδγδ ψαδ (3.0.14) C −1 γµ C = −γµT

¡

C = iγ0 γ2 = C ∗ = −C −1 = −C + = −C T

¢

(3.0.15)

63 With C, we see that (γµ C)αβ = (γµ C)βα (Σµν C)αβ = (Σµν C)βα

- 4 matrices - 6 matrices

µ

Σµν

i = [γµ , γν ] 2

¶

For a total of 10 symmetric 4 × 4 matrices. We can then have, 1 ψαβ (x) = mAµ (x) (γµ C)αβ + F µν (x) (Σµν C)αβ 2 Substitute this into (3.0.13), (3.0.14), and add; · ¸ 1 µν ν µ 0 = (iγµ ∂ − m)αδ mA (γν C)δβ + F (Σµν C)δβ 2 ¸³ · ´ ¡ ¢ ← − 1 µν ν + mA (γν C)αδ + F (Σµν C)αδ i γµT δβ ∂ µ − mδδβ 2

(3.0.16)

(3.0.17)

−1 Now, right multiplying by Cβ² (remember that CγµT C −1 = −γµ ). This gives,

¢ ¡ 0 = 2m [γµ , γν ] (∂ µ Aν ) + i [γµ , Σλσ ] ∂ µ F λσ − 2m2 γµ Aµ − mΣλσ F λσ

(3.0.18)

[γµ , γν ] = −2iΣµν [γµ , Σλσ ] = 2i (gµλ γσ − gµσ γλ )

(3.0.19) (3.0.20)

but recall that we know,

So, 0 = 2mΣµν ∂ µ Aν − 2 (gµλ γσ − gµσ γλ ) ∂ µ F λσ − 2m2 γµ Aµ − mΣµν F µν £ ¤ 0 = mΣµν [(∂ µ Aν − ∂ ν Aµ ) − F µν ] + γ µ −4∂λ Fµλ − 2m2 Aµ

Thus, we get the Proca Equations (Massive spin 1 equations). Fµν ∂µ F µν

= ∂µ Aν − ∂ν Aµ 2 = − m2 Aν

(3.0.21)

If we now set m = 0, we recover the free Maxwell equations: 0 = ∂µ Aν − ∂ν Aµ Fµν µν ∂µ F = 0

(3.0.22) (3.0.23)

64

CHAPTER 3. BARGMANN-WIGNER EQUATIONS

Where, recall that (3.0.22) gives ~ =0 , ∇·B

~ ~ + ∂B = 0 ∇×E ∂t

~ =0 , ∇·E

~ ~ − ∂E = 0 ∇×B ∂t

and (3.0.23) gives

with, F ij = ²ijk Bk , F 0i = E i . Oct. 15/99 Transformation of the field Aµ (x) under a Lorentz transformation. ½ ¾ i µν 0 0 c.f. ψ (x ) = exp − ω σµν ψ(x) 4 i → σ µν = [γ µ , γ ν ] 2 Since, {γµ , γν } = 2gµν → (+, −, −, −) ¤ ¡ ¢ ∴ σ µν , σ λρ = A g µλ σ νρ − g νλ σ µρ + g νρ σ µλ − g µρ σ νλ ¡ ¢ = −2i g µλ σ νρ − . . . £

(Aside - if µ = λ = 1 , ν = 2 , ρ = 3, ¤ £ LS = σ 12 , σ 13 £ ¤ = iγ 1 γ 2 , iγ 1 γ 3 ¡ ¢ = − γ 1γ 2γ 1γ 3 − γ 1γ 3γ 1γ 2 = −γ 2 γ 3 + γ 3 γ 2 £ ¤ = − γ2, γ 3 = 2iσ 23 RS = Ag 11 σ 23 = −Aσ 23 ⇒ A = −2i)

For Aµ , since it is a vector, under an infinitesimal Lorentz transformation, we have, x0µ = xµ + ω µν xν (ω µν = −ω νµ ) A0µ (x0 ) = Aµ (x)ωµν Aν (x) ¢ i ¡ λσ = Aµ (x) + ω Sλσ µν Aν (x) u Where (Sλσ )µν = 2i (gλµ gσν − gλν gσµ )

(3.0.24)

(3.0.25) (3.0.26)

65 But now, (Sλσ )(Sαβ ) − (Sαβ )(Sλσ ) ≡ (Sλσ )κµ (Sαβ )κν − (Sαβ )κµ (Sλσ )κν = −2i (gλα Sσβ − gσα Sλβ + gσβ Sλα − gλβ Sσα )µν (3.0.27) i.e. µ ½ ¾¶ i λσ 0 0 Aµ (x ) = exp ω Sλσ Aν (x0 ) (3.0.28) 4 µν for a finite transformation. ψ - spin 1/2 representation of SO(3, 1) (fundamental rep.) Aµ - spin 1 representation of SO(3, 1) (Adjoint representation). i.e. under a rotation, ½

¾ ~σ ψ → exp i · ω ~ ψ 2

where

hσ σ i σk i j , = i²ijk 2 2 2 ¶ µ ¶2 µ ~σ 1 1 +1 = 2 2 2

and A00 (x0 ) = A0 (x) i h ~ A0i (x0 ) = ei~ω·S Aj (x) ij

where 

     0 0 0 0 0 i 0 −i 0 S1 =  0 0 −i  , S2 =  0 0 0  , S3 =  i 0 0  0 i 0 −i 0 0 0 0 0

(3.0.29)

Note that,

1 ²ijk (Sk )mn 2 = −i²mij

(Si )mn =

[Si , Sj ] = i²ijk Sk ~ 2 = 1(1 + 1) S = s(s + 1) Massless Particles

(3.0.30) (3.0.31)

(3.0.32)

66

CHAPTER 3. BARGMANN-WIGNER EQUATIONS Dirac equation:

→ γ0 =

·

0 = (iγ µ ∂µ − m) ψ (3.0.33) ¸ ¸ · ¸ · i 0 σ −1 0 0 1 , γi = , γ5 = i 0 1 1 0 −σ 0 ¡ ∂ ¢ ¸· ¸ · φ ¡ ∂ −m ¢ i ∂t − ~σ · p~ , (~p = −i∇) (3.0.34) ∴0 = χ −m i ∂t + ~σ · p~

If m = 0, then we get the Decoupled (Weyl) equations; ∂ χ = ~σ · p~χ ∂t ∂ i φ = −~σ · p~φ ∂t

i

More generally, if we let p+ =

1+γ5 2

, p− =

1−γ5 2

p+ + p − (p+ )(p− ) (p+ )2 (p− )2

(3.0.35) (3.0.36)

→ where {γ5 , γµ } = 0 , γ52 = 1, = = = =

1 0 p+ p−

So, ·

¸ 0 0 p+ = 0 1 · ¸ 0 → p+ ψ = χ

·

1 0 p− = 0 0 · ¸ φ p− ψ = 0

¸

(3.0.37) (3.0.38)

If ψ± = p ± ψ

(3.0.39)

0 = iγ · ∂ψ+ − mψ− 0 = iγ · ∂ψ− − mψ+

(3.0.40) (3.0.41)

then

For massless particles, we can’t go to the rest frame, i.e. p2 = (p0 )2 − p~2 = m2 = 0 → there is no rest frame, so look at the frame where pµ = (0, 0, p, p) ;

(p1 = p2 = 0 , p3 = p0 = p)

(3.0.42)

67 In this frame, p+

p−

·

·

1 0 0 1

1 0 0 1

¸·

¸·

χ1 χ2

φ1 φ2

¸

·

¸· ¸ 1 0 χ1 = p+ χ2 0 −1 (first of Weyl equations) ¸ · χ1 χ = 0 (only one state of negative energy) ·

¸

1 0 0 −1

¸·

φ1 φ2

¸

= −p− ¸ · 0 φ = φ2 (only one state of positive energy)

(3.0.43)

(3.0.44)

(3.0.45) (3.0.46)

So, Weyl equations are: • Negative energy solutions: • Positive energy solutions:

∂ χ = ~σ · p~χ ∂t

(3.0.47)

∂ φ = −~σ · p~φ ∂t

(3.0.48)

i

i

Oct. 19/99 Can we define “spin” of a massless particle, seeing that there is no rest frame? - Employ the Pauli-Lubanski tensor. i Sµ = ²µνλσ σ λσ k ν (3.0.49) 2 i £ λ σ¤ γ ,γ → σ λσ = 2 In the rest frame of a massive (m) particle, k µ = (0, 0, 0, m) i Sµ = ²µ0νλ σ νλ m 2 (0 in ² index → µ, ν, λ all spatial.)

(3.0.50)

²0123 = +1 i Si = ²ijk σ jk m 2 · ¸ σi 0 = m 0 σi

(3.0.52)

(3.0.51)

(3.0.53)

68


Note: σij σi

· ¸ i σk 0 = ²ijk 0 σk 2 · ¸ 0 iσi = iσi 0

σ . Thus, in the rest frame 12 Sµ reduces to m~ 2 1 S generalizes spin to an arbitrary frame. For a massless particle moving along the 2 µ z-axis (with k 2 = 0, k µ = (0, 0, k, k)), then we have,

i ²1λσν k ν σ λσ 2 i = (²1λσ3 k + ²1λσ0 k) σ λσ 2 etc. for S2 , S3 , S4

S1 =

(3.0.54)

Now consider the two massless solutions to the Bargmann Wigner equations Wα(1) = δα1 2 δα2 2 . . . δα2s 2 1 α2 ...α2s

(3.0.55)

Wα(2) = δα1 3 δα2 3 . . . δα2s 3 1 α2 ...α2s

(3.0.56)

Generalize the Pauli Lubanski tensor, (Sµ )α1 β1 ,...,α2s β2s = (Sµ )α1 β1 (δα2 β2 . . . δα2s β2s ) + (Sµ )α2 β2 (δα1 β1 δα3 β3 . . . δα2s β2s ) ... ¡ ¢ + (Sµ )α2s β2s δα1 β1 . . . δα2s−1 β2s−1

(3.0.57)

It can now be shown that

S1 W (1) S2 W (1) S1 W (2) S2 W (2)

=0 =0 =0 =0

S3 W (1) S3 W (2) S4 W (1) S4 W (2)

= −2skW (1) = +2skW (2) = +2skW (1) = −2skW (2)

(3.0.58) (3.0.59) (3.0.60) (3.0.61)

But kµ = (0, 0, −k, k) and thus 12 Sµ = ±skµ . Hence for a massless particle, there are two solutions to the Bargmann-Wigner equations, and these two solutions correspond to spin states whose eigenvalues are ±S, and whose direction is in the direction of kµ . Thus there are only two polarizations for any massless particle of spin S. We lose states through the introduction of gauge symmetry. ex.

69 • massless spin 1/2 - 2¡states ¢ massive spin 1/2 - 2 21 + 1 = 2 states

• massless spin 1 - 2 states massive spin 1 - (2s + 1) = 3 states

• massless spin 3/2 - 2 states massive spin 3/2 - (2s + 1) = 5 states

70


Chapter 4 Gauge Symmetry and massless spin one particles Remember that Fµν = ∂µ Aν − ∂ν Aµ ∂µ F µν = −m2 Aν → 0 as m2 → 0

(4.0.1) (4.0.2)

Aµ → A µ + ∂µ Λ

(4.0.3)

∂µ F µν = 0 Fµν = ∂µ Aν − ∂ν Aµ

(4.0.4) (4.0.5)

If m2 → 0, there is the gauge invariance, We now have,

which are derived from the action µ ¶ Z 1 2 1 4 µν d x Fµν F → L = − F S=− 4 4

(4.0.6)

(the -ve sign is present to keep the energy +ve, and the magnitude 1/4 is convention). i.e. µ ¶ Z 1 δS 4 αβ δFαβ = − d s 2F δAµ (y) 4 δAµ (y) Z 1 δ = − d4 x F αβ (∂α Aβ − ∂β Aα ) 2 δAµ (y) Z ¢ ¡ 1 d4 x F αβ (x) 2∂α δ 4 (x − y) gµβ = − 2 ½ ¾ δAβ (x) 4 → = gµβ δ (x − y) ∂Aµ (y) = ∂α F αβ (x) = 0 71

72

CHAPTER 4. GAUGE SYMMETRY AND MASSLESS SPIN ONE PARTICLES

Just as with L = L (qi (t), q˙i (t)) and S = we have

Z

dt L (qi (t), q˙i (t))

Z

d4 x L(Aµ , ∂ν Aν ) ¶ µ Z Z ∂ 3 = dt d x L Aµ (~x, t), ∇Aµ (~x, t), Aµ (~x, t) ∂t | {z }

S =

(4.0.7)

(4.0.8)

L

qi (t) ↔ Aµ (~x, t), ∇Aµ (~x, t) ∂ q˙i (t) ↔ Aµ (~x, t) ∂t

(4.0.9) (4.0.10)

and, where previously we had i discrete, we now have a continuum (µ) number of degrees of freedom. So, just as pi =

∂L ∂L ¢ → momentum density ¡∂ −→ πµ (~x, t) = ∂ q˙i ∂ ∂t Aµ (~x, t) ¶ µ ∂ c.f. ∂µ = ∂xµ

(4.0.11)

Oct. 20/99 Now, as ∂Aµ (~x0 , t) = gµν δ 3 (~x − ~x0 ) ∂Aν (~x, t) ∂L ¡ ∂Aµ ¢ ∴ πµ (~x, t) = ∂ ∂t ¶ µ ∂ 1 µν ¡ µ ¢ − Fµν F = 4 ∂ ∂A ∂t ¡ ¢ 1 ∂ = − Fλν ¡ ∂Aµ ¢ F λν 2 ∂ ∂t ∂ = ∂0 → ∂t 1 ∂ = − Fλν F λν 2 ∂(∂0 Aµ ) But, → Fλν F λν = F0i F 0i + Fi0 F i0 + Fij F ij + F00 F 00 = 2F0i F 0i + Fij F ij | {z } No t dep.

(4.0.12)

73 ∂ F 0i µ ∂(∂0 A ) 0i → F = ∂ 0 Ai − ∂ i A0

(4.0.13)

∴ πµ = −F0i

(4.0.14)

Thus, π0 = 0 πj

(4.0.15)

∂ = −F 0i F0i ∂(∂0 Aµ ) ∂ = −F 0i (∂0 Ai − ∂i A0 ) ∂(∂0 Aj ) = −F 0j

(4.0.16)

Remember, ´ ³ ~ φ (A0 = φ, Ai = A) ~ A,

Aµ =

(4.0.17)

~ ~ = −∇φ − ∂ A E ∂t = −∂i A0 − ∂0 Ai = +∂ i A0 − ∂ 0 Ai = −F 0i ∴ πj

(4.0.18)

= E j = −F 0j = +F0j

(4.0.19)

(4.0.19) is Canonical momentum (Not mechanical momentum). Canonically conjugate to A3 . i.e. Ai has canonically conjugate momentum π j = E j . Ai A0

= −Ai = +A0

(4.0.20)

= 0

(4.0.21)

π0 i.e. A0 has no conjugate momentum. Define a Poison bracket, [qi , pj ] = δij ⇒ [A, B] =

X ½ ∂A ∂B i

∂A ∂B − ∂qi ∂pi ∂pi ∂qi

¾

(4.0.22)

Similarly [F (Aµ , πµ ), G(Aν , πν )] =

Z

3

dxg

µν

·

δF δG δG δF − µ µ ν ν δA (~x, t) δπ (~x, t) δπ (~x, t) δA (~x, t)

¸

(4.0.23)

74


4.1

Canonical Hamiltonian Density ∂ Aµ − L (re. H = pi q˙i − L) ∂t µ ¶ ¢ 1¡ i 0 0i ij = |{z} π ∂0 A0 + π ∂0 Ai − − 2F0i F + Fij F 4

H = πµ

(4.1.1) (4.1.2)

=0

(→ π 0 = primary constraint)   1 1 H = π i ∂0 Ai − ∂i A0 +∂i A0  + F0i F 0i + Fij F ij | {z } 2 4 F 0i

Remember that,

~ i = (∇ × A) ~ i = ²ijk (∂j Ak ) (B) ²imn Bi = ²imn ²ijk ∂j Ak → ²imn ²ijk = δmj δnk − δmk δnj So that ²imn Bi = (δmj δnk − δmk δnj )∂j Ak = ∂ m An − ∂ n Am = F mn ∴ Fmn

= ²mnl Bl

(4.1.3) (4.1.4)

Consequently: 1 1 H = π i |{z} F 0i +π i ∂i A0 + F0i F 0i + (²ijk Bk )(²ijl Bl ) {z } | {z } 2 4| i π

π i (−π i )

2Bi Bi

1 i i 1 π π + B i B i + π i ∂i A 0 ; π j = E j 2 2 1 ~2 ~2 H = ( E + B ) + π i ∂i A 0 2 =

(4.1.5)

Primary constraint (π 0 (~x, t) = 0. Thus, £ ¤ ∂ 0 π = 0 = π 0 (~x, t), H(~y , t) ∂t

(4.1.6)

But we have the fundamental poisson bracket

[Aα (~x, t), πβ (~y , t)] = gαβ δ 3 (~x = ~y )

(c.f. [qi , pj ] = δij )

(4.1.7)

4.1. CANONICAL HAMILTONIAN DENSITY ·

75

∂ π (~x, t), π (~y , t) i A0 (~y , t) ∂y ∂ 0 = π i (~y , t) i δ 3 (~x − ~y ) ∂y ~ 0 = +∇ · E(~y , t)δ 3 (~x − ~y )

∂ ∴ π0 = 0 = ∂t

0

i

¸

(4.1.8)

~ = 0 (Gauss’s Law). ∴ Secondary constraint ∇ · E Oct. 22/99 So, 1 L = − Fµν F µν 2 π i = −Ei = −F0i ¾ π0 = 0 0 First class Constraints π = 0 ∂i E i = 0 1 Bi = ²ijk Fjk 2 ~2 + B ~2 E H0 = + A 0 ∂i E i 2 Z d3 x H

H =

HT = H 0 + c i φi ~2 + B ~ 2 ) + A0 ∂i Ei + c1 Π0 + c2 ∂i Ei ; = 12 (E

can absorb c2 → A0

(4.1.9) (4.1.10) (4.1.11) (4.1.12) (4.1.13) (4.1.14)

(4.1.15)

∴ A0 - Lagrange multiplier for the constraint ∂i Ei = 0. We need two gauge conditions: • Coulomb gauge (usual choice) A0 = 0 ∂i A i = 0

(4.1.16) (4.1.17)

A0 = 0 A3 = 0 (say)

(4.1.18) (4.1.19)

• Axial gauge also possible

Now, replace Poisson Brackets by Dirac Brackets. The fundamental P.B. is: {Ai (~x, t), Πj (~y , t)} = δij δ 3 (~x − ~y ) Θi = {φi . . . γi }

(4.1.20) (4.1.21)

76


The Dirac Bracket is then: {A, B}∗ = {A, B} − where

d−1 ij

= {Θi , Θj }

X i,j

{A, Θi }dij {Θj , B}

(4.1.22)

So,

½

{Π0 (~x, t), A0 (~y , t)} = δ 3 (~x − ~y ) ¾ ¢ ∂ ∂ ∂ ∂ ¡ 3 −δ δ (~ x − ~ y ) Πi (~x, t), Aj (~y , t) = ij ∂xi ∂yi ∂xi ∂y i = ∇2x δ 3 (~x − ~y )

(4.1.23)

(4.1.24)

We now need (∇2 )−1 = G(~x − ~y ). i.e. 2

3

∇ G(~x − ~y ) = δ (~x − ~y ) = Z

Z

d3 k i~k·(~x−~y) e (2π)3

d3~k i~k·(~x−~y) ~ e g(k) (Fourier transf.) (2π)3 Z d3~k ~ 2 (−~k 2 g(~k)) eik·(~x−~y) ∴ ∇ G(~x − ~y ) = 3 (2π) | {z }

Write G(~x − ~y ) =

=1

Aside:

Z

Z

∞

δ(x) = 0 (x 6= 0) dx δ(x) = 1

−∞

r

r µr ¶ a −ax2 a π dx e = π π a −∞ = 1½ r 0 if x 6= 0 a −ax2 e = lim a→∞ ∞ if x = 0 π ∞

(4.1.25) (4.1.26) (4.1.27)

4.1. CANONICAL HAMILTONIAN DENSITY

77

Thus, δ(x) = So

Z

∞ −∞

dk ikx e = 2π = = = =

r

a −ax2 e a→∞ π Z ∞ dk ikx−ak2 lim e ak 2 is regulator a→0 −∞ 2π Z ∞ ¡ ¢ dk 2 ix lim exp{−a k − 2a } exp{− x4a } a→0 −∞ 2π Z ∞ 0 x2 dk −a(k0 )2 exp{− } lim e a→0 −∞ 2π 4a r 2 π 1 x lim exp{− } a→0 a 2π 4a δ(x) lim

end of aside. Now, recall (4.1.27) ∴ g(~k) = − Thus, G(~x − ~y ) =

Z

d3 k (2π)3

µ

1 ~k 2

−1 ~k 2

(4.1.28) ¶

~

eik·(~x−~y)

(4.1.29)

Say (~x − ~y ) is along k3 axis. µ ¶ Z 2π Z +1 Z ∞ dk 2 −1 ikr cos θ k e ; r = |~x − ~y | G(~x − ~y ) = dφ d(cos θ) (2π)3 k2 0 −1 0 Z 1 Z ∞ 1 dz dk eikrz ; (z = cos θ) = 2 (2π) −1 0 can extend k integral to (−∞) → int. over z makes even Z 1 Z ∞ −1 dz dk eik(rz) = 2(2π)2 −1 −∞ Z −1 1 = dz δ(rz) 4π −1 Z ∞ −1 = dz 0 δ(z 0 ) ; z 0 = rz 4πr −∞ −1 (4.1.30) = 4πr Thus, G(~x − ~y ) =

−1 4π|~x − ~y |

(4.1.31)

78


One can verify that ∇

2

and that Z

3

d ~x ∇

2

µ

−1 4π|~x|

¶

µ

−1 4π|~x − ~y |

¶

=0

, ~x 6= ~y

µ ¶ Z −1 −1 3 = d ~x ∇ · ∇ 4π |~x| by Gauss’s Law (Radius of Sphere = R) ¶ µ Z −1 1 2 = dΩ R n ˆ· ∇ lim 4π R=∞ Sphere |~x| = 1

(4.1.32)

(4.1.33)

Thus our Dirac Brackets are {F (~x, t), G(~y , t)}∗ = {F (~x, t), G(~y , t)} ¾ ¾ ½ ·½ Z ∂ ∂ 0 0 0 0 3 0 3 0 Aj (~y , t), G(~y , t) F (~x, t), 0i Πi (~x , t) G(~x − ~y ) − d ~x d ~y ∂x ∂yj0 ¾¸ ½ ¾ ½ ∂ ∂ 0 0 0 0 Aj (~y , t), F (~y , t) (4.1.34) − G(~x, t), 0i Πi (~x , t) G(~x − ~y ) ∂x ∂yj0 Hence; {Ai (~x, t), Πj (~y , t)}

∗

µ

¶ ∂ ∂ = δij δ(~x − ~y ) − G(~x − ~y ) ∂xi ∂xj µ ¶ ∂i ∂j 1 = δij − 2 δ(~x − ~y ) → 2 = G(~x − ~y ) ∇ ∇ {z } |

(4.1.35)

3 (~ δij⊥ x−~ y)

(4.1.36)

Consistency: ¶ µ ∂ ∂ ∂ ∂ 1 ∗ δ(~x − ~y ) {Ai (~x, t), Πj (~y , t)} = δij − ∂xi ∂xi ∂xi ∂xj ∇2 = 0 ∗ 3 {Ai (~x, t), Πj (~y , t)} = δij⊥ (~x − ~y )

(4.1.37) (4.1.38)

Oct. 26/99 H0 =

→ A˙ 0 = = ˙ → Ai = =

1 2 (Π + B2 ) + c1 Π0 + c2 ∂i Πi 2 [A0 , H] [A0 , c1 Π0 ] = c1 (Entirely arbitrary prior to fixing gauge) [Ai , H] [Ai , 21 Π2 + c2 ∂j Πj ] = Ai − ∂i c2

(4.1.39) (4.1.40) (4.1.41)


79

where the usual gauge invariance in A is recovered; Aµ → A µ + ∂µ Λ c1 = ∂ 0 Λ −∂i c2 = ∂i Λ

(4.1.42)

The Fourier expansion of A(x, t) in the Coulomb gauge (∂i Ai = 0) is

A(x, t) =

Z

2 X © ª d4 k 2 ε(k, λ)[a(k, λ)e−ik·x + a∗ (k, λ)e+ik·x ] δk 4 (2π) λ=1

(4.1.43)

1. δ(k 2 ) is the mass shell condition (i.e. ensures that ¤A = 0 → ∂µ F µν = 0 ⇒ ¤A = 0 if ∂i Ai = 0) 2. ε(k, λ) · k = 0

(λ = 1, 2) (Ensures ∂i Ai = 0). and

ε(k, λ) · ε(k, λ0 ) = δλλ0 k → ε(k, 1) × ε(k, 2) = |k| 3. A = A∗ ⇒ is real. Thus,

occurs.

ae−ik·x + a∗ eik·x

80


Note: Z ∞

−∞

2

Z

2

dx δ(x − a )f (x) =

a+² a−²

Z

dx δ((x + a)(x − a))f (x) −a+²

dx δ((x + a)(x − a))f (x) + −a−² Z −a+² Z a+² dx δ(−2a(x + a))f (x) dx δ(2a(x − a))f (x) + = −a−² a−² Z ∞ f (x) dx δ(cx)f (x) = but recall that |c| −∞ f (a) f (−a) + (4.1.44) = |2a| |2a| Hence δ(x2 − a2 ) = Z

d3 k (2π)3

Z

δ(x − a) + δ(x + a) |2a| 2

∞

X dk0 δ(k02 − k2 ) ε(k, λ)[ae−ik·x + a∗ eik·x ] ∴ A(x, t) = (2π) −∞ λ=1 √ ∗ 2 Rescaling a, a and setting ωk = k , εk,λ = ε(k, λ), etc., we get 1 ∴ A(x, t) = 2

Z

(4.1.45) (4.1.46)

2 £ d3 k 1 X √ ε ak,λ e−i(k·x+ωk t) + ak,λ e−i(k·x−ωk t) k,λ (2π)3/2 2ωk λ=1

¤ + a∗k,λ e+i(k·x+ωk t) + a∗k,λ e+i(k·x−ωk t) (4.1.47)

If k → −k, and if

ε−k,λ (λ = 1) −ε−k,λ (λ = 2) (−k) = |k|

εk,λ = ε−k,1 × ε−k,2

½

(4.1.48) (4.1.49)

then Z 2 2 1 X 1 X d3 k d3 k −i(k·x+ωk t) √ √ εk,λ ak,λ e = ε−k,λ a−k,λ e+i(k·x−ωk t) (2π)3/2 2ωk λ=1 (2π)3/2 2ωk λ=1 (4.1.50) n ∗ λ=1 ak,λ Now set a−k,λ = −a∗ λ=2 . Thus everything collapses to Z

k,λ

A(x, t) =

Z

2 ¤ £ d3 k 1 X ik·x −ik·x ∗ √ e ε a e + a k,λ k,λ k,λ (2π)3/2 2ωk λ=1

where k · x = k · x − ωk t

(4.1.51)


81

(we need not use lane wave basis; any complete set of functions will do). Recall the Klein-Gordon equation: 0 = (¤ + m2 )φ µ 2 ¶ ∂ 2 2 0 = −∇ +m φ ∂t2 Let θ =

∂φ ∂t

→∴

∂θ ∂t

= (∇2 − m2 )φ µ ¶ 1 i Ψ1 = √ φ + θ m 2

1 Ψ2 = √ 2

µ

(4.1.52) (4.1.53)

i φ− θ m

¶

(4.1.54)

Let Ψ = ∂Ψ = ∴i ∂t

·

·

Ψ1 Ψ2

¸

(4.1.55)

¸ (−i∇)2 (τ3 + iτ2 ) + mτ3 Ψ 2m {z } | H ¸ ¸ · · 1 0 0 −i τ3 = → τ2 = 0 −1 i 0

K.G. equation follows from an action ¸ · Z m2 2 1 µ 4 (∂µ φ)(∂ φ) − φ S= dx 2 2 {z } |

(4.1.56)

(4.1.57)

(4.1.58)

L

Now,

∂L = ∂0φ ∂ (∂0 φ) H = Π(∂0 φ) − L ¶ µ 1 1 2 2 2 2 (∂0 φ) − (∇φ) − m φ = Π · (∂0 φ) − 2 2 £ ¤ 1 2 = Π + (∇φ)2 + m2 φ2 2

→Π =

Fundamental P.B.

[φ(x, t), Π(y, t)] = δ 3 (x − y) (¤ + m2 )φ = j Describes Π, K, . . . ∂ψ = p2mψ → we’ve treated as single particle i~ ∂t µ 2 ¶ ∂ψ p i~ = + V ψ → in a classical potential ∂t 2m

(4.1.59)

(4.1.60)

(4.1.61) (4.1.62) (4.1.63) (4.1.64)

82


Dirac showed that the potential V itself, when it’s due to an E.M. field, should be quantized. Make relativistic by considering Dirac eq. i

∂ψ = (α · p + βm)ψ ∂t

(4.1.65)

• the above equation doesn’t describe just 1 particle anymore, but in fact now describes 4 • this one equation describes ψ particles (spin 21 ) • if you put c’s and ~’s back in, ~’s don’t all cancel, → ~ floating around Maxwell’s equations (Fiddle with Bargmann-Wigner (spin 1), and get these) ∂µ F µν = 0 F µν = ∂ µ Aν − ∂ ν Aµ

(4.1.66) (4.1.67)

~’s cancel, if m = 0 → classical equations. (2 particles/polarizations). i.e. in order to include relativity, you end up having to describe more than 1 particle. Wave function describing single particle → actually quantizing wave function itself (2nd quantization). Oct. 27/99

Chapter 5 (2nd) Quantization, Spin and Statistics 5.1

Harmonic Oscillator 1 H = (p2 + ω 2 q 2 ) ; 2

ω=

r

k m

(5.1.1)

Quantize: [q, p] = i ; (~ = 1) p˙ = [p, H] = −ω 2 q q˙ = [q, H] = p

(5.1.2) (5.1.3) (5.1.4)

Define: ωq + ip √ ; 2ω ωq − ip √ = ; 2ω

a = a†

( annihilation)

(5.1.5)

(creation)

(5.1.6)

Note that: [a, a† ] = 1 [a, a] = [a† , a† ] = 0 ¶ µ 1 † H = ω a a+ 2 and [H, a† ] = ωa† [H, a] = −ωa 83

CHAPTER 5. (2N D ) QUANTIZATION, SPIN AND STATISTICS

84 Suppose

H|ni = ωn |ni H(a† |ni) = ([H, a† ] + a† H)|ni = (ωa† |ni + a† ωn |ni) = (ωn + ω)(a† |ni) and similarly H(a|ni) = (ωn − ω)(a|ni)

(5.1.7)

(5.1.8) (5.1.9)

More generally, H(am |ni) = (ωn − mω)(am |ni)

(5.1.10)

For the energy to have a lower bound, there must be a state |0i such that a|0i = 0. In this case, µ ¶ 1 † H|0i = ω a a + |0i ; a|0i = 0 2 1 = ω|0i (Consistent with uncertainty principle). 2 ¢ ~2 1¡ 2 = p + ω 2 q 2 |0i → ∆p ∆q ≥ (5.1.11) 2 4

i.e. Vacuum state |0i is lowest energy state, not nothing. a† |0i ∝ |1i (a† )n |0i ∝ |ni µ ¶ 1 H|ni = n+ ω|ni 2 For hn|mi = δnm

(n = 0, 1, . . .)

(a† )n |0i |ni = p (n + 1)

For our real K.G. field:

¢ ¤ + m2 ψ has a solution Z ¢ d3 k 1 ¡ −ik·x √ φ(x) = ak e + a∗k eik·x 3/2 (2π) 2ωk 0 =

¡

(5.1.12)

(5.1.13)

(5.1.14)

(note that this is only a scalar field here, ∴ don’t need polarization vectors ε k,λ ). For a complex field, we have the same treatment for real and complex parts. → k · x = ωk t − k · x q with ωk = k 2 + m2 fk

e−ik·x p = (2π)3 2ωk

5.1. HARMONIC OSCILLATOR As

85 Z

then

Z

Remember that

d3 k ik·x e = δ 3 (x) (2π)3

d3 x fk∗ φ(x, t) =

(5.1.15)

¤ 1 £ ak + a∗−k e2iωk t 2ωk

∂φ(x, t) ∂t Z £ ¤ i ∴ d3 x fk∗ Π(x, t) = − ak − a∗−k e2iωk t 2

(5.1.16)

Π(x, t) =

Thus, ak = i

If we now quantize

Z

3

dx

·

fk∗ (x, t) (∂t φ(x, t))

|

µ

¶ ¸ ∂ ∗ f (x, t) φ(x, t) − ∂t k {z }

(5.1.17) (5.1.18)

(5.1.19)

↔

⇒fk∗ ∂ φ

ˆ t), Π(y, ˆ t)] = i~δ 3 (x − y) ( let ~ = 1) [φ(x,

(5.1.20)

[ˆ a(k), a ˆ(k 0 )] = 0 [ˆ a∗ (k), a ˆ∗ (k 0 )] = 0 ˆ∗ (k 0 )] = δ 3 (k − k 0 ) [ˆ a(k), a

(5.1.21) (5.1.22) (5.1.23)

then we can show that

and Z

·

m2 2 1 2 1 Π + (∇φ)2 + φ H = dx 2 2 2 · µ ¶¸ Z 1 ∗ 3 = d x ωk a (k)a(k) + 2 3

¸ (5.1.24)

→ Harmonic oscillator! (Note the 12 term above will diverge → at each point, get this vacuum P oscillation → over all points, get ∞.) States in the Hilbert space of these operators: |0k , 0k0 , 0k00 , . . .i = a†l

vacuum ≡ |0i

√ |0i = |0k , . . . , 1l , . . .i 2!

(5.1.25) (5.1.26)


86

i.e. the a†l “creates” a K.G. particle of momentum l. So, (a†l )2 (a†l0 ) √ √ |0i 3! 2!

(5.1.27)

is a state with one K.G. particle of momentum l 0 and two of momentum l. Notation: When we quantize, the ∗ goes to †; a∗k −→ a†k

(5.1.28)

As it is only the difference between energy levels that is observed, we take, :H: →

normal order of H Z :H: = d3 k ωk (a†k ak )

(5.1.29) (5.1.30)

(excitations above lowest energy state)

As 1 H= 2

Z

´ ³ d3 k ωk a†k ak + ak a†k

(5.1.31)

→ : H : is obtained by rewriting H so that all annihilation operators are on the right. More generally, : φ(x)φ(y) : - ordered so that, again, all annihilation operators are on the right hand side. As,

i

Z

↔

d3 xfk∗ ∂ fk0 = δ 3 (k − k 0 ) Z − φ = d3 x a†k fk∗ Z + φ = d 3 x ak fk

Oct. 29/99 ¡ ¢¡ ¢ : φ(x)φ(y) : = : φ− (x) + φ+ (x) φ− (y) + φ+ (y) : = φ− (x)φ+ (y) + φ− (y)φ+ (x) +φ− (x)φ− (y) + φ+ (x)φ+ (y) {z } | inversion

(5.1.32) (5.1.33) (5.1.34)

5.1. HARMONIC OSCILLATOR

87

¤ (φ+ (x) + φ− (x)), (φ+ (y) + φ− (y)) ·Z ³ ´ Z ³ ´¸ † ∗ † ∗ 3 3 0 = d k ak fk (x) + ak fk (x) , d k ak0 fk0 (y) + ak0 fk0 (y)   Z i h i h  d3 k d 3 k 0 † † −i(k·x−k 0 ·y) i(k·x−k0 ·y)   0 √ a , a + a , a e = 0 e k k k k  (2π)3 2ωk · ωk0 | {z } | {z }

[φ(x), φ(y)] =

£

δkk0

Z

3

−δkk0

£ −ik·(x−y) ¤ dk +ik·(x−y) e − e (2π)3 (2ωk ) Z 3 d k ik·(x−y) −i e = sin [ωk (x0 − y0 )] 3 (2π) ωk Z Z 3 dk → d4 k δ(k 2 − m2 ) 2ωk Z d4 k = δ(k 2 − m2 )ε(k0 )e−k·(x−y) (2π)3 ½ +1 k0 > 0 where ε(k0 ) = −1 k0 < 0 = i∆(x − y)

=

(5.1.35)

Properties: ¢ ¡ 0 = ¤x − m2 ∆(x − y) → expected, as (¤ − m2 )φ = 0 ∆(x − y) = −∆(y − x) → Reasonable, as [φ(x), φ(y)] = − [φ(y), φ(x)] Note: ∆(x − y, 0) = 0 i.e. →

Z

d4 k δ(k02 − k 2 − m2 )ε(k0 )e−ik·(x−y) = 0 (2π)3

Thus, ∆(x − y) = 0 if (x − y)2 < 0

(5.1.38)

[φ(x), φ(y)] = 0 if (x − y)2 < 0

(5.1.39)

Hence, as required by causality. If x, y are separated temporally, (i.e. within light cone), then the order of x, y (in time) is the same ∀ observers. If they are separated by spatial separation, (outside light cone), order (time order) not necessarily the same.


88

5.2

Feynman Propagator h0|T φ(x)φ(y)|0i

(5.2.1)

T φ(x)φ(y) = θ(x0 − y0 )φ(x)φ(y) + θ(y0 − x0 )φ(y)φ(x) · ½ ¾¸ 1 x>0 θ(x) = 0 x0 2πi λ − iε

(5.2.5)

→ λ → k0 Z ∞ dk0 eitk0 θ(t) = (5.2.6) −∞ 2πi k0 − iε Z d4 k e−ik·(x−y) T (φ(x)φ(y)) = i (5.2.7) (2π)4 k 2 − m2 + iε p R Note that k02 − k 2 − m2 + iε → poles at k0 = ± k 2 + m2 − iε. Hence for dk0 we have the contour Hence Z d4 k e−ik·(x−y) h0|T φ(x)φ(y)|0i = i (2π)4 k 2 − m2 + iε = i∆F (x − y) (Feynman Propagator) (5.2.8)

5.2. FEYNMAN PROPAGATOR

89

t0 →

no poles, ∴ θ(t) = 0 1 pole, ∴ residue θ(t) = 1

Note: (¤x + m2 )∆F (x − y) = −δ 4 (x − y)

(5.2.9)

(¤ + m2 )∆(x − y) = 0

(5.2.10)

Remember

We can repeat this procedure for the vector field. Classical P.B. [Ai (x, t), Πj (x, t)] = gij δ 3 (x − y) (i, j → spatial indices) ↓ i h ˆ j (x, t) = −iδij δ 3 (x − y) Aî (x, t), Π

(5.2.11) (5.2.12)

90


But, in order to eliminate the constraints, we need the classical Dirac Bracket ¶ µ £ ¤∗ ∂i ∂j Ai (x, t), Πj (y, t) = − δij − 2 δ 3 (x − y) (Coulomb gauge) ∇ which goes to the commutator, ¶ µ h i ∂ ∂ i j ˆ j (y, t) = −i δij − δ 3 (x − y) Aî (x, t), Π ∇2

We’ll work with this commutator. Nov. 2/99 Z 2 ¢ ¡ d3 k 1 X −ik·x † ik·x √ Ai (x, t) = λ)) a(k, λ)e + a (k, λ)e (ε(k, (2π)3/2 2ωk λ=1

(5.2.13)

(5.2.14)

(5.2.15)

Re: ε(k, 1) = ε(−k, 1) ε(k, 2) = −ε(−k, 2) ε(k, 1) × ε(k, 2) =

k |k|

These ε are appropriate for the Coulomb gauge - the gauge condition is taken care of by the two conditions with k → −k, above. Also, once again we have a superposition of harmonic oscillators making up A(x, t); [a(k, λ), a(k 0 , λ0 )] = 0 ¤ a† (k, λ), a† (k 0 , λ0 ) = 0 £ ¤ a(k, λ), a† (k 0 , λ0 ) = δ 3 (k − k 0 )δλλ0

£ Recall (~ = 1)

(5.2.18)

Z

1 d3 x : (E 2 + B 2 ) : 2 Z 2 X 3 = d kωk a† (k, λ)a(k, λ)

H =

(5.2.16) (5.2.17)

(5.2.19)

λ=1

(i.e. E = ~ωk Nk Einstein photoelectric effect) Z p = d3 x : E × B : = (~)

Z

3

dkk

2 X

a† (k, λ)a(k, λ)

(5.2.20)

λ=1

¡ p = ~kNk de Broglie)

(5.2.21)

Remember that, for a Harmonic Oscillator, if

N = a† a

(5.2.22)

5.2. FEYNMAN PROPAGATOR

91

then N |ni = n|ni

(5.2.23)

General expression for Aµ (x, t) and its time ordered product: Now Aµ (x, t) =

Z

So h0|T Aµ (x)Aν (y)|0i =

Z

2 ¢ ¡ 1 X d3 k √ (5.2.24) εµ (k, λ) a† (k, λ)eik·x + a(k, λ)e−ik·x 3/2 (2π) 2ωk λ=1 2 d4 k e−ik·(x−y) X εµ (k, λ)εν (k, λ) (2π)4 k 2 + iε λ=1

= iDFµν (x − y)

(5.2.25)

In the Coulomb gauge, if εµ (k, λ) = (0, ε(k, λ)) (k · ε = 0)

(5.2.26)

Let ηµ = (1, 0, 0, 0). In frame where εµ has the above form, kµ − k · η η µ k˜µ = p (k · η)2 − η 2

Note

k˜ · η = 0 k˜ · ε = 0

(5.2.27)

η·ε=0

(5.2.28)

kµ , εµ (1), εµ (2), ηµ are 4 orthonormal vectors, all defined with respect to frame where εµ (k, λ) = (0, ε(k, λ)). Now, 2 X λ=1

εµ (k, λ)εν (k, λ) = −gµν + ηµ ην − k˜µ k˜ν

(5.2.29)

Reduces to ki kj = δij − 2 in frame where εµ = (0, ε) k

(5.2.30)

Hence in the Coulomb gauge,

iDFµν (x − y) =

Z

d4 k e−ik·(x−y) (2π)4 k 2 + iε





2   −gµν + k ηµ ην + kµ kν − k · η(kµ ην + kν ηµ )  (5.2.31)   2 2 −k + (η · k) | {z } ∗

∗ is an artifact of using the Coulomb gauge and cannot affect any physical process.


92

5.3

Quantizing the Dirac Field

We have φ(x) → ak , a†k0 with states:

(a†k1 )nk1 (a†km )nkm |nk1 , nk2 , . . . , nkm i = p ··· p |0, . . . , 0i (nk1 + 1)! (nkm + 1)!

(5.3.1)

(5.3.2)

where nk1 , . . . , nkm = 1, 2, . . . , ∞. For electrons, Pauli suggested that no two electrons can be in the same state. Fermionic Harmonic Oscillator ¶ µ 1 † H = ω b b+ 2 © † †ª {b, b} = b , b = 0 © †ª b, b = 1 = bb† + b† b

where the above are now anti-commutator relations. From this, · µ ¶¸ 1 † [b, H] = b, ω b b + 2 † † = ω(bb b − b bb) ; bb† = 1 − b† b = ω(b − 2b† bb ; bb + bb = 0 = {b, b} = ωb − ωb† {b, b} = ωb

(5.3.3) (5.3.4) (5.3.5)

(5.3.6)

If H|ni = ωn |ni

(5.3.7)

H(b|ni) = (bH − ωb)|ni (where [b, H] = bH − Hb = ωb) = (ωn − ω)(b|ni)

(5.3.8)

So also, H(b† |ni) = (ωn + ω)(b† |ni)

(5.3.9)

For the energy of the system to be bounded below, b|0i = 0

(5.3.10)

5.3. QUANTIZING THE DIRAC FIELD

93

For this state; µ ¶ 1 † H|0i = ω b b + |0i 2 ³ ω´ ω |0i ω0 = = 2 2

Now,

(5.3.11)

b† |0i = |1i

(5.3.12)

H|1i = ω(b† b + 1/2)|1i = ω(b† b + 1/2)b† |0i (b† bb† = b† (bb† + b† b − b† b) = b† ) = ω(b† + (1/2)b† )|0i 3ω 3ω † (b |0i) = |1i = 2 2

(5.3.13)

But (b† )2 |0i = 0 as {b† , b† } = 0. Nov. 3/99 Consider, µ ¶ 1 † H = ω b b− 2 † {b, b } = 1 b|0i = 0 , b† |0i = |1i , (b† )2 |1i = 0

(5.3.14)

ω H|0i = − |0i 2 ω |1i H|1i = 2 H

0

µ

1 = ω −b b + 2  †

¶

= −H 

1 = ω |−b† b{z− bb}† +bb† +  2 =−1 · ¸ 1 † = ω bb − 2

(5.3.15)

There is a symmetry between b and b† ; b† |0i0 = 0 b|0i = |1i0

(5.3.16) (5.3.17)

94


Consequently though, H|0i

0

H|1i0

µ

1 = ω b b− 2 " †

¶

|0i0 #

1 = ω |b† b {z + bb}† −bb† − |0i0 2 1 ω 0 |0i = 2µ ¶ 1 † = ω b b− (b|0i0 ) 2 ω = − |1i02 2 |0i ↔ |1i0 |1i ↔ |0i0

(5.3.18)

(5.3.19) (5.3.20) (5.3.21)

with the operator H, b destroys a state |1i while this operator b becomes a creation operator for a system with Hamiltonian H 0 . Now for the electron field ψ. X Z d3 p r m £ ¤ ψ(x) = b(p, s)e−ip·x u(p, s) + d∗ (p, s)eip·x v(p, s) (5.3.22) 3/2 (2π) E p s=± (6 p − m)u = 0 = (6 p + m)v X

uα (p, s)¯ uβ (p, s) =

s=±

X

vα (p, s)¯ vβ (p, s) =

s=±

µ

µ

6p + m 2m 6p + m 2m

(5.3.23) ¶ ¶

(5.3.24) αβ

(5.3.25) αβ

u¯(p, s)v(p, s0 ) = 0 u¯(p, s)u(p, s0 ) = −¯ v (p, s)v(p, s0 ) = δss0 Ep δss0 u† (p, s)u(p, s0 ) = v † (p, s)v(p, s0 ) = m

(5.3.26) (5.3.27)

(i 6 ∂ − m)ψ = 0

(5.3.29)

(5.3.28)

Quantizing ψ This can be derived from the Lagrangian ¯ 6 ∂ − m)ψ L = ψ(i

(ψ¯ = ψ † γ0 )

(5.3.30)

5.3. QUANTIZING THE DIRAC FIELD

95

provided ψ and ψ¯ are treated as being independent. ∂L ¯ 0 )α = iψ † = ψ(iγ α ∂(∂0 ψα ) ∂L = =0 ∂(∂0 ψ¯α )

Πα =

(5.3.31)

¯α Π

(5.3.32)

Eventually the constraint formalism gives, ª ¤∗ £ © ψα (x, t), Πβ (y, t) quantum ↔ ψα (x, t), Πβ (y, t) Classical Dirac Bracket

(5.3.33)

The quantization of the E.M. field gives

{ψα (x, t), Πβ (y, t)}

= iδ 3 (x − y)

(5.3.34)

{ψα (x, t), ψβ† (y, t)}

= δ 3 (x − y)

(5.3.35)

As b, d† are now operators, we obtain,

{b(p, s), b† (p0 , s0 )} = δ 3 (p − p0 )δss0 {d(p, s), d† (p0 , s0 )} = δ 3 (p − p0 )δss0

)

Fermionic creation and annihilation operators

b† (p, s) (b(p, s)) creates (destroys) an electron with energy Ep = + momentum p.

q

d(p, s) (d† (p, s)) creates (destroys) a positron with energy −Ep = − momentum −p. or

(5.3.36)

p2 + m2 and spin s,

q

p2 + m2 , spin s and

q d(p, s) (d† (p, s)) annihilates (creates) a positron with energy Ep = + p2 + m2 , spin s and momentum +p. i.e. Now, µ ¶ Z ∂ψα 3 : H := d x : Πα −L : (5.3.37) ∂t ¡ ¢ Here : ψα ψβ := ψα+ ψβ+ + ψα− ψβ− + ψα+ ψβ− − ψβ+ ψα− We find that XZ ¡ ¢ H= d3 p Ep b† (p, s)b(p, s) + d† (p, s)d(p, s) ← Positive Definite Hamiltonian s=±

(5.3.39) where this is a positive definite Hamiltonian because the b(p, s) destroys an electron of energy Ep , and the d(p, s) destroys a positron of momentum Ep . If we had quantized using a commutator, we would have gotten; XZ ¡ ¢ H= d3 p Ep b† (p, s)b(p, s) − d† (p, s)d(p, s) (5.3.40) s=±

96


where the negative sign would mean there is no lower bound on the spectrum of H. Consider the current ¯ µ ψ Vector - eventually this will be the e.m. current. (5.3.41) jµ = ψγ ¯ 0ψ : : Q :=: j0 : = : ψγ = : ψ†ψ : Z ¡ ¢ = d3 p b† (p, s)b(p, s) − d† (p, s)d(p, s) (5.3.42)

with b† (p, s)b(p, s) the positive contribution from electrons, and d† (p, s)d(p, s) term the negative contribution from positrons. Nov. 6/99 {ψα (x), ψ¯β (y)} = i(i 6 ∂ + m)∆(x − y) [φ(x), φ(y)] = i∆(x − y) = 0 if (x − y)2 < 0

(5.3.43) (5.3.44)

If we used commutators for ψα (x, t) and Πβ (y, t) , [ψα (x), ψβ (y)] 6= 0 for (x − y)2 < 0. This would be inconsistent. So also, h0|T ψα (x)ψ¯β |0i = iSF αβ (x − y) Z d4 p e−ip·(x−y) (6 p + m) (Scalar case) SF (x − y) = (2π)4 p2 − m2 + iε

(5.3.45) (5.3.46)

Since (6 p + m)(6 p − m) =6 p2 − m2 = p2 − m2 , (last = due to anti-commutation relations) SF (x − y) =

Z

d4 p e−ip·(x−y) (2π)4 6 p − m + iε

(5.3.47)

For Fermi Dirac particles T a(t)b(t0 ) = θ(t − t0 )a(t)b(t0 ) − θ(t0 − t)b(t0 )a(t)

(5.3.48)

Chapter 6 Interacting Fields 1 λφ4 m2 2 L = (∂µ φ)(∂ µ φ) − φ − 2 } 4! {z |{z} |2

Interaction

Free field

The equation of motion is

∂ 2 φ + m2 φ +

(6.0.1)

λφ3 =0 6

(6.0.2)

(Most important type of interaction → Gauge).

6.1

Gauge Interaction

Schrodinger equation

∂ψ = Hψ ψ − ψ complex ∂t Probability density ψ ∗ (x, t)ψ(x, t) is invariant under a phase change i

ψ(x, t) → eiΛ ψ(x, t)

(6.1.1)

(6.1.2)

where Λ is a constant. If ψ(x, t) → eiΛ(x,t) ψ(x, t) is an invariant then ψ ∗ ψ is unaltered, but

∂ψ = Hψ is changed (6.1.3) ∂t Introduce Aµ (x, t) and replace ∂µ by Dµ = ∂ − ieAµ , and let Aµ → Aµ + 1e ∂µ Λ if ψ → eiΛ ψ. i.e. Dµ ψ → eiΛ Dµ ψ (6.1.4) i

Aµ is the electromagnetic potential; ψ becomes complex.

L = (∂µ ψ ∗ )(∂ µ φ) − m2 φ∗ φ − λ(φ∗ φ)2 ∂µ φ → D µ φ L = [(∂µ + ieAµ ) φ∗ ] [(∂ µ − ieAµ ) φ] − m2 φ∗ φ − λ(φ∗ φ)2 97

(6.1.5) (6.1.6)

98

CHAPTER 6. INTERACTING FIELDS

This is invariant under φ → eiΛ φ φ∗ → e−iΛ φ

1 A µ → A µ + ∂µ Λ e Also, add in 1 L = − Fµν F µν 4 where Fµν = ∂µ Aν − ∂ν Aµ i [Dµ , Dν ] (f ) (f = test function) = e i [(∂µ − ieAµ ) , (∂ν − ieAν )] (f ) = e = Aν,µ − Aµ,ν

(6.1.7)

(6.1.8)

We can also include spinors L = ψ¯ (i (∂µ − ieAµ ) γ µ − m) ψ

(6.1.9)

So also there is an interaction between spinors and real scalars: 2 4 ¯ ¯ 6 ∂ − m)ψ + 1 (∂µ φ)(∂ µ φ) − µ φ − λφ − ig ψψφ L = ψ(i 2 2 4!

(6.1.10)

where the last term is the yukawa interaction (the i is required for L = L † . (you must have a complex scalar field to interact with the electromagnetic field).

6.2

Heisenberg Picture of Q.M. AH = Ah (x, t) (Heisenberg operator) ∂ AH = [AH , H] µ ∂t ¶ ∂A 1 c.f. = [A, H]P B → [ ]Commutator ∂t i i

(6.2.1) (6.2.2) (6.2.3)

Solution Heisenberg states |ψiH

AH (x, t) = eiHt AH (x, 0)e−iHt ¢ ¡∂ |ψiH = 0 . Matrix elements ∂t

hφ|AH |ψiH = hφ|eiHt AH (x, 0)e−iHt |ψiH

(6.2.4)

(6.2.5)

6.2. HEISENBERG PICTURE OF Q.M.

99

Let AH (x, 0) = AS (x) ← Schrodinger operator |ψs i = e−iHt |ψiH = |ψ(t)iS ∂ Now As (x) = 0 ∂t i

∂ |ψ(t)iS = H|ψ(t)iS ∂t

(6.2.6)

(6.2.7)

Dirac’s interaction representation, Aip (x, t) H |a(t)i Thus hφH |AH |φH i

= = = = =

eiH0 t AS (x)e−iH0 t H0 + HI = eiH0 t e−iHt AH (x, t)eiHt e−iH0 t eiH0 t |a(t)iS hφS |AS |φS i hφip |Aip |ψip i

(6.2.8) (6.2.9) (6.2.10)

We note that

¤ ∂ £ iH0 t i∂Aip =i e AS e−iH0 t = [Aip , H0 ] ∂t ∂t (Aip (x, t) evolves as if there were no interactions). So, also i

¤ ∂ £ iH0 t ∂ e |a(t)iS |a(t)iip = i ∂t µ ·∂t ¶¸ ∂ iH0 t iH0 t i |a(t)iS = −H0 e |a(t)iS + e ∂t iH0 t iH0 t = −H0 e |a(t)iS + e (H0 + HI ) |a(t)iS ¡ iH0 t ¢ iH0 t −iH0 t = e HI e e |a(t)iS = HIip |a(t)iip

(6.2.11)

(6.2.12)

Nov. 9/99 Now, drop the “ip” (we’re always working in the interaction picture from now on). So, ∂ |a(t)i = HI |a(t)i i ∂t ∂ i ∂t A(t) = [A(t), H0 ]

(6.2.13)

is in the interaction picture. Suppose, |a(t)i = U (t, t0 )|a(t0 )i where U (t, t0 ) is an evolution operator. Note: 1. U (t, t) = 1 2. U (t, t0 )−1 = U (t0 , t)

(6.2.14)

100


3. By (6.2.13), i∂t U (t, t0 )|a(t0 )i = HI U (t, t0 )|a(t0 )i Notice |a(t0 )i has no t dependence ∂ ∴ i U (t, t0 ) = HI (t)U (t, t0 ) ∂t

(6.2.15)

Take the Hermitian conjugate −i Thus, from

¡

∂ † U (t, t0 ) = U † (t, t0 )HI ∂t

U † (6.2.15) + (6.2.16)U

(HI† = HI )

¢

¤ ∂ £ † U (t, t0 )U (t, t0 ) = 0 ∂t

and

(6.2.16)

(6.2.17)

U † (t, t0 )U (t, t0 ) = 1 (i.e. = constant, let constant = 1)

(6.2.18)

U −1 (t, t0 ) = U † (t, t0 ) (i.e. U is unitary)

(6.2.19)

So we have We can integrate (6.2.15) U (t, t0 ) = K − i

Z

t

dt0 HI (t0 )U (t0 , t0 )

(6.2.20)

t0

As U (t, t) = 1, ∴ K = 1. We can iterate this equation, U (t, t0 ) = 1 − i

Z

t 2

Z

t

Z

t1

dt1 HI (t1 ) + (−i) dt1 dt2 HI (t1 )HI (t2 ) t0 t0 t0 Z t Z t1 Z t3 3 + (−i) dt1 dt2 dt3 HI (t1 )HI (t2 )HI (t3 ) + . . . (6.2.21) t0

t0

t0

This satisfies (6.2.15). Examine: We can also write as: ·Z t ¸ Z t1 Z t Z t2 (−i)2 dt1 dt2 dt2 HI (t1 )HI (t2 ) + dt1 HI (t2 )HI (t1 ) 2 t0 t0 t0 t0

(6.2.22)

where the first region of integration is as in diagram above, and the second region of integration is the same diagram with the region reflected in the t1 = t2 line. But this is also, Z Z t (−1)2 t dt1 dt2 [T HI (t1 )HI (t2 )] (6.2.23) = 2 t0 t0

6.2. HEISENBERG PICTURE OF Q.M.

101

(−i)

2

Z

t

dt1 t0

Z

t1

dt2 HI (t1 )HI (t2 ) t0

So also (−i)

3

(−i)3 = 3!

Z

t

dt1 t0

Z

t

dt1 t0

Z

t1

dt2 t0

Z

t

dt2 t0

Z

t2

t Z 0t

dt3 HI (t1 )HI (t2 )HI (t3 ) dt3 T HI (t1 )HI (t2 )HI (t3 )

t0

In general, then, Z Z t (−i)2 t dt1 HI (t1 ) + T dt1 dt2 HI (t1 )HI (t2 ) 2! t0 t0 t0 Z t Z (−1)n t dtn HI (t1 ) . . . HI (tn ) + ... + T dt1 . . . n! t0 t0 ¸ · Z t dt1 HI (t1 ) (6.2.24) = T exp −i

(−i) U (t, t0 ) = 1 + T 1!

Z

t

t0

Consider now a scattering problem: • In this we go from a state |a(t = −∞)) to a state at (a(t = +∞)| • At t = ±∞, we have free particles (Adiabatic Approximation)

102


At t = −∞, we have a free particle state, |a(t = −∞)), and at t = +∞, we have a free particle state |b(t = +∞)), and what we want to compute is (b(∞)|a(∞)) - This gives the amplitude for |a(−∞)) evolving into |b(∞)). But, |a(∞)) = U (∞, −∞) |a(−∞))

(6.2.25)

(b(∞)| U (∞, −∞) |a(−∞)) = Sba → (S-matrix)

(6.2.26)

This means we want,

To evaluate this, we need, Z t Z t Z t (−i)n (b(∞)| T dt1 dt2 . . . dtn HI (t1 ) . . . HI (tn ) |a(−∞)) n! t0 t0 t0

6.3

(6.2.27)

Wick’s Theorem

This converts Time-ordered products to normal-ordered products. (actual theorem works for any two orderings of operators). Trivial Case: (n=1) T HI (t1 ) =: HI (t1 ) : (6.3.1) n=2 (let t1 > t2 ). ¡ ¢¡ ¢ T HI (t1 )HI (t2 ) = : HI+ (t1 ) + HI− (t1 ) HI+ (t2 ) + HI− (t2 ) : (let’s let HI (t1 ) = H1 , etc.) = H1+ H2+ + H1+ H2− + H1− H2+ +H1− H2− | {z }

(6.3.2)

Not N.O.’d

Now,

      − +  H − H + H − − H + H − (Bosonic)   H | 1 2 {z 2 1} | 2{z 1} N.O.’d H1− H2+ = [H1− ,H2+ ]   − +   H + H + H − − H + H − (Fermionic) H  | 1 2 {z 2 1} | 2{z 1}   − + N.O.’d

(6.3.3)

{H1 ,H2 }

Thus,

( £ ¤ H1− , H2+ ← (Bosonic) T H1 H2 =: H1 H2 : + © − + ª ← (Fermionic) H1 , H 2

(6.3.4)

But this (anti)-commutator is a c-number. In general these are just h0|T H1 H2 |0i. i.e. We have (t1 > t2 ), £ − +¤ H1 , H 2 = h0|H1− H2+ |0i = h0|T H1− H2+ |0i = h0|T H1 H2 |0i (6.3.5)

6.3. WICK’S THEOREM

103

For both t1 > t2 and t2 > t1 , we have T HI (t1 )HI (t2 ) =: HI (t1 )HI (t2 ) : +h0|T HI (t1 )HI (t2 )|0i

(6.3.6)

Nov. 10/99 Inductively, we can show that T χ(x1 ) . . . χ(xn ) = : χ(x1 )χ(x2 ) . . . χ(xn ) : X + (0| T χ(xi )χ(xj ) |0) : χ(x1 ) . . . χ(xi−1 )χ(xi+1 ) . . . χ(xj−1 )χ(xj+1 ) . . . χ(xn ) : i<j

X

+

all possible 2 pairs

+... +

X

(0| T χ(xi )χ(xj ) |0) (0| T χ(xk )χ(xl ) |0) : χ(x1 ) . . . χ(xn ) : {z } |

all possible n/2 pairs

χ(xijkl ) all excluded

(0| T . . . |0) . . . (0| T . . . |0) : χ :

This is useful, as (0| : χ(x1 ) . . . χxn : |0) = 0 Consider the case where HI =

Z

d3 x H(x)

¯ 5 ψφ ( Pseudo-scalar Yakawa coupling ) H(x) = iκψγ

(6.3.8) (6.3.9) (6.3.10)

i.e.

1 µ2 ¯ 6 ∂ − m)ψ − iκψγ ¯ 5 ψφ L = (∂µ φ)(∂ µ φ) − φ2 + ψ(i 2 2 (Note that φ(−x, t) = −φ(+x, t)). Hence: ½ Z ¾ ½ Z ¾ 4 T exp −i dt HI (t) = T exp −i d xHI (x) " Z ¯ = T 1 − i(iκ) d4 x (ψ(x)γ 5 ψ(x)φ(x)) (+κ)2 + 2! #

+ ...

Z

(6.3.11)

¯ 1 )γ5 ψ(x1 )φ(x1 )ψ(x ¯ 2 )γ5 ψ(x2 )φ(x2 ) d4 x1 d4 x2 ψ(x (6.3.12)

(6.3.7)

104


Chapter 7 Electron-Positron Scattering Now suppose |a(−∞)) consists of an electron of momentum p and polarization s, and a positron of momentum q and polarization t; and suppose |a(∞)) consists of an electron of momentum p0 and polarization s0 , and a positron of momentum q 0 and polarization t0 . Thus, we have |a(−∞)) = b†ps d†qt |0) −→ |0) = bps dqt |a(−∞))

|b(−∞)) = b†p0 s0 d†q0 t0 |0) −→ (0| bp0 s0 dq0 t0 = (b(−∞)|

(7.0.1) (7.0.2)

Thus, R

4

(b(∞)| T e−i d xH(x) |a(−∞)) µ ¶ Z Z κ2 4 4 4 ¯ = (b(∞)| T 1 + κ d x ψ(x)γ5 ψ(x)φ(x) + d x1 d x2 . . . |a(−∞)) (7.0.3) 2! Upon applying Wick’s theorem, the only surviving contributions to this matrix element will be those terms with destruction operators for e− and e+ acting on |a(−∞)) and creation operators for e− and e+ acting on (b(∞)| . So, (b(∞)| T e−i

R

d4 xH(x)

|a(−∞)) · Z ¯ = (b(∞)|a(−∞)) + κ (b(∞)| d4 x : ψ(x)γ 5 ψ(x)φ(x) : |a(−∞)) ¸ Z 4 + (b(∞)| d x (0| T ψ¯α (x)ψβ (x) |0) γαβ : φ(x) : |a(−∞))

= 0 i.e. ap |a(−∞)) = 0 ap → destruction op. for φ(x)

Term of order κ2 : The only surviving term. Z n h i o κ2 4 4 ¯ ¯ d x1 d x2 (b(∞)| (0| T φ(x1 )φ(x2 ) |0) : ψ(x1 )γ5 ψ(x1 )ψ(x2 )γ5 ψ(x2 ) : |a(−∞)) | {z } 2 i∆F (x1 −x2 )

(7.0.4)

105

106

CHAPTER 7. ELECTRON-POSITRON SCATTERING

The following (ψ + , ψ¯+ ) act on |a(−∞)) to give |0) by (7.0.1); Z r m X d3 p00 00 + ψ = b(p00 , s00 )u(p00 , s00 )e−ip ·x 3/2 (2π) Ep00 s00 =± Z r d3 p00 m X 00 ψ¯+ = d(p00 , s00 )¯ v (p00 , s00 )e−ip ·x 3/2 (2π) Ep00 s00 =±

(7.0.5) (7.0.6)

where the ψ + represents e− (b operator) and the ψ¯+ represents e+ (d operator). So also, by (7.0.2), Z r m X d3 p00 00 + b(p00 , s00 )u(p00 , s00 )e−ip ·x (7.0.7) ψ = 3/2 (2π) Ep00 s00 =± Z 3 00 r d p m X 00 + d(p00 , s00 )¯ v (p00 , s00 )e−ip ·x (7.0.8) ψ¯ = (2π)3/2 Ep00 s00 =± which act on (b(∞)| to give (0| . (ψ¯− has b− in it → e− .) Nov. 12/99 So, the only surviving term is: Z κ2 = dx1 dx2 (0| T φ(x1 )φ(x2 ) |0) (γ5 )αβ (γ5 )γδ · 2 £ · (b(∞)| ψ¯α− (x1 )ψβ− (x1 )ψ¯γ+ (x2 )ψδ+ (x2 ) + ψ¯− (x2 )ψ − (x2 )ψ¯+ (x1 )ψ + (x1 ) γ

δ

α

β

− ψ¯α− (x1 )ψδ− (x2 )ψ¯γ+ (x2 )ψβ+ (x1 ) ¤ − ψ¯− (x2 )ψ − (x1 )ψ¯+ (x1 )ψ + (x2 ) |a(−∞)) γ

β

α

δ

(7.0.9)

where the first two terms give identical contributions, and the last two terms give identical contributions (the minus sign on the last two terms is due to Fermi-Dirac statistics). ex. r XZ m † − 3 d (p, s)v(p, s)eip·x ψ (x) = dp E p s=± and so on . . . Recall |a(−∞)) = d†q0 t0 b†p0 s0 |0) ⇒ |0) = dq0 t0 bp0 s0 |a(−∞)) So also, i (0| T φ(x1 )φ(x2 ) |0) = (2π)4

Z

d4 k

e−ik·(x1 −x2 ) k 2 − µ2 + iε

(7.0.11)

107 Thus, Sba = κ2

Z

h

d 4 x1 d 4 x2

Z

4

−ik·(x1 −x2 )

d k ie (2π)4 (k 2 − µ2 + iε)

s

m4 · E p E q E p0 E q 0 0

0

· (¯ u(p0 , s0 )γ5 v(q 0 , t0 )) (¯ v (q, t)γ5 u(p, s)) ei(p ·x1 +q ·x1 −q·x2 −p·x2 ) 0

0

− (¯ u(p0 , s0 )γ5 u(p, s)) (¯ v (q, t)γ5 v(q 0 , t0 )) ei(p ·x1 −p·x1 +q ·x2 −q·x2 ) Integrate over x1 and x2 : →

Sba

Z

i

d4 x ik·x e = δ 4 (k) (2π)4

(7.0.13)

s

Z m2 1 = iκ · d4 k 2 E p E q E p0 E q 0 k − µ2 − iε £ · (¯ up0 s0 γ5 vq0 t0 )(¯ vqt γ5 ups )δ 4 (−k + p0 + q 0 )δ 4 (k − q − p) ¤ −(¯ up0 s0 γ5 ups )(¯ vqt γ5 vq0 t0 )δ 4 (−k + p0 − p)δ 4 (k + q 0 − q) s · £ 4 0 ¤ (¯ up0 s0 γ5 vq0 t0 )(¯ vqt γ5 ups ) m2 0 2 δ (p + q − p − q) = iκ 0 0 2 E p E q E p0 E q 0 (p + q ) − µ2 + iε ¸ (¯ up0 s0 γ5 ups )(¯ vqt γ5 vq0 t0 ) − (p0 − p)2 − µ2 + iε 2

(7.0.12)

(7.0.14)

Pictorial Representation

The relative minus sign in figure 7.0.2 comes from interchanging two fermions. We don’t actually observe the intermediate particles, and so the fact that they are off mass shell doesn’t effect the calculation. After integrating over x1 and x2 , and going to momentum space, we get

108

CHAPTER 7. ELECTRON-POSITRON SCATTERING

Figure 7.0.1: Electron Scattering

Figure 7.0.2: Electron Scattering (Momentum Space, after integration) The integral over x1 , x2 leads to momentum conservation at each vertex (total momentum incoming = total outgoing momentum). Nov 16/99

Chapter 8 Loop Diagrams 8.1

Feynman Rules in Momentum Space

Propagator for spinor

i 6 p − m + iε (Spin 0)

k2

i − µ2 + iε

(Spin 0)

−igµν + (gauge dependent) k 2 + iε (Spin 1)

109

110

CHAPTER 8. LOOP DIAGRAMS

−ie(γµ )αβ (Vertex)

−κ(γ5 )αβ (Vertex)

−iλ (Vertex)

• momentum is conserved at each vertex • for each loop, there is an overall factor of Now, take an example:

One more photon added: (i.e. many different possibilities) example:

R

d4 k (2π)4

→ (This leads to infinities).

8.1. FEYNMAN RULES IN MOMENTUM SPACE

111

Figure 8.1.1: we only ever observe the external particles →∴ don’t know where particles in loop are ∴ must sum over all possibilities → Uncertainty Principle.

(p + q 0 = q + p0 ). The External lines (spinors) give r

r r r m m m m u(p, s) u¯(p, s) v(p, s) v¯(p, s) Ep V Ep V Ep V Ep V | {z }| {z }| {z }| {z } incoming e− momentum p, spin s

where

1 V

outgoing e−

Outgoing e+

(8.1.1)

incoming e+

→ unit probability in a box of volume V .

• Factor of (−1) for each fermion loop • If we exchange fermions to go from one diagram to another, their relative phase is (−1). • Overall factor of (2π)4 δ(pi − pf ) For example, e− e+ → e− e+ (with respective momenta p, q → p0 , q 0 .) Consider one particular pair of diagrams, which are of order e4 .

112


Figure 8.1.2: For Diagram I

¶ ¸ µ ¸ ·r m m −igµν 0 0 · uβ (p, s) · u¯α (p , s ) · (−ie(γµ )αβ ) · (−1) E p0 V Ep V (p − p0 )2 + iε µ µ ¶ ¶ # i i · (−ieγν )γδ (−ieγλ )δ0 γ 0 · (6 k+ 6 p− 6 p0 ) − m + iε δδ0 6 k − m + iε γ 0 γ ¶r µ r m m −igλσ v¯α0 (q, t)(−ieγσ )α0 β 0 vβ 0 (q 0 , t0 ) · 0 2 (p − p ) + iε Eq V Eq 0 V Z d4 k 1 m2 4 0 0 p = (−1) (2π) δ(p + q − p − q ) [¯ u(p0 , s0 )γµ u(p, s)] · 2 (2π)4 V E p E p0 E q E q 0 ¶2 µ ¶ µ 1 1 1 Tr γµ γλ v¯(q, t)γλ v(q 0 , t0 ) (8.1.2) · 0 2 0 (p − p ) + iε (6 k+ 6 p− 6 p ) − m + iε 6 k − m + iε Z

d4 k (2π)4 "

·r

Note that (1) the (−1) is the fermion loop factor, and (2) the spinor propagator is 1 6 −m+iε p

Thus II becomes (−1)(−1)

Z

d4 k . . . (u(p, s) ↔ v(q 0 , t0 )) (2π)4

where the second (−1) will come from interchanging the spinors.

6p+m p2 −m2 +iε

⇒

(8.1.3)

8.2. COMBINATORIC FACTORS

113

Figure 8.1.3: For Diagram II

8.2

Combinatoric Factors

(Always (+1) in QED, as all particles entering a vertex are distinguishable).

¯ µ ψAµ −eψγ (3 distinct fields)

But, not so lucky in other theories - for example 1 µ2 λ L = (∂µ φ)2 − φ2 − φ4 |2 {z 2 } 4!

(8.2.1)

Scalar Propagator

with the 4! in the φ4 term to take into account the indistinguishability of the 4 φ’s. The Feynman diagrams are:

Consider the one-loop correction to φ(p)φ(q) → φ(p0 )φ(q 0 ), above. There are three possible diagrams that have the same topology; note that there is no factor of (−1) here, though, as these are scalars (bosons) here. Nov. 17/99 External Photons; recall, for electrons there was the factor r m − e − u(p, s) (8.2.2) Ep V

114


k2

i − µ2 + iε

Scalar Propagator

−iλ Vertex

Figure 8.2.1: 1-loop correction to φφ → φφ

Symmetry Factors recall the diagram for φφ → φφ, figure 8.2.1 above. As shown in figure 8.2.2, the p leg can attach to either vertex, on any one of the eight different legs, then the p0 leg can attach to the same vertex to any of the 3 remaining legs. After that the q leg can attach to any of the 4 legs of the remaining vertex, followed by the q 0 leg attaching to any of the 3 legs of that vertex; finally, one of the remaining legs on the first

8.3. CROSS SECTIONS FROM MATRIX ELEMENTS

115

εµ (k, λ) √ 2ωk V e− γ → e − γ

Figure 8.2.2: Symmetry Factor diagram (example) vertex can attach to either remaining leg on the other vertex, giving in the numerator of the symmetry factor (8 × 3) × (4 × 3) × (2). In the denominator, we must take into account that there are 4 possible legs on each vertex (4! × 4!), and that we can switch vertices (2!); thus, the symmetry factor is (8 × 3)(4 × 3)(2) 1 = = Symmetry factor 4! 4! 2! 2

8.3

(8.2.3)

Cross Sections From Matrix elements

Consider e− p → e− p , (p = proton). The proton is treated as a “heavy” electron with opposite charge. To lowest order,

116


Figure 8.3.1: Electron-Proton scattering

¶ ¶ µr ¶µ −igµν m m Sf i = (a(−∞)|b(∞)) = u¯(pf , sf ) (−ieγµ ) u(pi , si ) Ef V Ei V (pf − Pi )2 + iε Ãs ! Ãr ! M M · u¯(Pf , sf ) (+ieγν ) u(Pi , si ) Ef V Ei V µr

(2π)4 δ 4 (pi + Pi − pf − Pf )

(8.3.1)

(For e− e− → e− e− , we would have to account for the fact that we can exchange e− ’s (identical particles). - also note the “+00 in front of the second e above - this is because the proton is positive.) The probability of this transition occurring per unit volume of space per unit time is |Sf i |2 = Wf i VT

(8.3.2)

Note: £

¤2 δ 4 (pi + Pi − pf − Pf ) = δ 4 (0)δ 4 (pi + Pi − pf − Pf ) Z ∞ Z dx ip·x dx V As δ(p) = e ⇒ δ(0) = ∼ (2π) 2π −∞ (2π) £ ¤2 VT 4 ∴ δ 4 (pi + Pi − pf − Pf ) = δ (pi + Pi − pf − Pf ) (2π)4

(8.3.3)

8.3. CROSS SECTIONS FROM MATRIX ELEMENTS

117

Thus, Kinematics

Wf i

z

}|

{ 2 2 m M = (2π)4 δ 4 (pi + Pi − pf − Pf ) 4 · V E i E f Ei Ef ¯2 ¯ ¶ µ ¯ ¯ −i ¯ ¯ (¯ u γ u ) · ¯(−ie)(ie)(¯ upf sf γµ upi ,si ) Pf s f ν Pi s i ¯ 2 ¯ ¯ (pf − pi ) + iε | {z }

(8.3.4)

|Mf i |2

The number of states in a volume d3 pf , d3 Pf is going to be (V d3 pf )(V d3 Pf )Wf i

(8.3.5)

Divide through by Jinc V1 , where Jinc = flux of incident particles = # of particles per unit area that run by each other per unit time for collinear beams. Jinc =

|vi − Vi | V

(8.3.6)

1 V

is the number of target particles per unit volume. (as ψi is thus dσf i = V 2 d3 pf d3 Pf (Wf i )

So also,

mM mM =p Ei Ei |vi − Vi | (pi − Pi )2 − m2 M 2

d3 p = (2Ep )

Z

m EV

u(p, s) ). The cross section

V Jinc 1 m2 M 2 |Mf i |2 Ei Ef Ei Ef |vi − Vi |

(8.3.7)

(Lorentz invariant)

(8.3.8)

= d3 pf d3 Pf (2π)4 δ 4 (pi + Pi − pf − Pf ) Note that

p

dp0 δ(p2 − m2 )θ(p0 )d3 p → also Lorentz invariant

(8.3.9)

Thus dσ is Lorentz invariant. For unpolarized beams, • Average over the polarizations of the incoming particles • Sum over outgoing polarizations i.e.

1 X 1 X XX · dσf i 2 s 2 S s S i

i

f

f

(8.3.10)

118


Remember that, X

uα (p, s)¯ uβ (p, s) =

s=±

Ã

X

vα (p, s)¯ vβ (p, s) =

s=±

¶

µ

6p + m 2m

µ

6p − m 2m

(8.3.11) αβ

¶ !

(8.3.12)

αβ

Thus, · ¸ · ¸ 6 pf + m 6 pi + m 1 X 6 Pf + M 6 Pi + M e4 1 2 γµ γν Tr γµ γν |Mf i | = Tr 4 s ,S 4 2m 2m 2M 2M (pf − pi )4

(8.3.13)

i i sf ,Sf

Recall the relations: Tr[6 a 6 b] = 4a · b Tr[6 a 6 b 6 c] = 0 Tr[6 a 6 b 6 c 6 d] = 4 (a · bc · d − a · cb · d + a · db · c) And so 1 X 4

=

si ,Si sf ,Sf

(8.3.14) (8.3.15) (8.3.16)

¡ ¢ e4 Pf · pf Pi · pi + Pf · pi Pi · pf − m2 Pf · Pi − M 2 pf · pi + 2m2 M 2 2 2 2 2m M (pf − pi )

(8.3.17)

Nov. 19/99 dσ = p

mM (pi · Pi )2 − m2 M 2

(2π)4 δ 4 (pi + Pi − pf − Pf )

md3 pf M d3 Pf |Mf i |2 3 3 (2π) Ef (2π) Ef

(8.3.18)

Consider,

d3 pf = dΩ |pf |2 d|pf |

= dΩ |pf |2 Ef dEf

(8.3.19)

In the frame of reference where pµf = (pf , Ef )

(8.3.20)

pµi = (pi , Ei ) Pfµ ≈ Piµ = (0, M ) → Proton at rest during this scattering

(8.3.21) (8.3.22)

Note that if pi · pf = |pi ||pf | cos θ, then q 2 = (pf − pi )2

µ ¶ θ = −4Ei Ef sin 2 2

(8.3.23)

8.4. HIGHER ORDER CORRECTIONS Hence,

where α =

8.4

e2 4π

µ ¶µ ¶¸ · 1 q2 π 2 α2 θ 2 ¡ ¢ 1 + sin − −1 dσ = 2 2 2 M2 m Ei Ef sin4 2θ

119

(8.3.24)

if q 2 ¿ M 2 . The first term (the “1”) above is the classical Rutherford term.

Higher order corrections

Figure 8.4.1: Higher order corrections

Figure 8.4.2: Generates coulomb field → classically The loop (solid) is generated as the e− travels - it “Polarizes” - the e− e+ created in vacuum. The Loop contribution is

120


Figure 8.4.3: Loop contribution

I = = = =

· µ ¶¸ d4 q i i Tr (−ieγµ ) (−ieγν ) (−1) (2π)4 (6 k+ 6 q) − m + iε 6 q − m + iε · ¸ Z 4 γµ ((6 k+ 6 q) + m)γν (6 q + m) dq Tr −e2 4 (2π) ((k + q)2 − m2 )(q 2 − m2 ) · ¸ Z d4 q γµ (6 k+ 6 q)γν 6 q + m2 γµ γν 2 Tr −e (2π)4 ((k + q)2 − m2 )(q 2 − m2 ) ¾ ½ Z d4 q (k + q)µ qν − (k + q) · qgµν + (k + q)ν qµ + m2 gµν 2 −4e (2π)4 ((k + q)2 − m2 )(q 2 − m2 ) Z

(8.4.1)

Now, use the relation: 1 = ab ³

Z

1

dx 0

1 [ax + b(1 − x)]2

(8.4.2)

example: Z

dt = t(t + 1) = = = =

Z

Z

1

1 [tx + (t + 1)(1 − x)]2 0 Z Z 1 1 dt dx [t + (1 − x)]2 0 ¸ · Z 1 −1 +K dx (t + (1 − x)) 0 ¯1 ln [t + (1 − x)] ¯0 + K ´ ln(t) − ln(t + 1) + K dt

dx

(8.4.3)

8.4. HIGHER ORDER CORRECTIONS

121 Quadratically divergent

Z

}| {¤ d4 q £z 2 I = −4e2 (k + q) q + (k + q) q + (−(k + q) · q + m )g µ ν ν µ µν (2π)4 Z 1 1 · dx 2 2 [x((k + q) − m ) + (1 − x)(q 2 − m2 )]2 0 Z Z 1 d4 q 1 2 = −4e Nµν dx 2 4 (2π) [q + 2k · qx +k 2 x − m2 ]2 0 {z } | (q+kx)2 −k2 x2

Shift variable of integration (not straightforward) to q 0 = q + kx. Z 1 Z 1 d4 q 0 2 dx 02 [(k + q 0 − xk)µ (q 0 − xk)ν I = −4e 4 2 (2π) 0 [q + k x(1 − x) − m2 ]2 ¤ +(k + q 0 − xk)ν (q 0 − xk)µ + (−(k + q 0 − xk) · (q 0 − xk) + m2 )gµν

We can drop terms odd in q (because it’s integrating over odd terms over an area symmetric about origin). Z Z 1 d4 q 2qµ qν − 2x(1 − x)kµ kν + (−q 2 + x(1 − x)k 2 + m2 )gµν 2 I = −4e dx (2π)4 0 [q 2 + k 2 x(1 − x) − m2 ]2 where qµ qν = →

∴

Z

Z

1 gµν q 2 4

d4 q f (q 2 )qµ qν = Agµν → multiply by gµν

Z

d4 q f (q 2 )q 2 = 4A 4

2

d q f (q )qµ qν 1 = ab Z

Z

1

dx 0

gµν = 4

Z

d4 q f (q 2 )q 2

1 [ax + (1 − x)b]2

1

1 1 2 [xa + (1 − x)b] c 0 d take of (8.4.5) da Z 1 1 2y = dy 2 ab [ay + (1 − y)b]3 0 Z 1 Z 1 2y 1 dy dx ∴ = abc {y[ax + (1 − x)b] + (1 − y)c}3 0 0 1 = abc

(8.4.5)

dx

(8.4.6)

122


Nov. 23/99 R∞ In general (where Γ(x) = 0 dt tx−1 e−t ), 1 . . . Dnan

D1a1 D2a2

=

r(a1 + a2 + . . . + an ) · Γ(a1 )Γ(a2 ) . . . Γ(an ) Z 1 Z 1 Z 1 δ(1 − x1 − · · · − xn )xa11 −1 · · · xann −1 dxn dx1 dx2 . . . (8.4.7) · [x1 D1 + x2 D2 + . . . + xn Dn ]2 0 0 0

Recall the loop contribution from last time: figure 8.4.3, with equation I = −4e

2

Z

d4 q (2π)4

Z

1 0

[ 12 gµν q 2 − 2kµ kν x(1 − x) + (−q 2 + x(1 − x)k 2 + m2 )gµν dx [q 2 + x(1 − x)k 2 − m2 ]2

(8.4.8)

How do we make sense out of a divergent integral? We first insert some parameter into the theory to render this finite in a way consistent with the symmetries of the theory (Regularization). Then absorb this parameter into the quantities that characterize the theory. (mass, couplings, external wave functions, etc.) (Renormalization).

8.5

Renormalization

Recall the mass on a string problem (see figure 10.4.1):

Figure 8.5.1: Loop contribution The Lagrangian for this system is characterized by m, k, x0 ; 1 1 L = mx˙ 2 − k(x − x0 )2 2 2

(8.5.1)

8.6. REGULARIZATION

123

Now, if we turn on a gravitational field, mx˙ 2 k(x − x0 )2 − + mgx 2 2 · ¸ m 2 kh mg i2 k ³ mg ´2 2x0 mg = x˙ − x − x0 − + − 2 2 k 2 2 k {z } |

L =

Constant

2

k mx˙ 2 − (x − x00 ) + constant 2 2 mg x00 = x + k ! x0 has been renormalized by an amount mg k =

8.6

(8.5.2)

Regularization

1. Cut-off (Λ)

Z

4

dq→

Z

|q| Λ

(8.6.2)

2. Pauli Villars S → SP V =

Z

4

dx

·

1 ¯ − m)ψ + Ψ(iγM ¯ − Fµν F µν + ψ(iγ )Ψ 4

¸

(8.6.3)

(a) Bosonic spinor (b) M → ∞

(c) Still have gauge invariance

Now we must consider

(figure 8.6.1) Z 1 1 Z g q 2 − 2kµ kν x(1 − x) + (−q 2 + x(1 − x)k 2 + m2 )gµν d4 q 2 2 µν = −4e dx (2π)4 0 [q 2 + x(1 − x)k 2 − m2 ]2 1 g q 2 − 2kµ kν x(1 − x) + (−q 2 + x(1 − x)k 2 + M 2 )gµν 2 µν (8.6.4) [q 2 + x(1 − x)k 2 − M 2 ]2

124


Figure 8.6.1: Loop contribution This is Finite! (Divergence reappears as M 2 → ∞. The dependence on M 2 disappears when we renormalize. Combine terms using Z 1 1 1 1 dx − n = −n(a − b) n a b [xa + (1 − x)b]n+1 0 let y = x(a − b) + b · −n ¸a Z b y 1 dy/(a − b) 1 = −n(a − b) = −n = n − n = LHS n+1 y −n b a b a

(8.6.5)

(8.6.6)

Actually computing the integrals involves the following general integral. In n dimensions (n = 1, 2, 3, . . .) (in Minkowski space); Z

Γ dn q i (q 2 )a = (−1)a−b (m2 )n/2+a−b n 2 2 b n/2 (2π) (q − m ) (4π)

¡n 2

¢ ¡ ¢ + a Γ b − a − n2 Γ(a)Γ(b)

(8.6.7)

(q 2 )a (q 2 − m2 )b

(8.6.8)

Note:

q2 =

Z

∞

dq0

−∞ 2 q0 − q12

Z

−

∞

dq1 . . .

−∞ q22 −

Z

∞

dqn−1

−∞

2 . . . − qn−1

Let q0 = iqn (Wick Rotation - i.e. Cartesian co-ordinates in Minkowski space, giving Euclidean space) 2 qE = qn2 + q12 + . . . + qn−1

So (8.6.8) becomes =i

Z

∞ −∞

dqn . . .

Z

∞ −∞

dqn−1 = i(−1)

a−b

Z

d n qe

(qE2 )a (qE2 + m2 )b

(8.6.10)

8.6. REGULARIZATION

125

FigureR8.6.2: Integral R i∞ around contour is ∞ zero - −∞ dq0 ↔ −i∞ dq0

qn = q cos θ1 q1 = q sin θ1 cos θ2 q2 = q sin θ1 sin θ2 cos θ3 .. . qn−1 = q sin θ1 sin θ2 . . . sin θn−1 2 + qn2 q 2 = q12 + q22 + . . . qn−1 i.e. Z

n

d qE =

Z

dq q 0

n−1

Z

2π

dθ1 0

Z

φ

dθ2 sin θ2 0

Z

π 2

dθ3 sin θ3 . . . 0

Z

π

dθn−1 sinn−2 θn−1 (8.6.11) 0

Nov. 24/99 Z

(q 2 )a dn q (2π)n (q 2 − m2 )b Z N d qE (qE2 )a a−b = i(−1) (2π)n (qE2 + m2 )b Z ∞ Z 2π Z π a−b n−1 = i(−q) dq q dθ1 dθ2 sin(θ2 ) . . . 0 0 0 Z π (q 2 )a dθn−1 sinn−2 (θn−1 ) 2 (q + m2 )b 0

I =

(8.6.12)

126


But, √

¢ ¡ π Γ m+1 2 ¢ ¡ dθ sin θ = Γ m+2 0 2 Z ∞ tm−1 Γ(m)Γ(n − m) dt 2 = 2 n (t + a ) (a2 )n−m Γ(n) 0 ¢ ¡ ¢ ¡n n + a Γ Γ b − a − i 2¡ ¢ 2 I= (m2 )n/2+a−b (4π)n/2 Γ(b) Γ n2 Z

π

m

(8.6.13)

This is held to be true for n non-integer valued also. Substitute this into the integral for the Pauli Villars loop ( figure 8.6.1). Z 1 1 1 1 − n = −n(a − b) dx (8.6.14) n a b [xa + (1 − x)b]n+1 0 i.e. Formally, (n = 4, a = 1, b = 2) Z ¢ d4 q q2 i ¡ 2 2 2+1−2 Γ(2 + 1) Γ(2 − 1 − 2) = m − x(1 − x)k (2π)4 [q 2 + x(1 − x)k 2 − m2 ]2 (2π)2 Γ(2) Γ(2) (8.6.15) But Γ(−1) is at a pole of Γ(x). (Γ(x) has poles at x = 0, − 1, − 2, − 3, . . .) We can just let the number of dimensions be “n”, not 4. We have, Z dn k q2 IP V = (2π)n [q 2 + x(1 − x)k 2 − m2 ]2 ¡n ¢ ¡ ¢ n i 2 2 n/2+1−2 Γ 2 + 1 Γ 2 − 1 − 2 ¡ ¢ (m − x(1 − x)k ) = (4π)n/2 Γ n2 Γ(2) ¡n ¢ ¢ ¡ n ¡ 2 ¢ i 2 n/2−1 Γ 2 + 1 Γ 1 − 2 ¡ ¢ m − x(1 − x)k = (4π)n/2 Γ(2) Γ n2 ¡ ¡ ¢ ¡ ¢¢ Now let ε = 2 − n2 → 0 as n → 4 Γ n2 + 1 = n2 Γ n2 IP V =

i (m2 − x(1 − x)k 2 )−ε+1 (2 − ε)Γ(ε − 1) (4π)2−ε

(8.6.16)

But, "

Γ(1 + ε) = exp −γε +

∞ X (−1)n εn ζ(n) n=2

n

#

∞ X 1 → ζ(n) = kn k=1 Ã N ! X 1 Z N dt γ = ± lim − = 0.577 . . . N →∞ k t 1 k=1

(8.6.17)

8.6. REGULARIZATION

127

From this, 1 Γ(1 + ε) = − γO(ε) 2 ε

Γ(ε) =

(8.6.18)

aε = eε ln a ε2 ln2 a = 1 + (ε ln a) + + ... 2!

(8.6.19)

Thus our integral becomes IP V

" i (m2 − x(1 − x)k 2 ) (1 + ε ln(4π)) (1 − ε(m2 − x(1 − x)k 2 )) · = (4π)2 #µ ¶µ ¶ ³ 1 1 ε´ − γ + ... · 1− 2 ε−1 ε ¶ µ 1 1 −2i 2 2 2 2 + O(ε) (m − x(1 − x)k ) − γ + ln(4π) − ln(m − x(1 − x)k ) + = (4π)2 ε 2

Note that the first term of

1 ε

will be a pole as ε → 0 (other terms finite). This is only a portion

Combine this with the integral for

and will see that all terms of order

1 ε

will cancel.

128

CHAPTER 8. LOOP DIAGRAMS We note that as M 2 ≫ m2 , the combination of these two diagrams (recall figure 8.6.1)

is, 2iα Iµν (k) = (kµ kν − gµν k 2 ) π 2

Z

1 0

dx x(1 − x) ln

µ

M2 m2 − x(1 − x)k 2

¶

(8.6.20)

e (note that the ln still diverges as M 2 → ∞.) Note that the result is where α = 4π~c proportional to kµ kν − gµν k 2 .

Now, photon propagator is i(gλµ + gauge dependant parts) k 2 + iε

(8.6.21)

→ gauge dependant parts ∼ kλ kµ . But, k µ Iµν = k µ (kµ kν − gµν k 2 )f (k 2 ) = (k 2 kν − kν k 2 )f (k 2 ) = 0 Thus gauge dependant parts of the photon propagator don’t contribute to the physical process. This form of Iµν does not arise if we were to use cutoffs (i.e. cutoffs are not gauge invariant.) Note also that the dependance on M 2 is logarithmic. i.e.

∼

Z

d4 q[Agµν q 2 + (Bx(1 − x)k 2 + Cm2 )gµν ] [q 2 + x(1 − x)k 2 − m2 ]2

Would expect quadratic divergence ∼ M 2 . (M 2 terms all cancelled.) Quadratic divergences

8.6. REGULARIZATION

129

proportional to gµν M 2 all cancel. → still logarithmically divergent. Nove 26/99 Elimination of ln M 2 . Remember the photon propagator itself. (µ2 → some mass scale introduced →arbitrary (important)) Write · µ 2¶ µ 2 ¶¸ Z 1 ¢ M iα ¡ m − q 2 z(1 − x) 2 qµ qν − q gµν ln Iµν = −6 dz z(1 − z) ln 3π µ2 µ2 0 ¶· µ 2¶ ¸ µ M iα qµ qν ln + Π(q 2 , m2 µ2 ) (8.6.22) = − q 2 gµν − 2 3π q µ2 {z } | T gµν

(

µ µ 2¶ · ¶¸ µ ¶ −igµν −igµλ −iαq 2 T M −igλν ln +Π = (−ieγµ ) + g q2 q2 3π λσ µ2 q2 µ µ µ 2¶ ¶· ¶¸ −igµλ M −iαq 2 T + ln g +Π · q2 3π λσ µ2 ) ¶" ¶ ¶# µ ¶ µ T µ µ −iαq 2 gκρ −igρν M2 igσκ +Π + . . . (−ieγν ) (8.6.23) ln · − 2 q 3π µ2 q2 a This is a geometric series, a + ar + ar 2 + . . . = 1−r , |r| < 1. So, " µ ¶¶ µ µ 2¶ M −igµν −igµλ α T + − gλν ln +Π = (−ieγµ ) q2 q2 3π µ2 # µ µ 2 ¶¶¶2 µ M −igµλ α T + +Π + . . . (−ieγν ) − gλν ln q2 3π µ2   T −igµν 1 ³ ³ ´ ´  (−ieγν ) = (−ieγµ )  2 2 q 1 + α ln M + Π(q 2 , m2 , µ2 ) 3π µ2  

= ³ 1+

 −e2 ³ ´´ γµ   · α M2 ln 2 q2 q + 3π µ

m 3π³

α ln 1+ 3π

T −igµν ´ 2

M µ2

(Π(q 2 , m2 , µ2 ))

 ¸  γν

(8.6.24)

130


(neglecting terms of order α2 = e2R

³ 2 ´2 e 4π

). Now, Define:

e2

= 1+

α 3π

ln

³

M2 µ2

´ ≈e

2

µ

α 1− ln 3π

µ

M2 µ2

¶¶

(8.6.25)

(Renormalized Charge). Thus, continuing (8.6.24) ) ( T −ig µν ¢ γν ¡ = −e2R γµ αR 2 q 1 + 3π Π(q 2 , m2 , µ2 )

(8.6.26)

We now let M 2 → ∞.

• Note that there is still a dependence on µ2 . (Changes in µ2 give rise to corresponding changes in αR . • Also, in → (−ieR γµ ) shell. • Also

T gµν = gµν −

³

qµ qν q2

´ 2

qµ qν q2

(−ieγν ) gives zero provided the ends of the electron legs are on

2

2

Π(q , µ , m ) = −6

Z

1 0

dz z(1 − z) ln

is such that q 2 Π(q 2 , m2 , µ2 ) → 0 as q 2 → 0. Hence,

µ

m2 − q 2 z(1 − z) µ2

¶

still has a pole at q 2 = 0. Thus the photon is massless. In 2-D, if m2 = 0, Π(q 2 , µ2 ) is finite, and Π(q 2 ) = becomes T e2 γµ gµν γν ´ = − ³ e2 q 2 1 − πq 2

= −

T e2 γµ gµν γν

q2 −

e2 π

e2 , πq 2

then the above diagram sum

8.7. NOETHER’S THEOREM

131

Thus the photon develops a mass in 2-D due to the radiative corrections. (e has dimensions of mass in 2-D). i.e. Z S = d2 xL   2  Z  1  d2 x −  ∂µ Aν −∂ν Aµ  + ψ¯ ( 6 ∂ −i |{z} S = e 6 A)ψ  (8.6.29) |{z} |{z} |{z} |{z} |{z} 4 |{z} [µ]=0

8.7

−2

0

1

1 2

1

+1

Noether’s Theorem

For every infinitesimal change which leaves the action invariant there exists a conserved current which leads to a conserved charge. Change of type I L = L(φA (x), ∂µ φA (x)) xµ → x0µ = xµ φA (x) → φ0A (x) = φA + εAB φB (x)

(8.7.1) (8.7.2) (8.7.3)

with εAB → infinitesimal. (ex. we had a gauge transformation ψ(x) → eiθ(x) ψ(x) = lim

N →∞

µ

Now, S = =

Z

Z

iθ 1+ N

¶N

ψ(x) ≈ (1 + i θ(x))ψ(x) |{z} εAB

!

d4 x L(φA , ∂µ φA ) d4 x L(φ0A , ∂µ φ0A )

(8.7.4)

where both L(φA , ∂µ φA ), L(φ0A , ∂µ φ0A ) have the same functional dependence. (For example, f (x, y) = x2 + y 2 ; let x0 = x cos θ − y sin θ, y 0 = x sin θ + y cos θ, and we still get f (x0 , y 0 ) = x02 + y 02 (invariant) - whereas g(x, y) = xy 6= g(x0 , y 0 ) ). Nov 30/99 Suppose L is invariant under an infinitesimal translation xµ → xµ + εµ . Z S = d4 x L(φ(x), ∂µ φ(x)) (8.7.5) Z = d4 x0 L(φ(x0 ), ∂µ0 φ(x0 )) (8.7.6)

132


in the same way as S =

Z

4

d x L(x) =

→ d 4 x0 =

Z

d4 x0 L(x0 )

(8.7.7)

∂(x01 . . . x00 ) 4 d x = d4 x ∂(x1 . . . x0 )

Now, δL = L(x0 ) − L(x) = L(xµ + εµ ) − L(xµ ) ∂L = εµ µ ∂x or = L(φ(x0 ), ∂µ0 φ(x0 )) − L(φ(x), ∂µ φ(x))

(8.7.8) (8.7.9)

If now, δφ(x) = φ(x0 ) − φ(x) = φ(xµ + εµ ) − φ(x) = εµ

∂φ ∂xµ

(8.7.10)

then ∂L δL δL = δφ(x) + δ(∂µ φ(x)) ∂φ(x) δ(∂µ φ(x)) ¶ µ ¶ µ ∂L ∂ δL ν ∂φ µ ∂φ + µ ε = ε µ ν ∂x ∂φ(x) ∂x ∂x δ(∂µ φ(x)) However, we also have the equations of motion. Z δ d4 xL(φ(x), ∂µ φ(x)) = 0 → φ = φcl + δφ

(8.7.11)

(8.7.12) (8.7.13)

i.e. Z

d4 x [L(φcl + δφ, ∂φcl + ∂µ δφ)] · ¸ Z δL(φcl , ∂µ φcl ) δL(φcl , ∂µ φcl ) 4 = d x L(φcl , ∂µ φcl ) + δφ + (∂µ δφ(x)) + ... δφ ∂(∂µ φ) If δφ → 0 as xµ → ∞   =

Z

¶ µ   ∂ δL(φcl , ∂µ φcl ) δL(φcl , ∂µ φcl )   + . . . d x L(φcl , ∂µ φcl ) + δφ(x) − µ (8.7.14) δφ ∂x δ(∂µ φ(x))   | {z } 4

=0 if φcl is extremum of S


133

Hence as δφ is arbitrary, 0=

δL ∂ δL − µ δφ ∂x δ(∂φ)

(8.7.15)

on the classical path. Recall (8.7.11), ¶ ¶µ µ ¶ ∂ δL ∂ δL ∂φ ν ∂φ + µ ε δL = ε ν µ ν ∂x ∂x δ(∂µ φ) ∂x ∂x δ(∂µ φ) · ¸ ∂ ∂φ δL = εν ν µ ∂x ∂x δ(∂µ φ) µ

ν

(8.7.16)

(8.7.8) and (8.7.16) must be identical. As a result, εµ

Thus

∂L ∂ = (gµν εν L) ∂xµ ∂xµ · ¸ δL ∂ ν ∂φ = ε ∂xµ ∂xν δ(∂µ φ)

¸ · ∂ ∂φ δL 0 = µ −gµν L + ν ∂ ∂x δ(∂µ φ) | {z }

(8.7.17)

Tµν Energy-Momentum Tensor

Or,

∂ Tµν = 0 ∂xµ

(8.7.18)

Thus, there are 4 conserved currents, (µ = 1, 2, 3, 0), one for each invariance. This only holds if φ that it depends on is a solution to the classical equation. (ex. kµ (gµν k 2 − kµ kν ) = 0 always, but kµ (gµν m2 − kµ kν ) = 0 only if k 2 = m2 . ) There are 4 conserved currents; one fore each direction in which we have an infinitesimal invariance. For T00 = −g00 L +

∂φ δL ∂x0 δ(∂0 φ)

∂φ Π=H ∂t = −L + pq) ˙ = −L +

(H Now, if

∂µ j µ = 0

(8.7.19)

134


Then, Z

d 3 x ∂µ j µ = 0 · ¸ Z ∂ 0 3 0 = d x ∇·j+ j ∂t Z Z ∂ d3 x j 0 (x, t) ( by Gauss) = dS · j + ∂t R The first term is 0 if j → 0 as |x| → ∞. Let Q(t) = d3 x j 0 (x, t). ∴

∂ Q(t) = 0 (“Charge” conservation) ∂t

(8.7.20)

For ν = 0, Q(t) = =

Z

Z

d3 x T00 (x, t) d3 x H

(8.7.21)

(The “charge” here is the energy of the field.) For ν = i Z d3 xT0i (x, t) = Pi (t) → Mechanical momentum of field

(8.7.22)

Infinitesimal internal transformations xµ → x0µ = xµ L = L(φA (x), ∂µ φA (x))

(8.7.23)

φA → φ0A = φA + εAB φB

(8.7.24)

Invariance under (ex. Aµ → Aµ + ∂µ Λ, ψ → (1 + iΛ)ψ ⇒ Gauge transformation.) Z Z 4 S = d x L(φA , ∂µ φA ) = d4 x L(φ0A , ∂µ φ0A ) · ¸ Z δL δL 4 = d x L(φA , ∂µ φA ) + εAB φB + (∂εAB φB ) + ... δφA δ(∂µ φA ) Also if φA satisfies Hence,

δL δφA

(8.7.25)

. = ∂µ δ(∂δL µ φA ) Z

·

δL δL 0 = d x (εAB φB )∂µ + (∂µ εAB φB ) δ(∂µ φA ) δ(∂µ φA ) · ¸ Z δL ∂ εAB φB 0 = d4 x µ ∂x δ(∂µ φA ) 4

¸ (8.7.26)


135

Thus we have a conserved current, JµAB

·

δL = φB δ(∂µ φA )

¸

(8.7.27)

¯ µ ψ. ¯ 6 ∂− 6 A)ψ − 1 F 2 , then jµ = ψγ ex. if L = ψ(i 4 Dec. 1/99 x0µ = xµ + εµ φ0A (x) = φA (x) + εAB φB (x) Consider a general transformation in which, x0µ = xµ + δxµ ψ(x0 ) = ψ(x) + δψ(x)

(8.7.28) (8.7.29)

ex. consider a rotation of a vector field Ai (xi ) in 2-D. x0 y0 → δx → δy

= = = =

x cos θ − y sin θ ≈ x − θy for small θ x sin θ + y cos θ ≈ θx + y −θy ( small θ) θx ( small θ)

At the same time, A0x (x0 , y 0 ) = Ax (x, y) cos θ − Ay (x, y) sin θ A0y (x0 , y 0 ) = Ax (x, y) sin θ + Ay (x, y) cos θ δAx = A0x (x0 , y 0 ) − Ax (x, y) = A0x (x0 , y 0 ) − Ax (x0 , y 0 ) + Ax (x0 , y 0 ) − Ax (x, y) | {z } | {z } 0 0 ¯ = δAx (x , y ) + [Ax (x − θy, y + θx) − Ax (x, y)] · ¸ ∂ ∂ 0 0 ¯ = δAx (x , y ) + −θy Ax (x, y) + θx ∂x ∂y Invariance of the action S S = =

Z

Z

d4 x L(ψ(x), ∂µ ψ(x)) d4 x0 L(ψ 0 (x0 ), ∂µ0 ψ 0 (x0 ))

(8.7.30)

136


In general, d4 x0 = |J|d4 x, ¯ · ¸¯ ¯ ∂(x01 . . . x00 ) ¯¯ ¯ → |J| = ¯det ¯ ¯ ∂(x1 . . . x0 ) ¯  ∂x01  01 ¯ ¯ . . . ∂x ¯ ¯ ∂x1 ∂x0  .. ¯ ¯ . . = ¯det  . .  ¯¯ ¯ ∂x00 ∂x00 . . . ∂x0 1  ³∂x ´ ∂δx01 ∂δx01 1 + ∂x1 ... 2  ³ ∂x 02 ´ ¯  . ¯ ..  1 + ∂δx ¯ ∂x2  = ¯det  ... ¯  ³ 

So also, if we call

∂δx00 ∂x1

.. .

1+



¯ ¯ ¯ ¯ ¯  ´  00

∂δx ∂x0

¯ ¯ ¯ ¡ µ 2 ¢ ¯¯ ∂δxµ ¯ = ¯1 + + O (δx ) ¯ ¯ ¯ ∂xµ

(8.7.31)

L0 (x0 ) = L(ψ 0 (x0 ), ∂µ0 ψ 0 (x0 ))

L0 (x0 ) − L(x) = δL = L0 (x0 ) − L(x0 ) + L(x0 ) − L(x) {z } | {z } | ¯ δL(x)

Now,

(8.7.32)

δxµ

(8.7.33)

∂L(x) ∂xµ

¯ δL(x) = L0 (x0 ) − L(x0 ) = L(ψ 0 (x0 ), ∂µ0 ψ 0 (x0 )) − L(ψ(x0 ), ∂µ0 ψ(x0 )) 0 ¯ (Re: ψ 0 (x0 ) − ψ(x0 ) = δψ(x )) 0 0 0 0 0 ¯ ¯ )) − L(ψ(x0 ), ∂µ0 ψ(x0 )) = L(ψ(x ) + δψ(x ), ∂µ ψ(x ) + ∂µ0 δψ(x Hence, δL δL 0 0 ¯ = δψ(x ¯ ¯ δL ) ) + ∂µ0 δψ(x 0 0 δψ(x ) δ(∂µ ψ(x0 ))

(8.7.34)

By the equations of motion, ∂L ∂ δL 0 0 ¯ ¯ = δψ(x ¯ )) + (∂µ0 δψ(x δL ) µ 0 0 0 ∂x δ(∂µ ψ(x )) δ(∂µ ψ(x0 )) ¶ µ ∂ δL 0 ¯ = δψ(x ) ∂x0µ δ(∂µ0 ψ(x0 ))

(8.7.35)


137

So, putting together, Z Z 4 0 0 0 0 0 0 0 = d x L(ψ (x ), ∂µ ψ (x )) − d4 x L(ψ(x), ∂ψ(x)) ( subs. in (8.7.31), (8.7.33), (8.7.35)) ¸· ¸ Z · Z ∂δxµ ∂L µ 4 ¯ + δx L(ψ(x), ∂µ ψ(x)) + δL − d4 x L(ψ(x), ∂µ ψ(x)) = d x 1+ ∂xµ ∂xµ To leading order, ·

¶¸ µ ∂δxµ ∂ δL µ ∂L ¯ 0 = dx L + δx + δψ(x) ∂xµ ∂xµ ∂xµ δ(∂µ ψ(x)) ¶¸ · µ Z ∂ δL ¯ = d4 x δxµ L + δψ µ ∂x δ(∂µ ψ(x)) Z

4

¯ contain an arbitrary parameter, then As δxµ and δψ ∂µ j µ = 0 where

¯ j µ = δxµ L + δψ

(8.7.36) δL δ(∂µ ψ)

(8.7.37)

(Noether Current). Suppose δxµ = x0µ − xµ = Aµj (x)εj (translation: εµ ) δψA (x) = ψA0 (x0 ) − ψA (x) = BAj (x)εj But ¯ A (x) = ψ 0 (x) − ψA (x) δψ A = ψA0 (x0 ) − ψA (x) + ψA0 (x) − ψA0 (x0 ) ¶ µ 0 µ ∂ψA (x) = δψA + −δx ∂xµ ∂ψ 0 ≈ BAj εj − Aµj εj A ∂xµ ∂ψ ≈ BAj εj − Aµj εj µ ∂x (can replace ψ 0 → ψ because ψ 0 , ψ differ by a term which would lead to an over-all term ε2 → no good). Thus, µ ¶ δL µ j j j ∂ψA j = (Aµj ε )L + BAj ε − Aµj ε (8.7.38) µ ∂x δ(∂µ ψA )

138


For each εj there is a different current; jkµ

µ

∂ψA = Aµk L − BAk − Aµk µ ∂x

¶

δL δ(∂µ ψA )

(8.7.39)

(Translations: Aµk = gµk , BAk = 0. Angular momentum → rotations of space-time points and fields.) Dec. 3/99 Recall the spin 21 particle: ¯ 6 ∂ − m)ψ L = ψ(i (8.7.40) ψ = proton, neutron. Yukawa postulated that the strong force was mediated by spin-zero pseudo-particles −µr (mass µ, V (r) ∼ e r ). Heisenberg: postulated that the strong force treated the p and n exactly the same way and these two “nucleons” were just different states of the same “particle”, and they corresponded to spin up and spin down states of the electron. (Not really spin → but same mathematical structure, “Isosspin”). ·

¸ ψn → “Isospin doublet” ψp ψn , ψp → four component Dirac sponors ψ = ψiα ; i = 1, 2 (isospin) , α = 1, . . . , 4 ( Dirac).

1 µ2 λ L = ψ¯i (i 6 ∂ − m)ψi − ig ψ¯i γ5 (φ · τ ij )ψj − (∂µ φ)(∂ µ φ) − φ2 − (φ2 )2 2 2 4

(8.7.41)

(8.7.42)

with τ = Pauli matrices, and φ1 , φ2 , φ3 - 3 spin zero pseudo-scalars, “pions”. This is invariant under ψ → eiτ ·θ/2 ψ ¯ −iτ ·θ/2 ψ¯ → ψe τ · φ → eiτ ·θ/2 τ · φ e−iτ ·θ/2 where θ = (θ1 , θ2 , θ3 ) → constant 3 component isovector. ρ1µ , ρ2µ , ρ3µ → 3 vector particles, also interact with the nucleons. Postulate 1 µ2 ¯ µ ρi τ i ψ + ( Rest of terms from last L) L = − (∂µ ρiν − ∂ν ρiµ )2 − ρiµ ρiµ + ψγ µ 4 2

(8.7.43)


139

If ρiµ τ i → eiτ

For θi ≈ 0

=

µ

i θ i /2

ρiµ τ i e−iτ

iτ i θi 1+ 2



¶

i θ i /2

ρjµ τ j

(8.7.44)

µ

iτ i θi 1− 2

¶



¢ ¡ i   = ρjµ τ j + θi ρjµ τ i τ j − τ j τ i +O(θ2 ) 2 | {z }

(8.7.45)

2iεijk τ k

Thus,

ρiµ → ρiµ + εijk ρj θk

Conserved current arises in L=− Here;

¢2 µ2 1¡ i ∂µ ρν − ∂ν ρiµ − ρiµ ρiµ 4 2

∂L ∂L i δρ + δ∂ν ρiµ µ ∂ρiµ ∂(∂ν ρiµ ) ∂L ∂L = ∂ν δρiµ + δ∂ν ρiµ i i ∂(∂ν ρµ ) ∂(∂ν ρµ ) ¸ · ∂L = ∂ν δρi ∂(∂ν ρiµ ) µ | {z }

(8.7.46)

(8.7.47)

0 = δL =

(8.7.48)

jν

Thus as

¢ ¡ i ∂L 1 i = − (2)(2) ∂ ρ − ∂ ρ ν µ µ ν ∂(∂ν ρiµ ) 4

Thus,

(8.7.49)

i fνµ

¡z

}| ¢{ jν = − ∂ν ρiµ − ∂µ ρiν (εijk ρj θk )

i ijk j jνk = −fνµ ε ρµ

(8.7.50)

Couple jµk to Akµ ,

L

0

L

{ 1 i iµν µ2 i iµ = − fµν f − ρµ ρ +gjµk ρkµ 4 2 ρiµ → ρiµ + εijk ρjµ θk ; A new piece is added to jµk ¢2 µ2 1¡ = − ∂µ ρiν − ∂ν ρiµ − ρiµ ρiµ − g(∂ν ρiµ − ∂µ ρiν )εijk ρjµ ρkν 4 2 z

}|

(8.7.51)

140


i.e. Lnew = −gεijk (∂ν ρiµ − ∂µ ρiν )ρjµ ρkν ∂Lnew i δρ jνnew = ∂(∂ν ρiµ ) µ

(8.7.52)

= −2gεijk (ρjµ ρkν )(εimn ρm θn )

(8.7.53)

Again, as θ is arbitrary, n lji jµ jνnewi = −2gεlmn ρm µ ρν ε ρ

(8.7.54)

L” = L0 + gjνnewk ρkν µ2 1 n lji j iν = − (∂µ ρiν − ∂ν ρiµ )2 − ρkν ρkν − g(∂ν ρiµ − ∂µ ρiν )εijk ρjµ ρkν + g(−2gεlmn ρm µ ρν ε ρµ )(ρ ) 4 2 fourth term - only depends on ρ, not derivative - need no more terms 1 i iµν µ2 i iµ F − ρµ ρ (8.7.55) = − Fµν 4 2 i where Fµν = ∂ρiν − ∂ν ρiµ − 2gεijk ρjµ ρkν Rescale g by ? µ2 1 L = − (∂µ ρiν − ∂ν ρiµ + gεijk ρjµ ρkν )2 − ρiµ ρiµ 4 2 ijk j k i i is invariant under ρµ → ρµ + ε ρµ θ . This can be generalized to an SU (N ) gauge series. 1 µ2 L = − (∂µ Aν − ∂ν Aµ + gf ijk Ajµ Akν )2 − Aiµ Aiµ 4 2

(8.7.56)

(8.7.57)

with f ijk → structure functions for SU (N ) group.

Lspin-2 = L(hµν ) + Tµν hµν + T 0µν hµν + T 00µν hµν + . . . hµν (x) = hνµ

(8.7.58)

Define Tµν (stress tensor ) from L. Couple

hµν T µν ⇒ L + L0 0 Tµν → Tµν + Tµν

Sum Lspin-2 (∞), get

√ Lspin-2 = R g (Einstein- Lagrangian)

(8.7.59)

If µ2 = 0, the gauge invariance becomes local (allows for renormalizability). ρiµ (x) → ρiµ (x) + ∂µ

θi (x) + εijk ρjµ (x)θk (x) g

(8.7.60)

Chapter 9 Path Integral Quantization recall:

9.1

Heisenberg-Dirac dA(t) = [A(t), H]PB dt ¶ X µ ∂A ∂H ∂A ∂H ; = − ∂q ∂p ∂p ∂q i i i i i

(9.1.1) µ

∂H =0 ∂t

¶

(9.1.2)

With (9.1.1), we can go from a classical to a quantum mechanical (QM) operator, ˆ A(t) → A(t) (QM operator) ˆ dA(t) dt

i 1 hˆ ˆ A(t), H i~ i 1 hˆˆ ˆ Aˆ = AH − H i~ 1 [ , ] [ , ]PB → i~ h i ∂ Aˆ ˆ H ˆ = A, i~ ∂t

(9.1.3)

=

(9.1.4) (9.1.5)

(Note: will drop “ˆ” now, as we will deal with operators only from now on.) The solution to this is, A(t) = eiHt/~ A(0)e−iHt/~ = AH (t) ; (Heisenberg Variable - Time dependent) 141

(9.1.6) (9.1.7)

142

CHAPTER 9. PATH INTEGRAL QUANTIZATION

State vectors |ψH i are time independent hψH |AH (t)|ψH i = hψH |eiHt/~ A(0) e−iHt/~ |φH i | {z } | {z }

(9.1.8)

|φs (t)i = e−iHt/~ |φH i d → i~ |φs (t)i = H|φs (t)i Schrodinger eq. dt hφs (t)| = hφH |eiHt/~ As = A(0) → time independent.

(9.1.9)

hφs |

|φs i

We go to Schrodinger state vectors by;

(9.1.10) (9.1.11) (9.1.12)

Consider eigenstates of the Heisenberg operators. AH (t)|a(t)i = a|a(t)i e A(0)e−iHt/~ |a(t)i = a|a(t)i ¤ £ ¤ £ A(0) e−iHt/~ |a(t)i = a e−iHt/~ |a(t)i

(9.1.13)

A(0)|a(0)i = a|a(0)i

(9.1.15)

|a(t)i = eiHt/~ |a(0)i ha(t)| = ha(0)|e−iHt/~

(9.1.16a) (9.1.16b)

iHt/~

Let t = 0,

(9.1.14)

Hence

9.2

Wave Functions

Schrodinger wave function ψ(q, t) = hq|ψ(t)is

(9.2.1)

where i~ dtd |ψs (t)i = H|ψs (t)i, |ψ(t)i = e−iHt/~ |ψ(0)i. Thus ψ(q, t) = hq|e−iHt/~ |ψ(0)i | {z } (9.1.16b)

= hqH (t)|ψ(0)i

(here Q(t)|qH (t)i = q|qH (t)i → because we have a complete set of eigenfunctions). Note: Z dq(t) |qH (t)ihqH (t)| = 1 for any t

(9.2.2)

(9.2.3)

9.2. WAVE FUNCTIONS

143

This is the Completeness Relation. Let 0 < t1 < t2 < . . . < tn < t ∗ ∗. ψ(q, t) =

Z

For

0 0 (0)|ψ(0)i dq 0 (0) hqH (t)|qH (0)i hqH {z }| {z } |

(9.2.4)

ψ(q,0)=hq|ψ(0)i

K(t,0) → propagator

0 K1 (t, 0) = hqH (t)|qH (0)i Z = hqH (t)|qHn (tn )ihqHn (tn )|qH(n−1) (tn−1 )ihqH(n−1) (tn−1 )| . . . |qH1 (t1 )i ·

· hqH1 (t1 )|q 0 (0)idqn (tn ) dqn−1 (tn−1 ) . . . dq1 (t1 )

(9.2.5)

Consider one part of above: hqH(i+1) (ti+1 )|qHi (ti )i = hqs(i+1) (0)|e−iHti+1 /~ e+iHti /~ |qsi (0)i = hqs(i+1) (0)|e−iH(ti+1 −ti )/~ |qsi (0)i

(9.2.6)

Let ε = ti+1 − ti . If ε ≈ 0, then, µ

iHε hqH(i+1) (ti+1 )|qHi (ti )i ∼ hqs(i+1) (0)| 1 − ~

¶

|qsi (0)i

(9.2.7)

Remember [q, p] = i~

(9.2.8)

eipq/~ < q, p >= √ 2π~

(9.2.9)

Thus,

i.e. hq|ˆ p|ψi = −i~

∂ hq|ψi ∂q

(9.2.10)

which is satisfied if |ψi = |pi. So, µ

iHε hqs(i+1) (0)| 1 − ~

¶

|qi (0)i =

Z

dpi+1

i.e. used hq|ˆ p|ψ(t)i = −i~

µ

iHε dpi hqi+1 (0)|pi+1 ihpi+1 | 1 − | {z } ~ ip qi+1 /~ e i+1 √ 2π~

∂ ∂ hq|ψ(t)i − −i~ ψ(q, t) ∂q ∂q

¶

|pi i hpi |qi (0)i | {z } i qi /~ e−ip √ 2π~

(9.2.11)

(9.2.12)

144

CHAPTER 9. PATH INTEGRAL QUANTIZATION → hq| [ˆ q , pˆ] |ψ(t)i = i~hq|ψ(t)i = qhq|ˆ p|ψ(t)i − hq|ˆ pqˆ|ψ(t)i ¶ µ ¶ µ ∂ ∂ q |ψ(t)i = q −i~ hq|ψ(t)i − −i~ hq|ˆ ∂q ∂q ¶ µ ¶ ¸ · µ ∂ ∂ − −i~ q hq|ψ(t)i = q −i~ ∂q ∂q = i~hq|ψ(t)i [ˆ q , pˆ] = i~ · ¸ ∂ hq|ˆ p|ψ(t)i = −i~ + f (q) hq|ψ(t)i ∂q

(9.2.13) (9.2.14) (9.2.15)

→ See Dirac - f (q) can be absorbed into the phase of ψ(q, t). Jan. 13/2000 i.e. · ¾ · ¾ ¸ ½ Z q ¸ ½ Z q ∂ ∂ i i 0 0 0 0 −i~ + f (q) exp f (q )dq ψ(q, t) = −i~ ψ(q, t) exp f (q )dq ∂q ~ ∂q ~ (9.2.16) 0 0 Re: K(q , t ; q, t), Z 0 0 ψ(q , t ) = dq K(q 0 , t0 ; q, t)ψ(q, t) (9.2.17)

→ K(q 0 , t0 ; q, t) = h(q 0 , t0 )H |(q, t)H i Z = dq1 . . . dqn h(q 0 , t0 )H |(qn , tn )H ih(qn tn )H |(qn−1 tn−1 )H i . . . h(q1 t1 )H |(q, t)H i (9.2.18)

i1

i

→ h(q i+1 ti+1 )H |(q i ti )H i = hqsi+1 |eiH(t −t )/~ |qsi i µ ¶ i i+1 ≈ hqs | 1 − H(ˆ q , pˆ)ε |qsi i ~ ¶ µ Z iε i i+1 q , pˆ) |pi ihpi |q i i = dp hq | 1 − H(ˆ ~ hq|ˆ p|pi = −i~ ∴

Z

∂ hq|pi ∂q

→ hq|pi = Keipq/~ Z 0 0 dp hq |pihp|qi = dq eip·(q −q)/~ = δ(q 0 − q) 1 ⇒K = √ 2π~

(9.2.19)

(9.2.20)

(9.2.21) (9.2.22)

9.2. WAVE FUNCTIONS ∴ h(q

i+1 i+1

t

145

i i

)H |(q t )H i =

Z

¶ µ dpi iε i i i+1 √ q , pˆ) |pi ie−ip q /~ hq | 1 − H(ˆ ~ 2π~

↓ ¤ → hq |H(ˆ q , pˆ)|p i ≈ H(qi+1 , pi )hq i+1 |pi i ↓ ( Have to worry about ordering of pˆ0 s, qˆ0 s.) µ ¶ Z dpi ipi (qi+1 −qi )/~ iε i+1 i+1 i i √ ∴ h(q t )H |(q t )H i = 1 − H(qi+1 , pi ) e ~ 2π~ £

i+1

i

(9.2.23)

(The H term is important in curved space). Thus, K(q 0 , t0 ; q, t) is; Z Z dpn ip0 (q1 −q)/~ ip1 (q2 −q1 )/~ dp0 0 0 0 ... e e . . . eipn (q −qn )/~ · K(q , t ; q, t) = dq1 . . . dqn 2π~ 2π~ ¶µ ¶ µ ¶ µ iε iε iε 0 1 − H(q2 , p1 ) . . . 1 − H(q , pn ) · 1 − H(q1 , p0 ) ~ ~ ~ ( Z Z h dpn i dp0 ... exp p0 (q1 − q0 ) − εH(q1 , p0 ) ' dq1 . . . dqn 2π~ 2π~ ~ ) i 0 0 + p1 (q2 − q1 ) − εH(q2 , p1 ) + . . . + pn (q − qn ) − εH(q , pn ) Let

qi+1 −qi ε

≈ q(t ˙ i ). Thus,

K(q 0 , t0 ; q, t) =

Z

dq1 . . . dqn

Z

dp0 dpn ... exp 2π~ 2π~

(

) Riemann Sum n z }| { X i ε [pi (q˙i (ti )) − H(qi+1 , pi )] ~ i=0 (9.2.24)

See figure 9.2.1. As ε → 0,

Figure 9.2.1: Particle path, where the qn+1 = q 0 , q0 = q are fixed, and the other q’s (and p’s) are all integrated over.

K→

Z

q(t0 )=q 0 q(t)=q

Dq(t)

Z

) ( Z 0 i t Dp(t) exp dt [p(t)q(t) ˙ − H(q(t), p(t))] ~ t

(9.2.25)

146


Which is the Feynman Path Integral (Dirac/Feynman Papers → in a collection edited by Schwinger ). Suppose, H(q, p) = i.e.

- this is a Gaussian

³

Z

i

dp e 2π~

p2 + V (q) 2m

(9.2.26)

" Ã !# p2i i pi (qi+1 −qi )−ε ~ 2m

(9.2.27)

Recall: Z

∞

dx e

−ax−bx2

=

−∞

=

Z

∞

e

−b(x+

a 2 a2 + 2b ) 4b

−∞

r

π a2 /4b ´ e b

“Ignore” i in exponent of integral by putting in a convergence factor. i.e., Z ³ 2´ io n h dpi p (let δ → 0) exp ~i pi (qi+1 − qi ) − ε 2mi + iδp2 2π~ ³ ´2  −i(qi+1 −qi ) r ~ 1 π  ¡ iε ¢  = exp   iε 2π~ 2m~ 4 2m~ ( ) r im (qi+1 − qi )2 m = exp 2iπ~ε 2ε~

(9.2.28)

(9.2.29)

Plug this in ∀ pi ’s , ( n · ¶n+1 ¸) X m(qi+1 − qi )2 i m K(q 0 , t0 ; q, t) = dq1 . . . dqn exp − εV (qi+1 ) 2iπ~ε ~ i=0 2ε | {z } N ( ) µ ¶2 Z n i iε X h m qi+1 − qi = N dq1 . . . dqn exp V (qi+1 ) ~ i=0 2 ε {z } | Z

µr

(q˙i (ti ))2

q˙2 l2 L(q, q) ˙ = − V (q) → Dim. of Energy = mass × 2 2m t ( Z 0 ) Z q(t0 )=q0 t i dq1 . . . dqn exp ∴ K(q 0 , t0 ; q, t) = N dt L(q(t), q(t)) ˙ ~ t q(t)=q

(9.2.30)

9.3. FREE PARTICLE

147

Jan. 17/2000 As ~ → 0, the dominant contribution when integrating over all paths comes from the path that minimizes Z t0 dτ L(q, q) ˙ (9.2.31) S= t

i.e. δ

Z

t0

dτ L(q, q) ˙ =0

(9.2.32)

t

(Hamilton’s Principle of Least Action).

9.3

Free Particle

For a free particle, we mean V = 0. hq 0 , t0 |q, ti = N

Z

q(t0 )=q 0 q(t)=q

( Z Dq(τ ) exp i

(~ = m = 1) ( Z Z q(t0 )=q0 Dq(τ ) exp i = N q(t)=q

t0

dτ L(q(τ ˙ ), q(τ )) t

t0 t

q˙2 (τ ) dτ 2

)

)

(9.3.1)

But recall; ³ ´³ ´ p2 ˆ 0 ˆ hq 0 |e−iHt e+iHt |qi ; free particle → H = 2 Z 2 0 2 = dp hq 0 |e−iˆp t /2 |pihp|eiˆp t/2 |qi Z 2 0 = dp e−ip (t−t )/2 hq 0 |pi hp|qi | {z } | {z }

h(q 0 t0 )H |(qt)H i =

0

e√iq p 2π

=

Z

−iqp e√ 2π

dp −ip2 (t−t0 )/2 ip(q0 −q)−δp2 e e 2π

(9.3.2)

where δ → 0+ (inserted so integral converges). Now, recall the Gaussian: Z

∞

dx e

−ax2 −bx

=

−∞

=

Z

∞

dx e

¶2 µ b −a x+ b2 /4a 2a

e

−∞

r

π b2 /4a e a

(9.3.3)

148


Here we have a = δ +

i(t0 −t) , 2

b = −i(q 0 − q). So, s

 

0

2

 

(q − q) ´ − ³  4 δ + i(t0 −t)  δ+ 2 ½ 0 ¾ 2 i(q − q) 1 exp = p 2(t0 − t) 2πi(t0 − t)

∴ h(q 0 , t0 )H |(q, t)H i =

Remember that, 0

0

1 2π

ψ(q , t ) = where

Z

π

i(t0 −t) 2

exp

dq K(q 0 , t0 ; q, t)ψ(q, t)

K(q 0 , t0 ; q, t) = h(q 0 , t0 )H |(q, t)H i

but also

∂ 1 i ψ(q, t) = − ∂t 2

Hence, i

µ

∂2 ∂q 2

¶

ψ(q, t)

(9.3.4)

(9.3.5) (9.3.6) (9.3.7)

1 ∂2 ∂ 0 0 K(q , t ; q, t) = − K(q 0 , t0 ; q, t) ∂t0 2 ∂q 02

(9.3.8)

lim K(q 0 , t0 ; q, t) = δ(q 0 − q)δ(t0 − t)

(9.3.9)

And t0 →t

If we start putting in potentials, it’s awfully difficult to work out, especially when you put in boundary conditions. It’s easier to consider vacuum to vacuum transitions (in order to eliminate having to consider q(t) = q and q(t0 ) = q 0 → i.e. the B.C.’s). Examine, Z q(t0 )=q0 R t0 ˙ )q(τ )] 0 0 J Dq(t) ei t dτ [L(q,q)+J(τ (9.3.10) h(q t )H |(qt)H i = q(t)=q

(where J = “source”) Look at, (now dropping subscript “H” → always in Heisenberg eigenstates); Z 0 0 J h(Q , T )H |(Q, T )H i = dq 0 dq hQ0 T 0 |q 0 t0 iJ hq 0 t0 |qtiJ hqt|QT iJ (9.3.11) Now ˆ

0

ˆ

0

hQ0 T 0 |q 0 t0 i = hQ0 |e−iHT eiHt |q 0 i ˆ If H|ni = En |ni X 0 0 = hQ0 |e−iEn T |nihn|eiEn t |q 0 i n

If hq|ni = φn (q) X 0 0 = e−iEn (T −t ) φn (Q0 )φ∗n (q 0 ) n

(9.3.12)

9.3. FREE PARTICLE

149

and similarly for hq, t|Q, T iJ . Z X 0 0 0 0 e−iEn (T −t ) e−iEn0 (t−T ) φn (Q0 )φ∗n (q 0 )φ∗n0 (Q)φn0 (q)hq 0 t0 |qtiJ ∴ hQ T |QT i = dq dq 0 n,n0

(9.3.13) Let T → −i∞, T → +i∞ → The vacuum becomes the dominant contribution. (i.e. only the vacuum state |0i survives - projects only the vacuum state) - holds for any t, t 0 . Z 0 0 (−t0 +t) 0 ∗ −iE0 (T 0 −T ) ∗ 0 (9.3.14) hQ T |QT i = dq dq 0 e|−iE0{z φ0 (q )hq 0 t0 |qtiφ0 (q) } φ0 (Q )φ0 (Q)e {z } | 0

∗

∗

(“ * ” → involves integral over q, q 0 ). If φ0 (q)e−iE0 t = φ0 (q, t), Z hQ0 T 0 |QT iJ → dq dq 0 φ∗0 (q 0 , t0 )hq 0 t0 |qtiJ φ0 (q, t) = 0lim T →−i∞ φ∗ (Q, T )φ0 (Q0 , T 0 ) 0 T →+i∞ = h0, t0 |0, ti

(9.3.15) (9.3.16)

(with hQ0 T 0 |QT iJ = Path integral, and φ∗0 (Q, T )φ0 (Q0 , T 0 ) = Normal factor N). So, ( Z 0 ) Z Q(T 0 )=Q0 h i T ˙ ), Q(τ )) + J(τ )Q(τ ) dτ L(Q(τ DQ(τ ) exp i h0, ∞|0, −∞iJ = 0lim N T →−i∞ T →+i∞

T

Q(T )=Q

(9.3.17)

Jan. 19/2000 Consider now, hq 0 , t0 |ˆ q (tn )ˆ q (tn−1 ) . . . qˆ(t1 )|q, ti = hq 0 , t0 |T qˆ(t1 )ˆ q (t2 ) . . . qˆ(tn )|q, ti

(9.3.18)

where T is the time ordering operator (largest times to the left), and where t 0 > tn > . . . > t1 > t. Proceed as we did with the QM Path Integral. i.e. hq 0 t0 |ˆ q (tn ) . . . qˆ(t1 )hqt| =

Z

qn |qn tn i

qn−1 |qn−1 tn−1 i

z }| { z }| { dqn . . . dq1 hq 0 , t0 | qˆ(tn )|qn tn ihqn tn | qˆ(tn−1 )|qn−1 tn−1 i · q1 |q1 t1 i

z }| { hqn−1 tn−q | . . . qˆ(t1 )|q1 t1 ihq1 t1 |qti

Thus, just as hq 0 t0 |qti =

Z

q(t0 )=q 0 q(t)=q

( Z Dq(t) exp i

(9.3.19)

t0

dτ L(q(τ ˙ ), q(τ )) t

)

(9.3.20)

in this case we obtain hq 0 t0 |T qˆ(t1 ) . . . qˆ(tn )|qti =

Z

q(t0 )=q 0 q(t)=q

( Z Dq(t) q(t1 )q(t2 ) . . . q(tn ) exp i

t0

dτ L(q(τ ˙ ), q(τ )) t

)

(9.3.21)

150


(where t0 > ti > t ∀ i = 1, . . . , n). Add J(τ )q(τ ) to L: hq 0 t0 |T qˆ(t1 ) . . . qˆ(tn )|qti = =

Z

q(t0 )=q 0 q(t)=q

)¯ ¯ ¯ dτ [L(q(τ ˙ ), d(τ )) + J(τ )q(τ )] ¯ ¯ t J=0 ( Z t0 Z q(t0 )=q0 dτ [L(q(τ ˙ ), q(τ )) Dq(t) exp i

( Z Dq(t) q(t1 ) . . . q(tn ) exp i

δ δ ... iδJ(t1 ) iδJ(tn )

Thus, h0, ∞|T q(t1 ) . . . (tn )|0, −∞i =

t

q(t)=q

)¯ ¯ ¯ +J(τ )q(τ )] ¯ ¯

t0

(9.3.22) J=0

δ δ ... lim N hQ0 T 0 |QT i iδJ(t1 ) iδJ(tn ) TT0 →−i∞ →i∞

(9.3.23)

With Q = Q0 = 0, h0, ∞|T q(t1 ) . . . q(tn )|0, −∞i ( Z 0 Z Q(T 0 )=0 h T δ δ ˙ ), Q(τ )) ... N 0lim = dτ L(Q(τ DQ(T ) exp i |{z} iδJ(t1 ) iδJ(tn ) TT→−i∞ T Q(T )=0 →i∞ ∗ ) i (9.3.24) + J(τ )Q(τ ) “ ∗ ” Oscillates? Q˙ 2 →L= − V (Q) = 2 ∗ →

1 2 Z

= i

µ

d Q dτ

ic∞

Z0

¶2

− V (Q)

dx eix ix = −y +∞

dy d−y

(well defined)

(9.3.25)

0

Now let τ = −iτE , (E = Euclidean). So, we get h0, ∞|T q(t1 ) . . . q(tn )|0, −∞i µ ¶ δ δ 1 ... N· = n i δJ(t1 ) δJ(tn ) ( Z Z q(∞)=0 · lim Dq(t) exp + 0 T →∞ E TE →−∞

q(−∞)=0

0 TE

TE

dτE

"

Q˙ 2 − − V (Q) + J(τ )Q(τ ) 2

#) ¯ ¯ ¯ ¯(9.3.26) ¯ J(τ )=0

9.3. FREE PARTICLE

151

The same procedure can be applied in QFT. Here,

δ δ 1 ... N h0|T φ(x1 ) . . . φ(xn )|0i = n i δJ(x1 ) δJ(xn )

Z

Dφ(x) e

R

d4 x

E

(where φ(x) → 0 as x → ±∞.) If

LM (∂µ φ, φ) =



¯ ¯ [LE (∂µ φ,φ)+J(x)φ(x)] ¯ ¯ ¯ J=0 (9.3.27)



λ 1 (∂0 φ)2 − (∇φ)2 −m2 φ2  − φ4 {z } 2 | 4!

(9.3.28)

A

where A = (∂µ φ)(∂ µ φ) → (+, −, −, −) (Bj. and Drell), or = −(∂µ φ)(∂ µ φ) → (−, +, +, +). (and LM is the Minkowski lagrangian). And,

LE =

¢ λ 1¡ −(∂4 φ)2 − (∇φ)2 − m2 φ2 − φ4 2 4!

(9.3.29)

4 2 4 2 2 with |{z} x0 = −ix | {z } → xE = (x ) + (x) . (M )

(E)

Let

ZE (J) = N

Z

½ ¸¾ · 1 E 2 µ2 φ2 λφ4 4 Dφ exp d xE − (∂µ φ) − − + J(x)φ(x) 2 2 4!

(9.3.30)

(Generating Functional) - (call the mass “µ” now). So,

¯ ¯ 1 δ δ ¯ ˆ ˆ h0|T φ(x1 ) . . . φ(xn )|0i = n ... ZE (J)¯ ¯ i δJ(x1 ) δJ(xn )

J=0

Jan. 21/2000

(9.3.31)

152


Consider, ¯ Z ∞ ¯ 2 2 4 −a x +jx−λx ¯ dx e ¯ ¯ −∞

=

j=0

=

Z

∞

∞ X n=0

=

dx

−∞

∞ X n=0

∞ X (−λx4 )n n=0

(−λ) n!

n

µ

d dj

(−λ)n d4n n! dj 4n

−a2 x2 +jx

e n! ¶4n Z ∞

dx e

−∞

Z

∞ −∞

¯ ¯ ¯ ¯ ¯

j=0

−a2 x2 +jx

¯ ¯ ¯ ¯ ¯

j=0

(

dx exp −a2

µ

j x− 2 2a

¶2

+

µ

j2 4a2

j → Let x0 = x − 2 2a ¯ Z ∞ ¯ ∞ 4n n X (−λ) d 2 02 2 2 0 −a x j /(4a ) ¯ dx e = e ¯ ¯ n! dj 4n −∞ n=0 j=0 | {z } √

¶) ¯¯ ¯ ¯ ¯

j=0

π a

Note:

¯ √ X ∞ π (−λ)n d4n j 2 /(4a2 ) ¯¯ = e ¯ ¯ a n=0 n! dj 4n j=0   ¯ ¯ √ 2 8 ¯ 4 ¯ π −λ d ¯ (−λ) d ¯ 2 2 = 1+ ¯ + ¯ + . . . ej /(4a ) 4 8 a 1! dj ¯ 2! dj ¯ j=0 j=0 µ ¶ ¸ √ · π −λ 3 1+ + ... = a 1! 4a2 2j 2 2 d 2 2 → ej /(4a ) = 2 ej /(4a ) dj 4a . . . and so

° " ¶2 # µ ° d2 2 2j 2 2 2 2 ° ej /(4a ) = + ej /(4a ) ° 2 2 2 ° dj 4a 4a ¯ 8 4 3 d4 j 2 /(4a2 ) ¯¯ e = + = 4 ¯ 4 4 4 ¯ dj 16a 16a 4a

(9.3.32)

j=0

Thus, Z ∞

 ¯ ¯ (−λ) d ¯ π −λ 2 2 4  ej 2 /(4a2 ) (9.3.33) + + . . . dx e−a x +jx−λx = 1+ ¯ a 1! 2! dj 8 ¯ −∞ j=0 j=0 ¡1¢ with the (−λ) representing a dot (vertex), and a2 being the loops. (see figure 9.3.1) For · ½Z ¸¾ Z 1 µ µ2 2 λφ4 4 d x − (∂ φ) (∂µ φ) − φ − Z(J) = Dφ exp + Jφ 2 2 4! ¶k µZ ¶k Z ¸¾ ½ · ∞ µ X ¢ λ δ4 1 ¡ 2 2 4 4 − = φ ∂ − µ φ + Jφ(9.3.34) dx Dφ exp d x 4 (x) 4! δJ 2 k=0 √



µ

¶

¯ d ¯¯ ¯ dj 4 ¯ 4

2

8

9.3. FREE PARTICLE

153

Figure 9.3.1: Simple Feynman Diagram example

But now, · ¸ Z ¢ 1 ¡ 2 4 2 → dx φ ∂ − µ φ + Jφ (Now, complete square) 2 µ µ µ ¶¶ ¶ ¶ ¶ ¸ µ · µ Z ¢ ¡ 2 1 1 1 1 1 2 4 φ+J φ+ J − J J ∂ −µ = dx 2 ∂ 2 − µ2 ∂ 2 − µ2 2 ∂ 2 − µ2 (9.3.35) ¶ µ 1 J Let φ0 = φ + 2 ∂ − µ2 So, Z(J) =

¶ µZ ∞ µ X −λ k=0

Here,

1 ∂ 2 −µ2

4!

δ4 dx δJ 4 (x) 4

¶k Z

0

Dφ (x) e

R

d4 x

h

1 0 φ 2

(∂ 2 −µ2 )φ0 − 12 J

Also, if,

1 ∂ 2 −µ2

´ i J

(9.3.36)

1 ⇒ (∂ 2 − µ2 ) (∂ 2 −µ 2 ) = 1. So,

1 = G(x, y) → (∂x2 − µ2 )G(x, y) = δ 4 (x − y) 2 2 ∂ −µ Thus,

³

1 J 2 J= ∂ − µ2

Z

G(x, y) =

(9.3.37)

d4 x d4 y J(x)G(x, y)J(y)

(9.3.38)

Z

(9.3.39)

d4 k g(k)eik·(x−y) (2π)4

Then using (9.3.37) , Z Z ¡ 2 ¢ d4 k ik·(x−y) d4 k ik·(x−y) 2 e g(k) = e ∂ −µ (2π)4 (2π)4 | {z } δ(x−y)

2

2

(−k − µ )g(k) = 1 Z → G(x, y) =

d4 k eik·(x−y) (2π)4 (−k 2 − µ2 )

(Non-singular integral) (9.3.40)

154


Note that this provides meaning of the last term in the exponent of Z(J). (also, k 2 = k12 + k22 + k32 + k42 ). For the other part of the exponent, Z i R 4 h 1 0 2 d x − φ (∂ −µ2 )φ0 2 (9.3.41) Dφ e Consider, Z dn x e−xMx

MT = M (M∗ = M)  λ1  −1 ... → O MO = D =  ;

O

−1

=O

T

 λn

  (λi = eigenvalues of M)

Let y = O−1 x ¡ ¢ dn y = det O−1 dn x = (1)dn x

(9.3.42) (9.3.43)

(det(O) = Product of eigenvalues of O). So, Z Z n −xT Mx d xe ⇒ dn y e−yDy Z ∞ Z ∞ Z ∞ 2 −λ1 y 21 −λ2 y 22 = dy 1 e dy 2 e ... dy n e−λn yn −∞ r−∞ r r −∞ π π π = ... λ1 λ2 λn n/2 π = det1/2 (M) So,

Z

and so,

Dφ0 ed

4x

[− 21 φ0 (∂ 2 −µ2 )φ0 ] = N det1/2 (∂ 2 − µ2 )

(9.3.45)

R

(9.3.46)

1

Z0 (J) = N det1/2 (∂ 2 − µ2 )e− 2 Thus, Z(J) =

µZ ∞ X (−λ)n n=0

n!

δ4 dw δJ 4 (w) 4

(9.3.44)

¶n

d4 x d4 y J(x)G(x,y)J(y)

1

N det−1/2 (∂ 2 − µ2 )e+ 2

R

˜ d4 x d4 y J(x)G(x,y)J(y)

(9.3.47)

9.3. FREE PARTICLE ˜ y) = where G(x, Jan. 24/2000 Recall;

R

155

d4 k eik·(x−y) (2π)4 k2 +µ2

Z[J] =

Z

= −G(x, y).

Dφ exp

Z0 [J] = N e(1/2)

R

½Z

·

1 1 λφ4 d x (∂µ φ)2 − µ2 φ2 − + Jφ 2 2 4! 4

d4 xd4 yJ(x)G(x,y)J(y)

ˆ 1 ) . . . φ(x ˆ n )|0i (i)n h0|T φ(x ¯ ¯ δ δ ¯ ... Z[J]¯ = ¯ δJ(x1 ) δJ(xn )

(λ = 0)

J=0

µ ¶j µZ ¶j ∞ R X λ δ4 δ 1 δ (1/2) d4 xd4 y 4 − N e dz ... = δJ(x1 ) δJ(xn ) j=0 (j!) 4! δJ 4 (z)

For example, say n = 4 (and let

δ δJ1

=

δ , δJ(x1 )

¸¾

(9.3.48) (9.3.49)

¯ ¯ J(x)J(y)G(x,y) ¯ ¯ ¯

J=0

(9.3.50)

etc.),

ˆ 1 ) . . . φ(x ˆ 4 )|0i (i)n h0|T φ(x ( ) µ ¶Z µ ¶2 Z 4 Z 4 4 δ δ δ δ λ 1 λ δ = d4 z1 4 d4 z 4 + ... 1+ − − dz2 4 + . . . · δJ1 δJ4 4! δJz 2! 4! δJz1 δJz2 ¯ ¾¯ ½ Z 1 ¯ 4 4 d xd yJ(x)J(y)G(x, y) ¯ · exp ¯ 2 J=0

= [G(x1 − x2 )G(x3 − x4 ) + G(x1 − x3 )G(x2 − x4 ) + G(x1 − x4 )G(x2 − x3 )](−λ)0 | {z } ∗ · ¸ Z 4 + (−λ) G(0) [ ∗ ] + d z G(x1 − z)G(x2 − z)G(x3 − z)G(x4 − z) + . . . + . . .

156


or, diagrammatically, this is



  =   

     



 +(−λ)  +

+

+

+

+

+

+

+ ... +

+

+ 



 + (−λ)2   +

+ 

 + . . . + . . .

(9.3.51)

where,

Ã G(x, y)

(9.3.52)

Ã (−λ)

9.4. FEYNMAN RULES

9.4

157

Feynman Rules

= −λ

(9.4.1)

= G(x − y)

(9.4.2)

• Integrate over all internal points z • Take into account symmetry factors and number of diagrams of a given topology R

4

4

Z0 [J] = N e(1/2) d x d y J(x)G(x−y)J(y) ¶j X 1 µ −λ Z δ 4 4 Z[J] = d z Z0 [J] 4 j! 4! δJ z j

(9.4.3) (9.4.4)

h i δ δ It can be shown (c.f. Cheng and Li) . . . Just as δJ(x · · · Z[J] gives rise to all N δJ(xn ) 1) point Feynman diagrams, so also for the connected diagrams, ˆ 1 ) . . . φ(x ˆ n )|0iconnected = δ . . . δ W [J] (i)n h0|T φ(x δJ1 δJn

(9.4.5)

W [J] = ln (Z[J])

(9.4.6)

where For example, in n = 4, ˆ 1 ) . . . φ(x ˆ 4 )|0iconnected (i)4 h0|T φ(x   

 =  

+

  +   +

+

+ 

 + . . .  + ...

(9.4.7)

158


(i.e. we don’t get diagrams with separate “disconnected” parts) ⇒ (*) and disconnected diagrams can be shown to contribute only to the phase of the Green’s Function (G.F.). We thus only need the connected G.F. to calculate physical cross sections. Futhermore, if δW [J] (9.4.8) Φ(x) = δJ(x) and Γ[Φ] = W [J] − then

Z

d4 x J(x)Φ(x)

δ δ ... Γ[φ] δΦ(x1 ) δΦ(xn )

(9.4.9)

(9.4.10)

gives rise to the one-particle irreducible (1PI) connected Green’s Functions. → i.e. can’t cut diagrams into 2 parts by cutting a single internal line. For example, → Not 1PI →

is 1PI

→ 1PI

→

Not 1PI

This is useful in calculations because we can compute each 1PI part and connect the two to get non-1PI parts. For example, = (−λ)

2

Z

d4 z1 d4 z2 G(x1 − z1 )G(x2 − z2 )G3 (z1 − z2 )

(9.4.11)

→ # = 1 (one such diagram). Symmetry factor (# of ways we can connect these lines) S = 16 . (Recall, for the symmetry factor: for the numerator, x1 can go to 8 points (four points per vertex), then x2 can go to 4 points (four points on the second vertex), and then one of the remaining three lines on one of the vertices can go to 3 possible lines on the other vertex, then one of the remaining two lines can go to 2 possible lines on the other, and the

9.4. FEYNMAN RULES

159

Figure 9.4.1: Momentum Space diagram - note momentum conserved at each vertex

final line can go to the 1 line on the other vertex - for the denominator, there are two 4! factors (four points per vertex, two vertices), and we can switch the vertices, so there is also a 2!; thus, 8×4×3×2×1 4! 4! 2! 1 = 6

S =

φ4 per vertex is accounted for in the symmetry factor, as is the Note that the 4! in −λ 4! would arise from the sum (see (9.4.4))). Convert to p-space, G(x − y) =

In momentum space

µ

for

ˆ 1 )φ(x ˆ 2 )|0i = 1 (−λ)2 h0|T φ(x 6 µ · (see figure 9.4.1)

¶

Z

d4 k eik·(x−y) (2π)4 k 2 + µ2

1 2!

that

(9.4.12)

the number of integrals ≡ number of loops.

· µ ¶ Z 1 1 d4 p ip·(x1 −x2 ) 4 4 d k1 d k2 · e (2π)4 (p2 + m2 )2 k12 + m2 ¶ ¸ 1 1 (9.4.13) k22 + m2 (k1 + k2 + p)2 + m2 Z

160

CHAPTER 9. PATH INTEGRAL QUANTIZATION In Minkowski space, we have the same diagrams, with,

= −iλ

(9.4.14)

= i∆(x − y) =

Note: k 2 = k02 − k 2 ,

∴ pole at k0 = ±

p

eik·(x−y) d4 k (2π)4 k 2 − µ2 + iε

(9.4.15)

k 2 + µ2 .

ˆ 1 ) . . . φ(x ˆ n )|0iMinkowski = h0|T φ(x

9.5

Z

δ δ ... Zm [J] iδJ(x1 ) iδJ(xn )

(9.4.16)

Path Integrals for Fermion Fields ¯ 6 ∂ − m)ψ L = ψ(i

(9.5.1)

¯ = i~δ( ). In the path integral, ψ and ψ¯ are Grassmann Canonical Quantization → {ψ, ψ} Variables. i.e. ¾ ½ Z Z £ ¤ 4 ¯ ¯ (9.5.2) Z(η, η¯) = DψD ψ exp i d x L + η¯ψ + ψη

It’s important that ψ, ψ¯ are treated independently. (Note, though, if ψ = ψc = C¯ ψ¯T (majorana), then ψ, ψ¯ are no longer independent).

9.6

Integration over Grassmann Variables

Recall, θ1 θ2 = −θ2 θ1 (θi2 = 0) ¶ µ d dθm dθn (θm θn ) = θn + θ m − dθa dθa dθa = δma θn − δna θm

(9.6.1)

(9.6.2)

If we have θ1 . . . θn , then p(x, θ1 , θ2 , . . . , θn ) = p0 (x) + pi (x)θi + pij θi θj + . . . + pi1 ,...,in (x)θi , . . . , θin

(9.6.3)

9.6. INTEGRATION OVER GRASSMANN VARIABLES

161

(note no (n + 1) term due to θi2 = 0) where pij = −pji , pijk = −pjik , etc.. We have,

→

Z

p(x, θ) = p0 + p1 θ Z dθ p(x, θ) = dθ [p0 + p1 θ] Z = dθ [p(x, θ + α)] (α → a displacement)

(9.6.4)

For this to hold, Z

Z Thus, (9.6.4) is =

Z

dθ θ = 1

(9.6.5)

(1)dθ = 0

(9.6.6)

dθ [p0 + p1 · (θ + α)]

(9.6.7)

R d (i.e. are the same operator). Also, note that the only term that will survive Thus, dθ ⇔ dθ in the integral will be the one with n θ’s (all other terms won’t have enough θ’s). So, Z Z dθ1 . . . dθn [p(x, θ1 , . . . , θn )] = dθn . . . dθ1 [p0 + pi θi + pij θi θj + . . . + pi1 ,...in θi1 . . . θin ] Z = dθn . . . dθ1 pi1 ...in θi1 . . . θin = ²i1 ...in pi1 ...in | {z } n! terms

= n!p123...n

(9.6.8)

(the ²i1 ...in is present because order of θ’s may not be i1 , i2 , . . .). If we now change variables of integration, θ˜i = aij θj (9.6.9) then, Z

dθi p(x, θi ) = = = =

Z

Z

Z

Z

dθ˜i p(x, θ˜i ) (expect) dθ˜i pi1 ...in θ˜i1 . . . θ˜in dθ˜i (pi1 ...in )ai1 j1 . . . ain jn θj1 . . . θjn

(9.6.10)

dθ1 . . . dθn (pi1 ...in )θi1 . . . θin

(9.6.11)

162


and (9.6.10) must be equal to (9.6.11). Thus, Z Z dθ1 . . . dθn p1...n θ1 . . . θn = dθ˜1 . . . dθñ ²i1 ...in ai1 j1 . . . ain jn p1...n θ1 . . . θn

(9.6.12)

Hence dθ˜1 . . . dθñ = dθ1 . . . dθn ²i1 ...in ai1 j1 . . . ain jn {z } | =det(a)

i.e.

= det(a)dθ1 . . . dθn # " ˜1 . . . θñ ) ∂( θ dn θ dn θ˜ = det ∂(θ1 . . . θn ) {z } |

(9.6.13)

∗

where “*” is the reciprocal of the normal jacobian; recall · ¸ Z Z ∂(x1 . . . xn ) n n d x f (x) = d x˜ det f (˜ x) ∂(˜ x1 . . . xñ ) · ¸ ∂(x1 . . . xn ) n n i.e. d x = det d x˜ ∂(˜ x1 . . . xñ )

(9.6.14) (9.6.15)

Also note, Z

¯ dθ1 . . . dθn dθ¯1 . . . dθ¯n eθi Cij θj (expand exponential) ¸ · Z 1 ¯ n ¯ ¯ (θi Cij θj ) = dθ1 . . . dθn dθ1 . . . dθn n! 1 = ²i ...i ²j ...j Ci j . . . Cin jn n! 1 n 1 n 1 1 1 (n! det(C)) = n! = det(C)

³

(9.6.16)

Recall, Z

dθ1 . . . dθn eθi Cij θj

= but

det1/2 (C) Z ∞ dx1 . . . dxn d¯ x1 . . . d¯ xn e−¯xi Cij xj −∞

If λi > 0 for Cij then √ πn πn ´ 0 p → x¯/x s distinct, ∴ not = detC det(C)

9.6. INTEGRATION OVER GRASSMANN VARIABLES

163

For n = 2, Z

½ · ¸· ¸¾ · · ¸· ¸¸ Z 0 γ θ1 0 γ θ1 dθ1 dθ2 exp [θ1 , θ2 ] = dθ1 dθ2 1 + [θ1 , θ2 ] −γ 0 θ2 −γ 0 θ2 Z = dθ1 dθ2 [θ1 γθ2 − θ2 γθ1 ] = γ

= det

(1/2)

½·

0 γ −γ 0

¸¾

(9.6.17)

Hence, for, Z(η, η¯) = =

= =

¾ ½ Z £ ¤ 4 ¯ p − m)ψ + η¯ψ + ψη ¯ DψD ψ¯ exp i d x ψ(6 ½ Z ·µ µ ¶¶ µµ ¶ ¶ Z 1 1 4 DψD ψ¯ exp i d x ψ¯ + η¯ (6 p − m) η+ψ 6p − m 6p − m ¶ ¸¾ µ 1 η ( recall p ≡ i∂) −¯ η 6p − m ½ Z · ¶ ¸¾ µ Z 1 4 ¯ ¯ DψD ψ exp i d x ψ(6 p − m)ψ + η¯ η (ψ = ψ(xµ )) 6p − m ½ Z · µ ¶ ¸¾ 1 4 η (9.6.18) det(6 p − m) exp i d x η¯ | {z } 6p − m Z

normalization

³

Recall: 1 exp Z[J] = det(1/2) (p2 − m2 )

½ Z µ ¶ ¾´ i 1 dx J J 2 p2 − m 2

Only fallout from having to take into account that η¯, η are Grassmann. i Propagator 6 p − m + iε

(9.6.20)

→ Closed loop of fermions → factor of (−1)

(9.6.21) (9.6.22)

Jan. 27/2000

164

9.7


Gauge Invariance

(1st Principle) U (1) ¶ 1 2 2 S = d x − φ(∂ + µ )φ 2 Z ¡ ¢ 1 (∂µ φ)2 − µ2 φ2 ( 1 degree of freedom φ) = d4 x 2 Z

4

µ

If φ → complex, Z £ ¤ S = d4 x (∂µ φ∗ )(∂µ φ) − µ2 φ∗ φ ( 2 degrees of freedom φ∗ , φ)

(9.7.1)

(9.7.2)

We demand

φ → eiΛ(x)φ

(9.7.3)

L = (∂µ φ∗ )(∂ µ φ) − m2 φ∗ φ

(9.7.4)

be an invariance. Invariant under a global transformation, φ → eiΛ φ (Λ → const.)

(9.7.5)

If Λ = Λ(x), 1 L = − (∂µ Aν − ∂ν Aµ )2 + [(∂µ + ieAµ ) φ∗ ] [(∂ µ − ieAµ ) φ] − m2(9.7.6) φ∗ φ 4 where 1 A µ → A µ + ∂µ Λ (9.7.7) e Dµ = ∂ − ieAµ → covariant derivative (9.7.8) Dµ φ → U D µ φ (9.7.9) (Dµ Dν − Dν Dµ )φ ⇒ U (Dµ Dν − Dν Dµ )φ = −U (ie)Fµν φ ; Fµν = ∂µ Aν − ∂ν Aµ (9.7.10) Advantage of looking at E.M. this way → can generalize, · ¸ p = N → Heisenberg postulate n N = Nαi

where α is the Dirac index, and i = 1, 2 (1 ≡ ↑, 2 ≡ (forgetting Dirac index); N i → U ij (Λ)N j

(9.7.11) (9.7.12)

↓. The invariance is of the form (9.7.13)

9.7. GAUGE INVARIANCE

165

where and 1 τ= 2

½·

¡ ¢ij U ij (Λ) = eiτ ·Λ/2 0 1 1 0

¸

,

·

0 −i i 0

¸

,

(9.7.14) ·

1 0 0 −1

¸¾

(9.7.15)

¯ i (i 6 ∂δ ij − mδ ij ) N j is globally SU (2) invariant. Suppose Λ = Λ(xµ ). Add in Aij N µ = a a ij Aµ (τ ) . So, ¶ ¸ · µ i µ ¯ 6 −m N ; A 6 = Aaµ (τ a )ij γαβ → N i 6∂ − g A 2 · µ ¸ ¶ i µ µ µ ij i ij a a ij ¯α i ∂ γ δ − gAµ (τ ) γ = N − mδ δαβ Nβj αβ αβ 2

(9.7.16)

¯ →N ¯ U −1 , then, If N → U N , N µ

· ¶ ¸ ig ig −1 ¯i 6∂ − A ¯ iU N 6∂ − A 6 N → N 6 UN 2 2 ¸ · ig −1 µ −1 ¯ γ i U ∂µ − U A µ U N = N 2   ∂(U N ) }| { ig z ¯ U −1 γ µ (∂µ U ) + (U ∂) − Aµ U  N = iN 2 ¸ · ig 0 µ ¯ = iN [γ ] ∂µ − Aµ N 2

(9.7.17)

(Notation: note that (∂µ A) = A,µ i.e. “,” ≡ derivative wrt. µ.) where, igA0µ ig −1 U Aµ U = − 2 2 ig ig U −1 U,µ − U −1 Aµ U = − A0µ 2 2

U −1 (∂µ U ) −

(9.7.18)

For Λa ≈ 0, U ij ≈ δ ij + iΛa (τ a )ij U −1 = 1 − iΛ · τ ¡ ¡ ¢ ig ¢ ig − A0µ · τ = (1 − iτ · Λ) iτ · Λ,µ − (1 − iτ · Λ) Aµ · τ (1 + iτ · Λ) 2 2 ¤ ig £ −iτ · Λτ · Aµ + iτ · Aµ τ · Λ = iτ · Λ,µ − 2

(9.7.19) (9.7.20)

(9.7.21)

166


But, ¡ ¢ τ · Λ τ · A − τ · A τ · Λ = Λa Abµ τ a τ b − τ b τ a 1 = i²abc Λa Abµ τ c (τ = a) 2 = iΛ × A · τ So,

Hence,

¤ ig £ ig − A0µ · τ = iτ · Λ,µ − −i(2i)(Λ × Aµ ) · τ 2 2 ³g ´ ¡ ¢a ig − A0µ = iΛa,µ + i Aµ × Λ 2 2 2 ∴ A0a = − Λa,µ − ²abc Abµ Λc µ g

(9.7.22)

(9.7.23)

(9.7.24)

In other words, we have 3 photon fields, 1 for each spin matrix. (Aµ → Aµ + 1e ∂µ Λ). Consider three scalars φa (a = 1, 2, 3). L=

¤ 1£ (∂µ φa )(∂ µ φa ) − m2 φa φa 2

For an invariance under φa → φa + g²abc φb Λc (x), for Λa ≈ 0, then, ¢ ¤ ¢ ¤ £¡ £¡ L = ∂µ δ ab + g²acb Acµ φb ∂µ δ ap + g²aqp Aqµ φp − m2 φa φa

(9.7.25)

(9.7.26)

with

a abc b c a Aµ Λ A0a µ = Aµ + ∂µ Λ + g²

(9.7.27)

(again, SU (2), but SU (2) rep. that is isomorphic to SU (3)). Define: Dµab = ∂µ δ ab + g²amb Am µ

(9.7.28)

(Analogous to Dµ = ∂µ − ieAµ → do the same thing Dµ Dν − Dν Dµ = ieFµν ), L = (Dam φm )(Dan φn ) − m2 φa φa = Aaµ + Dµab Λb A0a µ

(9.7.29) (9.7.30)

φ0a = φa + g²abc φb Λc

(9.7.31)

Dµab (A)φ → (δ ab + g²abc Λc )(Dbm φm )

(9.7.32)

Now, under the gauge transformations. Consider,

→ £ ¤ (Dµab Dνbc − Dνab Dµbc )φ (9.7.32) (δ ab + g²abc Λc ) (Dµbm Dνmn − Dνbm Dµmn )φn

(9.7.33)

9.7. GAUGE INVARIANCE

167

But now, Dµab Dνbc − Dνab Dµbc = (∂µ δ ab + g²apb Apµ )(∂ν δ bc + g²bqc Aqν )

− (∂ν δ ab + g²apb Apν )(∂µ δ bc + g²bqc Aqµ ) ¡ ¢ n = g²abc (∂µ Abν ) − (∂ν Abµ ) + g²bmn Am µ Aν

(c.f. ²abc ²bmn = −δ am δ cn + δ an δ cm ) b = g²abc Fµν

Lagrangian for Aaµ

(Gauge invariant). Jan. 31/2000

(9.7.34)

Fµν → Fµν (U (1) case) a b Fµν → (δ ab + ²abc Λc )Fµν

(9.7.35) (9.7.36)

1 a aµν F L = − Fµν 4

(9.7.37)

168


Chapter 10 Quantizing Gauge Theories We could look at canonical quantization → Aaµ (x); Define: π aµ =

δL δ(∂µ Aaµ )

(10.0.1)

• Identify constraints • insert gauge conditions for each of the 1st class constraints • Form Dirac Brackets • From these, determine commutators

Problems with this in practice are;

1. Manifest Lorentz invariance is lost (Important) 2. For the Coulomb Gauge (for example), where ∂i Aai (x, t) = 0, in forming the Dirac Brackets → will involve £ ¤−1 ∂i (Dµab ) (10.0.2) where Dµab = ∂i δ ab + g²apb Api (“ * ” part drops out in abelian case). | {z } ∗

10.1

Quantum Mechanical Path Integral

Use the Q.M Path Integral (QMPI). 1 a (10.1.1) (A)F aµν (A) + Jµa Aaµ L = − Fµν 4 where for now, we will eliminate the second term Jµa Aaµ → purely classical L. Note that we have full gauge invariance Aaµ → Aaµ + ∂µ Λa + g²abc Abµ Λc . Consider, ½ Z ¾ Z ¡ ¢ a 4 a aµ Z[J] = DAµ exp i d x L + Jµ A (10.1.2) 169

170

CHAPTER 10. QUANTIZING GAUGE THEORIES L ( Z " z }|0 ³ ¡ ¢{ ¡ ¢ 1 a 4 a a 2 Z[J] = DAµ exp i d x − ∂µ Aν − ∂ν Aµ +2g ∂µ Aaν − ∂ν Aaµ ²abc Abµ Acν 4 #) ´ a aµ n (10.1.3) +g 2 ²amn Apµ Aqν Am µ Aν + Jµ A

Z

but, L0 = 12 Aaµ (∂ 2 g µν − ∂ µ ∂ ν ) Aaν , and, ¸¾ ½ Z · Z £ ¤ 1 a 2 µν µ ν a a aµ a 4 ∼ det(−1/2) δ ab (g µν ∂ 2 − ∂ µ ∂ ν ) DAµ exp i d x Aµ (∂ g − ∂ ∂ )Aν + Jµ A 2 (10.1.4) and, ¡ 2 µν ¢ ∂ g − ∂ µ ∂ ν (∂ν Λ) = 0 (10.1.5) → ∂ ν Λ is an eigenvector with vanishing eigenvalue! Hence, £ ¤ det ∂ 2 g µν − ∂ µ ∂ ν = 0

(10.1.6)

(Dirac put in − 12 [∂µ Aaµ ]2 into (10.1.2), (in exponential integral) → eliminates ∂ µ ∂ ν → but he found inconsistencies). To see the Faddeev-Popov procedure, consider, Z π n/2 n −xT N x d xe (N ) = (10.1.7) det1/2 ) → det(N ) = λ1 . . . λn R Only if λi > 0∀i (10.1.8) 2 → dx1 . . . dxn e−x1 λ1 . . . e−x2n λn Suppose,

N m(x)λ0 = 0 m(x)λ0 → eigenvector corresponding to eigenvalue zero. Note that, Z dn y δ(Ay + b)f (y) Let z = Ay + b, so dn y = Z

1 dn z. detA

(10.1.9)

(10.1.10)

So,

Z 1 d y δ(Ay + b)f (y) = dn z δ n (z)f (A−1 (z − b)) detA f (−b) = det(A) n

Thus, δ(Ay + b) =

δ(y + A−1 b) det(A)

(10.1.11)

(10.1.12)

10.1. QUANTUM MECHANICAL PATH INTEGRAL (c.f. 1-D → δ(ax) =

Insert this into Z T dn x e−x N

171

1 δ(x)). |a|

Consider Z 1 = dλ0 δ [L(x + m(x)λ0 )] det(L m(x))

x

=

Z

n

d x

Z

(10.1.13)

TN

dλ0 δ [L(x − m(x)λ0 )] det [L m(x)] e−x

Let x → x − m(x)λ0 . Then, Z Z Z T −xT N x n n = d x dλ0 δ(L x) det [L m(x)] e−x N d xe

x

(10.1.14)

x

T

(note no change in e−x N x because N (m λ0 ) = 0) Z Z T = dλ0 dn x δ(L x) det (L m(x)) e−x N x (10.1.15)

R (The dλ0 contains the ∞ occurring in the integral over the eigenvector with vanishing eigenvalue). Note: R T 1. det [L m(x)] = dcd¯c e¯c (L m(x))c (where c, ¯c → Grassmann vectors). p 2 2. δ(L x) = limα→0 απ e−α(L x)

So,

Z

n

d xe

−xT N x

= =

Z

Z

n

r

d x lim α→0 r π dλ0 lim α→0 α | {z }

dλ0 Z

Z 2 π −xT N x T e dc d¯c e¯c L m(x)c−α(L x) α Z T T T ] dn x dc d¯c e[−x (N +αc L)x+¯c L m(x)c(10.1.16)

absorb into Normalizing factor

• Choose L so that (N + αLT L) is invertible (i.e. (N + αLT L) has no vanishing eigenvalue). • ¯cT L m(x)c cancels the contribution of eigenvectors with vanishing eigenvalues (“ghost fields”). Feb. 2/2000 The condition is N (m(x)λ0 ) = 0 Z 1 = dλ0 δ(L(x − m(x)λ0 ))det(L m(x))

x → x − m(x)λ0 Z Z T = dλ0 dn x δ(L x) det(L m(x)) e−x N | {z } | {z } A

B

x

172

CHAPTER 10. QUANTIZING GAUGE THEORIES q π −(L x)2 /α lim e α α→∞ Z T B → dcd¯ c e−¯c L m(x)c A →

10.2

Gauge Theory Quantization

Recall the path integral Z

If J = 0,

DAaµ (x)e

R

d4 x

µ

−

a (x))2 (Fµν +Jµa Aaµ 4

¶

Aaµ → Aaµ + Dµab (A)Ωb (x)

(Analogy x ↔

With Dµab (A) =

Aaµ (x) from above ∂µ δ ab + g²apb Abµ

(10.2.1)

(10.2.2) ¢

Insert a factor of “1”. 1 =

Z

=

Z

dΩa (x) δ( ∂µ (Aaµ (x) + Dµab (A)Ωb ))det(∂µ Dµab (A)) |{z} | {z } | {z } | {z } L

x

m(x)λ0

(10.2.3)

Lm(x)

µ ab a dΩa (x) δ(∂ µ AΩ µ (x))det(∂ Dµ (A))

Where, −1 a (Ω)(Aaµ + ∂µ )U (Ω) ; → AΩ µ (x) = U

≈ Aaµ + Dµab (A)Ωb

U (Ω) ≈ δ ab + ²apb Ωp

a a Let AΩ µ → Aµ (reverse gauge transf. → −Ω) Z Z R 4 1 a 2 a dAaµ δ(∂ µ Aaµ )det(∂ µ Dµab (A)ei d x (− 4 (Fµν ) ) 1 = dΩ r Z ½ Z · ¸¾ Z Z π 1 a aµν 1 a a a a 4 aµ 2 a µ ab b = lim dΩ dAµ dc d¯ c exp i d x − Fµν F − (∂µ A ) − c¯ ∂ Dµ (A)c α→0 2α 4 2α {z } | Absorb into Normalization

(10.2.4)

Recall that the c’s are grassmann. Also, note that, in the last term, Dµab (A) = ∂µ δ ab + g²apb Apµ , and so ghost fields are important (ghosts are free fields) → Mathematical constructs, Longitudinal mode of vector particle. Note that in the U (1) case (gauge group), the Ghost Lagrangian is c¯∂µ ∂ µ c = c¯∂ ν c.

10.3. FEYNMAN RULES

10.3

173

Feynman Rules

Here are the Feynman Rules.

−iδ

ab

·

k k

(gµν − µ 2 ν ) k k2

+

αkµ kν k4

¸

Where the last term is the α dependance - (longitudinal part of propagator).

ab

− iδk2

g²abc pµ

−g²abc {(p − q)ν gλµ + (q − r)λ gµν + (r − p)µ gνλ }

174

CHAPTER 10. QUANTIZING GAUGE THEORIES

−ig 2

©

f abc f cde (gλν gµσ − gλσ gµν )

f abe f bde (gλµ gνσ − gλσ gµν ) ª f ade f cbe (gλν gµσ − gλµ gσν )

Feb. 3/2000

10.4

Radiative Corrections L=

1 m2 2 λφ4 (∂µ φ)2 − φ − 2 2 4!

(10.4.1)

and recall the Feynman rules are, i p2 −m2

−iλ

Recall that, in general, ¢ ¡ ¢ ¡ Z dn ` i(−1)a−b Γ a + n2 Γ a − b − n2 (m2 )b−a+n/2 (`2 )a ¡ ¢ = (2π)n (`2 − m2 )b (4π)n/2 Γ(b)Γ n2 Z dn ` i(−1) ³ n ´ 2 n/2−1 i → (m ) = Γ 1 − (2π)n (`2 − m2 ) (4π)n/2 | {z 2 } pole at n=4

(10.4.2) (10.4.3)

10.4. RADIATIVE CORRECTIONS

175

Also, in general, Γ(α + β) 1 = α β a b Γ(α)Γ(β)

Z

1

dx 0

xα−1 (1 − x)β−1 [xa + (1 − x)b]α+β

(10.4.4)

The 1PI contribution to the 2-pt. function is

i = 2 + p − m2

µ

i 2 p − m2

¶

1 (−iλ) 2

Z

i dn ` n 2 (2π) (` − m2 )

µ

i 2 p − m2

¶

(10.4.5)

Which can be easily integrated. Another loop integral is (dropping external leg contribution for now);

= = = =

=

Z n n (−iλ)2 d kd ` 1 2n 2 2 2 2 6 (2π) (k − m )(` − m )[(p − k − `)2 − m2 ] Z Z 1 1 (−iλ)2 dn k d n ` 1 dx 2n 2 2 2 2 6 (2π) (` − m ) 0 [(1 − x)(k − m ) + x((p − k − `)2 − m2 )]2 Z Z 1 1 (−iλ)2 dn k d n ` 1 dx 6 (2π)2n (`2 − m2 ) 0 [k 2 − 2xk(p − `) + x(p − `)2 − m2 ]2 Z Z 1 1 dn k d n ` 1 (−iλ)2 dx 2n 2 2 2 6 (2π) (` − m ) 0 [(k − x(p − `)) + x(1 − x)(p − `)2 − m2 ] {z } | k0 Z n 0 n Z 1 d k d ` 1 (−iλ)2 1 dx 02 2n 2 2 6 (2π) (` − m ) 0 [k + x(1 − x)(p − `)2 − m2 ]2

The integral over k 0 can be evaluated; recall (10.4.2) (Note: General integral correct in

176


(10.4.2), but check signs on a’s and b’s). ¢ ¡ Z Z Γ 2 − n2 dn ` (−iλ)2 1 i dx = 6 (2π)n (4π)n/2 (`2 − m2 )(m2 − x(1 − x)(p − `)2 )2−n/2 0 ¢ ¡ Z Z (−iλ)2 1 1 dn ` iΓ 2 − n2 1 = dx (−x(1 − x))−2+n/2 ³ ´2−n/2 n n/2 2 2 6 (2π) (4π) (` − m ) m2 0 2 (` − p) − x(1−x)

Now, recall (10.4.4). (−iλ)2 = 6 Z

Z

1

dx 0

¢ ¢ ¡ ¡ n dn ` iΓ 2 − n2 −2+n/2 Γ 1 + 2 − 2 ¢· ¡ (−x(1 − x)) (2π)n (4π)n/2 Γ(1)Γ 2 − n2

Z

y 1−1 (1 − y)2−n/2−1 ³ í3−n/2 m2 y(`2 − m2 ) + (1 − y) (` − p)2 − x(1−x) ¢ ¡ Z 1 Z 1 Z iΓ 3 − n2 (−x(1 − x))−2+n/2 (1 − y)1−n/2 dn ` dy dx = i3−n/2 h (2π)n m2 (1−y) 0 0 n/2 2 2 2 (4π) ` + y(1 − y)p − m y − x(1−x) Z 1 Z 1 ³ ³ i n´ n n´ i −3+n/2 −2+n/2 1−n/2 = dx dy Γ 3 − (−1) Γ 3 − − (−x(1 − x)) (1 − y) · (4π)n/2 2 (4π)n/2 2 2 0 0 · ¸n/2+n/2−3 m2 (1 − y) 2 2 −y(1 − y)p + m y + x(1 − x) ¶2 ³ Z 1 Z 1 µ n´ i Γ 3− Γ(3 − n)(−1)−5+n (x(1 − x))−2+n/2 (1 − y)1−n/2 · = dx dy n/2 (4π) 2 0 0 · ¸n−3 2 m (1 − y) 2 2 −y(1 − y)p + m y + x(1 − x) dy h

Now, the integrals over x, y are problems. But, if m2 → 0, µ

¶2 Z

1

Z

1

³ dyΓ 3 − 0 0 ¡ ¢n−3 n−3 (−1) y(1 − y)p2 µ ¶2 Z 1 Z 1 ³ i dx dyΓ 3 − = (4π)n/2 0 0 =

i (4π)n/2

dx

n´ Γ(3 − n)(−(−1)n )(x(1 − x))−2+n/2 (1 − y)1−n/2 · 2 n´ Γ(3 − n)(x(1 − x))−2+n/2 y n−3 (1 − y)n/2−2 (p2 )n−3 2

Now, using (10.4.4), we have a = b = 1, so #" ¡ ¢# " ¡ ¢# µ ¶2 " ³ Γ(n − 2)Γ n2 − 1 Γ2 n2 − 1 i n´ ¡ ¢ = (p2 )n−3 Γ 3− Γ(3 − n) (4π)n/2 2 | {z } Γ(n − 2) Γ 3n − 3 2 Pole at n=4


177

The (p2 )n−3 can be expanded as (1 + (n − 3) ln(p2 ) + . . .), and recall, Γ(ε) =

1 − γE + ln(4π) + . . . ε

(10.4.6)

Thus, the 2-loop integral is =

¢ A ¡ + B ln(p2 ) + C + O(ε) 2

(10.4.7)

(ln(p2 ) is singular if p2 = m2 = 0). At one loop,

∝ =

Z

dn ` 1 n 2 (2π) (` − m2 )

A +B 2

(10.4.8)

B is independent of p2 (this is a fluke). Feb. 7/2000 © µν λρ ª ¡ ¢ σ ,σ = −2 g µρ g λν − g µλ g ρν + 2i²µνλρ γ5

(10.4.9)

where σ µν = 2i [γ µ , γ ν ], {γ µ , γ ν } = 2g µν , and γ5 = Let

1 µνλρ ² γµ γν γλ γρ 4!

(10.4.10)

1 µ2 φ 2 λ0 φ 4 L = (∂µ φ0 )2 − 0 0 − 2 2 4!

(10.4.11)

= −iΣ(p)

= =

+ + Sum of all 2pt 1PI diagrams.

+ ...

178


Let

h0|T φ0 (x)φ0 (y)|0iconnected =

i i i + (−iΣ) 2 2 p2 − µ0 + iε p2 − µ0 + iε p2 − µ20 + iε i i i (−iΣ) 2 (−iΣ) 2 + ... + 2 2 2 p − µ0 + iε p − µ0 + iε p − µ20 + iε = a + ar + ar 2 + . . . a = 1 − r2

=

=

i p2 −µ20 +iε

1 − (−iε) p2 −µi 2 +iε 0

i = 2 2 p − µ0 − Σ(p2 ) + iε = i∆(p2 )

(10.4.12)

Eliminate the divergence (at least to one loop order). Follow treatment by Lurié “on-shell” Renormalization scheme → µ2 is the mass of the field φ(x) (p2 = µ2 → “on shell”) Σ(p2 ) = Σ(µ2 ) +(p2 − µ2 ) Σ0 (µ2 ) + Converging terms {z } | {z } | | {z } When p2 =µ2 diverges

where Σ0 (p2 ) =

d Σ(p2 ). dp2

log divergence

(10.4.13)

Σc (p2 )

Thus,

i∆(p2 ) =

2

p − |

i − Σ(µ ) −(p2 − µ2 )Σ0 (µ2 ) − Σc (p2 ) {z } 2

µ20

(10.4.14)

∗

* - let µ2 = µ20 +Σ(µ2 ) (µ2 → finite, µ20 , Σ(µ2 ) diverge if n = 4). This is mass renormalization, eliminates quadratic divergences. (c.f. recall figure 10.4.1, from previous notes from QFTI). Thus,


179

Figure 10.4.1: Recall mass on string example. x0 gets renormalized by mg . x00 = x0 + mg k i∆(p2 ) =

=

' Let Zφ =

i (p2 − µ2 )(1 − Σ0 (µ2 )) − Σc (p2 ) → Pole occurs at p2 = µ2 ³ ´ 1 i 1−Σ0 (µ2 ) 2

Σc (p ) p2 − µ2 − 1−Σ 0 (µ2 ) ³ ´ i 1−Σ10 (µ2 )

to lowest order in λ0

p2 − µ2 − Σc (p2 )

1 ≈ 1 + Σ0 (µ2 ) (log divergent) 0 2 1 − Σ (µ ) iZφ p2 − µ2 − Σc (p2 ) = F.T.{h0|T φ0 (x)φ0 (y)|0iconnected }

(10.4.15)

(10.4.16)

→ i∆(p2 ) =

(10.4.17)

(F.T. = Fourier Transf.). Now let −1/2

φ(x) = Zφ

φ0 (x)

(re-scaling of φ0 field in L). Hence: F.T.h0|T φ(x)φ(y)|0i =

p2

−

µ2

i (Finite) − Σc (p2 )

(10.4.18)

(10.4.19)

What diagrams do diverge? The superficial degree of divergence = D. For any diagram, 1 1 ; (g # ) → coupling const. ∼ (g)# d4 `1 . . . d4 `L 2 ... 2 2 2 `1 − µ `L − µ = 4L − 2I (only if g is dimensionless) (10.4.20)

180


• L = # of loops • I = # of internal lines • V = # of vertices

= h0|T φ1 . . . φ4 |0i

4V = 2I + E (E = # of external legs)

(10.4.21)

(10.4.22)

The 4 is in front of the V because each vertex has 4 lines going out. Check: (figure 10.4.2) In figure 10.4.2, L = 2, I = 3, V = 2. Thus,

Figure 10.4.2: E = 2, I = 3, V = 2 4(2) = 2(3) + 2 8 = 6 + 2 Check

(10.4.23)

Also, L = I − (V − 1) = I −V +1

(10.4.24)

Hence, D = 4L − 2I 4V = 2I + E → 2I = 4V − E →∴ 4I − 8V = −2E ∴ D = 4L − (4V − E) = 4(I − V + 1) − 4V + E = 4I − 8V + 4 + E = −2E + 4 + E D = 4 − E → Depends only on E!! Thus, D = 4 − E as λ is dimensionless.

(10.4.25)


181

• D = 4 if (E = 0) → not possible • D = 3 if (E = 1) → not possible φ → −φ, ∴ L even in φ. There must be a symmetry, ∴ E= 6 1 • D = 2 if (E = 2) → Quadratic divergence + Logarithmic • D = 0 if (E = 4) → Log divergence • D < 0 if (E > 4) → Converges (Higher pt. diagrams are ok).

Now, consider for example a φp theory: Use the following scalar field theory in field theory in four dimensions 1 1 (10.4.26) L = − (∂φ)2 − m2 φ2 − λφp 2 2 (where p ≥ 3 is a positive integer) to explain the difference between theories that are: renormalizable, nonrenormalizable and super-renormalizable. (Simply consider the (superficial) degree of divergence of the Feynman diagrams). The superficial “degree of divergence” is given by D = 4L − 2I

where

(10.4.27)

• L = # of loops • I = # of internal lines • E = # of external lines • V = # of vertices

Each loop momentum k has a volume element d4 k associated with it, and each (scalar) internal line is associated with a propagator, which for large | 6 k| behaves like | 6 k| −2 . Now, each vertex has p lines emerging from it, and each internal line removes two of these, so and the number of loops is given by

E = pV − 2I

(10.4.28)

L=I −V +1

(10.4.29)

(only V − 1 conservation constraints, due to overall momentum conservation). So, subs. (10.4.29) into (10.4.27), D = = = = = The conditions are:

4(I − V + 1) − 2I 4I − 4V + 4 − 2I 2I − 4V + 4 ; Subs. in (10.4.28) pV − E − 4V + 4 4 − E + (p − 4)V

(10.4.30)

182


• If D < 0 → super-renormalizable • if D = 0 → renormalizable • if D > 0 → non-renormalizable So, as can be seen from (10.4.30), if p < 4, then D will become negative very quickly as V increases, so p = 3 theories would be super-renormalizable; for p = 4, we get the usual φ 4 theory, which is renormalizable; and for p > 4, the final term will fast outweigh the first two, D will be positive as more and more vertices are added, and the theory will be (superficially) non-renormalizable. Feb. 9/2000 Recall: iZφ F.T.h0|T φ0 (x)φ0 (y)|0i = 2 (10.4.31) p − µ2 − Σc (p2 ) where

Σc (µ2 ) = 0 φ0 = F.T.h0|T φ(x)φ(y)|0i =

(10.4.32)

1/2 Zφ φ

p2

−

µ2

(10.4.33) i − Σc (p2 )

(10.4.34)

The four-point function:

F.T.h0|T φ0 (x1 ) . . . φ0 (x4 )|0i = = −iλ0 + Γ(t) + Γ(s) + Γ(u) + . . .

1 = (−iλ0 )2 2

Z

dn ` (2π)n

¡ ¢ = Γ (p1 − p3 )2 = Γ(t) (recall the Mandelstem Variables:) • s = (p1 + p2 )2 • t = (p1 − p3 )2

µ

i 2 ` − µ20

¶µ

i (` + p1 − p3 )2 − µ20

(10.4.35)

¶

(10.4.36)


183

• u = (p1 − p4 )2 These variables satisfy: s + t + u = p21 − 2p1 p3 + p23 + p21 + 2p1 p2 + p22 + p21 − 2p1 p4 + p24 But p1 + p1 = p3 + p4 , p2i = µ2 = 6µ2 + 2(−p1 p3 + p1 p2 − p1 p4 ) = 6µ2 + 2[−p1 p3 + p1 p2 − p1 (p1 + p2 − p3 )] = 6µ2 − 2p21 = 4µ2

(10.4.37)

The “truncated” Fourier transform of h0|φ0 (x1 ) . . . φ0 (x4 )|0i means the external legs are removed. Hence, Truncated F.T.h0|φ0 (x1 ) . . . φ0 (x4 )|0i = −iλ0 + Γ(s) + Γ(t) + Γ(u) where Γ(s) ∼

Z

d4 ` ∼ `4

Z

dx → log divergent x

(10.4.38)

(10.4.39)

(c.f. D = 4 − E . . . E = 4 → D = 0). Somehow must get rid of divergences in Γ(s), Γ(t), 2 2 2 Γ(u). Expand Γ(s), Γ(t), Γ(u) about s = 4µ3 , t = 4µ3 , u = 4µ3 . µ

4µ2 3

¶

˜ ˜ + Γ(u) ˜ + Γ(s) + Γ(t) | {z } Finite µ 2¶ 4µ −1 → Now let − iZλ λ0 = −iλ0 + 3Γ 3 −1 ˜ ˜ ˜ = −iZλ λ0 + Γ(s) + Γ(t) + Γ(u) (10.4.40)

Truncated F.T.h0|φ0 (x1 ) . . . φ0 (x4 )|0i = −iλ0 + 3Γ

What we want to compute is h0|T φ(x1 ) . . . φ(x4 )|0i (φ are Renormalized fields)

(10.4.41)

But −1/2

φ(x) = Zφ

φ0 (x)

(10.4.42)

Recall F.T.h0|T φ0 (x)φ0 (y)|0i =

p2

iZφ − − Σc (µ2 ) µ2

(10.4.43)

184


Now: G0 (x1 , . . . , x4 ) = F.T.h0|φ0 (x1 ) . . . φ0 (x4 )|0i

=

=

(where the circle in the line in the second diagram = n-pt. unrenormalized G.F.

−iZφ . p2i −µ2

→ Γ0 (x1 , . . . , xn ) = Truncated

G(x1 , . . . , xn ) = h0|T φ(x1 ) . . . φ(xn )|0i −n/2

= Zφ

−n/2

= Zφ

G0 (x1 . . . xn )

(10.4.44)

Γ(x1 , . . . , xn ) = Zφ Γ0 (x1 , . . . , xn )

n/2

(10.4.45)

˜ ˜ + Γ(u) ˜ F.T. Γ0 (x1 , . . . , xn ) = −iZλ−1 λ0 + Γ(s) + Γ(t)

(10.4.46)

But then We have Thus,

h0|T φ0 (x1 ) . . . φ0 (xn )|0i

h

O(λ2 )

}|0 {i z ˜ ˜ + Γ(u) ˜ + Γ(t) − iZλ−1 λ0 + Γ(s)

F.T. Γ(x1 , . . . , xn ) = Zφ2 ¢2 ¡ (recall that Zφ2 ∼ 1 + λ20 ). To order λ20 ,

˜ ˜ + Γ(u) ˜ F.T. Γ(x1 , . . . , x4 ) ≈ −iZφ2 Zλ−1 λ0 + Γ(s) + Γ(t)

(10.4.47)

(10.4.48)

10.5. DIVERGENCES AT HIGHER ORDERS

185

Let λ = Zφ2 Zλ−1 λ0 (Renormalized Coupling - where Zφ2 arises from 2 pt. function, and Zλ−1 arises from 4 pt. function - Zφ2 → ∞, Zλ−1 → ∞ (cancel)). So, ˜ ˜ + Γ(u) ˜ F.T. Γ(x1 , . . . , x4 ) = −iλ + Γ(s) + Γ(t) (Finite)

(10.4.49)

aside i p2 − µ2 − Σc (p2 ) = F.T. G(x1 , x2 ) p2 − µ2 − Σc (p2 ) F.T. Γ(x1 , x2 ) = i

F.T.h0|T φ(x1 )φ(x2 )|0i =

(10.4.50) (10.4.51)

end aside

10.5

Divergences at Higher orders

10.5.1

Weinberg’s Theorem

A Feynman integral is convergent if its degree of divergence is negative and the degree of divergence of any integral associated with a subdiagram is also negative. Recall: φ 4 : D =4−E

Figure 10.5.1: Convergence

Figure 10.5.2: Subdiagram is divergent: ∴ whole diagram is divergent

In figure 10.5.3, 1. All subdiagrams have D < 0

186


Figure 10.5.3:

2. D < 0 for entire diagram W.Th. ⇒ Overall diagram convergent! (i.e. no unanticipated divergences will appear). Thus, all divergences can be eliminated by an iterative procedure (i.e. first eliminate divergences in subdiagrams, and then in the diagrams as a whole (if there are any)). Feb. 10/2000 Consider 1 m2 2 λφ6 L = (∂µ φ)2 − φ − (4 − D) (10.5.1) 2 2 6! What goes wrong? Consider:

∼

Z

d4 ` (2π)4

µ

1 2 ` − m2

¶µ

1 (` + p)2 − m2

¶

∼ divergent

(10.5.2)

8

We can add in − λ88!φ to L to absorb this divergence. Once we have λφ8 , we can examine the 12-point function:

∼

Z

d4 ` (2π)4

12

µ

1 `2 − m 2

¶µ

1 (` + p)2 − m2

¶

(10.5.3)

We now need − λ1212!φ in order to absorb this new divergence. This goes on and on - in total, there will be an infinite number of vertices to absorb all of these divergences. Consider a diagram with ni → # of vertices of type i bi → # of Bosons in the ith vertex fi → # of Fermions in the ith vertex di → # of derivatives in the ith vertex B → # of external Bosons IB → # of internal Bosons


187

F → # of external Fermions IF → # of internal Fermions L → # of loops D → degree of divergence ¯ µ ψ∂µ φ → b = 1 ψγ f = 2 d = 1 ¯ µν ψFµν → b = 1 ψσ f = 2 d = 1 (where Fµν = (∂µ Aν − ∂ν Aµ )) So, 2IB + B =

X

ni b i

i

2IF + F =

X

ni f i

i

# of loops L = IB + IF −

X

ni + 1

i

(where the “1” is present in L because one of the δ-functions is superfluous). X D = 4L − 2IB − IF + ni d i i

Now eliminate L, IB , IF .

X 3 D =4−B− F + ni δ i 2 i

(10.5.5)

where δi = bi + 32 fi + di − 4. The ith vertex: Z LI = g |{z} d 4 x ψ 1 . . . ψ f i ∂ µ 1 . . . ∂ µ d i φ 1 . . . φ bi | {z } | {z } | {z } p−4

p3fi /2

pd i

(10.5.6)

p bi

This means that the dimension of g must be p−δi in order to keep overall dimensionless. (recall: Z d4 x ψ¯ 6 ∂ ψ |{z} |{z} |{z} |{z} p−4

p3/2

p+1

p3/2

188


→ dimensionless). With a dimensionless coupling constant, δi = 0 and 3 D =4−B− F 2

(10.5.8)

and then there are only a certain number of divergent diagrams. For δi > 0, the degree of divergence increases with more vertices of type ni (BAD). (ex. 6 λφ in 4 dimensions). → NON-RENORMALIZABLE THEORY! δi < 0; ex. λφ3 in 4 dimensions - δi = 3 − 4 = −1 < 0 → Only a finite # of divergent diagrams. In fact:

are the only fundamental divergent diagrams. → SUPERRENORMALIZABLE. This (above) treatment is a bit “primitive” - we are only counting powers of momenta in diagrams to determine divergence. In Gauge theories, there is a cancellation of divergences between 2 different diagrams. (See for example Figure 10.5.4) Figure 10.5.4: Example of gauge theories where (here) the divergences in last 3 diagrams cancel For example 2 ¯ p − m)ψ + 1 (∂µ φ)2 − M φ2 − g ψψφ ¯ − λ φ4 L = ψ(6 2 2 4!

(10.5.9)

(with g dimensionless). There is still a diagram that is divergent:

Z

d4 ` ∼ log divergent (6 `)4

(10.5.10)


189

We require − 4!λ φ4 term to absorb this. The following diagram

¯ 2 − κ(ψψ)

(10.5.11)

³R 4 ´ d ` is convergent. `6 For massive vectors, the lagrangian contribution is 1 m2 L = − (∂µ Vν − ∂ν Vµ )2 − Vµ V µ 4 2

(10.5.12)

Note that ³ the ´last term isn’t gauge invariant, and so we’re stuck with a longitudinal polarµ kν ) in the propagator; The propagator is: ization ( km 2 −

³ i gµν −

kµ kν m2

´

k 2 − m2 + iε

(10.5.13)

0

The integral goes like ∼ (k 2 ) ∼ constant. This causes divergences. ¯ µ ψV µ + g(ψψ) ¯ 4 LI = ψγ

(10.5.14)

Then we could have

Z

(Higgs Mechanism)

d4 ` ∼ log divergent (6 `)4

(10.5.15)

190

10.6


Renormalization Group

There is an arbitrariness inherent in renormalization. ex. Self-energy in φ 4 :

i∆(p) =

p2

−

µ20

i = − Σ(p2 )

(10.6.1)

If we’re just interested in eliminating the infinities, we could expand about some point κ 2 . Let 1 00 Σ(κ2 ) +(p2 − κ2 ) Σ0 (κ2 ) − (p2 − κ2 )Σ (κ2 ) + . . . | {z } | {z } |2! {z } Quadratic log

Σ(p2 ) =

divergence

=

p2

−

µ20

=Σc (p2 ,κ2 ) (converges)

divergence

+

Σ(κ2 )

−

(p2

1 − κ2 )Σ0 (κ2 ) − Σc (p2 , κ2 )

Wave function renormalization is done first: i (1−Σ0 (κ2 ))

= "

p2 −

³

µ20 +Σ(κ2 )−κ2 1−Σ0 (κ2 )

Let µ2 = i∆(p) =

p2

−

µ2

µ20

´ 2

Σc (p2 , κ2 ) − (1 − Σ0 (κ2 )) {z } | ≈Σc (p2 ,κ2 )

+ Σ(κ ) − κ2 1 − Σ0 (κ2 )

iZφ ; − Σc (p2 , κ2 )

#

Zφ =

1 ∼ 1 + Σ0 (κ2 ) 1 − Σ0 (κ2 )

(10.6.2)

where κ → arbitrary. Note 1. µ2 → no longer the physical mass of φ (Σc (µ2 , κ2 ) 6= 0, physical mass, given by µ2p − µ2 − Σc (µ2p , κ2 ) = 0

→ Σc (κ2 , κ2 ) = 0). µ2p is the (10.6.3)

2. How does something “physical” depend on κ? See “New methods for the renormalization group”, J.C. Collins and A.J. Macfarlane, Phys. Rev. D, Vol 10, Number 4, 15 August 1974. Feb. 14/2000 Momentum subtraction → is difficult to work out in practice.

10.6. RENORMALIZATION GROUP

191

“Minimal Subtraction” in conjunction with dimensional regularization is the easiest way to renormalize. c.f. Collins and MacFarlane. m2 λ0 1 (∂µ φ0 )2 − 0 φ20 − φ40 2 2 4! 2 m 1 (∂µ φ0 )2 − φ2 (m → renormalized) = 2 2 0 (m2 − m2 ) 2 λ0 4 = − 0 φ0 − φ0 2 | {z } 4!

(10.6.4)

L = L(2) LI Again, we have,

(10.6.5) (10.6.6)

F.T.h0|T φ0 φ0 |0iTruncated =

1 (−iλ0 ) µ4−n −iΣ(p ) = |{z} 2 2

∗

Z

i dn ` + n 2 (2π) (` − m2 )

µ

−i 2

¶

(m2 − m20(10.6.7) )

* → thus λ0 = µ4−2n+n = µ4−n . Keep in mind, we’re working in n dimensions. The Action is:   Z 1 λ0 S = |{z} dn x  ( ∂µ φ0 )2 − φ40  (10.6.8) 2 |{z} 4! a

b

Where the dimensions are: [a] = µ−n , [b] = µ+2 . Since the dimension of the action must be zero, this means that the dimension of [φ0 ] ∼ µ−(2+n)/2 . Note: as expected, if n = 4, λ0 is dimensionless. Z i 1 −i dn ` 2 4−n −iΣ(p ) = (−iλ0 )µ + (m2 − m20 ) n 2 2 2 (2π) (` − m ) 2 ´ ³ 4−n 1 µ λ0 n Σ(p) = (m2 )−1+n/2 + (m2 − m20 ) Γ 1− n/2 2(4π) 2 2 µ ¶ −2+n/2 ³ ´ n m2 1 λ0 m 2 Γ 1 − + (m2 − m20 ) = n/2 2 (4π) 2 µ 2 Let n → 4 ¢ µ ¶−2+n/2 ¡ Γ 2 − n2 m2 λ0 m 2 1 ¡ ¢ = + (m2 − m20 ) n n/2−2+2 2 (4π) µ 2 1− 2

192


Let ε = 2 −

n 2

(ε → 0 as n → 4). µ

¶ ¡ ¢ 1 − γE + O(ε) (−1) 1 + ε + O(ε2 ) · ε | {z } {z } | (4π)−n/2+2 n −1 (1− 2 ) Γ(2− n 2) ¶ µ µ 2¶ 1 m + O(ε) + (m2 − m20 ) 1 − ε ln 2 µ 2 | {z }

λ0 m 2 = (1 + ε ln(4π) + . . .) {z } (4π)2 |

λ0 m 2 (−1) = (4π)2 ³ Let m20 = m2 1 −

2λ0 (4π)2 ε

µ

³

m2 µ2

´ −2+n/2

1 + ln(4π) − γE − ln ε

µ

m2 µ2

¶¶

1 + (m2 − m20 ) + O(ε) 2

´ .

· µ 2 ¶¸ λ0 m ln(4π) − γE − ln = − 2 (4π) µ2 = Finite (where the ln

³

m2 µ2

´

(10.6.10)

is a residual dependence on renormalization).

We could also have absorbed the ln(4π), γE in to m20 , or µ2 . Let µ02 = µ2 eγE −ln(4π) . → ln

µ

m2 µ2

¶

= ln

µ

m2 µ02

¶

¢ ¡ + ln e(γE −ln(4π))

(10.6.11)

So we’d get λ0 Σ(p) = − (4π)2

µ

− ln

µ

m2 µ02

¶¶

(10.6.12)

(→ modified minimal subtraction: gets rid of ln(4π) and γE .) So also, for the four-point function:

+ |

{z

−iλ0 µ−4+n

}

(10.6.13) |

{z

absorb terms of

1 ε

into λ0

}


193

Net Result: ∞ X aν (λ, m, µ)

λ0 µ−4+n = λ +

Ã ν=1

m0 = m 1 + Zφ = 1 +

(10.6.14)

(n − 4)ν

∞ X bν (λ, m, µ) ν=1

(n − 4)ν

!

= mZν

(10.6.15)

∞ X cν (λ, m, µ)

(10.6.16)

(n − 4)ν

ν=1

Note: aν , bν , cν in minimal subtraction are independent of m2 and µ2 in practice. Note also: µ ∞ 1 ¶ X i = # of powers of n−4 j aij λ (10.6.17) ai (λ) = j = # of loops j=i+1 i.e.   

|

{z

# λ2 (n−4)





}

|



  +

(and the sum goes higher, etc.). So also, bi = ci =

λ3

∞ X

j=i+1 ∞ X

³

{z

0 # + # 3 (n−4)2 (n−4)

´

 

(10.6.18)

}

bij λj

(10.6.19)

cij λj

(10.6.20)

j=i+1

Note that

−λ0 φ40 4!

→ λ0 occurs in Feynman rules, but isn’t finite. ex: see figure 10.6.1. Figure 10.6.1: Overall divergence comes only partially from the diagram → λ0 also has poles due to calculated graph at 1-loop order.

Net result: ˜ R (p, m(n), λ(n), µ, n) ΓR (p, m, λ, µ) = lim Γ n→4

(10.6.21)

194


But now, £ ¡ ¢ ¤ −1 ˜ R (p, m(n), λ(n), µ, n) = Γ ˜ R p, λ λ0 µn−4 , n m0 Zm →Γ (λ(λ0 )) , µ, n −N/2

= Zφ

Γ0 (pi , m0 , λ0 , n)

(10.6.22)

(N is the # of external legs - Note µ dependence vanishes. ∴ can’t be µ dependent on L.H.S.) Feb. 16/2000 or ˜ R = ZΓ (λ0 µn−4 , n)Γ0 (p, λ0 (n), m0 (n), n) Γ (10.6.23) N/2

with ZΓ = Zφ . →µ

d Γ0 = 0 dµ

i d h −1 ˜ n−4 n−4 = µ lim Z ΓR (p, λ(∂0 µ , n), m(λ0 µ , n), µ, n) dµ n→4 Γ · ¸ −1 ∂ ∂ ∂λ ∂ ∂Zm ∂ZΓ −1 ˜ ΓR 0 = µ +µ + Zm mµ −µ Z ∂µ ∂µ ∂λ ∂µ ∂m ∂µ Γ ´ ´ ³ ´ ³ ³ −1 ∂ZΓ −1 ∂Zm ∂λ Now set β(λ) ≡ µ ∂µ , −γm m ≡ Zm mµ ∂µ , −γΓ ≡ −µ ∂µ ZΓ , and let n → 4. µ ¶ ∂ ∂ ∂ 0= µ + β(λ) − γm m − γΓ Γ (10.6.24) ∂µ ∂λ ∂m

Note: as Γ is finite, β, γm and γΓ are finite. Suppose we just look at the “engineering dimension” (depends only on the # of external legs), DΓ , of Γ = Γ(p, λ, m, µ).

DΓ = −2

DΓ = +2

DΓ = 0

Consider pµ = κpµ0 (pµ0 → reference momentum).


10.6.1

195

Euler’s Theorem d f (λx, λy) = λD f (x, y) → at λ = 1 dλ ¶ µ ∂ ∂ f = DλD−1 F x +y ∂x ∂y

(10.6.25) (10.6.26)

From this, if Γ = Γ(κp0 , λ, m, µ), then ¶ µ ∂ ∂ ∂ Γ = DΓ Γ κ +m +µ ∂κ ∂m ∂µ

(10.6.27)

∂ Now, eliminate µ ∂µ from (10.6.24), (10.6.27): · ¸ ∂ ∂ ∂ 0= κ − β(λ) + (1 + γm )m + (γΓ − DΓ ) Γ(κp0 , λ, m, µ) ∂κ ∂λ ∂m

(10.6.28)

• Can trade in dependence on mass scale parameter µ for dependence on scale in front of momentum p0 . • By rescaling momentum (changing κ) → will effectively change value of coupling constant/mass parameter (β, m). −1

∂λ • β(λ) = µ ∂µ , −γm = µZm ∂Z∂µm .

Formal Solution

Γ(κp0 , λ, m, µ) = κ

DΓ

µ Z exp −

¶ 0 dκ ¯ )) ¯ γλ (λ(κ Γ(p0 , λ(κ), m(κ), ¯ µ) (10.6.29) κ0 1 ´ £ ¤ ¯ = − 1 + γm (λ(κ)) m(κ) ¯ and with boundary conditions κ

0

´ ³ ³ ¯ ¯ ¯ = β( λ(κ)) , κ ∂ m(κ) with κ ∂ λ(κ) ∂κ ∂κ ¯ λ(1) = λ, m(1) ¯ = m. For example, consider the 2-pt. function:    +  



 +  |

|



    +   | }

{z A

+

+ {z C

+



{z

}

B



 + . . . +... }

 

196


The net result of renormalization will be: µ µ 2¶ ¶ p + B1 A = λ A1 ln µ2 µ 2¶ µ 2¶ p p 2 + B2 ln + C2 B = A2 ln 2 µ µ2 µ 2¶ µ 2¶ µ 2¶ p p p 3 2 C = A3 ln + B3 ln + C3 ln + D3 2 2 µ µ µ2

(10.6.30) (10.6.31) (10.6.32)

¯ → we’re just taking the leading order terms, (A0 s), → all terms subsumed into λ(κ). Thus, κ

¯ ∂ λ(κ) ¯ = β(λ(κ)) ∂κ = β1 λ2 +β2 λ3 + β3 λ4 + . . . |{z}

(10.6.33)

lowest order

¯ For perturbation theory to make sense, λ(κ) must be small. • If,

¯ lim λ(κ) =0

κ→∞

(10.6.34)

we have asymptotic freedom → (Yang-Mills Theory) • If

¯ lim λ(κ) =0

κ→0

(10.6.35)

we have infrared freedom → (almost everything else). Feb. 17/2000 Recall ¯ ∂ λ(κ) ¯ = β(λ(κ)) ∂κ £ ¤ ∂ m(κ) ¯ ¯ κ = − 1 + γm (λ(κ)) m(κ) ¯ ∂κ κ

(10.6.36) (10.6.37)

In Dimensional Regularization

λ0 µn−4 = λ +

∞ X aν (λ) (n − 4)ν ν=1

The µn−4 was introduced so that λ is dimensionless. λ0 → µ4−n ∂λ β(λ) = µ ∂µ

(10.6.38)


197

Now, how do we get β(λ)? β(λ) = x0 + x1 (n − 4) + x2 (n − 4)2 + . . . # " ∞ X ∂ ∂ ∂λ a (λ) ν µ (λ0 µn−4 ) = µ λ+ ∂µ ∂µ ∂λ (n − 4)ν ν=1 |{z}

(10.6.39) (10.6.40)

β(λ)

Subs. in β(λ)

"

∞ X a0ν (λ) = x0 + x1 (n − 4) + x2 (n − 4)2 + . . . 1 + (n − 4) λ0 µn−4 | {z } (n − 4)ν ν=1 subs. (10.6.38) # " # " ∞ ∞ X X a0ν (λ) aν (λ) = [x0 + x1 (n − 4) + . . .] 1 + (n − 4) λ + (n − 4)ν (n − 4)ν ν=1 ν=1

£

¤

#

Thus,

(n − 4)λ + a1 +

£ ¤ a2 a3 2 + . . . = x + x (n − 4) + x (n − 4) + . . . + + 0 1 2 (n − 4) (n − 4)2 · ¸ x0 a01 0 0 + + x1 a1 + x2 a1 (n − 4) + . . . (n − 4) · ¸ x1 a02 x0 a02 0 + + + x 2 a2 + . . . + . . . (n − 4)2 (n − 4)

We set x2 = x3 = x4 = . . . = 0. (n − 4)λ + a1 +

Matching powers of (n − 4): ( x1 = λ x0 + x1 a01 = a1 ½

x0 a01 + x1 a02 = a2 .. . x0 a0ν + x + 1a0ν+1

a2 + . . . = x1 (n − 4) + (x0 + x1 a01 ) (n − 4) 1 = +(x0 a01 + x1 a02 ) (n − 4) µ ¶2 1 0 0 +(x0 a2 + x1 a3 ) (n − 4)

(10.6.41)

)

x1 = λ → β(λ) = [a1 − λa01 ] + λ(n − 4) x0 = a1 − λa01

)

constraint eq’s that fix a2 , a3 , . . . , aν in terms of a1

(Note as n → 4, β(λ) = (a1 − λa01 )). Thus, λ0 µ

n−4

∞ X aν (λ) =λ (n − 4)ν ν=1

(10.6.42)

198


• only a1 is needed for β(λ) • a1 fixes a2 , a3 , . . . Remember that λφ4 in 4-D has the form: a1 (λ) =

a12 λ2 + a13 λ3 + a14 λ4 + . . . |{z} |{z} |{z}

From 1-loop diagrams

Recall that:

aν =

2-loop

∞ X

(10.6.43)

3-loop

aνj λj

(10.6.44)

j=ν+1

ex: for

a2 = x + 0a01 + x1 a02 £¡ a23 λ3 + a24 λ4 + . . . = a12 λ2 + a13 λ3 + . . . ¶ ¸ ¢ d ¡ ¢ d ¡ 2 3 2 3 a12 λ + a13 λ + . . . a12 λ + a13 λ + . . . −λ dλ dλ · ¸ ¢ d ¡ 3 4 a23 λ + a24 λ + . . . +λ dλ

Here, a2j can be solved for in terms of a1j . Similarly, we can show that, if Zm = 1 +

∞ X ν=1

bν (n − 4)ν

,

Zφ = 1 +

∞ X ν=1

cν (n − 4)ν

(10.6.45)

Then (***)

10.6.2

λb01 m = λc01 (c.f. Collins & MacFarlane)

γm (λ) =

(10.6.46)

γΓ

(10.6.47)

Explicit Calculations 1 1 λ0 (∂µ φ)2 − m20 φ2 − φ4 2 2 4! 1 1 2 2 2 = (∂µ φ) − m φ 2 2 1 2 λ0 φ 4 = − (m0 − m2 )φ2 − 2 4!

L = L(2) LI

(10.6.48) (10.6.49) (10.6.50)


199

Two point function 

+ |

{z

}

A



 +

|

{z

}

B

+

    +    

|



{z

λ0 (1+one contribution to λ0 )∗

       

}

 

+

(* - because it is itself renormalized). We find that, to 1-loop order,

SF−1

1-loop contrib.

B

A

}| { z µ }| ¶{ z z }| { 2 2 m λ b λ 11 − + finite +(two-loop) = (p2 − m2 ) − m2 (n − 4) 16π 2 (n − 4)

i.e. m20 = m2

Ã

∞ X bν (λ) 1+ (n − 4)ν ν=1

!

(10.6.51)

(10.6.52)

(we just have to take lowest order term. (b2 (λ) = b11 λ + b12 λ2 + . . .). → SF−1 will be finite if we choose · ¸ −λ b11 = (10.6.53) 16π 2 Note, no divergence ∝ p2 (could be, in principle, but it just doesn’t occur in φ4 theory). ∴ Zφ = 1 to one-loop order For the 4-pt. function

Γ =

+ |

{z A

λ20 = λ0 + (n − 4)

} µ

3 16π 2

+ (two-loop)

¶

+ (finite)

(10.6.54)

200


but, a12 λ2 + ... (n − 4) a12 λ2 3 λ2 Γ = λ+ + + ... (n − 4) (n − 4) 16π 2

λ0 = λ +

→ arrange for pole from A and from 1-loops to cancel · ¸ 3 ⇒ a12 = − 16π 2

(10.6.55)

Feb. 28/2000 Recall:

+

= −m SF−1

2

µ

¶ ³ ´ b11 λ b12 λ2 + + ... + ... (n − 4) (n − 4)

µ ¶ b11 λ m2 λ − =p −m 1+ + finite + O(λ2 ) 2 (n − 4) 16π (n − 4) 2

2

(10.6.56)

1 Z = 1. where b11 = − 16π 2 ,

Γ =

+ O(λ3 )

3λ2 λ2 a12 + finite + O(λ3 ) + = λ+ 2 (n − 4) (n − 4)16π

(10.6.57)

3 with ⇒ a12 = − 16π 2 . Higher order contributions to the 2-pt. function are:

µ ¶ ³ a12 λ2 n ´ 2 (n/2−1) −4+n 1 1 λ+ Γ 1− (m ) µ = 2 (4π)n/2 2 (n − 4) ¸ · 2a12 λ m2 λ 2 + (γ − 1)a12 λ + + finite (10.6.58) = O(λ) + 32π 2 (n − 4)2 (n − 4)


201

(where the vertex of the loop with the propagator is λ0 = λ +

µ8−4n = − 4(2π)2n

Z

(k 2

1 1 2 2 − m ) (` − m2 )2

m 2 λ2 1 (4γ − 2)m2 λ2 + + finite (16π 2 )2 (n − 4)2 (32π 2 )2 (n − 4)

=

λ2 m2 a11 = O(λ) + 16π 2 Ã

dn k d n `

a12 λ2 ). (n−4)

µ

1 (n − 4)2

1 Recall L = − (m20 − m2 )φ2 + . . . 2

¶

+

(10.6.59)

γb11 λ2 m2 + finite (10.6.60) 32π 2 (n − 4)

!

(where the cross at the top of the loop is due to the mass insertion).

= ( Collins )

=





λ2 m2 m 2 λ2 1   p2 finite + − + (γ − 1)m2  +(10.6.61)  2 2 2 2 2 (16π ) (n − 4) (16π ) (n − 4) |{z} 12 2 ∗

(∗ ⇒ wave function renormalization) All together (to 2-loop order - only showing two loop order here): " · ¶ ¸ µ 2 λ λ2 1 1 −1 2 2 Sp = p 1 − −m 1+ b12 − 12(16π 2 )2 (n − 4) (n − 4) 2(16π 2 )2 | {z } would think 1 b12 = 2(16π 2 )2

λ2 + (n − 4)2

¶# 2 b22 − (16π 2 )2 | {z } µ

(10.6.62)

and here 2 b22 = (16π 2 )2

First!! Remember that

FT h0|T φφ|0i →

(p2

−

Zφ − finite )

m2

(10.6.63)

202


Thus,

£ ¤ S −1 = Zφ−1 p2 − m2 − finite

Thus,

Zφ−1 = 1 − So,

(10.6.64)

λ2 1 2 2 12(16π ) (n − 4)

(10.6.65)

£ ¤ S −1 = Zφ−1 p2 − m2 − finite µ µ ¶ ¶" µ λ2 λ2 1 1 1 2 2 = 1− b12 − p −m 1+ + 12(16π 2 )2 (n − 4) (n − 4) 2(16π 2 )2 12(16π 2 )2 µ ¶¶ # 2 λ2 (10.6.66) b22 − + (n − 4)2 (16π 2 )2 Thus,

b22 = b12 =

2 (16π 2 )2 5 12(16π 2 )2

)

Zφ = 1 +

λ2 1 2 12(16π ) (n − 4)

m20

2

λ0 = µ Thus,

4−n

= m

·

·

(10.6.67)

µ

5λ2 −λ + 16π 2 12(16π 2 )2 ¶ ¸ 2λ2 1 3 + + O(λ ) (16π 2 )2 (n − 4)2

1 1+ (n − 4)

3λ2 1 + O(λ3 ) λ− 2 16π (n − 4)

3λ2 + O(λ3 ) a1 (λ) = − 2 16π µ ¶ ∂ β(λ) = 1−λ a1 (λ) ∂λ 3λ2 = 16π 2

¸

(10.6.68)

(10.6.69)

(10.6.70)

(10.6.71)

Thus the running coupling is the solution to: ¯ ∂ λ(κ) ¯ κ = β(λ(κ)) ∂κ Z κ 0 Z λ(κ) ¯ ¯ dκ dλ = 0 ¯ β(λ) κ0 κ λ

(10.6.72)

10.6. RENORMALIZATION GROUP Ã

203

originally µ

∂ ∂ µ + β(λ) ∂µ ∂λ µ

¶

Γ(p, µ, λ)

=

0(?)

κ

→

¯ Γ(κp0 , µ, λ) → Γ(p0 , µ, λ(κ)) !

∂ ∂ + β(λ) κ ∂κ ∂λ

¶

?

z}|{ Γ = 0

(10.6.73)

¯ 0 ) = λ. So, The boundary conditions are: λ(κ ¯κ ¯ ¯ ln (κ0 ) ¯ ¯

=

κ0

ln

3 ln 16π 2

µ

µ

κ κ0

¶

Z

¯ λ(κ) λ

16π 2 = 3

¶

¯ dλ ³

µ

¯2 3λ 16π 2

1 −¯ λ

´

¯ ¶ ¯¯λ(κ) ¯ ¯ ¯ λ

1 1 + = −¯ λ(κ) λ 1 ¯ ³ ´ λ(κ) = 3 1 − 16π2 ln κκ0 λ κ κ0

λ

¯ λ(κ) =

1−

3λ 16π 2

ln

³ ´≈ κ κ0

½

0− as κ → ∞ ( unstable ) 0+ as κ → 0+ ( stable *)

¾

(10.6.74)

∗ → i.e. perturbation theory acceptable for small momenta. Thus, as κ → 0+ , perturbation theory reliable. κ

¯ ∂ λ(κ) ¯ = β(λ(κ)) ∂κ

(10.6.75)

The plot of this can be seen in Figure 10.6.2. If we include higher orders, we get Figure 10.6.2. Mar. 1/2000 ¯ 3 ¯2 ∂ λ(κ) = λ (κ) → κ ∂κ 16π 2

Z

¯ λ(κ) λ

¯ 0 (κ) dλ 3 = λ02 (κ) 16π 2

λ

¯ λ(κ) = 1−

3λ 16π 2

ln

³ ´ κ κ0

Z

κ κ0

dκ0 κ0

(10.6.76)

(10.6.77)

204


Figure 10.6.2: plot

Figure 10.6.3: Beta function Beta function plot 2 - note β = 0!! - depends ¯ on where you start. (λ(κ) → λfixed as κ → ∞)

¯ 0 ) = λ. B.C. λ(κ ¯ We often let λ = ∞ at κ0 = Λ. i.e. λ(Λ) = ∞. Thus, ¯κ Z λ(κ) ¯ ¯ dλ0 3 0 ¯ = ln(κ ) ¯ ¯ λ02 16π 2 ∞ Λ ¯λ(κ) ¯ ³κ´ 3 1 ¯¯ = ln − 0¯ λ¯ 16π 2 Λ ∞ µ ¶ 3 Λ = − ln 16π 2 κ 16π 2 ¯ ¡ ¢ λ(κ) = 3 ln Λκ

(10.6.78)

Note that:

¯ • κ → 0, λ(κ) → 0+

¯ • κ → ∞, λ(κ) → 0− (unacceptable)

¯ • κ ≈ Λ, λ(κ) → ∞ → perturbation breaks down. In QCD ¡ ¢2 1 ¡ a ¢2 ¯ Fµν + ψ (i 6 ∂ − gτ a 6 Aa ) ψ + φ∗i ∂µ − iegT a Aaµ ij φj (10.6.79) 4 · ¸ 11 4 1 g3 − C2 (v) + C2 (F ) + C2 (S) (10.6.80) β(g) = 16π 2 3 3 3 £ a b¤ T ,T = cabc T c (10.6.81) L = −


205

Figure 10.6.4: Beta func- Figure 10.6.5: Beta function plot for pure Yang- tion plot for pure YangMills - two-loop order. Mills. Then, C2 (v)δab = camn cbmn = n for SU (n) (10.6.82) a b δab C2 (F ) = Tr(τ τ ) 1 for quarks in QCD (10.6.83) = 2 C2 (S)δ ab = Tr(T a T b ) 1 = for complex scalars in the fundamental rep. for SU (3) (10.6.84) 2 For pure SU (n) Yang-Mills theory, β(g) = −

11 g 3 n 3 16π 2

(10.6.85)

whose graph looks like Figure 10.6.4 Thus, κ

(B.C. g¯(Λ) = ∞)

g 3 (κ) d¯ g (κ) 11 n¯ = − dκ 3 16π 2 d¯ g < 0 as κ → ∞ → asymptotic freedom dκ Z g¯(κ) 0 Z κ 0 dg dκ 11n = − g 03 3(4π)2 Λ κ0 ∞

¯g¯(κ) ³κ´ 11n 1 ¯¯ = − − 02 ¯ ln 2g ¯ 3(4π)2 Λ ∞ ) 3(4π)2 κ → ∞, g¯2 → 0 ¡κ¢ g¯2 = κ < Λ, g¯2 (κ) < 0 (Not acceptable) 22n ln Λ

(10.6.86) (10.6.87) (10.6.88)

(10.6.89)

206


For Nf flavours of quarks, (see Figure 10.6.5) g3 β(g) = 16π 2

"

SU (3)

µ ¶ 11 z}|{ 4 1 Nf − (3) + 3 3 2 | {z }

−11+

2Nf 3

0 - Figure 11.0.1) or a negative mass squared (m2 < 0 - Figure 11.0.2). For the first case, m2 φ2 λφ4 + 2 4! For the second, we get dV dφ

λφ3 6 φ=0q

(11.0.6)

= m62φ + = 0 if

φ=±

−6m2 λ

207

)

(2 possible vacua)

(11.0.7)

208

CHAPTER 11. SPONTANEOUS SYMMETRY BREAKING 5

4

4

3

3 y 2

y2 1

1

–3

–2

–1

0

–4 1 phi 2

–2

2 phi

4

–1

3

–1

–2

Figure 11.0.1: Positive Figure 11.0.2: Negative 2 2 mass (m > 0) plot. mass (m < 0) plot Let’s consider excitations above φ0 = +

q

−6m2 . λ

0

φ=φ +

r

−

6m2 λ

(11.0.8)

The Lagrangian for φ0 is: L =

1 m2 0 λ 2 2 4 [∂µ (φ0 + φ0 )] − (φ + φ0 ) − (φ0 + φ0 ) 2 2 4!   



2

=

1 m  02 λ (∂µ φ0 )2 − φ + 2φ0 φ0 +φ20  − φ04 + 4φ03 φ0 + 6φ02 φ20 + 4φ0 φ30 +φ40  | {z } | {z } 2 2 4! ∗

∗

∗0 s cancel using φ0 definition Ã r ¶! µ 2 2 2 1 m λ 6m 6m + const (∂µ φ0 )2 − φ02 − φ04 + 4φ03 − + 6φ02 − = 2 2 4! λ λ r µ ¶ 2 2 1 6m2 m 3m λ λ 2 = (∂µ φ0 ) + − + φ02 − φ04 − φ03 − 2 2 2 4! 3! λ r 1 1 λ 6m2 03 λ 04 2 = (∂µ φ0 ) − (−2m2 )φ02 − − φ − φ 2 2 3! λ |{z} 4!

(11.0.9)

†

→ mass of φ0 is (−2m2 ) > 0. † breaks the φ ↔ −φ symmetry. This can be seen in Figure 11.0.2 - the bottom of the two wells will have ground states |0− i, |0+ i - this ground state doesn’t respect symmetry. For example, Fˆ φFˆ −1 = −φ (suppose) Fˆ |0i = |0i if (m2 > 0) Fˆ |0+ i = |0− i if (m2 < 0)

209 ³

´ → symmetry restored . Mar. 2/2000 Heisenberg (1-D) Ferromagnet: |0− i+|0+ i √ 2

1 2 3 4 5 6 7 8 → → ← ← → ← → → (where σ = +1 for positions (1 + 2), σ = −1 for positions (3 + 4),. . . ). X H = − κσi σi+1

(11.0.10)

(11.0.11)

i

£ ¤ = −κ (+1)2 + (+1)(−1) + (−1)2 + . . .

There is an “up-down” symmetry in this Hamiltonian (no preference for up/down). The lowest energy states are “all up” or “all down”. → Generalize to an O(N ) model. φ = (φ1 , φ2 , . . . , φN )

(11.0.12)

φ → Rφ (RT = R−1 )

(11.0.13)

m2 λ 1 φ · φ − (φ · φ)2 L = (∂µ φ)(∂ µ φ) − 2 2 4!

(11.0.14)

L is invariant under R is a global orthogonal matrix.

The potential for this Lagrangian is

Figure 11.0.4: O(N ) model Figure 11.0.3: O(N ) model (m2 < 0) plot “Mexican for m2 > 0. Hat” potential

V (φ) =

m2 2 λ φ + ((φ)2 )2 2 4!

(11.0.15)

210

CHAPTER 11. SPONTANEOUS SYMMETRY BREAKING

(where λ > 0 for stability). This can be plotted - see figures 11.0.3 (for m 2 > 0) and 11.0.4 (for m2 < 0). ∂V λ 2 = m φ + (2φ2 )(2φi ) i ∂φi 4! µ ¶ λ 2 2 = m + φ φi 6 = 0 at φi = 0 or at |φ| =

r

−6m2 λ

(11.0.16)

Suppose, φ0 = V δi1 ( in the “one” direction) φ = (V + φ01 , φ02 , . . . , φ0N ) L =

¤ λ£ ¤ m2 £ 1 02 2 (∂µ φ0 )(∂ µ φ0 ) − (V + φ01 )2 + φ02 + . . . + φ0N − (V + φ01 )2 + φ02 2 + . . . + φN 2 2 | {z } 4! =V 2 +2V φ01 +φ02 1 +... =V 2 +2V φ01 +φ02

= = = = =

1 (∂µ φ0 )2 − 2 1 (∂µ φ0 )2 − 2 1 (∂µ φ0 )2 − 2 1 (∂µ φ0 )2 − 2 1 (∂µ φ0 )2 − 2

¤ λ£ 2 ¤2 m2 £ 2 V + 2V φ01 + φ02 − V + 2V φ01 + φ02 2 4! ¤ λ £ 2 02 2V φ + 4V 2 φ02 1 + interaction terms 4! µ ¶ ¢ m2 02 λ −6m2 ¡ 02 2φ + 4φ02 φ − 1 2 4! λ ¶ µ 2 1 02 m 02 2 02 φ +m φ + φ1 2 2 1 (−2m2 )φ02 1 2

(11.0.17)

Thus • φ01 → mass (−2m2 ) • φ02 , . . . , φ0N → massless (N − 1 massless Goldstone Bosons). In general, if a system of N scalars has an invariance under a group G, and if the ground state of the theory has an invariance under a group G0 ⊂ G, then there is a massless excitation above this ground state for each generator of G0 . → Not realized in Nature. Here we have G → O(N ) (Orthogonal group, N-dimensional) (11.0.18) (see figures 11.0.5,11.0.6). Figure 11.0.7 → No massless scalars (spin 0) particles observed

211

Figure 11.0.5: So G0 → O(N − 1)

Figure 11.0.6: Excitation where you just roll around bottom of well (Correspond to Goldstone Bosons)

Figure 11.0.7: m2 > 0 - (No massless excitations). in nature. 1 2 m2 2 L = − Fµν − A (Massive spin-1 particle) 4³ 2 ´µ µ kν i gµν − km 2 ( whole term is transverse) iDµν = − k 2 − m2

(11.0.19) (11.0.20)

µ kν where the km term comes from the longitudinal polarization (Destroys Renormalizability) 2 - the gauge invariance is ruined by m2 . In the Higgs Mechanism, the massless Goldstone Bosons are absorbed by the longitudinal polarization of the massless vectors to which they are coupled to give massive vector particles. However, gauge invariance is not sacrificed, so renormalizability is retained. The massive scalars left over are all the “Higgs” particles.

212


Mar. 6/2000

11.1

O(2) Goldstone model: ¢¡ ¢ m2 i i λ i i 2 1¡ ∂µ φ i ∂ µ φ i − φ φ − (φ φ ) 2 ·2 1 ¸ ¸· 1 ¸ ·4! φ φ cos θ − sin θ i 1 2 → φ = (φ , φ ) → 2 φ sin θ cos θ φ2

L =

Let φ =

φ1√ +iφ2 , 2

φ∗ =

(11.1.1)

φ1√ −iφ2 . 2

λ L = (∂µ φ∗ )(∂ µ φ) − m2 φ∗ φ − (φ∗ φ)2 6 iθ φ→e φ

(11.1.2)

Local gauge invariance φ → eiθ(x) φ

(11.1.3)

Introduce a gauge field 1 L = − Fµν F µν + [(∂ + ieAµ )φ∗ ] [(∂ µ − ieAµ )φ] − m2 φ∗ φ − g(φ∗ φ)2 4

(11.1.4)

If m2 < 0 - (see figure 11.1.1).

Figure 11.1.1: m2 < 0, → V (φ) = m2 φ∗ φ + g(φ∗ φ)2

∂V = 0 ⇒ |φ| = ∂φ

s

−m2 2g

(11.1.5)

11.1. O(2) GOLDSTONE MODEL:

213

Choose φmin =

s

−m2 (Real) 2g

φ =

s

−m2 + φ0 2g

V = √ + φ0 2

(11.1.6)

(11.1.7)

Now, · µ ¶¸ · µ ¶¸ V 1 2 V µ µ 0∗ 0 (∂ − ieA ) √ + φ L = − Fµν + (∂µ + ieAµ ) √ + φ 4 2 2 µ ¶µ ¶ ·µ ¶µ ¶¸2 V V V V 2 0∗ 0 0∗ 0 √ +φ −g √ +φ √ +φ −m √ +φ 2 2 2 2 V 1 2 = − Fµν + [(∂µ + ieAµ )φ0∗ ] [(∂µ − ieAµ )φ0 ] + √ [(ieAµ (∂ µ φ0 ) − (∂µ φ0∗ )(ieAµ )] 4 2 · ¸ · 2 ¸2 V V V2 2 V2 V 0∗ 0 2 µ 0 0∗ 0 0∗ 0∗ 0 −g e Aµ A −m φ φ + √ (φ + φ ) + + √ (φ + φ ) + φ φ + 2 2 2 2 |2 {z } ∗

(Note that it would appear, from the * term, that Aµ Aµ has a mass V 2 e2 ). Now, let Dµ = ∂µ − ieAµ . e2 V 2 ieV 1 2 + (Dµ φ0 )∗ (Dµ φ0 ) + √ [Aµ ∂ µ φ0 − (∂µ φ0∗ )Aµ φ0 ] + Aµ Aµ − m2 φ0∗ φ0 L = − Fµν 4 2 2 # " µ 2¶ 2 2V V V (φ0 + φ0∗ )2 + 2 φ∗ φ0 + √ (φ0 + φ0∗ )φ0∗ φ0 + (φ0∗ φ0 )2 + const.(11.1.8) −g 2 2 2 {z } | appears that Im(φ0 ) is massless

To determine the actual degrees of freedom, set µ ¶ η(x) + V √ φ(x) = eiξ(x)/V 2

(11.1.9)

where η(x), ξ(x) are real. Then, let φv (x) = e−iξ/V φ(x) ¶ µ 1 ξ(x) B µ = A µ − ∂µ e V

(11.1.10) (11.1.11)

214


where (11.1.10,11.1.11) are a gauge transformation. Then, φv (x) =

η(x) + V √ 2

(11.1.12)

Now we find that L =

1 1 1 1 (∂µ η)2 − (−2m2 )η 2 − (∂µ Bν − ∂ν Bµ )2 + (eV )2 Bµ B µ 2 2 4 2 1 2 g 4 µ 2 2 2 + e Bµ B (η + 2V η) − gV η − η 2 4

(11.1.13)

• ξ(x) → disappeared (Goldstone boson has vanished) • η(x) → mass (−2m2 ) → Higgs • Bµ → massive vector (3 polarizations) - mass (eV )2

q 2 There are still only 3 parameters that characterize the theory, η, e, g. (recall V = −m ). 2g This is the form of the Lagrangian in the “U ” (unitary) gauge - physical degrees of freedom are apparent. Renormalizability is only manifest in “other” gauges (f-gauges) (Note, though, that since the “original theory” was renormalizable, this one is also). In general, for a non-Abelian Field, L= If we say

¤ 1 a aµν 1£ (∂µ φ∗i + igTija Aaµ φ∗j )(∂ µ φi − igTilb Abµ φl ) − V (φi ) − Fµν F 2 4

(11.1.14)

• n → # of scalar fields φi (2n if φi → complex). and if •

∂V ∂φi

= 0 if φi =

Vi √ 2

6= 0 (i = 1, . . . , n0 ) (n0 < n)

then there are then n0 massive Higgs and (n − n0 ) massive vector particles. Electroweak SU (2) × U (1) model φi ⇒ i = 1, 2 (Complex doublet of SU (2)) (n = 4), • n0 = 1 → One Higgs particle • → 4 vectors, (4-1) = 3 are massive, and 1 is massless (photon). Mar. 8/2000

(11.1.15)

11.2. COLEMAN WEINBERG MECHANISM

11.2

215

Coleman Weinberg Mechanism

L =

1 1 λ (∂µ φ)2 − m2 φ2 − φ4 2 2 4! 2 m 2 λ 4 φ + φ →V = 2 4!

Vtot =

(11.2.1)

+ |



  +  |

{z

}

Classical

+ {z

One-loop



   + . . . +   | }



{z

Two-loop

+ . . .

= Sum of 1PI (irreducible) diagrams with an arbitrary number of external legs where momenta = 0

}

Remember (Generator function of all diagrams) Z[J] =

Z

dφ e(i/~)

R

dx (L+Jφ)

= eiW (J)/~

(11.2.2) (11.2.3)

where W (J) is the generating functional of all connected diagrams. Φc (x) =

δW (J) δJ(x)

→ Legendre Transformation Z Renormalized → Γ(Φc ) = W (J) − dx J(x)Φc (x)

(11.2.4) (11.2.5)

(where Φc is the 1PI generating functional). Define the effective potential to be ∞ X 1 V (φ) = − Γ(0, . . . , 0)φn n! n=0

• We first of all renormalize Γ(φ)

(11.2.6)

216


• We next define m2R λR

¯ d V ¯¯ = ¯ dφ2 ¯ ¯φ=κ d4 V ¯¯ = ¯ dφ4 ¯ 0 2

(11.2.7)

(11.2.8)

φ=κ

(mR = Renormalized mass). Thus,

V = V (φ, λR , m2R )

(11.2.9)

We can now determine if there is Spontaneous Symmetry breaking. Note that even if Vcl has φmin = 0 (i.e. no spontaneous symmetry breakdown), it is conceivable that Vtot does generate the breakdown. If there is spontaneous symmetry and the Higgs mechanism has generated massive vectors, then there will be no divergences in this theory with vectors. • Heuristic argument for the renormalizability of models with the Higgs mechanism for generating massive vectors.

One loop Effective Potential in λφ4 model

11.3

+

+

φc − constant field (external momenta = 0) First loop:

th

n

Loop:

Z

i d4 k 4 2 (2π) (k − m2 + iε)

Sn

Z

d4 k (−iλ)n (2π)4

µ

i 2 (k − m2 + iε)

2n legs →

(2n)! (2n)2n

¶n

(Sn → Symmetry factor)

11.3. ONE LOOP EFFECTIVE POTENTIAL IN λφ4 MODEL

217

(where, in the denominator, the 2n is because we can start anywhere in the circle, and the 2n is because the legs are interchangeable). Thus, ¶n µ Z (2n)! d4 k λ Γ(0, . . . , 0) = n (11.3.2) | {z } 2 (2n) (2π)4 k 2 − m2 + iε 2n zero momenta

n = 0 (constant term)

(11.3.3)

Therefore, ∞ X

1 Γ(0, . . . , 0)φ2n c (2n)! n=1 ¶n µ Z ∞ X 1 (2n)! λφ2c 4 = − dk n (2n) (2n)! 2 k 2 − m2 + iε n=1 Ã !n 2 Z ∞ λ φ2c d4 k 1 X 1 = − (2π)4 2 n=1 n k 2 − m2 + iε

V (φ) = −

(n = 1, 2diverges) Ã ! Z λφ2c d4 k 1 2 ln 1 − 2 = − (2π)4 2 k − m2 + iε

x2 x3 (as ln(1 − x) = x + + + . . .) 2 3   ¶ µ Z 2 4 ¡ ¢ λφ d k 1 ln k 2 − m2 − c − ln k 2 − m2 + iε  = − 4 (2π) 2 2 | {z }

(11.3.4)

constant

The minus sign means R 4 we√must still renormalize. Note that the second (constant) term is a problem in G.R. - d x g∞ k ← Cosmological constant = 0 actually. Alternative Derivation ½ Z ¾ Z i 4 Z(J) = dφ exp d x (L(φ) + J(φ)) (11.3.5) ~ Let φ(x) = φc + h(x) = (constant field) + (quantum fluctuations about φc ). ½ Z ¾ Z i 4 Z(J) = dh exp [L(φc + h) + Jh] d x ~ Consider only diagrams with external φc ’s and are 1PI to get V (φc ). But, L = L(φc + h) 1 = (∂µ h)2 − 2 1 (∂µ h)2 − = 2

m2 λ (φc + h)2 − (φ2c + 2φc h + h2 )2 2 4! ¢ m2 2 λ¡ 4 φc + 4φ3c h + 6φ2c h2 + O(h3 ) (φc + 2φc h + h2 ) − 2 4!

(11.3.6)

218


The only terms that will contribute to V (φc ) at one-loop order are O(h2 ). i.e. One-Loop m2 2 1 (∂µ h)2 − h ⇒ 2 2

(< hh >)

(11.3.7)

λ − φ2c h2 ⇒ 4

(11.3.8)

One loop

Two-loop 



   3  h φc   

      

,

(11.3.9)

Then, ½ Z · m2 2 i 1 4 Zˆ = dh exp dx (∂µ h)2 − h − ~ 2 2 ½ Z · µ Z i 1 4 = dh exp dx ln −p2 − m2 − ~ 2 µ ¶ λφ2c −1/2 2 2 = det p +m + κ 2 Z

¸¾ λ 2 2 φh 4 c ¶ ¸¾ λφ2c h 2 (11.3.10)

Mar. 9/2000 µ ¶ ∞ 1 λφ2 d4 k 1 X 1 2 V (φ) = i (2π)4 2 n=1 n k 2 − m2 + iε µ ¶ Z d4 k 1 2 i 2 2 ln k − m − λφ + iε = 2 (2π)4 2 Z

(11.3.11)

11.3. ONE LOOP EFFECTIVE POTENTIAL IN λφ4 MODEL Recall we can get the same thing from: ½ Z ¶ ¾ µ Z 1 λφ2c i 4 2 2 Z(J) = dh exp d x h(x) p − m − h(x) ~ 2 2 µ ¶ λφ2c −1/2 2 2 p −m − = det 2 (1-loop 1PI contribution to Z if φc is a constant) = eiW

219

(11.3.12)

where we define µ

µ

¶¶ λφ2c p −m − iW = ln det 2 ¶¶ µ µ λφ2c i 2 2 ln det p − m − W = 2 2 Z = Γ(φc ) − dx φc J | {z } −1/2

2

2

(11.3.13)

=0

Thus,

Note:

¶¶ µ µ i λφ2c 2 2 Γ(φc ) = ln det p − m − 2 2

(11.3.14)

ln(det(O)) = ln(det(U † OU )) ; O = O †    λ1    ... = ln det   λn = ln (λ1 . . . λn ) = ln(λ1 ) + . . . + ln(λn )

(11.3.15)

and Tr(ln(O)) = Tr(ln(U † OU ))   λ1   ... = Tr ln 

 λn

= ln(λ1 ) + . . . + ln(λn )

 

(11.3.16)

Then, Tr(ln(O)) = ln(det(O))

(11.3.17)

220


So Γ(φc ) = = = =

µ µ ¶¶ λφ2c i 2 2 ln det p − m − 2 2 µ µ ¶¶ λφ2c i 2 2 Tr ln p − m − 2 2 µ ¶ Z i λφ2c 2 2 dp hp| ln p − m − |pi 2 2 µ ¶ Z d4 k λφ2c i 2 2 ln k − m − 2 (2π)4 2

(11.3.18)

Now, to compute V (φc ) in closed form.

11.4

Dimensional Regularization µ ¶ dn k λφ2c 2 2 ln k − m − + iε (2π)2 n 2! Z ∞ ¢ d(it) ¡ i(b+ia)t e − ei(a+iε) → ln(a + iε) − ln(b + iε) = it 0 i V (φc ) = 2

i.e.:

Z

Z ∞ d(it) i(a+iε)t d d ln(a + iε) = e (−1) da da 0 it Z ∞ 1 = − d(it) ei(a+iε)t a + iε 0 ¯∞ i(a+iε)t ¯ e ¯ = − ¯ a + iε ¯

(11.4.1) (11.4.2)

0

1 = a + iε

fL (a, b) ∂fL ∂a ∂fL ∂b ∂k ∂a

= fR (a, b) + k ∂fR = ∂a ∂fR = ∂b ∂k = =0 ∂b

(11.4.3)

(11.4.4) (11.4.5) (11.4.6) (11.4.7)

11.4. DIMENSIONAL REGULARIZATION

221

· µ ¸ ¶ ¡ 2 ¢ λφ2c dn k 2 2 2 ln k − m − + iε − ln k − m + iε (2π)n 2 Ã ¶ ! µ Z Z 2 2 −m2 − λφc +iε t i k d(it) dn k i 2 2 2 = ei(k −m +iε)t − e 2 (2π)n it ¶ µ Z ∞ Z 2 i d(it) i k2 −m2 − λφ2 c +iε t dn k = (11.4.8) e 2 (2π)n 0 it µ ¶ Z ∞ Not true strictly speaking Will often 1 d(it) i(H+iε)t =− e → it (11.4.2) is true see written H + iε 0

i V (φc ) = 2

Z

However, −s

a Γ(s) = Hence,

Z

dn k (2π)n

∞

dt ts−1 e−at

2

=

No angular comp.

Z

(11.4.9)

0

e|i(k{z+iε)t}

Thus, one loop i −V (φc ) = − 2

Z

d4 x Γ(0 . . . 0)

i (4πit)n/2

(11.4.10)

φn n!

µ ¶ Z λφ2 i −m2 − 2 c +iε t i ∞ d(it) i e = − 2 0 it (4πit)n/2 → Let it = u µ ¶ Z λφ2 − m2 + 2 c u 1 1 ∞ du e = 2 0 u1+n/2 (4π)n/2 ¶n/2 ³ n´ µ 1 λφ2c 2 = Γ − m + 2(4φ)n/2 2 2

(11.4.11)

¡ ¢ This diverges as n → 4 (Γ − n2 → Γ(−2) → Γ diverges at negative integers). So, let ε = 2 − n2 . ³ n´ Γ − = Γ(−2 + ε) 2 Γ(−1 + ε) = −2 + ε 1 Γ(ε) = (−2 + ε) (−1 + ε) (Γ(x + 1) = xΓ(x)) 1 → Γ(ε) = − γE + . . . ε

(11.4.12)

222

CHAPTER 11. SPONTANEOUS SYMMETRY BREAKING µ ¶µ ¶µ ¶2−ε ¶µ 1 1 1 1 1 λφ2c 2 V (φc ) = − γE + . . . m + 2 (4π)2−ε 2 − ε 1−ε ε 2 ¸· · ¸· ¸ ¢ 1 1 1 ¡ 1 2 1 + ε ln(4π) + O(ε ) = − γE + . . . 2 (4π)2 2 − 3ε + . . . ε ¸2 µ µ ¶ ¶ · λφ2c λφ2c 2 2 1 − ε ln m + + ... m + 2 2 µ µ ¶2 · ¶ ¸ 1 1 λφ2c 1 3 λφ2c 2 2 = m + + − γE + ln(4π) − ln m + + O(ε) (11.4.13) 4 (4π)2 2 ε 2 2

So, VTOT

Ã ! 2 4 m2 φ2c λφ2c (−1) λ φ c m4 + m2 λφ2c + + + · = 2 | {z } 2 4 64π 4 | {z } |{z} | {z } ∗

·

∗

∗∗

∗∗ µ ¸ ¶ 2 1 λφc 3 2 − ln m + + ln(4π) − γE + + O(ε) ε 2 2

(11.4.14)

So, to one loop order, µ ¶µ 2 ¶ −1 mλ m2R m2 = + 2 2 64π 2 ε ¶µ ¶ µ 2 1 λR −1 λ λ = + 2 4! 4! 4 64π ε

+

+

∴

finite VTot

+

λ2 φ4c = 256π 2

(11.4.15) (11.4.16)

+

µ µ 2¶ ¶ λφc ln +k µ2

+ ldots

(11.4.18)

(lim m2 → 0). Thus, as m2 → 0, V

(See Figure 11.4.1)

µ µ 2¶ ¶ λφc λ2 φ4c ln +k (φc ) = 256π 2 µ2 µ µ 2¶ ¶ λφ4c λ2 φ4c λφc Tot V = ln +k + 4! 256π 2 µ2

1 loop

(11.4.19) (11.4.20)

11.4. DIMENSIONAL REGULARIZATION

223

5 4 3 y 2

0

Figure 11.4.1: V c (φc ) =

1

–3

–2

–1

0

1 phi 2

λφ4c 4!

3

–1

µ µ 2¶ ¸ · ¶ λφc dV Tot (φc ) λφ3c λ2 3 4 1 4φc ln + k + φc = + dφc 6 256π 2 µ2 φc · µ µ 2¶ ¶ ¸ 2 λ λφc λ = φ3c 4 ln +k +1 + 6 256π 2 µ2 · µ 2 ¶¸ φc λ2 3 λ = φc + ln 2 6 64π µ02    if φc = 0  2 02 −(64π 2 /6λ) or φc = µ e = 0   2 = µ02 e−(32π /3λ) → Changes in µ are compensated for by changing λ. Mar. 13/2000 Recall:

Vtot

µ µ 2¶ ¶ f λ2 4 λφ4 +k + f ln = 4! 256π 2 µ2 µ 2¶ f λf 4 λ2 f 4 = + ln (f = φc from last time) 2 4! 256π µ02

µ µ 2 ¶¶ µ ¶ λ λ2 λ2 f 00 1 f ∂Vtot 3 = 4f + ln + ∂f 4! 256π 2 µ2 256π 2 f µ µ ¶ ¶ f2 λ 4λ2 λ2 3 + ln 0 = f + 6 256π 2 µ2 256π 2

(11.4.21)

224


If f = 0, or ln

µ

f2 µ2

¶

= = = = =

→ ln

µ

f2 µ02

¶

=

2 fmin =

At this value of fmin , we have,

µ ¶ λ 256π 2 λ2 − − 6 256π 2 4λ2 2 2 −42 3 π 1 − 4λ 4 −21 31 π 2 1 − 2λ 4 64π 2 1 − − 6λ 4 32π 2 1 − − 3λ 4 2 32π − 3λ ¾ ½ 32π 2 02 µ exp − 3λ

(11.4.22)

VTREE ¿ V1−loop

(11.4.23)

1 L = − Fµν F µν + (Dµ φ)∗ (Dµ φ) − m2 φ∗ φ − λ(φ∗ φ)2 4

(11.4.24)

Thus we are beyond the region where perturbation theory can be trusted. Note: For scalar electrodynamics (See Coleman and Weinberg).

VClassical = m2 φ∗ φ + λ(φ∗ φ)2 , Dµ = ∂µ + ieAµ

VTot = Vcl + V1−loop = Vtot (m2 , λ, e, f )

Here,

∂VTot = 0 at f = fmin (11.4.25) ∂f This fixes λ in terms of e, but introduces a new parameter fmin into the resting effective potential. “Dimensional Transmutation” (Coleman and Weinberg). Vtot to higher loop order             + + + +    +... +

Ã

+ ...

!



+



+ . . . + . . .

(11.4.26)

11.5. SPONTANEOUS SYMMETRY BREAKING IN GAUGE THEORIES

225

The loop expansion is an expansion in powers of ~. 1 m2 2 λφ4 L = (∂µ φ)2 − φ − 2 2 4! Z R 4 Z = Dφe(i/~) d x L Vertices ∼

1 → ~

Propagator ∼ ~

→−

(11.4.27) (11.4.28)

iλ ~

µ

∂ 2 + m2 ~

¶−1

For a given diagram ∼ (~)I−V But in a given diagram,

=

~ 2 | − {zm } k2

Mom. space

I → # of internal lines V → # of vertices L = # of loops = I − (V − 1) L = I −V +1

(11.4.29)

(11.4.30)

(Where the V is present because the δ-function at each vertex imposes a restriction, but the (−1) in the second line is because one of the δ-functions is superfluous due to “overall” δ-function). Thus, a given diagram ∼ (~)L−1 (11.4.31)

So, for example, the expression (11.4.26) will have the following powers of ~: µ ¶ ¡ ¢ ¡ ¢ 1 + ~0 + . . . + (~) + ~2 + . . . ~

11.5

(11.4.32)

Spontaneous Symmetry Breaking in Gauge Theories

Choice of gauge 1 2 L = − Fµν + (Dµ φ)∗ (Dµ φ) − m2 φ∗ φ − λ(φ∗ φ)2 4 φ(x) = V + φ0 (x) (V = V ∗ )

(11.5.1)

226


The kinetic energy for scalar: K.E. = [∂µ φ0∗ − ieAµ (V + φ0∗ )] [∂µ φ0 + ieAµ (V + φ0 )] = ∂φ0∗ ∂φ0 + e2 V 2 Aµ + ie(∂µ φ0∗ − ∂µ φ0 )V Aµ +O(φ3 ) | {z }

(11.5.2)

leads to mixed propagator

Lgf = −

1 (∂A)2 (gf = gauge fixing) 2α

(11.5.3)

(Feynman gauge). We can eliminate the mixed propagator by working in the Rξ gauge.

(recall previously, φ = by gauge transf.). L

(2)

+ Lgf

³

1 (∂ · A − iξeV (φ0∗ − φ0 ))2 2ξ If φ0 = φ1 + iφ2 1 = − (∂ · A − 2ξeV φ2 )2 2ξ

Lgf = −

(11.5.4)

Lgf

(11.5.5)

V√+η 2

´

eiρ/

√

2V

, where the imaginary (exponential) part is eliminated

³ ´ 1 2 2 2 2 2 µ = − (Fµν ) + (∂µ φ1 ) + (∂µ φ2 ) + e V Aµ A From (11.5.2) 4 1 + 2eV Aµ (∂µ φ2 ) − (∂ · A)2 + 2m(∂µ Aµ )φ2 2ξ 2 2 2 2 − 2ξ(eV ) φ2 + 2m φ1 (11.5.6)

The Aµ − φ2 propagator has disappeared. φ1 Propagator

i k 2 + 2m2

φ2 Propagator

i 2 k − ξm2

Aµ Propagator

³ −i gµν −

(11.5.7)

·

(eV )2 M = 2 2

kµ kν (1−ξ) k2 −ξM 2

k2 − M 2

´

¸

(11.5.8)

(11.5.9)

Mar. 15/2000 Recall: φ(x) = V + φ1 + iφ2

(11.5.10)

11.5. SPONTANEOUS SYMMETRY BREAKING IN GAUGE THEORIES i k 2 − 2µ2 i = 2 k − ξM 2 ´ ³ ν (1−ξ) −i gµν − kkµ2k−ξM 2 = 2 2 k −M

227

i∆1 =

(11.5.11)

i∆2

(11.5.12)

i∆µν Rξ gauge

Lgf = −

(11.5.13)

1 (∂ · A + ξM φ2 )2 2ξ

(11.5.14)

• In the limit ξ → 0 i∆1 =

k2

i (Higgs Field) − 2µ2

(11.5.15)

i (Serves to act as the longitudinal mode of the vector)(11.5.16) k2 ³ ´ kµ kν −i gµν − k2 (Purely Transverse) (11.5.17) = k2 − M 2

i∆2 = i∆µν

This is the “renormalized” (R) gauge. • If ξ → ∞, i∆1 =

k2

i∆2 = 0 i∆µν =

i − 2µ2

³ −i gµν −

(11.5.18) (11.5.19) kµ kν M2

k2 − M 2

´

µ

1 m2 − Fµν F µν − Aµ Aµ 4 2

¶

(11.5.20)

→ Unitary (“U ”) gauge. • ξ = 1 (Feynman-’tHooft gauge). i − 2µ2 i = 2 k − M2 −igµν = 2 k − M2

i∆1 = i∆2 i∆µν → Easy to calculate with.

Note that Physical effects are independent of ξ.

k2

(11.5.21) (11.5.22) (11.5.23)

228


Chapter 12 Ward-Takhashi-Slavnov-Taylor Identities The WTST identities are relations between different Green’s Functions that follow from gauge invariance. (We’ll look at Slavnov’s approach). Z 2 (12.0.1) I(a, b) = d3 x e−ax +b·x If

x → x + ε × x (ε ≈ 0) x2 → x2

Let

x = x0 + ε × x0 d3 x = d3 x0 , x02 = x2

(12.0.2a) (12.0.2b)

So, I(a, b) = ≈

Z

Z

d3 x0 e−ax

02 +b·(x0 +ε×x0 )

d3 x0 e−ax

02 +b·x0

(Re: ε ∼ 0)

(1 + b · ε × x0 )

(12.0.3)

(relabel x0 → x, and note that the exponent times the “1” in the last line = I(a, b).) Z 2 0 = d3 x e−ax +b·x (b · ε × x) Z 2 → 0 = b × d3 x xe−ax +b·x ¶ µ Z ∂ 3 −ax2 +bm xm ²ijk bj d x xk e ⇒0 = ∂ba Z 2 0 = d3 x (²iak xk + ²ijk bj xk xa ) e−ax +b·x (12.0.4) 229

230

CHAPTER 12. WARD-TAKHASHI-SLAVNOV-TAYLOR IDENTITIES

Perform analogous operation in the functional integral. 1 ¯ 6∂ − e A ¯ − 1 (∂ · A)2(12.0.5) L = − Fµν F µν + ψ(i 6 − m)ψ + Jµ Aµ + η¯ψ + ψη 2 Z4 R 1 2 ¯ (∂·A) i dx L +J·A+¯ η ψ+ ψη− ) ( cl 2α Z(Jµ , η¯, η) = dψ dψ¯ dAµ e (12.0.6) Use transformation analogous to (12.0.2a,12.0.2b). 1 A µ → A µ + ∂µ Ω e −iΩ ψ → e ψ ¯ iΩ ψ¯ → ψe

(12.0.7a)

(12.0.7b) (12.0.7c) 02 2 Lcl → Lcl (analogous to x → x ) (12.0.7d) ¯ ¯ dAµ dψ dψ → dAµ dψ dψ (Note demonstrating this NOT trivial) (12.0.7e) Thus,

Z(J, η¯, η) =

Z

( Z " ¯ iΩ η dAµ dψ dψ¯ exp i dx Lcl + J · (A + dΩ) + η¯e−iΩ ψ + ψe 1 − (∂A + ∂ 2 Ω)2 2α

#)

(12.0.8)

For Ω ≈ 0 (i.e. expanding exponent) #) ( Z " Z 1 ¯ − · (∂ · A)2 = dAµ dψ dψ¯ exp i dx Lcl + J · A + η¯ψ + ψη 2α ( Ã Z ¯ · 1 + dx iψ(x)Ω(x)η(x) − i¯ η (x)Ω(x)ψ(x) i − (∂ · A(x))(∂ 2 Ω(x) + iJµ (x)∂ µ Ω(x)) |α {z }

!)

=0

Let

Ω(x) =

Z

d4 yDF (x − y)κ(y)

∂ 2 Ω(x) = κ(x)

(12.0.9) (12.0.10)

231 Thus,

0 =

Z

dAµ dψ dψ¯ ei

R

¯ η ψ+J·A− 1 (∂·A)2 ] dx [Lcl +ψη+¯ 2α

Z

"

¯ d4 x iψ(x)η(x)D F (x − y)

i ∂ −i¯ η (x)ψ(x)DF (x − y) + iJµ (x) µ DF (x − y) − (∂ · A(x))δ 4 (x − y) ∂x α " µ ¶ µ ¶ Z δ δ 4 0 = dx i η(x)DF (x − y) − i¯ η (x) DF (x − y) iδη(x) iδ η¯(x) # ¶ µ ∂ ∂ δ −i µ δ 4 (x − y) + Jµ (x) µ DF (x − y) · Z(J µ , η¯, η) ∂x iδJµ (x) ∂x

³ ψ→

δ iδ η¯(x)

#

´ ³ ´ δ . Acto on this with iδJν (x) and then let Jµ = η = η¯ = 0.

0 =

Z i

"

∂ d x (−i) µ ∂x 4

µ

δ

δ iδJµ (x) iδJν (x) #

∂ DF (x − y)Z[0, 0, 0] ∂xν

¶

δ(x − y)Z[J, 0, 0] + (12.0.11)

Mar. 16/2000 From last time:

Z

=

Aµ → ψ → ψ¯ →

¶¾ ½ Z µ 1 2 ¯ + η¯ψ − dAµ dψ dψ¯ exp i dx Lcl + J · A + ψη (∂ · A) (12.0.12) 2α 1 ¯ (i 6 ∂ − e A A µ + ∂µ Ω Lψψ 6 − m) ψ ¯ = ψ e e−iΩ ψ ¯ iΩ ψe (12.0.13) Z

232 0 =

0 =

0 =

0 = But,

CHAPTER 12. WARD-TAKHASHI-SLAVNOV-TAYLOR IDENTITIES · R ¯ η ψ− 1 (∂·A)2 ) 1 i dx(Lcl +J·A+ψη+¯ ¯ − iΩ¯ ¯ 2α J · ∂Ω + iΩψη ηψ dAµ dψdψe e µ ¶¸ 1 1 2 − (∂ · A) ∂ Ω α e Z → Ω(x) = d4 yDF (x − y)κ(y) · Z Z 1 ∂ ¯ Jµ (x) µ DF (x − y) + iψ¯α (x)ηα (x)DF (x − y) dAµ dψdψ dx e ∂x ¸ R 1 ∂ −i¯ ηα (x)ψα (x)DF (x − y) − Aµ (x)δ(x − y) ei dx (Lcl +...) eα ∂xµ · δ ∂ δ 1 Jµ (x) µ DF (x − y) + i ηα (x)DF (x − y) − i¯ ηα (x) DF (x − y) e ∂x iδηα (x) iδ η¯α (x) ¸ δ 1 ∂ δ(x − y) Z − eα ∂xµ iδJµ (x) δ → × by and let J = η = η¯ = 0 δJν (z) ¯ ¸ · Z ¯ 1 ∂ 4 δ 1 ∂ δ ¯ Z[J, 0, 0] δ (x − y) gµν δ(x − z) µ DF (x − y) − dx ¯ ¯ e ∂x eα ∂xµ iδJµ (x) δJν (z) Z

¯ ¯ δ ¯ Z[J, 0, 0]¯ h0|T Aµ (x)Aν (y)|0i = ¯ iδJµ (x) iδJν (y)

J=0

δ

J=0

µ ¸ · ¶ Z 1 1 ∂ ∂ 4 4 ih0|T Aµ (x)Aν (y)|0iδ (x − y) − 0 = − dx + δ(x − z) ν DF (x − y) α ∂xµ α ∂x i ∂ ∂ 0 = − h0|T Aµ (y)Aν (z)|0i + ν DF (z − y) µ α ∂y ∂z ∂ i ∂ h0|T Aµ (y)Aν (z)|0i + ν DF (y − z) (12.0.15) 0 = α ∂y µ ∂y Thus if Z

d4 k ik·(y−z) e iΠµν (k) (2π)4 Z d4 k eik·(y−z) DF (y − z) = (2π)4 k 2 + iε

h0|T Aµ (y)Aν (z)|0i =

Hence 0=−

kµ (ikν ) Πµν (k) + 2 α k

(12.0.16) (12.0.17)

(12.0.18)

233

h0|T Aµ (y)Aν (z)|0iF T =

+

+

+

+ ...

= iΠµν (k)

(12.0.19)

Here, = Hence, 0=− ³ but kµ gµν −

And so,

kµ kν k2

´



³

³ −i gµν −

kµ kν (1−α) k2

k 2 + iε

kµ kν (1−α) k2

kµ  −i gµν − α k 2 + iε

´

(12.0.20)

´

 + ikν k2

= 0. Thus, 

 0 = kµ 

+



 + . . .

+

+

+ ...

must be proportional to gµν −

kµ kν k2

(12.0.23)

i.e.

+

+ . . . = (k 2 gµν − kµ kν )Π(k 2 ) = k 2 gµν Π1 (k 2 ) − kµ kν Π2 (k 2 ) + m2 Π3 (k 2 )gµν Must have → Π1 (k 2 ) = −Π2 (k 2 ) Π3 (k 2 ) = 0

234


Thus, all divergences are proportional to (gµν k 2 − kµ kν ). → No divergences proportional to gµν alone. If, 1 1 2 = (kµ Aν (k) − kν Aµ (k))2 Lcl = − Fµν 4 4 1 = Aν (k 2 gµν − kµ kν )Aν 2 → divergences and Lcl ∼ k 2 gµν − kµ kν . Thus all divergences in the vacuum polarization can be removed by a wave function renormalization. 1/2 Abare = Z3 ARenormalized (12.0.24) µ µ i.e. if Π1 6= −Π2 , Π3 6= 0, quadratic

log div. }| { z div. z}|{ ¢ z}|{ − kµ kν Π1 + (Π2 + Π1 ) kµ kν + m2 Π3 gµν {z } | {z } {z } | log div.

¡ = k 2 gµν |

divergence can be absorbed into Aµ (gµν k2 −kµ kν )Aν

• Π1 , Π2 → Log. div.

need Aµ ∂µ ∂ν Aν

(12.0.25)

m2R Aµ Aν

• Π3 → quadratic Recall: = (−1)(−ie)

2

Z

¸ · d4 ` i(6 `+ 6 k + m) i(6 ` + m) Tr 2 γµ γν (12.0.26) (2π)4 ` − m2 + iε (` + k)2 − m2 + iε

= appears to be quadratic divergent

(12.0.27)

Gauge invariance overrides simple power counting arguments. • Quadratic divergence disappears • Automatically satisfied in dimensional Regularization. Mar. 20/2000

12.1 Z

Dimensional Regularization with Spinors dn k i (k 2 )a ab 2 (n/2)+a−b Γ = (−1) (m ) (2π)n (k 2 − m2 )b (4π)n/2

¡n 2

¢ ¢ ¡ + a Γ b − a − n2 ¡ ¢ Γ n2 Γ(b)

(12.1.1)

12.1. DIMENSIONAL REGULARIZATION WITH SPINORS

235

with spinors, {γ µ , γ ν } = 2g µν ;

α η

γµ γ γ

= = = = = =

12.1.1

0 =

Take

−

"

g µµ = n

(12.1.2)

·

¸ γµ γ µ + γ µ γµ α − γ γµ γ µ 2gµα 2γ α − γ α γµ γ µ ¶ µ 1 µ µ α α (γµ γ + γ γµ ) 2γ − γ 2 1 2γ α − γ α (2g µµ ) 2 (2 − n)γ α →n=4 −2γ α

(12.1.3a) (12.1.3b)

Spinor Self-Energy

µ ¶ Z Z 1 ∂ δ δ δ ¯ − + ie dx ξ(x) DF (x − y) D (x − y) + ie dx ξ(x) F ¯ α ∂y µ iδJ µ (y) iδξ(x) iδ ξ(x) # Z ∂ ¯ (12.1.4) + dx Jµ (x) µ DF (x − y) Z[Jµ , ξ(x), ξ(x)] ∂x

δ δ ¯ δξ(w) δ ξ(z)

of (12.1.4) ξ = ξ¯ = Jµ = 0. We get,

h i 1 ∂ (0) (0) ¯ ¯ h0|T Aµ (x)ψ(y)ψ(z)|0i = e DF (x − y) − DF (x − z) h0|T ψ(y)ψ(z)|0i {z } | α ∂xµ

(12.1.5)

SF (y−z)

(0)

(Note that ¤DF (x − y) = δ(x − y)). In F.T. space, if ¯ h0|T Aµ (x)ψ(y)ψ(z)|0i = e

Z

dx0 dy 0 dz 0 DF (x − x0 )SF (y − y 0 )Λµ (x0 , y 0 , z 0 )SF (z 0(12.1.6) − z)

F → ∂µx Dµν (x − x0 ) = α

∂ (0) DF (x − x0 ) ∂xν

236


→ amputating external legs:

=

→ (p − p0 )Λµ (p, p0 ) =

In the limit p → p0 , Λµ (p, p) =

¡

¢ ¡ ¢ SF−1 (p0 ) − SF−1 (p)

∂ −1 S (p) ∂pµ F

Ward Identity

(12.1.7)

At tree level, let Λµ = γµ SF−1 = 6 p − m

→

∂ −1 S = γµ = Λµ ∂pµ F

at tree level

(trivial at tree level to get Ward identity). One loop: Λµ =

Z

d4 k (−ieγλ ) (2π)4

µ

i 6 p+ 6 k − m

¶

(−ieγµ )

µ

i 0 6p − 6k − m

¶

(−ieγσ )

µ

¶ −ig λσ (12.1.8) k 2 − iε

SF = µ ¶ µ ¶ d4 k i −ig λσ (−ieγλ ) = (−ieγσ ) (2π)4 6 p+ 6 k − m k 2 − iε → Bj. and Drell Z

(12.1.9)

12.1. DIMENSIONAL REGULARIZATION WITH SPINORS

237

(One loop order, evident (12.1.7) holds also). In general, as 6 p → m, SF (p) → And vertex function,

Z2 6 p − m − Σc (p) ¡ ¢ Σc (p) → 0 as p2 → m2

u¯(p)Λµ (p, p)u(p) = Z1−1 u¯(p)γµ u(p)

(12.1.10)

(12.1.11)

Thus our Ward identity gives Z1 = Z2 . Tells us, if we compute a self-energy diagram,

+

+

1/2

ψBare = Z2 ψRenormalized ,

1/2

ABare = Z3 ARen. µ µ

(12.1.13)

And look at all vertex functions,

Z1 :

+

+

(12.1.14)

Z1 eRenormalized 1/2 Z3 Z2

(12.1.15)

(Z1 parameterizes ∞’s in diagrams). eBare =

¢ ¡ ¢ 1¡ B 2 ¯B 6 p − mB ψ − eB AB ψ¯B γ µ ψB + ψ ∂µ A B − ∂ A ν µ ν µ 4 ¡ ¢ ¡ ¢ 1 R 2 ¯R 6 p − mR + δm ψ R = − Z3 ∂ µ A R − ∂ A + Z ψ ν 2 ν µ 4 Z1 1/2 ¯R µ R − 1/2 eR (Z3 AR µ )Z2 ψ γ ψ Z3 Z2 (What we would expect from renormalization)

L = −

(12.1.16)

We also have Z1 = Z 2 Z1 eR eB = 1/2 Z3 Z2 −1/2 R

= Z3

e

(12.1.17)

(12.1.18)

238


Suppose we have 2 fermions (e− , proton): L = −

¢ ¡ ¢ 1¡ B 2 ¯B 6 p − mB ψ − eB AB ψ¯B γ µ ψB + ψ ∂µ A B − ∂ A ν µ ν µ 4 ¢ B ¡ µ B B B B ¯ B γ ΨB ¯ 6 p − M Ψ − e Aµ Ψ +Ψ | {z }

(12.1.19)

eB for proton

Renormalization of proton will be different. eB for proton → Ward identity insures Z1 = Z2 . (Lurie). i.e. eB same for e− , proton, → Ward identity ensures eR will be same { Z1 for e− 6= Z1 for p, but Z2 for e− 6= Z2 for p either}. This has the effect that if 2 different charged fields have the same bare charge, then they will have same renormalized charge.

12.2

Yang-Mills Theory

Mar. 22/2000 Z(Jµ ) =

Z

½ Z µ ¶¾ 1 4 2 DAµ ∆F (Aµ ) exp i d x LY M − (∂ · A) + J · A (12.2.1) 2α

∆F (A) = det(∂ · D(A)) , Dµab = ∂µ δ ab + gf apb Apµ Z −1 a ab b ∆F (A) = dΩ δ(∂AΩ ) , AaΩ µ = Aµ + Dµ (A)Ω Z 0 −1 Ω0 dΩ δ(∂ · AΩ Ω ) ∆F (A ) = 0

00

Ω But, AΩ = AΩ µ µ and

R

dΩ → integrate over all gauge transformations. Z 00 −1 Ω0 ∆F (A ) = dΩ00 δ(∂ · AΩ ) = ∆−1 F (Aµ )

(12.2.2) (12.2.3) (12.2.4)

(12.2.5)

Thus, ∆F (Aµ ) = ∆f (AΩ µ)

(12.2.6)

Aµ → A µ + Dµ Ω DAµ → DAΩ µ = DAµ

(12.2.7) (12.2.8)

In the functional integral for Z, let,

Thus, Z[J] =

Z

½ Z µ ¶¾ 1 Ω 2 Ω (∂ · A ) + J · A DAµ ∆F (Aµ ) exp i dx LY M − 2α

(12.2.9)

12.2. YANG-MILLS THEORY

239

For small Ω,

¶¾ ½ Z µ 1 Ω 2 (∂ · A ) + J · A · Z[J] = DAµ ∆F (A) exp i dx LY M − 2α ¸ · µ ¶ −1 (∂ · A)(∂ · DΩ) + J · (DΩ) · 1+2 2α " κb Z Z z }| { R 1 1 2 0 = DAµ ∆F (A)ei dx [LY M − 2α (∂·A) +J·A] d4 x − (∂µ Aµa (x)) (∂ν Dνab (A)Ωb (x)) α # Z

+ Jµa (x)Dabµ (A) Ωb (x) | {z }

(12.2.10)

M bc κc

If ∂µ Dabµ (A)M bc (x, y) = δ µ (x − y)δ ac

(12.2.11)

and

a

κ (x) =

Z

d4 y M ab (x, y)Ωb (y)

(12.2.12)

⇒ ∂µ Dabµ κb (x) = Ωa (x) Then, as Aaµ (x) →

0=

(Z

d4 x

·

1 − α

δ 1 , i δJµa (x)

µ

we have,

∂ δ µ ∂x iδJµa (x)

¶

¸

κa (x) +

Z

d4 x Jµa (x)Dabµ

µ

δ iδJ

¶

)

(M bc )−1 (x, y) κb (y) Z | {z } Messy

(12.2.13) From the resulting Ward identities, (WTST identities), it was proven that the coupling constant g could be renormalized by considering any vertex into which it enters, and the same result would arise.

240


i.e.

+

+

+ . . . (A − A − A vertex)

+

+ . . . (A − A − A − A vertex)

+

+ . . . (c − c¯ − A vertex)

+

+ . . . (ψ − ψ¯ − A vertex)

If gB = Zg gR , Zg is the same for all 4 vertices ⇒ Gauge invariance of Lagrangian has not been altered. → Renormalization doesn’t break gauge invariance.

12.3

BRST Identities

WSTS → Were derived by using a local, linear transformation on Aaµ that left LY M and ∆F (A) invariant. BRST identities involve global non-linear transformations which leave L Y M + Lghost + Lgf invariant (gf = gauge fixing).

Z[Jµa , η a , η¯a ] =

Z

( Z Ã 1 a DAaµ Dca D¯ ca exp i d4 x − (Fµν (A))2 − c¯a Dµab (A)∂ µ cb 4 !) 1 (12.3.1) (∂ · Aa )2 + Jµa Aaµ + c¯a η a + η¯a ca − 2α

Remarkably, Lef f = LY M + Lgh + Lgf

(12.3.2)

12.3. BRST IDENTITIES

241

is invariant under Aaµ → Aaµ + Dµab (A)cb ε (ε, c → Grassmann) 1 ca → ca − f abc cb cc ε 2 1 c¯a → c¯a + (∂ · Aa )ε α

(12.3.3) (12.3.4) (12.3.5)

(The Proof involves the Jacobi identity). To derive the BRST identities, let A → A + (Dc)ε,etc., in the functional integral, and to leading order in ε,

Z[J] =

Z

Ã ½ Z ¾" Z 4 dA dc d¯ c exp i dx (Lef f + J · A + η¯c + c¯η) 1 + d x Jµa Dabµ (A)cb ε

!# 1 a abc b c 1 a a + (∂ · A )c + η¯ f c c ε α 2 " ¶µ ¶ ¶ µ µ Z δ 1 ∂ δ δ δ 4 a ab 0 = d x Jµ (x)D + iδJ iδ η¯b (x) α ∂xµ iδJµa iδ η¯a (x) # µ ¶µ ¶ δ δ 1 abc a Z[j, η, η¯] − f η¯ (x) 2 iδ η¯b (x) iδ η¯c (x)

(12.3.6)

This can be streamlined: Any Green’s Fn.

δBRST Consider

z }| { h0|T O1 . . . ON |0i = 0

(12.3.7)

0 = δBRST h0|T Aaµ (x)¯ cb (y)|0i ¡ ¢ 0 = h0|T Dµac (A)cc (x)ε¯ cb (y) |0i µ ¶ 1 ∂ b a +h0| T Aµ (x) A (y)ε |0i α ∂y ν ν

1 ∂ h0|T Aaµ (x)Abν (y)|0i = h0|T Dµab (A)cc (x)¯ cb (y)|0i ν α ∂y

Typical of BRST identity → Relates two Green’s Functions. Note: h0|T Aaµ (x)Abν (y)|0i is transverse only if α = 0 (Landau gauge). Mar. 23/2000

(12.3.8)

242

12.4


Background Field Quantization L = Z[j] =

m2 2 λ 4 1 (∂µ φ)2 − φ − φ 2 2 4! ½ Z Z dφ exp i

(12.4.1)

d4 x (L(φ) + Jφ)

¾

(12.4.2)

write φ(x) = f (x) + h(x), (f (x) = Classical Background, and h(x) = Quantum Correction). ½ Z ¾ Z 4 ˜ J) = Dh exp i d x (L(f + h) + Jh) Z(f, (12.4.3) 1 m2 λ (∂µ (f + h))2 − (f + h)2 − (f + h)4 2 2 4! 2 m 2 λ 4 1 (∂µ f )2 − f − f + f (. . .) = 4! } | {z } |2 {z2

L =

+

With

Classical Lagrangian m2 h2 1 (∂µ h)2 −

|2

{z A

2 }

Can be neglected

−

λ λ (6f 2 h2 ) + (4f h3 + |{z} h4 ) | {z } |4! {z } 4! C

B

i − m2

A =

=

B =

→ 4 - pt.

p2

(12.4.4)

D

C =

D =

(12.4.5)

12.4. BACKGROUND FIELD QUANTIZATION

243

4 3 y2

Figure 12.4.1: φ = V + φ0 → V − background - (Just 2 2 4 some const. previously). Where V (φ) = m 2φ + λφ 4!

1

–4

–2

2 phi

4

–1 –2

Aside: Recall the Figure 12.4.1: If we consider only diagrams with external fields, we will reproduce Z(J). Let h → h − f ˜ in Z. ½ Z ¾ Z ˜ Z(f, J) = Dh exp i dx (L(h) + J(h − f )) R

= e−i dx Jf Z(J) ¾ W (J) = −i ln(Z(J)) ˜ (f, J) = −i ln(Z(f, ˜ J)) Generating functional for connected diagrams W

(12.4.6)

(12.4.7)

Thus,

³ R 4 ˜ W (f, J) = −i ln e−i d x

Jf

´

Z(J) ³ R 4 = −i ln (Z(J)) − i ln e−i d x Z = W (J) − d4 x Jf

δW (J) δJ Z Γ(Φ) = W (J) − d4 x JΦ

Jf

´

Φ ≡

(12.4.8)

(12.4.9) (12.4.10)

= Generating functional for 1PI diagrams

So also, ˜ ˜ ≡ δ W (J) Φ δJ Z ˜ ˜ ˜ Γ(Φ) = W (J) − d4 x J Φ µ ¶ Z δ W − dx Jf δJ = Φ−f ˜ +f ∴Φ = Φ

(12.4.11) (12.4.12)

˜ = Φ

(12.4.13)

244


So also Z ˜ ˜ ˜ Γ(f, Φ) = W (f, J) − dx J φ˜ Z Z 4 ˜ = W (J) − d x Jf − d4 x J Φ Z = W (J) − d4 x Φ = Γ(Φ)

(12.4.14)

˜ f˜, Φ) ˜ Γ(Φ) = Γ( or ˜ = Γ(f, ˜ Φ) ˜ Γ(f + Φ) ˜ =0 → if Φ ˜ 0) Γ(f ) = Γ(f,

(12.4.15)

Hence,

Γ(Φ)

˜ Φ) ˜ Γ(

L = L(φ)

˜ column, If we make the identification φ → f , we can ignore the diagrams on the left in Γ


245

and get the diagram on the right. Applying this to YM theory, 1 L = − (Fµν (V ))2 ½ Z ¾ Z4 Z[J] = dV ∆F exp i dx (L + Lgf + JA)

(12.4.16) (12.4.17)

Vµa (x) = Aaµ (x) + Qaµ (x) ½ Z ¾ Z a a ˜ DQ exp i dx (L(A + Q) + Lgf (A + Q) + J · Q) (12.4.18) Z[Aµ , Jµ ] = δVµa = Dµab (V )Ωb = ∂µ Ωa + gf abc Vµb Ωc This can become: ½ ¾ δAaµ = 0 (i) Choose Lgf so that this gauge invariance is broδQaµ = ∂µ Ωa + gf abc (Abµ + Qbµ )Ωc ken. ¾ ½ δAaµ = ∂µ Ωa + gf abc Aaµ Ωc Leave this intact. (ii) δQaµ = gf abc Qbµ Ωc Homerkemp gauge, 1 (Dµab (A)Qbµ )2 2α Leaves type (ii) gauge invariance unbroken. Lgf = −

(12.4.19)

˜ aµ , Q ˜ aµ ) = Γ(Aaµ + Aãµ ) Γ(A

(12.4.20)

˜ Φ) ˜ = Γ(f + Φ). in same way as in scalar case → Γ(f, ˜ → Let Q = 0. Thus, ˜ a , 0) Γ(Aaµ ) = Γ(A | {zµ }

(12.4.21)

∗

* - Invariant under

Mar. 27/2000 Recall: if we have

Aaµ

→

Aaµ

+

Dµab (A)Ωb .

Thus,

£ a ¤ £ a ¤ ˜ a ) = Γ(Tr ˜ Γ(A (Fµν (A))2 + Tr (Fµν (A))3 ) µ Lgf = −

1 (Dab (A)Qbµ )2 2α µ

(12.4.22)

(12.4.23)

246


then Z[Aaµ ] ⇒ invariant under δAaµ = Dµab (A)Ωb . This gauge fixing (Lgf ) breaks the invariance. δAaµ = 0 δQaµ

=

(12.4.24)

Dµab (A

+ Q)Ω

b

(12.4.25)

If we follow through with the standard way of deriving the ghost Lagrangian, we find that, ⇒ Lghost = −¯ ca Dµab (A)Dµbc (A + Q)cc (12.4.26) a a (A), etc. (i.e. another (A)Fµν (where c¯, c are Grassmann ghost fields). Z depends only on Fµν abc a b c would be ² Fµλ Fλν Fνµ → must construct Z out of gauge invariant quantities). But divergences, by simple power counting, can arise only in 2, 3 and 4 point functions. Z kµ kν ∼ quadratic divergent d4 k 2 2 (12.4.27) k k (kµ , kν at each vertex)

∼ Linear

+

Z

d4 k

kµ kµ kµ k2k2k2

(12.4.28)

∼ log

(12.4.29)

∼ log

(12.4.30)

The divergences in all three diagrams can only serve to renormalize a a Fµν F aµν → Fµν = ∂µ Aaν − ∂ν Aaµ + g²abc Abµ Acν

(12.4.31)

We can get divergences in (12.4.28,12.4.29) diagrams automatically if we know the divergences in (12.4.27). Hence, the divergences in Z(A) are of the form, ∞ X aj j=1

εj

a a Fµν (A)Fµν (A)

(12.4.32)


247

ex: ¢ ¢ ¡ ¡ a b c ²abc Fµλ Fλν Fνµ → ²abc ∂µ Aaν − ∂ν Aaµ (∂λ Aaν − ∂ν Aaλ ) ∂ν Aaµ − ∂µ Aaν ppp

Z

1 dk 2 2 2 k k k 4

µ

¶ Not divergent (12.4.33) no renormalization req.

a Fµν F aµν → one power of external momentum associated with it.

Z

Z kµ kν kµ kν kλ → d k 2 2 2 pλ → Could have d4 k 2 2 2 k k k k k k {z } {z } | | 4

∼ log div.

(12.4.34)

∼ linearly div.

From the two point function, Aaµ

Bare

1/2

= ZA Aaµ

Ren.

(12.4.35)

From the three point function, g Bare = Zg g Ren.

(12.4.36)

Thus Z[A] has a contribution ¢2 ∂µ Aaν Bare − ∂ν Aaµ Bare + g Bare ²abc Abµ Bare Acν Bare h i´2 ³ 1/2 1/2 = ZA ∂µ Aaν Ren. − ∂ν Aaµ Ren. + ZA Zg g Ren. ²abc Abµ Ren. Acν Ren. ¡

(12.4.37)

In order for this to remain gauge invariant, 1/2

ZA Zg = 1

(12.4.38)

1/2

So, ZA and Zg are not independent in background field quantization! But ZA can be determined from the 2 point function alone. Note also - the relation between 2, 3 and 4 point functions implied by the gauge invariance of Z(A) is the Ward identities in the context of background field quantization.

248

CHAPTER 12. WARD-TAKHASHI-SLAVNOV-TAYLOR IDENTITIES The Feynman rules are:

For ZA to one loop order   

+

→ Have the divergent parts ig 2 δ ab c2 10 2 ig 2 ab 1 2 (p g − p p ) + δ c2 (p gµν − pµ pν ) µν µ ν (4π)2 3ε (4π)2 3ε 11 c2 3 → β(g) = − g 3 (4π)2

(Details: See Abbott - Nuclear Physics 1981).

   (12.4.39) (12.4.40)

Chapter 13 Anomalies A theory has an anomaly if a symmetry of the classical Lagrangian is broken by quantum effects. “Most important one”: → Chiral anomaly for massless fermions. 1 1 L = − Fµν (A)F µν (A) − Fµν (B)F µν (B) + ψ¯ (i 6 ∂− A 6 −B 6 γ5 ) ψ 4 4

(13.0.1)

There are two invariances in this Lagrangian, 1. ψ(x) → eiΩ(x) ψ(x) provided Aµ → Aµ − ∂µ Ω(x) 2. ψ(x) → eiΩ5 (x)γ5 ψ(x) provided Bµ → Bµ − ∂µ Ω5 (x) i.e. Fµν (B) = ∂µ Bν − ∂ν Bµ → ∂µ (Bν − ∂ν Ω5 ) − ∂ν (Bµ − ∂µ Ω5 ) | {z } | {z } ∗

(13.0.2)

∗

* - cancel, provided derivatives commute with each other. ψ(x) → eiΩ5 (x)γ5 ψ(x) †

† −iΩ5 (x)γ5†

(13.0.3) † −iΩ5 (x)γ5

=ψ e ψ (x) → ψ e † † −iΩ5 (x)γ5 ¯ ¯ iΩ5 (x)γ5 ψ = ψ γ0 → ψ e γ0 = ψe

(13.0.4) (13.0.5)

Thus, ¯ iΩ5 γ5 [i 6 ∂ − (6B − 6 ∂Ω5 )γ5 ] eiΩ5 γ5 ψ ψ¯ (i 6 ∂− B 6 γ5 ) ψ → ψe ¡ iΩ5 γ5 ¢ −iΩ5 γ5 ¯ = ψe e [i (6 ∂ + i 6 ∂Ω5 γ5 ) − (6B − 6 ∂Ω5 ) γ5 ] ψ = ψ¯ (i 6 ∂− B 6 γ5 ) ψ Note:

¡ iΩ5 γ5 ¢ ¡ iΩ5 γ5 ¢ ¯ → ψe ¯ ¯ ψψ e ψ 6= ψψ

∴ Chiral gauge invariance is not respected by the mass term in a Lagrangian. 249

(13.0.6) (13.0.7)

250

CHAPTER 13. ANOMALIES In Euclidean space,

γµ = γµ† → {γµ , γν } = 2δµν (13.0.8) ¯ 6 ∂ − 6 B γ5 )ψ is not chirally gauge So ψ¯ = ψ † (not ψ¯ = ψ † γ 0 as in Minkowski), and so ψ(i ¯ invariant, but ψψ is. Mar. 29/2000 1 1 ¯ 6∂ − e A 6 −gB 6 )ψ L = − F 2 (A) − F 2 (B) + ψ(i 4 4 ψ → eieΩ(x) ψ Aµ → Aµ − ∂µ Ω ψ → eigΩ5 (x) ψ Bµ → Bµ − ∂µ Ω5 ∂µ j µ (x) = 0 ∂µ j5µ (x) = 0

¯ µψ j µ = ψγ ¯ µ γ5 ψ j5µ = ψγ

(13.0.9) (13.0.10) (13.0.11) (13.0.12) (13.0.13)

The Green’s Functions are external in the fields Aµ , Bµ . i.e. Gµ1 ,...µn ,ν1 ,...,νm (x1 . . . xn , y1 . . . ym ) = h0|T Aµ1 (x1 ) . . . Aµn (xn )Bν1 (y1 ) . . . Bνm (ym (13.0.14) )|0i ∂G ∂G = 0= (13.0.15) ∂xµi ∂yνj → Remembera CAµ C −1 = −Aµ → n is even , CBµ C −1 = +Bµ → m is even or odd. Examine h0|T Aµ (x1 )Aν (x2 )Bσ (y)|0i (13.0.16) In momentum space at one-loop order, 



        +       

or can draw

               

(13.0.17)

½ i(6 k+ 6 p) i 6k i(6 k− 6 p) d4 k Tr (−iγσ γ5 ) (−iγν ) 2 (iγµ ) Rµνσ (p, q) = (−1)(−ie) (−ig) 4 2 (2π) (k + q) k (k − p)2 µ ¶ ¾ i 6k i(6 k+ 6 p) i(6 k− 6 q) +(−iγσ γ5 ) (−iγµ ) (−iγν ) (13.0.18) 2 2 (k + p) k (k − q)2 2

Z

251 (the first term in the trace refers to the second diagram, and the second term to the first diagram). By using 6 a 6 b 6 c = a · b 6 c − a · c 6 b + b · c 6 a + i²αβγδ aα bβ cγ γ δ γ5 ¢ ¡ 0123 ² = +1 , γ5 = iγ 0 γ 1 γ 2 γ 3

→ can show that the two traces are identical. Examine: q ν Rµνσ (p, q). Note: µ ¶ ¶ µ (6 k+ 6 q) ν 6 k 1 ν 1 qν γ γ = qν (k + q)2 k 2 (6 k+ 6 q) 6 k 1 1 = ((6 k+ 6 q)− 6 k) (6 k+ 6 q) 6k µ ¶ 6k 1 = 1− (6 k+ 6 q) 6 k µ ¶ 1 1 = − 6 k (6 k+ 6 q)

(13.0.19)

We finally obtain ½ 1 1 1 1 d4 k Tr γσ γ5 γµ − γ σ γ5 γµ q Rµνσ (p, q) = −e g 4 (2π) 6 k (6 k− 6 p) (6 k+ 6 q) (6 k− 6 p) ¾ 1 1 1 1 γµ + γ σ γ5 γµ −γσ γ5 (13.0.20) (6 k+ 6 p) 6 k (6 k+ 6 p) (6 k− 6 q) ν

2

Z

Now, k → k + p in the first term in the trace, and k → k + p − q in the second term ? q ν Rµνσ (p, q) = 0 upon this shifting of variables. But, it is quadratically divergent (actually just linearly divergent because of ² αβγδ appearing upon taking the trace), so care must be taken when shifting variables. ex.

Figure 13.0.1: Plot of tan−1 (x + a) and tan−1 (x)

I=

Z

∞ −∞

¡ ¢ ? dx tan−1 (x + a) − tan−1 (x) = 0(shifting x → x − a) =

but (see figure 13.0.1) Z Λ+ ¡ ¢ lim dx tan−1 (x + a) − tan−1 (x)

Λ+ →∞ Λ− →−∞

Λ−

252

CHAPTER 13. ANOMALIES

Or tan−1 (x + a) = tan−1 (x) + a

a2 d 2 d tan−1 (x) + tan−1 (x) + . . . dx 2! dx2

(13.0.21)

Substitute this back into I · ¸ Z ∞ d a2 d 2 −1 −1 −1 −1 I = dx tan (x) + a tan (x) + tan (x) + . . . − tan (x) dx 2! dx2 −∞ ¸∞ · a3 d 2 a2 d −1 −1 −1 tan (x) + . . . tan (x) + = a tan (x) + 2! dx 3! dx2 −∞ ³³ π ³ π ´´ ´ = a − − + 0 + ... 2 2 = πa (13.0.22) R 4 In general, if d k f (k 2 ) is linearly divergent, then, Z Z £ ¤ ∂ 4 2 2 λ f (k 2 ) d k f ((k + a) ) − f (k ) = a d4 k ∂k λ ¯ Z ¯ ¯ = i dΩk k 3 a · k n f (k 2 )¯ by Gauss ¯2 k →∞ ¯ ¯ ˆ (k 2 )¯¯ = 2π 2 ia · kf (13.0.23) ¯2 k →∞ R R ( dΩk = 2π 2 in 4-d, recall in 3-d, dΩk = 4π). Go back to ½ Z 1 1 1 d4 k 2 Tr γσ γ5 γν γµ Rµνσ (a) = −e g 4 (2π) (6 k+ 6 a1 + 6 q) (6 k+ 6 a1 ) (6 k+ 6 a1 − 6 p) ¾ 1 1 1 γµ γν γσ γ5 (6 k+ 6 a2 + 6 p) (6 k+ 6 a2 ) (6 k+ 6 a2 − 6 q) · 2 1 1 1 2 (2π i) Rµνσ (a) − Rµνσ (0) = −e g lim Tr aλ1 k 2 k λ γσ γ5 γν γµ 4 2 (2π) k →∞ (6 k+ 6 q) 6 k (6 k− 6 p) ¸ 1 1 1 λ 2 λ + a2 k k γσ γ5 γµ γν (6 k+ 6 p) 6 k (6 k− 6 q) .. . e2 g = ²κµνσ (a1 − a2 )κ (13.0.24) π2 (13.0.25) aκ1 − aκ2 = Apκ + Bq κ ⇒ Rµνσ (a) = Rµνσ (0) +

e2 g ²κµνσ (Apκ + Bq κ ) π2

(13.0.26)

So, (by same analogy), by surface terms, q ν Rµνσ (p, q) =

¢ e2 g ¡ α β ² q p ναµβ π2

(13.0.27)

253 (does not appear to be = 0). Mar. 30/2000 Recall:

Rµνσ (a) − Rµνσ (0) =

e2 g ²κµνσ (Apκ + Bq κ ) π2

´ h ³γ γ γ γ γ (2π 2 i) σ 5 α µ β α β λ 2 q Rµνσ (0) = −e g lim (−p )k kλ Tr p k (2π)4 k2 →∞ k4 ³γ γ γ γ γ í σ 5 α µ β λ 2 α β α β − (p − q) k kλ Tr (q k − k p ) k4 e2 g = − 2 ²σαµβ q α pβ (unsure of sign - see (13.0.34)) π ν

2

In the same way, pµ Rµνσ (0) = −

e2 g ²σανβ pα q β π2

(13.0.28)

(13.0.29)

And lastly the axial vertex, (p + q)σ Rµνσ (0) = 0 Maybe can choose A, B so we get conservation of momentum at all 3 vertices. ¸ · e2 q κ κ qν Rµνσ (a) = qν Rµνσ (0) + 2 ²κµνσ (Ap + Bq ) π 2 £ ¤ eg ²σαµβ q α pβ + ²βµασ q α Apβ = 2 π ¤ e2 g £ α β = ² (q p )(1 + A) σαµβ π2

(13.0.30)

(13.0.31)

So also,

¸ e2 g κ κ pµ Rµνσ (a) = pµ Rµνσ (0) + 2 ²κµνσ (Ap + Bq ) π 2 eg e2 g = − 2 ²σανβ pα q β + 2 ²βανσ pα q β B π π e2 g = ²σανβ pα q β (−1 − B) (unsure of signs - see (13.0.34)) (13.0.32) π2 ·

254

CHAPTER 13. ANOMALIES

And, σ

(p + q) Rµνσ (a) = (p + q) = 0+

σ

·

e2 g Rµνσ (a) + 2 ²κµνσ (Apκ + Bq κ ) π

e2 g ²κµνσ (Apκ q σ + Bq κ pσ ) π2

e2 g = ²κµνσ (A − B)pκ q σ 2 π It should be: If

q ν Rµνσ (a) = 0 pµ Rµνσ (a) = 0

¸

(13.0.33)

(13.0.34)

then

e2 g ²µνλσ pλ q σ (13.0.35) 2 2π So, if we do have conservation of vector current, then axial current not conserved. i.e. If, (p + q)σ Rµνσ (a) = −

then ∂µ jsµ = e2

∂µ j µ = 0

(13.0.36)

²µνλσ F µν (A)F λσ (A) (4π)2

(13.0.37)

(Note: anomaly independent of mass). This is the modification of ∂ · j5 = 0 that is required to account for (p + q)σ Rµνσ 6= 0. ¯ µ γ5 ψ = 0 (13.0.38) ∂ µ ψγ Note: The presence of the anomaly accounts for the decay of the π 0 (π 0 → γγ). i.e. If ψ were massive,

But,

¯ µ γ5 ψ = 2mψγ ¯ 5ψ ∂ · ψγ ¤ £ ¯ µ γ5 ψ = ψ¯ [∂ µ γµ γ5 ψ] + ∂µ ψγ (i 6 ∂− A 6 −B 6 γ5 )ψ − mψ = 0 1 6 ∂ψ = [(6A+ B 6 γ5 )ψ + mψ] i

(13.0.39)

Similarly, ¤ 1£ † † ψ (6A + B 6 † γ5 ) + mψ † i £ ¤ ← 1 ¯ A− B 6 γ5 ) + mψ † ψ¯ 6 ∂ = − ψ(6 i ←

ψ† 6 ∂ = −

(13.0.40)

13.1. ANOMALIES

255 ¯ 5 ψ (pseudoscalar) π 0 ∼ ψγ 1 ¯ µ γ5 ψ ∂µ ψγ = 2m 1 = ∂µ j µ5 2m

→

(13.0.41)

→ 0 if ∂ · j5 = 0

(13.0.42)

(i.e. says π 0 cannot go to γ 0 γ 0 ). But the anomaly says this 9 0 if ∂ · j5 =

e2 ²µνλσ F µν F λσ (4π)2

(13.0.43)

If j5µ couples to a vector Bµ , and j5 is not conserved because of the anomaly, then renormalizability is lost. ψ¯1 (i 6 ∂)ψ1 + ψ¯2 (i 6 ∂)ψ2 − g ψ¯1 γµ γ5 ψ1 B µ + g ψ¯2 γµ γ5 ψ2 B µ ;

Bµ → Bµ − ∂µ Ω5

(13.0.44)

Anomalous contributions to ∂µ j5µ cancel in this case, and renormalizability is no longer a problem. In the standard model, having equal numbers of quarks and leptons ensures the cancellation of anomalies. Apr. 3.2000

13.1

Anomalies

1. “Adler Bardeen” theorem → Anomalies are exclusively one-loop effects. i.e. h0|T Vµ (x)Vν (y) Aσ (z) |0i | {z } | {z } Transverse

Anomaly

+ |

{z A

}

(13.1.1)

(13.1.2)

+ |

{z B

→ ∂µ j5µ =

e2 ²µνλσ F µν F λσ 16π 2

} (13.1.3)

256

CHAPTER 13. ANOMALIES All of the anomaly can be shown to be from the A graph only. No anomalous contribution comes from B, though they are needed in order to ensure conservation of e4 momentum at all three vertices. So, no (4π) 4 correction for B.

2. ½ Z ¾ i 4 Z(J, K) = DAµ DBµ DψD ψ¯ exp d x [Lcl + Lgf + A · J + B · J5 ] (13.1.4) ~ 1 1 ¯ p−m−eA Lcl = ψ(6 6 −gB 6 γ5 )ψ − F 2 (A) − F 2 (B) 4 4 Z

Aµ → A µ + ∂µ Ω , Bµ → Bµ + ∂µ Ω5 ,

ψ → eieΩ(x) ψ ψ → eigΩ5 γ5 ψ | {z }

(13.1.5) (13.1.6)

∗

* - this gauge transformation does affect measure → there is a non-trivial Jacobian associated with this gauge transformation. (Fujikawa showed that this Jacobian gives the “anomalous” term in the WTST identity for the axial current.) → anomaly arises in trying to make this gauge transformation in this infinite measure ¯ → integrating over all possibilities. → (DψD ψ)

Chapter 14 Instantons Instantons are generated in a 2 stage process. 1. Make Wick rotation to Euclidean space 2. Solve resulting equations of motion (“Instanton”)

14.1

Quantum Mechanical Example

ex: (see figure 14.1.1)

Figure 14.1.1: Classically, if in the lowest energy state at the bottom (one of the V (q) = 0 → ±q0 positions), particle would just sit there.

“Vacuum tunnelling”

V (q) = (q 2 − q02 )2

(14.1.1)

|q(t → −∞)i = |−q0 i |q(t → ∞)i = |+q0 i

(14.1.2) (14.1.3)

Compute transition (Q.M.) → Path integral. hqf |e−iHt |qi i 257

(14.1.4)

258

CHAPTER 14. INSTANTONS

is the amplitude for being in state |qi i at t = 0 and being in state |qf i at t = t. Path integral: ½ Z t ¾ Z q(t)=qf i −iHt Dq(t) exp hqf |e |qi i = dτ L(q(τ )) ~ 0 q(0)=qi 1 2 q˙ (τ ) − V (q(τ )) → L(q(τ )) = 2 1 2 = q˙ (τ ) − (q 2 − q02 )2 2 Let τ → −iτ . ½ Z t ¶¾ µ 2 Z q(t)=qf 1 q˙ −iHt 2 2 2 Dq(t) exp hqf |e |qi i = dτ − − (q − q0 ) ~ 0 2 q(0)=qi ½ ¾ Z q(t)=qf 1 Dq(t) exp − SE = ~ q(0)=qi µ 2 ¶ Z t q˙ 2 2 2 dτ + (q − q0 ) = SEuclidean SE = 2 0 Dominant path: path that satisfies Now then, if the

δSE δq(τ )

(14.1.6)

= 0 with the appropriate boundary conditions.

ti → −∞ tf → +∞

q(t) → −q0 q(t) → +q0

The solution to the equations of motion that satisfies the above is √ qcl (t) = q0 tanh( 2q0 t) The more general solution is

(14.1.5)

√ qcl (t) = qo tanh( 2q0 (t − t0 )) |{z}

(14.1.7) (14.1.8)

∗

* - localized at an “instant” t0 → “instanton”. We will now consider

hq(+∞) = +q0 |q(−∞) = −q0 i =

Z

D(δq) e−[SE (qcl +δq)]

δq = 0 at t → ±∞ → expand about qcl . ½ · Z Z hq(+∞) → +q0 |q(−∞) → −q0 i = D(δq) exp −SE (qcl ) − dτ (δq(τ ))2 = e−SE (qcl ) [1 + O(~)] √ q3 here SE (qcl ) = 4 2 0 3~ ( ¶ ¸) · µ 2 D S(q ) cl + ... = e−SE (qcl ) 1 + det (Dq(τ ))2

(14.1.9)

D 2 SE + ... (Dδq(τ ))2

¸¾

(14.1.10)

(14.1.11)

14.1. QUANTUM MECHANICAL EXAMPLE

259

Suppose we have 2 spinors X, Y . Consider two functions f0 and f1 . fi : X → Y (i = 0, 1)

(14.1.12)

f0 (x), f1 (x) are homotopic if there exists a function F (x, t) such that F (x, 0) = f 0 (x) and F (x, 1) = f1 (x) and F (x, t) is continuous in t in (0 ≤ t ≤ 1). Suppose we took X: Y :

0 < θ ≤ 2π Z; Z ∈ C ,

|Z| = 1

(14.1.13) (14.1.14)

Consider fα (θ) = ei(θ+α) fβ (θ) = ei(θ+β)

(14.1.15) (14.1.16)

Homotopic → F (x, t) = ei(θ+tβ+(1−t)α) → Homotopy connecting fα , fβ . But can also write two functions not homotopic. fα (θ) = ei(θ+α) fβ (θ) = ei(2θ+β)

(14.1.17) (14.1.18)

Not homotopic (still functions that map X → Y , but can’t continuously deform one into the other). We have in general different classes of functions characterized by integers “n”. These are of the form, fαn (θ) = ei(nθ+α) ;

n = ±1, ±2, . . . (sign indicates direction of winding.)

(14.1.19)

In general, if two functions have different n’s, Not homotopic. n → Pontryagin index (“Winding number”) Z 2π f 0 (θ) n = −i dθ f (θ) 0

(14.1.20)

Apr. 5/2000 Recall: X : → θ ∈ [0, 2π) Y : → Z, |Z| = 1 F (n) (θ) = einθ n = ±1, ±2, . . . Z 2π f 0 (θ) dθ n = −i (counts # of times each pt. in X mapped onto Y) f (θ) 0

260


Figure 14.1.2: X :→ (−∞, ∞)

We can let X be (−∞, ∞) provided +∞ and −∞ are mapped on to the same point. (See figure (14.1.2)). Consider, ¾ ½ iπx (14.1.21) f (x) = exp √ x2 + λ 2 This maps −∞ < x < ∞ onto Z where |Z| = 1. Z ∞ 1 f 0 (x) (Winding number for this mapping) n=− dx 2π −∞ f (x)

(14.1.22)

Two functions with different winding numbers have different homotopies. Consider the mapping from X = S3 to Y = SU (2). i.e. S3 → Set of all points with x21 + x22 + x23 + x20 = 1 = x20 + x2 (4-d sphere). SU (2) → set of 2 × 2 matrices U with U U † = 1. f (x0 , x) = x0 + ix · σ

(14.1.23)

→ (x0 + ix · σ)† (x0 + ix · σ) = (x0 − ix · σ)(x0 + ix · σ) = 1 (if x20 + x2 = 1) Winding number for this mapping is, Z 1 n=− dθ1 dθ2 dθ3 Tr {²ijk Ai Aj Ak } 24π 2

(14.1.24)

• Where θi are the polar angles characterizing S3 • and Ai = f −1 (θ) ∂θ∂ i f (θ) (Coleman’s lectures on Instantons). We now extend X to be the entire three dimensional Euclidean space with all points at ∞ identified with a single point. Here, ) ( iπx · π (14.1.25) f1 (x) = exp p 2 x0 + x2 + λ2

14.2. CLASSICAL SOLUTIONS TO SU (2) YM FIELD EQUATIONS IN EUCLIDEAN SPACE261 is such a mapping. The winding number is Z 1 d3 x ²ijk Tr {Ai Aj Ak } n=− 24π 2 where Ai (x) = f −1

∂ f ∂xi

(14.1.26)

(14.1.27)

Recall: Gauge theories, Aµ (x) → U −1 (Aµ (x) + ∂µ )U

(14.1.28)

(if Aµ originally 0 → would get pure gauge function → very similar to above).

14.2

Classical Solutions to SU (2) YM field equations in Euclidean Space

SY M

¶ 1 a aµν = d x − Fµν (A)F (A) 4 i i Fµν = ∂µ Aν − ∂ν Aiµ + ²ijk Ajµ Akν Z

4

µ

(14.2.1)

Equations of motion: j Dµij (A)Fµν (A) = 0

→ Dµij = ∂µ δ ij + ²imj Am µ

(14.2.2)

Bit of a short-cut to solve this (not in every case). “Bianchi Identity”. 1 i ²µνλσ Fλσ 2 = 0

i ∗Fµν =

i → Dµij ∗Fµν

(Dual)

(14.2.3)

Thus, if i i Fµν = ± ∗Fµν

(14.2.4)

(ex: in E.M. → would be analogous to requiring E = B) then, j j = ±Dµij ∗Fµν Dµij Fµν = 0

(14.2.5)

Let Aim = (²imk ∂k ∓ δmi ∂0 ) ln[f (x0 , xm )] Ai0 = ±∂i ln[f (x0 , xm )]

(14.2.6) (14.2.7)

262


(m → spacial index, i → group index). If F = ± ∗F , then f −1 ∂ 2 f = 0 n X f = i=1

(14.2.8) λ2i (x − xi )2

(“n Instanton solution”)

→ Size → λi position → xi

Perform a gauge transformation on the “one instanton” solution. A0µ = U −1 (Aµ + ∂µ )U

(14.2.9)

with ¾ ix · σ θ U = exp √ x2 + λ 2 ¶ µ nπ x0 → tan(θ) = √ − 2 x2 + λ 2 ½

(14.2.10) (14.2.11)

(n is integer, not the n from above). Note, as |x| → ∞ , A0i0 → 0 (not easy to show) ∂ Ωn x0 → +∞ , A0im → Ω−1 n ∂xm ∂ x0 → −∞ , A0im → Ω−1 Ωn−1 n−1 ∂xm h n oin −iπx·σ √ where Ωn = exp . x2 +λ2

(14.2.12)

Figure 14.2.1: I, II, III are all 3d spaces. In all 3 spaces, Aim → U −1 ∂m U (i.e. Aim is pure gauge function).

14.2. CLASSICAL SOLUTIONS TO SU (2) YM FIELD EQUATIONS IN EUCLIDEAN SPACE263 To show this, consider Fµν (A) ; Fµν → 0 as |xµ | → ∞. Consider U as a mapping from 3-d Euclidean space onto SU (2). Z 1 d3 x ²ijk Tr {Ai Aj Ak } (14.2.13) n = 24π 2 ∂ → Ai = U −1 i U ∂x In region III (figure 14.2.1); Z 1 nIII = d3 x ²ijk Tr {Ai Aj Ak } 2 24π III where i, j, k has one index “0” because we’re on one side of sphere = 0 (on III) (14.2.14) But, from (14.2.12), nI

Z 1 d3 x ²ijk Tr {Ai Aj Ak } = 24π 2 I = n ((i, j, k) are a mixture of 1,2,3 (no 0) in integral)

(14.2.15)

and nII

1 = 24π 2 = n−1

Z

II

d3 x ²ijk Tr {Ai Aj Ak } (14.2.16)

See figure 14.2.2.

Figure 14.2.2:

Figure 14.2.3:

264


Figure √ 14.2.4: q(t) q0 tan( 2q0 t) → instanton interpolated from −∞ → +∞.

= →

Apr. 6/2000 For completeness, One-instanton: x2 x4 ± ix · σ √ Ω−1 ∂µ Ω ; Ω = 2 2 x +λ x2 −1 → Ω ∂µ Ω (pure gauge) as |x| → ∞ Z 1 m= d3 x ²ijk Tr(Ai Aj Ak ) ; Ai = f −1 ∂i f 2 24π Aµ (x) =

Note: Kµ

∂µ K µ

¶ 2 (not gauge invariant) = 4²µνλσ Tr Aν ∂λ Aσ + Aν Aλ Aσ 3 . . . satisfies 1 = 2Tr {Fµν ∗Fµν } ; ∗Fµν = ²µνλσ Fλσ 2 µ

(14.2.17)

(14.2.18)

(14.2.19)

(14.2.20)

Thus, Tr

½Z

4

d x Fµν ∗Fµν

¾

Z

1 d 4 ∂µ K µ 2 Z 1 d 3 σµ Kµ = 2 S∞

=

(By Gauss’ Theorem)

(14.2.21)

But, If Aµ = Ω−1 ∂µ Ω, then,

Aµ → Ω−1 ∂µ Ω as |x| → ∞

© ª 4 Kµ = ²µνλσ Tr (Ω−1 ∂ν Ω)(Ω−1 ∂λ Ω)(Ω−1 ∂σ Ω) 3

Thus, ½Z ¾ Z © ª 4 1 4 d x Fµν ∗Fµν = d3 σµ ²µνλσ Tr (Ω−1 ∂ν Ω)(Ω−1 ∂λ Ω)(Ω−1 ∂σ Ω) Tr 2 S∞ 3 {z } | m (above)

(14.2.22)

(14.2.23)

(14.2.24)

14.2. CLASSICAL SOLUTIONS TO SU (2) YM FIELD EQUATIONS IN EUCLIDEAN SPACE265 Tr

½Z

4

d x Fµν ∗Fµν

¾

1 = 2

µ ¶ ¢ 4 ¡ 24π 2 m 3

= 16π 2 m

(14.2.25)

m is instanton #. Sectors of the vacuum are labelled by the winding number n. |ni

(14.2.26)

T1 |ni = |n + 1i

(14.2.27)

[T1 , H] = 0

(14.2.28)

Perform a gauge transformation But,

as H is gauge invariant. Thus the vacuum must be an eigenstate of T1 . Consider the “θ vacuum”. +∞ X |θi = einθ |ni

(14.2.29)

n=−∞

i.e.

T1 |θi =

∞ X

n=−∞

= e−iθ

einθ |n + 1i

∞ X

n=−∞ iθ

ei(n+1)θ |n + 1i

= e |θi

(14.2.30)

Thus we have a vacuum labelled by this parameter θ. Vacuum-vacuum transition. hθ0 |e−iHt |θi = δ(θ 0 − θ)hθ0 |e−iHt |θi ∞ ∞ X X 0 = hm|e−imθ e−iHt einθ |ni

(14.2.31)

m=−∞ n=−∞

=

X m,n

0

ei(nθ−mθ ) hm|e−iHt |ni

θ = θ0 from (14.2.31) Z B ∞ X R 4 iνθ e DAµ ei d x = ν=−∞

A

LY M

(m = n + ν)

(14.2.32)

266


−1 ([A = Aµ → Ω−1 n ∂µ Ωn at t = −∞] ⇒ n, [B = Aµ → Ωm ∂µ Ωm at t = +∞] ⇒ m) The dominant instanton for this transition has instanton number ν. ½Z ¾ 1 4 Tr d x Fµν ∗Fµν ν= 24π 2

hθ0 |e−iHt |θi =

∞ Z X

ν=−∞

m n

( Z Ã DAµ exp i d4 x LY M +

1 → LY M = − Fµν Fµν 4

θ Tr {Fµν ∗Fµν } 2 {z } |24π

(14.2.33) !)

(14.2.34)

(Violates Parity, CP)

θ → leads to effective term in action, which leads to the Parity, CP violation. θ < 10 −9 (electric dipole moment of neutron).

Quantum Field Theory

Quantum Field Theory

Quantum Field Theory