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- A. Zee

Solutions Manual for A. Zee, Quantum Field Theory in a Nutshell, 2nd Ed. Yoni BenTov

PRINCETON UNIVERSITY PRESS PRINCETON AND OXFORD

c Copyright 2012 by Princeton University Press Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540. In the United Kingdom: Princeton University Press, 6 Oxford Street, Woodstock, Oxfordshire OX20 1TW. press.princeton.edu All Rights Reserved ISBN (pbk.) 978-0-691-15040-6 The publisher would like to acknowledge the author of this volume for providing the cameraready copy from which this book was printed. Printed in the United States of America.

Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I

Motivation and Foundation I.2 Path Integral Formulation . . . . . . . . . . . . . I.3 From Mattress to Field . . . . . . . . . . . . . . . I.4 From Field to Particle to Force . . . . . . . . . . I.5 Coulomb and Newton: Repulsion and Attraction . I.6 Inverse Square Law and the Floating 3-Brane . . I.7 Feynman Diagrams . . . . . . . . . . . . . . . . . I.8 Quantizing Canonically . . . . . . . . . . . . . . . I.9 Disturbing the Vacuum . . . . . . . . . . . . . . . I.10 Symmetry . . . . . . . . . . . . . . . . . . . . . . I.11 Field Theory in Curved Spacetime . . . . . . . .

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1 1 6 11 14 16 17 21 26 32 37

II Dirac and the Spinor II.1 The Dirac Equation . . . . . . . . . . . . . . . . . . . . . . . . II.2 Quantizing the Dirac Field . . . . . . . . . . . . . . . . . . . . II.3 Lorentz Group and Weyl Spinors . . . . . . . . . . . . . . . . II.4 Spin-Statistics Connection . . . . . . . . . . . . . . . . . . . . II.5 Vacuum Energy, Grassmann Integrals, and Feynman Diagrams II.6 Scattering and Gauge Invariance . . . . . . . . . . . . . . . . . II.7 Diagrammatic Proof of Gauge Invariance . . . . . . . . . . . . II.8 Photon-Electron Scattering and Crossing . . . . . . . . . . . .

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39 39 41 42 55 56 57 68 71

III Renormalization and Gauge Invariance III.1 Cutting Off Our Ignorance . . . . . . . . . . . . . . . III.3 Counterterms and Physical Perturbation Theory . . . III.5 Field Theory without Relativity . . . . . . . . . . . . III.6 The Magnetic Moment of the Electron . . . . . . . . . III.7 Polarizing the Vacuum and Renormalizing the Charge III.8 Becoming Imaginary and Conserving Probability . . .

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74 74 75 78 80 81 90

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100 100 102 113 119 122 136

IV Symmetry and Symmetry Breaking IV.1 Symmetry Breaking . . . . . . . . . IV.3 Effective Potential . . . . . . . . . . IV.4 Magnetic Monopole . . . . . . . . . IV.5 Nonabelian Gauge Theory . . . . . . IV.6 Anderson-Higgs Mechanism . . . . . IV.7 Chiral Anomaly . . . . . . . . . . .

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V Field Theory and Collective Phenomena V.1 Superfluids . . . . . . . . . . . . . . . . . . . . . . V.2 Euclid, Boltzmann, Hawking, and Field Theory at V.3 Landau-Ginzburg Theory of Critical Phenomena . V.4 Superconductivity . . . . . . . . . . . . . . . . . . V.6 Solitons . . . . . . . . . . . . . . . . . . . . . . . V.7 Vortices, Monopoles, and Instantons . . . . . . . .

. . . . . . . . . . . . Finite Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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150 150 151 155 156 158 158

VI Field Theory and Condensed Matter 169 VI.1 Fractional Statistics, Chern-Simons Term, and Topological Field Theory . . 169 VI.2 Quantum Hall Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 VI.4 The σ Models as Effective Field Theories . . . . . . . . . . . . . . . . . . . 185 VI.5 Ferromagnets and Antiferromagnets . . . . . . . . . . . . . . . . . . . . . . 187 VI.6 Surface Growth and Field Theory . . . . . . . . . . . . . . . . . . . . . . . 194 VI.7 Disorder: Replicas and Grassmannian Symmetry . . . . . . . . . . . . . . . 194 VI.8 Renormalization Group Flow as a Natural Concept in High Energy and Condensed Matter Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 VII Grand Unification VII.1 Quantizing Yang-Mills Theory and Lattice Gauge Theory VII.2 Electroweak Unification . . . . . . . . . . . . . . . . . . . VII.3 Quantum Chromodynamics . . . . . . . . . . . . . . . . . VII.4 Large N . . . . . . . . . . . . . . . . . . . . . . . . . . . VII.5 Grand Unification . . . . . . . . . . . . . . . . . . . . . . VII.6 Protons Are Not Forever . . . . . . . . . . . . . . . . . . VII.7 SO(10) Unification . . . . . . . . . . . . . . . . . . . . .

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206 206 213 220 226 240 245 257

VIII Gravity and Beyond 271 VIII.1 Gravity as a Field Theory and the Kaluza-Klein Picture . . . . . . . . . . 271 VIII.3 Effective Field Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 VIII.4 Supersymmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 IX Part N 302 IX.1 N.2 Gluon Scattering in Pure Yang-Mills Theory . . . . . . . . . . . . . . . . 302 IX.2 N.3 Subterranean Connections in Gauge Theories . . . . . . . . . . . . . . . 311 X Appendix E: Dotted and Undotted Indices

318

Preface

These are the solutions to the problems in Quantum Field Theory in a Nutshell, 2nd Ed., that are not already solved in the back of the book. For most of the problems I provide a detailed solution, while for others I sketch the solution and provide a reference to the literature for further details. Some problems are intentionally open-ended, serving as more of an introduction to the literature rather than as a homework assignment. The goal of Zee’s text is not only to teach quantum field theory but also to facilitate the transition from student to researcher. I thank my colleagues and teachers for helpful discussions in preparing these solutions. In particular, I have benefitted from talking to Jen Cano, Gavin Hartnett, Kurt Hinterbichler, Josh Ilany, Yonah Lemonik, Eugeniu Plamadeala, Yinbo Shi, Joe Swearngin, Benson Way and Chiu-Tien Yu. I am grateful to Joshua Feinberg for providing some of the solutions. I am also indebted to the faculty at the University of California at Santa Barbara for their patience in answering my questions. In particular, I thank David Berenstein, Andreas Ludwig, Ben Monreal, Joe Polchinski, Mark Srednicki, and my thesis adviser, Tony Zee. Finally, I thank my editor Ingrid Gnerlich at Princeton University Press for providing the opportunity to undertake this project. I also thank my friends and family for their support during the completion of this work. Yoni BenTov Apr. 9, 2012

v

I I.2

Motivation and Foundation Path Integral Formulation

1. Verify (5) hqF |e

−iHT

Z |qI i =

Dq e i

RT 0

dt[ 12 mq˙2 −V (q)]

(5)

Solution: ˆ = 1 pˆ2 + V (ˆ q ). Let T = N , with N → ∞, → 0, T Start with the Hamiltonian H 2m fixed. In this way, split up the time evolution operator into N pieces: ˆ

ˆ

ˆ

ˆ

H e−iHT = e|−iH e−i{z ... e−iH} N copies ˆ

Use this decomposition in the transition amplitude hqF |e−iHT |q0 i, and insert one copy of the ˆ identity matrix between each pair of e−iH s: ˆ

ˆ

ˆ

hqF |e−iHT |q0 i = hqF |e−iH e−iH ... e−iH |q0 i ˆ

ˆ

ˆ

ˆ

ˆ

= hqF |e−iH 1(N −1) e−iH 1(N −2) e−iH ... e−iH 1(1) e−iH |q0 i The superscript is just a label to keep track of the fact that we have inserted N − 1 identity matrices. It is convenient to resolve each identity matrix in a complete set of position eigenstates: Z ∞

1(i) =

dqi |qi ihqi | −∞

ˆ 1 2 pˆ + V (ˆ q ) + O(2 ) between To see why, consider the matrix element of e−iH = 1 − i 2m two position eigenstates, |qi i and |qj i: 1 2 ˆ −iH hqi |e |qj i = hqi |1 − i pˆ + V (ˆ q ) + O(2 )|qj i 2m 1 2 = hqi |1 − i pˆ + V (qj ) + O(2 )|qj i 2m 1 2 = hq |e−i( 2m pˆ +V (qj )) |q i j

i

−iV (qj )

=e

1 2 −i 2m pˆ

hqi |e

|qj i

The re-exponentiation is justified by the Baker-Campbell-Hausdorff formula: 1

eA eB = eA+B+ 2 [A,B]+(higher commutators) 2

pˆ Set A = −i 2m and B = −iV (ˆ x) in the above to get: p ˆ2

p ˆ2

e−i 2m e−iV (ˆx) = e−i 2m −iV (ˆx)+O( 1

2)

So to O(2 ) → 0 we can perform the above manipulations. 1

2

Now insert an identity matrix to the right of e−i 2m pˆ , but this time resolve it in a complete set of momentum eigenstates: Z ∞ dp 1= |pihp| −∞ 2π This gives, for the above matrix element, the following: ˆ

1

2

hqi |e−iH |qj i = e−iV (qj ) hqi |e−i 2m pˆ 1 |qj i Z ∞ 1 2 dp −iV (qj ) −i 2m pˆ =e hqi |e |pihp| |qj i −∞ 2π Z ∞ 1 2 dp −iV (qj ) hqi |e−i 2m pˆ |pihp|qj i =e 2π Z−∞ ∞ dp −i 1 p2 e 2m hqi |pi hp|qj i = e−iV (qj ) | {z } | {z } −∞ 2π eiqi p e−iqj p Z ∞ dp −i 1 p2 +ip(qi −qj ) = e−iV (qj ) e 2m −∞ 2π Now do all of this for the original transition amplitude. This will require N − 1 resolutions of the identity in position eigenstates, as already indicated, and it will require N resolutions of the identity in momentum eigenstates. ˆ

hqF |e−iHT |q0 i = ˆ

ˆ

ˆ

hqF |e−iH 1(N −1) e−iH 1(N −2) ... 1(1) e−iH |q0 i = Z ∞ Z ∞ Z ∞ ˆ ˆ ˆ −iH −iH dq1 |q1 ihq1 | e−iH |q0 i hqF |e dqN −1 |qN −1 ihqN −1 | e dqN −2 |qN −2 ihqN −2 | ... −∞ −∞ −∞ Z ∞ ˆ ˆ ˆ ˆ dqN −1 ...dq1 hqF |e−iH |qN −1 ihqN −1 |e−iH |qN −2 ihqN −2 | ... |q2 ihq2 |e−iH |q1 ihq1 |e−iH |q0 i = −∞

2

∞

dpN −1 −i p2N −1 +ipN −1 (qF −qN −1 ) e 2m 2π −∞ Z ∞ dpN −2 −i p2N −2 +ipN −2 (qN −1 −qN −2 ) ˆ −iH −iV (qN −2 ) • hqN −1 |e |qN −2 i = e e 2m 2π −∞ ... Z ∞ dp1 −i p21 +ip1 (q2 −q1 ) ˆ −iH −iV (q1 ) e 2m • hq2 |e |q1 i = e 2π Z−∞ ∞ dp0 −i p20 +ip0 (q1 −q0 ) ˆ • hq1 |e−iH |q0 i = e−iV (q0 ) e 2m −∞ 2π ˆ −iH

• hqF |e

|qN −1 i = e

−iV (qN −1 )

Z

Note that we need N resolutions into momentum eigenstates instead of just N − 1: after inserting N − 1 resolutions of the identity into position eigenstates, we need one set of momentum eigenstates for each position “ket,” |q0 i, ..., |qN −1 i, including the initial position q0 over which we do not integrate. In any case, the amplitude is now: ˆ

hqF |e−iHT |q0 i = 2 2 Z ∞ p p0 −p0 (q1 −q0 ) −i 2m dp0 dpN −1 −i N2m−1 −pN −1 (qF −qN −1 ) dq1 ...dqN −1 ... e ... e e−iV (qN −1 ) ...e−iV (q0 ) 2π 2π −∞ Completing the square for each term in brackets will yield one-dimensional Gaussian integrals: 2m p2i 2 − pi (qi+1 − qi ) = pi (qi+1 − qi ) p − 2m 2m i 2 m2 m 2 = pi − (qi+1 − qi ) − 2 (qi+1 − qi ) 2m 2 2 m m qi+1 − qi pi − (qi+1 − qi ) = − 2m 2 Therefore, each of the momentum integrals gives: 2 Z ∞ Z pi 2 −pi (qi+1 −qi ) m dpi −i 2m 1 +i m2 qi+1−qi 2 ∞ e e dpi e−i 2m (pi − (qi+1 −qi )) = 2π −∞ 2π −∞ r 1 +i m2 qi+1−qi 2 π = e 2π (i/(2m)) r q 2 i+1 −qi m +i m 2 = e 2πi Again, we have N copies of this integral, so the transition amplitude is: r N Z ∞ q −q 2 F N −1 m q1 −q0 2 m ˆ +i m −iHT 2 hqF |e |q0 i = dq1 ...dqN −1 e ... e+i 2 ( ) e−iV (qN −1 ) ... e−iV (q0 ) 2πi −∞ r N PN −1 m qj+1 −qj 2 Z i −V (q ) m j j=0 2 e = dN −1 q 2πi 3

Remember that the limits N → ∞ and → 0 with T fixed are supposed to be implied throughout. With this in mind, define the integral over paths and go to a continuum notation: Z

Z

q(0) = q0 q(T ) = qF

N −1 X j=0

"

m 2

Dq ≡

qj+1 − qj

lim N →∞ →0 T fixed #

2

d

N −1

Z

T

− V (qj ) →

r

m 2πi

N

"

2

q

dt 0

m 2

dq dt

# − V (q)

In the above, qN ≡ qF . This yields the desired result: Z RT 1 2 −iHT hqF |e |q0 i = Dq ei 0 dt ( 2 mq˙ −V (q)) with q(t = 0) = q0 , q(t = T ) = qF

2. Derive (24) hxi xj ...xk x` i =

X

(A−1 )ab ...(A−1 )cd

(24)

Wick

Solution: Define the following N -dimensional integral: Z 1 T 1 ~T −1 ~ ~ Z[J~ ] ≡ dN x e− 2 ~x M~x+J·~x = Ce+ 2 J (M )J In the above, ~x and J~ are N -dimensional vectors, M is an N × N symmetric matrix, the superscript T denotes the transpose, and C is a constant that we set equal to 1. If you don’t like that, then assume that all expectation values in what follows are implicitly divided by the constant C. It is just a statistical normalization factor, the equivalent of ensuring that the sum of all probabilities equals 1 rather than some other number. With this in mind, the expectation value of x1 is: Z 1 T hx1 i = dN x e− 2 ~x M~x x1 Z 1 T ∂ ~ dN x e− 2 ~x M~x+J·~x |J=0 = ~ ∂J1 ∂ ~ |~ = Z[J] J=0 ∂J1

4

In general, for an n-point function, we have: hxi1 xi2 ... xin i =

∂ ∂ ∂ ~ |~ ... Z[J] J=0 ∂Ji1 ∂Ji2 ∂Jin

Take the case n = 4 to see how this works. To clean up the notation, define G ≡ M −1 and use the following shorthand: J · G · J ≡ J~ T GJ~ =

N X N X

Jα Gαβ Jβ ≡ Jα Gαβ Jβ

α=1 β=1

First, we have the result from doing the Gaussian integral: 1

Z[J] = e+ 2 J·G·J

Now take a derivative with respect to Ji4 : 1 ∂ + 1 J·G·J 1 ∂ e 2 = e+ 2 J·G·J (Jα Gαβ Jβ ) ∂Ji4 2 ∂Ji4 1 1 = e+ 2 J·G·J (δi4 α Gαβ Jβ + Jα Gαβ δβi4 ) 2 + 21 J·G·J Gi4 α4 Jα4 =e

The last line follows from the symmetry Gαβ = Gβα . Also, we relabeled the dummy index to α4 just to associate it with i4 for later convenience. Before proceeding, note that setting J = 0 here gives zero, which means that hxi4 i = 0. Now take another derivative, this time with respect to Ji3 : 1 ∂ + 1 J·G·J ∂2 Gi4 α4 Jα4 e 2 e+ 2 J·G·J = ∂Ji3 ∂Ji4 ∂Ji3 1 1 ∂ + 21 J·G·J ∂ Jα Gαβ Jβ Gi4 α4 Jα4 + e+ 2 J·G·J Gi4 α4 Jα =e ∂Ji3 2 ∂Ji3 4 1

= e+ 2 J·G·J (Gi3 α3 Jα3 Gi4 α4 Jα4 + Gi4 i3 ) Again, before proceeding try setting J = 0. This time, a non-zero piece is left over. We see therefore that hxi3 xi4 i = Gi4 i3 = (M −1 )i4 i3 . Now differentiate with respect to Ji2 : i 1 ∂ h + 1 J·G·J ∂3 e+ 2 J·G·J = e 2 (Gi3 α3 Jα3 Gi4 α4 Jα4 + Gi4 i3 ) ∂Ji2 ∂Ji3 ∂Ji4 ∂Ji2 1

1

= e+ 2 J·G·J (Gi2 α2 Jα2 ) (Gi3 α3 Jα3 Gi4 α4 Jα4 + Gi4 i3 ) + e+ 2 J·G·J (Gi3 i2 Gi4 α4 Jα4 + Gi3 α3 Jα3 Gi4 i2 ) 5

As in the case for just one derivative, setting J = 0 gives 0, which means that hxi2 xi3 xi4 i = 0. You can now see a general result for this generating function: hxi1 xi2 ... xin i = 0 if n is odd Having taken enough derivatives to see how this works, while taking the next and final derivative, with respect to Ji1 , keep only the terms that have no powers of J left over, so that you don’t have to bother keeping track of terms that will go to zero anyway. This gives: 1 ∂4 e+ 2 J·G·J |J=0 = Gi2 i1 Gi4 i3 + Gi3 i2 Gi4 i1 + Gi3 i1 Gi4 i2 ∂Ji1 ∂Ji2 ∂Ji3 ∂Ji4

On the right-hand side, all possible pairings of the indices {i1 , i2 , i3 , i4 } appear exactly once.

I.3

From Mattress to Field

1. Verify that D(x) decays exponentially for spacelike separation. Solution: The free-field propagator in (3+1) spacetime dimensions is Z d4 k e ik·x D(x) = (2π)4 k 2 − m2 + iε where k · x = k 0 x0 − ~k · ~x. Rewriting this in the form 0

D(x , ~x) =

Z

d3 k −i~k·~x e (2π)3

Z

∞

−∞

0 0

dk 0 e ik x , Ek ≡ 2π (k 0 )2 − (Ek − iε)2

q ~k 2 + m2

we see that the integral over k 0 can be obtained by the contour integral I I≡ C

0

dz e izx 2π z 2 − (E − iε)2

with simple poles at z = ±(E − iε). For x0 > 0 we can use a semicircular contour in the upper half plane to pick up the residue at z = −E + iε and obtain ! Z ∞ 0 0 0 0 dk e ik x e izx e−iEx I= = i lim = −i , x0 > 0 . 0 2 2 z→ −E z−E 2E −∞ 2π (k ) − (E − iε)

6

For x0 < 0 we can use a semicircular contour in the lower half plane to pick up the residue at z = +E − iε and, being careful to take into account the orientation of the contour, we obtain ! Z ∞ 0 0 0 0 e ik x e izx e+iEx dk = −i , x0 < 0 . I= = (−1)i lim 0 )2 − (E − iε)2 z→+E 2π (k z + E 2E −∞ Therefore the propagator is Z

0

D(x , ~x) = −i

d3 k 1 −iE|x0 |−i~k·~x e , E= (2π)3 2E

q ~k 2 + m2 .

Writing d3 k = 2π dk k 2 d cos θ we can perform the integral over cos θ: Z 1 Z 1 1 −i~k·~ x d cos θ e−ikr cos θ = d cos θ e = e−ikr − e+ikr −ikr −1 −1 where r ≡ |~x |. We have 1 D(x , ~x) = 2 8π r 0

Z

∞

dk √

0

√ k 0 2 2 e−i k +m |x | e−ikr − e+ikr k 2 + m2

Note that the integrand is even in k. Therefore Z ∞ √ k 1 −ikr +ikr 0 −i k2 +m2 |x0 | √ dk e − e . D(x , ~x) = e 16π 2 r −∞ k 2 + m2 √ Let k ≡ m sinh t so that k 2 + m2 = cosh t. We have Z ∞ m −im|x0 | cosh t −imr sinh t +imr sinh t 0 dt sinh t e e − e D(x , ~x) = . 16π 2 r −∞ From now on consider the spacelike coordinate vector xµ = (0, ~x) =⇒ xµ xµ = −r2 < 0. In other words, consider D(r) ≡ D(0, ~x). The above integral with x0 = 0 is almost in the form of the modified bessel functions, where the order-α modified bessel function of the second kind is defined as Z 1 −i 1 απ ∞ dt e−αt−ix sinh t , Kα (x) ≡ e 2 2 −∞ which is a real function as long as x is real and positive. In our case, we have x = mr and therefore m D(r) = −i 2 K1 (mr) . 4π r For mr 1, the function K1 (mr) behaves as r π K1 (mr) ≈ e−mr 2mr showing that D(r → ∞) ∼ e−mr → 0. Alternatively, for m → 0 we have K1 (mr) ≈

1 1 =⇒ D(r) ≈ −i 2 2 . mr 4π r

7

2. Work out the propagator D(x) for a free field theory in (1+1)-dimensional spacetime and study the large x1 behavior for x0 = 0. Solution: The propagator in (1+1) spacetime dimensions is Z d2 k e ik·x . D(x) = (2π)2 k 2 − m2 + iε Performing the integral over k 0 and setting x0 = 0 as in problem I.3.1 gives Z ∞ 1 1 dk √ e−ikx . D(0, x) = −i 2 2 4π −∞ k +m The substitution k ≡ m sinh t gives 1 D(0, x) = −i 4π

Z

∞

dt e−imx sinh t = −i

−∞

1 K0 (mr) 2π

where we have used the definition of the modified bessel function from problem I.3.1. For mr 1, we obtain r π −mr 1 D(x, 0) ≈ −i e ∼ e−mr 2π 2mr as in (3+1) dimensions. In contrast to (3+1) dimensions, for m → 0 we have K0 (mr) ≈ − ln(mr/2) − γ where γ ≈ 0.577.

3. Show that the advanced propagator defined by Z d4 k e ik(x−y) Dadv (x − y) = (2π)4 k 2 − m2 − i sgn(k0 )ε is nonzero only if x0 > y 0 . In other words, it only propagates into the future. [Hint: Both poles of the integrand are now in the upper half of the k0 -plane.] Incidentally, some authors prefer to write (k0 − iε)2 − ~k 2 − m2 instead of k 2 − m2 − isgn(k0 )ε in the integrand. Similarly, show that the retarded propagator Z d4 k e ik(x−y) Dret (x − y) = (2π)4 k 2 − m2 + i sgn(k0 )ε propagates into the past. 8

Solution: The advanced propagator is: Z µ eikµ x d4 k Dadv (x) = (2π)4 kν k ν − m2 − i sgn(k 0 )ε Z Z ∞ 0 0 0 ~ d3 k eik x e−ik·~x dk = (2π)3 −∞ 2π (k 0 )2 − |~k|2 − m2 − i sgn(k 0 )ε Z Z 0 0 eik x d3 k −i~k·~x ∞ dk 0 e = 0 0 (2π)3 k 0 + (ω~k + i sgn(k 0 )ε) −∞ 2π k − (ω~k + i sgn(k )ε) q

In the above, ω~k ≡ |~k|2 + m2 , O(ε2 ) terms are set to zero, and as explained in the book we take ε to be a generic positive infinitesimal, so for instance we write “2ε = ε” instead of bothering to define new symbols. Now evaluate this integral using the complex plane. Define the complex variable z ≡ k 0 + i β, where β is real, and consider the following integral: 0

Z

I(x ) ≡ c

0

eizx dz 2π [z − (ω + i sgn(Re{z}))] [z + (ω + i sgn(Re{z}))]

So far the contour c is unspecified.R The intention is to provide a contour such that integration along the real axis will yield the dk 0 that we’re actually interested in evaluating, and that the contributions along the imaginary parts of the complex plane will evaluate to zero. In anticipation of an appropriate choice of contour, consider the behavior of the integrand in I(x0 ) when z is pure-imaginary: z → iβ, where β is real. The exponential factor in the 0 numerator becomes e−βx . If x0 > 0, then this exponential goes to zero for β → +∞. If x0 < 0, then this exponential goes to zero for β → −∞. The denominator of the integrand is a polynomial, so if the exponential goes to zero then the whole integrand will go to zero. So now we’ve distinguished the two cases: for x0 > 0, we must close the contour in the upper half-plane, and for x0 < 0, we must close the contour in the lower half-plane. The relevant contours for this problem, which will yield the desired real-valued integral, are shown below:

9

Im z C>

Re z

Im z

Re z

C< c> is what we use for x0 > 0, and c< is what we use for x0 < 0. If you evaluate the integrand for the piece of either c> or c< that is on the real axis, you will recover the integral you are actually interested in evaluating (Re{z} = k 0 ). The ×s in the above diagram indicate the locations of the poles of the integrand. A pole is where the integrand goes to infinity, which in this case corresponds to a zero of the denominator. The pole for Re{z} > 0 is z = ω +i sgn(Re{z})ε = ω +iε. The pole for Re{z} < 0 is z = −ω −i sgn(Re{z})ε = −ω +iε. So both of the poles have positive imaginary parts and are therefore in the upper half-plane. The residue theorem says that if you choose a contour that encircles the locations of the poles of the integrand, then you will get some nonzero number for the integral. If, however, your contour does not encircle any of the poles, then you will get zero. For the purposes of this problem all you have to do is decide whether the integral is zero. From the above discussion, we have determined two facts: 1. All poles of the integrand are in the upper half-plane. 2. If x0 > 0, we close the contour in the upper half-plane. If x0 < 0, we close in the lower half-plane. Therefore, the integral is nonzero if x0 > 0, and the integral is zero if x0 < 0. Therefore, the advanced propagator is only nonzero if x0 > 0.

10

For the retarded propagator, everything is the same except the poles of the integrand are in the lower half-plane. So the retarded propagator is only nonzero if we close the contour in the lower half-plane, which we can only do when x0 < 0.

I.4

From Field to Particle to Force

1. Calculate the analog of the inverse square law in a (2+1)-dimensional universe, and more generally in a (D+1)-dimensional universe. Solution: Taking the analog of equation (1) on page 24 of the book, we have in (d + 1) dimensions: Z Z 1 iW [J] dd+1 x dd+1 y J(x)D(x − y)J(y) Z[J] ∝ e , W [J] = − 2 The free-field propagator D(x − y) is: Z D(x − y) =

dd+1 k −eik(x−y) (2π)d+1 −k 2 + m2 − i

As in the book, consider J(x0 , ~x) = δ (d) (~x − ~x1 ) + δ (d) (~x − ~x2 ), and look only at the cross term ∼ δ (d) (~x − ~x1 )δ (d) (~x − ~x2 ) in W [J]: W [J]cross Z Z Z Z Z d Z 0 0 0 ~ d k dk 0 −eik (x −y ) e−ik·(~x−~y) (d) 0 d 0 d δ (~x − ~x1 )δ (d) (~x − ~x2 ) = − dx d x dy d y d 0 2 2 2 ~ (2π) 2π −(k ) + |k| + m − i Z Z Z Z 0 0 0 ~ dk 0 −eik (x −y ) e−ik·(~x1 −~x2 ) dd k 0 0 = − dx dy (2π)d 2π −(k 0 )2 + |~k|2 + m2 − i Z Z Z Z 0 0 ~ −eik x e−ik·(~x1 −~x2 ) dd k dk 0 0 0 −ik0 y 0 = − dx dy e 0 2 2 ~2 (2π)d 2π {z } −(k ) + |k| + m − i | 2πδ(k 0 ) Z Z ~ dd k −e−ik·(~x1 −~x2 ) 0 =− dx (2π)d |~k|2 + m2 − i Z ~ dd k e−ik·(~x1 −~x2 ) = +T (2π)d |~k|2 + m2 Since iW = −iET , we have an expression for the potential energy: Z E=−

~

dd k e−ik·(~x1 −~x2 ) (2π)d |~k|2 + m2 11

Now set m to zero and use dimensional analysis: Z Z Z ikr du u d−3 iu 1 d−1 e d−3 ikr E ∼ dk k = dk k e = e ∼ d−2 2 k r r r The notation is intended to extract the dimensional information only; for instance, ~k · ~x = kr cos θ, etc, but for the purposes of this problem the angular information is not important. For d = 3, the above gives E ∼ 1/r, as expected. However, for d = 2 this gives E ∼ r0 = 1, which may naively suggest that the force between the sources is zero, but that is incorrect. Define r ≡ |~x1 − ~x2 | and consider the force between the sources: Z ~ d2 k e−i k·(~x1 −~x2 ) d dE =+ F =− dr dr (2π)2 |~k |2 Z 2π Z ∞ e−ikr cos θ 1 d dθ k dk = 2 4π dr 0 k2 Z ∞ Z 2π 0 1 dk = 2 dθ e−ikr cos θ (−ik cos θ) 4π 0 k 0 Z ∞ Z 2π −i = 2 du dθ e−iu cos θ cos θ 4π r 0 0 1 ∼ r The force is not zero; it goes as 1/r. The potential energy is therefore E ∼ ln r. However this is still not quite right, since the argument of the log must be dimensionless. So really we get E ∼ ln(r/r0 ) for some cutoff r0 , which explains our naive estimate that the potential energy is dimensionless. Let’s study the properties of this cutoff. Explicitly, the potential energy is: 1 E(r) = − 2 4π

Z 0

∞

dk k

Z

2π

dθ e−ikr cos θ

0

R 2π 1 The integral 2π dθ e−ikr cos θ = J0 (kr), where J0 (z) is the Bessel function of the first 0 kind of order 0 as a function of z. We may now use Mathematica’s Bessel function input R BesselJ[n, z] ≡ Jn (z) to plot the integral. Plotting the dimensionless function du J0 (u)/u gives:

12

Ù âu J0 HuLu

0.1

u 2

4

6

8

10

-0.1

-0.2

Identifying u = kr, we see that the integral over k is uneventful at high k but diverges as k → 0. This suggests we regularize the integral by imposing a cutoff kmin ≡ 1/a on the lower bound of integration. So really, the potential energy is: Z ∞ Z dk 2π 1 dθ e−ikr cos θ E(r) = − 4π 1/a k 0 Remember that in our units of ~ = c = 1, momentum has units of inverse length. If we take the potential energy as written above and change variables to u = kr, then the potential energy is: Z ∞ Z du 2π 1 dθ e−iu cos θ E(r) = − 4π r/a u 0 Now we see that the expression for the potential energy is more sensible: the reason we naively guessed that E does not depend on r is that we had improperly chosen the lower bound of the k integral to be 0. We also now see what the r0 is in E ∼ ln(r/r0 ): r0 = a = 1/kmin is the inverse of the minimum momentum, or equivalently the maximum separation between the particles that we are willing to consider. Finally, the double integral in the force is between the particles is

R∞

du

0

F =−

R 2π 0

1 . 2πr

13

dθ e−iu cos θ cos θ = −2π i, so the force

I.5

Coulomb and Newton: Repulsion and Attraction

P (a) (a) 1. Write down the most general form for a εµν (k)ελσ (k) using symmetry repeatedly. For example, it must be invariant under the exchange {µν ↔ λσ}. You might end up with something like AGµν Gλσ + B(Gµλ Gνσ + Gµσ Gνλ ) + C(Gµν kλ kσ + kµ kν Gλσ ) + D(kν kλ Gνσ + kµ kσ Gνλ + kν kσ Gµλ + kν kλ Gµσ ) + Ekµ kν kλ kσ P (a) (a) with various unknown A, ..., E. Apply k µ a εµν (k)ελσ (k) = 0 and find out what that implies for the constants. Proceeding in this way, derive (13). (13)

X a

(a)

ε(a) µν (k)ελσ (k) =

(Gµλ Gνσ + Gµσ Gνλ ) − 32 Gµν Gλσ k 2 − m2

Solution: The ingredients we have are kµ and Gµν (k) ≡ gµν − m12 kµ kν . G has the properties that Gµν = Gνµ and k µ Gµν = k µ gµν − m12 kµ kν = kν − m12 (m2 )kν = 0. (a)

Each polarization tensor εµν is symmetric under interchange of µ and ν, so we need fµνλσ = fνµλσ and fµνλσ = fµνσλ . This leads us to write: fµνλσ ⊃ A Gµν Gλσ + C Gµν kλ kσ + C˜ kµ kν Gλσ However, because the components of the polarization tensors are just ordinary numbers, we know that εµν ελσ = ελσ εµν , which means that f should be invariant under the interchange of pairs of indices, (µν) ↔ (λσ): fµνλσ = fλσµν . This sets C˜ = C. Another thing we can do is stagger the indices across Gs, by which I mean something involving the term Gµλ Gνσ . If we take that and interchange µ and ν, we get Gνλ Gµσ . We therefore have another term to add to f : fµνλσ ⊃ B (Gµλ Gνσ + Gνλ Gµσ ) We can also use the same idea but instead of using two Gs, use two ks and one G, such as kµ kλ Gνσ . If we take that and interchange µ and ν, we get kν kλ Gµσ . If we then interchange λ and σ, we get kν kσ Gµλ . Interchanging µ and ν again gives us something different: kµ kσ Gνλ . These four terms should be included in f : fµνλσ ⊃ D (kµ kλ Gνσ + kν kλ Gµσ + kν kσ Gµλ + kµ kσ Gνλ ) Finally, there is the possibility of multiplying together four momenta to yield a term fµνλσ ⊃ 14

E kµ kν kλ kσ . We therefore have: fµνλσ = A Gµν Gλσ + B (Gµλ Gνσ + Gνλ Gµσ ) + C (Gµν kλ kσ + kµ kν Gλσ ) + D (kµ kλ Gνσ + kν kλ Gµσ + kν kσ Gµλ + kµ kσ Gνλ ) + E kµ kν kλ kσ Now we need to fix the constants A, B, C, D and E. To do that, we use the fact that k µ εµν (k) = 0, which implies that k µ fµνλσ = 0. Recalling that k µ Gµν = 0 and k µ kµ = m2 gives the following: k µ fµνλσ = A(0) + B(0 + 0) + C(0 + m2 kν Gλσ ) + D(m2 kλ Gνσ + 0 + 0 + m2 kσ Gνλ ) + Em2 kν kλ kσ = 0 =⇒ Ckν Gλσ + D(kλ Gνσ + kσ Gνλ ) = 0 , E = 0 Multiplying by k ν implies C = 0, and multiplying by k λ implies D = 0. We are now left with: fµνλσ = A Gµν Gλσ + B (Gµλ Gνσ + Gνλ Gµσ ) To proceed we recall the other trace-free condition on the polarization tensor, namely that g µν εµν = 0. In anticipation of multiplying fµνλσ by g µν , we first compute the following:

µν

g Gµλ

1 =g gµλ − 2 kµ kλ m 1 = δλν − 2 k ν kλ m 1 =⇒ g µν Gµν = δνν − 2 k ν kν = 4 − 1 = 3 m µν

Therefore, the second trace-free condition implies: 1 ν 1 ν µν ν ν g fµνλσ = A(3)Gλσ + B δλ − 2 k kλ Gνσ + Gνλ δσ − 2 k kσ m m 1 ν 1 1 1 ν ν ν = 3A Gλσ + B δλ − 2 k kλ gνσ − 2 kν kσ + gνλ − 2 kν kλ δσ − 2 k kσ m m m m " # 2 1 1 1 = 3A Gλσ + 2B gλσ − 2 kλ kσ − 2 kσ kλ + (m2 )kλ kσ 2 m m m = 3A Gλσ + 2B Gλσ = 0 =⇒ A =

−2B 3

We now have the answer up to an overall normalization constant: 15

fµνλσ

2 = B − Gµν Gλσ + (Gµλ Gνσ + Gλµ Gσν ) 3

As suggested in the book, fix the constant B by imposing the normalization f1212 = 1 for all k. If this is supposed to hold for all k, then try k = 0. Since Gij (k = 0) = 1 for i, j = 1, 2, 3, we get: 1 = f1212 (0) = B G11 (0)G22 (0) = B =⇒ B = 1 We now have the desired result: 2 fµνλσ = − Gµν Gλσ + (Gµλ Gνσ + Gλµ Gσν ) 3

I.6

Inverse Square Law and the Floating 3-Brane

1. Putting in the numbers, show that the case n = 1 is already ruled out. 2 MPl(n+3+1) =

2 MPl

[MPl(n+3+1) R]n

Solution: To clarify what this question is asking: The original point of this compact extra dimension setup is to propose that the scale of quantum gravity in the (n+3+1)-dimensional space is actually similar to the weak scale, so that our guessing that the Planck mass is so high is just an artifact of measuring in (3+1) dimensions. So, really, we are interested in answering the question: “Can one extra dimension fix the hierarchy problem?” With that in mind, we set n = 1 and try MPl(5) ∼ 102 GeV. Using MPl ∼ 1019 GeV, we predict a value for the characteristic length scale of the proposed 5th extra dimension: 1 = MPl(5) R

MPl(5) MPl

2

2

∼ 10 GeV

102 1019

2

= 10−32 GeV = 10−23 eV ∼ 10−42 J

Remembering that ~c ∼ 10−26 J·m gets us from ~ = c = 1 units to SI units, we get: R→

R 1042 1042 ∼ =⇒ R ∼ 10−26 J · m = 1016 m ∼ 1 parsec ~c J J

We conclude that adding only one extra dimension is not a valid way to fix the hierarchy problem. 16

I.7

Feynman Diagrams

1. Work out the amplitude corresponding to figure I.7.11 in (24). Solution: Since this is a 3-loop amplitude, we will just write down the answer using the Feynman rules and then get the symmetry factor by looking at the diagram. Using the propagator i∆(p) = p2 −mi 2 +iε and the vertex −iλ, we have M= Z 1 d4 p d4 q d4 r [−iλ]4 [i∆(p)][i∆(k1 +k2 −p)][i∆(q)][i∆(p−q−r)][i∆(k1 +k2 − r)][i∆(r)] 4 4 4 S (2π) (2π) (2π) where S is the symmetry factor. Up to the symmetry factor, this matches (24). Symmetry factors arise in loop diagrams for the following reason. Consider the pictorial representation of the vertices and of the propagators with sources: 1

A

4

2

B

3

We normalize the vertex as L = − 4!1 λϕ4 so that the permutations of the labels 1, 2, 3, 4 on the “hands” contribute 4! terms and cancel out the 4!1 in the Lagrangian. We normalize the kinetic term as L = − 12 ϕ(∂ 2 + m2 )ϕ so that swapping the labels A, B on the “blobs” contribute 2 terms and cancel out the 12 in the Lagrangian. Suppose we have a diagram in which hands 1 and 2 eat blobs A and B respectively:

17

1,A

4 2,B

3

From our previous argument, this diagram takes into account the 2 terms from swapping hands 1 and 2, and also the 2 terms from swapping blobs A and B, leading to a total of 4 terms. But swapping 1 and 2 and simultaneously swapping A and B gives exactly the same diagram with which we started, so this bookkeeping overcounts by a factor of S = 2. Now instead of rearranging individual hands on a given vertex, and individual blobs on a given propagator, consider rearranging the vertices themselves and the propagators themselves. Consider the path integral Z R 4 Z(J) = Dϕ e i d x(L+Jϕ) " Z 4 #V X Z P ∞ ∞ X 1 1 δ 1 1 4 4 4 1 . −i d x λ i d x d y J(x)∆(x − y)J(y) = V ! 4! i δJ(x) P ! 2 P =0 V =0 The quantity in the first square brackets raised to the power V is the vertex diagram drawn above, and the quantity in the second square brackets raised to the power P is the barbell diagram drawn above. Thus we see (as explained in the main text) that a general diagram is given by pasting together a bunch of different vertices with a bunch of different propagator lines. The blobs δ (the J(x)) on the propagator lines get eaten by the hands (the δJ(x) ) on the vertices. Just as before, we might think that the V vertices can be arranged in V ! ways, and the P propagators can be arranged in P ! ways, so that the V !P ! cancel out of the path integral. But if exchanging two vertices and simultaneously exchanging two propagators gives you the same diagram you started with, then counting those two swaps separately would overcount the contribution of that term to the sum in Z(J). The mismatch between the naive counting the correct counting of matching derivatives to sources is the symmetry factor. The last thing to take note of is that symmetry factors 18

coming from external blobs are canceled once those blobs are given fixed external values. In other words, tree diagrams do not have symmetry factors. Now let us apply this reasoning to figure I.7.11, which we repeat below for convenience:

W X 4

3

1 C A D B

E G 2

F H

5

6 Y Z

Here we omit the momentum arrows and labels, since they play no role in counting the symmetry factor.1 We label internal lines by numbers, and “hands” on the vertices by letters. Suppose we swap the vertices labeled by (ABCD) and (EF GH). Can we swap the internal lines in such a way that we recover the original diagram? The answer is yes: swap lines 3 ↔ 4 and 5 ↔ 6, and then also the hands W ↔ X and Y ↔ Z. This contributes a factor of 2 to S. This is also the same as keeping the vertices fixed, but swapping the hands A ↔ B and E ↔ F , while also swapping the lines 1 ↔ 2. This contributes another factor of 2 to S. There is nothing else we can do, so we find S = 2 × 2 = 4. 1

In theories with complex scalar fields, the arrows can denote the flow of a conserved U (1) charge. In such cases, the arrows do play a role in determining the symmetry factor since the U (1) charge must remain conserved.

19

3. Draw all the diagrams describing two mesons producing four mesons up to and including order λ2 . Write down the corresponding Feynman amplitudes. Solution: The Lagrangian density is: L=

1 1 (∂ϕ)2 − m2 ϕ2 − λ ϕ4 2 4!

This yields the following propagator and vertex:

p =

−i −p2 +m2 −i

= −iλ The 2 → 4 scattering diagrams up to O(λ2 ) are:

1’ 1 p

2’ 3’ 4’

2

1’

1

2’ 3’

p 2

4’

There are also the diagrams with different permutations of the outgoing lines, which correspond to different momenta flowing in the internal propagator. Reading the first diagram from right to left, the corresponding amplitude is: −i λ2 iMfirst = [−iλ] [−iλ] = +i −p2 + m2 − i −p2 + m2 − i 20

In the first diagram, the propagator’s momentum is p = p1 + p2 − p40 . The amplitude for the second diagram has the same form as the first one, except p = p1 − (p10 + p20 ). So the total amplitude is, suppressing the iεs: 0

0

0

4 4 X 4 X X 1 1 + O(λ4 ) . M=λ + 2 2 2 2 0 0 0 −(p + p − p ) + m −(p − p − p ) + m 1 2 n 1 i j 0 j=10 n = 10 2

i=1 i0 0 only Z Z ∞ 1 dk 0 δ(k 0 − ωk ) f (k 0 , ~k ) = d3 k 2ωk −∞ Z 1 = d3 k f (ωk , ~k ) 2ωk

b. Consider Lorentz transformations that do not change the signRof the time component of a vector. The matrices Λ ≡ ∂x/∂x0 have determinant +1, so d4 k δ(k 2 − m2 )θ(k 0 ) is unchanged under a Lorentz transformation. The function f (k) has no Lorentz indices and therefore does not transform under Lorentz transformations. Therefore the whole integral R 4 2 d k δ(k − m2 )θ(k 0 )f (k) is Lorentz invariant. R If the integral d3 k 2ω1k f (ωk , ~k ) is to be Lorentz invariant, and if the function f (ωk , ~k ) is Lorentz invariant, then the only conclusion we can make is that the measure d3 k/(2ωk ) is also Lorentz invariant. c. The point is to find the new commutation relations if you do not keep the square root in the definition of the Fourier coefficients a~k and a~†k . Assume the relation [ak , a†p ] = δ(k − p) p then shift ak → ak / (2π)D 2ωk to get: [a~k , ap~ ] = (2π)D 2ω~k δ (D) (~k − p~) If you like, we can compute this explicitly as well. Using the square-root integration measure, the Fourier expansion of ϕ(~x, t) is: Z h i dD k ~ ~ p ϕ(~x, t) = a~k e−i(ωk t−k·~x ) + a~†k e i(ωk t−k·~x ) (2π)D 2ωk The canonical commutation relations are between ϕ and π ≡ ∂L/∂ ϕ˙ = ϕ, ˙ so we also need the Fourier expansion of π: 22

Z π(~x, t) = −i

dD k p (2π)D

r

i ωk h † i(ωk t−~k·~ −i(ωk t−~k·~ x) x) a~k e − a~k e 2

Now that we’ve taken the time derivative we may as well set t = 0 to simplify the calculation. We need to change basis from (ϕ, π) to (a, a† ), and to do that we can invert the Fourier transform: Z

D

−i~ p·~ x

d xe

dD k

Z ϕ(~x, 0) =

Z

p

i h ~ ~ dD x a~k ei(k−~p)·~x + a~†k e−i(k+~p)·~x

(2π)D 2ωk s Z i (2π)D h (D) ~ = dD k a~k δ (k − p~) + a~†k δ (D) (~k + p~) 2ωk s (2π)D = ap~ + a†−~p 2ωp

p Note that ωp = |~p |2 + m2 , so the sign of p~ does not affect ωp . The algebra goes through in exactly the same way for π except with the relative minus sign: r Z (2π)D ωp † D −i~ p·~ x d xe π(~x, 0) = −i ap~ − a−~p 2 Rearranging these for clarity, we have: s

Z 2ωp dD x e−i~p·~x ϕ(~x, 0) ap~ + = (2π)D s Z 2 † ap~ − a−~p = i dD x e−i~p·~x π(~x, 0) (2π)D ωp a†−~p

Organized in this way it’s pretty clear that what we want to do is to add the two equations and divide by 2. Once we have ap~ we can just Hermitian conjugate it to get ap†~ . Therefore: Z 1 ap~ = p dD x e−i~p·~x [ωp ϕ(~x, 0) + iπ(~x, 0)] (2π)D 2ωp The non-zero equal-time canonical commutation relation is [ϕ(~x, t), π(~y , t)] = iδ (D) (~x − ~y ). The other commutators are zero. Therefore we can compute [ap~ , a~k ]:

23

1 [ap~ , a~k ] = √ D (2π) 2 ωp ωk 1 = √ D (2π) 2 ωp ωk

Z

D

−i~ p·~ x

Z

d xe

Z

D

−i~ p·~ x

d xe

Z

~

dD y e−ik·~y [ ωp ϕ(~x) + iπ(~x) , ωk ϕ(~y ) + iπ(~y ) ]

~ dD y e−ik·~y 0 + iωp [ϕ(~x), π(~y )] +iωk [π(~x), ϕ(~y )] +0 | {z } | {z } (D) (D) i δ (~x − ~y ) −i δ (~x − ~y )

Z 1 ~ = dD x e−i(~p+k)·~x (−ωp + ωk ) √ (2π)D 2 ωp ωk 1 δ (D) (~k + p~) (−ωp + ωk ) = 0 since ω−~p = ω+~p = √ 2 ω p ωk To get a non-zero result, we’re going to have to change the sign on one term but not the other; but that’s exactly what Hermitian conjugation will do for us. So, without spending much extra effort we can change the sign in the eipx factor and the sign of ωp to get: 1 [ap~ , a~†k ] = √ δ (D) (~k − p~)(+ωp + ωk ) = δ (D) (~k − p~) 2 ωp ωk Finally, [ap†~ , a~†k ] = 0 by Hermitian conjugating [ap~ , a~k ] = 0.

3. For the complex scalar field discussed in the text calculate h0|T [ϕ(x)ϕ† (0)]|0i. Solution: We want to calculate h0|T [ϕ(~x, t)ϕ(~0, 0)]|0i, h0|T [ϕ(~x, t)ϕ† (~0, 0)]|0i and h0|T [ϕ† (~x, t)ϕ† (~0, 0)]|0i. For convenience, let’s write the Fourier expansions for ϕ and ϕ† again: d3 k † i(ω~k t−~k·~ −i(ω~k t−~k·~ x) x) a e + b e ~ ~k (2π)3 2ω~k k Z d3 k † +i(ω~ t−~k·~x) † −i(ω~k t−~k·~ x) k ϕ (~x, t) = a e + b~k e (2π)3 2ω~k ~k Z

ϕ(~x, t) =

The vacuum state is defined by a~k |0i = b~k |0i = 0. Since a commutes with b† , we see immediately that the two-point function of ϕ with itself is zero. Similarly, the two-point function of ϕ† with itself is zero. So we just have to calculate the two point function for ϕ with its Hermitian conjugate. For t > 0, we have:

24

Z 3 d k d3 p † † i(ω~k t−~k·~ x) x) −i(ω~k t−~k·~ a + b + b e h0|ϕ(~x, t)ϕ (~0, 0)|0i = h0| a e ~k p ~ |0i p ~ ~k (2π)3 2ωk (2π)3 2ωp Z Z d3 p d3 k ~ = h0|a~k ap†~ |0i e−i(ω~k t−k·~x) 3 3 (2π) 2ωk (2π) 2ωp | {z } 3 (2π) 2ωp δ (3) (~k − p~) Z d3 k ~ = e−i(ωk t−k·~x) 3 (2π) 2ωk †

Z

We also have, for the opposite ordering: Z Z d3 k d3 p † i(ω~k t−~k·~ † x) x) † ~ −i(ω~k t−~k·~ |0i + b e h0|ϕ (0, 0)ϕ(~x, t)|0i = h0| a + b a e ~k p ~ ~k p ~ (2π)3 2ωk (2π)3 2ωp Z Z d3 k d3 p ~ = ei(ω~k t−k·~x) h0|bp~ b~†k |0i 3 3 (2π) 2ωk (2π) 2ωp | {z } 3 (2π) 2ωp δ (3) (~k − p~) Z d3 k ~ = e i(ωk t−k·~x) 3 (2π) 2ωk So the time ordered product of the two fields is: h0|T [ϕ(~x, t)ϕ† (~0, 0)]|0i ≡ θ(t)h0|ϕ(~x, t)ϕ† (~0, 0)|0i + θ(−t)h0|ϕ† (~0, 0)ϕ(~x, t)|0i Z d3 k −i(ωk t−~k·~ x) i(ωk t−~k·~ x) θ(t)e + θ(−t)e = (2π)3 2ωk Z µ d4 k e ikµ x =i (2π)4 k 2 − m2 + iε

25

I.9

Disturbing the Vacuum

1. Choose the damping function g(v) = 1/(1 + v)2 instead of the one in the text. Show that this results in the same Casimir force. [Hint: To P sum the resulting series,Rpass to an integral P ∞ ∞ R∞ −(1+nξ)t = 0 dt e−t /(1 − eξt ). representation H(ξ) = − ∞ 1/(1 + nξ) = − n=1 n=1 0 dt e Note that the integral blows up logarithmically near the lower limit, as expected.] Solution: The function f (d) is f (d) =

∞ ∞ π X nπ 1 ∂ X nπ ng a = h a 2d n = 1 d 2 ∂a n = 1 d

R If g(v) = (1 + v)−2 , then h(v) = dv g(v) = −(1 + v)−1 up to some constant that we can set equal to zero. Define ε ≡ πa/d. Using the hint, we can rewrite the function f (d) as an integral: ∞ ∞ 1 ∂ X 1 1 ∂ X h (nε) = − f (d) = 2 ∂a n = 1 2 ∂a n = 1 1 + nε Z ∞ 1 ∂ X ∞ =− dt e−(1+nε)t 2 ∂a n = 1 0 Z ∞ ∞ X n 1 ∂ −t =− dt e e−εt 2 ∂a 0 n=1 Z ∞ −t 1 ∂ e =− dt 2 ∂a 0 1 − e−εt

Now we can Taylor expand the denominator, being careful to keep higher order terms: e−εt = 1 − εt +

1 1 (εt)2 − (εt)3 + O(ε4 ) 2 6

So, the denominator is: 1 1 1 − e−εt = +εt − (εt)2 + (εt)3 + O(ε4 ) 6 2 1 1 2 3 = εt 1 − εt + (εt) + O(ε ) 2 6 We’d like to get that series out of the denominator, so use the expansion (1 + δ)−1 = 1 − δ +

26

δ 2 + O(δ 3 ) with δ = − 12 εt + 16 (εt)2 + O(ε3 ): " # 2 Z ∞ 1 ∂ e−t 1 1 1 f (d) = − dt 1 − − εt + (εt)2 + O(ε3 ) + − εt + O(ε2 ) + O(ε3 ) 2 ∂a 0 εt 2 6 2 Z ∞ 1 ∂ e−t 1 1 =− dt 1 + εt + (εt)2 + O(ε3 ) 2 ∂a 0 εt 2 12 Z ∞ 1 1 1 ∂ −t 1 2 + + εt + O(ε ) dt e =− 2 ∂a 0 εt 2 12 Z ∞ 1 ∂ d 1 πat −t 2 =− dt e + + + O(a ) 2 ∂a 0 πat 2 12 d Z 1 ∞ d πt −t =− − 2 +0+ + O(a) dt e 2 0 πa t 12 d Notice that differentiating with respect to a kills the second term and turns the O(a) term into a finite piece; that is why we needed to goR to third order in the original expansion. There R∞ ∞ −t is one integral we have to do: 0 dt e−t t = R ∞0 dt−te = 1. The other integral is just some number that will cancel out, so call it I ≡ 0 dt e /t and proceed: f (d) = +

π Id − 2 2πa 24 d

We have taken the limit a → 0 to kill the finite a-dependent terms. Now we’re in business. Differentiate with respect to d to get: f 0 (d) =

π I + 2 2πa 24 d2

Just as in the book for the case of the exponential damping factor, the infinite term is independent of the distance d, so subtracting off f 0 (L − d) removes this term and leaves only the finite contribution. Remembering that the force is minus the finite piece of the above derivative, we get the same Casimir force as before: π F (d) = −f 0 (d) + lim f 0 (L − d) = − L→∞ 24 d2

2. Show that with the regularization used in the appendix, the 1/d expansion of the force between two conducting plates contains only even powers. Solution: Using the regularization given in on p. 74, we have: Z π X cα ∞ 1 −t f (d) = − dt e −1 2 α bα 0 1 − e−bα t/d 27

where bα =

π Λα

→ 0 regulate the series. The constants cα and bα are subject to the constraints X cα α

bα

=0,

X cα α

b2α

=0,

X

cα = 1 .

α

The force is F = −[f 0 (d) − (d → L − d)] with L → ∞, where in the limit L → ∞ the second term serves simply to cancel off the leading divergent piece. The derivative of f (d) is Z ∞ e bα t/d π X dt e−t t cα f (d) = − 2 2 . 2d α (e bα t/d − 1) 0 0

The function

x2 e x g(x) ≡ x = 12 x2 csch2 x 2 (e − 1)

is even in x. Therefore its Taylor series contains only even powers of x. The function we are interested in is x−2 g(x) with x = bα t/d, with the first term to be canceled by subtracting f 0 (L − d). Therefore the series is even in 1/d.

3. Show off your skill in doing integrals by calculating the Casimir force in (3+1)-dimensional spacetime. For help, see M. Kardar and R. Golestanian, Rev. Mod. Phys. 71: 1233,1999; J. Feinberg, A. Mann, and M. Revzen, Ann. Phys. 288:103, 2001. Solution: As given in the book, the energy per unit plate-area is given by adding up the energies of all the individual modes: X ε(d) = ω(kx , ky , kz ) kx ,ky ,kz

For a massless scalar field quantized in the x-direction, the dispersion relation is: r q nπ 2 + ky2 + kz2 ω(kx , ky , kz ) = |~k |2 = d Since ky and kz are continuous, the discrete sum indicated above should actually be an R y P integral: ky → dk , and the same for kz . Therefore, the energy per unit area between the 2π plates at x = 0 and x = d is: r Z ∞ Z ∞ X dky ∞ dkz nπ 2 ε(d) = + ky2 + kz2 2π 2π d −∞ n=1 −∞ As in the (1 + 1)-dimensional case, we’ve made a mistake in writing down this formula: 28

we haven’t accounted for the important physical fact that we can’t contain arbitrarily high frequencies within the plates. Regularizing with the function F (ω, a) = e−a ω , the energy per unit area between the plates at x = 0 and x = d is: r Z q ∞ Z ∞ X 2 dky ∞ dkz nπ 2 −a ( nπ +ky2 +kz2 2 2 d ) + ky + kz e ε(d) = lim a→0 2π −∞ 2π d n=1 −∞ First let’s do the integrals. Since the integrand depends only on the combination k 2 ≡ ky2 +kz2 , change to polar coordinates: r Z q ∞ Z 2π X 2 dφ ∞ dk k nπ 2 −a ( nπ +k2 2 d ) ε(d) = lim +k e a→0 2π 2π d n=1 | 0 {z } 0 =1

Let c ≡ nπ/d for convenience, and while doing the integral over k don’t forget that c is a function of n. Change integration variables to x2 ≡ c2 + k 2 =⇒ k dk = x dx: ε(d) = lim

a→0

∞ Z X n=1

∞

c(n)

∞

X dx x −a x 1 xe = lim 2π 2π a→0 n=1

Z

∞

dx x2 e−a x

c(n)

The integral can be done via integration by parts twice: Z ∞ Z ∞ −1 2 −a x ∞ 2 −a x −a x dx x e = x e |c − 2 dx x e a c c Z ∞ 2 −1 2 −ac −a x ∞ −a x −c e + x e |c − dx e = a a c −1 2 1 −ac 2 −ac −ac −c e + −c e − e = a a a 2 +1 2 2c + 2 e−ac = c + a a a To summarize, the energy per unit area is now: ∞ 1 X 2c(n) 2 nπ 1 2 c(n) + ε(d) = lim + 2 e−ac(n) , where c(n) ≡ a→0 2π a n=1 a a d The parameter a has dimensions of length, so to Taylor expand we should define a dimensionless ratio, α ≡ a/D, where D ≡ d/π. The π was included in the α purely for notational

29

convenience. In terms of these variables, c = n/D and a = αD, so the energy is: ∞ 2 1 1 X 2 2n n + + 2 e−αn ε(d) = lim 3 α→0 2πD α n=1 α α ∞ 2 1 2 X −αn 1 2 = ∂α − ∂α + 2 e lim 2π(d/π)3 α→0 α α α n=1 ∞ π2 1 X −αn 1 1 2 1 = 3 lim ∂ − ∂α + 2 e d α→0 α 2 α α α n=1 The sum is an infinite geometric series: ∞ X

−αn

e

∞ X

=

n=1

(e

−α n

) =

n=1

∞ X

(e−α )n − 1

n=0

1 e−α 1 = − 1 = = α −α −α 1−e 1−e e −1 Therefore, the energy is: 1 π2 ε(d) = 3 lim α→0 d α

1 2 1 1 ∂α − ∂α + 2 2 α α

1 α e −1

The tedious partial derivatives are: 1 −eα ∂α α = α e −1 (e − 1)2 1 e2α + eα ∂α2 = eα − 1 (eα − 1)3

The energy is now: π2 1 ε(d) = 3 lim α→0 d α

e2α + eα eα 1 + + 2 α α 3 α 2 2(e − 1) α(e − 1) α (e − 1)

The denominators go to zero as α → 0, but hidden within the divergent expression is a finite, physically meaningful result. To see that there is in fact a finite piece hidden in there, compute the Taylor series in α about α = 0 for the energy (don’t forget the 1/α prefactor). The answer is: π2 3 1 π2 2 ε(d) = lim 3 − − + O(α ) 3 α→0 d | {z } α4 2α3 720 d | {z } = 0 when α→0 independent of α

30

Recalling the definition α = a/d gives: 3 d4 π2 d3 − − a4 2 a3 720 d3 3d π2 1 − = lim π 2 − a→0 a4 2 a3 720 d3

π2 ε(d) = lim 3 a→0 d

The force per unit area between the pairs of plates at x = 0 and x = d and at x = d and x = L as a function of the location x = d of the middle plate is: ∂ ε(d) ∂ ε(L − d) ∂ εT (d) =− − ∂d ∂d ∂d ∂ ε(d) ∂ ε(L − d) ∂(L − d) =− − ∂d ∂(L − d) ∂d ∂ ε(d) ∂ ε(d) + =− ∂d ∂d

f (d) = −

d→L−d

The derivative of ε with respect to the distance d is: 2 3π ∂ ε(d) π4 = lim + a→0 ∂d a4 240 d4 The divergent piece is independent of the distance d; when we subtract the contribution to the force from the plate at x = L, the infinity arising from lima→0 1/a4 will cancel out: 2 3π ∂ ε(d) π2 = lim + ∂d d→L−d a→0 a4 240 (L − d)4 2 3π d π2 = lim + 1 + O a→0 a4 240 L4 L 2 3π → lim for L → ∞ a→0 a4 The force per unit area between two parallel conducting plates a distance d apart is: f (d) = −

π2 240 d4

The force is attractive, meaning that it takes effort to pull apart two conducting plates. Restoring the factors of ~ and c to give an idea for how large this force is gives: f (d) = −

m 4 π 2 ~c −27 ≈ − 1.3 × 10 N · m−2 240 d4 d 31

So the force (per unit area) between plates of size ∼ 1 m2 separated a distance d ∼ 1 mm = 10−3 m apart is ∼ 10−15 N. Since 1 N is about the gravitational force exerted by the Earth on an apple, the Casimir force is extremely small, as expected from the fact that it is an effect of quantum origin. Note: Each harmonic oscillator mode contributions 12 ω to the vacuum energy, so for a real scalar field, which has only one component, this answer should be divided by 2. We keep it as is because it is the usual result quoted for the electromagnetic field, which has 2 polarizations. Also, if we decide to use a complex scalar field instead of a real scalar field, then we have a theory with two real scalar fields and for that case the answer would be correct as written.

I.10

Symmetry

1. Some authors prefer the following more elaborate formulation of Noether’s theorem. Suppose that the action does not change under an infinitesimal transformation δϕa (x) = θA VaA [with θA some parameters labeled by A and VaA some function of the fields ϕb (x) and possibly also of their first derivatives with respect to x.] It is important to emphasize that when we say the action S does not change we are not allowed to use the equations of motion. After all, the Euler-Lagrange equations of motion follow from demanding that δS = 0 for any variation δϕa subject to certain boundary conditions. Our scalar field theory example nicely illustrates this point, which is confused in some books: δS = 0 merely because S is constructed using the scalar product of O(N ) vectors. Now let us do something apparently a bit strange. Let us consider the infinitesimal change written above but with the parameters θA dependent on x. In other words, we now consider δϕa (x) = θA (x)VaA . Then of course there is no reason for δS to vanish; but, on the other hand, know that since δS does vanish when θA is constant, δS must have the form R 4 we µ δS = d x J (x)∂µ θA (x). In practice, this gives us a quick way of reading off the current J µ (x); it is just the coefficient of ∂µ θA (x) in δS. Show how all this works for the Lagrangian in (3). (3)

L=

1 1 (∂ ϕ ~ )2 − m2 ϕ ~ 2 − λ(~ ϕ 2 )2 2 4

Solution: Consider the following real scalar field Lagrangian with global O(N ) symmetry: 1 1 1 L = ∂µ ϕα ∂ µ ϕα − m2 ϕα ϕα − λ(ϕα ϕα ) 2 2 4 The field index α runs from 1 to N . The global O(N ) symmetry is: O(N ) : ϕα → Rαβ ϕβ with RT R = 1 32

N (N −1)/2

Let {T A }A=1 be the generators of the Lie algebra of the group O(N ) in the N -dimensional (“defining”) representation. Given these generators, the N × N orthogonal matrix R can be written as: h A Ai Rαβ = eθ T

α

β

= δα β + θA (T A )αβ + O(θ2 )

The goal is to find the conserved current corresponding to the O(N ) symmetry by promoting the symmetry to a local one and finding the coefficient of ∂µ θA in the variation of the action. Afterwards, we’ll compare the result of this method to the result of the formal procedure for generating the Noether current and see that they are equivalent. The action is: Z 1 2 1 1 µ α α α 4 ∂µ ϕα ∂ ϕ − m ϕα ϕ − λ(ϕα ϕ ) S[ϕ] = d x 2 2 4 Consider a first-order change in the fields:

S[ϕ + δϕ] = Z 1 1 2 1 µ α α α 4 d x ∂µ (ϕ + δϕ)α ∂ (ϕ + δϕ) − m (ϕ + δϕ)α (ϕ + δϕ) − λ((ϕ + δϕ)α (ϕ + δϕ) ) 2 2 4 Z = S[ϕ] + d4 x ∂µ ϕα ∂ µ (δϕα ) − m2 ϕα δϕα − λϕβ ϕβ ϕα δϕα The O(N ) symmetry transformation is: ϕα → Rαβ ϕβ = ϕα + θA (T A )αβ ϕβ + O(θ2 ) Therefore, the corresponding first-order variation in the field is: δϕα = θA (T A )αβ ϕβ The symmetry is global, but we can find the conserved current by imagining that the symmetry were local and finding a term of the form: Z δS = d4 xj µ ∂µ θ If we use the explicit expression for δϕ in the first-order change in the action, we get:

33

Z

δS ≡ S[ϕ + δϕ] − S[ϕ] = d4 x ∂µ ϕα ∂ µ (δϕα ) + ... Z = d4 x ∂µ ϕα ∂ µ θA (T A )αβ ϕβ + ... Z = d4 x ∂µ ϕα (T A )αβ ϕβ ∂ µ θA We know the other terms must be zero because the action is invariant under global O(N ) transformations. Therefore, the conserved current corresponding to the global O(N ) symmetry is: jµA = ∂µ ϕα (T A )αβ ϕβ You might worry that the current doesn’t look hermitian, since (T A )T = −T A , but once you transpose everything you will also switch the order of the ∂ϕ and the ϕ, which means you have to integrate by parts and therefore generate a second minus sign to get back to the original form. So jµA is hermitian. To verify the result using Noether’s theorem, we compute the current directly: ∂L δϕα = ∂µ ϕα δϕα ∂(∂ µ ϕα ) = ∂µ ϕα θA (T A )αβ ϕβ

θA jµA =

So there it is, the same answer as obtained previously: jµA = ∂µ ϕα (T A )αβ ϕβ 4. Add a Lorentz scalar field η transforming as a vector under SO(3) to the Lagrangian in exercise I.10.3, maintaining SO(3) invariance. Determine the Noether currents in this theory. Using the equations of motion, check that the currents are conserved. Solution: The Lagrangian is L = Lϕ + Lη + Lϕη , where: 1 Lϕ = Tr ∂µ ϕ ∂ µ ϕ − m2 ϕϕ − λ [Tr(ϕϕ)]2 2 1 Lη = ∂µ η α ∂ µ ηα − µ2 η α ηα − g(η α ηα )2 2 Lϕη = κ ηα ηβ ϕαβ = κ ~η Tϕ ~η . 34

The index α runs from 1 to 3, and the indices are contracted with usual identity matrix.2 ϕ transforms under the 5-representation of SO(3), which can either be written as a 5-dimensional column vector or as a 3 × 3 symmetric traceless matrix. For this problem, we choose the latter description and write: A C D 1 1 E ϕ = ϕαβ = (ϕαβ + ϕβα ) − (ϕγδ δ γδ ) δαβ = C B 2 3 D E −(A + B) Note that the term Tr(ϕϕϕϕ) is not included, because in this case it is proportional to [Tr(ϕϕ)]2 . For further details about ϕ, see the solution to problem 9.3 in the book. The SO(3) transformation acts on ϕ and on η by multiplying each index by a 3 × 3 orthogonal matrix with determinant 1: ϕαβ → Rαγ Rβ δ ϕγδ ηα → Rαβ ηβ RT R = 1 , det R = 1 To find the Noether current, we need an infinitesimal version of the above transformations. If T A are the generators of the 3-dimensional representation of SO(3), where A runs from 1 to 3(3 − 1)/2 = 3, then the matrix R can be written: h A Ai β β β Rα = eθ T = 1 + θA T A + O(θ2 ) α = δα β + θA (T A )αβ + O(θ2 ) α

Therefore, the infinitesimal versions of the SO(3) transformations on ϕ and η are: i h ϕαβ → δα γ + θA (T A )αγ + O(θ2 ) δβ δ + θB (T B )β δ + O(θ2 ) ϕγδ h i = δα γ δβ δ + θA (T A )αγ δβ δ + δα γ (T A )β δ + O(θ2 ) ϕγδ h i A A γ A δ = ϕαβ + θ (T )α ϕγβ + (T )β ϕαδ +O(θ2 ) {z } | A = δϕαβ

ηα → [δα β + θA (T A )αβ + O(θ2 )] ηβ = ηα + θA (T A )αβ ηβ +O(θ2 ) | {z } = δηαA 2

In SO(3) there is no distinction between upper and lower indices. For this problem we use them interchangeably.

35

The Noether current is: ∂L ∂L δϕA δη A αβ + µ ∂(∂ ϕαβ ) ∂(∂ µ ηα ) α α A = ∂µ ϕαβ δϕA αβ + ∂µ η δηα h i = ∂µ ϕαβ (T A )αγ ϕγβ + (T A )β δ ϕαδ + ∂µ η α (T A )αβ ηβ

jµA =

Now we have to check that this current is in fact conserved, or in other words that ∂ µ jµA = 0. The equations of motion are: 2 ∂ + m2 + λ Tr(ϕϕ) ϕαβ + κ ηα ηβ = 0 2 ∂ + µ2 + g (~η · η~ ) δαβ + 2κ ϕαβ ηβ = 0 The important point is that ∂ 2 ϕαβ = a(ϕ) ϕαβ − κ ηα ηβ and ∂ 2 ηα = b(η) ηα − 2κ ϕαβ ηβ , where a and b are SO(3)-scalar functions of ϕ and η, respectively. Therefore, the 4-divergence of the current is: h i µ A αβ A γ αβ A δ ∂ jµ = a(ϕ) ϕ (T )α ϕγβ + ϕ (T )β ϕαδ + b(η) η α (T A )αβ ηβ − 2κ ηα [(T A )αβ ϕβγ + ϕαβ (T A )βγ ]ηγ Each of the three terms is zero individually by symmetry. To see why, look at the first term: ϕαβ (T A )αγ ϕγβ = (T A )αγ ϕγβ ϕαβ = (T A )αγ ϕγβ ϕβα = (T A )αγ (ϕϕ)γ α Now the symmetry argument comes in: {T A }3A=1 are the generators of 3-dimensional rotations (ϕϕ) is a symmetric matrix, so and are therefore antisymmetric matrices. Meanwhile, Tr (T A )(ϕϕ) = 0. Similarly, η α (T A )αβ ηβ = Tr (T A )(ηη) = 0. For the third term, ~η T (T A ϕ + ϕT A )~η , we have ϕ = ϕT and T A = −(T A )T . Therefore: ~η T T A ϕ ~η = ηα (T A )αβ ϕβγ ηγ = ηγ ϕβγ (T A )αβ ηα = ηγ [+ϕγβ ][−(T A )βα ]ηα = −~η Tϕ T A ~η . Therefore, ∂ µ jµA = 0. The current is conserved.

36

I.11

Field Theory in Curved Spacetime

1. Integrate by parts to obtain for the scalar field action Z √ √ 1 1 µν 2 4 ∂µ −g g ∂ν + m ϕ S = − d x −g ϕ √ 2 −g and write the equation of motion for ϕ in curved spacetime. Discuss the propagator of the scalar field D(x, y) (which is of course no longer translation invariant, i.e., it is no longer a function of x − y). Solution: The equation of motion that follows from the above action is √ 1 √ ∂µ −g g µν ∂ν + m2 ϕ = 0 . −g We will specialize to the maximally symmetric de Sitter spacetime, which has scalar curvature R = n(n − 1)/a2 in n spacetime dimensions. Here a is the de Sitter length. We will follow B. Allen and T. Jacobson, “Vector Two-Point Functions in Maximally Symmetric Spaces,” Commun. Math. Phys. 103, 669-692 (1986). Although the scalar propagator D(x, y) is no longer a function of x − y, it is still a function of x and y only through the geodesic distance between the points x and y: 1

Z µ(x, y) = 0

dX µ (λ) dX ν (λ) dλ gµν dλ dλ

1/2

where X µ (0) = xµ and X µ (1) = y µ . Therefore the defining differential equation for the free scalar propagator in maximally symmetric curved space can also be written as an ordinary differential equation in one variable, µ. Writing D(x, y) = D(µ(x, y)), the equation of motion for ϕ implies D00 (µ) + (n − 1)A(µ)D0 (µ) − m2 D(µ) = 0 for x 6= y. Here 0 =

d dµ

and A(µ) =

1 a

cot(µ/a). Change variables to µ z = cos 2a 2

to put the differential equation in the form d2 d z(1 − z) 2 + [c − (a + b + 1)z] − ab D(z) = 0 dz dz

37

where h i p n − 1 + (n − 1)2 − (2ma)2 h i p 1 2 2 b = 2 n − 1 − (n − 1) − (2ma) a=

1 2

c = 12 n . This is the defining equation of the hypergeometric function 2 F1 (a, b, c, z). Since a + b + 1 − 2c = 0, the differential equation is invariant under z → 1 − z. Two independent particular solutions are therefore 2 F1 (a, b, c, z) and 2 F1 (a, b, c, 1 − z), and the general solution is a linear combination of these: D(µ) = C(1) 2 F1 (a, b, c, z(µ)) + C(2) 2 F1 (a, b, c, 1 − z(µ)) . To fix the coefficients, we need to consider the µ → 0 and µ → ∞ behavior of D(µ), as well as the location of singular points and branch cuts. For further discussion, consult the reference.

38

II II.1

Dirac and the Spinor The Dirac Equation

7. Show explicitly that (25) violates parity. L = G (ψ¯1L γ µ ψ2L )(ψ¯3L γµ ψ4L )

(25)

Solution: Parity acts on a Dirac spinor as ψ → iγ 0 ψ. Define PL ≡ 21 (1 − γ 5 ) and PR ≡ 12 (1 + γ 5 ) as usual. Then parity acts on a left-handed spinor as: ψL ≡ PL ψ → PL iγ 0 ψ = iγ 0 PR ψ = iγ 0 ψR . Thus parity transforms left-handed spinors into right-handed spinors, and so the Lagrangian in (25) violates parity.

11. Work out the Dirac equation in (1+1)-dimensional spacetime. See VII.7.7.

12. Work out the Dirac equation in (2+1)-dimensional spacetime. Show that the apparently innocuous mass term violates parity and time reversal. [Hint: The three γ µ s are just the three Pauli matrices with appropriate factors of i.] Solution: Consider the Dirac equation in (2+1) spacetime: (i6 ∂ − m)ψ = 0 =⇒ (iγ 0 ∂0 + iγ 1 ∂1 + iγ 2 ∂2 − m)ψ(x) = 0 Multiply the whole equation by γ 2 and anticommute it through to the right to get: (−iγ 0 ∂0 − iγ 1 ∂1 + iγ 2 ∂2 − m)γ 2 ψ(x) = 0 Divide by −1 to get: (iγ 0 ∂0 + iγ 1 ∂1 − iγ 2 ∂2 + m)γ 2 ψ(x) = 0 Defining the parity transformation P : x ≡ (x0 , x1 , x2 ) → x0 ≡ (x0 , x1 , −x2 ) puts the above equation into the form: (iγ 0 ∂00 + iγ 1 ∂10 + iγ 2 ∂20 + m)γ 2 ψ(x) = 0 39

Therefore the spinor ψ 0 (x0 ) ≡ γ 2 ψ(x) satisfies the Dirac equation with the wrong sign for the mass term: (i6 ∂x0 + m)ψ 0 (x0 ) = 0. At this point we should ask why the same argument doesn’t hold in (3+1) dimensions. After all, if we parity transform using P : x ≡ (x0 , x1 , x2 , x3 ) → x0 ≡ (x0 , x1 , −x2 , x3 ), we get the exact same result as above: (i6 ∂x0 + m)ψ 0 (x0 ) = 0. Where does the argument fail? There are two answers. One answer is that in (odd+1) dimensions, the operation of flipping one spatial coordinate is related by a rotation to the operation of flipping all of the spatial coordinates, while in (even+1) dimensions that is not true. See the footnote on page 98 of the book for more details. The second answer is that in (3+1) dimensions, we can multiply (i 6 ∂x0 + m)ψ 0 (x0 ) = 0 on the left by the γ 5 matrix, which anticommutes with the other 4 gamma matrices, to get (−i6 ∂x0 + m)γ 5 ψ 0 (x0 ) = 0 =⇒ (i6 ∂x0 − m)γ 5 ψ 0 (x0 ) = 0, and the field redefinition ψ → γ 5 ψ leaves the path integral unchanged. Therefore we conclude that the sign of the Dirac mass term does not matter in (3+1) dimensions. However, this operation is not possible in (2+1) dimensions. As noted in the partial solution to this question in the back of the book, in (2+1) dimensions we can define the gamma matrices to be the Pauli matrices: γ 0 = σ 3 , γ 1 = iσ 2 , γ 2 = −iσ 1 . Now try to define γ 5 : γ 5 ≡ iγ 0 γ 1 γ 2 = iσ 3 (iσ 2 )(−iσ 1 ) = iσ 3 (−iσ 3 ) = I In (2+1) dimensions, γ 5 is just the identity matrix. In other words, the Dirac spinor representation of SO(2, 1) is irreducible. Therefore, in (2+1) dimensions, the sign of the Dirac mass term does matter and therefore the mass term violates parity. Time reversal is going to work the same way. Take the Dirac equation and multiply by γ 0 , then anticommute γ 0 all the way to the right. This yields (iγ 0 ∂0 − iγ i ∂i − m)γ 0 ψ(x) = 0 Divide by −1 to get (−iγ 0 ∂0 + iγ i ∂i + m)γ 0 ψ(x) = 0 Define the time reversed coordinates x0 ≡ (−x0 , ~x) to write the above as (+iγ 0 ∂00 + iγ i ∂i0 + m)γ 0 ψ(x) = 0 Therefore the spinor ψ 0 (x0 ) ≡ γ 0 ψ(x) satisfies the Dirac equation with the wrong sign for the mass term. Again, in (2+1) dimensions the Dirac spinor is irreducible.

40

II.2

Quantizing the Dirac Field

2. Quantize the Dirac field in a box of volume V and show thatR the vacuum energy E0 is indeed proportional to V . [Hint: The integral over momentum d3 p is replaced by a sum over discrete values of the momentum.] Solution: P This problem was worded incorrectly in the book, since the vacuum energy is E0 = (#) p ωp with no factor of volume (as required by dimensional analysis). The intention is to place the Dirac field in a box and to calculate its contribution to the energy of the vacuum, and show that the factors of volume work themselves out. The solution proceeds according to the discussion on pp. 111 and 139, so we outline the salient steps: • When quantizing in a box, momentum integrals are replaced with sums as: Z 1 X d3 p → (2π)3 V p ~

• Each harmonic oscillator for a Lorentz-scalar field contributes + 12 ωp~ to the vacuum energy. • Each harmonic oscillator for a Lorentz-spinor field contributes − 21 ωp~ to the vacuum energy. • The Dirac field contains two Lorentz-spinor fields, for example the electron and the positron, so the above gets multiplied by 2. • Each Lorentz-spinor field contains two spin states, so the above gets multiplied by yet another factor of 2. P Therefore, a Dirac field contributes a total of −2 p~ ωp~ to the energy of the vacuum.

41

II.3

Lorentz Group and Weyl Spinors

1. Show by explicit computation that ( 12 , 21 ) is indeed the Lorentz vector. Solution: It is sufficient to compute the infinitesimal transformation properties, but it is pedagogically instructive to show that it all works out for the finite group transformation. Let χ transform under the ( 12 , 12 ) representation of the Lorentz group: χac˙ , where a {1, 2} ˙ 2} ˙ denotes the twodenotes the two-dimensional representation of SU (2)L and c˙ {1, dimensional representation of SU (2)R . ~ and of the boosts, The Lorentz group consists of the rotations, which are generated by J, ~ Let A transform under the 2-dimensional representation of which are generated by K. SU (2)L , which means that A carries one SU (2)L 2-index: Aa . The rotations and the boosts act on this object A as: i b h i b h iθ n ˆ ·( ~σ2 ) iθ n ˆ ·J~ Ab Rotation through angle θ about axis n ˆ : Aa → e Ab = e a a i h i b h b ~ σ ~ Boost through β ≡ Arctanh(v) about n ˆ : Aa → eiβ nˆ ·K Ab = eiβ nˆ ·( 2i ) a

a

Since the Pauli matrices have the property {σ i , σ j } = 2δ ij , these transformations simplify: θ θ iθ n ˆ ·( ~σ2 ) + in ˆ · ~σ sin e = 12×2 cos 2 2 θ θ cos 2 0 0 x 0 1 y 0 −i z 1 = +i n +n +n sin θ 1 0 i 0 0 −1 0 cos 2 2 θ θ θ z − cos 2 + i n sin in 2 sin z 2 , n± ≡ nx ± i ny = θ θ + i n sin 2 cos 2 − i n sin 2θ Using cos(−i x) = cosh(x) and i sin(−i x) = sinh(x), the boost transformation matrix can be obtained from the above with θ → −iβ: β β − cosh β2 + nz sinh n sinh βn ˆ ·( ~σ2 ) 2 2 = e n+ sinh β2 cosh β2 − nz sinh β2 As indicated on page 114 of the book, the transformations for SU (2)R are the same as those for SU (2)L except that the boosts get an extra minus sign from the replacement β → −β. We can now compute explicitly how the four components of χac˙ transform under rotations and under boosts. First specialize to the case n ˆ = (1, 0, 0), for which the transformations on χ are:

42

Rotation through angle θ about xˆ-axis: h i bh i ~ ~ c˙ χac˙ → eiθ nˆ ·J eiθ nˆ ·J χb e˙ a e˙ b c˙ c is c is = χb e˙ is c a is c e˙ b 1˙ e˙ T c˙ 2˙ c is χ1 χ1 c is = i s c a χ2 1˙ χ2 2˙ i s c e˙ b 2 1˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ c χ1 − s2 χ2 2 + i cs(χ1 2 + χ2 1 ) c2 χ1 2 − s2 χ2 1 + i cs(χ1 1 + χ2 2 ) = 2 1˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ c χ2 − s2 χ1 2 + i cs(χ1 1 + χ2 2 ) c2 χ2 2 − s2 χ1 1 + i cs(χ1 2 + χ2 1 ) θ θ , s ≡ sin c ≡ cos 2 2 Some trigonometric identities will be useful. Since cos(2x) = 2 cos2 x − 1 = 1 − 2 sin2 x and sin(2x) = 2 sin x cos x, we have c2 = 12 (1 + cos θ), s2 = 12 (1 − cos θ) and cs = 12 sin θ. With these, the above transformations simplify to: 1 1 i ˙ ˙ ˙ ˙ ˙ χ1 1 → (1 + cos θ)χ1 1 − (1 − cos θ)χ2 2 + sin θ(χ1 2 + χ2 1 ) 2 2 2 1 1˙ 1 i ˙2 ˙1 ˙2 ˙ ˙ = (χ1 − χ2 ) + cos θ(χ1 + χ2 ) + sin θ(χ1 2 + χ2 1 ) 2 2 2 1 1 i ˙2 ˙2 ˙1 ˙ ˙ χ1 → (1 + cos θ)χ1 − (1 − cos θ)χ2 + sin θ(χ1 1 + χ2 2 ) 2 2 2 1 i 1 2˙ ˙1 ˙2 ˙1 ˙ ˙ = (χ1 − χ2 ) + cos θ(χ1 + χ2 ) + sin θ(χ1 1 + χ2 2 ) 2 2 2 1 1 i ˙1 ˙1 ˙2 ˙ ˙ χ2 → (1 + cos θ)χ2 − (1 − cos θ)χ1 + sin θ(χ1 1 + χ2 2 ) 2 2 2 1 i 1 2˙ ˙1 ˙1 ˙ ˙2 ˙ = − (χ1 − χ2 ) + cos θ(χ1 + χ2 ) + sin θ(χ1 1 + χ2 2 ) 2 2 2 1 1 i ˙2 ˙ ˙2 ˙1 ˙ χ2 → (1 + cos θ)χ2 − (1 − cos θ)χ1 + sin θ(χ1 2 + χ2 1 ) 2 2 2 1 1˙ 1 i ˙2 ˙1 ˙2 ˙ ˙ = − (χ1 − χ2 ) + cos θ(χ1 + χ2 ) + sin θ(χ1 2 + χ2 1 ) 2 2 2 These all appear in ± pairs, so add and subtract some of these transformations: ˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

χ1 2 − χ2 1 → χ1 2 − χ2 1 χ1 1 − χ2 2 → χ1 1 − χ2 2 χ1 1 + χ2 2 → cos θ(χ1 1 + χ2 2 ) + i sin θ(χ1 2 + χ2 1 ) χ1 2 + χ2 1 → cos θ(χ1 2 + χ2 1 ) + i sin θ(χ1 1 + χ2 2 ) ˙

˙

˙

˙

˙

˙

˙

˙

Define the symbols (v t , ~v ) by v t ≡ χ2 1 −χ1 2 , v x ≡ χ2 2 −χ1 1 , v y ≡ −i(χ1 1 +χ2 2 ), v z ≡ χ1 2 +χ2 1 . 43

In terms of these new labels, then above transformation laws are: t t 1 0 0 0 v v 0 1 v x v x 0 0 y → 0 0 cos θ sin θ v y v vz vz 0 0 − sin θ cos θ That is exactly how a 4-vector v µ transforms under rotations about the x-axis. c s iθ n ˆ ·(~ σ /2) For n ˆ = (0, 1, 0), the transformation matrix is e = , so the components −s c of χ transform as: ˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

χ1 1 → c2 χ1 1 + s2 χ2 2 + cs(χ1 2 + χ2 1 ) 1 1 1 ˙ ˙ ˙ ˙ = (1 + cos θ)χ1 1 + (1 − cos θ)χ2 2 + sin θ(χ1 2 + χ2 1 ) 2 2 2 1 1˙ 1 1 ˙2 ˙1 ˙2 ˙ ˙ = (χ1 + χ2 ) + cos θ(χ1 − χ2 ) + sin θ(χ1 2 + χ2 1 ) 2 2 2 χ1 2 → c2 χ1 2 − s2 χ2 1 − cs(χ1 1 − χ2 2 ) 1 1 ˙ ˙ = (1 + cos θ)χ1 2 − (1 − cos θ)χ2 1 − 2 2 1 1 2˙ ˙1 ˙ ˙ = (χ1 − χ2 ) + cos θ(χ1 2 + χ2 1 ) − 2 2 ˙

˙

˙

˙

˙

1 ˙ ˙ sin θ(χ1 1 − χ2 2 ) 2 1 ˙ ˙ sin θ(χ1 1 − χ2 2 ) 2

χ2 1 → c2 χ2 1 − s2 χ1 2 − cs(χ1 1 − χ2 2 ) 1 1 1 ˙ ˙ ˙ ˙ = (1 + cos θ)χ2 1 − (1 − cos θ)χ1 2 − sin θ(χ1 1 − χ2 2 ) 2 2 2 1 1 1 ˙ ˙ ˙ ˙ ˙ ˙ = − (χ1 2 − χ2 1 ) + cos θ(χ1 2 + χ2 1 ) − sin θ(χ1 1 − χ2 2 ) 2 2 2 ˙

˙

˙

˙

˙

χ2 2 → c2 χ2 2 + s2 χ1 1 − cs(χ1 2 + χ2 1 ) 1 1 1 ˙ ˙ ˙ ˙ = (1 + cos θ)χ2 2 + (1 − cos θ)χ1 1 − sin θ(χ1 2 + χ2 1 ) 2 2 2 1 1˙ 1 1 ˙ ˙ 2˙ 1˙ 2˙ = (χ1 + χ2 ) − cos θ(χ1 − χ2 ) − sin θ(χ1 2 + χ2 1 ) 2 2 2 Adding and subtracting these gives: ˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

χ1 1 + χ2 2 → χ1 1 + χ2 2 ˙

˙

˙

˙

˙

˙

˙

χ1 1 − χ2 2 → cos θ(χ1 1 − χ2 2 ) + sin θ(χ1 2 + χ2 1 ) ˙

˙

χ1 2 − χ2 1 → χ1 2 − χ2 1 ˙

χ1 2 + χ2 1 → cos θ(χ1 2 + χ2 1 ) − sin θ(χ1 1 − χ2 2 ) 44

˙

˙

˙

˙

˙

˙

Using the same definitions from before, v t ≡ χ2 1 −χ1 2 , v x ≡ χ2 2 −χ1 1 , v y ≡ −i(χ1 1 +χ2 2 ), v z ≡ ˙ ˙ χ1 2 + χ2 1 , we get: t t v 1 0 0 0 v v x 0 cos θ 0 − sin θ v x y → v 0 0 1 0 v y vz 0 sin θ 0 cos θ vz That is exactly how a 4-vector v µ transforms under rotations about the y-axis. iθ/2 c + is 0 e 0 iθ n ˆ ·(~ σ /2) , For n ˆ = (0, 0, 1), the transformation matrix is e = = 0 c − is 0 e−iθ/2 so the transformation of χ is easy: ˙

˙

χ1 1 → (cos θ + i sin θ)χ1 1 ˙

˙

˙

˙

χ1 2 → χ1 2 χ2 1 → χ2 1 ˙

˙

χ2 2 → (cos θ − i sin θ)χ2 2 Rearrange these to make them look like the previous two cases: ˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

˙

χ1 2 − χ2 1 → χ1 2 − χ2 1 χ1 2 + χ2 1 → χ1 2 + χ2 1 χ1 1 + χ2 2 → cos θ(χ1 1 + χ2 2 ) + i sin θ(χ1 1 − χ2 2 ) χ1 1 − χ2 2 → cos θ(χ1 1 − χ2 2 ) + i sin θ(χ1 1 + χ2 2 ) ˙

˙

˙

˙

˙

˙

So, again, with the same definitions v t ≡ χ2 1 − χ1 2 , v x ≡ χ2 2 − χ1 1 , v y ≡ −i(χ1 1 + χ2 2 ), v z ≡ ˙ ˙ χ1 2 + χ2 1 , these transformations are: t t v 1 0 0 0 v v x 0 cos θ sin θ 0 v x y → v 0 − sin θ cos θ 0 v y vz vz 0 0 0 1 That is exactly how a 4-vector v µ transforms under rotations about the z-axis. Therefore, we know that the object χac˙ , which transforms under the ( 12 , 12 )-representation of the Lorentz group, can be repackaged into an object v µ , which transforms under the 4-dimensional (“vector”) representation of the Lorentz group. Before proceeding to check the boosts, let us reflect on the repackaging of the components of

45

χ. The definitions we used are: ˙

˙

˙

˙

v t ≡ χ2 1 − χ1 2 v x ≡ χ2 2 − χ1 1 ˙

˙

v y ≡ −i(χ1 1 + χ2 2 ) ˙

˙

v z ≡ χ1 2 + χ2 1 Note the following very important are raised and lowered with point: since SU (2) indices 0 1 0 −1 the antisymmetric tensors ab = and ab = , the diagonal components −1 0 +1 0 of χac˙ correspond to the off-diagonal components of χac˙ and vice versa, and all sorts of minus ˙ ˙ ˙ ˙ ˙ ˙ signs appear. For example, χ2 1 = 2a χa1 = 21 χ11 = −12 χ11 = +12 χ11 = χ11 . Raising all of the indices in that way results in the following for the definitions of (v t , ~v ): ˙

˙

˙

˙

˙

˙

˙

˙

v t ≡ χ2 1 − χ1 2 = +χ11 + χ22 = Iaa˙ χaa˙ v x ≡ χ2 2 − χ1 1 = +χ12 + χ21 = σaxa˙ χaa˙ ˙

˙

˙

˙

v y ≡ −i(χ1 1 + χ2 2 ) = +iχ21 − iχ12 = σaya˙ χaa˙ ˙

˙

˙

˙

v z ≡ χ1 2 + χ2 1 = −χ22 + χ11 = σaza˙ χaa˙ ˙ 12, ˙ 21, ˙ 22} ˙ of SU (2)L ⊗ SU (2)R We therefore have discovered a way to turn the pair aa˙ {11, indices into the single SO(3, 1) 4-vector index, µ {0, 1, 2, 3}: for any value of µ {x, y, z}, use the corresponding Pauli matrix to contract the aa˙ indices; for µ = 0, use the identity matrix. This is often written as σaµa˙ , where σ 0 ≡ I. Back to the problem at hand, we still have to check that the boosts work, but now we have two options as to how to go about doing that. We could proceed as before, by writing out the boost matrices explicitly, calculating the transformation of each component of χ, repackaging them into a 4-vector, and showing that it’s the same 4-vector that we got before. However, we now have a second option: if it is true that we can use the machine σaµa˙ to change the ( 21 , 12 ) indices to one 4-index, then it is already manifest that the object χµ ≡ σaµa˙ χaa˙ transforms as a 4-vector under rotations and boosts, and therefore that the two representations are equivalent. A valid way to approach this problem is simply to check that the symbol σaµa˙ is invariant under simultaneous Lorentz transformations for all of its indices. It’s easiest to work with an object that has one lower SU (2)L index and one upper SU (2)R index, so define the quantity sµa c˙ ≡ −σaµa˙ a˙ c˙ whose components are, numerically: sta c˙ = (−iσ y )ac˙ sxa

c˙

= (σ z )ac˙

sya

c˙

= (−iσ t )ac˙

sza

c˙

= (−σ x )ac˙ 46

Let’s take a boost in the x-direction, meaning n ˆ = (1, 0, 0), and apply it to sµa c˙ : h β x i b h β x ic˙ e− 2 σ Λµν sνb e˙ sµa c˙ → e 2 σ a e˙ µ cosh β − sinh β 0 0 b c˙ − sinh β cosh β 0 0 cosh β2 sinh β2 cosh β2 − sinh β2 sνb = β β β β 0 0 1 0 sinh 2 cosh 2 a − sinh 2 cosh 2 e˙ 0 0 0 1 ν Do one index at a time. For µ = t, we have: Λt ν sbν

e˙

= cosh β(iσ y )b e˙ − sinh β(σ z )b e˙ e˙ e˙ 0 −1 1 0 = cosh β − sinh β 1 0 b 0 −1 b e˙ − sinh β − cosh β = cosh β sinh β b

Therefore, sta c˙

c˙ cosh β2 sinh β2 − sinh β − cosh β cosh β2 − sinh β2 → cosh β sinh β sinh β2 cosh β2 − sinh β2 cosh β2 a c˙ 0 −1 = = (−iσ y )a c˙ = sta c˙ X 1 0 a

For µ = x, we have: Λxν sbν

e˙

= − sinh β(−iσ y )b e˙ + cosh β(σ z )b e˙ e˙ e˙ 0 −1 1 0 = − sinh β + cosh β 1 0 b 0 −1 b e˙ cosh β sinh β = − sinh β − cosh β b

Therefore, sxa c˙

c˙ cosh β sinh β cosh β2 − sinh β2 cosh β2 sinh β2 → − sinh β − cosh β sinh β2 cosh β2 − sinh β2 cosh β2 a c˙ 1 0 = = (σ z )a c˙ = sxa c˙ X 0 −1 a

For µ = y, we have: Λ

y

ν e˙ ν sb

t

e˙

= (−iσ )b = 47

−i 0 0 −i

e˙

b

e˙

Therefore, sya c˙

c˙ −i 0 cosh β2 − sinh β2 cosh β2 sinh β2 → 0 −i sinh β2 cosh β2 − sinh β2 cosh β2 a c˙ −i 0 = = (−iσ t )a c˙ = sya c˙ X 0 −i a

For µ = z, we have: Λ

z

ν e˙ ν sb

x

e˙

= (−σ )b =

e˙ 0 −1 −1 0 b

Therefore, sza c˙

c˙ cosh β2 sinh β2 0 −1 cosh β2 − sinh β2 → −1 0 sinh β2 cosh β2 − sinh β2 cosh β2 a c˙ 0 −1 = = (−σ x )a c˙ = sza c˙ X −1 0 a

Since ab is invariant under SU (2)L (if it weren’t, we couldn’t use it to raise and lower SU (2)L indices), the invariance of sµa c˙ implies the invariance of σaµc˙ . We have checked only the case of a Lorentz boost in the x-direction, but since we have already checked for rotational invariance, we can just argue by symmetry that boosts in the y- and z- directions also leave the symbol σaµc˙ invariant. Therefore, the ( 21 , 12 )-representation of SU (2)L ⊗ SU (2)R is equivalent to the 4-vector representation of SO(3, 1).

2. Work out how the six objects contained in the (1, 0) and (0, 1) transform under the Lorentz group. Recall from your course on electromagnetism how the electric and mag~ and B ~ transform. Conclude that the electromagnetic field in fact transforms as netic fields E (1, 0)⊕(0, 1). Show that it is parity that once again forces us to use a reducible representation. Solution: A spin-j object transforms under the (2j + 1)-dimensional representation of SU (2), so a spin-1 object transforms under the 3-dimensional representation of SU (2). Let a {1, 2} be an index for the 2-dimensional representation of SU (2). An object X transforming under the 2 ⊗ 2-dimensional representation of SU (2) therefore carries two indices: X = Xab . The 3-dimensional representation of SU (2) can be obtained from the symmetrized tensor product of two 2-dimensional representations, or in fancy notation, 3 = 2 ⊗S 2. Therefore, the quantity X(ab) ≡ 21 (Xab + Xba ) transforms under the 3-dimensional representation of SU (2). 48

Now we can find out how the object X(ab) in the 3-dimensional representation of SU (2) transforms, since each index gets its own transformation matrix. Explicitly, for a rotation through an angle θ about an axis n ˆ , the transformation is: i h i d h c iθ n ˆ ·( ~σ2 ) iθ n ˆ ·( ~σ2 ) e X(cd) X(ab) → e a b h i h iT d c iθ n ˆ ·( ~σ2 ) iθ n ˆ ·( ~σ2 ) = e X(cd) e a b c + inz s in− s X11 X(12) c + inz s in+ s = in+ s c − inz s X(12) X22 in− s c − inz s θ θ , s ≡ sin where c ≡ cos 2 2 For a boost through β about n ˆ , the transformation is: ch + n z s h n− sh X11 X(12) ch + n z s h n+ sh X(ab) → n+ sh ch − n z s h X(12) X22 n− sh ch − n z s h β β ch ≡ cosh , sh ≡ sinh 2 2 We’ve just been referencing SU (2), but remember that we’re talking about the Lorentz group, SO(3, 1) ∼ = SU (2)L ⊗ SU (2)R . The discussion above applies for SU (2)L , so the transformations we found are those for an object in the (1, 0)-representation, meaning that said object transforms under the 3-dimensional representation of SU (2)L and is unaffected by SU (2)R . What about the (0, 1)-representation? As for the case of (0, 21 ) versus ( 12 , 0) from the previous problem, the (0, 1) transforms in ~ (1,0) = +~σ /(2i) exactly the same way as the (1, 0) except for the critical difference that K ~ (0,1) = −~σ /(2i). In practice that means we can copy the results from above except whereas K with β → −β, which implies ch → ch and sh → −sh . ˙

Let Y transform under the (0, 1) representation of the Lorentz group: Y = Y (a˙ b) . The Lorentz group transformations for Y are: Rotation through θ about n ˆ: 1˙ 1˙ ˙˙ c + inz s in− s Y Y (12) c + inz s in+ s ˙ (a˙ b) Y → ˙˙ ˙˙ in+ s c − inz s in− s c − inz s Y (12) Y 22 θ θ , s ≡ sin c ≡ cos 2 2 Boost through β about n ˆ: 1˙ 1˙ ˙˙ ch − nz sh −n− sh Y (12) ch − nz sh −n+ sh Y ˙ (a˙ b) Y → ˙˙ ˙˙ −n+ sh ch + nz sh −n− sh ch + nz sh Y (12) Y 22 49

β β , sh ≡ sinh ch ≡ cosh 2 2 Now consider a quantity Φ that transforms under the (1, 0) ⊕ (0, 1) representation of the Lorentz group defined by the following: Φ Φ 0 0 1 3 Φ3 Φ2 0 0 X(ab) 02×2 ≡ Φ≡X ⊕Y = ˙ 0 0 Φ4 Φ6 02×2 Y (a˙ b) 0 0 Φ6 Φ5 This quantity Φ is a 4 × 4 matrix with 6 independent components. Using the above work, it transforms in the following way under boosts: Φ→

ch +nz sh n− sh 0 0 Φ1 + z Φ3 n sh ch −n sh 0 0 z − 0 0 0 ch −n sh −n sh + z 0 0 −n sh ch +n sh 0

Φ3 Φ2 0 0

0 0 Φ4 Φ6

0 ch +nz sh n+ sh 0 0 − z 0 n s c −n s 0 0 h h h z + Φ6 0 0 ch −n sh −n sh − z Φ5 0 0 −n sh ch +n sh

To understand what’s going on, specialize to a boost in the x-direction: n ˆ = (1, 0, 0). For that case, the field Φ transforms as: ch s h 0 0 Φ1 Φ3 0 0 ch sh 0 0 Φ3 s h s 0 0 Φ 0 0 c 0 0 h ch 2 h Φ→ 0 0 ch −sh 0 0 Φ4 Φ6 0 0 ch −sh 0 0 −sh ch 0 0 Φ6 Φ5 0 0 −sh ch

Therefore, each component of Φ transforms as: Φ1 Φ2 Φ3 Φ4 Φ5 Φ6

→ ch (ch Φ1 + sh Φ3 ) + sh (ch Φ3 + sh Φ2 ) → ch (ch Φ2 + sh Φ3 ) + sh (ch Φ3 + sh Φ1 ) → ch (ch Φ3 + sh Φ2 ) + sh (ch Φ1 + sh Φ3 ) → ch (ch Φ4 − sh Φ6 ) − sh (ch Φ6 − sh Φ5 ) → ch (ch Φ5 − sh Φ6 ) − sh (ch Φ6 − sh Φ4 ) → ch (ch Φ6 − sh Φ5 ) − sh (ch Φ4 − sh Φ6 )

Now use some hyperbolic trigonometric identities. Recall that ch ≡ cosh(β/2) and sh ≡ sinh(β/2), so that c2h = 21 (cosh β + 1), s2h = 12 (cosh β − 1), and 2ch sh = sinh β. Using these, we get: Φ1 → c2h Φ1 + s2h Φ2 + 2sh ch Φ3 1 1 = (cosh β + 1)Φ1 + (cosh β − 1)Φ2 + sinh β Φ3 2 2 1 1 = cosh β (Φ1 + Φ2 ) + (Φ1 − Φ2 ) + sinh β Φ3 2 2 50

Φ2 → c2h Φ2 + s2h Φ1 + 2sh ch Φ3 1 1 = (cosh β + 1)Φ2 + (cosh β − 1)Φ1 + sinh β Φ3 2 2 1 1 = cosh β (Φ1 + Φ2 ) − (Φ1 − Φ2 ) + sinh β Φ3 2 2 Φ3 → (c2h + s2h )Φ3 + ch sh (Φ1 + Φ2 ) 1 = cosh β Φ3 + sinh β (Φ1 + Φ2 ) 2 The fields appear in the combinations Φ± ≡ 12 (Φ1 ±Φ2 ) and Φ3 . From adding and subtracting the transformations for Φ1 and Φ2 , we get: Φ+ → cosh β Φ+ + sinh β Φ3 Φ3 → cosh β Φ3 + sinh β Φ+ Φ− → Φ− Now do the same thing for Φ4 , Φ5 and Φ6 : Φ4 → c2h Φ4 + s2h Φ5 − 2ch sh Φ6 1 1 = cosh β (Φ4 + Φ5 ) + (Φ4 − Φ5 ) − sinh β Φ6 2 2 Φ5 → c2h Φ5 + s2h Φ4 − 2ch sh Φ6 1 1 = cosh β (Φ4 + Φ5 ) − (Φ4 − Φ5 ) − sinh β Φ6 2 2 Φ6 → (c2h + s2h )Φ6 − ch sh (Φ4 + Φ5 ) 1 = cosh β Φ6 − sinh β (Φ4 + Φ5 ) 2 ˜ ± ≡ 1 (Φ4 ± Φ5 ) so that the transformations are: As before, define Φ 2 ˜ + → cosh β Φ ˜ + − sinh β Φ6 Φ ˜+ Φ6 → cosh β Φ6 − sinh β Φ ˜− → Φ ˜− Φ To collect and rearrange the results, boosting in the x-direction with speed v = tanh β

51

transforms the components of Φ as follows: Φ− → Φ− Φ+ → cosh β Φ+ + sinh β Φ3 ˜+ Φ6 → cosh β Φ6 − sinh β Φ ˜− → Φ ˜− Φ ˜ + → cosh β Φ ˜ + − sinh β Φ6 Φ Φ3 → cosh β Φ3 + sinh β Φ+ ˜ −, By ≡ Φ ˜ + and B z ≡ Φ3 , we get: With the relabeling E x ≡ Φ− , E y ≡ Φ+ , E z ≡ Φ6 , B x ≡ Φ Ex → Ex E y → cosh β E y + sinh β B z E z → cosh β E z − sinh β B y Bx → Bx B y → cosh β B y − sinh β E z B z → cosh β B z + sinh β E y ~ and a magnetic field B ~ These are precisely the transformation rules of an electric field E under a Lorentz boost in the x-direction with speed v = tanh β. We see that the electromagnetic field transforms under the reducible representation (1, 0) ⊕ (0, 1) of the Lorentz group. That fact is typically repackaged in a different way. A spin-1 representation corresponds to the 3-dimensional representation of the rotation group, which corresponds to the 4-dimensional representation of the Lorentz group, denoted by the index µ {0, 1, 2, 3}. Therefore, an object that transforms under two copies of spin-1 implies that the object gets two such indices, µ and ν. But a matrix M µν has 4 × 4 = 16 independent components, where as the object Φ has only 6 independent components. However, the antisymmetric matrix F µν ≡ 12 (M µν − M νµ ) has 4 × (4 − 1)/2 = 6 independent components. Such a matrix F transforms under a Lorentz boost with speed v = tanh β in the x-direction as: T cosh β sinh β 0 0 0 F 01 F 02 F 03 cosh β sinh β 0 0 sinh β cosh β 0 0 −F 01 0 F 12 F 13 sinh β cosh β 0 0 → 02 12 23 0 0 1 0 0 1 0 −F −F 0 F 0 03 13 23 0 0 0 1 −F −F −F 0 0 0 0 1

F µν

So the 6 components transform as: 52

F 01 F 02 F 03 F 23 F 13 F 12

→ F 01 → cosh β → cosh β → F 23 → cosh β → cosh β

F 02 + sinh β F 12 F 03 + sinh β F 13 F 13 + sinh β F 03 F 12 + sinh β F 02

~ and B ~ implies F 01 = E x , F 02 = Comparing with the transformations for the components of E E y , F 03 = E z , F 23 = B x , F 13 = −B y and F 12 = B z , or more compactly, E i = F 0i and B i = 21 ijk Fjk . We know that parity interchanges SU (2)L with SU (2)R . How can we get an object that transforms as spin-1 under the rotation group (that is, as a “3-vector” under spatial rotations) that is also invariant under parity? As discussed, an object that transforms as (1, 0) of SU (2)L ⊗ SU (2)R does in fact transform as spin-1 under the rotation group, but a parity operation would change such an object to (0, 1). If we want the theory of the electromagnetic field to preserve parity and to be Lorentz invariant, then it must transform as (1, 0) ⊕ (0, 1) under Lorentz transformations.

3. Show that e

~ 0 +i~ ~ −i~ ϕ·K ϕ·K

γ e

=

0

e+~ϕ·~σ

e−~ϕ·~σ 0

and e ϕ~ ·~σ = cosh ϕ + ~σ · ϕˆ sinh ϕ with the unit vector ϕˆ ≡ ϕ ~ /ϕ. Identifying p~ = mϕˆ sinh ϕ, derive the Dirac equation. Show that 0 σi i γ = . −σi 0 Solution: If D is a diagonal matrix, then eD = diag(eD11 , eD22 , ...). Page 117 tells us that ~σ 0 0 I 0 2 ~ and γ = iK = I 0 0 − ~σ2

53

so by direct computation we have

~

~

e−~ϕ·(iK) γ 0 e+~ϕ·(iK)

~σ /2 0 ~σ /2 0 +~ ϕ· 0 −~σ /2 γ 0 e 0 −~σ /2 =e −~ϕ·(~σ/2) +~ϕ·(~σ/2) e 0 0 I e 0 = I 0 0 e+~ϕ·(~σ/2) 0 e−~ϕ·(~σ/2) −~ϕ·(~σ/2) 0 e−~ϕ·(~σ/2) e 0 = e+~ϕ·(~σ/2) 0 0 e+~ϕ·(~σ/2) −~ ϕ·~ σ 0 e = +~ϕ·~σ e 0 −~ ϕ ·

ˆ σ) Next, compute ei(θ ϕ·~ = cos(θ ϕˆ · ~σ ) + i sin(θ ϕˆ · ~σ ) by Taylor expansion:

cos(θ ϕˆ · ~σ ) =

∞ X

(−1)n (θ ϕˆ · ~σ )n n! n = 0,2,4,6...

Since {σ i , σ j } = 2δ ij I, any even power of (ϕˆ · ~σ ) is equal to the identity matrix: 1 ˆ =I (ϕˆ · ~σ )2 = ϕˆi ϕˆj σ i σ j = ϕˆi ϕˆj {σ i , σ j } = ϕˆi ϕˆi I = ϕˆ · ϕI 2 Therefore we have the simple result cos(θ ϕˆ · ~σ ) = I cos θ. By the same argument, any odd power of (ϕˆ · ~σ ) is equal to itself. This allows us to write another simple expression: sin(θ ϕˆ · ~σ ) = ϕˆ · ~σ sin θ. Therefore, we have just proved that ˆ σ) ei(θ ϕ·~ = I cos θ + iϕˆ · ~σ sin θ

Since sin(ix) = i sinh x, we can let θ = −iϕ to get the desired result eϕ~ ·~σ = I cosh ϕ + ϕ ~ · ~σ sinh ϕ Next, we start with the form of the boosted Dirac equation written above equation (17) on p. 118: ~ ~ e−i~ϕ·K γ 0 e+i~ϕ·K − I4 ψ(p) = 0 Using what we just derived in the first part of this problem gives −I2 e−~ϕ·~σ ψ(p) = 0 e+~ϕ·~σ −I2 Using the second part of this problem, the above becomes −I2 I2 cosh ϕ − ϕ ~ · ~σ sinh ϕ ψ(p) = 0 I2 cosh ϕ + ϕ ~ · ~σ sinh ϕ −I2 54

p p Identifying p~ = ϕm ˆ sinh ϕ gives cosh ϕ = 1 + |~p |2 /m2 = m2 + |~p |2 /m = p0 /m, so after multiplying through by m, the above equation becomes −m I2 p0 − ~σ · p~ ψ(p) = 0 I2 p0 + ~σ · p~ −m Define the 4-vectors σ µ ≡ (I2 , ~σ ) and σ ¯ µ ≡ (I2 , −~σ ) (the bar is part of the name and does not denote any sort of complex conjugation operation) to rewrite the above as −m σ µ pµ ψ(p) = 0 σ ¯ µ pµ −m 0 σµ µ Defining γ = gives the Dirac equation: σ ¯µ 0 (γ µ pµ − m)ψ(p) = 0 This shows that γ i =

II.4

i

0 σ , which completes the problem. −σ i 0

Spin-Statistics Connection

1. Show that we would also get into trouble if we quantize the Dirac field with commutation instead of anticommutation rules. Calculate the commutator [J 0 (~x, 0), J 0 (~0, 0)]. Solution: ¯ 0 ψ = ψ † ψ. Consider the commutation relations The charge density is J 0 = ψγ [ψα (~x, t), ψβ† (~0, t)]ζ = δ (3) (~x )δαβ and [ψα (~x, t), ψβ (~0, t)]ζ = 0 where [A, B]ζ ≡ AB − ζBA denotes the commutator for ζ = +1 and the anticommutator for ζ = −1. From now on all times t will be set to zero, so for notational convenience drop the vector arrow over the spatial coordinates. The commutator of charge densities is [J 0 (x),J 0 (0)] = [ψ † (x)ψ(x), ψ † (0)ψ(0)] = ψα† (x)ψα (x)ψβ† (0)ψβ (0) − ψβ† (0)ψβ (0)ψα† (x)ψα (x) = ψα† (x)ψα (x)ψβ† (0)ψβ (0) − ψβ† (0)[ζψα† (x)ψβ (0) + δ(x)δαβ ]ψα (x) = ψα† (x)ψα (x)ψβ† (0)ψβ (0) − ζψβ† (0)ψα† (x)ψβ (0)ψα (x) − ζδ(x)ψα† (0)ψα (x) = ψα† (x)ψα (x)ψβ† (0)ψβ (0) − ζ[ζψα† (x)ψβ† (0)][ζψα (x)ψβ (0)] − ζδ(x)ψα† (0)ψα (x) = ψα† (x)ψα (x)ψβ† (0)ψβ (0) − ζψα† (x)[ζψα (x)ψβ† (0) + δ(x)δαβ ]ψβ (0) − ζδ(x)ψα† (0)ψα (x) = ψα† (x)ψα (x)ψβ† (0)ψβ (0) − ψα† (x)ψα (x)ψβ† (0)ψβ (0) − δ(x)[ψα† (x)ψα (0) + ζψα† (0)ψα (x)] = −δ(x)[ψα† (x)ψα (0) + ζψα† (0)ψα (x)] The delta function is only nonzero when x = 0, so we have [J 0 (x), J 0 (0)] = −δ(x)ψα† (0)ψα (0)(1 + ζ) . If ζ = −1, meaning fermion anticommutation relations, then [J 0 (x), J 0 (0)] = 0 as desired. But if ζ = +1, meaning fermion commutation relations, then indeed we run into trouble. 55

II.5

Vacuum Energy, Grassmann Integrals, and Feynman Diagrams for Fermions

1. Write down the Feynman amplitude for the diagram in figure II.5.1 for the scalar theory (19). The answer is given in chapter III.3. Solution: The vertices and propagators are: = if k

=

p =

−i −k −i(

2

2

p +m)

−p 2 + m 2

Reading the diagram from left to right gives: Z −i −i (6 k+6 p + m) d4 k (if ) (if ) u(p, s) (Fig. II.5.1) = u¯(p, s) (2π)4 −k 2 + µ2 − i −(k + p)2 + m2 − i Z d4 k 1 (6 k+6 p + m) 2 = f u¯(p, s) u(p, s) (2π)4 k 2 − µ2 (k + p)2 − m2 Compare this with page 180 in the book for verification.

2. Applying the Feynman rules for the vector theory (22) show that the amplitude for the diagram in figure II.5.3 is given by Z d4 k 1 kµ kν 6 p +6k + m µ 2 2 − gµν u¯(p)γ ν γ u(p) . (26) (ie) i 4 2 2 2 (2π) k − µ µ (p + k)2 − m2

56

Solution: The vertices and propagators are:

k

=

−i −k 2

2

( −g + k 2k )

The fermion propagator is the same as for the previous question. Again, reading the diagram from left to right gives: (Fig. II.5.3) = Z 4 −i kµ kν −i (6 k+6 p + m) dk µ (−ieγ ) −gµν + 2 (−ieγ ν ) u(p, s) u¯(p, s) 4 2 2 2 2 (2π) −k + µ − i µ −(k + p) + m − i Z 4 1 kµ kν 6 k+6 p + m dk γ ν u(p, s) −gµν + 2 u¯(p, s)γ µ = e2 4 2 2 (2π) −k + µ µ −(k + p)2 + m2 Z d4 k 1 kµ kν 6 k+6 p + m 2 =e γ µ u(p, s) −gµν + 2 u¯(p, s)γ ν 4 2 2 (2π) k − µ µ (k + p)2 − m2

II.6

Scattering and Gauge Invariance

1. Show that the differential cross section for a relativistic electron scattering in a Coulomb potential is given by dσ α2 = 2 2 4 (1 − v 2 sin2 (θ/2)) dΩ 4~p v sin (θ/2) Solution: The amplitude is given in equation (4) on p. 133 as M = −i

4πα 0 0 u¯ γ u k2

where 4πα ≡ e2 , and we have abbreviated u0 ≡ u(P, S) and u ≡ u(p, s). Taking the magnitude squared gives 2 2 4πα 4πα 0 0 0 0 2 |M| = (¯ u γ u)(¯ uγ u ) = tr γ 0 (u¯ u)γ 0 (u0 u¯0 ) 2 2 k k 57

Summing over spins as

P

s

u¯ u=

1 (6 p 2m

+ m) gives

1X 1 M ≡ |M|2 = 2 s,S 8m2 2

4πα k2

2

tr γ 0 (6 p + m)γ 0 (6 P + m)

The trace over gamma matrices simplifies as tr γ 0 (6 p + m)γ 0 (6 P + m) = tr(γ 06 pγ 06 P + m2 I) = tr(−6 p6 P + 2p0 γ 06 P + m2 I) = 4(−pµ P µ + 2p0 P 0 + m2 ) = 4(p0 P 0 + p~ · P~ + m2 ) Therefore the spin-summed amplitude is 2 1 4πα 2 M = (p0 P 0 + p~ · P~ + m2 ) 2m2 k2 q p 0 0 2 2 where p = |~p | + m and P = |P~ |2 + m2 , and the scattering angle θ is defined as p~ · P~ = |~p ||P~ | cos θ. The differential scattering cross section is 1 d3 P 2π δ(p0 − P 0 ) M2 γ|~v | (2π)3 (P 0 /m) p where the relativistic factor γ ≡ 1/ 1 − |~v |2 is included in the denominator since we calculate in the lab frame. The energy conserving delta function implies |~p | = |P~ |, as expected: only the direction of the 3-momentum should change while the magnitude stays fixed. Choosing not to integrate over the solid angle but integrating over |P~ |, we get ! p ! |~p |2 + m2 1 4πα 2 dσ m |~p |2 p = (|~p |2 +m2 )+|~p |2 cos θ+m2 2 2 2 dΩ 2(2π) γ|~v | |~p | m k |~p |2 + m2 2 |~p | 4πα 2 2 = 2 |~ p | (1 + cos θ) + 2m 8π γ|~v |m k 2 dσ =

Using k 2 = −4|~p |2 sin2 (θ/2) from p. 133 along with cos θ = 1 − 2 sin2 (θ/2), we get 2 |~p | πα dσ 2 |~p |2 (1 − sin2 (θ/2)) + m2 = 2 2 2 dΩ 8π γ|~v |m |~p | sin (θ/2) 1 2 α2 2 2 2 = |~ p | + m − |~ p | sin (θ/2) 4 4γ|~v ||~p |2 sin (θ/2) m|~p | p The relativistic 3-momentum is p~ = γm~v , with γ ≡ 1/ 1 − |~v |2 as before. From now on, since there are no more 4-vectors, let v ≡ |~v | and p ≡ |~p |. The quantity in square brackets 58

simplifies as p2 + m2 − p2 sin2 (θ/2) = γ 2 m2 v 2 + m2 − γ 2 m2 v 2 sin2 (θ/2) 2 v v2 2 2 =m +1− sin (θ/2) 1 − v2 1 − v2 m2 2 2 1 − v sin (θ/2) = 1 − v2 = γ 2 m2 [1 − v 2 sin2 (θ/2)] The cross section is therefore 1 dσ α2 = γ 2 m2 [1 − v 2 sin2 (θ/2)] 4 2 dΩ m(γmv) 4γvp sin (θ/2) α2 = 2 2 4 [1 − v 2 sin2 (θ/2)] . 4v p sin (θ/2) 2. To order e2 the amplitude for positron scattering off a proton is just minus the amplitude (3) for electron scattering off a proton. Thus, somewhat counterintuitively, the differential cross sections for positron scattering off a proton and for electron scattering off a proton are the same to this order. Show that to the next order this is no longer true. Solution: Consider the 1-loop corrections to the propagator of the exchanged photon. These will all contribute with the same sign to the amplitudes for e+ p → e+ p and e− p → e− p. Since the tree level amplitudes differ by a sign, the amplitudes including 1-loop effects are different.

3. Show that the trace of a product of an odd number of gamma matrices vanishes. Solution: σαµα˙ The gamma matrices can be written in the Weyl basis as γ = . In this ˙ σ ¯ µαα 0 representation, it is immediately clear that the trace of an odd number of gamma matrices is zero simply because such a quantity cannot be written down. For example, µ ν αβ σαα˙ σ ¯ ˙ 0 tr(σ µ σ ¯ν ) 0 µ ν µ ν =⇒ tr(γ γ ) = γ γ = ˙ 0 σ ¯ µαα σαν β˙ 0 tr(¯ σµσν )

µ

but γ µγ ν γ ρ =

˙ 0 σαµα˙ σ ¯ ν αβ σβρ γ˙ ˙ ˙ σ ¯ µαα σαν β˙ σ ¯ ρβγ

!

cannot be traced over, since the spinor indices cannot be contracted. 59

0

6. For those who relish long calculations, determine the differential cross section for electronelectron scattering without taking the relativistic limit. Solution: R P For this problem, we use s us (~p )¯ us (~p ) =6 p + m for the spin sum and measure. The relevant Feynman rules are: electron propagator: i

p2

for the

6p + m − m2 + iε

−i ηµν p2 + iε

photon propagator:

d3 p (2π)3 2ωp~

(Feynman gauge)

vertex: = −ieγ µ An incoming electron gets a us (~p ) and an outgoing electron gets a u¯s0 (~p 0 ). We will use the notation ui ≡ usi (~pi ) and primes for the outgoing particles. There are two diagrams that contribute to M at tree level: 1

1’

1

q’

q

2

2’

2’

2

1’

Remembering the relative minus sign between these diagrams gives the amplitude: −iηµν −iηµν M = (¯ u10 ieγ µ u1 ) 2 (¯ u20 ieγ ν u2 ) − (¯ u20 ieγ µ u1 ) 0 2 (¯ u10 ieγ ν u2 ) + O(e4 ) q + iε (q ) + iε µ µ 0 0 0 0 (¯ u2 γ u1 )(¯ u1 γµ u2 ) u1 γ u1 )(¯ u2 γµ u2 ) 2 (¯ = ie − + O(e4 ) 2 2 (p1 − p10 ) + iε (p1 − p20 ) + iε Now we need |M|2 . Remember that M is a Lorentz-scalar (it is just a number), so the conjugation can be carried out as M∗ = M† . However, we are used to working with u and u¯ rather than u and u† , as required by Lorentz invariance. We have defined a barred spinor as ψ¯ ≡ ψ † γ 0 , so for consistency we define a barred matrix as M ≡ γ 0 M † γ 0 . For instance, (γ µ ) = γ 0 (γ µ )† γ 0 = γ 0 (γ 0 γ µ γ 0 )γ 0 = γ µ .

60

Just as (AB)† = B † A† , barring also reverses the order. So to compute the conjugate of the amplitude, we write |M|2 = MM : |M|2 /e4 =

A B C +D + − [(p1 − p10 )2 ]2 [(p1 − p20 )2 ]2 (p1 − p10 )2 (p1 − p20 )2

A = (¯ u10 γ µ u1 )(¯ u20 γµ u2 )(¯ u1 γ ν u10 )(¯ u2 γµ u20 ) µ ν B = (¯ u20 γ u1 )(¯ u10 γµ u2 )(¯ u1 γ u20 )(¯ u2 γν u10 ) µ ν C = (¯ u10 γ u1 )(¯ u20 γµ u2 )(¯ u1 γ u20 )(¯ u2 γν u10 ) µ ν D = (¯ u1 γ u10 )(¯ u2 γµ u20 )(¯ u20 γ u1 )(¯ u10 γν u2 ) Each set of parentheses denotes a Lorentz-scalar, meaning that all spinor indices are contracted. To figure out how the Dirac traces work, we write out the indices explicitly and rearrange everything in matrix multiplication order. Denoting a = 1, 2, 3, 4 as a Dirac 4spinor index, we rearrange as follows: A = u¯a10 (γ µ u1 )a u¯b20 (γµ u2 )b u¯c1 (γ ν u10 )c u¯d2 (γν u20 )d = (γ µ u1 u¯1 )ac (γ ν u10 u¯10 )ca (γµ u2 u¯2 )bd (γν u20 u¯20 )db = Tr [(γ µ u1 u¯1 )(γ ν u10 u¯10 )] Tr [(γµ u2 u¯2 )(γν u20 u¯20 )] B = u¯a20 (γ µ u1 )a u¯b10 (γµ u2 )b u¯c1 (γ ν u20 )c u¯d2 (γν u10 )d = (γ µ u1 u¯1 )ac (γ ν u20 u¯20 )ca (γµ u2 u¯2 )bd (γν u10 u¯10 )db = Tr[(γ µ u1 u¯1 )(γ ν u20 u¯20 )] Tr[(γµ u2 u¯2 )(γν u10 u¯10 )] C = u¯a10 (γ µ u1 )a u¯b20 (γµ u2 )b u¯c1 (γ µ u20 )c u¯d2 (γν u10 )d = (γ µ u1 u¯1 )ac (γ ν u20 u¯20 )cb (γµ u2 u¯2 )bd (γν u10 u¯10 )da = Tr[(γ µ u1 u¯1 )(γ ν u20 u¯20 )(γµ u2 u¯2 )(γν u10 u¯10 )] D = u¯a1 (γ µ u10 )a u¯b2 (γµ u20 )b u¯c20 (γ ν u1 )c u¯d10 (γν u2 )d = (γ µ u10 u¯10 )ad (γν u2 u¯2 )db (γµ u20 u¯20 )bc (γ ν u1 u¯1 )ca = Tr[(γ µ u10 u¯10 )(γν u2 u¯2 )(γµ u20 u¯20 )(γ ν u1 u¯1 )] P Now we would like to sum over all spins and use s ui u¯i = (6 pi + m). To compute the amplitude, we want to average over the initial spins, which gives a factor of (1/2) for each incoming electron. We will multiply by this factor of (1/2)2 = 1/4 later. For now, we simply define the spin-summed version of each of the above pieces: X X X X A , β≡ B , γ≡ C , δ≡ D α≡ s1 ,s2 ,s10 ,s20

s1 ,s2 ,s10 ,s20

s1 ,s2 ,s10 ,s20

61

s1 ,s2 ,s10 ,s20

We now simplify the above with the following gamma matrix identities: Tr[6 a6 b] = 4 (ab) Tr[6 a6 b6 c6 d] = 4 [(ab)(cd) + (ad)(bc) − (ac)(bd)] Tr[odd number of γs] = 0 γµ γ µ = 4 = η µν ηµν γµ6 aγ µ = −26 a γµ6 a6 bγ µ = 4(ab) γµ6 a6 b6 cγ µ = −26 c6 b6 a The notation (ab) = aµ bµ is used above. It is also useful to define a further condensed notation: (ij) ≡ pi µ pµj . We will only use the latter notation when there is no risk of confusion. Let us now simplify the first term: α = Tr[γ µ (6 p1 + m)γ ν (6 p10 + m)]Tr[γµ (6 p2 + m)γν (6 p20 + m)] = Tr[(γ µ6 p1 + mγ µ )(γ ν6 p10 + mγ ν )]Tr[(γµ6 p2 + mγµ )(γν6 p20 + mγν )] = Tr[γ µ6 p1 γ ν6 p10 + m2 γ µ γ ν ]Tr[γµ6 p2 γν6 p20 + m2 γµ γν ] = 4[(pµ10 pν1 + pν10 pµ1 ) − (p1 p10 )η µν + 4m2 η µν ]4[(p2µ p20 ν + p2ν p20 µ ) − (p2 p20 )ηµν + 4m2 ηµν ] The parentheses on the first term inside each [...] is there just to show that the term is symmetric in µ and ν, and it is thus useful to organize the calculation using this grouping. Continuing: α = 16[(pµ10 pν1 + pν10 pµ1 )(p2µ p20 ν + p2ν p20 µ ) − 2(p2 p20 )(p10 p1 ) + 4m2 × 2(p1 p10 ) − 2(p1 p10 )(p2 p20 ) + (p1 p10 )(p2 p20 ) η µν ηµν −4m2 (p1 p10 )η µν ηµν + 4m2 × 2(p2 p20 ) | {z } = +4

2

µν

4 µν

− 4m (p2 p20 )η ηµν + 16m η ηµν ] = 16[(10 2)(120 ) + (10 20 )(12) + (10 20 )(12) + (10 2)(120 ) − 2(220 )(10 1) + 8m2 (10 1) − 2(110 )(220 ) + 4(110 )(220 ) − 16m2 (110 ) + 8m2 (220 ) − 16m2 (220 ) + 64m4 ] = 16[2(10 2)(120 ) + 2(10 20 )(12) − 8m2 (110 ) − 8m2 (220 ) + 64m4 ] = 32[(10 2)(120 ) + (12)(10 20 ) − 4m2 (110 ) − 4m2 (220 ) + 32m4 ]

62

A quick check from dimensional analysis: Each (ij) has dimensions of m2 , so each term has dimensions of m4 , so the result is at least dimensionally correct. The next term is: β = Tr[γ µ (6 p1 + m)γ ν (6 p20 + m)]Tr[γµ (6 p2 + m)γν (6 p10 + m)] = Tr[(γ µ6 p1 + mγ µ )(γ ν6 p20 + mγ ν )]Tr[(γµ6 p2 + mγµ )(γν6 p10 + mγν )] = Tr[γ µ6 p1 γ ν6 p20 + m2 γ µ γ ν ]Tr[γµ6 p2 γν6 p10 + m2 γµ γν ] = 16[(pµ1 pν20 + pν1 pµ20 ) − (p1 p20 )η µν + 4m2 η µν ][(p2µ p10 ν + p2ν p10 µ ) − (p2 p10 )ηµν + 4m2 ηµν ] = 16[(pµ1 pν20 + pν1 pµ20 )(p2µ p10 ν + p2ν p10 µ ) − 2(120 )(210 ) + 4m2 × 2(120 ) − 2(120 )(210 ) + 4(120 )(210 ) − 4m2 (120 ) × 4 + 8m2 (210 ) − 4m2 (210 ) × 4 + 16m4 × 4] = 16[2(12)(10 20 ) + 2(110 )(220 ) − 2(120 )(10 2) + 8m2 (120 ) − 2(120 )(10 2) + 4(120 )(10 2) − 16m2 (120 ) + 8m2 (10 2) − 16m2 (10 2) + 64m4 ] = 16[2(12)(10 20 ) + 2(110 )(220 ) − 8m2 (10 2) − 8m2 (120 ) + 64m4 ] = 32[(12)(10 20 ) + (110 )(220 ) − 4m2 (10 2) − 4m2 (120 ) + 32m4 ] Now for the longer terms, the first of which is:

63

γ = Tr[γ µ (6 p1 + m)γ ν (6 p20 + m)γµ (6 p2 + m)γν (6 p10 + m)] = Tr[(γ µ6 p1 + mγ µ )(γ ν6 p20 + mγ ν )(γµ6 p2 + mγµ )(γν6 p10 + mγν )] = Tr[(γ µ6 p1 γ ν6 p20 + mγ µ6 p1 γ ν + mγ µ γ ν6 p20 + m2 γ µ γ ν ) × (γµ6 p2 γν6 p10 + mγµ6 p2 γν + mγµ γν6 p10 + m2 γµ γν )] = Tr[ γ µ6 p1 γ ν6 p20 γµ 6 p2 γν6 p10 + 0 + 0 + m2 γ µ6 p1 γ ν6 p20 γµ γν | | {z } {z } ν ν −26 p20 γ 6 p1 −26 p20 γ 6 p1 + 0 + m2 γ µ6 p1 γ ν γµ 6 p2 γν + m2 γ µ6 p1 γ ν γµ γν6 p10 + 0 | {z } | {z } 4pν1 4pν1 + 0 + m2 γ µ γ ν6 p20 γµ 6 p2 γν + m2 γ µ γ ν6 p20 γµ γν6 p10 + 0 | {z } | {z } ν 4p20 4pν20 + m2 γ µ γ ν γµ6 p2 γν 6 p10 + 0 + 0 + m4 γ µ γ ν γµ γν ] | {z } | {z } 4p2µ −2γ ν = Tr[−26 p20 γ ν6 p16 p2 γν 6 p10 − 2m26 p20 γ ν6 p1 γν +4m26 p26 p1 | {z } | {z } 4(p1 p2 ) −26 p1 + 4m26 p16 p10 + 4m26 p206 p10 + 4m26 p26 p10 − 2m4 γ ν γν ] |{z} 4 = −8(p1 p2 )Tr[6 p206 p10 ] + 4m2 Tr[6 p206 p1 +6 p26 p1 +6 p16 p10 +6 p26 p20 +6 p206 p10 +6 p26 p10 ] − 8m4 Tr[I4 ] = −32(p1 p2 )(p10 p20 )+16m2 [(p20 p1 )+(p2 p1 )+(p1 p10 )+(p2 p20 )+(p10 p20 )+(p2 p10 )] − 32m4

64

One more: δ = Tr[γ µ (6 p10 + m)γν (6 p2 + m)γµ (6 p20 + m)γ ν (6 p1 + m)] = Tr[(γ µ6 p10 + mγ µ )(γν6 p2 + mγν )(γµ6 p20 + mγµ )(γ ν6 p1 + mγ ν )] = Tr[(γ µ6 p10 γν6 p2 + mγ µ6 p10 γν + mγ µ γν6 p2 + m2 γ µ γν ) × (γµ6 p20 γ ν6 p1 + mγµ6 p20 γ ν + mγµ γ ν6 p1 + m2 γµ γ ν )] = Tr[ γ µ6 p10 γν6 p2 γµ 6 p20 γ ν6 p1 + 0 + 0 + m2 γ µ6 p10 γν6 p2 γµ γ ν {z } | {z } | −26 p2 γν6 p10 −26 p2 γν6 p10 + 0 + m2 γ µ6 p10 γν γµ6 p20 γ ν + m2 γ µ6 p10 γν γµ γ ν6 p1 + 0 | {z } | {z } 4p10 ν 4p10 ν + 0 + m2 γ µ γν6 p2 γµ 6 p20 γ ν + m2 γ µ γν6 p2 γµ γ ν6 p1 + 0 | {z } | {z } 4p2ν 4p2ν + m2 γ µ γν γµ6 p20 γ ν 6 p1 + 0 + 0 + m4 γ µ γν γµ γ ν ] | {z } | {z } 4p20 µ −2γν = Tr[−26 p2 γν6 p106 p20 γ ν 6 p1 − 2m26 p2 γν6 p10 γ ν | {z } | {z } 4(p10 p20 ) −26 p10 + 4m2 (6 p206 p10 +6 p106 p1 +6 p206 p2 +6 p26 p1 +6 p206 p1 ) − 2m4 γν γ ν ] |{z} 4 = −8(p10 p20 )Tr[6 p26 p1 ] + 4m2 Tr[6 p206 p10 +6 p106 p1 +6 p206 p2 +6 p26 p1 +6 p206 p1 ] − 8m4 Tr[I4 ] = −32(p10 p20 )(p1 p2 ) + 16m2 [(p10 p2 ) + (p10 p20 ) + (p1 p10 ) + (p2 p20 ) + (p1 p2 ) + (p1 p20 )] − 32m4 We see that this equals the previous term: γ = δ. Now we have all the terms we need to compute the amplitude-squared. Define theP amplitude-squared after averaging over initial 1 2 spins and summing over final spins: M ≡ 4 {all spins} |M|2 . We have: e4 α β 2δ 2 + − M = 4 [(p1 − p10 )2 ]2 [(p1 − p20 )2 ]2 (p1 − p10 )2 (p1 − p20 )2 α = 32[(10 2)(120 ) + (12)(10 20 ) − 4m2 {(110 ) + 220 )} + 32m4 ] β = 32[(12)(10 20 ) + (110 )(220 ) − 4m2 {(10 2) + (10 2)} + 32m4 ] δ = −16[2(12)(10 20 ) − m2 {(12) + (10 2) + (10 20 ) + (110 ) + (220 ) + (120 )} + 2m4 ] where (ij) ≡ pi µ pjµ 65

Until now, we have not used any information about the actual kinematics of the scattering process. Remembering that all external particles are on-shell with equal masses, we know p2i = m2 , where m is the of the electron. We now define the Mandelstam variables s, t, u Pmass 2 such that s + t + u = i mi = 4m2 : 1 s ≡ (p1 + p2 )2 = (p21 + p22 + 2(12)) = 2m2 + 2(12) =⇒ (12) = (s − 2m2 ) 2 = (p10 + p20 )2 =⇒ (10 20 ) = (12) 1 t ≡ (p1 − p10 )2 = (p21 + p210 − 2(110 )) = 2m2 − 2(110 ) =⇒ (110 ) = − (t − 2m2 ) 2 = (p2 − p20 )2 =⇒ (220 ) = (110 ) 1 u ≡ (p1 − p20 )2 = (p21 + p220 − 2(120 )) = 2m2 − 2(120 ) =⇒ (120 ) = − (u − 2m2 ) 2 = (p2 − p10 )2 =⇒ (210 ) = (120 ) Therefore, the pieces of the amplitude-squared are: α = 32[(10 2)2 + (12)2 − 8m2 (110 ) + 32m4 ] β = 32[(12)2 + (110 )2 − 8m2 (10 2) + 32m4 ] δ = −16[2(12)2 − 2m2 {(120 ) + (12) + (110 )} + 2m4 ] Now that we have gotten this far, we will drop the mass of the electron to check with the results in the text. Setting m = 0 simplifies the dot products: (12) = +s/2 , (110 ) = −t/2 , (120 ) = −u/2, where now s + t + u = 0. The pieces of the amplitude-squared are now: 1 1 α = 32[ u2 + s2 ] = 8(s2 + u2 ) 4 4 1 2 1 2 β = 32[ s + t ] = 8(s2 + t2 ) 4 4 1 2 δ = −16[ s ] = −8s2 2 The amplitude-squared is therefore: e4 α β 2δ 2 M = + − 4 t2 u2 tu e4 8(s2 + u2 ) 8(s2 + t2 ) 2(−8s2 ) = + − 4 t2 u2 tu 2 2 2 2 2 s +u s +t 2s = 2e4 + + 2 2 t u tu 66

2 From the definition of s, we see that s = ECM , where ECM is the total energy in the centerof-mass frame. We now calculate the differential cross section, whose formula is given in Appendix C. The differential cross section in the center-of-mass frame is:

dσCM =

d3 p10 d3 p20 √ M2 (2π)4 δ 4 (p1 + p2 − p10 − p20 ) (2π)3 2p010 (2π)3 2p020 4|~p1 |CM s 1

p where p0i = |~pi |2 + m2 = |~pi | for m = 0. For 4 identical particles with zero mass, the prefactor can be simplified: q √ 1 1 |~p1 |CM s = s2 − 2(m21 + m22 )s + (m21 − m22 )2 = s 2 2 Now, the 4-dimensional delta function is a product of a one dimensional energy-conserving delta function and a 3-dimensional momentum-conserving delta function. As discussed, the √ 0 0 total energy of the incoming electrons in the center-of-mass frame is p1 + p2 = s . Furthermore, the center-of-mass frame is defined such that p~1 + p~2 = ~0, so the differential cross section is: dσCM =

=

d3 p20 1 2 d3 p10 M (2π)4 δ 4 (p1 + p2 − p10 − p20 ) 2s (2π)3 2p010 (2π)3 2p020 M2 d3 p10 d3 p20 √ δ( s − |~p10 | − |~p20 |) δ 3 (~p10 + p~20 ) 3 2 2 (2π) s |~p10 | |~p20 |

We see that the 3-dimensional delta function simply sets p~20 = −~p10 : Z d3 p20 f (~p10 , p~20 ) δ 3 (~p10 + p~20 ) = f (~p10 , −~p10 ) Therefore, integrating over p~20 along with the fact that δ(2x) = 12 δ(x) gives: dσCM =

M2 d3 p1 0 √ δ( s − |~p10 |) 24 (2π)2 s |~p10 |2

Now, using spherical coordinates for p~10 gives d3 p10 = dΩ dp p2 , where Ω is the solid angle over which you are not asked to integrate, and p ≡ |~p10 |. The differential cross section per solid angle is now a 1-dimensional integral: √ Z ∞ dσ 1 s 2 = dp M δ −p dΩ CM 24 (2π)2 s 0 2 √ The delta function just sets p = s /2, so all we have to do now is figure out how these integrals over delta functions affected M2 . We can do this by looking at the Mandelstam

67

variables: t = (p1 − p10 )2 = −2p1µ pµ10 = 2(−p01 p010 + p~1 · p~10 ) = 2(−|~p1 ||~p10 | + |~p1 ||~p10 | cos θ) √ = −2|~p1 ||~p10 |(1 − cos θ) = − s |~p10 |(1 − cos θ) √ √ s s (1 − cos θ) = − (1 − cos θ) =− s 2 2 2 = −s sin (θ/2) Using s + t + u = 0, the other kinematic variable is s s s u = −s − t = −s + (1 − cos θ) = − + s cos θ = − (1 + cos θ) = −s cos2 (θ/2) 2 2 2 The amplitude-squared is therefore: 2 s + u2 s2 + t2 2s2 2 4 + + M = 2e t2 u2 tu 1 + cos4 (θ/2) 1 + sin4 (θ/2) 2 4 = 2e + + cos4 (θ/2) sin4 (θ/2) cos2 (θ/2) sin2 (θ/2) {z } | ≡ f (θ) Putting it all together:

dσ dΩ

CM

2e4 f (θ) M2 = = = 4 2 (2π)2 s 24 (2π)2 s

e4 25 π 2

f (θ) = s

e2 4π

2

f (θ) 2 2ECM

2 Finally, ECM = E1 + E2 = 2E =⇒ ECM = 4E, so we get the result:

II.7

dσ dΩ

= CM

e2 4π

2

f (θ) 8E 2

Diagrammatic Proof of Gauge Invariance

1. Extend the proof to cover figure II.7.1c. [Hint: To get oriented, note that figure II.7.1b corresponds to n = 1.] Solution: We will work out the case n = 2, which will make the generalization to arbitrary n clear. There are now (2 + 1)! = 6 diagrams, corresponding to the different ways in which the photon lines can cross. We will compute them in groups of three. First consider the three diagrams given below: 68

p’ 2

k2

p’−k 2

1

p’

k2

2

k

p’−k 2

k2

p’−k k1

k1

k

1

k1 p+k1

p+k

p

p’

k

p+k2 1

2

p

p

where we choose the convention in which the internal photon momenta always flow towards the left. The three ks are related by p0 − p = k + k1 + k2 . The amplitude given by adding (bose statistics) these three diagrams is: 1 1 1 1 1 1 λ2 λ1 λ1 λ1 λ2 0 λ2 6k + γ γ γ 6k γ + 6k 0 γ M1 = u¯ γ u. 6 p 0 − 6 k2 6p +6k 6 p 0 − 6 k2 6 p + 6 k1 6p −6k 6 p + 6 k2 Here we have dropped the fermion masses since they play no role in the proof. We have also suppressed all irrelevant factors in the amplitude, retaining only the kµ term in the photon propagator, as explained in the text. In the numerator of the first term, write 6 k = (6 p +6 k)−6 p, and in the third term write 6 k = −(6 p 0 − 6 k)+ 6 p 0 . The Dirac equation is now 6 pu = 0 and 6 p 0 u0 = 0, so that the 6 p gives zero acting to the right, and 6 p 0 gives zero acting to the left. We have 1 1 1 1 λ1 λ1 λ2 λ2 λ1 0 λ2 γ +γ 6k γ −γ γ u M1 = u¯ γ 6 p 0 − 6 k2 6 p 0 − 6 k2 6 p + 6 k1 6 p + 6 k2 1 1 1 1 λ1 λ1 λ2 0 λ2 λ1 λ2 γ +γ 6k γ −γ γ = u¯ γ u 6 p + 6 k+ 6 k1 6 p + 6 k+ 6 k1 6 p + 6 k1 6 p + 6 k2 where in the second line we have written p 0 − k2 = p + k + k1 . In the numerator of the middle term, write k = (p + k1 + k) − (p + k1 ) to get: 1 1 1 1 λ1 λ2 λ1 λ2 λ1 λ1 λ2 0 λ2 γ +γ γ −γ γ −γ γ u M1 = u¯ γ 6 p + 6 k+ 6 k1 6 p + 6 k1 6 p + 6 k+ 6 k1 6 p + 6 k2 1 1 0 λ2 λ1 λ1 λ2 = u¯ γ γ −γ γ u 6 p + 6 k1 6 p + 6 k2 where we have canceled the first and third terms from the first line. We see that the remaining two terms do not cancel; indeed, this is one way to discover additional diagrams in case you had forgotten them. That is, it is obvious that there are at least 3 diagrams but it may have escaped your attention that there are actually a total of 3! = 6 diagrams. The remaining three diagrams are displayed below: 69

p’

p’ 1

k1 p’−k 1

k2

p’ 1

k1

p’−k 1

k p’−k

k

2

p+k k

2

p+k1

p+k2 k2

1

2

p

k2

p

k1

p

As before, two of the resulting terms will cancel each other and the remaining terms will cancel the leftover terms in M1 . The amplitude from these three new diagrams is: 1 1 1 1 1 1 λ1 λ1 λ2 λ2 λ2 λ1 M2 = u¯ γ 6k + γ γ γ 6k γ + 6k 0 γ u. 6 p 0 − 6 k1 6p +6k 6 p 0 − 6 k1 6 p + 6 k2 6p −6k 6 p + 6 k1 In the numerator of the first term, write k = (p + k) − p and in the third term write k = −(p 0 − k) + p0 . Again 6 p acting to the right gives zero, and 6 p 0 acting to the left gives zero. The amplitude is 1 1 1 1 λ2 λ2 λ1 λ1 λ2 λ1 γ +γ 6k γ −γ γ u M2 = u¯ γ 6 p 0 − 6 k1 6 p 0 − 6 k1 6 p + 6 k2 6 p + 6 k1 1 1 1 1 λ2 λ2 λ1 λ1 λ1 λ2 γ +γ 6k γ −γ γ = u¯ γ u 6 p + 6 k+ 6 k2 6 p + 6 k+ 6 k2 6 p + 6 k2 6 p + 6 k1 where we have used p0 − k1 = p + k + k2 . In the numerator of the middle term, write k = (p + k + k2 ) − (p + k2 ) to get: 1 1 1 1 λ1 λ2 λ2 λ2 λ1 λ1 λ1 λ2 M2 = u¯ γ γ +γ γ −γ γ −γ γ u 6 p + 6 k+ 6 k2 6 p + 6 k2 6 p + 6 k+ 6 k2 6 p + 6 k1 1 1 λ2 λ2 λ1 λ1 γ −γ γ = u¯ γ u = −M1 6 p + 6 k2 6 p + 6 k1 Thus the contribution from the offending term kµ kν /µ2 in the photon propagator contributes a total of M = M1 + M2 = 0 to the amplitude. In general, the photon with momentum k can connect to the left-most line in (n + 1) ways, which results in (n + 1)! different diagrams, whose constituent terms will cancel in pairs.

2. You might have worried whether the shift of integration variable is allowed. Rationalizing the denominators in the first integral Z d4 p 1 1 1 ν σ λ tr γ γ γ (2π)4 6 p2 − m 6 p1 − m 6 p − m 70

in (13) and imagining doing the trace, you can convince yourself that this integral is only logarithmically divergent and hence that the shift is allowed. This issue will come up again in chapter IV.7 and we are anticipating a bit here. Solution: Rationalizing the denominators gives Z Z d4 p 1 1 d4 p 1 N νσλ σ λ ν γ γ = tr γ I≡ (2π)4 6 p2 − m 6 p1 − m 6 p − m (2π)4 (p22 − m2 )(p21 − m2 )(p2 − m2 ) where the numerator is N νσλ = tr γ ν (6 p2 + m)γ σ (6 p1 + m)γ λ (6 p + m) and p1 ≡ p + q1 and p2 ≡ p + q2 . The key is that after suitably combining denominators and shifting integration variables, the cubic terms in the numerator will vanish by Lorentz invariance. For large loop momentum p the integral behaves as Z Z Z 2 p2 1 4 3 p I ∼ d p 2 3 ∼ dp p 6 ∼ dp ∼ ln p (p ) p p which depends logarithmically on the cutoff.

II.8

Photon-Electron Scattering and Crossing

1. Show that averaging and summing over photon polarizations amounts to replacing the 0 square bracket in (9) by 2[ ωω + ωω0 − sin2 θ]. [Hint: We are working in the transverse gauge.] Solution: Let us first derive a general Lorentz-covariant definition of the sum over photon polarizations. For a circularly polarized3 photon moving in the +ˆ z direction, the polarization vectors can be taken as 0 0 1 1 1 1 µ . √ , ε = εµ1 = √ 2 2 +i 2 −i 0 0 Explicitly computing the sum over polarizations 0 X 0 εµa εν∗ a = 0 a = 1,2 0 3

gives 0 1 0 0

0 0 1 0

0 0 . 0 0

We are using circular polarization to show an example with complex polarization vectors.

71

We would like to write this in terms of Lorentz-covariant notation. The 4-momentum of the photon is k µ = |~k |(1, 0, 0, 1)T , so that 1 0 0 1 0 0 0 0 kµkν . = 2 0 0 0 0 ~ |k | 1 0 0 1 Define the spacelike unit vector in the direction of propagation of the photon: nµ = (0, 0, 0, 1)T . Then we have 0 0 0 1 k µ nν + k ν nµ 0 0 0 0 . = 0 0 0 0 ~ |k | 1 0 0 2 Furthermore, we can write |~k | = −k · n. Therefore, using the Lorentz-invariant metric η µν = diag(+1, −1, −1, −1) we can write X

εµa (k)ενa (k)∗ = −η µν +

a = 1,2

kµkν k µ nν + k ν nµ + . k·n (k · n)2

This is valid for any reference frame. We may now proceed with the problem. We can use the above result to write: " #" # X X S≡ εµa (k)ενa (k)∗ εa0 µ (k 0 )εa0 ν (k 0 )∗ = a0 = 1,2

a = 1,2

2 (k · n)(k 0 · n0 )

(k 0 · n) (k · n0 ) 1 (k · k 0 )2 (k · n )(k · n)+(k · k ) (n · n )+ 0 0 + + . (k · n ) (k · n) 2 (k · n)(k 0 · n0 ) 0

0

0

0

Now let nµ = (0, zˆ)T , k µ = ω(1, zˆ)T and sin θ 1 0 k 0µ = ω 0 , nµ = , n ˆ0 = 0 . 0 n ˆ n ˆ0 cos θ We have: k · k 0 = ωω 0 (1 − cos θ) , n · n0 = − cos θ k · n = −ω , k 0 · n0 = −ω 0 k · n0 = −~k · n ˆ 0 = −ω cos θ , k 0 · n = −~k 0 · n ˆ = −ω 0 cos θ .

72

Therefore (let c ≡ cos θ): −ω 0 c −ωc 1 [ωω 0 (1 − c)]2 2 0 0 (ω c)(ωc) + ωω (1 − c) −c + + + S= ωω 0 −ω 0 −ω 2 [−ω][−ω 0 ] 2 = 2 c + (1 − c)c + 12 (1 − c)2 = 2{c2 + c − c2 + 21 (1 − 2c + c2 )} = 2{c + 21 − c + 12 c2 } = 1 + c2 = 2 − sin2 θ . The problem wants us to take equation (9) on p. 154 and replace it with X X XX 2 1 M2 ≡ 12 |M| 2 a = 1,2 a0 = 1,2

= = = =

e4 (2m)2 e4 (2m)2 e4 (2m)2 e4 (2m)2

0 ω 2 ω 0 ω 2 ω 0 ω 2 ω 0 ω 2 ω

+ + + +

ω 1 − 2 + 4 × 2S ω0 ω 2 − 4 + 2(2 − sin θ) ω0 ω 2 − 2 sin θ ω0 ω 2 − sin θ . ω0

That is the result we were asked to prove.

2. Repeat the calculation of Compton scattering for circularly polarized photons. Solution: The explicit forms of the photon polarization vectors never entered into the derivation of the Klein-Nishina formula on p. 155; the only assumption was that the polarization vectors were real. For circularly polarized photons, everything goes through as before except that the εµa are complex. The result is equation (12) on p. 155 with the replacement (εε0 )2 → |ε∗ ε0 |2 .

73

III III.1

Renormalization and Gauge Invariance Cutting Off Our Ignorance

1. Work through the manipulations leading to (9) without referring to the text. Solution: Start with equation (3): M = −iλbare +

iCλ2bare

log

Λ2 s

+ log

Λ2 t

+ log

Λ2 u

+ O(λ3bare )

Here we write the physical coupling as λ and the unphysical bare coupling as λbare . We measure the physical coupling λ by performing some experiment at a particular centerof-mass energy and so on, and thus obtain equation (4): 2 2 2 Λ Λ Λ 2 −iλ = −iλbare + iCλbare log + log + log + O(λ3bare ) s0 t0 u0 We now pass to the notation of equations (5) and (6), where the sum of logs in the square brackets is denoted simply by L: (5) (6)

M = −iλbare + iCλ2bare L + O(λ3bare ) − iλ = −iλbare + iCλ2bare L0 + O(λ3bare )

Eliminate λbare by solving for it in terms of λ: −iλbare = −iλ − iCλ2bare L0 + O(λ3bare ) 2 = −iλ + iC −iλ + O(λ2 ) L0 + O(λ3 ) = −iλ − iCλ2 L0 + O(λ3 ) Now plug this into (5) to get: 2 M = −iλ − iCλ2 + O(λ3 ) − iC −iλ + O(λ2 ) L + O(λ3 ) = −iλ − iCλ2 L0 + iCλ2 L + O(λ3 ) = −iλ + iCλ2 (L − L0 ) + O(λ3 ) u t0 s0 0 2 + log + log = −iλ + iCλ log s t u Note that all dependence on the arbitrary cutoff has disappeared.

74

III.3

Counterterms and Physical Perturbation Theory

1. Show that in (1 + 1)-dimensional spacetime the Dirac field ψ has mass dimension 12 , and hence the Fermi coupling is dimensionless. Solution: R The action S = dd x L is dimensionless, so L has dimensions of (mass)d , denoted by [L] = d. ¯ µ ∂µ ψ gives: The derivative ∂µ has mass dimension +1, so the kinetic term ψγ d−1 L = iψ6¯ ∂ψ =⇒ [L] = [∂] + 2[ψ] =⇒ d = 1 + 2[ψ] =⇒ [ψ] = 2 Therefore, for d = 1 + 1 = 2, we have [ψ] = 1/2. ¯ ψψ). ¯ The Fermi interaction has the form L = G (ψψ)( As above, this implies: d−1 [L] = [G] + 4[ψ] =⇒ d = [G] + 4 =⇒ [G] = 2 − d 2 In d = 3 + 1 = 4, [G] = −2, as discussed in the text. In d = 1 + 1 = 2, we have [G] = 0, so the Fermi coupling is dimensionless and thus the theory is renormalizable.

2. Derive (11) and (13). Solution: Consider first the Yukawa theory, defined by the Lagrangian density: ¯ ∂ − m)ψ + 1 [(∂ϕ)2 − µ2 ϕ2 ] −λ ϕ4 + f ϕψψ ¯ L = ψ(i6 | {z } 2 ≡ Lint As in the book, we are really doing physical (“dressed”) perturbation theory, in which case we should have renormalizing factors for each term or, equivalently, we should add a series of counterterms. For this problem, all that matters is the structure of the Lagrangian, so this issue is not important. Introduce the sources J for ϕ, η for ψ¯ and η¯ for ψ and expand the path integral with sources in a Taylor series, as in equation (4) on page 44 and as described below equation (20)

75

on page 128: Z

¯

¯ DψDψDϕ e iS[ψ,ψ,ϕ]+i

Z(η, η¯, J) = =e ∼

i

R

R

¯ d4 x(Jϕ+¯ η ψ+ψη)

¯ d4 x Lint (ϕ→−iδJ ,ψ→−iδη ,ψ→+iδ η ¯)

∞ X Vλ = 0

[δJ4 ]Vλ

X

[δJ δη δη¯]Vf

∞ X BI = 0

Vf = 0

Z

¯

¯ DψDψDϕ e iS[ψ,ψ,ϕ]

[J Gϕ J]BI

∞ X

[¯ η Gψ η]FI

FI = 0

The tilde means we are being schematic and ignoring all numerical factors, which are irrelevant for this problem. Gϕ is the propagator for ϕ, and Gψ is the propagator for ψ, both of which appear from performing the Gaussian integrals as usual. Vλ , Vf , BI and BF are, at this level, simply dummy summation variables. These names are chosen for a reason: Vλ counts the number of λϕ4 vertices, Vf counts ¯ vertices, BI counts the number of scalar propagators (“internal boson the number of f ϕψψ lines”), and FI counts the number of fermion propagators (“internal fermion lines”). Moreover, define BE ≡ # Js killed, and FE ≡ # ηs killed + # η¯s killed. How many Js are killed? Each δJ kills one, and there are 4Vλ + Vf of them in a given term, as can be seen directly from the terms [δJ4 ]Vλ and [δJ ...]Vf . How many Js persist? Each propagator term goes with two Js, so 2BI persist, as can be seen from the term [J...J]BI . Thus BE = 4Vλ +Vf −2BI , which matches equation (10) on page 163. Similarly, each vertex kills Vf ηs, as can be read off from the term [...δη ...]Vf , and each fermion propagator goes with one η, as can be read from [...η]FI , so the number of ηs killed is Vf − FI . The number of η¯s killed is the same, so FE = 2(Vf − FI ). If the theory is renormalizable, then D should be independent of Vλ and Vf , so in anticipation of this let’s solve for Vλ and Vf in terms of the other quantities: 1 Vf = F I + F E 2 1 4Vλ = BE + 2BI − Vf = BE + 2BI − FI − FE 2 Now we need to count R powers of momentum. As explained on page 162, the number of loops L is the number of d4 k we have to perform, so the superficial degree of divergence D gets a +4L. R Each internal Rline carries a d4 k, but each vertex carries a δ 4 (k) that cancels those, so the number of d4 k that we actually have to perform is L = BI + FI − (Vλ + Vf − 1), where the −1 accounts for the overall momentum conserving delta function. Each boson propagator goes like 1/k 2 , and each fermion propagator goes like 1/k, so D gets a contribution of −2BI − FI . All together, we have: D = 4L − 2BI − FI = 4[BI + FI − (Vλ + Vf − 1)] − 2BI − FI = 2BI + 3FI − 4(Vλ + Vf ) + 4 76

Plugging in the values for Vf and Vλ gives: 1 1 D = 2BI + 3FI − BE + 2BI − FI − FE − 4 FI + FE + 4 2 2 1 = (2 − 2)BI + (4 − 4)FI − BE + FE − 4FI − 2FE + 4 2 3 = −BE − FE + 4 2 Now consider the Fermi theory in (1+1) dimensions: ¯ ∂ − m)ψ − G (ψψ) ¯ 2 L = ψ(i6 The path integral with sources is ∞ X

Z∼

[(δη δη¯)2 ]V

V =0

∞ X

[¯ η Gψ η]FI

FI = 0

Same procedure asRabove: FE ≡ 2×(# ηs killed) = 2(2V − FI ) =⇒ 2V = 21 FE + FI . This time, each loop is d2 k, so D = 2L − FI , as stated above equation (13) on page 165. The number of loops is L = FI − (V − 1) = FI − 12 ( 12 FE + FI ) + 1 = 12 FI − 14 FE + 1. So the superficial degree of divergence is: 1 1 1 FI − FE + 1 − FI = − FE + 2 D=2 2 4 2 That is equation (13) on page 165.

5. Show that the result P = L − 1 holds for all the theories we have studied. Solution: Take the QED Lagrangian rescaled by L → L/~ ¯ µ ψ − 1 Fµν F µν ¯ 1 (i6 ∂ − m)]ψ + 1 eAµ ψγ L = ψ[ ~ ~ 4~ Each fermion propagator contributes ~, each photon propagator contributes ~, and each fermion-photon vertex contributes 1/~. Therefore the total power of ~ in a diagram is P = FI + AI − V , where FI is the number of internal fermion lines, AI is the number of internal photon lines, and V is the number of vertices. R But the number of loops L is the number of d4 k/[(2π)4 ] we have to do, which is equal to the number of momenta flowing through each P internal line minus the number of vertices, each of which conserves momentum with a δ 4 ( k). So L = FI + AI − V , or in other words P = L − 1 again. It is clear that this happens for any theory. 77

III.5

Field Theory without Relativity

1. Obtain the Klein-Gordon equation for a particle in an electrostatic potential (such as that of the nucleus) by the gauge principle of replacing (∂/∂t) in (2) by ∂/∂t − ieA0 . Show that in the nonrelativistic limit this reduces to the Schr¨odinger’s equation for a particle in an external potential. Solution: To add a photon to the relativistic free field theory L = (∂Φ† )(∂Φ) − m2 Φ† Φ, replace the partial derivatives with covariant derivatives as ∂µ Φ → Dµ Φ ≡ (∂µ − ieAµ )Φ. (We also need the photon kinetic term − 41 Fµν F µν , but it will play no role in what follows, so we drop it.) The Lagrangian is L = (Dµ Φ)† Dµ Φ + m2 Φ† Φ = ∂µ Φ† ∂ µ Φ + m2 Φ† Φ + ieAµ (Φ† ∂µ Φ − Φ∂µ Φ† ) + e2 Φ† ΦAµ Aµ with a purely electric potential independent of time: Aµ (~x, t) = V (~x)δ µ0 . Now take the non-relativistic limit as in the text: 1 1 e−imt ϕ(~x, t) =⇒ ∂t Φ = √ (−im ϕ + ∂t ϕ)e−imt Φ(~x, t) = √ 2m 2m Therefore Φ† ∂t Φ = √

1 1 +imt † 1 e ϕ √ (−im ϕ + ∂t ϕ)e−imt = (−im ϕ† ϕ + ϕ† ∂t ϕ) 2m 2m 2m

and so we have 1 1 (−im ϕ† ϕ + ϕ† ∂t ϕ) − (+im ϕ† ϕ + ϕ∂t ϕ† ) 2m 2m 1 = −i ϕ† ϕ + (ϕ† ∂t ϕ − ϕ∂t ϕ† ) . 2m The term linear in the potential is therefore 1 µ † † † † † ieA (Φ ∂µ Φ − Φ∂µ Φ ) = ieV (~x) −i ϕ ϕ + (ϕ ∂t ϕ − ϕ∂t ϕ ) 2m ie = eV (~x)ϕ† ϕ + V (~x)(ϕ† ∂t ϕ − ϕ∂t ϕ† ) 2m ie = eV (~x)ϕ† ϕ + V (~x)ϕ† ∂t ϕ + (total time derivative) m To this we add the term quadratic in the potential, Φ† ∂t Φ − Φ∂t Φ† =

e2 Φ† ΦAµ Aµ =

e2 † ϕ ϕV (~x)2 2m

Adapting the previously obtained Lagrangian from the book (equation (6) on p. 191), we obtain (up to total derivatives) 1 2 ie e2 † 2 L = ϕ i∂t + ∇ + V (~x)∂t + eV (~x) + V (~x) ϕ 2m m 2m 78

Set

δS δϕ†

= 0 to get the equation of motion: h i i e 1 2 h e 1 + V (~x) i∂t + ∇ + 1+ V (~x) eV (~x) ϕ(~x, t) = 0 m 2m 2m

In the non-relativistic limit m ∂t ϕ ϕ, and neglecting the term quadratic in the potential, we obtain 1 2 i∂t + ∇ + eV (~x) ϕ(~x, t) = 0 2m 2

∇ which is precisely the Schr¨odinger equation (i∂t + H)ϕ = 0 with H = 2m + eV the correct non-relativistic Hamiltonian for a particle in the presence of an external potential eV .

3. Given a field theory we can compute the scattering amplitude of two particles in the nonrelativistic limit. We then postulate an interaction potential U (~x) between the two particles and use nonrelativistic quantum mechanics to calculate the scattering amplitude, for example in Born approximation. Comparing the two scattering amplitudes we can determine U (~x). Derive the Yukawa and the Coulomb potentials this way. The application of this method to the λ(Φ† Φ)2 interaction is slightly problematic since the delta function interaction is a bit singular, but it should be all right for determining whether the force is repulsive or attractive. Solution: Given a scattering amplitude M, the non-relativistic potential between distinguishable particles of masses m1 and m2 is Z d3 q 1 M(~q ) e i~q·~x . U (~x ) = − 4m1 m2 (2π)3 The Yukawa potential is derived in the text. Here we consider the U (1)-invariant scalar Lagrangian λ L = ∂Φ∗ ∂Φ − m2 Φ∗ Φ − (Φ∗ Φ)2 . 4 We would like to compute the potential between a Φ meson and an anti-meson. The scattering amplitude is just the constant iM = −iλ, so the potential is Z 1 d3 q +λ 3 i~ q ·~ x U (~x ) = − 2 (−λ) e = δ (~x ) . 4m (2π)3 4m2 For λ > 0, this is a repulsive contact interaction. If you would like to consider a less singular Lagrangian, then write r 1 λ L = ∂Φ∗ ∂Φ − m2 Φ∗ Φ + (∂χ)2 − M 2 χ2 + iM χΦ∗ Φ 2 2 where χ is a real scalar field. Then consider the potential due to exchange of the field χ, which will reduce to the result obtained previously in the limit of taking the χ momentum to zero. 79

III.6

The Magnetic Moment of the Electron

3. By Lorentz invariance the right hand side of (7) has to be a vector. The only possibilities are u¯γ µ u, (p + p0 )µ u¯u, and (p − p0 )µ u¯u. The last term is ruled out because it would not be consistent with current conservation. Show that the form given in (7) is in fact the most general allowed. iσ µν qν 0 0 µ 0 0 2 µ 2 F2 (q ) u(p, s) , q ≡ p0 − p (7) hp , s |J (0)|p, si = u¯(p , s ) γ F1 (q ) + 2m Solution: First, for notational convenience, write u(p, s) = u and u(p0 , s0 ) = u0 . Lorentz invariance dictates that if we have one free µ index on the left, we need one free µ index on the right. The only such vectors we have are γ µ , q µ and (p + p0 )µ . However, since Lorentz invariance also dictates that we have no free spinor indices on the right, any vector we write down is going to be sandwiched between u¯0 and u. Therefore, we may use the Gordon decomposition rearranged as: u¯0 (p + p0 )µ u = u¯0 (2m γ µ − iσ µν qν ) u So between the spinors, (p + p0 ) is a linear combination of γ and q. Therefore anytime we write a function of (p + p0 ) we may as well write it as a function of q. Equation (7) follows immediately.

4. In chapter II.6, when discussing electron-proton scattering, we ignored the strong interaction that the proton participates in. Argue that the effects of the strong interaction could be included phenomenologically by replacing the vertex u¯(P, S)γ µ u(p, s) in (II.6.1) by iσ µν qν 2 µ µ 2 F2 (q ) u(p, s) (17) hP, S|J (0)|p, si = u¯(P, S) γ F1 (q ) + 2m Careful measurements of electron-proton scattering, thus determining the two proton form factors F1 (q 2 ) and F2 (q 2 ), earned R. Hofstadter the 1961 Nobel Prize. While we could account for the general behavior of these two form factors, we are still unable to calculate them from first principles (in contrast to the corresponding form factors for the electron.) See chapters IV.2 and VII.3. Solution: This form follows from parity and gauge invariance. Consult the literature, for example L. N. Hand, D. G. Miller and R. Wilson, “Electric and magnetic form factors of the nucleon,” Rev. Mod. Phys, Vol. 35, No. 2, Apr. 1963 and J. D. Bjorken and E. A. Paschos, “Inelastic Electron-Proton and γ-Proton Scattering and the Structure of the Nucleon,” Phys. Rev. Vol. 185, No. 3, 25 Sep. 1969. For a modern development regarding the form factors for the proton, see V. Barger, C. W. Chiang, W. Y. Keung and D. Marfatia, “Proton Size Anomaly,” Phys. Rev. Lett. 106:153001, 2011 (arXiv:1011.3519v2 [hep-ph]). 80

III.7

Polarizing the Vacuum and Renormalizing the Charge

1. Calculate Πµν (q) using dimensional regularization. Solution: For this problem, we will use the metric signature g = (−, +, +, +). The scheme of dimensional regularization is to calculate everything in d dimensions, then write d = 4 − ε and take ε → 0+ . The divergent piece will show up as a term proportional to 1/ε. The polarization tensor is:

µν

dd p Tr [(−6 k−6 p + m) γ µ (−6 p + m) γ ν ] (2π)d [(k + p)2 + m2 ] [p2 + m2 ] Z dd p Tr [(6 k+6 p)γ µ6 pγ ν + m2 γ µ γ ν ] =− (2π)d [(k + p)2 + m2 ] [p2 + m2 ] Z

iΠ (k) = −

The simplification followed from remembering that the trace of an odd number of gamma matrices is zero. In d dimensions, the gamma matrices have dimension 2d/2 , so the trace identities become: Tr[γ µ γ ν ] = −2d/2 g µν Tr[(6 k+6 p)γ ν6 pγ ν ] = 2d/2 [(k + p)µ pν + (k + p)ν pµ − g µν (k + p) · p ]

So the polarization tensor is now: Z dd p (k + p)µ pν + (k + p)ν pµ − g µν (k + p) · p − m2 g µν µν d/2 iΠ (k) = −2 (2π)d [(k + p)2 + m2 ] [p2 + m2 ] Play with the denominator using Feynman’s formula: Z 1 1 1 dx = AB [xA + (1 − x)B]2 0 Identifying A = (k + p)2 + m2 and B = p2 + m2 , the polarization tensor is:

81

µν

iΠ (k) = −2

= −2

= −2

= −2

d/2

d/2

d/2

d/2

Z

Z

Z

Z

dd p (2π)d

Z

dd p (2π)d

Z

dd p (2π)d

Z

dd p (2π)d

Z

1

dx

(k + p)µ pν + (k + p)ν pµ − g µν (k + p) · p − m2 g µν {x [(k + p)2 + m2 ] + (1 − x)(p2 + m2 )}2

dx

(k + p)µ pν + (k + p)ν pµ − g µν (k + p) · p − m2 g µν [x(k 2 + p2 + 2k · p + m2 ) + p2 + m2 − xp2 − xm2 ]2

dx

(k + p)µ pν + (k + p)ν pµ − g µν (k + p) · p − m2 g µν [p2 + 2x k · p + xk 2 + m2 ]2

dx

(k + p)µ pν + (k + p)ν pµ − g µν (k + p) · p − m2 g µν [(p + xk)2 + x(1 − x)k 2 + m2 ]2

0

1

0

1

0

1

0

R R Now interchange the order of integration so that the dd p comes before the dx, and shift the integration variable to q ≡ p + xk =⇒ p = q − xk. For notational convenience, also define D ≡ x(1 − x)k 2 + m2 . We get: iΠµν (k) = Z 1 Z d d q [q+(1−x)k]µ (q−xk)ν +(µ ↔ ν)−g µν [q+(1−x)k]·(q−xk)−m2 g µν d/2 −2 dx (2π)d [q 2 + D]2 0 Z 1 Z dd q 2q µ q ν − 2x(1 − x)k µ k ν − g µν [q 2 − x(1 − x)k 2 ] − m2 g µν = −2d/2 dx (2π)d [q 2 + D]2 0 We have dropped terms in the numerator that are in q, because now the denominator R linear 2 d µ is a function of the Lorentz-invariant q , so that d q q /(q 2 + D)2 = 0. On a related note: Z

d

µ ν

2

µν

Z

d q q q f (q ) = C g dd q q 2 f (q 2 ) by Lorentz invariance Z Z d µ ν 2 µν d 2 2 gµν d q q q f (q ) = C g d q q f (q ) Z Z dd q q 2 f (q 2 ) = C Tr(Id ) dd q q 2 f (q 2 ) 1=Cd Z =⇒

1 d q q q f (q ) = g µν d d

µ ν

2

Z

dd q q 2 f (q 2 )

Therefore, in the numerator of our integral we can substitute q µ q ν → g µν q 2 /d to get: 82

iΠµν (k) = Z 1 Z dd q d2 q 2 g µν − 2x(1 − x)k µ k ν − g µν [q 2 − x(1 − x)k 2 ] − m2 g µν d/2 dx = −2 (2π)d [q 2 + D]2 0 Z 1 Z dd q 2( d1 − 1) q 2 g µν − 2x(1 − x)k µ k ν + g µν x(1 − x)k 2 − m2 g µν dx = −2d/2 (2π)d [q 2 + D]2 0 Z 1 2 µν µ ν µν 2 2 µν d/2 dx − 1 g I1 + −2x(1 − x)k k + g x(1 − x)k − m g I0 = −2 d 0 Γ(2 − n − d2 )Γ(n + d2 ) −(2−n− d ) dd q (q 2 )n i 2 = D (2π)d [q 2 + D]2 (4π)d/2 Γ( d2 ) ε/2 4π i Γ(n + 2 − 2ε ) ε n = Γ(−n + ) D ε 2 16π Γ(2 − 2 ) 2 D Z

In ≡

As you can see, we have started to plug in d = 4 − ε. Before continuing, note a subtlety. In d = 4 spacetime dimensions, the gauge field Aµ has mass dimension +1 and the Dirac spinor ψ has mass dimension +3/2, so the gauge coupling e is dimensionless. But we are no longer working in d = 4 but rather in d = 4 − ε, so we need to find out how e changes with dimension. Taking typical terms in a Lagrangian, we get: d−2 2 d − 1 L = iψ¯ 6 ∂ψ =⇒ [ψ] = 2 ¯ µ ψ =⇒ [e] = 2 − d = ε L = e Aµ ψγ 2 2 L = −(∂A)2 =⇒ [A] =

To make this mass dimension explicit, we rewrite the gauge coupling as e → e˜ µε/2 , where the parameter µ ˜ has dimensions of mass, so that e remains dimensionless for all d. The important point here is that since Π appears with an e2 , this consideration sends Π → µ ˜ε Π. ε/2 2 ε/2 So in the expression for the integral In , the (4π/D) term is changed to (4π µ ˜ /D) . Here are a few more intermediate steps4 : 4

Keeping ε 6= 0 in 2d/2 = 4 × [1 + O(ε)] does not yield any new poles and is therefore not strictly necessary.

83

ε 2d/2 = 22−ε/2 = 4 × 2−ε/2 = 4 × 1 − ln 2 + O(ε2 ) 2 2 1 ε 1 ε 2 −1= −1= 1 + + O(ε2 ) − 1 = − + + O(ε2 ) d 4−ε 2 4 2 8 ε ε Γ(3 − 2 ) ε Γ(2 − 2 ) ε = 2− =2− ε ε Γ(2 − 2 ) 2 Γ(2 − 2 ) 2 ε 2 Γ −1 + = − + γ − 1 + O(ε) 2 ε ε 2 Γ = + − γ + O(ε) 2 ε 2 ε/2 4π µ ˜ 4π µ ˜2 ε + O(ε2 ) = 1 + ln D 2 D

Let’s put these into the expressions for I1 and I0 :

I1 = = = = = =

ε/2 i ε 4π µ ˜2 ε 2− Γ −1 + D 16π 2 2 2 D 4π µ ˜2 ε −2 ε i 2 1− + γ − 1 + O(ε) D 1 + ln + O(ε ) 8π 2 4 ε 2 D i −2 1 ε 4π µ ˜2 2 + γ − 1 + + O(ε) D 1 + ln + O(ε ) 8π 2 ε 2 2 D i −2 4π µ ˜2 D − ln + γ − 1 + O(ε) 8π 2 ε D i −2 4π µ ˜2 1 γ D − ln + ln e − + O(ε) 8π 2 ε D 2 2 −i 2 4π µ ˜ 1 D + ln + + O(ε) 2 γ 8π ε e D 2

I0 = = = =

i 16π 2 i 16π 2 i 16π 2 i 16π 2

ε/2 ε 4π µ ˜2 Γ 2 D 2 ε 4π µ ˜2 2 − γ + O(ε) 1 + ln + O(ε ) ε 2 D 2 4π µ ˜2 + ln − γ + O(ε) ε D 2 4π µ ˜2 + ln + O(ε) ε eγ D 84

Define µ2 ≡ 4π µ ˜2 /eγ for convenience. After all, µ ˜ was defined arbitrarily anyway, just as long as it has dimensions of mass. Notice that now everything has an overall factor of i, so divide by that factor and plug in all of the results so far into the expression for the polarization tensor: Πµν (k) = Z 1 2 µν µ ν µν 2 2 µν d/2 dx − 1 g I1 /i + −2x(1 − x)k k + g x(1 − x)k − m g I0 /i −2 d 0 2 Z 1 ε −1 ε −1 2 µ 1 2 µν dx + g D + ln + + O(ε) = −[4 − ln 2 + O(ε )] 2 2 8 8π ε D 2 0 ε − [4 − ln 2 + O(ε2 )]× 2 Z 12 1 2 µ µ ν µν 2 2 µν dx −2x(1 − x)k k + g x(1 − x)k − m g + ln + O(ε) 16π 2 ε D 0 2 iZ 1 1 h µ ε 1 1 1 1 2 µν = + 2 1 − ln 2 + O(ε ) dx g D − − ln − + + O(ε) 2π 8 ε 2 D 4 4 0 Z 1 h i ε 1 2 µ ν µν 2 2 µν 2 1 − ln 2 + O(ε ) dx −2x(1 − x)k k + g x(1 − x)k − m g − 4π 2 8 ε 2 Z0 1 h i ε 1 µ 2 µ ν µν 2 2 µν 1 − ln 2 + O(ε ) dx −2x(1 − x)k k + g x(1 − x)k − m g ln − 2 4π 8 D 0 + O(ε)

Now group the terms in orders of ε: Πµν ε−1 (k) = Z 1 µν 1 µ ν µν 2 2 µν 1 dx −g D + 2x(1 − x)k k − g x(1 − x)k + m g 2π 2 0 ε Z 1 1 1 = 2 dx −g µν [x(1 − x)k 2 + m2 ] + 2x(1 − x)k µ k ν − g µν x(1 − x)k 2 + m2 g µν 2π 0 ε Z 1 1 1 = 2 dx x(1 − x) k µ k ν − g µν k 2 π 0 ε 1 1 = 2 k µ k ν − g µν k 2 6π ε The divergent piece is indeed transverse. What about the finite piece?

85

Πµν ε0 (k) = 2 Z 1 Z µ ln 2 µν 1 1 −1 µν ln + dx g D g dx D 2π 2 0 2 D 16π 2 0 Z 1 ln 2 + dx −2x(1 − x)k µ k ν + g µν x(1 − x)k 2 − m2 g µν 2 16π 0 2 Z 1 1 µ µ ν µν 2 2 µν − 2 dx −2x(1 − x)k k + g x(1 − x)k − m g ln 4π 0 D 2 Z 1 µ 1 1 2 µν µ ν 2 2 2 1 µν 1 µν x(1−x)k + m + x(1−x)k k − 2 g x(1−x)k + m g ln = 2 dx − 2 g 2π 0 2 D 1 2 1 1 ln 2 g µν k + m2 − k µ k ν + g µν k 2 − m2 g µν + 2 16π 6 3 6 2 Z 1 µ ν 1 µ 2 µν = 2 dx x(1 − x) k k − k g ln 2π 0 D ln 2 2 µν + k g − kµkν 2 48π 1 = 2 k µ k ν − k 2 g µν 2π

Z

1

dx x(1 − x) ln

0

µ2 D

ln 2 + 24

The finite piece is also transverse. We have shown that dimensional regularization preserves the transverse structure of the polarization tensor and therefore preserves gauge invariance. We have shown that Πµν (k) = (k µ k ν − k 2 g µν )Π(k 2 ), where 2 Z 1 1 µ ln 2 1 2 dx x(1 − x) ln +A + Π(k ) = 2 + 2 6π ε 2π 0 D 24 with D ≡ x(1 − x)k 2 + m2 , and with A being the counterterm. That is exactly the form of equation (13) on page 187 from using Pauli-Villars regularization. To proceed with any physical calculation, we need to choose a renormalization scheme. Choose the “on-shell” renormalization scheme, in which the renormalized photon propagator has a pole at k 2 = 0 with residue 1, just like the tree-level propagator.R In other words, we de1 1 mand Πrenormalized (k 2 = 0) = 0. This fixes A = − 6πε − 2π1 2 ln(µ2 /m2 ) 0 dx x(1 − x) − ln 2/24, and therefore: Z 1 1 m2 2 Πrenormalized (k ) = 2 dx x(1 − x) ln 2π 0 x(1 − x)k 2 + m2 86

2. Study the modified Coulomb’s law as determined by the Fourier integral e2 Π(~q 2 )]}.

R

d3 q e i~q·~x /{~q 2 [1+

Solution: The modified Coulomb potential is V = Vcoul + δV , where Z e i~q·~r e2 e2 (as usual) = Vcoul (~r) = 2 d3 q 2π |~q |2 r δV (~r) = −

e2 2π 2

2 Z

e i~q·~r dq |~q |2 3

Z

1

0

m2 dx x(1 − x) ln m2 + x(1 − x)|~q |2 − iε

We have used the on-shell renormalized result for Π(k 2 ) from the previous problem, setting k 0 = 0 and ~k = ~q. First notice that the log has a branch cut at |~q |2 /4 + m2 < 0, which causes the log to have a nonzero imaginary piece, which means that with enough energy a virtual electronpositron pair can become real. This can happen if the energy is greater than 2m, which is precisely the branch point of the log. This is enough to show that there is a characteristic length scale of 1/(2m) in the potential. Now let us compute the integrals explicitly. First evaluate the angular part of the Fourier integral: Z ∞ Z 1 Z i~ q ·~ r 2 2 1 2 3 e f (|~q | ) = 2π dq q 2 f (q ) dη e iqrη dq |~q |2 q 0 −1 Z ∞ iqr e − e−iqr = 2π dq f (q 2 ) iqr 0 Therefore we have δV = − Z I≡

e i~q·~r dq ln |~q |2 3

e2 2π 2

2 R

1 0

dx x(1 − x) I, where

m2 m2 + x(1 − x)|~q |2 − iε

2πi =− r

Z 0

∞

e+iu − e−iu du ln u

1 1 + βu2 − iε

with β ≡ x(1 − x)/(mr)2 . Now we will evaluate the integral over u by the method of contour integration. Consider the integral I e iz J≡ dz ln(1 + βz 2 ) z C over the complex variable z, with the contour C = Γ2 + C3 + γ1 + C2 + γ2 + C4 + Γ1 + C1 as shown in the picture: 87

Im z

C4

2

C2

C3

1

2m

C1 −R

2

1

+R

Re z

As stated √ previously and as shown in the diagram, the log has a branch cut starting at z = i/ β extending along the imaginary axis to z → +i∞. Taking the radius R of the contour to infinity and the radius of the small semicircle C1 to zero gives Z Z ∞ e iu du ln(1 + βu2 ) . = u Γ1 +Γ2 −∞ R The residue of the pole at C1 is ln 1 = 0, so C1 = 0. Along γ2 , the log function can be written ln z = ln |z| + iθ, where θ = π/2 is the location of the cut. The line γ1 is on the other side of the cut, so the angle picks up an extra factor of 2π and hence on γ1 , we write ln z = ln |z| + i(θ + 2π). Furthermore,√along γ2 the complex variable of integration is z = 0− + iy, where y goes from R to 1/ β . The zero indicates that we take γ2 arbitrarily close √ to the left-hand side of the cut. Along γ1 , we have z = 0+ + iy, where now y goes from 1/ β to R. Therefore, the only contribution to the integral from γ1 + γ2 is the part along the cut, since the rest cancels out. We have Z Z ∞ e−y = −2πi √ dy y γ1 +γ2 1/ β where we have again taken R → ∞. Since the contour does not enclose any residues, the whole contour integral J equals zero. Therefore: Z ∞ Z ∞ e iu e−y 2 du ln(1 + βu ) = 2πi √ dy . u y −∞ 1/ β We also need Z ∞ Z −∞ Z ∞ e−iu e+iu e iu 2 2 du ln(1 + βu ) = (−du) ln(1 + βu ) = − du ln(1 + βu2 ) u (−u) u −∞ +∞ −∞ 88

and

Z

∞

0

so therefore:

e+iu − e−iu 1 du f (u2 ) = u 2

Z

∞

du −∞

e+iu − e−iu f (u2 ) u

∞

e iu − e−iu du ln(1 + βu2 ) = 2πi u −∞

Z

Thus we find

+2πi I= r

Z 2πi

∞

√ 1/ β

e−y dy y

Z

∞

dy

√ 1/ β

4π 2 =− r

Z

e−y . y

∞

√ 1/ β

dy

e−y . y

The 1-loop correction to the Coulomb potential is now written as two nested integrals over the real parameters x and y: Z ∞ Z 1 e4 e−y δV (r) = + 2 dy dx x(1 − x) π r 0 y a(x) p √ where a(x) ≡ 1/ β = mr/ x(1 − x) . In the short-distance limit, we can Taylor expand around mr = 0 to get Z ∞ e−y 1 mr dy = − ln(mr) + ln[x(1 − x)] − γ + p + O(mr)2 y 2 x(1 − x) a(x) where γ ≡ −

R∞ 0

dx e−x ln x ≈ 0.577. Using the integrals

Z 0

and

R1 0

1

5 , dx x(1 − x) ln[x(1 − x)] = − 18

Z

1

dx

p

x(1 − x) =

0

π 8

dx x(1 − x) = 16 , we have: Z 0

1

∞

e−y 1 3π 5 2 dx x(1 − x) dy =− mr + O(mr) ln(mr) + + γ − y 6 6 4 a(x) Z

Therefore, the correction to the Coulomb potential in the limit mr 1 is 5 3π e4 δV (r) ≈ − 2 ln(mr) + + γ − mr . 6π r 6 4 It is customary to express loop corrections in terms of α ≡ e2 /(4π), so we can rewrite the potential as e2 2α 5 3π V (r) ≈ 1− ln(mr) + + γ − mr . r 3π 6 4

89

III.8

Becoming Imaginary and Conserving Probability

1. Evaluate the imaginary part of the vacuum polarization function, and by explicit calculation verify that it is related to the decay rate of a vector particle into an electron and a positron. Solution: We first evaluate the imaginary part of the vacuum polarization function by the method leading to equation (8) on p. 211. The vacuum polarization function is Z 1 h z i e2 2 dx x(1 − x) ln 1 − x(1 − x) 2 Π(z) = − 2 M 2π m 0 where we have included a factor of e2 from using the Lagrangian L = 14 Fµν F µν instead of L = 4e12 Fµν F µν , and we have included a factor of M 2 since we are shifting the pole from k 2 = M 2 rather than from k 2 = 0. The log goes imaginary when its argument is negative. Since ln(−1) = iπ, we find Z 1 e2 σ 2 ImΠ(σ + iε) = − 2 M dx x(1 − x)(−π) θ[x(1 − x) 2 − 1] 2π m Z 0x+ 2 e dx x(1 − x) = + M2 2π x− e2 2 2 2 3 x+ 1 2 2 = M (x − 3 x ) x− = M x+ − x2− − 23 x3+ − x3− 4π 4π p where x± ≡ 12 1 ± 1 − 4m2 /σ are the two roots of the quadratic equation x2 −x+m2 /σ = p 2 2 0 found by setting the argument of the step function to zero. We have x −x = 1 − 4m2 /σ + − p and x3+ − x3− = 1 − 4m2 /σ (1 − m2 /σ), and so r 4m2 1 2m2 2 2 3 3 2 x+ − x− − 3 (x+ − x− ) = 1− 1+ . 3 σ σ The decay rate is given on p. 212 as Γ = ImΠ(M 2 )/M , so we find r 4m2 2m2 e2 M 1− 1+ . Γ= 12π M2 M2 Now we will calculate the decay rate of a vector particle into a fermion-antifermion pair using Feynman diagrams. Consider the Lagrangian for quantum electrodynamics with a massive photon: ¯ + 1 M 2 Aµ Aµ − 1 Fµν F µν L = ψ¯ i D 6 ψ − mψψ 2 4 The Dirac field ψ represents the electron with mass m and electric charge −e, as specified by the covariant derivative Dµ ψ = ∂µ ψ − ie(−1)Aµ ψ = ∂µ ψ + ieAµ ψ. This implies a vertex −ieγ µ and an amplitude iM(γ → e+ e− ) = ∗µ (p0 )¯ u2 [−ieγ µ ]v1 . (Notation: 90

0 = photon, 1 = outgoing positron, 2 = outgoing electron). The magnitude squared is u2 γ µ v1 )(¯ v1 γ ν u2 ) = e2 ∗µ (p0 )ν (p0 )tr[γ µ (v1 v¯1 )γ ν (u2 u¯2 )]. |M|2 = e2 ∗µ (p0 )ν (p0 )(¯ P Next we sum over the spins of the outgoing fermions using u = (6 p + m)/(2m) and s u¯ P v = (6 p − m)/(2m), and average over the three spin states of the incoming massive s v¯ photon using X pµ pν ∗(a)µ (p)(a)ν (p) = −ηµν + . M2 a P The spin-summed amplitude squared M2 ≡ 13 a,s1 ,s2 |M|2 is 2 h 1 p0µ p0ν i ηµν − tr [γ µ (6 p1 − m)γ ν (6 p2 + m)] M = −e 3 2m M2 e2 =− tr [γ µ (6 p1 − m)γµ (6 p2 + m)] − tr γ 0 (6 p1 − m)γ 0 (6 p2 + m) 2 12m 2

21

where we have used pµ0 = (M, ~0) in the rest frame of the decaying photon. Now we need some of the identities in Appendix D. Using {γ µ , γ ν } = 2η µν =⇒ γ µ γµ = 4I, tr(γ µ γ ν ) = 4η µν and γ µ6 pγµ = −2 6 p, we have tr[γ µ (6 p1 − m)γµ (6 p2 + m)] = tr[(−2 6 p1 − 4m)(6 p2 + m)] = tr[−2 6 p1 6 p2 − 4m2 ] = −8[p1 · p2 + 2m2 ] . Also using (γ 0 )2 = I and tr[6 a 6 b 6 c 6 d] = 4[(a · b)(c · d) + (a · d)(b · c) − (a · c)(b · d)] we obtain tr[γ 0 (6 p1 − m)γ 0 (6 p2 + m)] = tr[γ 0 6 p1 γ 0 6 p2 − m2 I] = 4[2p01 p02 − p1 · p2 − m2 ] . Subtracting these two gives tr [γ µ (6 p1 − m)γµ (6 p2 + m)] −tr γ 0 (6 p1 − m)γ 0 (6 p2 + m) = −8(p1 · p2 + 2m2 ) − 4(2p01 p02 − p1 · p2 − m2 ) = −4[2(p1 · p2 + 2m2 ) + 2p01 p02 − p1 · p2 − m2 ] = −4[p1 · p2 + 2p01 p02 + 3m2 ] = −4[3p01 p02 − p~1 · p~2 + 3m2 ]. Therefore the squared amplitude is M2 =

e2 3p01 p02 − p~1 · p~2 + 3m2 2 3m 91

where p0i =

p

|~pi |2 + m2 .

The differential decay rate is dΓ =

1 d3 p2 d3 p1 (2π)4 δ 4 (p0 − p1 − p2 )M2 2M (2π)3 p01 /m (2π)3 p02 /m

where δ 4 (p0 − p1 − p2 ) = δ(M − p01 − p02 )δ 3 (~p1 + p~2 ) in the rest frame of the parent photon. Integrating over the 3-dimensional delta function sets p~2 = −~p1 , for which the squared amplitude becomes e2 [3(p01 )2 + p~12 + 3m2 ] 3m2 e2 = [3(~p12 + m2 ) + p~12 + 3m2 ] 3m2 e2 = [4~p 2 + 6m2 ] . 3m2 1

M2 =

The decay rate is Z q 1 m2 d3 p1 Γ= δ(M − 2 p~12 + m2 ) M2 |p~2 =−~p1 2M (2π)2 p~12 + m2 Z q d3 p1 1 1 e2 δ(M − 2 p~12 + m2 ) (4~p12 + 6m2 ) = 2 2 2 2M 4π 3 p~1 + m 2 Z p e p2 = dp 2 δ(M − 2 p2 + m2 ) (2p2 + 3m2 ) 2 3πM p +m where p ≡ |~p1 |. The delta function is p M δ(M − 2 p2 + m2 ) = δ(p − p∗ ) 4p∗ where p∗ =

p

(M/2)2 − m2 . The decay rate is e2 p2∗ M Γ= (2p2∗ + 3m2 ) 2 2 3πM p∗ + m 4p∗ e2 p∗ 3 2 2 = p∗ + m 6π p2∗ + m2 2 2 2 4p∗ M e 3 2 2 = −m + m 6π M 2 4 2 2 2 e 2m = p∗ 1 + 6π M2 r e2 4m2 2m2 = M 1− 1+ . 12π M2 M2 92

This agrees with the result obtained previously. 2. Suppose we add a gϕ3 term to our scalar ϕ4 theory. Show that to order g 4 there is a “box diagram” contributing to meson scattering p1 + p2 → p3 + p4 with the amplitude Z 1 d4 k 4 I=g 4 2 2 2 2 (2π) (k − m )[(k + p2 ) − m ][(k − p1 )2 − m2 ][(k + p2 − p3 )2 − m2 ] (where m2 above is understood as m2 −iε. Note that this is a typo in the problem in the text.) Calculate the integral explicitly as a funtion of s = (p1 +p2 )2 and t = (p3 −p2 )2 . Study the analyticity property of I as a function of s for fixed t. Evaluate the discontinuity of I across the cut and verify Cutkosky’s cutting rule. Check that the optical theorem works. What about the analyticity property of I as a function of t for fixed s? And as a function of u = (p3 −p1 )2 ? Solution: We add the term L = − 3!1 gϕ3 to the Lagrangian, which generates the cubic vertex −ig. The diagram in question is displayed below:

p1

k−p 1

k

p2

p4

k + p 2 −p 3

k+p2

p3

Each of the external momenta is on-shell, meaning p2i = m2 for i = 1, 2, 3, 4. Using the scalar propagator i i∆(p) = 2 p − (m2 − iε) we obtain the amplitude Z d4 k 4 [i∆(k)][i∆(k + p2 )][i∆(k − p1 )][i∆(k + p2 − p3 )] I = (−ig) (2π)4 Z d4 k 1 1 1 1 4 =g 4 2 2 2 2 2 2 (2π) [k − m ] [(k + p2 ) − m ] [(k − p1 ) − m ] [(k + p2 − p3 )2 − m2 ] where m2 is understood as m2 − iε.

93

Note that the integral has four powers ofR k in the numerator R and 4 × 2 = 8 powers of k in the denominator, so the integral goes as dk k 3 /[(k 2 )4 ] ∼ dk/k 5 , which converges at high energy and thus does not need to be regularized. Now that we have the integral, we should be clear as to what exactly this problem is asking. The steps are as follows: First compute the integral using Feynman parameters, and evaluate the discontinuity of I across the cut at s = 4m2 . Second, “verify Cutkosky’s cutting rule” by returning to the original integral, replacing the propagators with on-shell delta functions with positive frequency, evaluating the resulting integral and showing that we get the same result for the imaginary part as we got before. Third, to “check the optical theorem,” we are to compute the tree-level scattering amplitude obtained by cutting the box diagram down the middle, and plug it into the right-hand side of equation (19) on p. 215, which we repeat below for convenience: ! X Y 1 (2π)4 δ (4) (Pni ) F(i → n) [F(f → n)]∗ ImF(i → f ) = 21 2 ρn n n where in our case F(i → f ) = −iI. By evaluating the right-hand side explicitly, we are to observe that it equals the left-hand side, namely the imaginary part of the original box diagram that we computed previously in two different ways. Finally, all results as a function of s for fixed t should be the same as those obtained as a function of t for fixed s. To evaluate the integral directly, one introduces Feynman parameters and integrates over the loop momentum to put the integral in the form P Z δ( 4i = 1 xi − 1)θ(x1 )θ(x2 )θ(x3 )θ(x4 ) g4 4 . I= dx 16π 2 [m2 (1 + x1 x2 + x2 x3 + x3 x4 + x4 x1 ) + sx1 x3 + tx2 x4 ] This is invariant under the interchange s ↔ t. Proceeding in this way, eventually one arrives at an expression in terms of logarithms and dilogarithms, with branch cuts in the appropriate kinematic channels (s > 4m2 for fixed t, and t > 4m2 for fixed s). For explicit details, see G. ’t Hooft and M. Veltman, “Scalar One-Loop Integrals,” Nucl. Phys. B153 (1979) 365-401 and A. Denner, U. Nierste and R. Scharf, “A Compact Expression for the Scalar One-Loop Four-Point Function,” Nucl. Phys. B 367 (1991) 637-656. We will now verify Cutkosky’s cutting rule and the optical theorem. We compute the imaginary part of the integral by replacing the cut internal propagators with momentum-conserving delta functions and thereby verify Cutkosky’s cutting rule, as described on p. 215. For convenience, we collect the results from the example on pp. 216-217. 94

Given the amplitude Π = +ig

2

Z

d4 k [i∆(µ) (k)][i∆(m) (q − k)] (2π)4

where the subscript on the scalar propagator denotes the location of the pole in momentumsquared (that is, the mass), Cutkosky tells us that twice the imaginary part of the amplitude is given by replacing the propagators i∆ with on-shell delta functions with positive frequency: Z 2 d4 k 0 1 2 . θ(k )2πδ(k 2 − µ2 ) θ (q − k)0 2πδ (q − k)2 − m2 Im Π = 2 g 4 (2π) As explained on p. 215, in our case we identify I with iΠ. We first shift the loop momentum as k → k + p1 to obtain Z d4 k 1 4 I=g 4 2 2 2 2 (2π) [k − m ][(k + p1 ) − m ][(k + p4 )2 − m2 ][(k + p1 + p2 )2 − m2 ] where we have used p1 + p2 − p3 = p4 . From this point on, we follow5 P. van Nieuwenhuizen, “Muon-Electron Scattering Cross Section to Order α3 ,” Nucl. Phys. B28 (1971) 429-454. We will cut the diagram through the propagators labeled by momenta k and k + p1 + p2 . Cutkosky tells us that this entails replacing i∆(k) → θ(k 0 )2πδ(k 2 −m2 ) and i∆(k+p1 +p2 ) → θ[−(k 0 + p01 + p02 )]2πδ[(k + p1 + p2 )2 − m2 ]. Note the minus sign in the second theta function; we are supposed to ensure that the momentum flows in one direction through the loop when we cut it. The imaginary part of the diagram I ≡ Im(−iI) is Z 2 2 0 0 2 2 0 0 1 g4 4 θ(k )δ(k − m ) θ[−(k + p1 + p2 )]δ[(k + p1 + p2 ) − m ] . d k I= 2 (2π)2 [(k + p1 )2 − m2 ][(k + p4 )2 − m2 ] We can perform the integral over k 0 by using q where ωk ≡ |~k |2 + m2 . Therefore: g4 I= 2(2π)2

Z

g4 = 2(2π)2

Z

R

dk 0 θ(k 0 )δ(k 2 − m2 )f (k 0 , ~k) =

1 f (ωk , ~k) 2ωk

d3 k θ[−(k 0 + p01 + p02 )]δ[(p1 + p2 )2 + 2k · (p1 + p2 ))] 0 2ωk [ p21 + 2k · p1 ][p24 + 2k · p4 ] k =ωk √ √ d3 k θ[−(ωk + s )]δ[s + 2ωk s ] √ √ 2ωk [ m2 + ωk s − 2~k · p~1 ][ m2 + ωk s − 2~k · p~4 ]

where we have defined s ≡ (p1 + p2 )2 and have chosen to work in the center-of-mass frame: µ µ 1√ 1√ p1 = ( 2 s , +~p ), p2 = ( 2 s , −~p ) =⇒ p~1 + p~2 = 0. The reference uses the notation qm = p1 , qe = p2 , qe0 = p3 , qm = p4 , ∆ = k and P = p1 + p2 in Eq. (3.9). We treat the case for which all internal masses are equal, so M = λ = m. Also their metric is mostly positive, whereas ours is mostly negative. 5

95

Now let us define coordinates carefully in order to perform the integral. First fix the direction of the incoming momentum p1 to point along the zˆ-axis: √ p~1 = |~p1 |ˆ z , |~p1 | = 12 s − 4m2 . Then fix the scattering plane as the (x, z)-plane, and define the scattering angle θ0 in the center-of-mass frame as follows: √ p~4 = |~p4 | (ˆ x sin θ0 + zˆ cos θ0 ) , |~p4 | = |~p1 | = 21 s − 4m2 . We will perform the integral in spherical coordinates, so that the 3-vector to be integrated over is ~k = |~k | (ˆ x sin θ cos ϕ + yˆ sin θ sin ϕ + zˆ cos θ) . In these coordinates, we have ~k · p~1 = |~k ||~p1 | cos θ ~k · p~4 = |~k ||~p4 | (sin θ0 sin θ cos ϕ + cos θ0 cos θ) . Note that θ and ϕ are to be integrated over, while θ0 is fixed. Next use ωk =

√

k 2 + m2 to rewrite the integral over k ≡ |~k | as one over ω ≡ ωk : √ d3 k = dΩ dω ω ω 2 − m2

where dΩ ≡ d cos θ dϕ. This will facilitate integration over the remaining delta function. With these coordinates, the integral becomes: √ √ √ Z 4 ω 2 − m2 θ[−(ω + s )] 2√1 s δ( 12 s + ω) g √ √ I= dΩ dω 4(2π)2 [ m2 + ω s − 2kp cos θ ][ m2 + ω s − 2kp(sin θ0 sin θ cos ϕ + cos θ0 cos θ)] where we have defined k ≡ |~k | =

√

ω 2 − m2

and

p ≡ |~p1 | = |~p4 | =

1 2

√

s − 4m2 .

√ √ The delta function sets ω = − 12 s , so that k = 21 s − 4m2 = p, and therefore kp = 1 (s − 4m2 ). The integral is 4 r g4 4m2 I= 1 − 8(2π)2 s Z 1 × dΩ . [s − 2m2 + (s − 4m2 ) cos θ][s − 2m2 + (s − 4m2 )(sin θ0 sin θ cos ϕ + cos θ0 cos θ)] The integral over angles is in the standard form6 Z 1 2π dΩ = √ ln (a + b cos θ)(A + B cos θ + C sin θ cos ϕ) X 6

√ ! aA − bB + X √ aA − bB − X

See Appendix A of W. Beenakker and A. Denner, “Infrared Divergent Scalar Box Integrals with Applications in the Electroweak Standard Model,” Nucl. Phys. B338 (1990) 349-370.

96

where a = s − 2m2 b = s − 4m2 A = s − 2m2 = a B = (s − 4m2 ) cos θ0 = b cos θ0 C = (s − 4m2 ) sin θ0 = b sin θ0

and X = (aA − bB)2 − (a2 − b2 )(A2 − B 2 − C 2 ). Defining t ≡ (p1 − p4 )2 as usual implies cos θ0 = 1 + 2t/(s − 4m2 ) = 1 + 2t/b, so that for example B = b + 2t. The result can be brought into the form: 1 1 − Q(s, t, m) g4 ln I= 4π P (s, t, m) 1 + Q(s, t, m) where s 4m2 3m2 2 P (s, t, m) ≡ t 1− s − 4m 1 − t t s Q(s, t, m) ≡

4m2 1− t

s − 3m2 s − 4m2

.

Finally, let us verify the optical theorem. If we cut the box diagram vertically, we arrive at the amplitudes ML and MR for the left and right halves respectively:

k = p1+ k1

ML =

p2

We have ML = −i

p4

k1

k1

p1

MR =

k’ = k1 + p 4

k2

k2

p3

g2 g2 , M = −i . R (k1 + p1 )2 − m2 (k1 + p4 )2 − m2

We can also cut the diagram horizontally. After suitably shifting the internal momenta, this second cut will contribute an identical term to the imaginary part of the amplitude. 97

P Define F ≡ −iM as on p. 215. The sum n over intermediate states becomes Z 3 Z 3 X d k1 d k2 → 3 (2π) (2π)3 n and the product over normalization factors is Y 1 1 → 2 ρn (2ω1 )(2ω2 ) n q where ωi ≡ |~ki |2 + m2 . The optical theorem states that the imaginary part of the amplitude is given by Z 3 Z 3 1 d k1 d k2 1 −g 2 −g 2 4 4 I = 2× (2π) δ (k1 +k2 −p1 −p2 ) . 2 (2π)3 (2π)3 (2ω1 )(2ω2 ) (k1 + p1 )2 − m2 (k1 + p4 )2 − m2 where the factor of accounts for the other possible cut, as discussed. In the center-of-mass frame, we have p~1 + p~2 = 0 and therefore √ δ 4 (k1 + k2 − p1 − p2 ) = δ(ω1 + ω2 − s ) δ 3 (~k1 + ~k2 ) . We can therefore do the integral over ~k2 and get Z 3 √ πg 4 1 dk 1 I= δ(2ωk − s ) 2 3 2 2 4(2π) ωk [(k + p1 ) − m ] [(k + p4 )2 − m2 ] q √ where ωk ≡ |~k |2 + m2 . As before, we can write d3 k = dΩ dω ω ω 2 − m2 and integrate √ √ over the remaining delta function using δ(2ω − s ) = 12 δ(ω − 12 s ). We have: √ Z ω 2 − m2 1 1 dΩ 2 2 ω [(k + p1 ) − m ] [(k + p4 )2 − m2 ] r Z g4 4m2 1 1 = 1− dΩ . 2 2 2 8(2π) s [(k + p1 ) − m ] [(k + p4 )2 − m2 ]

g4 I= 8(2π)2

This is exactly the expression we got when applying the Cutkosky cutting rule.

3. Prove (28). R R 4 [Hint: Do unto dP x e iqx h0|[O(x), O(0)]|0i what we did to d4 x e iqx h0|T (O(x)O(0))|0i, namely insert 1 = n |nihn| (with |ni a complete set of states) between O(x) and O(0) in the commutator. Now we don’t have to bother with representing the step function.]

98

Solution: Z Z 4 iq·x d x e h0|[O(x), O(0)]|0i = d4 x e iq·x (h0|O(x)O(0)|0i − h0|O(0)O(x)|0i) Z = d4 x e iq·x h0|e+iP ·x O(0)e−iP ·x O(0)|0i − h0|O(0)e+iP ·x O(0)e−iP ·x |0i Z = d4 x e iq·x h0|O(0)e−iP ·x O(0)|0i − h0|O(0)e+iP ·x O(0)|0i ! ! ! Z X X |nihn| e−iP ·x O(0)|0i − h0|O(0) |nihn| e+iP ·x O(0)|0i = d4 x e iq·x h0|O(0) n

Z =

n

! d4 x e iq·x

X

h0|O(0)|nihn|O(0)|0i e−iPn ·x −

n

=

X

X

h0|O(0)|nihn|O(0)|0i e+iPn ·x

n 2

|h0|O(0)|ni|

Z

4

d xe

i(q−Pn )·x

Z −

4

d xe

i(q+Pn )·x

n

= (2π)4

X

|O0n |2 δ (4) (q − Pn ) − δ (4) (q + Pn )

(†)

n

Equation (26) on p. 217 is Z X 1 4 iq·x Im i d x e h0|T (O(x)O(0))|0i = (2π)4 |O0n |2 δ (4) (q − Pn ) + δ (4) (q + Pn ) 2 n We are told to prove equation (28) on p. 218, which is Z Z 1 4 iq·x d4 x e iq·x h0|[O(x), O(0)]|0i . Im i d x e h0|T (O(x)O(0))|0i = 2 We should now check equation (†) for each case on the right-hand side, namely for whether q 0 is positive or negative. For q 0 > 0, only the first delta function is nonzero and equation (28) follows immediately. For q 0 < 0, we have to track down a sign. For this we use translation invariance7 in the form h0|[O(x), O(0)]|0i = h0|[O(0), O(−x)]|0i to get Z Z 4 iq·x d x e h0|[O(x), O(0)]|0i = − d4 x e iq·x h0|[O(0), O(x)]|0i (commutator) Z = − d4 x e−iq·x h0|[O(0), O(−x)]|0i (x → −x) Z = − d4 x e−iq·x h0|[O(x), O(0)]|0i (translation invariance) . Thus for q satisfying the delta function δ (4) (q+Pn ), equation (28) also follows. This concludes the problem. Explicitly, use O(x) = e iP ·x O(0)e−iP ·x and Pµ |0i = 0 to get h0|[O(x), O(0)]|0i = h0|O(x)O(0)|0i − h0|O(0)O(x)|0i = h0|O(0)e−iP ·x O(0)|0i − h0|O(0)e+iP ·x O(0)|0i = h0|O(0)O(−x)|0i − h0|O(−x)O(0)|0i = h0|[O(0), O(−x)]|0i. 7

99

IV IV.1

Symmetry and Symmetry Breaking Symmetry Breaking

2. Construct the analog of (2) with N complex scalar fields and invariant under SU (N ). Count the number of Nambu-Goldstone bosons when one of the scalar fields acquires a vacuum expectation value. L=

1 1 (∂ ϕ ~ )2 − m2 ϕ ~ 2 − λ(~ ϕ 2 )2 2 4

(2)

Solution: Let ϕ be a complex scalar field that transforms under the N -dimensional representation of SU (N ). Its Lagrangian is λ L = ∂µ ϕ†a ∂ µ ϕa + µ2 ϕ†a ϕa − (ϕ†a ϕa )2 2 The index a runs from 1 to N . Note that this Lagrangian has U (N ) symmetry, rather than just SU (N ) symmetry. Actually, this Lagrangian secretly has an even larger symmetry group that is at this level opaque because of our choice of field coordinates. Decompose each field into its real and imaginary parts: ϕa = invariant scalar product is

√1 2

(χa + iη a ). The U (N )-

1 1 ϕ†a ϕa = (χa − iηa )(χa + iη a ) = (χa χa + ηa η a ) 2 2 Repackage the N χs and the N ηs into a 2N -dimensional column vector: χ ~ φA ≡ ~η Since the U (N )-invariant Lagrangian above depends only on this scalar product, it is actually invariant under the symmetry group O(2N ) (for more on this point, see p. 407): 2N X λ 1 L= ∂ µ φA ∂ µ φA + µ 2 φA φA − 2 8 A=1

2N X

!2 φA φA

A=1

The group O(2N ) has 2N (2N − 1)/2 = N (2N − 1) generators, as compared with the N 2 generators of U (N ). Now the problem is identical to IV.1.1, in which you are asked to show that O(N ) breaks to O(N − 1) and yields N − 1 Nambu-Goldstone bosons. Before computing the Lagrangian explicitly, we first count the number of Nambu-Goldstone bosons from group theory, as described in the section “Counting Nambu-Goldstone bosons” on p. 199. We start with the group O(2N ), which has 2N (2N −1)/2 = N (2N −1) generators. We break 100

it to O(2N − 1), which has (2N − 1)(2N − 2)/2 = (2N − 1)(N − 1) generators. Therefore, the number of Nambu-Goldstone bosons is N (2N − 1) − (N − 1)(2N − 1) = 2N − 1. Alternatively, if we break U (N ) to U (N − 1), we get N 2 − (N − 1)2 = N 2 − (N 2 − 2N + 1) = 2N − 1 Nambu-Goldstone bosons, which is the same number. Now let us compute the Lagrangian explicitly. It is more straightforward to work with the explicitly O(2N )-invariant theory, repeated below λ 1 L = (∂µ φA ∂ µ φA + µ2 φA φA ) − (φA φA )2 2 8 We have written indices up and down just for convenience; since O(N ) has the invariant tensor δAB , up and down make no difference. (Contrast with SU (N ), which does not have the invariant tensor δAB but only the invariant tensors εA1 ...AN , εA1 ...AN and δ AB .) Use the O(2N ) freedom to rotate the vacuum expectation value (VEV) of φ into the 2N th component: hφA i = vδA,2N . For clarity, let us define a new index i = 1, ..., 2N − 1 to single out the component with nonzero VEV. Let us also denote the shifted value of the last field by h(x), so that we write: φA = 2N = v + h(x) , φA6=2N = φi The Lagrangian is: λ 2 µ2 1 (v + h)2 + φi φi − (v + h)2 + φi φi L = (∂µ h∂ µ h + ∂µ φi ∂ µ φi ) + 2 2 8 We have: (v + h)2 = v 2 + 2vh + h2 [(v+h)2 + φi φi ]2 = (v + h)4 + 2(v + h)2 φi φi + (φi φi )2 = (v 2 + 2vh + h2 )2 + 2(v 2 + 2vh + h2 )φi φi + (φi φi )2 = v 4 + 2v 2 (2vh + h2 ) + (2hv + h2 )2 + 2v 2 + 4vhφi φi + 2h2 φi φi + (φi φi )2 = v 4 + 4v 3 h + 2v 2 h2 + 4v 2 h2 + 4vh3 + h4 + 2v 2 φi φi + 4vhφi φi + 2h2 φi φi + (φi φi )2 = v 4 + 4v 3 h + 6v 2 h2 + 4vh3 + h4 + 2v 2 φi φi + 4vhφi φi + 2h2 φi φi + (φi φi )2 We display this explicitly because the numerical factors are critical to get the right answer. The Lagrangian is:

101

v2 λv 2 1 2 L= µ − + (∂µ h∂ µ h + ∂µ φi ∂ µ φi ) 2 4 2 2 2 2 µ λ 3 µ λ 2 µ λ 2 2 + 2v − (4v ) h + − (6v ) h + − (2v ) φi φi 2 8 2 8 2 8 λ − 4vh3 + h4 + 4vhφi φi + 2h2 φi φi + (φi φi )2 8 v2 λv 2 1 2 = µ − + (∂µ h∂ µ h + ∂µ φi ∂ µ φi ) 2 4 2 2 λv 1 2 3λv 2 2 1 2 λv 2 2 +v µ − h+ µ − h + µ − φi φi 2 2 2 2 2 λv 3 λv λ λ λ − h − hφi φi − h2 φi φi − h4 − (φi φi )2 2 2 4 8 8 You see now what is going to happen: minimizing with respect to h at the point h = 0 and φi = 0, along with v 6= 0, yields the condition µ2 = λv 2 /2. (Alternatively, you can minimize the Lagrangian with respect to v, set h = φi = 0, then solve for v.) This also sets the coefficient of the term quadratic in φi to zero, resulting in 2N − 1 massless Nambu-Goldstone bosons. As you can see, the resulting theory has O(2N − 1) symmetry, along with cubic interaction terms, indicating that an h-particle can decay into two Nambu-Goldstone bosons.

IV.3

Effective Potential

2. Study Veff in (1+1)-dimensional spacetime. Solution: The 1-loop effective potential for massive ϕ4 theory in d = 1 + 1 dimensions is 1 2 n Z ∞ λϕ 1 4 i d2 k X 1 1 2 2 2 . Veff (ϕ) = (m + B)ϕ + λϕ + 2 2 2 4! 2 Λ (2π) n = 1 n k − m2 We will take m2 → 0 later. We have included a counterterm 12 Bϕ2 to cancel a quadratic dependence on the cutoff Λ, but we have not included a counterterm 4!1 Cϕ4 , because there is no quartic dependence on the cutoff. Only the n = 1 integral requires a finite UV cutoff: 2 Z d2 k 1 1 Λ = −i ln . 2 2 2 4π m2 Λ (2π) k − m In writing the right-hand side, we have dropped terms small compared to Λ, which we take arbitrarily large. For n > 1, we have Z d2 k 1 m2 (−1)n = +i . (2π)2 (k 2 − m2 )n 4π (n − 1)(m2 )n 102

Thus the regularized 1-loop effective potential is 2 1 4 1 1 2 Λ 1 2 2 2 λϕ ln −m S Veff (ϕ) = (m + B)ϕ + λϕ + 2 4! 8π 2 m2 where we have defined the sum 2 n ∞ X λϕ λϕ2 (−1)n = (1 + x) ln(1 + x) − x , x ≡ S≡ . n(n − 1) 2m2 2m2 n=2 We now require renormalization conditions. We would like to follow p. 241 in the text and 00 (0) = m2 . This fixes impose “on-shell” renormalization conditions, for which Veff 2 λ Λ B=− ln . 8π m2 Putting this B back into the effective potential, we find 2 λϕ 1 2 2 1 4 1 2 2 2 Veff (ϕ) = m ϕ + λϕ + λϕ − (λϕ + 2m ) ln +1 . 2 4! 16π 2m2 Taking m2 arbitrarily small but nonzero, we have λϕ2 . 2m2 2 1 2m 2 For ϕ → 0, the potential behaves as Veff (ϕ) → + 16π > 0. The Z2 : ϕ → −ϕ 2 λϕ ln λϕ2 1 1 Veff (ϕ) = λϕ4 − λϕ2 ln 4! 16π 2

symmetry of the classical potential is not spontaneously broken.8 Now consider the case for which we start with a massless theory, for which the 1-loop effective potential is n Z ∞ i 1 4 1 d2 k X 1 λϕ2 2 Veff (ϕ) = λϕ + Bϕ + 4! 2 2 (2π)2 n = 1 n 2k 2 Z 2 1 d kE λϕ2 1 4 1 2 = λϕ + Bϕ + ln 1 + 2 4! 2 2 (2π)2 2kE where in the second line we have summed the series and rotated to Euclidean momentum. Regularizing the integral with an arbitrarily large UV cutoff Λ, we have 2 Z 2 d kE λϕ2 1 2Λ 2 ln 1 + 2 = λϕ ln 2 (2π) 2kE 8π λϕ2 where we have dropped terms that go to zero for Λ → ∞. The regularized effective potential is 2 1 4 1 1 2Λ 2 2 Veff (ϕ) = λϕ + Bϕ + . λϕ ln 4! 2 16π λϕ2 8

Actually the ϕ → −ϕ symmetry is not broken in (3 + 1) dimensions either. See the addendum at the end of this section.

103

The second derivative of this has a log divergence at ϕ = 0. Instead, define the curvature at 00 (µ) ≡ m2 (µ). This fixes an arbitrary field point ϕ = µ as Veff 2 λ 2Λ 2 2 B = m (µ) − ln + 4πµ − 3 . 8π λϕ2 Note that in (1+1) dimensions, ϕ is dimensionless so µ is dimensionless, and λ has dimensions of mass-squared. Putting B back into the effective potential gives 2 2 1 µ 1 4 1 2 2 2 1 λϕ ln +3 . Veff (ϕ) = λϕ + m (µ) − 2 λµ ϕ + 4! 2 16π ϕ2 9 This is now independent of Λ, as it should be by renormalizability. At this point, if we take 2 λ ϕ2 ln ϕµ2 > 0 . However, in this limit the log becomes large, ϕ → 0 we find Veff (ϕ) → + 16π

and since the expansion parameter is λ ln(µ2 /ϕ2 ), perturbation theory is not trustworthy. To obtain an improved perturbation expansion, we use the renormalization group. The effective potential must be independent of the arbitrary point µ, so differentiating the equation for Veff (ϕ) with respect to µ and setting the result to zero gives a flow equation for the parameter m2 (µ): d 2 1 2 µ m (µ) = + µ − λ. dµ 4π In (1 + 1) dimensions, λ does not run and is therefore a constant with respect to µ. Thus this equation can be readily integrated (from an arbitrary point µ0 ): 1 µ 1 2 2 2 2 ln m (µ) = m (µ0 ) + λ (µ − µ0 ) − . 2 4π µ0 Parametrize the RG integration constant as µ20 = ξϕ2 . Then Veff (ϕ) = 21 m2eff (ξ) ϕ2 + 4!1 λeff (ξ) ϕ4 , where: λ • m2eff (ξ) ≡ m2 (µ = ξ 1/2 ϕ) + (3 + ln ξ) , 8π • λeff (ξ) ≡ (1 − 12ξ)λ . The idea is to approach the origin ϕ = 0 carefully by keeping ξ small but nonzero, taking ϕ → 0 and then taking ξ → 0. 9

Note also that, as in d = (3 + 1) dimensions, all logs of λ have disappeared after renormalization. In Appendix A.2 of the Coleman-Weinberg paper, it is shown that by rescaling the loop momenta k → λ1/2 k, d and therefore dd k → λd/2 dd k, the contribution to the effective potential is of the form ϕ4 f (ϕ/M )λV + 2 L−I , where V is the number of vertices, L is the number of loops, and I is the number of internal lines. For d = 4, the contribution goes as λV +2L−I = λL+1 , and thus the 1-loop correction goes as λ2 , as shown in the text. For the present case of d = 2, the contribution goes as λV +L−I = λ, which is independent of the number of loops L. This explains our result that the 1-loop correction to massless ϕ4 theory in d = (1 + 1) dimensions goes as λ.

104

00 We are interested in “massless” ϕ4 theory, so we want m2 (µ → 0) → 0. (i.e., Veff (0) = 0.) From the RG equation, we have

m2 (ϕ) = m2 (ξ 1/2 ϕ) + λ[ 12 (1 − ξ)ϕ2 + Consider ξ > 0, but ϕ → 0. Then m2 (ϕ) ≈ m2 (ξ 1/2 ϕ) + 3 3 m2 (ϕ) + 8π λ. If m2 (ϕ → 0) → 0, then m2eff (ξ) → 8π λ > 0.

1 ln ξ] . 8π λ 8π

ln ξ, which implies m2eff (ξ) ≈

1 in massless ϕ4 theory implies that Veff (ϕ) ≈ So near the origin ϕ → 0, taking 0 < ξ 12 1 λ ϕ4 . The Z2 symmetry is not spontaneously broken. 4!

5. Consider the electrodynamics of a complex scalar field 1 L = − Fµν F µν + [(∂ µ + ieAµ )ϕ† ][(∂µ − ieAµ )ϕ] + µ2 ϕ† ϕ − λ(ϕ† ϕ)2 . 4 In a universe suffused with the scalar field ϕ(x) taking on the value ϕ independent of x as in the text, the Lagrangian will contain a term (e2 ϕ† ϕ)Aµ Aµ so that the effective mass squared of the photon field becomes M (ϕ)2 ≡ 2e2 ϕ† ϕ. Show that its contribution to Veff (ϕ) has the form 2 Z k − M (ϕ)2 d4 k ln . (2π)4 k2 Compare with (14) and (26). [Hint: Use the Landau gauge to simplify the calculation.] If you need help, I strongly urge you to read S. Coleman and E. Weinberg, Phys. Rev. D7: 1883, 1973, a paragon of clarity in exposition. Solution: In the Landau gauge (see p. 267 with ξ = 0), computing the photon contribution to the effective potential constitutes summing up the one-loop diagrams depicted below: +

+

+

+ ...

The external lines are scalar lines with no external momenta. The diagrams (from left to right) contain n = 1, 2, 3, 4, ... powers of the photon propagator. The reason for choosing Landau gauge is that otherwise there would be diagrams with both photons and scalars in the internal lines. The massive photon propagator is (suppressing the i): −i kµ kν i∆µν (k) = 2 ηµν − 2 k − M (ϕ)2 k = −i∆(k)Pµν (k) 105

where we have recognized the scalar propagator i∆(k) =

k2

i − M (ϕ)2

and we have defined a projection operator Pµν (k) ≡ ηµν −

kµ kν . k2

This is a projection operator in the sense that Pµν (k)P νρ (k) = Pµ ρ (k), so that tr(P n ) = trP = 3 for any positive integer n. Because of this, and since every diagram in the above one-loop expansion contains an equal number of e2 ϕ† ϕAµ Aµ → 2ie2 η µν vertices and photon propagators i∆µν (k) = −i∆(k)Pµν (k), we can simply replace the series by an equivalent series with internal scalar lines instead of photon lines, as long as we include a minus sign in the vertex and multiply the sum by an overall factor of 3. In terms of the scalar coupling λ, the replacement is λ → 2e2 . In other words, we can immediately jump to equation (14) in the text with the replacement V 00 (ϕ) = M (ϕ) and an overall factor of 3 multiplying the integral. Therefore, the photon results in an effective potential 2 Z 4 2 d k k − M (ϕ) ln . Veff (ϕ) = V (ϕ) − 3 × 21 i (2π)4 k2 The photon contribution is equal to (+3) real scalar contributions, which makes sense given the 3 polarization states of a massive vector boson (the photon obtains an effective mass in the ϕ background). Now compare this to the contribution of a fermion (equation (26) on p. 243, with slight notational adjustments): 2 Z k − M (ϕ)2 d4 k 1 ln . Veff (ϕ) = V (ϕ) + 4 × 2 i (2π)4 k2 The Dirac fermion contribution is equal to (−4) real scalar contributions to the effective potential.

Addendum 1: Effective Potential Revisited On p. 242 of the text, it is shown that the one-loop correction +ϕ4 ln ϕ2 overwhelms the classical +ϕ4 potential near ϕ = 0. However, the conclusion to draw from this is not that quantum fluctuations break the discrete symmetry ϕ → −ϕ but rather that perturbation 106

theory is not valid near ϕ = 0. The issue is not whether ~ is small (indeed we have set it to 1), but rather, as explained on p. 242, that the expansion parameter is λ ln ϕ rather than just λ. For small ϕ, the second term in the expansion becomes larger than the first term, and the third term will be larger than second term, and so forth, which means that the perturbation expansion breaks down near ϕ = 0 and cannot tell us whether ϕ = 0 is a maximum or a minimum of the potential. We now use the renormalization group to obtain a perturbative expansion that is valid near ϕ = 0 and show that the ϕ → −ϕ symmetry is not spontaneously broken by one-loop effects.10 The one-loop beta function for the quartic coupling λ is (equation (19) on p. 242) M

3 dλ(M ) = [λ(M )]2 + O[λ(M )]3 . 2 dM 16π

The solution to this equation, integrated from some arbitrary scale M0 , is λ(M0 )

λ(M ) = 1−

3 λ(M0 ) ln 32π 2

M2 M02

.

In the text we found the effective potential 2 1 1 ϕc 25 4 2 4 [λ(M )] ϕc ln . Veff (ϕc ) = λ(M )ϕc + − 4! 256π 2 M2 6 The integration constant M0 was arbitrary; let us choose the value M0 = ϕc . Then the solution of the renormalization group equation for λ gives ( 2 2 2 ) M 3 M λ(ϕc ) ln . λ(M ) = λ(ϕc ) 1 + + O λ(ϕc ) ln 2 2 32π ϕc ϕ2c Putting this into Veff , we find 1 25 Veff (ϕc ) = 1− λ(ϕc ) λ(ϕc ) ϕ4c . 4! 64π 2 25 Since 64π 2 ≈ 0.04, we find that for λ(ϕc ) small and positive Veff (ϕc ) looks like an ordinary 4 +ϕ potential with a modified (but still positive) quartic coupling.

The question now is whether we can trust perturbation theory near ϕc = 0. Rearranging the solution of the flow equation for λ, we find 2 3 M λ(M ) = λ(ϕc ) 1 + . λ(ϕc ) ln 2 32π ϕ2c 10

This argument follows section V of S. Coleman and E. Weinberg, “Radiative Corrections as the Origin of Spontaneous Symmetry Breaking,” Phys. Rev. D, Vol. 7 No. 6, 15 Mar 1973 and Section 18.2 in S. Weinberg, Volume II.

107

We are interested in taking ϕc → 0, so that we may consider M close to but larger than ϕc and thereby take ln(M 2 /ϕ2c ) > 0. For perturbation theory to be valid, we need 2 3 M λ(ϕ ) ln 1 c 32π 2 ϕ2c to keep the second term in the expansion smaller than the first term. Assuming that this is true by taking M sufficiently close to ϕc (so that the log is not large) and by taking λ(ϕc ) sufficiently small, then the solution to the flow equation implies that λ(M ) and λ(ϕc ) have the same sign. For vacuum stability, we take λ(M ) positive. Therefore, the effective potential is 1˜ 4 Veff (ϕc ) = λ(ϕ c ) ϕc 4! with 25 ˜ λ(ϕc ) ≡ 1 − λ(ϕc ) λ(ϕc ) > 0 . 64π 2 The minimum of the effective potential is at ϕc = 0, and the symmetry ϕc → −ϕc of the classical potential is not broken. For a scalar theory in which a classical symmetry truly is broken by one-loop effects, see problems IV.3.5 and IV.6.9.

108

Addendum 2: Effective Potential using Dimensional Regularization It is pedagogically instructive to repeat the calculation of the effective potential using dimensional regularization, in contrast to the cutoff regularization used in the text and in the original paper by Coleman and Weinberg. Useful formulas:

" # n X 1 (−1)n 2 −γ+ + O(ε) Γ(−n + ε/2) = n! ε k k=1 Z

dd kE (kE2 )a Γ(b − a − d/2)Γ(a + d/2) −(b−a−d/2) D = 2 d b (2π) (kE + D) (4π)d/2 Γ(b)Γ(d/2)

Metric signature: η = (+, −, −, −). Lagrangian: 1 1 1 µ ε ϕ4 L = Z(∂ϕ)2 − Zm m2 ϕ2 − Zλ λ˜ 2 2 4! 1-loop effective potential: 1 1 V (ϕc ) = Zm m2 ϕ2c + Zλ λ˜ µε ϕ4c + i 2 4! Specialize to d = 4 − ε =⇒

d 2

= 2 − 2ε , 1 −

d 2

Z

1 ε 2 n ∞ λ˜ µ ϕc dd k X 1 2 (2π)d n = 1 2n k 2 − m2 + iε

= −1 + 2ε , 2 −

d 2

= + 2ε .

Only the n = 1 and n = 2 integrals diverge, and their divergences will be absorbed in Zm and Zλ . The relevant integrals are the following:

109

•

n = 1 (note k 2 = −kE2 so an overall (−1) gets factored out of the integral): Z d Z d kE dd k 1 1 ε 2 1 1 ε 2 λ˜ µ ϕc = λ˜ µ ϕc (−i) 2 d 2 2 d 2 (2π) k − m + iε 2 (2π) kE + m2 Γ(1 − d2 ) 2 −(1− d ) 1 ε 2 2 = λ˜ µ ϕc (−i) (m ) 2 (4π)d/2 Γ(−1 + 2ε ) 4π ε/2 2 1 ε 2 = λ˜ µ ϕc (−i) m 2 (4π)2 m2 ε/2 m2 ˜2 1 2 ε 4π µ = λϕc (−i) Γ(−1 + ) 2 (4π)2 2 m2 1 2 m2 ε 4π µ ˜2 2 = λϕc (−i) + O(ε2 )] (−1)[ − γ + 1 + O(ε)][1 + ln 2 (4π)2 ε 2 m2 1 2 m2 2 4πe−γ µ ˜2 = λϕc (+i) + 1 + O(ε)] [ + ln 2 (4π)2 ε m2

So in total we have 1 ε 2 Z λ˜ µ ϕc dd k 1 4πe−γ µ ˜2 m2 2 2 2 i λϕ + ln =− +1 . (2π)d 2 k 2 − m2 + iε 4(4π)2 c ε m2 •

n = 2 (since the denominator is squared this time, the (−1) doesn’t matter): Z Z d dd k 1 1 ε 2 2 d kE 1 1 ε 2 2 = ( λ˜ µ ϕc ) µ ϕc ) (+i) ( λ˜ 2 d 2 2 2 d 2 (2π) (k − m + iε) 2 (2π) (kE + m2 )2 Γ(2 − d2 ) 2 −(2− d ) 1 ε 2 2 2 = ( λ˜ µ ϕc ) (+i) (m ) 2 (4π)d/2 Γ( 2ε ) 4π ε/2 1 ε 2 2 = ( λ˜ µ ϕc ) (+i) 2 (4π)2 m2 ε/2 1 ε 4π µ ˜2 ε 1 2 2 Γ( ) =µ ˜ ( λϕc ) (+i) 2 (4π)2 2 m2 1 2 ε 4π µ ˜2 2 2 ε 1 =µ ˜ ( λϕc ) (+i) [ − γ + O(ε)][1 + ln + O(ε2 )] 2 (4π)2 ε 2 m2 1 2 4π e−γ µ ˜2 ε 1 2 2 =µ ˜ ( λϕc ) (+i) [ + ln + O(ε)] 2 (4π)2 ε m2

So in total we have 1 ε 2 2 Z λ˜ µ ϕc dd k 1 µ ˜ε 4π e−γ µ ˜2 2 4 2 2 i =− λ ϕc + ln . (2π)d 2 · 2 k 2 − m2 + iε 16(4π)2 ε m2

110

The 1-loop effective potential is (define µ2 ≡ 4π e−γ µ ˜2 ): V (ϕc ) = 1 ε 2 1 ε 2 2 Z Z λ˜ µ ϕc λ˜ µ ϕc dd k 1 dd k 1 1 1 2 2 ε 4 2 2 Zm m ϕc + Zλ λ˜ µ ϕc + i +i d 2 2 d 2 2 4! (2π) 2 k − m + iε (2π) 2 · 2 k − m2 + iε 1 ε 2 n Z ∞ λ˜ µ ϕc dd k X 1 2 +i d 2 (2π) n = 3 2n k − m2 + iε 2 2 1 λ 2 µ 1 λ 2 µ 2 2 = Zm − + ln + 1 m ϕc + Zλ − + ln λ˜ µε ϕ4c 2 2 2 2 2 64π ε m 4! 256π ε m Z n ∞ 1 ε 2 d λ˜ µ ϕc d k X 1 2 +i (2π)d n = 3 2n k 2 − m2 + iε At this point we are free to take µ ˜ε → 1 everywhere. The regularized 1-loop effective potential is: Vreg (ϕc ) = 2 2 λ µ λ µ 1 2 1 2 2 2 Zm − + ln Zλ − + ln + 1 m ϕc + λϕ4c 2 2 2 2 2 64π ε m 4! 256π ε m 1 2 n Z 4 ∞ n X λϕc (−1) d kE 2 − 2 4 (2π) n = 3 2n kE + m2 We have Wick rotated the integral to Euclidean momentum and set d = 4 in all of the convergent integrals. We can now perform the integrals and then resum the series. For each n ≥ 3, the integral is: Z 4 1 Γ(n − 2) 2 −(n−2) m4 1 1 1 d kE = (m ) = 2 4 2 n 2 2 (2π) (kE + m ) (4π) Γ(n) 16π (n − 1)(n − 2) (m2 )n Thanks to Mathematica, we also have the series: ∞ X n=3

(−1)n

1 1 xn = x(3x + 2) − 2(1 + x)2 ln(1 + x) n(n − 1)(n − 2) 4

So defining x ≡ λϕ2c /(2m2 ), we have 1 2 n Z 4 ∞ λϕc d kE X (−1)n 1 m4 1 2 = x(3x + 2) − 2(1 + x)2 ln(1 + x) 2 4 2 2 (2π) n = 3 2n kE + m 2 16π 4 " 2 # m4 λϕ2c 3λϕ2c λϕ2c λϕ2c = +2 −2 1+ ln 1 + 128π 2 2m2 2m2 2m2 2m2 111

Therefore, the dimensionally regularized and resummed 1-loop effective potential is (now dropping the subscript on the field): Vreg (ϕ) = 2 2 λ 2 µ 1 λ 2 µ 1 2 2 Zm − + ln +1 m ϕ + Zλ − + ln λϕ4 2 2 2 2 2 64π ε m 4! 256π ε m " # 2 m4 λϕ2 3λϕ2 λϕ2 λϕ2 − ln 1 + + 2 − 2 1 + 128π 2 2m2 2m2 2m2 2m2 Now we are ready to pick renormalization conditions. We choose the “on-shell” renormalization scheme defined by: 4 d V (ϕ) d2 Vreg (ϕ) reg 2 ≡ m , ≡ λ(µ) dϕ2 ϕ = 0 dϕ4 ϕ = µ This is the same as the scheme chosen in Coleman and Weinberg’s paper and in the text. Imposing these renormalization conditions fixes: λ 2 µ2 Zm = 1 + + 1 + ln 2 64π 2 ε m 2 2 λ µ λµ µ2 2 2 Zλ = 1 + + ln 2 − 24 ln + 1 − 48 2 λ + O(λ ) . 256π 2 ε m 2m2 m Here we are dropping O(λ3 ) and higher terms, but not Taylor expanding the logs. The reason is that in the m → 0 limit, the λ 1 in the numerator is offset by m → 0 in the denominator, so that the log is not well approximated by any finite order in its Taylor series. Putting these back into Vreg (ϕ) gives the renormalized effective potential: 2 3 2 λϕ 1 2 m ln + 1 + 12ϕ m2 Vren (ϕ) = { 24 8π 2 2m2 2 1 λϕ 3 2 2 4 + ln +1 − m ϕ +ϕ λ 8π 2 2m2 2 2 3 4 λϕ + 2m2 3 2 ϕ ln − λ + O(λ3 )} . + 2 2 2 32π λµ + 2m 2 In the limit m → 0, this becomes: 1 λ(µ)2 4 ϕ2 25 4 Vren (ϕ) = λ(µ)ϕ + ϕ ln 2 − 24 256π 2 µ 6 which matches equation (20) on p. 242. This is a useful check on the arithmetic: although we have regularized the integrals differently, if we choose the same renormalization scheme then we must get the same answer.

112

IV.4

Magnetic Monopole

2. Show by writing out the components explicitly that dF = 0 expresses something that you are familiar with but disguised in a compact notation. Solution: dF = 0 =⇒ ∂[µ Fνρ] = 0. (The brackets denote complete antisymmetrization of the ~ ·B ~ = 0 imindices.) Contract that equation with ε0µνρ and recognize B i = 12 εijk Fjk to get ∇ λµνρ mediately (Note that choosing one index of e to be 0 fixes the other indices to be purely 1 spatial.) Now expand ∂[µ Fνρ] = 3! (∂µ Fνρ + ∂ρ Fµν + ∂ν Fρµ − ∂ρ Fνµ − ∂µ Fρν − ∂ν Fµρ ) = 0. Since Fµν = −Fνµ , the last three terms just combine with the first three terms. ∂µ Fνρ + ∂ρ Fµν + ∂ν Fρµ = 0 Choose µ = i, ρ = j, ν = 0 to get ∂i F0j + ∂j Fi0 + ∂0 Fij = 0 The electric field is Ei = F0i , so the above is ∂i Ej − ∂j Ei + ∂0 Fij = 0 =⇒ 2∂[i Ej] + ∂0 Fij = 0 Contract this equation with ([~a × ~b ]i = εijk aj bk ) to get

1 kij ε 2

and recognize the definition of the curl of two vectors ~ ×E ~ + ∂t B ~ =0 ∇

Introducing the Hodge star operation (∗ M )µ1 ...µp =

1 εµ1 ...µp µp+1 ...µn M µp+1 ...µD p!

it is possible to repeat this exercise with d(∗ F ) = 0 as follows. The electromagnetic action on a D-dimensional spacetime M is: Z 1 ∗ ∗ S= − F F +A j 4 M A is the 1-form potential, F = dA is its 2-form field strength, and j is a 1-form current that couples to the potential A. If j is a 1-form, then ∗ j is a (D − 1)-form, so that the term A ∗ j is a D-form and thus can be integrated over the space M . Varying the action with respect to A as δS ≡ S[A + δA] − S[A] = 0 implies d(∗ F ) = ∗ j. So d(∗ F ) = 0 should be the source-free ~ ·E ~ = 0 and ∇ ~ ×B ~ − ∂t E ~ = 0. Maxwell equations ∇ Now specialize to D = 4 and compute: ∗

1 F = εµνρσ F ρσ dxµ dxν 2 113

Now take the derivative:

1 d(∗ F ) = εµνρσ ∂λ F ρσ dxλ dxµ dxν 2 ∗ The operation is defined such that operating twice gives you back the original thing (hence the name “dual”), so the equation d(∗ F ) = ∗ j can be written ∗ d(∗ F ) = j, which is more convenient. The left-hand side is: 1 µνρσ λ 1 ∗ ∗ ε ∂ Fρσ dxα d( F ) = εαµνλ 2 2 Using εµναλ εµνρσ = 2(δαρ δλσ − δλρ δασ ) we get: 1 d(∗ F ) = (∂ σ Fασ − ∂ ρ Fρα )dxα = ∂ µ Fαµ dxα 2 ∗ ∗ So the equation d( F ) = j in components becomes just ∗

∂µ F αµ = j α Writing the 4-current as j α = (ρ, J~ ), setting α = 0 and defining again E i ≡ F 0i in the above ~ ·E ~ = ρ. (In verifying this remember F 00 = 0 by antisymmetry.) gives ∂i F 0i = j 0 =⇒ ∇ Now instead of α = 0, set α = i to get: ∂µ F iµ = ∂0 F i0 + ∂j F ij = j i =⇒ −∂t E i + ∂j F ij = J i Recalling the definition B i = 21 εijk Fjk consider the following: 1 k kij kij `m ~ × B] ~ ≡ ε ∂i Bj = ε ∂i [∇ εj`m F 2 1 i k i − δm δ` )∂i F `m = ∂m F km = (δ`k δm 2 Therefore the term with the F leftover is exactly curl B, which is what we expect. Therefore, we have ~ ×B ~ − ∂t E ~ = J~ ∇ D = 4 is special because F and ∗ F are both 2-forms, so there is an electric-magnetic duality that swaps the geometric identity dF = 0 with the equation of motion d(∗ F ) = 0.

3. Consider F = (g/4π)d cos θ dϕ. By transforming to Cartesian coordinates show that this describes a magnetic field pointing outward along the radial direction. Solution:

114

The coordinate transformation from spherical to cartesian coordinates is x = r sin θ cos ϕ , y = r sin θ sin ϕ , z = r cos θ so cos θ =

z y z =p , tan ϕ = r x x2 + y 2 + z 2

dz z z 1 − 2 dr = (dz − dr) r r r r p 1 (x dx + y dy + z dz) dr = d x2 + y 2 + z 2 = p x2 + y 2 + z 2 1 z x dx + y dy + z dz 1 z2 z d cos θ = dz − = 1 − 2 dz − 2 (x dx + y dy) r r r r r r d cos θ =

dϕ = 1 d cos θ dϕ = 2 1− (x + y 2 )r 1 = 2 1− (x + y 2 )r

1 (−y dx + x dy) x2 + y 2 z2 dz(−y dx + x dy) − r2 z2 dz(−y dx + x dy) − r2

z 2 2 (x dx dy − y dy dx) r2 z 2 2 (x + y )dx dy r2

z2 1 1 = 2 (r2 − z 2 ) = 2 (x2 + y 2 ) 2 r r r 1 d cos θ dϕ = 3 [dz(−y dx + x dy) − z dx dy] r 312 dz dx = ε dy = +dy , dz dy = ε321 dx = −dx , dx dy = ε123 dz = +dz 1 1 d cos θ dϕ = 3 (−y dy − x dx − z dz) = − 3 (x dx + y dy + z dz) r r 1 The unit vector in the r direction is precisely dr = r (x dx + y dy + z dz), so 1−

F =

g g d cos θ dϕ = − dr 4π 4πr2

That is a radial magnetic field from a point charge −g.

4. Restore the factors of ~ and c in Dirac’s quantization condition. Solution (due to J. Feinberg): To do this, we will consider the Lagrangian of a nonrelativistic particle in an electromagnetic 115

potential. The generalization to a relativistic particle is straightforward and only affects kinetic terms, and that is the first quantized version of field theory. If we are still working in “natural” units, with ~ = c = 1, the Lagrangian is 1 ~ x) L = mx˙ 2 − eφ(x) + e~x˙ · A(~ 2 ~ + ~v × B) ~ → with e the charge of the particle. This would give the familiar Lorentz force e(E ~ ~ e(E + (~v /c) × B), where we have restored c in Heaviside-Lorentz units (in SI units, we would instead change the dimensions of the magnetic field). Therefore, we need to take the vector potential coupling to ~ x) . ~ x) → e ~x˙ · A(~ e~x˙ · A(~ c In the case of the monopole, the needed gauge transformation is eg ~→A ~ − 1 ∇Λ, ~ φ, A Λ= e 2π so our Lagrangian for the particle goes to L→L−

1˙ ~ ~x · ∇Λ . c

The text states that we must have e iΛ (2π) = e iΛ (0) for the gauge transformation to make sense. Physically, this comes from the fact that the wavefunction of our particle undergoes ψ → e−iΛ ψ in natural units, and the wavefunction must remain single-valued. We should now check how gauge transformations alter the wavefunction in Heaviside-Lorentz units. Restoring ~, the amplitude to go from initial wavefunction ψ1 (~xi ) to a final wavefunction ψ2 (~xf ) in path integral notation is Z Z Z ~xf R i ∗ Dx e ~ dt L hψ2 |ψ1 i ≡ dxf ψ2 (~xf ) dxi ψ1 (~xi ) ~ xi

where L is the one with c restored as discussed, and Dx is the path integral measure. Doing the gauge transformation shifts the phase in the path integral by Z 1 ~ = − 1 [Λ(~xf ) − Λ(~xi )] dt ~x˙ · ∇Λ − ~c ~c ~ = ∂t Λ(~x(t)). The amplitude is therefore left invariant if the wavefunctions because ~x˙ · ∇Λ transform as ψ → e−iΛ/(~c) ψ . The rest follows as in the text, and we find the condition g=

hcn . e

116

where h = 2π~ as usual. Note also that this is consistent with comparing equations (2) and (3) on p. 307.

5. Write down the reparametrization-invariant current J µνλ of a membrane. Solution: The generalization of equation (12) on p. 251 is immediate: Z µνλ J (x) = dX µ dX ν dX λ δ (D) (x − X) .

6. Let g(x) be the element of a group G. The 1-form v = gdg † is known as the Cartan-Maurer form. Then trv N is trivially closed on an N -dimensional manifold since it is already an N R form. Consider Q = S N trv N with S N the N -dimensional sphere. Discuss the topological meaning of Q. These considerations will become important later when we discuss topology in field theory in chapter V.7. [Hint: Study the case N = 3 and G = SU (2).] Solution: Let’s consider the case for which G is some group whose manifold has no pathologies, for instance S N , so that infinitesimal deviations from any point in G are sufficient to determine the global structure of object Q. So consider letting g → g + δg. Then v changes as v → (g + δg)d(g −1 + δg −1 ) = v + δgdg −1 + gd(δg −1 ) + O(δg 2 ) Since gg −1 = I (the identity element on G), we have: (g + δg)(g −1 + δg −1 ) = gg −1 + δgg −1 + gδg −1 + O(δg 2 ) = I =⇒ δg −1 = −g −1 δg g −1 So v transforms as v → v + δg dg −1 − gd(g −1 δg g −1 ) = δg dg −1 − d(δg g −1 ) − dg g −1 δg g −1 = δg dg −1 − d(δg)g −1 + δg dg −1 − dg g −1 δg g −1 = −d(δg)g −1 + dg g −1 δg g −1 = −d(gg −1 δg)g −1 + dg g −1 δg g −1 = −g d(g −1 δg)g −1

117

So under g → g + δg, we have δv ≡ v[g + δg] − v[g] = −g d(g −1 δg)g −1 , and therefore: Z Z Z Z δQ = tr δv v...v + tr vδv...v + ... + tr vv...δv = N tr vv...v δv Z = −N tr vv...v gd(g −1 δg)g −1 Z = −N tr (gdg −1 )(gdg −1 )...(gdg −1 )g d(g −1 δg)g −1 Z = −N tr dg −1 gdg −1 ... gdg −1 gd(g −1 δg) (by cyclicity of trace) Z = +N tr dg −1 g dg −1 g ...gdg −1 (gg −1 )dg d(g −1 δg) (by dg −1 g = d(g −1 g)−g −1 dg = 0−g −1 dg) Z = +N tr dg −1 g dg −1 g ...gdg −1 dg d(g −1 δg) (since g −1 g = I) Z N −1 = (−1) N tr dg dg ... dg d(g −1 δg) (by repeating the above another (N − 1) times) Z N −1 = (−1) N d tr dg dg ... dg g −1 δg (since d2 = 0) =0 So the quantity Q calculated at a point g on the group space G and the quantity Q calculated at a point g + δg on G are the same. Therefore, barring any unforeseen pathologies on the space G, we can compound the infinitesimal transformations and conclude that we can calculate Q using any point on the group manifold and get the same answer. Thus Q is a topological quantity, which depends only on the particular group G we pick. Now that we know Q is topological, let’s try to figure out what it means. First consider the case G = U (1) and N = 1. Then g(x) = einθ(x) , so v = gdg −1 = neiθ d(e−iθ ) = −nidθ, and the Cartan-Maurer form is Z Z Q= v = −in dθ = −i2πn , n Z S1

S1

In conclusion, for this case the quantity Q/(−2πi) counts the number of times the spatial circle wraps around the group circle: Q Π1 (S 1 ) −2πi For any N , the object Q properly normalized counts the number of times the spatial N -sphere wraps around the group manifold G: Q ΠN (G) where Q is suitably normalized. In particular, for G = S N , ΠN (S N ) = Z. This mathematical fact, that Q is proportional to an integer determined purely by topology, tells us that the chiral anomaly does not get renormalized (see Chapter IV.7).

118

IV.5

Nonabelian Gauge Theory

3. In 4 dimensions εµνλρ trFµν Fλρ can be written as trF 2 . Show that dtrF 2 = 0 in any dimensions. Solution: Using differential forms notation, we have d trF 2 = tr(dF F + (−1)2 F dF ) = 2 tr dF F (cyclic property of trace) = −2 tr[A, F ]F = 0 where in the last step we have used the Bianchi identity DF ≡ dF + [A, F ] = 0. This result is equation (A.15) in B. Zumino, Wu Yong-Shi, A. Zee, “Chiral Anomalies, Higher Dimensions, and Differential Geometry,” Nucl. Phys. B239:477 (1984)

5. For a challenge show that trF n , which appears in higher dimensional theories such as string theory, are all total divergences. In other words, there exists a (2n − 1)-form ω2n−1 (A) n such R 1 that trF = dω2n−1 (A). [Hint: A compact representation of the form ω2n−1 (A) = dt f2n−1 (t, A) exists.] Work out ω5 (A) explicitly and try to generalize knowing ω3 and 0 ω5 . Determine the (2n − 1)-form f2n−1 (t, A). For help, see B. Zumino et al., Nucl. Phys. B239:477, 1984. Solution: The Bianchi identity, DF ≡ dF + [A, F ] = 0 will be useful. d (tr F n ) = n tr dF F n−1 = −n tr [A, F ]F n−1 n

(Bianchi identity) n−1

= −n tr AF − F AF = −n tr (AF n − AF n ) =0

(cyclicity of trace)

Therefore tr F n is locally a total divergence: tr F n = d(something) ≡ dω2n−1 (A) Now let us find this (2n − 1)-form ω2n−1 (A). We follow the reference B. Zumino, Wu YongShi, A. Zee, “Chiral Anomalies, Higher Dimensions, and Differential Geometry,” Nucl. Phys. B239:477 (1984).

119

Given a Yang-Mills potential A, define a 1-parameter class of potentials As ≡ sA and their associated field strength tensors Fs = dAs + A2s = s dA + s2 A2 . Now differentiate: dFs n−1 d n (tr Fs ) = n tr F ds ds s d 2 2 n−1 = n tr (s dA + s A )Fs ds = n tr (dA + 2sA2 )Fsn−1 = n tr (dA + 2As A)Fsn−1 The commutator of a Yang-Mills 1-form As with a p-form X is: [As , X] = As X − (−1)p XAs So letting X = A and thus p = 1 gives [As , A] = As A − (−1)AAs = As A + AAs But As = sA, so I can take the s from one A and put it in the other A to give [As , A] = 2As A So we have d (tr Fsn ) = n tr (dA + [As , A])Fsn−1 ds = n tr Ds A Fsn−1 Ds = d + [As , · ] is the covariant derivative associated with the potential As . Using the Bianchi identity Ds Fs = 0, the covariant derivative can act on the product AFsn−1 : d (tr Fsn ) = n tr Ds A Fsn−1 ds = n d tr AFsn−1 + n tr As , AFsn−1 Fs is a 2-form, so Fsn−1 is a 2(n − 1)-form. A is a 1-form, so AFsn−1 is a (2n − 1)-form, which is odd for all n. So the commutator gives a plus sign: tr [As , AFsn−1 ] = tr As AFsn−1 + AFsn−1 As It is tempting to use the cyclicity of the trace to have those two terms add, but we have to be careful. Let Ω be a (2n − 1)-form. Consider the following trace: tr (ΩA) = tr (Ωµ1 ...µ2n−1 Aµ ) dxµ1 ...dxµ2n−1 dxµ Cyclicity of the trace indeed lets us move the matrix Aµ to the left of the matrix Ωµ1 ...µ2n−1 . But to repackage the expression into forms notation, we have to move the dxµ to the left of all of the other dxµi , with each exchange picking up a minus sign. Since there are (2n − 1) 120

exchanges, we pick up an overall minus sign. So tr (ΩA) = −tr (AΩ), for Ω being an arbitrary (2n − 1)-form. Since AFsn−1 is also a (2n − 1)-form, we have tr [As , AFsn−1 ] = 0. Therefore: d (tr Fsn ) = n d tr AFsn−1 ds Now, remember what we want is tr F n = lims→1 tr Fsn , and note that lims→0 tr Fsn = 0. So if we integrate the above expression from s = 0 to 1, we get the desired result: Z 1 n ds tr A(s dA + s2 A2 ) n−1 tr F = dω2n−1 (A) , ω2n−1 (A) = n 0

Now to figure out what all this means, recall IV.4.6 whose result was that, for some R question −1 element g of the group G, the object Q = S N tr (g dg) suitably normalized is an element of the homotopy class ΠN (G). If A(x S N ) = g −1 dg (that is, if A approaches the gauge transformation of 0 on the spatial S N ) then: dA = d(g −1 dg) = dg −1 dg and AA = g −1 dg g −1 dg = −dg −1 gg −1 dg = −dg −1 dg So dA = −A2 = −(g −1 dg)2 . Therefore, we have dA + sA2 = −(1 − s)A2 and Z 1 n−1 ds sn−1 (1 − s)n−1 tr A2n−1 ω2n−1 (A) = (−1) n 0

Meanwhile, the integral over the parameter s evaluates to Z 1 π 1/2 Γ(n) ds sn−1 (1 − s)n−1 = 2n−1 2 Γ(n + 1/2) 0 So remembering that nΓ(n) = Γ(n + 1), we have the result: ω2n−1 (A) = (−1)n−1

π 1/2 Γ(n + 1) tr (g −1 dg)2n−1 22n−1 Γ(n + 1/2)

Thus for g G, and for 2n-dimensional Euclidean space (E2n ) whose boundary can be taken as spherical, we have the result: Z Z Z Z n tr F = dω2n−1 = ω2n−1 ∝ tr (g −1 dg)2n−1 Π2n−1 (G) E2n

E2n

S 2n−1

S 2n−1

The quantity on the left is the nonabelian generalization of the chiral anomaly, and the quantity on the right is an integer determined purely by topology.

121

IV.6

Anderson-Higgs Mechanism

1. Consider an SU (5) gauge theory with a Higgs field ϕ transforming as the 5-dimensional representation: ϕi , i = 1, ..., 5. Show that a vacuum expectation value of ϕ breaks SU (5) to SU (4). Now add another Higgs field ϕ0 , also transforming as the 5-dimensional representation. Show that the symmetry can either remain at SU (4) or be broken to SU (3). Solution: Consider an SU (N ) gauge theory with a Higgs field ϕ in the N -dimensional (“defining”) representation. SU (N ) is the collection of transformations that leave the norm (ϕ, ϕ)N ≡

N X

ϕ † i ϕi

i=1

invariant. Suppose that the SU (N ) symmetry is broken by ϕ obtaining a vacuum expectation value (VEV). We can always choose a coordinate system (in field space) such that this VEV points in the N th direction: hϕi = (0, ..., 0, v). Writing the field as a fluctuation about this classical value ϕ1 ϕ2 ϕi = ... ϕN −1 v + ϕN changes the norm (ϕ, ϕ) into: (ϕ, ϕ)N =

N −1 X

ϕ† i ϕi + (v + ϕN )† (v + ϕN )

i=1

P †i If v = 0, then this just gets repackaged into the sum N i = 1 ϕ ϕi as before. If v 6= 0, then there is a term linear in ϕN , which clearly not invariant under a full SU (N ) transformation. PisN −1 However, the piece (ϕ, ϕ)N −1 ≡ i = 1 ϕ† i ϕi is invariant under SU (N − 1) transformations, since it is precisely the SU (N − 1)-invariant norm. Since the SU (N )-invariant Lagrangian (before spontaneous symmetry breaking) depends only on the SU (N ) norm (ϕ, ϕ)N , after −1 spontaneous symmetry breaking the Lagrangian’s dependence on {ϕi }N i = 1 will be completely in terms of the SU (N − 1) norm (ϕ, ϕ)N −1 . Therefore, the Higgs VEV breaks SU (N ) to SU (N − 1). Now consider an SU (N ) gauge theory with two Higgs fields in the N -dimensional representation. SU (N ) is the collection of transformations that leave the norm (ϕ, ϕ0 )N ≡

N X i=1

122

ϕ† i ϕ0i

invariant. Suppose again that the SU (N ) symmetry is broken by ϕ obtaining a vacuum expectation value (VEV). As before, we can choose a coordinate system in field space such that this VEV points in the N th direction: hϕi = (0, ..., 0, v). The question is, in which direction does the VEV of the other Higgs field, ϕ0 , point? This direction is in principle given to us by the potential. If this VEV also happens to be in the N th direction, then hϕ0 i = (0, ..., 0, v 0 ). If this VEV happens to be perpendicular to the N th direction, then we can still choose coordinates such that hϕ0 i is aligned along one of the other directions, say the (N − 1)th direction: hϕ0 i = (0, ..., 0, v 0 , 0), where now there are (N − 2) zeros to the left of v 0 rather than (N − 1) zeros as in the previous case. It may be the case that this VEV is somewhere in between, neither parallel nor perpendicular to the first VEV. We therefore parameterize the most general case as: ϕ01 ϕ1 ϕ02 ϕ2 .. .. 0 . ϕ= . , ϕ = 0 ϕ N −2 ϕN −1 v 0 cos β + ϕ0N −1 v + ϕN v 0 sin β + ϕ0N When β = 0, the second VEV is perpendicular to the first, and when β = π/2, the second VEV is parallel to the first. Putting this parameterization into the SU (N )-invariant norm gives 0

(ϕ, ϕ )N =

N −2 X

ϕ† i ϕ0i + ϕ†N −1 (v 0 cos β + ϕ0N −1 ) + (v + ϕN )† (v 0 sin β + ϕ0N )

i=1

= (ϕ, ϕ0 )N −2 + ϕ†N −1 ϕ0N −1 + ϕ†N ϕ0N + v 0 cos β ϕ†N −1 + v † ϕ0N + v 0 sin βϕ†N + v † v 0 sin β For generic β 6= π/2, there are terms linear in ϕN −1 and in ϕN , which are not invariant under SU (N ) or SU (N − 1) transformations; this expression is only invariant under SU (N − 2), as expressed by the SU (N − 2)-invariant norm (ϕ, ϕ0 )N −2 . For the special case β = π/2, the term linear in ϕN −1 drops out, so we can repackage the ϕN −1 into an SU (N − 1)-invariant norm. So for the particular case N = 5, having two Higgs fields transforming as a 5 generically breaks SU (5) to SU (3). In the special case for which the two VEVs happen to be aligned, SU (5) instead breaks to SU (4).

123

2. In general, there may be several Higgs fields belonging to various representations labeled by α. Show the mass squared matrix for the gauge bosons generalizes immediately to P that 2 ab 2 a (µ ) = α g (Tα vα · Tαb vα ), where vα is the vacuum expectation value of ϕα and Tαa is the ath generator represented on ϕα . Combine the situations described in exercises IV.6.1 and IV.6.2 and work out the mass spectrum of the gauge bosons. Solution: Let ϕ transform under the R-representation of some gauge group G. Then Dµ ϕ = ∂µ ϕ − P G a a a th igAµ ϕ, where Aµ = dim Lie ala = 1 Aµ TR is the matrix-valued gauge field, and TR is the a † µ gebra generator of P the R-representation of the group Lagrangian L = (Dµ ϕ) D ϕ = P G. 1The 2 † a b a bµ 2 † 2 † a b a bµ + ... implies, g ϕ AAϕ + ... = + ... = a,b g ϕ TR TR ϕ Aµ A a,b 2 g ϕ {TR , TR }ϕ Aµ A upon spontaneous symmetry breaking ϕ = V + ... the gauge boson mass-squared matrix (µ2 )ab = g 2 V † {TRa , TRb }V We can also write the complex vector V as V =

√1 2

v eiθ to get:

1 (µ2 )ab = g 2 v T {TRa , TRb }v 2 Now suppose there are α = 1, ..., N Higgs fields, each of which transforms under a representation Rα of the group G. Each has its own covariant derivative Dµ ϕα = ∂µ ϕα − igAaµ TRaα ϕα , so the kinetic term L = (Dϕ)† Dϕ yields L=

N X

(Dϕα )† Dϕα =

α=1

N dim X XG

g 2 Vα† TRaα TRb α Vα Aaµ Abµ + ...

α = 1 a,b = 1

Immediately we have the mass-squared matrix 2 ab

(µ ) = g

2

N X

Vα† {TRaα , TRb α }Vα

α=1

Now let’s specialize to the situation from problem IV.6.1. We have N = 2 Higgs fields, both of which transform under the 5-representation of G = SU (5). [The group SU (n) has n2 − 1 generators, so dim G = 52 − 1 = 24.] Taking the vacuum expectation value of the first field ϕ to be hϕi = V (0, 0, 0, 0, 1)T , we can without loss of generality take the vacuum expectation value of the second field ϕ0 to lie in the (ϕ4 , ϕ5 )-plane: 0 0 0 0 hϕ i = V 0 cos β sin β This gives (µ2 )ab = g 2 c2 V 02 (tab )44 + 2csV 02 (tab )45 + (V 2 + s2 V 02 )(tab )55 124

where we have defined c ≡ cos β, s ≡ sin β and a set of symmetric matrices tab with components: (tab )ij ≡ {T5a , T5b }ij where T5a is the ath generator of the 5-dimensional representation of SU (5). As explained in IV.6.1, the SU (5) symmetry can be broken down to either SU (4) or SU (3), depending on whether hϕi and hϕ0 i point along the same axis. The above mass matrix is another manifestation of that statement; we see that if cos β = 0, then (µ2 )ab = g 2 (V 2 + V 02 )(tab )55 so that whichever gauge bosons were massless with just the first Higgs field remain massless with the addition of the second field as long as the two vacuum expectation values point in the same direction. Alternatively, if sin β = 0 then (µ2 )ab = g 2 V 02 (tab )44 + V 2 (tab )55 which shows that if the vacuum expectation values are perpendicular, each Higgs breaks a separate direction in SU (5) and thereby gives mass to the corresponding gauge bosons.

4. In chapter IV.5 you worked out an SU (2) gauge theory with a scalar field ϕ in the I = 2 representation. Write down the most general quartic potential V (ϕ) and study the possible symmetry breaking patterns. Solution: The “I = 1” (or “spin-1”) representation of SU (2) is the symmetric 2 × 2 matrix, which has 3 components and thus can be written as a 3-component vector under SO(3). The “I = 2” (or “spin-2”) representation of SU (2) can be written as a symmetric traceless 3 × 3 tensor under SO(3). Let i = 1, 2, 3 denote the 3-vector under SO(3). Then the object ϕij ≡ ϕ(ij) − 13 tr(ϕ)δij transforms under the “spin-2” representation of SU (2). The most general renormalizable potential for ϕ in d = 3 + 1 spacetime dimensions is 1 V (ϕ) = m2 tr(ϕ2 ) + λ[tr(ϕ2 )]2 + λ0 tr(ϕ4 ) . 2 But you have already encountered this potential way back in problem I.10.3, whose solution (in the book) points out that tr(ϕ4 ) and [tr(ϕ2 )]2 are actually proportional to each other. We can therefore set λ0 = 0 with no loss of generality, and thus observe that the potential actually has an accidental SO(5) symmetry. We can repackage the five independent numbers ~ = (φ1 , ..., φ5 )T , where we will ϕ11 , ϕ12 , ϕ13 , ϕ22 , ϕ23 into a 5-dimensional column vector φ not actually display the explicit relations between the {φA }5A = 1 and the ϕij . The potential can be rewritten as ~ ) = 1 m2 φ ~·φ ~ + λ(φ ~·φ ~ )2 V (φ 2 125

~ → M φ, ~ with M any 5-by-5 which is manifestly invariant under the SO(5) transformation φ orthogonal matrix. Suppose that m2 < 0 and λ > 0 so that the potential exhibits spontaneous symmetry breaking. By SO(5) invariance, we can choose field coordinates for which the vacuum expec~ points purely in the fifth direction: hφi ~ = v(0, 0, 0, 0, 1)T . To study small tation value of φ oscillations about this vacuum, write χ1 χ2 ~ φ= χ3 χ4 v+H and study the SO(5) invariant norm ~·φ ~=χ φ ~ ·χ ~ + (v + H)2 where the vector arrow over χ runs over only 4 indices, χ ~ = (χ1 , ..., χ4 )T . The analysis is just ~·φ ~ is invariant only under SO(4) transformations on as in IV.6.1: if v 6= 0, then the norm φ the χ ~ fields. With only one scalar field in the theory, any negative mass squared instability will break SO(5) to SO(4). The situation is more complicated with additional scalar fields. If we stick with SO(5), ~ and φ ~ 0 both transforming as 5-vectors under SO(5), then the analysis meaning introduce φ proceeds analogously to that in problem IV.6.1: the SO(5) symmetry can break either to SO(4) or to SO(3). If instead we insist on starting with traceless symmetric tensors ϕij and ϕ0ij of SO(3), then we have to worry about cross terms of the form V (ϕ, ϕ0 ) = λ1 [tr(ϕ2 )][tr(ϕ02 )] + λ2 tr(ϕ2 ϕ02 ) + λ3 tr(ϕϕ0 ϕϕ0 ) + λ4 tr(ϕ3 ϕ) + λ5 tr(ϕϕ03 ) in which case the analysis is more complicated.

5. Complete the derivation of the Feynman rules for the theory in (3) and compute the amplitude for the physical process χ + χ → B + B. Solution: The Lagrangian is

where Bµν

1 1 1 1 L = − Bµν B µν + M 2 Bµ B µ + (∂χ)2 − m2 χ2 + Lint 4 2 2 2 √ = ∂µ Bν − ∂ν Bµ , M = ev, m = 2λ v and 1 1 Lint = eM χBµ B µ + e2 χ2 Bµ B µ − λvχ3 − λχ4 2 4 126

where we have dropped an additive constant. The propagator for χ (solid line) is i∆(q) =

q2

i − m2 + iε

and the propagator for the massive photon Bµ (wavy line) is i qµ qν i∆µν (q) = 2 −η + . µν q − M 2 + iε M2 The cubic χBB vertex comes from L = eM χB 2 = diagrammatically as:

1 (2eM η µν )χBµ Bν 2

and is represented

Similarly, the quartic χχBB vertex comes from L = 21 e2 χ2 B 2 = 14 (2e2 η µν )χ2 Bµ Bν and is represented diagrammatically as:

= i2e 2

The cubic and quartic self-couplings for χ come from L = −λvχ3 − 41 λχ4 = 1 (−3!λ)χ4 , which are represented diagrammatically as: 4!

=

1 (−3!λv)χ3 3!

+

v

=

P The tree-level amplitude for χχ → BB involves four diagrams: M = 4i = 1 Mi , where the Mi are defined as follows. The first, M1 , is the contribution from the quartic χ2 B 2 vertex: p 1’

p1 M1 = p2

p 2’

127

This gives M1 = ε10 µ ε20 ν (i2e2 )η µν . Next we have a contribution from s-channel exchange of a χ: p1

p 1’ k

M2 = p 2’

p2

This gives M2 = ε10 µ ε20 ν (−i6λv)i∆(k)(+i2e2 vη µν ) 1 = iε10 µ ε20 ν (12λM 2 ) η µν s − M2 where we have defined k = p1 + p2 = p10 + p20 and s ≡ k 2 , and we have used M = ev. The third contribution to the amplitude is from a t-channel exchange of a Bµ : p 1’

p1

M3 =

q p 2’

p2

This gives M3 = ε10 µ ε20 ν [(i2eM η µρ )i∆ρσ (q)(i2eM η σν )] 1 qµqν 2 2 µν = iε10 µ ε20 ν (4e M ) η − t − M2 M2 where q = p1 − p10 = p20 − p2 and t ≡ q 2 = (p1 − p10 )2 = (p20 − p2 )2 . We can use gauge invariance in the form of ε10 ·p10 = 0 and ε20 ·p20 = 0 to write ε10 ·q = ε10 ·p1 and ε20 ·q = −ε20 ·p2 , so that this part of the amplitude becomes pµ1 pν2 1 2 2 µν M3 = iε10 µ ε20 ν (4e M ) η + . t − M2 M2 The fourth is the crossed version of M3 , which results in 1 pµ2 pν1 2 2 µν M4 = iε10 µ ε20 ν (4e M ) η + u − M2 M2 128

where u ≡ (p20 − p1 )2 = (p2 − p10 )2 . P The full amplitude M = 4i = 1 Mi is therefore (recall v = M/e): 1 pµ1 pν2 6λv 2 µν 2 2 µν 2 η + (4e M ) η + + (1 ↔ 2) M = iε10 µ ε20 ν 2e 1 + s − M2 t − M2 M2

6. Derive (14). [Hint: The procedure is exactly the same as that used to obtain (III.4.9).] Write L = 12 Aµ Qµν Aν with Qµν = (∂ 2 + M 2 )g µν − [1 − (1/ξ)]∂ µ ∂ ν or in momentum space Qµν = −(k 2 − M 2 )g µν + [1 − (1/ξ)]k µ k ν . The propagator is the inverse of Qµν . −i kµ kν (14) gµν − (1 − ξ) 2 k 2 − M 2 + iε k − ξM 2 + iε Solution: The hint gave us the Fourier transform, so all we have to do is to find the matrix inverse of Qµν in the Lorentz-index space. That is, solve Qµν (Q−1 )νρ = δρµ for (Q−1 )νρ . Lorentz invariance tells us that (Q−1 )νρ = a gνρ + b kν kρ , so plug this into the matrix inverse equation: Qµν (Q−1 )νρ = (−k 2 + M 2 )g µν + (1 − 1/ξ)k µ k ν [a gνρ + b kν kρ ] = (−k 2 + M 2 )aδρµ + (−k 2 + M 2 )bk µ kρ + (1 − 1/ξ)ak µ kρ + (1 − 1/ξ)bk 2 k µ kρ = (−k 2 + M 2 )aδρµ + (1 − 1/ξ)a + −k 2 + M 2 + (1 − 1/ξ)k 2 b k µ kρ ≡ δρµ . The term multiplying δρµ must equal 1, and the term multiplying k µ kρ must equal zero. This tells us that a = 1/(−k 2 + M 2 ) and (1 − 1/ξ)a + [−k 2 + M 2 + (1 − 1/ξ)k 2 ]b = 0 [k 2 − M 2 − (1 − 1/ξ)k 2 ]b = +(1 − 1/ξ)a [−M 2 + (1/ξ)k 2 ]b = +(1 − 1/ξ)a [k 2 − ξM 2 ]b = +(ξ − 1)a 1 −1 1 a = (ξ − 1) 2 . b = (ξ − 1) 2 k − ξM 2 k − ξM 2 k 2 − M 2 Therefore, we have −1

(Q )νρ

−1 1 −1 = gνρ + (ξ − 1) 2 kν kρ k2 − M 2 k − ξM 2 k 2 − M 2 kν kρ −1 gνρ + (ξ − 1) 2 . = 2 k − M2 k − ξM 2

129

7. Work out the (...) in (13) and the Feynman rules for the various interaction vertices. (13)

L=

1 1 1 µ4 1 − Fµν F µν + M 2 Aµ Aµ − M Aµ ∂ µ ϕ2 + [(∂ϕ01 )2 − 2µ2 (ϕ01 )2 ] + (∂ϕ2 )2 + ... 4λ 4 2 2 2

Solution: We start with the U (1)-invariant scalar QED Lagrangian in Cartesian notation ϕ = iϕ2 ) as given in equation (12) of the text, on page 240:

√1 2

(ϕ1 +

1 1 µ2 λ L = − F 2 + [(∂ϕ1 + eAϕ2 )2 + (∂ϕ2 − eAϕ1 )2 ] + (ϕ21 + ϕ22 ) − (ϕ21 + ϕ22 )2 4 2 2 4 Now it’s a matter of algebra: (∂ϕ2 − eAϕ1 )2 = (∂ϕ2 )2 − 2eAϕ1 ∂ϕ2 + e2 A2 ϕ21 (∂ϕ1 + eAϕ2 )2 = (∂ϕ1 )2 + 2eAϕ2 ∂ϕ1 + e2 A2 ϕ22 So before spontaneous symmetry breaking, we have 2 1 2 X 1 1 2 2 1 2 2 2 λ 2 (∂ϕi ) + µ ϕi + e A ϕi − L=− F + 4 2 2 2 4 i=1

2 X

!2 ϕ2i

+ eAµ (ϕ2 ∂ µ ϕ1 − ϕ1 ∂ µ ϕ2 )

i=1

Now break the symmetry by writing ϕ1 = v + h (h ≡ ϕ01 ) and performing some more algebra: ϕ21 + ϕ22 = (v + h)2 + ϕ22 = v 2 + 2vh + h2 + ϕ22 (ϕ21 + ϕ22 )2 = (v 2 + 2vh + h2 )2 + 2(v 2 + 2vh + h2 )ϕ22 + ϕ42 = v 4 + 2v 2 (2vh + h2 ) + (2vh + h2 )2 + 2v 2 + 4vhϕ22 + 2h2 ϕ22 + ϕ42 = v 4 + 4v 3 h + 6v 2 h2 + 4vh3 + h4 + 2v 2 + 4vhϕ22 + 2h2 ϕ22 + ϕ42 Also, ϕ1 ∂ µ ϕ2 = v∂ µ ϕ2 + h∂ µ ϕ2 . The Lagrangian is: 1 1 µ2 + e2 A2 2 1 (v + 2vh + h2 + ϕ22 ) L = − F 2 + (∂h)2 + (∂ϕ2 )2 + 4 2 2 2 λ 4 − v + 4v 3 h + 6v 2 h2 + 4vh3 + h4 + 2v 2 ϕ22 + 4vhϕ22 + 2h2 ϕ22 + ϕ42 4 + eAµ (ϕ2 ∂ µ h − h∂ µ ϕ2 ) − evAµ ∂ µ ϕ2 Minimizing the potential with respect to v at the point for which all the fields are zero gives ∂L = µ2 v − λv 3 = 0 =⇒ µ2 = λv 2 ∂v

130

2

2

Therefore, µ2 2vh − λ4 4v 3 h = 0, which kills the terms linear in h, and µ2 h2 − λ4 6v 2 h2 = √ √ 1 1 2 2 2 2 2λ v) h , so the mass of the physical h particle is m = 2λ v. The (1 − 3)λv h = − ( h 2 2 Lagrangian is: 1 1 1 λ 1 L = − F 2 + (ev)2 A2 + (∂h)2 − m2h h2 − (4vh3 + h4 + 4vhϕ22 + 2h2 ϕ22 + ϕ42 ) 4 2 2 2 4 1 1 + (∂ϕ2 )2 − evA∂ϕ2 + eA(ϕ2 ∂h − h∂ϕ2 ) + e2 A2 (2vh + h2 + ϕ22 ) 2 2 As explained on p. 240, we add the gauge fixing term Lgf = − 2ξ1 (∂A+ξM ϕ2 )2 , where M = ev is the mass of the photon after spontaneous symmetry breaking. Since 1 [(∂A)2 + 2ξM ∂Aϕ2 + ξ 2 M 2 ϕ22 ] 2ξ 1 1 = − (∂A)2 − M ∂Aϕ2 − ξM 2 ϕ22 2ξ 2 1 1 = − (∂A)2 + M A∂ϕ2 − ξM 2 ϕ22 + (total derivative) 2ξ 2

Lgf = −

the A∂ϕ2 term cancels from the Lagrangian (by design), and we get: 1 1 1 L = − F 2 − (∂A)2 + M 2 A2 ← (yields massive vector boson propagator in Rξ gauge) 4 2ξ 2 1 1 + (∂h)2 − m2h h2 ← (yields scalar propagator for h) 2 2 1 1 + (∂ϕ2 )2 − ξM 2 ϕ22 ← (yields scalar propagator for Goldstone boson ϕ2 ) 2 2 1 1 − λv h3 − λ h4 − λϕ42 ← (cubic and quartic self-interactions for h, quartic for ϕ2 ) 4 4 1 + eM hAµ Aµ + e2 (h2 + ϕ22 )Aµ Aµ ← (photon-scalar interactions) 2 1 − λv hϕ22 − λ h2 ϕ22 ← (h-ϕ2 interactions) 2 + e(ϕ2 ∂µ h − h∂µ ϕ2 )Aµ ← (photon coupling to scalar electromagnetic current) Using solid lines for h, dashed lines for ϕ2 and wavy lines for Aµ , the interaction vertices in momentum space are:

131

v

= +i(2e 2

= +i(2e 2

=

v

k2 = + e ( k 1 −k 2 ) k1 where in the last diagram, the arrows indicate that both momenta flow into the vertex.

8. Using the Feynman rules derived in exercise IV.6.7 calculate the amplitude for the physical process ϕ01 + ϕ01 → A + A and show that the dependence on ξ cancels out. Compare with the result in exercise IV.6.5. [Hint: There are two diagrams, one with A exchange and the other with ϕ2 exchange.] Solution: As for IV.6.5, the amplitude is a sum of diagrams: M = M1 + M2 + M5 + M05 , where M1 and M2 are identical to those of IV.6.5 because the vertices are identical. The next two diagrams are the ones mentioned in the hint, one with t-channel ϕ2 exchange and one with t-channel Aµ exchange; we define both of these as contained within M5 . Finally, we define M05 as the crossed version of M5 . The four diagrams contained in M5 + M05 must conspire to give the same result as the sum M3 + M4 did in IV.6.5, since the two scattering processes should be identical. The gauge boson propagator in Rξ gauge is qµ qν −i ηµν − (ξ − 1) 2 i∆µν (q) ≡ wavy line = 2 q − M 2 + iε q − ξM 2 + iε 132

and the ϕ2 propagator is i∆µν (q) ≡ dashed line =

i . q 2 − ξM 2 + iε

The partial amplitude M5 is (again q = p1 − p10 = p20 − p2 ): M5 = ϕ2 -exchange + Aµ -exchange = ε10 µ ε20 ν [e(p1 + q)µ i∆(q) e(p2 − q)ν + (i2eM η µρ )i∆ρσ (q)(i2eM η σν )] (2p1 − p10 )µ (2p2 − p20 )ν 1 qµqν 2 µν 2 = ie ε10 µ ε20 ν η + (ξ − 1) 2 + (2M ) 2 q 2 − ξM 2 q − M2 q − ξM 2 pµ1 pµ2 M 2 µν M2 = 4ie2 ε10 µ ε20 ν η + 1 − (ξ − 1) t − M2 t − M 2 t − ξM 2 where t ≡ q 2 = (p1 − p10 )2 = (p20 − p2 )2 , and again we have used gauge invariance in the form of ε10 · p10 = 0 and ε20 · p20 = 0 to write ε10 · q = ε10 · p1 and ε20 · q = −ε20 · p2 . Now simplify the quantity in parentheses: 1 − (ξ − 1)

t − M 2 − (ξ − 1)M 2 t − ξM 2 M2 = = . t − M2 t − M2 t − M2

The ξ-dependent numerator cancels the t − ξM 2 in the denominator of pµ1 pν2 , so we find M2 pµ1 pµ2 2 µν M5 = i4e ε10 µ ε20 ν η + . t − M2 M2 The crossed diagrams can be obtained by switching p1 ↔ p2 to get pµ2 pµ1 M2 µν 0 2 η + M5 = i4e ε10 µ ε20 ν u − M2 M2 where u ≡ (p20 − p1 )2 = (p2 − p10 )2 . Thus we find M = M1 + M2 + M3 + M4 = M1 + M2 + M5 + M05 , as we must since the two theories are equivalent.

133

9. Consider the theory defined in (12) with µ = 0. Using the result of exercise IV.3.5 show that 1 ϕ2 1 4 25 2 4 4 + ... Veff (ϕ) = λϕ + (10λ + 3e )ϕ ln 2 − 4 64π 2 M 6 where ϕ2 = ϕ21 + ϕ22 . This potential has a minimum away from ϕ = 0 and thus the gauge symmetry is spontaneously broken by quantum fluctuations. In chapter IV.3 we did not have the e4 term and argued that the minimum we got there was not to be trusted. But here we can balance the λϕ4 against e4 ϕ4 ln(ϕ2 /M 2 ) for λ of the same order of magnitude as e4 . The minimum can be trusted. Show that the spectrum of this theory consists of a massive scalar boson and a massive vector boson, with m2scalar 3 e2 = . m2vector 2π 4π For help, see S. Coleman and E. Weinberg, Phys. Rev. D7: 1888, 1973. Solution: The Lagrangian is L = − 41 Fµν F µν + 12 (∂ϕ1 − eAϕ2 )2 + 21 (∂ϕ2 + eAϕ1 )2 − 4!1 λ(ϕ21 + ϕ22 )2 + Lct where Lct includes the counterterms. Since the result will depend only on ϕ2c = ϕ21c + ϕ22c , we need only to consider diagrams whose external lines are ϕ1 . The diagrams we need to consider are of the form

+

+

+

+ ...

where the internal dotted lines are either ϕ1 , ϕ2 , or Aµ . The text already calculated the scalar contributions to the effective potential. In problem IV.3.5, we showed that the photon contribution is the same as the scalar contribution, except with the replacement λ → 2e2 and an overall factor of 3. Thus the photon contribution to Veff is 1 ϕ2c 25 (A) 2 2 4 Veff (ϕc ) = 3 × [2e ] ϕc ln 2 − 256π 2 M 6 4 2 ϕ 3e 4 25 = ϕc ln c2 − . 2 64π M 6 Summing up all contributions, we find 1 4 5λ2 3e4 ϕ2c 25 4 Veff (ϕc ) = λϕc + + ϕc ln 2 − 4! 1152π 2 64π 2 M 6 134

where the couplings are renormalized at the scale M . Now treat λ as a coupling of order e4 , so that the term of order λ2 = O(e8 ) should be dropped. Then we have 3e4 4 ϕ2c 25 1 4 ϕ ln 2 − Veff (ϕc ) = λϕc + 4! 64π 2 c M 6 where the couplings are renormalized at the scale M . 0 (v) = 0. Let the value ϕc = v be the location of the minimum of Veff (ϕc ), or in other words Veff 0 Furthermore, let us choose M = v. Then the equation Veff (v) = 0 implies

λ=

33e4 . 8π 2

Putting this back into the effective potential results in ϕ2c 3e4 4 1 ϕ ln 2 − 2 . Veff (ϕc ) = 64π 2 c v The squared mass of the scalar is given by 00 m2S ≡ Veff (v) =

3e4 2 v . 8π 2

The mass of the vector after spontaneous symmetry breaking can be read off from the Lagrangian as mV = ev . Dividing these, we arrive at the relation m2S 3α = 2 mV 2π where α ≡ e2 /(4π).

135

IV.7

Chiral Anomaly

1. Derive (11) from (9). The momentum factors k1λ and k2σ in (9) become the two derivatives in Fµν Fλσ in (11). qλ ∆λµν (a, k1 , k2 ) = ∂µ J5µ =

i µνλσ ε k1λ k2σ 2π 2

e2 µνλσ ε Fµν Fλσ (4π)2

(9) (11)

Solution: We will derive this relation by comparing the matrix elements of both operators in the canonical formalism, and finding that they are the same up to numerical factors. We are interested in processes for which an operator causes two photons to pop out of the vacuum, or in other words we want matrix elements of the form hk, a; k 0 , b|O(0)|0i where a labels the polarization of the outgoing photon with momentum k, and b labels the polarization of the outgoing photon with momentum k 0 . By translation invariance, 0 hk, a; k 0 , b|O(x)|0i = e−i(k+k )x hk, a; k 0 , b|O(0)|0i, so it is sufficient to consider the operator at the origin x = 0. The first matrix element we will consider is for the operator O(x) = −i∂µ J5µ (x), for which all of the work has already been done in the chapter: hk, a; k 0 , b|(−i)∂µ J5µ (0)|0i = εaµ (k)∗ εbν (k 0 )∗ e2 (−i)qλ ∆λµν (a, k, k 0 ) = εaµ (k)∗ εbν (k 0 )∗

e2 µνλσ ε kλ kσ0 . 2 2π

The factor of −i came from the Fourier replacement ∂µ → iqµ . Now consider the matrix element for O(x) = εµνρσ ∂µ Aν (x)∂ρ Aσ (x). We will need the plane-wave expansion for the photon field: XZ d3 p a −ipx † a ∗ +ipx Aν (x) = a ε (p) e + a ε (p) e . p ν p ν (2π)3 2ωp a The spacetime derivative at the origin is XZ a d3 p † a ∗ (−ip ) a ε (p) − a ε (p) . ∂µ Aν (0) = p µ ν p ν (2π)3 2ωp a We will need the following properties of the creation and annihilation operators acting on the vacuum state: ak |0i = 0 ak a†k0 |0i = (2π)3 (2ωk )δ 3 (~k − ~k 0 ) hk, k 0 |0i = h0|ak0 ak |0i = 0 hk, a; k 0 , b|p, c; p0 , di = h i 3 3 ac bd 3 ~ 3 ~0 0 ad bc 3 ~ 0 3 ~0 [(2π) 2ωk ][(2π) 2ωk0 ] δ δ δ (k−~p )δ (k −~p )+δ δ δ (k−~p )δ (k −~p ) 136

3

In what follows, define (dp) ≡ (2π)d 3p2ωp and suppress the polarization labels on the outgoing states to save room. Now compute the matrix element: hk, k 0 |εµνρσ ∂µ Aν (0)∂ρ Aσ (0)|0i i XZ h µνρσ =ε (dp)(dp0 )(−ipµ )(−ip0ρ )hk, k 0 | ap εcν (p) − a†p εcν (p)∗ ap0 εdσ (p0 ) − a†p0 εdσ (p0 )∗ |0i c,d 2 µνρσ

= (−1) ε

XZ

(dp)(dp0 )pµ p0ρ hk, k 0 || ap εcν (p) − a†p εcν (p)∗ a†p0 |0iεdσ (p0 )∗

c,d

= − εµνρσ

XZ

(dp)(dp0 )pµ p0ρ hk, k 0 |p, p0 iεcν (p)∗ εdσ (p0 )∗

c,d

= −2 ε

µνρσ

kµ kρ0 εaν (k)∗ εbσ (k 0 )∗

= −2 ερµσν kρ kσ0 εaµ (k)∗ εbν (k 0 )∗ = −2 (−1)3 εµνρσ kρ kσ0 εaµ (k)∗ εbν (k 0 )∗ = +2 εµνρσ εaµ (k)∗ εbν (k 0 )∗ kρ kσ0 Recall the definition of the field strength tensor Fµν = ∂µ Aν − ∂ν Aµ . Then εµνρσ Fµν Fρσ = 4εµνρσ ∂µ Aν ∂ρ Aσ so in total we have hk, a; k 0 , b|εµνρσ Fµν (0)Fρσ (0)|0i = +8 εµνρσ εaµ (k)∗ εbν (k 0 )∗ kρ kσ0 . Therefore hk, a; k

0

, b|∂µ J5µ (0)|0i

e2 e2 = 2 εµνλσ εaµ (k)∗ εbν (k 0 )∗ kλ kσ0 = 2 2π 2π

1 0 µνρσ hk, a; k , b|ε Fµν (0)Fρσ (0)|0i . 8

We therefore have the operator equation ∂µ J5µ

e2 µνρσ ε Fµν Fρσ . = 16π 2

2. Following the reasoning in chapter IV.2 and using the erroneous (10) show that the decay amplitude for the decay π 0 → γ + γ would vanish in the ideal world in which the π 0 is massless. Since the π 0 does decay and since our world is close to the ideal world, this provided the first indication historically that (10) cannot possibly be valid. Solution: The decay π 0 → γγ has an amplitude A(π 0 → γγ) ∝ kµ hγ1 γ2 |J5µ (0)|π 0 (k)i 137

where k µ is the momentum of the decaying neutral pion. This amplitude has one initial hadron and no final hadrons. By translation invariance, we have hγ1 γ2 |J5µ (x)|π 0 (k)i = hγ1 γ2 |J5µ (0)|π 0 (k)ie−ik·x As in chapter IV.2, we have hγ1 γ2 |J5µ (0)|π 0 (k)i = f0 k µ and so kµ hγ1 γ2 |J5µ (0)|π 0 (k)i = f0 k 2 = f0 m2π0 . Since hγ1 γ2 |∂µ J5µ (x)|π 0 (k)i = −ikµ hγ1 γ2 |J5µ (0)|π 0 (k)ie−ik·x = −if0 m2π0 e−ik·x , we deduce (incorrectly) that ∂µ J5µ (x) = 0 if and only if f0 m2π0 = 0. In other words, mπ0 → 0 would seem to imply kµ hγ1 γ2 |J5µ (0)|π 0 (k)i → 0. Since A(π 0 → γγ) is proportional to that, we deduce incorrectly that the decay π 0 → γγ cannot occur.

¯ µ ∂µ − m)ψ. 3. Repeat all the calculations in the text for the theory L = ψ(iγ Solution (due to J. Feinberg): The momentum space amplitude for massive fermions is Z d4 p 1 1 1 λµν λ 5 ν µ ∆ (k1 , k2 ) = i tr γ γ γ γ + (µ, k1 ↔ ν, k2 ) , (2π)4 6 p− 6 q − m 6 p− 6 k1 − m 6 p − m which is linearly divergent by power counting. Therefore, we have to specify how we will regulate and evaluate it. Following the text, we will set up all divergent integrals as integrals of total derivatives, using the formula Z d4 p[f (p + a) − f (p)] = lim i aµ P µ (2π 2 P 2 )f (P ) P →∞

evaluated symmetrically over the 3-sphere at infinity (so that P µ P ν /P 2 → 41 g µν , etc.). Furthermore, we will exchange (µ, k1 ↔ ν, k2 ) at the last step in any calculation, and we define the momentum space amplitude as the amplitude given above with loop momentum p shifted by a chosen so that the vector currents are conserved: k1µ ∆λµν = 0 and k2ν ∆λµν = 0. Technically, also, when P → ∞, we should be taking that limit in Euclidean space, so we don’t get complications from null vectors, and we can always take P 2 m2 . So let us find the difference of the shifted and unshifted amplitudes, as in the text: ∆λµν (a) − ∆λµν (0) =

i 2 ρ 2N (i2π )a lim P P ρ P →∞ (2π)4 D

where the numerator is N = tr γ λ γ 5 (6 P − 6 q + m)γ ν (6 P − 6 k1 + m)γ µ (6 P + m) 138

and the denominator is D = [(P − q)2 − m2 ][(P − k1 )2 − m2 ][P 2 − m2 ] . To this we also add (µ, k1 ↔ ν, k2 ) as in the text. Only the terms in the numerator with six powers of P survive, so we need tr[γ λ γ 56 P γ ν6 P γ µ6 P ] = 2P µ tr[γ λ γ 56 P γ ν6 P ] − P 2 tr[γ λ γ 56 P γ ν γ µ ] using the gamma-matrix anticommutator. Then tr[γ 5 γ σ γ ν γ µ γ λ ] = 4iεσνµλ , so the first term vanishes. Then 4i ∆λµν (a) − ∆λµν (0) = 2 aρ εσνµλ lim Pρ Pσ + (µ, k1 ↔ ν, k2 ) P →∞ 8π i = 2 εσνµλ aσ + (µ, k1 ↔ ν, k2 ) . 8π Now we need to find the shift variable a so that the vector currents are conserved. The only independent momenta available in the problem are k1 and k2 , and Bose symmetry implies that we take a = β(k1 − k2 ). Thus iβ λµνσ ε (k1 − k2 )σ 4π 2 as in the text. Now we can enforce vector current conservation. Using 6 k1 = (6 p − m)− (6 p−6 k1 − m) = (6 p− 6 k2 − m) − (6 p−6 q − m), we can write Z 1 1 1 1 d4 p λ 5 ν λ 5 ν λµν tr γ γ γ −γ γ γ k1µ ∆ (0) = i (2π)4 6 p− 6 q − m 6 p− 6 k1 − m 6 p− 6 q − m 6 p − m Z 1 1 1 1 d4 p λ 5 ν λ 5 ν tr γ γ γ −γ γ γ +i (2π)4 6 p− 6 q − m 6 p − m 6 p− 6 k2 − m 6 p − m ∆λµν (a) − ∆λµν (0) =

Z

d4 p ∂ 1 1 λ 5 ν tr γ γ γ (2π)4 ∂pρ 6 p− 6 k2 − m 6 p − m

=

−ik1ρ

=

4k1ρ εσντ λ

Z

(p − k2 )σ pτ d4 p ∂ (2π)4 ∂pρ [(p − k2 )2 − m2 ][p2 − m2 ]

=−

Pρ Pτ 4i ρ k1 k2σ ελνστ lim 2 P →∞ P 2 8π

=+

i λντ σ ε k1τ k2σ . 8π 2

In the third equality, we have used the fact that traces of γ 5 with fewer than four other gamma matrices vanish. Putting this all together, we obtain: k1µ ∆λµν (a) = k1µ ∆λµν (0) + =

iβ λµνσ ε k1µ k2σ 4π 2

i (1 + 2β)ελµνσ k1µ k2σ 2 8π 139

Demanding vector current conservation means setting this quantity to be zero, so as in the text we need to set β = −1/2. Finally, we can calculate the divergence of the axial current. We use 6 q = (6 p − m) − (6 p − 6 q + m) + 2m where the difference in sign of m in the second term is due to anticommuting the slashed momenta with γ 5 . After some algebra, Z 1 1 d4 p ∂ ρ ν µ λµν 5 γ γ + (µ, k1 ↔ ν, k2 ) + 2m∆µν qλ ∆ = −ik1 tr γ 4 ρ (2π) ∂p 6 p− 6 k2 − m 6 p − m where

Z

µν

∆ (k1 , k2 , m) = i

d4 p 1 1 1 ν µ 5 γ γ . tr γ (2π)4 6 p− 6 q − m 6 p− 6 k1 − m 6 p − m

The first two terms evaluate just as in the book to give i 4i ρ Pσ Pτ k1 (−k2σ )εσντ µ lim + (µ, k1 ↔ ν, k2 ) = 2 εµνστ k1σ k2τ . 2 2 P →∞ P 8π 4π Thus, our properly regulated amplitude is qλ ∆λµν (a) =

i µνστ ε k1σ k2τ + 2m∆µν (k1 , k2 , m) . 2π 2

Now we just need to calculate ∆µν (k1 , k2 , m). We’ll work on the first integral and then perform the Bose swap. We have Z d4 p tr[γ 5 (6 p− 6 q + m)γ ν (6 p− 6 k1 + m)γ µ (6 p + m)] i . (2π)4 [(p − q)2 − m2 ][(p − k1 )2 − m2 ][p2 − m2 ] From the trace, the only terms that don’t vanish are linear in m, or m tr γ 5 γ ν (6 p− 6 k1 )γ µ 6 p + γ 5 (6 p− 6 q)γ ν γ µ 6 p + γ 5 (6 p− 6 q)γ ν (6 p− 6 k1 )γ µ , which become after some algebra and a bunch of cancellations −4i m εµνστ k1σ k2τ . Thus ∆µν (k1 , k2 , m) = 4m εµνστ k1σ k2τ [I(q, k1 , m) + I(q, k2 , m)] where

Z I(q, k, m) =

1 d4 p 4 2 2 (2π) [(p − q) − m ][(p − k)2 − m2 ][p2 − m2 ]

140

is superficially convergent, so we can just evaluate it and shift variables as desired. Using Feynman parameters and shifting variables, we get Z Z 1 d4 p δ(x + y + z − 1) dx dy dz I(q, k, m) = 2 4 2 2 (2π) [p − m − 2p · qx − 2p · ky + q 2 x + k 2 y]3 0 Z Z 1 d4 ` δ(x + y + z − 1) dx dy dz =2 (2π)4 [`2 − m2 + q 2 x(1 − x) + k 2 y(1 − y) + 2q · kxy]3 0 for ` ≡ p − qx − ky. Integrating gives Z 1 δ(x + y + z − 1) i dx dy dz 2 . I(q, k, m) = − 2 2 16π 0 m − q x(1 − x) − k 2 y(1 − y) − 2q · kxy It is relatively easy to put the expression back together again. The extra term is just what ¯ 5 ψJ µ J ν i. we would expect from hψγ

6. Discuss the anomaly by studying the amplitude h0|T [J5λ (0)J5µ (x1 )J5ν (x2 )]|0i given in lowest orders by triangle diagrams with axial currents at each vertex. [Hint: Call 5 2 the momentum space amplitude ∆λµν 5 (k1 , k2 ).] Show by using (γ ) = 1 and Bose symmetry that 1 λµν ∆λµν (a, k1 , k2 ) + ∆µνλ (a, k2 , −q) + ∆νλµ (a, −q, k1 )]. 5 (k1 , k2 ) = [∆ 3 Now use (9) to evaluate qλ ∆λµν 5 (k1 , k2 ). Solution (due to J. Feinberg): The momentum space amplitude, by analogy with the one studied in the text, is Z d4 p 1 λµν λ 5 1 ν 5 µ 51 tr γ γ γ γ γ γ + (µ, k1 ↔ ν, k2 ) . ∆5 = i (2π)4 6 p−6 q 6 p−6 k1 6p We have only the two terms because there are only two circular permutations of the 3 vertices. This is of course a divergent integral, so we must specify how we will regulate it. First off, by Bose symmetry, since the 3 currents are identical, we must choose a regulator which is invariant under (λ, −q ↔ µ, k1 ) and (λ, −q ↔ ν, k2 ) as well as (µ, k1 ↔ ν, k2 ). (The sign on q is because q is incoming as opposed to outgoing.) A simple way to implement this type λµν of regularization is to define ∆λµν above but with the integration 5 (a) as the expression ∆5 variable p shifted to p + a and then define our regulated amplitude as 1 νλµ ∆λµν = [∆λµν (0) + ∆µνλ 5 5 (k1 ) + ∆5 (q)] 3 5 141

with the permutation of indices indicating that we have also used the cyclic property of the trace. Using the second or third terms would be equivalent to evaluating the Feynman diagram starting at a different vertex. From looking at the integral expression for ∆λµν above, we can anticommute the second 5 5 γ matrix through 1/(6 p−6 k1 ) and 1/(6 p−6 k2 ) and then γ µ and γ ν , and use (γ 5 )2 = 1 to see that λµν µνλ νλµ ∆λµν (k1 , k2 ) , ∆µνλ (k2 , −q) , ∆νλµ (−q, k1 ) . 5 (0) = ∆ 5 (k1 ) = ∆ 5 (q) = ∆

We should now use the appropriately shifted versions of the ∆ amplitudes that conserve vector currents, so −1 −1 1 (k1 − k2 ), k1 , k2 ) + ∆µνλ (a = (k2 + q), k2 , −q) ∆λµν = [∆λµν (a = 5 3 2 2 1 + ∆νλµ (a = (q + k1 ), −q, k1 )] . 2 We see that 1 i qλ ∆λµν = qλ ∆λµν (a, k1 , k2 ) = 2 εµνλσ k1λ k2σ . 5 3 6π The anomaly is spread evenly over the three vertices.

7. Define the fermionic measure Dψ in (16) carefully by going to Euclidean space. Calculate the Jacobian upon a chiral transformation and derive the anomaly. [Hint: For help, see K. Fujikawa, Phys. Rev. Lett. 42: 1195, 1979.] Solution: We will proceed somewhat differently from Fujikawa’s original paper. First, we will use the two-component spinor notation instead of the 4-component Dirac notation. Second, we will stay in Minkowski signature for most of the discussion and rotate to Euclidean space only to compute an integral. For this problem we use the metric convention η = (−, +, +, +). Consider the Lagrangian for one massless 2-component spinor ψ charged under a U (1) gauge symmetry: 1 ˙ L = iψa†˙ σ ¯ µaa (∂µ − igAµ )ψa − Fµν F µν 4 The covariant derivative Dµ ψ ≡ (∂µ − igAµ )ψ is present to preserve gauge invariance. For later convenience, we define the following notation: ˙ ˙ vaa˙ ≡ σaµa˙ vµ , v¯aa ≡σ ¯ µaa vµ for any Lorentz vector v µ

Consider the U (1) gauge transformation on ψ: Z 0 iθ(x) ψa (x) = e ψa (x) = d4 y e iθ(x) δ 4 (x, y)δa b ψb (y) 142

We thus read off the Jacobian that defines this transformation: Ja b (x, y) ≡

δψa0 (x) = e iθ(x) δ 4 (x, y)δa b δψb (y)

R R For commuting numbers, the coordinate transformation x0 = Jx implies dx0 = dx det J. R R For anticommuting numbers, x0 = Jx implies dx0 = dx (det J)−1 . So the above change of variables gives Z Z Z 0 −1 Dψ = Dψ (det J) = Dψ e−tr ln J Imagine putting the field theory on a lattice, so that you are comfortable with δ 4 (x, y) being simply the identity matrix in spacetime. The log is then: ln J = ln e iθ I = ln [(1 + iθ + ...)I] = iθ I + O(θ2 ) R Putting this into the trace, we run into the immediate problem that tr I = d4 x δ 4 (x, x) = ∞, but this is really no surprise at all. When we evaluate the path integral for a free scalar field theory, for example, we get a determinant: Z Z −1 ∗ 4 ∗ 2 2 DφDφ exp i d x φ ∂ − m φ ∝ det ∂ 2 − m2 The determinant is a product of eigenvalues, and the eigenvalues can be arbitrarily large, which causes the determinant to diverge. When we can treat the determinant as an overall factor for the path integral, we don’t have to worry about this. When we need to extract physical content from the determinant, we need to regulate it. Phrased in this way, the natural way to regulate the determinant is to impose an upper cutoff on the possible size of the eigenvalues. We are now tempted to apply the above reasoning for the spinor field: when we evaluate the path integral over the fermion ψ, we get a determinant: Z Z † 4 † ¯ ¯ DψDψ exp i d x iψ Dψ ∝ det D ¯ does not have eigenHowever, continuing this line of thought fails, because the operator D values. Recall that an eigenvalue equation for the operator O is O~v = λ~v , where λ is just a number. That is, there exists a vector ~v for which applying the matrix O returns something proportional to that same vector ~v . ˙ ˙ ¯ defined as D ¯ aa Look at the operator D, ≡ σ ¯ µaa Dµ . The fact that one index is dotted ¯ acts on a vector while the other isn’t tells you everything you need to know: the operator D in SU (2)L but returns a vector in SU (2)R – it can’t possibly have an eigenvalue equation.

We remedy this problem by constructing operators that can have eigenvalue equations. Namely, consider the operators ˙ ˙ ¯ c ≡ Daa˙ D ¯ ac ¯ a˙ ≡ D ¯ aa (DD) and (DD) Dac˙ a c˙

143

¯ acts on a vector in SU (2)L and returns As you can see from the indices, the operator DD ¯ acts on a vector in SU (2)R and returns a a vector in SU (2)L . Similarly, the operator DD vector in SU (2)R . We are thus free to regulate these operators by imposing upper cutoffs on their eigenvalue spectra. R Back to the matter at hand: we wish to compute tr ln J = tr iθ I = d4 x iθ(x) tr δ 4 (x, x)I2 , where Ja b (x, y) = eiθ(x) δ 4 (x, y)δa b is the Jacobian for the transformation ψa0 (x) = eiθ(x) ψa (x). ¯ which is an We wish to regulate this by cutting off the eigenvalues of the operator DD, ¯ with eigenvalue λ, operator that acts purely within SU (2)L . Let |λi be an eigenstate of DD ¯ meaning (DD)|λi = λ|λi. Let us now regulate: ! X X δ 4 (x, x) = lim δ 4 (x, y) ≡ lim hx|yi = lim hx| |λihλ| |yi = lim hx|λihλ|yi y→x

regulate

y→x

→ lim

X

y→x

y→x

hx|λihλ|yi e

−λ/Λ2

y→x

λ

= lim

X

y→x

λ

hx| e

¯ 2 −DD/Λ

λ

|λihλ|yi

λ

! ¯

2

X

= lim hx|e−DD/Λ y→x

¯

2

2

y→x

λ

= lim hx|e−DD/Λ y→x

¯

|λihλ| |yi = lim hx|e−DD/Λ |yi

Z d4 k d4 k ¯ 2 |kihk| |yi = lim hx|e−DD/Λ |kihk|yi 4 4 y→x (2π) (2π)

The momentum states |ki in the position basis are hy|ki = e+iky . For any function f , we know how to write D acting on it in the position basis: ∂ − igAµ (x) f (x) hx|Dµ |f i = ∂xµ We therefore know how to write any function of D acting on any function in the position basis, so in particular: ¯ 2 ¯ 2 hx|e−DD/Λ |ki = e−(DD/Λ )(x) e ikx ¯ is there to remind you that these are derivatives The argument (x) for the operator DD with respect only to x, not y, so we can move the hk|yi = hy|ki∗ = e−iky to the left of the ¯ 2 hx|e−DD/Λ |ki. Taking the limit y → x, we arrive at the regulated delta function: Z d4 k −ikx −DD/Λ ¯ 2 +ikx 4 e e e δ (x, x)reg = 4 (2π) ¯ into a nicer form. Introduce the usual notation M(µν) ≡ Now let’s put the operator DD 1 1 (Mµν + Mνµ ) and M[µν] ≡ 2 (Mµν − Mνµ ), so that any matrix can be written as the sum 2 of its symmetric and antisymmetric parts: Mµν = M(µν) + M[µν] . For any function f (x), we have: (µ ν)ab [µ ν]ab ν ab ˙ ¯ ab f (x) = σ µ σ (DD) Dµ Dν f (x) = (σaa˙ σ ¯ ˙ + σaa˙ σ ¯ ˙ )Dµ Dν f (x) aa˙ ¯ (µ

˙ By Lorentz invariance, we know σaa˙ σ ¯ ν)ab = Cδa b η µν , where C is some constant. For µ = ν = 0, the left-hand side is just the identity matrix δa b , while the right-hand side equals Cδa b η 00 .

144

So the constant C simply cancels the 00 piece of the metric: η 00 = −1 =⇒ C = −1. To summarize: (µ ν)ab σaa˙ σ ¯ ˙ = −δa b η µν The antisymmetric piece has no neat simplification as-is, since it is just proportional to the generator of rotations. Recalling that the commutator of covariant derivatives defines the field strength, [Dµ , Dν ] = −igFµν , we proceed: [µ

ν]ab ˙ ¯ ab f (x) = −δa b η µν Dµ Dν f (x) + σ σ Dµ Dν f (x) (DD) aa˙ ¯ 1 [µ ν]ab = −δa b Dµ Dµ f (x) + σaa˙ σ ¯ ˙ [Dµ , Dν ]f (x) 2 1 [µ ν]ab b 2 ˙ = −δa D − ig σaa˙ σ ¯ Fµν f (x) 2

Since this expression appears exponentiated, you may expect that we’ll need to raise it to various powers. Actually we will only need the square of the term with the Fµν . For later convenience, let’s simplify that term now. By Lorentz invariance, we know: (σ [µ σ ¯ ν] σ [ρ σ ¯ σ] )ab = δa b A η ρ[µ η ν]σ + B εµνρσ ¯ 1 − σ1σ ¯ 0 ). Numerically, in the Weyl basis, Try µ = ρ = 0 and ν = σ = 1. σ [0 σ ¯ 1] = 21 (σ 0 σ 1 0 0 1 1 [0 1] 1 σ =σ ¯ = I, and σ ¯ = −σ , so we have σ σ ¯ = 2 (−σ −σ 1 ) = −σ 1 . Therefore, the left-hand side of the equation is (−σ 1 )2 = I. Since ε0101 = 0, the right-hand side of the equation is (suppressing the 2 × 2 identity matrix δa b ): 1 1 A η 0[0 η 1]1 = A η 00 η 11 − η 01 η 10 = − A 2 2 So comparing the two sides of the equation, we deduce A = −2. Now we need to solve for B. For that, consider the case µ = 0, ν = 1, ρ = 2, σ = 3. The right-hand side is just B ε0123 = B, and the left-hand side is: 1 1 0 1 (σ σ ¯ − σ1σ ¯ 0 )(σ 2 σ ¯ 3 − σ3σ ¯ 2 ) = (−σ 1 − σ 1 )(−σ 2 σ 3 + σ 3 σ 2 ) 2 2 4 1 1 2 3 1 1 1 = + σ [σ , σ ] = σ (+2iσ ) = iI 2 2

σ [0 σ ¯ 1] σ [2 σ ¯ 3] =

We therefore deduce that B = i. We have therefore arrived at the expression (σ [µ σ ¯ ν] σ [ρ σ ¯ σ] )ab = δa b −2 η ρ[µ η ν]σ + i εµνρσ We remind you that what we want to calculate is Z d4 k −ikx −DD/Λ ¯ 2 +ikx 4 δ (x, x)reg = e e e 4 (2π) ¯ can be written as where the operator DD ν]ab ˙ ¯ ab = −δa b D2 − 1 ig σ [µ σ (DD) Fµν aa˙ ¯ 2

145

The property f (∂)e ikx = e ikx f (∂ +ik) implies D2 e ikx = e ikx (D+ik)2 = e ikx (D2 +2ikD−k 2 ). Any series in powers of D2 will inherit the same property, so we get: Z d4 k −ikx −DD/Λ ¯ 2 +ikx 4 e e e δ (x, x)reg = 4 (2π) Z d4 k −ikx + 12 (I D2 + 12 igσ[µ σ¯ ν] Fµν ) +ikx = e e e Λ (2π)4 Z d4 k −ikx +ikx + 12 (I (D+ik)2 + 12 igσ[µ σ¯ ν] Fµν ) = e e e Λ (2π)4 Z d4 k −k2 /Λ2 12 (D2 +2ik·D+ 21 igσ[µ σ¯ ν] Fµν ) = e eΛ (2π)4 Z d4 k −k2 12 (D2 +2iΛk·D+ 12 igσ[µ σ¯ ν] Fµν ) 4 =Λ e eΛ (2π)4 In the last line I have simply rescaled k → Λk to facilitate the limit Λ → ∞. When expanding the exponential, we see that terms of order 1/Λ>4 will go to zero, so we know we only need to keep finitely many terms of the series. Before worrying about the terms that don’t go to zero (and in fact look divergent), let’s ¯ so in the path integral we remind ourselves that the original Lagrangian was L = iψ † Dψ, † need to integrate over ψ in addition to ψ. All of the previous work was to regulate the Jacobian J from the transformation ψa0 (x) = e iθ(x) ψa (x) =⇒ Ja c (x, y) = e iθ(x) δ 4 (x, y)δa c But we also need to include the Jacobian J˜ from the conjugate transformation 0 ψ † a˙ (x) = e−iθ(x) ψ †a˙ (x) =⇒ J˜a˙c˙ (x, y) = e−iθ(x) δ 4 (x, y)δ a˙ c˙

This leads to a total change in the path integral measure: Z Z Z ˜ 0 0 † † −1 −1 ˜ Dψ D(ψ ) = DψDψ (det J) (det J) = DψDψ † e−tr ln J−tr ln J There are two things to notice here. First, most of the manipulations carry through for J˜ exactly as for J, so the opposite sign for the phase kills most of the terms upon the addition ˜ The second thing to notice is that the opposite sign does not kill all of the tr ln J+tr ln J. terms when adding the two Jacobians. ¯ which acts purely within How can this be? To regularize J, we needed the operator (DD), ˜ we need the operator (DD), ¯ SU (2)L . To regularize J, which acts purely within SU (2)R . ¯ led to the Everything carries through in exactly the same way, except that squaring DD expression (σ [µ σ ¯ ν] σ [ρ σ ¯ σ] )ac = δa c (−2η ρ[µ η ν]σ + i εµνρσ ) ¯ leads to the expression whereas squaring DD (¯ σ [µ σ ν] σ ¯ [ρ σ σ] )a˙ c˙ = δ a˙ c˙ (−2η ρ[µ η ν]σ − i εµνρσ ) 146

The minus sign is critical: tracing and subtracting the two leaves a nonzero piece +4i εµνρσ . Therefore, subtracting the two regularized delta functions (recall there is a trace over each set of spinor indices sitting out front) gives: tr δ 4 (x,x)reg,J − tr δ 4 (x, x)reg,J˜ = Z i 1 1 1 1 2 [µ ¯ ν] F 2 d4 k −k2 h σ [µ σ ν] Fµν ) 4 µν ) 2 (D +2iΛk·D+ 2 igσ σ 2 (D +2iΛk·D+ 2 ig¯ Λ Λ tr e e − tr e Λ (2π)4 The first trace is over SU (2)L indices, and the second trace is over SU (2)R indices, which of course is the way it must be for the expression to make any sense. Now you see what is happening here. The lowest order term (just the identity matrix) drops out. The first order term also drops out, except for the subtraction σ [µ σ ¯ ν] − σ ¯ [µ σ ν] . ¯ ν − σν σ ¯ µ ), the cyclic property of the trace over SU (2)L indices causes Since σ [µ σ ¯ ν] = 12 (σ µ σ that term to be zero. Similarly, the trace over SU (2)R indices yields tr(¯ σ [µ σ ν] ) = 0. So expanding the exponentials yields terms that will cancel each other out, trace to zero, or will go to 0 for Λ → ∞. As discussed, the only exception is: tr (σ [µ σ ¯ ν] σ [ρ σ ¯ σ] ) − tr (¯ σ [µ σ ν] σ ¯ [ρ σ σ] ) = δa a iεµνρσ + δ a˙ a˙ iεµνρσ = 4i εµνρσ So tracing over the delta functions and subtracting them, and then taking the cutoff to infinity, gives 2 Z d4 k −k2 1 ig 4 4 e 4i εµνρσ Fµν Fρσ tr δ (x, x)reg,J − tr δ (x, x)reg,J˜ = (2π)4 2 2 Z 1 2 µνρσ d4 k −k2 Fµν Fρσ e = − ig ε 2 (2π)4 Now Wick rotate to evaluate the integral: k 0 ≡ −ikE0 =⇒ k 2 ≡ kµ k µ = −(k 0 )2 + ~k 2 = R R +(kE0 )2 + ~k 2 ≡ +kE2 and d4 k = −i d4 kE . We obtain for the integral Z

d4 k −k2 = −i e (2π)4

Z

d4 kE −kE2 e = −i (2π)4

Z

∞

−∞

dk −k2 e 2π

4

= −i

1 √ 2 π

So our final expression for the delta functions is 4

4

tr δ (x, x)reg,J − tr δ (x, x)reg,J˜ R

g 2 µνρσ = − ε Fµν Fρσ 32π 2

d4 x iθ(x) tr δ 4 (x, x)reg , so Z g 2 µνρσ 4 ˜ tr ln J + tr ln J = i d x θ(x) − ε Fµν Fρσ 32π 2

Now recall tr ln J =

147

4 =

−i 16π 2

We have finally managed to compute the change in the path integral measure. Under the change of integration variables 0

ψa0 (x) = e iθ(x) ψa (x) , ψ † a˙ (x) = e−iθ(x) ψ †a˙ (x) the path integral measure changes as Z Z R 4 1 2 µνρσ 0 0 † Dψ D(ψ ) = DψDψ † e i d x θ(x)(+ 32π2 g ε Fµν Fρσ ) We see that the measure is not invariant under this change of variables (in other words, the Jacobian is not 1). Recall the original Lagrangian: 1 ˙ L = i(ψ 0 )†a˙ σ ¯ µaa (∂µ − igAµ )ψa0 − Fµν F µν 4 where now we have primed the fields to follow the order in which we have defined the change of variables all along. Under the infinitesimal version of this change of variables, the Lagrangian becomes L0 = L + δL, where ˙ ˙ δL = iψa†˙ σ ¯ µaa (i ∂µ θ)ψa = +θ(x)∂µ ψa†˙ σ ¯ µaa ψa + (total derivative) Finally, we are ready to state the result. The infinitesimal transformation ψ → ψ + δψ, ψ † → ψ † + δψ † with δψa (x) = iθ(x) ψa (x), δψ †a˙ = −iθ(x) ψ †a˙ changes the path integral as: Z Z R 4 R 4 R † µaa 1 2 µνρσ ˙ † i d4 x L DψDψ e → DψDψ † e i d x L+i d x θ(x)[∂µ (ψa˙ σ¯ ψa )+ 32π2 g ε Fµν Fρσ ] R∞ R † Just as the integral −∞ dx f (x) is not a function of x, the path integral DψDψ † e iS(ψ,ψ ) is not a function of ψ or ψ † . Therefore the path integral should remain completely unchanged under the infinitesimal transformation described above. Moreover, since this transformation is for an arbitrary infinitesimal function of spacetime θ(x), we conclude that ˙ ∂µ ψa†˙ σ ¯ µaa ψa = −

1 2 µνρσ g ε Fµν Fρσ 32π 2

This entire analysis was for one complex Weyl spinor. Suppose we introduce a second complex Weyl spinor ξ. The Lagrangian density is 1 ˙ ˙ L = iψa†˙ σ ¯ µaa (∂µ − igAµ )ψa + iξa†˙ σ ¯ µaa (∂µ − igAµ )ξa − Fµν F µν 4 This Lagrangian exhibits a classical U (1) gauge symmetry iθ(x) ψa e ψa → ξa e iθ(x) ξa 148

with the associated Noether current ˙ ˙ j µ = ψa†˙ σ ¯ µaa ψa + ξa†˙ σ ¯ µaa ξa

Classically Noether’s theorem says ∂µ j µ = 0, but we have just derived that ψ and ξ each contributes a nonzero contribution to this classical conservation law, so that at the quantum level we have 1 2 µνρσ g ε Fµν Fρσ ∂µ j µ = − 16π 2 Thus this theory does not actually have the U (1) gauge symmetry, even though it appears as though it did at the classical level. That is the statement of the chiral anomaly. The Lagrangian also exhibits a different classical U (1) gauge symmetry iθ(x) ψa e ψ → −iθ(x) a ξa e ξa with the associated Noether current ˙ ˙ j µ = ψa†˙ σ ¯ µaa ψa − ξa†˙ σ ¯ µaa ξa

Note the relative minus sign between the two terms: the contributions to ∂µ j µ cancel exactly. Thus the theory truly does have this U (1) gauge symmetry, even quantum mechanically. To compare this to the usual treatment in terms of Dirac spinors, let us package the 2component fields ψa and ξa into a 4-component Dirac spinor: ψ Ψ ≡ †aa˙ ξ The U (1) transformation iθ(x) iθ(x) ψa e ψa ψa e ψ → =⇒ → −iθ(x) †aa˙ ξa ξ †a˙ e iθ(x) ξa e ξ is the transformation Ψ → e iθ(x)γ5 Ψ. The left-handed and right-handed components of the Dirac spinor Ψ are transformed oppositely, which is why this U (1) transformation is called chiral. On the other hand, the U (1) transformation iθ(x) iθ(x) ψa e ψa ψa e ψa → −iθ(x) =⇒ → ξa ξ †a˙ e ξa e iθ(x) ξ †a˙ is the transformation Ψ → e iθ(x) Ψ. The current associated with this transformation is the usual electromagnetic current, which is indeed conserved.

149

8. Compute the pentagon anomaly by Feynman diagrams in order to check remark 6 in the text. In other words, determine the coefficient c in ∂µ J5µ = ... + cεµνλσ trAµ Aν Aλ Aσ . Solution: R The pentagon diagram has 5 fermion propagators, so the integral goes as Λ d4 k 6k15 ∼ Λ1 , which is convergent as Λ → ∞ and thereby naively appears not to contribute to the anomaly. However, gauge invariance relates the pentagon diagram to the triangle and box diagrams, so that only the set of all three is physically meaningful. Demanding that the box diagram conserves vector currents then generates a pentagon anomaly in the form 1 µνλσ 1 ε tr{(T5a )[ 14 vµν vλσ + 12 aµν aλσ ] 4π 2 + i 23 (aµ aν vλσ + vµν aλ aσ + aµ vνλ aσ ) − 83 aµ aν aλ aσ ]} .

∂µ J5aµ = ... +

Here we have split up the gauge field into vector and axial vector components, 6 A =6 v + γ 56 a, and defined the corresponding field strengths vµν = ∂µ vν − ∂ν vµ − i[vµ , vν ] − i[aµ , aν ] and aµν = ∂µ aν − ∂ν aµ − i[vµ , aν ] − i[aµ , vν ]. The matrix T5a is the generator associated with the axial coupling to fermions: aµ = aaµ T5a . See W. A. Bardeen, “Anomalous Ward Identities in Spinor Field Theories,” Phys. Rev. Vol. 184 No. 5, 25 Aug 1969 for further details.

V V.1

Field Theory and Collective Phenomena Superfluids

2. To confine the superfluid in an external potential W (~x) we would add the term −W (~x)ϕ† (~x, t)ϕ(~x, t) to (1). Derive the corresponding equation of motion for ϕ. The equation, known as the Gross-Pitaevski equation, has been much studied in recent years in connection with the Bose-Einstein condensate. L = iϕ† ∂0 ϕ −

1 ∂i ϕ† ∂i ϕ − g 2 (ϕ† ϕ − ρ¯)2 2m

(1)

Solution: Adding the term Lpot = −W (~x)ϕ† ϕ implies the addition of a term −W (~x)ϕ to the equation of motion found from varying L with respect to ϕ† .

150

V.2

Euclid, Boltzmann, Hawking, and Field Theory at Finite Temperature

1. Study the free field theory L = 12 (∂ϕ)2 − 21 m2 ϕ2 at finite temperature and derive the Bose-Einstein distribution. Solution: One way to derive the Bose-Einstein distribution is to show that average quantities are computed with the correct probability distribution.11 We first compute the thermodynamic partition function for the relativistic free Bose gas, given by: Z Rβ R 3 1 1 Z= Dφ e− 0 dτ d x LE , LE = ∂µ φ ∂µ φ + M 2 φ2 . 2 2 φ(0,~ x)=φ(β,~ x) The path integral is Gaussian and therefore can be computed exactly: 1

2

Z = [det(−∂E2 + M 2 )]−1/2 = e− 2 tr ln(−∂E +M

2)

.

Here we have dropped an overall constant, which corresponds to an additive constant in the argument of the exponent. The trace is the sum of eigenvalues of the operator −∂E2 + M 2 , which is diagonalized in Matsubara frequency / momentum space. So we have12 Z 3 ∞ 1 X d kV 2πn ln Z = − ln(ωn2 + ~k 2 + M 2 ) , ωn = . 3 2 n = −∞ (2π) β Consider the sum f (x) ≡

∞ X

ln(n2 + x2 ) .

n=1

While this sum diverges, its derivative is finite and can be computed: 0

f (x) =

∞ X n=1

Using f (x) =

R

n2

2x 1 2π = −π − + . 2 +x x 1 − e−2πx

dx f 0 (x), we find: ∞ X

ln(n2 + x2 ) = 2πx + 2 ln 1 − e−2πx .

n = −∞ 11

We follow p. 394 of “Lattice Gauge Theories - An Introduction,” 3rd Ed, by Heinz J. Rothe. We have implicitly put the system into a box to quantize the spatial momenta and then taken the R d3 k P continuum limit k → (2π) 3 V , where V is the volume of the box. This treatment misses the formation of the Bose-Einstein condensate. 12

151

Therefore we can compute the sum over Matsubara frequencies to obtain, up to an irrelevant additive constant, q Z d3 k −βε(~k) ln 1 − e , ε(~k) ≡ ~k 2 + M 2 . ln Z = −V (2π)3 ∂ ln Z, so that the average energy per unit volume The average energy is given by hEi = − ∂β is given by: Z E d3 k ∂ −βε(~k) = ln 1 − e V ∂β (2π)3 Z d3 k ~ 1 . = ε(k) ~ 3 βε( k) (2π) e −1

Thus the average energy is computed with the probability distribution n(ε) =

1 e βε − 1

which is the Bose-Einstein distribution.

2. It probably does not surprise you that for fermionic fields the periodic boundary condition (6) is replaced by an antiperiodic boundary condition ψ(~x, 0) = −ψ(~x, β) in order to reproduce the results of chapter II.5. Prove this by looking at the simplest fermionic functional integral. [Hint: The clearest exposition of this satisfying fact may be found in appendix A of R. Dashen, B. Hasslacher, and A. Neveu, Phys. Rev. D12: 2443, 1975.] Solution: Consider the case of free field theory in (0 + 1)-spacetime dimensions (that is, the quantum mechanics of a fermionic harmonic oscillator). ¯ The Lagrangian is L = ψ(t)(i∂ t − m)ψ(t). The system has two possible single-particle energy states, ε0 and ε1 , corresponding to whether or not a fermion is present. The energy difference ε1 − ε0 = m is fixed, while the energies themselves are determined only up to an overall additive constant: ε0 = E0 − 12 m and ε1 = E0 + 21 m. We can now compute the partition function: Z ≡ tr(e

−iHT

)=

1 X

−iεn T

e

n=0

−iE0 T

= 2e

cos

mT 2

.

Any path integral representation of the partition function must reproduce this basic result.

152

The standard tricks give us the path integral representation Z RT ¯ Z= DψDψ¯ e i 0 dt ψ(t)(i∂t −m)ψ(t) ±BC

where the ±BC indicates periodic or antiperiodic boundary conditions, respectively: ψ(t + T ) = ±ψ(t). The integral is gaussian and may be performed explicitly: Z = C det(i∂t − m) where C is a normalization factor that is determined by the overall additive constant in the energy: Z RT ¯ −iH(m=0)T tr(e )= DψDψ¯ e i 0 dt ψ(t)i∂t ψ(t) = C det(i∂t ) ≡ 2e−iE0 T ±BC

=⇒ C =

2e−iE0 T . det(i∂t )

Therefore, the path integral calculation gives a partition function Z = 2e−iE0 T

det(i∂t − m) . det(i∂t )

The determinant of an operator is the product of its eigenvalues. The eigenvalue equation (i∂t − m)ϕn (t) = −ωn (m)ϕn (t) subject to ± boundary conditions implies the eigenvalues (n = 0, ±1, ±2, ...): 2nπ +m (periodic) T ωn (m) = (2n+1)π +m (antiperiodic) T Since

∞ Y ωn (m) det(i∂t − m) = det(i∂t ) ωn (0) n = −∞

we have a choice between two infinite products. Since ∞ Y 1+ n = −∞

x 2n + 1

= cos

πx 2

we deduce that only antiperiodic boundary conditions will give us the correct answer: ∞ Y det(i∂t − m) = det(i∂t ) −BC n = −∞

(2n+1)π + T (2n+1)π T

m

∞ Y

=

n = −∞

153

mT 1+ (2n + 1)π

= cos

mT 2

.

By the usual analytic continuation T = −iβ, we conclude that fermions at finite temperature obey antiperiodic boundary conditions. 3. It is interesting to consider quantum field theory at finite density, as may occur in dense astrophysical objects or in heavy ion collisions. (In the previous chapter we studied a system of bosons at finite density and zero temperature.) In statistical mechanics we learned to go from the partition function to the grand partition function Z = tr e−β(H−µN ) , where a chemical potential µ is introduced for every conserved particle number N . For example, for ¯ 6 ∂ − m)ψ + µψγ ¯ 0 ψ. noninteracting relativistic fermions, the Lagrangian is modified to L = ψ(i Note that finite density, as well as finite temperature, breaks Lorentz invariance. Develop the subject of quantum field theory at finite density as far as you can. Solution: ¯ µψ First let us clarify that finite density breaks Lorentz invariance because the term ψγ is a Lorentz 4-vector, and we add to the Lagrangian only the µ = 0 component of that term, ¯ 0 ψ = ψ † ψ. ψγ To work with a theory at nonzero temperature T ≡ β −1 , take the Minkowski theory and compactify the imaginary time direction: t → iτ , 0 ≤ τ ≤ β. The gamma matrices of SO(3, 1) satisfy the relation {γ µ , γ ν } = 2η µν , so that (γ 0 )2 = +1 while (γ 1 )2 = (γ 2 )2 = (γ 3 )2 = −1. The gamma matrices of SO(4) satisfy {γEµ , γEν } = 2δ µν , so that (γE0 )2 = (γE1 )2 = (γE2 )2 = (γE3 )2 = +1, where the subscript E stands for “Euclidean.” We can therefore write γE0 = γ 0 , γEi = iγ i for the Euclidean gamma matrices13 . The Dirac operator in Euclidean space is 6 ∂ = γ µ ∂µ = γ 0 ∂t + γ i ∂i = γE0 (−i∂τ ) + (−iγEi )∂i = −i(γE0 ∂τ + γEi ∂i ) ≡ −i6 ∂E . R The theory we will work with has the Euclidean action SE = β d4 x LE , where we have R Rβ R defined β d4 x ≡ 0 dx0 d3 x, and the Lagrangian ¯ ∂E + m)ψ + µψγ ¯ 0ψ . LE = ψ(6 E ¯ The thermal two-point function S(x) ≡ hψ(x)ψ(0)i is defined as the inverse of the operator 0 6 ∂E + m + µγE : 6 ∂E + m + µγE0 S(x) = Iδ 4 (x) where I is the R4 × 4 identity matrix in Dirac spinor space. Multiply this equation by e−ipµ xµ and integrate β d4 x to get (+i 6 pE + m + µγE0 )S(p) = I 13

We continue to label the spacetime indices by 0,1,2,3 in Euclidean spacetime, instead of the often-used convention 1,2,3,4.

154

where we have defined the Fourier transform Z S(p) ≡ d4 x e−ipµ xµ S(x) . β

We therefore have the thermal two-point function Z ∞ X d3 p −i(ωn + iµ)γE0 − ipi γEi + m +i(ωn τ +~p·~x) 1 e S(x) = 3 2+p 2 + m2 β (2π) (ω + iµ) ~ n n = −∞ where we have defined p0 = ωn ≡ βπ (2n + 1) and summed instead of integrated because the τ ≡ x0 direction is compact. Everything proceeds as in the case for which µ = 0, except with the replacement p0 → p0 + iµ.

V.3

Landau-Ginzburg Theory of Critical Phenomena

1. Another important critical exponent γ is defined by saying that the susceptibility χ ≡ (∂M/∂H)|H = 0 diverges as ∼ 1/|T − Tc |γ as T approaches Tc . Determine γ in Landau~ ·M ~ to (1) Ginzburg theory. [Hint: Instructively, there are two ways of doing it: (a) Add −H ~ and H ~ constant in space and solve for M ~ (H). ~ (b) Calculate the susceptibility function for M χij (x − y) ≡ [∂Mi (x)/∂Hj (y)]|H=0 and integrate over space.] Solution: ~ ·M ~ to (1) gives the free energy a) Adding −H ~2−H ~ ·M ~ )V + O[(M ~ 2 )2 ] G = (aM ~ so the susceptibility is χ = ∂M/∂H = ~ = 1 H, This is minimized for M a which implies γ = 1 in Landau-Ginzburg theory. b) Starting from (3) on p. 293, we have the susceptibility function χij (~x ) ≡

∂Mi (~x ) ∂Hj (~y )

√

H =0

e− a |~x−~y | = δij 4π|~x − ~y |

Stripping off the δij and integrating over space gives the susceptibility Z χ=

√

e− a |~x | dx = 4π|~x | 3

Z

∞

0

Again we find γ = 1.

155

dr r e−

√ a r

=

1 1 ∼ a T − Tc

1 a

∼ 1/(T − Tc ),

V.4

Superconductivity

1. Vary (1) to obtain the equation for A and determine the London penetration length more carefully. b 1 (1) F = Fij2 + |Di ϕ|2 + a|ϕ|2 + |ϕ|4 + ... 4 2 Solution: Expanding out the field strength Fij = ∂i Aj − ∂j Ai gives 1 1 Fij Fij = Ai −δij ∂~ 2 + ∂i ∂j Aj 4 2

(i)

and the covariant derivative Di ϕ = ∂i ϕ − 2ieAi ϕ implies (Di ϕ)† Di ϕ = ∂i ϕ† ∂i ϕ + 2ieAi (ϕ† ∂i ϕ − ϕ∂i ϕ† ) + (2e)2 ϕ† ϕAi Ai .

(ii)

Varying equation (i) with respect to Ai gives 1 δ Fij Fij = δAi −δij ∂~ 2 + ∂i ∂j Aj 4 and varying equation (ii) with respect to Ai gives δ (Di ϕ)† Di ϕ = δAi 2ie(ϕ† ∂i ϕ − ϕ∂i ϕ† ) + 2(2e)2 ϕ† ϕAi . The equation of motion found from setting δF = F[A + δA] − F[A] = 0 is therefore m2γ † 2 −δij ∂~ + ∂i ∂j + † ϕ ϕ δij Aj = −2ie(ϕ† ∂i ϕ − ϕ∂i ϕ† ) hϕ ϕi where we have defined the mass of the photon inside the superconductor via m2γ ≡ 2(2e)2 hϕ† ϕi. The extra factor of two is there because the mass term for a real vector boson is F = 1 2 m AA. 2 A i i Choose the gauge ∂j Aj = 0. When ϕ assumes its vacuum expectation value, we have (−∂~ 2 + m2γ )Ai = 0 . The physically relevant solution is the decaying exponential. Sub~ stitute the form Ai (~x) = C e−k·~x to get (−~k 2 + m2γ )C = 0 =⇒ |~k | = mγ . We find an exponential decay Ai (r) ∼ e−mγ r . The London penetration depth is defined as the length scale over which the field penetrates into the material, meaning `L = where v ≡

p

1 1 1 =p = √ mγ 2 2 ev 2(2e)2 v 2

hϕ† ϕi .

156

2. Determine the coherence length more carefully. Solution: Use the free energy from problem V.4.1, but this time vary ϕ† to get an equation of motion for ϕ. The relevant part of the free energy is F = ∂i ϕ† ∂i ϕ + 2ieAi (ϕ† ∂i ϕ − ϕ∂i ϕ† ) +

m2γ † b ϕ ϕAi Ai + aϕ† ϕ + (ϕ† ϕ)2 . 2 2v 2

If we choose the gauge ∂i Ai = 0, then we can integrate by parts to move the derivative off of ϕ† and onto ϕ. The variation δF = F[ϕ† + δϕ† ] − F [ϕ† ] immediately gives the equation of motion for ϕ: m2γ 2 2 † ~ ~ ~·∂+ ~ + a + bϕ ϕ ϕ = 0 . −∂ + 4ie A A 2v 2 Deep inside the superconductor (or more accurately, at depths much greater than `L derived in the previous problem), we can set the vector potential to zero. Assuming that the selfinteraction term governed by the parameter b is negligible, we obtain (−∂~ 2 + a)ϕ = 0 Recall that for this entire discussion to work, we must have a < 0 (the analog of the “negative mass squared” instability from chapter IV.) We therefore determine an oscillating solution ϕ(r) ∼ e±ir/`ϕ , where 1 `ϕ = √ −a is the coherence length, defined as the characteristic length scale over which ϕ varies.

157

V.6

Solitons

3. Compute the mass of the kink by the brute force method and check the result from the Bogomol’nyi inequality. Solution: The mass of the kink is given on p. 305 as " # Z 2 2 1 df 1 2 µ2 f −1 + M = µ dy λ 2 dy 4 where f (y) ≡ ϕ(x)/v and y ≡ µx. Setting δM ≡ M [f + δf ] − M [f ] = 0 gives the equation f 00 (y) = [f (y)2 − 1]f (y) . Mathematica can solve this equation using the command DSolve[{f 00 [y] == (f[y]2 - 1)f[y], f 0 [∞] == 0}, f[y], y] which yields the solution

y f (y) = tanh √ . 2 Putting this solution into M and integrating from −∞ < y < ∞, we get the result √ 2 2 µ2 M= µ. 3 λ The Bogomol’nyi inequality is

√ 2 2 µ2 M≥ µ|Q| 3 λ so the kink is at precisely the lower bound of the inequality. This implies |Q| ≤ 1, which is another way of showing that the kink has topological charge Q = 1 (using the convention for which the antikink has charge Q = −1).

V.7

Vortices, Monopoles, and Instantons

1. Explain the relation between the mathematical statement Π0 (S 0 ) = Z2 and the physical result that there are no kinks with |Q| ≥ 2. Solution: The quantity Π0 (S 0 ) counts the number of topologically inequivalent mappings of spatial S 0 into the group manifold M = S 0 . Since Π0 (S 0 ) = Z2 , and since the group Z2 has only two elements, there are only two topologically inequivalent kink configurations. These are the kink and antikink, with Q = +1 and Q = −1 respectively. Therefore, there is no solution with a nonzero Q for which |Q| 6= 1.

158

2. In the vortex, study the length scales characterizing the variation of the fields ϕ and A. Estimate the mass of the vortex. Solution: The mass of a time-independent configuration of the theory is Z 1 2 † † 2 2 M = d x (Di ϕ) Di ϕ + λ(ϕ ϕ − v ) + Fij Fij 4 where Di ϕ = ∂i ϕ−ieAi ϕ. For future reference, let us collect the mass dimensions of all quantities in this expression. In (2+1) spacetime dimensions, we have: [ϕ] = [v] = [A] = [e] = + 12 and [λ] = +1. To check this, note that [∂] = +1 must have the same dimension as [eA] = [e] + [A] = + 21 + 12 = +1 X. Also, [F ] = [∂A] = +1 + 12 = + 32 , so that [F 2 ] = +3, which is indeed the mass dimension of the Lagrangian density in (2+1) dimensions. Finally, [λϕ4 ] = [λ]+4[ϕ] = +1+2 = +3, which is again the correct mass dimension of the Lagrangian. Propose the ansatz 1 ϕ(r, θ) = ve iθ f (r) , Ai (r, θ) = ∂i θ g(r) e with f (r) and g(r) dimensionless functions that equal 1 at r = ∞. Using the relations ∂i r =

xi 1 , ∂i θ = (cos θ δiy − sin θ δix ) r r

we can now perform the requisite algebra to obtain each part of the Lagrangian. The derivative of the scalar is v ∂i ϕ = e iθ [i(cos θ δiy − sin θ δix )f (r) + xi f 0 (r)] . r The field strength Fij = ∂i Aj − ∂j Ai is Fij =

g 0 (r) [(cos θ δjy − sin θ δjx ) xi − (cos θ δiy − sin θ δix ) xj ] . er2

We find: f (r)2 0 2 ∂i ϕ ∂i ϕ = v + [f (r)] r2 v2 Ai ϕ† ∂i ϕ − h.c. = 2i 2 f (r)2 g(r) er v 2 A i Ai ϕ † ϕ = f (r)g(r) er 2 † 2 2 (ϕ ϕ − v ) = v 4 f (r)2 − 1 2 1 1 g 0 (r) Fij Fij = 4 2 er †

2

159

Therefore, the mass of the soliton is ( 0 2 ) Z 2 1 g (r) v M = d2 x 2 [g(r) − 1]2 f (r)2 + r2 [f 0 (r)]2 + λv 2 r2 [f (r)2 − 1]2 + r 2 ev Since [v 2 ] = +1, each term in the square brackets must have mass dimension +2. The reader may verify using the formulas collected previously that this is true. Note that the combination ξ ≡ evr is dimensionless. Let us now switch to dimensionless variables. Define the functions of dimensionless variables F (ξ) ≡ f (r) and G(ξ) ≡ g(r), so that f 0 (r) = ev F 0 (ξ) and g 0 (r) = ev G0 (ξ). Also define the dimensionless coupling β ≡ λ/e2 . In terms of these, the mass of the vortex is given by Z evR 1 0 1 2 2 2 0 2 2 2 2 2 [G(ξ) − 1] F (ξ) + ξ [F (ξ)] + β ξ [F (ξ) − 1] + [G (ξ)] M = 2πv dξ ξ 2 eva where a is the size of the vortex and R is the size of the spatial region. (We are using the notation of the discussion about the Kosterlitz-Thouless phase transition on p. 310.) We now vary with respect to the functions F (ξ) and G(ξ) to find the equations of motion. Minimizing with respect to F (ξ) gives d d 1 ξ F (ξ) = βξ[F (ξ) − 1] + [G(ξ) − 1]2 F (ξ) dξ dξ ξ and minimizing with respect to G(ξ) gives d 1 d G(ξ) = 2[G(ξ) − 1]F (ξ)2 . ξ dξ ξ dξ Recall that the boundary conditions on F (ξ) and G(ξ) are that they go to 1 for ξ → ∞. With this in mind, consider the ansatze F (ξ) = 1 − e−αF ξ and G(ξ) = 1 − e−αG ξ . Since ξ = evr, the quantities `ϕ ≡ 1/(evαF ) and `A ≡ 1/(evαG ) characterize the length scales over which the fields ϕ and A, respectively, vary. With the above forms, the equations of motion become αF (1 − ξαF ) = −βξ +

1 −2αG ξ +αF ξ e [e − 1] ξ

1 (αG − )αG = 2[1 − e−αF ξ ]2 ξ

ξ→∞

→

ξ→∞

→

αF2 = β

2 αG =2

Thus, recalling that β = λ/e2 , we obtain the lengths 1 `ϕ = √ λv

and 160

1 `A = √ . 2 ev

(provided that αF < 2αG )

Note that the gauge coupling e has dropped out of the length scale of ϕ, and that the length scale of A does not depend on the quartic coupling λ. We could have arrived at this conclusion simply on the basis of dimensional analysis: both length scales are determined by ∼ 1/v, but v has mass dimension +1/2 and so another power of mass dimension +1/2 is needed √ for each field. Since λ has mass dimension +1 and is associated with ϕ, we guess `ϕ ∼ 1/( λ v). Since e has mass dimension +1/2 and is associated with A, we guess `A ∼ 1/(ev). Note also that the condition αF < 2αG implies λ < 8e2 or λ < 32πα, where α ≡ e2 /(4π) as usual. This implies that the configuration makes sense only if the quartic coupling for the scalar field is less than ∼ 100α. √ As for√ the mass of the vortex, we take the above forms for F (ξ) and G(ξ) with αF = β and αG = 2 , and perform the integral. As discussed on p. 310, we need to cut off the integral at the low end by the size a of the vortex. On the other hand, the boundary conditions discussed previously allow us to take R → ∞. Asking Mathematica to perform the integral, we obtain: # " √ 2! √ β + 2 2 ) ( 1 1 √ + ln − γ + + O(eva) M = 2πv 2 ln √ eva 2 16( β + 2 ) where γ ≈ 0.577. Perhaps a more useful quantity to compute is the mass difference between a vortex of size a0 and one of size a: a 0 2 M (a ) − M (a) = 2πv ln 0 a which shows the amount of energy required to increase the size of the vortex from a size a to a size a0 > a. In any case, the mass of the vortex is estimated to be of order M ∼ 2πv 2 . Recall the remark on p. 309 that the mass of the topological monopole in (3 + 1) dimensions comes out to be of order M ∼ MW /α, where α ≡ e2 /(4π). In our case, we have 2 2 2 2 2 MW ≡ `−1 A ∼ ev =⇒ v ∼ MW /e and thus in (2 + 1) dimensions we find M ∼ MW /e , 1 which is dimensionally correct since e now has mass dimension + 2 .

3. Consider the vortex configuration in which ϕ(r → ∞) → ve iνθ with ν an integer. Calculate the magnetic flux. Show that the magnetic flux coming out of an antivortex (for which ν = −1) is opposite to the magnetic flux coming out of a vortex. Solution: Consider the field configuration at spatial infinity ϕ∞ = v e inθ

161

where n is an integer. The gauge field at infinity is † −i ϕ∞ ∂i ϕ∞ −i v(v in∂i θ) n A∞i = = = ∂i θ 2 2 e |ϕ∞ | e v e Everything R 2 proceeds H iprecisely as in the text, except multiplied by the integer n. The flux Φ ≡ d x F12 = dx Ai is 2π Φ=n = nΦ0 e where Φ0 ≡ 2π is the fundamental unit of flux in equation (2) on p. 307. As discussed, a e vortex corresponds to n = +1, while an antivortex corresponds to n = −1. Immediately we have Φvortex = Φ0 = −Φantivortex .

6. Display explicitly the map S 2 → S 2 , which wraps one sphere around the other twice. Verify that this map corresponds to a magnetic monopole with magnetic charge 2. Solution: An arbitrary element g 0 1 T ≡ 0 0

~

of SO(3) can be parametrized as g(~x) = e ~x·T , where the matrices 0 0 0 0 1 0 −1 0 0 −1 , T 2 ≡ 0 0 0 , T 3 ≡ 1 0 0 1 0 −1 0 0 0 0 0

satisfy [T a , T b ] = εabc T c (with ε123 ≡ +1) and thereby generate SO(3), and the coordinates x1 = sin θ cos(nφ) , x2 = sin θ sin(nφ) , x3 = cos θ satisfy ~x ·~x = 1 and thereby manifestly parameterize the surface of a two-sphere. The number n is an integer whose significance we will derive shortly. The function g(~x) thereby constitutes a map S 2 → S 2 . We now compute the winding number of this configuration, namely the quantity Z 1 tr[(g −1 dg)3 ] νg ≡ − 8π R3

162

~ where we are using forms notation. Using the notation g = e~x·T , we have g −1 dg = d~x · T~ . We therefore compute the integral: Z Z −1 3 tr[(g dg) ] = dxa dxb dxc tr(T a T b T c ) 3 3 R ZR = dxa dxb dxc 21 tr([T a , T b ]T c ) 3 ZR dxa dxb dxc 12 εabd tr(T d T c ) = 3 ZR = dxa dxb dxc 21 εabd (−2δ dc ) ← [check explicitly from the above T a ] 3 RZ dxa dxb dxc εabc =− 3 ZR =− d(xa dxb dxc )εabc ← (since d2 = 0) 3 ZR xa dxb dxc εabc ← (Stokes’ theorem) =− S2 Z 2π Z π =− dφ dθ εAB xa ∂A xb ∂B xc εabc ← [coordinate basis (θ, φ)] 0

0

Therefore we have 1 νg = 8π

Z

2π

Z

dφ 0

π

dθ εabc εAB xa ∂A xb ∂B xc .

0

Using our expressions for xa given above, we compute the partial derivatives: ∂θ x1 = cos θ cos(nφ) , ∂θ x2 = cos θ sin(nφ) , ∂θ x3 = − sin θ ∂φ x1 = −n sin θ sin(nφ) , ∂φ x2 = +n sin θ cos(nφ) , ∂φ x3 = 0 . Next we expand out the epsilon tensor εabc corresponding to the SO(3) algebra directions: εabc xa ∂A xb ∂B xc = x1 ∂A x2 ∂B x3 + x3 ∂A x1 ∂B x2 + x2 ∂A x3 ∂B x1 − x3 ∂A x2 ∂B x1 − x1 ∂A x3 ∂B x2 − x2 ∂A x1 ∂B x3

163

Now we compute: εabc xa ∂θ xb ∂φ xc = 0 + [cos θ][cos θ cos(nφ)][n sin θ cos(nφ)] + [sin θ sin(nφ)][− sin θ][−n sin θ sin(nφ)] − [cos θ][cos θ sin(nφ)][−n sin θ sin(nφ)] − [sin θ cos(nφ)][− sin θ][n sin θ cos(nφ)] − 0 = n[cos2 θ sin θ cos2 (nφ) + sin2 θ sin θ sin2 (nφ) + cos2 θ sin θ sin2 (nφ) + sin2 θ sin θ cos2 (nφ)] = n[cos2 θ sin θ + sin2 θ sin θ] = n sin θ . The other term contributes equally, so finally we have: Z 2π Z π 1 νg = dθ n sin θ = n . dφ 4π 0 0 Thus the integer n is itself the winding number for the gauge field configuration. If we desire the map that wraps one sphere around the other twice, then we choose n = 2 in the definition of the SO(3) algebra coordinates xa . Next we have to compute the magnetic charge of the monopole described by this field configuration. Equation (7) on p. 308 provides us with the definition of the gauge-invariant electromagnetic field: 1 Fij ≡ F~ij · ~x − εabc xa (Di x)b (Dj x)c e b a a abc b c where (Di x) = ∂i x + e ε Ai x and Ai = ∂i xb as before, and we have used xa xa = 1. (We are labeling the field variables by xa – as we have been throughout this problem – to emphasize that the parameters on the group manifold should be thought of as coordinates like any other. We use i, j, k to denote spatial R3 , and A, B to denote the S 2 of spatial infinity.) The gauge-covariant field strength is Fija = ∂i Aaj − ∂j Aa + e εabc Abi Acj = e εabc ∂i xb ∂j xc , so F~ij · ~x = e εabc xa ∂i xb ∂j xc . Next we have to contract two covariant derivatives with an epsilon. This will require using εabc εade = δ bd δ ce − δ be δ cd , which follows from SO(3) invariance and checking with b = d = 2 and c = e = 3. We compute:

164

xa εabc (Di x)b (Dj x)c = xa [εabc ∂i xb ∂j xc + e∂i xb (εabc εcpq )∂j xp xq + e(εabc εbef )∂i xe xf ∂j xc + e2 (εabc εbef εcpq )∂i xe xf ∂j xp xq ] = xa [εabc ∂i xb ∂j xc + e∂i xb (δ ap δ bq − δ aq δ pb )∂j xp xq + e(−δ ae δ cf + δ af δ ec )∂i xe xf ∂j xc + e2 (δ ap εqef − δ aq εpef )∂i xe xf ∂j xp xq ] = εabc xa ∂i xb ∂j xc + e [(∂i xb ∂j xa xb xa − ∂i xb ∂j xb ) + (−xa ∂i xa xb ∂j xb + ∂i xa ∂j xa )] {z } | cancel

2

qef

f

q

e

a a

pef

e f

p

+ e [(|ε {z x x} ∂i x ∂j x x ) − ( ε ∂i x x ∂j x )] | {z } =0 abc a b = −ε x ∂i x ∂j xc = (1 + e2 )εabc xa ∂i xb ∂j xc Therefore, the gauge-invariant field strength is 1 Fij = e εabc xa ∂i xb ∂j xc − [(1 + e2 )εabc xa ∂i xb ∂j xc ] e 1 abc a b c = − ε x ∂i x ∂j x . e The magnetic field is 1 1 Bi ≡ εijk Fjk = − εijk εabc xa ∂i xb ∂j xc . 2 2e The magnetic flux through a 2-sphere at infinity is Z Z π 1 2π g=− dφ dθ εabc εAB xa ∂A xb ∂B xc 2e 0 0 1 = − (8πn) 2e 4π =− n. e As explained in the book’s solution for V.7.5, the fundamental unit of charge is actually 21 e, so that we have Dirac’s quantization condition g=−

2π n (e/2)

as expected. As in V.7.5, the sign just corresponds to which we call the monopole and which one we call the antimonopole.

165

9. Discuss the dyon solution. Work it out in the BPS limit. [B. Julia and A. Zee, Phys. Rev. D11: 2227, 1975.] Solution: We follow M. K. Prasad and C. M. Sommerfield, “Exact classical solution for the ’t Hooft monopole and the Julia-Zee dyon,” Phys. Rev. Lett. Vol. 35, No. 12, 22 Sep 1975. The Lagrangian density is 1 a aµν 1 F + (Dµ φ)a (Dµ φ)a − V (φ) L = − Fµν 4 2 1 a a 1 a a 1 1 = − Fij Fij + F0i F0i − (Di φ)a (Di φ)a + (D0 φ)a (D0 φ)a − V (φ) 4 2 2 2 with potential V (φ) = − 21 µ2 (φa φa ) + 14 λ(φa φa )2 (µ2 > 0) and covariant derivative (Dµ φ)a = a = ∂µ Aaν − ∂ν Aaµ + εabc Abµ Acν . The equations of ∂µ φa + eεabc Abµ φc . The field strength is Fµν motion are (Dµ Fµν )a + eεabc φb (Dν φ)c = 0 (Dµ Dµ φ)a − µ2 φa + λ(φb φb )φa = 0 . There is also a constraint (Dµ Fµ0 )a + eεabc φb (D0 φ)c = 0. Choose the ansatz φa =

j xa xa a a aij x H(r) , A [1 − K(r)] , A J(r) = ε = i 0 er2 er2 er2

and plug into the equations of motion (including the constraint) to obtain: r2 K 00 = K(K 2 − 1) + K(H 2 − J 2 ) r2 J 00 = 2JK 2 λ r2 H 00 = 2HK 2 + 2 (H 2 − C 2 r2 )H e where C ≡ µe/λ1/2 . The boundary conditions are: H(0) = J(0) = 0 , K(0) = 1 K(r → ∞) = 0 J(r) lim =M r→∞ r H(r) lim = C cosh γ r→∞ r where M is the characteristic inverse length of Aa0 , and γ is an arbitrary constant.

166

The equations of motion admit the solution Cr sinh(Cr) J = sinh γ [Cr coth(Cr) − 1] H = cosh γ [Cr coth(Cr) − 1] . K=

Given the electromagnetic field (see equation (7) on p. 308 of the book) Fµν =

1 a a 1 abc a φ Fµν − ε φ (Dµ φ)b (Dν φ)c |φ| e|φ|3

we find the electric field Ei ≡ −F0i to be xi 1 − K2 xi d J(r) = sinh γ . Ei = r dr er r er2 This can be used to find the electric charge: Z Z 8π ∞ JK 2 4π 3 Q = d x ∂i Ei = dr = sinh γ . e 0 r e The case γ = 0 corresponds to the monopole solution discussed in the text. See both references for further discussion.

10. Verify explicitly that the magnetic monopole is rotation invariant in spite of appear~ are covariant ances. By this is meant that all physical gauge invariant quantities such as B b under rotation. Gauge variant quantities such as Ai can and do vary under rotation. Write down the generators of rotation. Solution: First a small but important typographical error: All physical gauge invariant quantities are invariant under rotation, not covariant. Covariant means transforms as a vector, while invariant means does not transform. Gauge transformations (rotations within the gauge group) are conceptually nothing but changing coordinates (in field space), and physics must be invariant under a change of coordinates. The electromagnetic field from the magnetic monopole is Fµν = where |ϕ| =

a a Fµν ϕ εabc ϕa (Dµ ϕ)b (Dν ϕ)c − |ϕ| e|ϕ|3

√ a a ϕ ϕ , (Di ϕ)a = ∂i ϕa + eεabc Abi ϕc and v 1 ϕa (r) = δ ai xi f (r) , Aai (r) = 2 εaij xj g(r) , Aa0 = 0 r er 167

with f (r) and g(r) yet unspecified functions of r. Since we are concerned only with the transformation properties of the various terms under SO(3) gauge transformations, we do not need to specify the functions f (r) and g(r) other than saying that they are scalars under gauge transformations, as indicated by their lack of SO(3) indices. Since the symbols δ ab and εabc are invariant under SO(3), the electromagnetic field Fµν is manifestly gauge invariant. We will now verify this explicitly using infinitesimal SO(3) transformations. An arbitrary infinitesimal SO(3) transformation can be parametrized as U (n, θ) = I +θna T a , where na na = 1 and the generators T a are 0 0 0 0 0 1 0 −1 0 T 1 = 0 0 −1 , T 2 = 0 0 0 , T 3 = 1 0 0 0 1 0 −1 0 0 0 0 0 and satisfy the commutation relations [T a , T b ] = εabc T c , where ε123 ≡ +1. (By the way, this immediately shows that εabc is invariant under SO(3), since the structure constants of a Lie algebra are invariant symbols of the corresponding group.) To show explicitly that Fµν is invariant under SO(3), we need to show explicitly that δ ab and εabc are invariant under simultaneous SO(3) transformations on all of their indices, since the first term of Fµν is constructed from δ ab and the second term is constructed from εabc (and the magnitude |ϕ| is also constructed from δ ab ). Note that the indices a, b, c label the adjoint representation of SO(3). First, transform δ ab . Under an infinitesimal SO(3) transformation, this symbol transforms as δ ab → U ac U bd δ cd = [I + iθne T e ]ac [I + iθnf T f ]bd δ cd = δ ab + iθne (T e )ac δ cb + iθnf (T f )bd δ ad + O(θ2 ) = δ ab + iθne (T e )ab + iθnf (T f )ba = δ ab X since (T a )bc = −(T a )cb , as can be seen above by the explicit construction of the matrices {T a }3a = 1 . Now transform εabc : εabc → U ad U be U cf εdef = [I + iθ~n · T~ ]ad [I + iθ~n · T~ ]be [I + iθ~n · T~ ]cf εdef = εabc + iθ~n · T~ ad εdbc + T~ be εaec + T~ cf εabf + O(θ2 ) abc ad dbc bd adc cd abd ~ ~ ~ = ε + iθ~n · T ε + T ε + T ε 168

Specialize to the case (abc) = (123), since all elements of εabc are defined in terms of ε123 . We have ε123 → ε123 + iθ~n · T~ 1d εd23 + T~ 2d ε1d3 + T~ 3d ε12d = ε123 + iθ~n · T~ 11 ε123 + T~ 22 ε123 + T~ 33 ε123 = ε123 X since all diagonal entries of the T a are equal to zero. This completes the calculation.

VI

Field Theory and Condensed Matter

VI.1

Fractional Statistics, Chern-Simons Term, and Topological Field Theory

1. In a nonrelativistic theory you might think that there are two separate Chern-Simons terms, εij ai ∂0 aj and εij a0 ∂i aj . Show that gauge invariance forces the two terms to combine into a single Chern-Simons term εµνλ aµ ∂ν aλ . For the Chern-Simons term, gauge invariance implies Lorentz invariance. In contrast, the Maxwell term would in general be nonrelativistic, consisting of two terms, f0i2 and fij2 , with an arbitrary relative coefficient between them (with fµν = ∂µ aν − ∂ν aµ as usual). Solution: The Lagrangian is L = c1 εij ai ∂0 aj + c2 εij a0 ∂i aj . A gauge transformation aµ → aµ + ∂µ λ implies L → L + (2c1 + c2 )εij ∂0 λ∂i aj =⇒ c2 = −2c1 . Up to total derivatives, εµνρ aµ ∂ν aρ = −ε0ij (ai ∂0 aj − 2a0 ∂i aj ). Therefore L = −c1 εµνρ aµ ∂ν aρ .

2. By thinking about mass dimension, convince yourself that the Chern-Simons term dominates the Maxwell term at long distances. This is one reason that relativistic field theorists find anyon fluids so appealing. As long as they are interested only in long distance physics they can ignore the Maxwell term and play with a relativistic theory (see exercise VI.1.1). Note that this picks out (2+1)-dimensional spacetime as special. In (3+1)-dimensional spacetime the generalization of the Chern-Simons term εµνλσ fµν fλσ has the same mass dimension as the Maxwell term f 2 . In (4+1)-dimensional space the term ερµνλσ aρ fµν fλσ is less important at long distances than the Maxwell term f 2 . Solution: 169

There are two standard ways to assign mass dimensions to gauge fields, which of course yield equivalent results. If we normalize the gauge field by L = − 4g12 f 2 with the covariant derivative Dµ ψ = (∂µ − iaµ )ψ (for some matter field ψ), then [a] = [∂] = +1 for any choice of the spacetime dimension d. The Chern-Simons term εµνρ aµ ∂ν aρ therefore has mass dimension [a∂a] = 2[a] + [∂] = 3. Meanwhile, the field strength fµν = ∂µ aν − ∂ν aµ has dimension [f ] = [∂a] = 2, so the Maxwell term f 2 has dimension 4. Therefore [a∂a] < [f 2 ]. Instead suppose we normalize the gauge field by L = − 41 f 2 with the covariant derivative Dµ ψ = (∂µ − igaµ )ψ. Since the Lagrangian always has dimension [L] = d in d spacetime R d dimensions to make the action S = d x L dimensionless, we have [f 2 ] = 2[∂a] = 2(1+[a]) = d =⇒ [a] = (d − 2)/2. The Chern-Simons term has dimension [a∂a] = 2[a] + 1 = d − 2 + 1 = d − 1. Since the Maxwell term f 2 has the same mass dimension as that of the Lagrangian L, namely d, we see again that [a∂a] < [f 2 ].

3. There is a generalization of the Chern-Simons term to higher dimensional spacetime different from that given in exercise IV.1.2. We can introduce a p-form gauge potential (see chapter IV.4). Write the generalized Chern-Simons term in (2p + 1)-dimensional spacetime and discuss the resulting theory. Solution: First a typo: The problem meant to contrast this with the previous exercise, VI.1.2. In (2p + 1)-dimensional spacetime, the Chern-Simons term is a (2p + 1)-form. If we introduce a p-form gauge potential A, then F ≡ dA is a (p + 1)-form. The quantity AF is a (p + p + 1) = (2p + 1)-form, so the Chern-Simons Lagrangian is LCS = γ AF , where γ is a coupling constant.

4. Consider L = γaε∂a − (1/4g 2 )f 2 . Calculate the propagator and show that the gauge boson is massive. Some physicists puzzled by fractional statistics have reasoned that since in the presence of the Maxwell term the gauge boson is massive and hence short ranged, it can’t possibly generate fractional statistics, which is manifestly an infinite ranged interaction. (No matter how far apart the two particles we are interchanging are, the wave function still acquires a phase.) The resolution is that the information is in fact propagated over an infinite range by a q = 0 pole associated with a gauge degree of freedom. This apparent paradox is intimately connected with the puzzlement many physicists felt when they first heard of the Aharonov-Bohm effect. How can a particle in a region with no magnetic field whatsoever and arbitrarily far form the magnetic flux know about the existence of the magnetic flux?

170

Solution: The Lagrangian is 1 fµν f µν + γεµνρ aµ ∂ν aρ 2 4g µ 2 ν = 2a (−ηµν ∂ + ∂µ ∂ν )a , we have 1 µ 1 2 λ L= a (ηµν ∂ − ∂µ ∂ν ) + 2γεµλν ∂ aν 2 g2 L=−

Since fµν f µν

The propagator G is the inverse of the thing in brackets, so 1 2 λ (ηµν ∂ − ∂µ ∂ν ) + 2γεµλν ∂ Gνρ (x) = δµρ δ 3 (x) 2 g R Multiplying by e−ikx , integrating d3 x and moving derivatives around via integration by parts gives 1 2 λ ˜ νρ − 2 (ηµν k − kµ kν ) + 2iγεµλν k G (k) = δµρ g ˜ νρ now we will run into trouble. To see why, use Lorentz invariance If we try to solve for G to write ˜ νρ (k) = A(k 2 )η νρ + B(k 2 )k ν k ρ + C(k 2 )ενρσ kσ . G ˜ νρ gives Putting this into the equation for G A C 2 2 ρ ρ − 2iγC k δµ − k kµ + 2iγA + 2 k εµρλ kλ = −δµρ 2 g g which has no solution. (Note that B dropped out.) This is just the usual gauge fixing ˜ νρ : problem, which we can resolve by adding a term 2ξ1 ηµν to the equation for G 1 1 2 λ ˜ νρ (k) = δ ρ − 2 (ηµν k − kµ kν ) + 2iγεµλν k + ηµν G µ g 2ξ ˜ = Aη + Bkk + Cεk results in the conditions Now plugging in the form G g2 A A B C g2 2 2 2 k − − 2iγCk = −1 , 2 + − 2iγC = 0 , 2iγA + 2 k − =0. 2ξ g 2 g 2ξ g 2ξ Solving these gives A= B= C=

−(k 2 − (k 2 −

g2 2 ) 2ξ

g2 )g 2 2ξ

− (2γg 2 )2 k 2

−g 2 (k 2 −

g2

(k 2 −

g2

2ξ

)2 − (2γg 2 )2 k 2 2iγg 4

2ξ

−g 2 → k2 − (2γg 2 )2

ξ→∞

)2 − (2γg 2 )2 k 2

2ξ 1 − 2 k 2 − (2γg 2 )2 g

2iγg 4 → k2 (k2 − (2γg 2 )2 ) .

ξ→∞

171

2ξ → k2

ξ→∞

The propagator for the gauge boson has a simple pole at k 2 = (2γg 2 )2 which persists in the limit ξ → ∞. Therefore, the gauge boson has a mass m = 2γg 2 . The propagator also has a simple pole at k 2 = 0; notice that the terms involving the pole at k 2 = m2 are independent of the gauge parameter ξ when taking ξ → ∞, while the pole at k 2 = 0 comes with a factor of ξ in the numerator in the coefficient B. This is what is meant by the k 2 = 0 pole being associated with a gauge degree of freedom. Even though the physical gauge boson has mass m = 2γg 2 6= 0, there is still a long-range interaction from the nontrivial topology of the gauge theory.

5. Show that θ = 1/4γ. There is a somewhat tricky factor of 2. So if you are off by a factor of 2, don’t despair. Try again. [X. G. Wen and A. Zee, J. de Physique, 50: 1623, 1989.] Solution: As instructed at the top of p. 318, one approach is to take the Lagrangian La = γεµνλ aµ ∂ν aλ + aµ j µ and integrate out aµ to get the nonlocal Hopf Lagrangian. In anticipation of fixing transverse gauge ∂µ aµ = 0, add a term − 2ξ1 (∂a)2 to the Lagrangian to get, after integration by parts and ignoring a total derivative, the gauge-fixed Lagrangian Lξa = 21 aµ (G−1 )µλ aλ , where (G−1 )µλ = 2γεµνλ ∂ν + 1ξ ∂ µ ∂ λ . The effective action SHopf is computed by performing the gauge-fixed path integral over aµ : Z Z R i d3 x Lξa SHopf = −i ln Da e = d3 x d3 y j λ (x)Gλρ (x − y)j ρ (y) where G is the operator inverse of G−1 defined by the equation 1 (2γεµνλ ∂ν + ∂ µ ∂ λ )Gλρ (x) = δρµ δ 3 (x) . ξ Multiplying by

R

d3 x e−ikx gives the momentum-space equation 1 ˜ λρ (k) = δρµ (+2γεµνλ ikν − k µ k λ )G ξ

R ˜ ˜ in Lorentz-index space. By where G(k) ≡ d3 x e−ikx G(x). We now need to solve for G ˜ λρ (k) as Lorentz invariance, we can write G Gλρ (k) = A(k 2 )ηλρ + B(k 2 )ελρσ k σ + C(k 2 )kλ kρ . ˜ Plugging this form into the above equation for G(k) and using εµνλ ελρσ = δρµ δσν − δσµ δρν results in the three conditions: 1 2iγBk 2 = 1 , 2iγAεµνρ kν = 0 , 2iγB + Ck 2 = 0 . ξ 172

Therefore A = 0 and

1 ξ , C=− 2 2 . 2 2iγk (k )

B=

The momentum-space propagator is therefore ξ→0

˜ λρ (k) = G

1 ξ ελρσ k σ − 2 2 kλ kρ → 2 2iγk (k )

Now the goal is to evaluate SHopf =

R

1 ελρσ k σ . 2 2iγk

d3 x d3 y j λ (x)Gλρ (x − y)j ρ (y) with Z

Gλρ (x − y) =

d3 k e+ik(x−y) ελρσ k σ (2π)3 2iγk 2

and the current j λ (x) = j1λ (x) + j2λ (x) describing the exchange of two particles. Let particle 1 with charge q1 be sitting at rest at the origin: j1λ (x) = q1 δ0λ δ 2 (~x ). We want particle 2 with charge q2 to move in a half-circle of radius R around particle 1: j2λ (y) = q2 uλ (y)δ 2 (~y − ~r(t)), where π x cos(ωt) + yˆ sin(ωt)), uλ (y) = (1, ~y˙ ). t ≡ y 0 [0, ], ~r(t) = R(ˆ ω The speed ~y˙ can be taken arbitrarily small to ignore the relativistic factor (1 − |~y˙ |2 )−1/2 . cross We want the cross term SHopf =

cross SHopf

Z =

3

Z

dx

dy

Z = q1 q2

3

dx

0

Z

R

d3 x d3 y j1λ (x)Gλρ (x − y)j2ρ (y). Plugging in gives

q1 δ0λ δ 2 (~x ) π/ω

d3 k e+ik(x−y) ελρσ k σ (2π)3 2iγk 2

Z

0

0

q2 uλ (y)δ 2 (~y − ~r(t))

~

d3 k e+ik (x −t)+ik·~r(t) ε0ij k j ui (t, ~r(t)) 3 0 2 2 ~ (2π) 2iγ[(k ) − |k | ]

Z dt

0

R∞ 0 +ik0 x0 Since = 2πδ(k 0 ), we can perform the integral over x0 and then the integral R 0 −∞ dx e dk , leaving behind the above expression with k 0 set to zero: cross SHopf

Z = q1 q2

π/ω

Z dt

0

~

d2 k e+ik·~r(t) ε0ij k j r˙ i (t) . (2π)2 −2iγ|~k |2

The arbitrary 2-momentum k j can be written in polar coordinates as ~k = k(ˆ x cos ϕ + yˆ sin ϕ), ~ so that k · ~r(t) = kR[cos(ωt) cos ϕ + sin(ωt) sin ϕ] = kR cos(ωt − ϕ). Also, since ~r˙ (t) = ωR(−ˆ x sin(ωt) + yˆ cos(ωt)) we have ε0ij r˙ i (t)k j = ωkR[− sin(ωt) sin ϕ + (−1) cos(ωt) sin ϕ] = −ωkR cos(ωt − ϕ)

173

R R 2π R∞ where the (−1) comes from ε021 = −ε012 = −1. In polar coordinates, d2 k = 0 dϕ 0 dk k, so all of the factors of k = |~k | cancel out in the integrand. We have cross SHopf

Z π/ω Z 2π Z dϕ ∞ dk ikR cos(ωt−ϕ) q1 q2 dt = (−ωR) e cos(ωt − ϕ) −2iγ 2π 0 2π 0 0 Z 2π Z Z ∞ dϕ 1 dk ∂ ikR cos(ωt−ϕ) q1 q2 ωR π/ω dt e = iγ 2π iR 0 2π ∂k 0 0 Z π/ω Z 2π q1 q2 ω dϕ 1 ikR cos(ωt−ϕ) ∞ = dt e k=0 (−1)2γ 0 2π 2π 0 Z π/ω Z 2π q1 q 2 ω dϕ ikR cos(ωt−ϕ) =− dt − 1 lim e k→∞ 4πγ 0 2π 0 | {z } =0 q1 q 2 ω π q1 q2 =− − =+ 4πγ ω 4γ

Therefore the statistics parameter θ = SHopf for two particles of charge q1 = q2 = +1 is θ = 1/(4γ).

6. Find the nonabelian version of the Chern-Simons term ada. [Hint: As in chapter IV.6 it might be easier to use differential forms.] Solution: The new feature in a non-abelian theory is that a3 6= 0. The Lagrangian will be of the form L = γ tr(ada + βa3 ) where the coefficient β should be fixed by gauge invariance (up to a total derivative). For a gauge transformation a → a + δa, we get δL = tr[(2 da + 3β a2 )δa], where we have dropped a total derivative. A non-abelian gauge transformation δa = [θ, a]−dθ with matrix-valued infinitesimal parameter θ(x) implies δL = γ tr[(2 − 3β)a2 dθ], where again we have dropped total derivatives. Demanding that δL = 0 implies β = 23 , so the non-abelian Chern-Simons Lagrangian is L = γ tr ada + 32 a3 .

7. Using the canonical formalism of chapter I.8 show that the Chern-Simons Lagrangian leads to the Hamiltonian H = 0. Solution: The Chern-Simons Lagrangian is L = γεµνρ aµ ∂ν aρ = 2γ a0 B + γε0ij a˙ i aj 174

where in the second equality we have integrated by parts, dropped a total derivative and defined B ≡ 21 ε0ij Fij . Notice that a0 has no time derivatives and is therefore constant; it acts as a Lagrange multiplier to enforce the constraint B = 0: ∂L = 0 =⇒ B = 0 . ∂a0 The Lagrangian is now L = γε0ij a˙ i aj which is linear in time derivatives. The conjugate momenta to ai are πi ≡

∂L = γε0ij aj ∂ a˙ i

and therefore H ≡ π i a˙ i − L = 0 .

8. Evaluate (6). Z

d3 p tr (2π)3

γν

1 1 γµ 6 p +6q − m 6 p − m

(6)

Solution: First we need gamma matrix identities in 3 dimensions. As pointed out in problem II.1.12, in (2+1) dimensions we can take the gamma matrices to be γ 0 = σ 3 , γ 1 = iσ 2 and γ 2 = −iσ 1 , where σ i are the usual 2-by-2 Pauli matrices. Let aµ , bµ , cµ and dµ be arbitrary 4-vectors. From the defining relation of the Clifford algebra {γ µ , γ ν } = 2η µν I, where here I is the 2 × 2 identity matrix, and using Lorentz invariance we get the following gamma matrix relations: tr(6 a6 b) = 2(ab) where (ab) ≡ a · b ≡ η µν aµ bν tr(6 a6 b6 c) = −2i εµνρ aµ bν cρ tr(6 a6 b6 c6 d) = 2 ((ab)(cd) + (ad)(bc) − (ac)(bd)) Note that unlike in four dimensions, the trace of three gamma matrices is not zero. Next notice that this integral is linearly divergent, so we will have define it properly such that gauge invariance is preserved (recall problem IV.7.3). We will adapt the method on p. 273 for d = 2 + 1 spacetime dimensions. For d = 3 + 1 dimensions, we had Z Pµ 4 µ f (P )2π 3 P 3 d p[f (p + a) − f (p)] = lim ia P →∞ P where the expression on the right-hand side is to be averaged over the sphere at infinity, for example replacing P µ P ν → 14 η µν P 2 . Adapting this to d = 2 + 1 dimensions we have Z Pµ 3 µ d p[f (p + a) − f (p)] = lim ia f (P )4πP 2 . P →∞ P 175

Replacing f (p) with the integrand f (p) ≡ tr γ ν µν

we find that the integral I µν (a) ≡

R

1 1 γµ 6 p+ 6 q − m 6 p − m

d3 p µν f (p (2π)3

+ a) satisfies the relation i µ Pµ µν µν f µν (P )P 2 . I (a) = I (0) + lim a 2 P →∞ 2π P

In this expression, f µν (P ) simplifies to tr(γ ν 6 P γ µ 6 P ) 2P µ P ν − η µν P 2 µν ν 1 µ 1 f (P ) = tr γ γ = = 6P 6P (P 2 )2 (P 2 )2 2( 1 η µν P 2 ) − η µν P 2 1 µν = 3 =− η . 2 2 (P ) 3P 2 Therefore, our integral satisfies i λ a I (a) = I (0) − lim P →∞ 6π 2 µν

µν

Pλ P

η µν .

The strategy now is to evaluate I µν (0) with a regulator in place, then to choose the vector aµ such that I µν (a) is gauge invariant, meaning qµ I µν (a) ≡ 0 and qν I µν (a) ≡ 0. Move the gamma matrices into the numerator using 1/(6 p − m) = (6 p + m)/(p2 − m2 ) to get Z N µν d3 p µν , N µν = tr [γ µ (6 p + m)γ ν (6 p + 6 q + m)] . I (0) = (2π)3 [p2 − m2 ][(p + q)2 − m2 ] Using the gamma matrix identities and tracing, we find N µν = 2[pµ (p + q)ν + pν (p + q)µ − η µν p · (p + q) − im eµνλ qλ + m2 η µν ] . The denominator can be simplified using the Feynman trick Z 1 1 1 dx = AB [xA + (1 − x)B]2 0 with A = p2 − m2 and B = (p + q)2 − m2 . After some arithmetic, this leads to Z 1 1 1 = dx 2 2 2 2 2 [p − m ][(p + q) − m ] (` + D)2 0 where ` ≡ p + (1 − x)q and D ≡ x(1 − x)q 2 − m2 . The denominator depends on ` only as `2 , so we can shift the integration variable from p to `, ignore terms linear in ` in the numerator, and make the replacement `µ `ν → 31 η µν `2 . This leads to 1 µν 2 µν µν 2 µ ν µνλ 2 µν N = 2 − η ` + x(1 − x)(η q − 2q q ) − im ε qλ + m η . 3 176

R R Now rotate to Euclidean momentum: d3 ` = i d3 `E and `2 = −`2E . Also define DE ≡ −D = x(1 − x)qE2 + m2 but leave q as Minkowski in the numerator. In total, we have i I (0) = 3 4π µν

Z

1

Z dx

d3 `E

0

+ 13 η µν `2E + K µν (`2E + DE )2

where K µν ≡ x(1 − x)(η µν q 2 − 2q µ q ν ) − im εµνλ qλ + m2 η µν does not depend on `E . The relevant integrals over `E are Z 1 π2 d3 `E 2 = 1/2 (`E + DE )2 DE and

Z

`2E 1/2 = 2πΛ − 3π 2 DE d `E 2 2 (`E + DE ) 3

p where we have regulated the second integral by imposing a cutoff Λ ≡ ( qE2 )max . We now have I µν (0) = Z 1 Z 2Λ µν µν 1 i −1/2 +1/2 µν 2 µ ν dx x(1−x)DE dx DE + (η q −2q q ) { η −η 4π 3π 2 0 0 Z 1 −1/2 dx DE } . + (−im εµνλ qλ + m2 η µν ) 0

p Now we need the integrals over x. In terms of β ≡ 2m/ qE2 , the relevant integrals are: Z

1

m [(1 + β 2 ) cot−1 β + β] 2β Z0 1 β −1/2 dx DE = cot−1 β m Z0 1 β −1/2 dx x(1 − x) DE = [(1 − β 2 ) cot−1 β + β] 8m 0 +1/2

dx DE

=

We therefore have i 2Λ µν β µν 2 −1 µ ν µνλ −1 µν I (0) = η +η R− [(1 − β ) cot β + β]q q − iε qλ β cot β 4π 3π 2 4m where R≡−

m β [(1 + β 2 ) cot−1 β + β] + q 2 [(1 − β 2 ) cot−1 β + β] + mβ cot−1 β . 2β 8m

The term R better simplify in such a way that q µ q ν and q 2 η µν appear only in the combination q µ q ν − q 2 η µν if the integral is to be gauge invariant. Indeed, this will be the case. Since 177

qE2 = 4m2 /β 2 = −q 2 , we have q 2 = −4m2 /β 2 so 2 4m β m 2 −1 [(1 − β 2 ) cot−1 β + β] + mβ cot−1 β R = − [(1 + β ) cot β + β] − 2β β2 8m m m = − [(1 + β 2 ) cot−1 β + β] − [(1 − β 2 ) cot−1 β + β] + mβ cot−1 β 2β 2β m 2 −1 = − [(1 + β ) cot β + β + (1 − β 2 ) cot−1 β + β − 2β 2 cot−1 β] 2β m = − [2 cot−1 β + 2β − 2β 2 cot−1 β] 2β m = − [(1 − β 2 ) cot−1 β + β] β Therefore we have 2 i 2Λ µν β µν 2 −1 µ ν µν 4m µνλ −1 I (0) = [(1 − β ) cot β + β][q q + η η − ] − iε qλ β cot β . 4π 3π 2 4m β2 Since β 2 = −4m2 /q 2 , we have i β 2Λ µν µν 2 −1 µ ν µν 2 µνλ −1 I (0) = η − [(1 − β ) cot β + β][q q − η q ] − iε qλ β cot β . 4π 3π 2 4m The finite part of I µν (0) therefore satisfies qµ I µν (0) = 0 and qν I µν (0) = 0. As discussed previously, we want i λ Pλ µν µν a I (a) = I (0) − lim η µν P →∞ 6π 2 P to be gauge invariant. We therefore choose aµ such that Λ Pµ µ lim a = P →∞ P π which makes β 2 −1 µ ν µν 2 µνλ −1 [(1 − β ) cot β + β][q q − η q ] + iε qλ β cot β 4m p satisfy qµ I µν (a) = 0 and qν I µν (a) = 0. Finally, let β = iα with α ≡ 2m/ q 2 to write the integral in terms of the Minkowskian momentum q µ . Since 1 α+1 −1 cot (iα) = −i ln 2 α−1 i I (a) = − 4π µν

we have

1 α+1 2 β[(1 − β ) cot β + β] = α (1 + α ) ln −α 2 α−1 and 1 α+1 −1 β cot β = α ln . 2 α−1 p Finally, in terms of the variable α ≡ 2m/ q 2 , the gauge invariant integral is α 1 α+1 α+1 µν 2 µ ν µν 2 µνλ I (a) = −i (1 + α ) ln − 2α q q − η q + iε qλ ln . 8π 4m α−1 α−1 2

−1

178

VI.2

Quantum Hall Fluids

1. To define filling factor precisely, we have to discuss the quantum Hall system on a sphere rather than on a plane. Put a magnetic monopole of strength G (which according to Dirac can be only a half-integer or an integer) at the center of a unit sphere. The flux through the sphere is equal to Nφ = 2G. Show that the single electron energy is given by E` = ( 12 ~ωc )[`(`+1)−G2 ]/G with the Landau levels corresponding to ` = G, G+1, G+2, ..., and that the degeneracy of the `th level is 2` + 1. With L Landau levels filled with noninteracting electrons (ν = L) show that Nφ = ν −1 Ne − S, where the topological quantity S is known as the shift. Solution: We follow X. G. Wen and A. Zee, “Shift and Spin Vector: New Topological Quantum Numbers for the Hall Fluids,” Phys. Rev. Lett., Vol. 69, No. 6, 10 Aug 1992. The point of this problem is to recognize that since a curved space has a connection 1form ω (see p. 443), it is possible to write down a Chern-Simons-type interaction between ω and the gauge potential aµ of the Hall fluid, which we remind the reader is defined as the po1 µνλ tential for the conserved electromagnetic current: J µ = 2π ε ∂ν aλ . Thus to the Lagrangian of equation (3) on p. 325, we add the interaction term Ls = s ωµ J µ =

s ωµ εµνλ ∂ν aλ 2π

where s is a real number. For example, on the sphere the connection 1-form is ω ab = −εab cos θ dϕ, where ε12 = +1. (See p. 444 in the text.) Augmenting the discussion on pp. 326-327 with this new term, we find the electromagnetic current 1 µνλ ∂Leff µ =+ JEM =− ε ∂ν (Aλ − sωλ ) ∂Aµ 2πk and the spin current Jsµ =

s µνλ ∂Leff = ε ∂ν (−Aλ + sωλ ) . ∂ωµ 2πk

The zero component of each is a number density of each type. Along the R 2current R 2 with 0 0 number of electrons Ne = d xJEMR and the number of spin quanta Ns = d xJs , define R 1 1 the number of flux quanta Nφ ≡ 2π F and the number of curvature quanta NR ≡ 2π R. ij 2 ij 2 Upon recognizing ε ∂i Aj d x = dA = F and ε ∂i ωj d x = dω = R, integrating the µ = 0 components of the currents implies 1 1 −s Ne Nφ = . 2 Ns NR k −s s Inverting this matrix equation shows that Nφ = kNe + sNR . 179

In the text it was shown that the filling factor is ν = 1/k (p. 327), so we have obtained Nφ = ν −1 Ne − S, with a shift Z R . S = −sNR = −s 2π R 1 The Gauss-Bonnet theorem says that 2π R = 2(1 − g), where g is the genus of the manifold over which we integrate. In particular, the sphere has genus zero so the shift is S = −2s.

2. For a challenge, derive the effective field theory for Hall fluids with filling factor ν = m/k with k an odd integer, such as ν = 25 . [Hint: Pm You have to introduce m gauge potentials aIλ µ µνλ and generalize (2) to J = (1/2π)ε ∂ν I=1 aIλ . The effective theory turns out to be m m X 1 X L= aI KIJ ε∂aJ + aIµ˜j Iµ + ... 4π I,J = 1 I =1

with the integer k replaced by a matrix K. Compare with (13).] (13)

L=

X 1 KIJ aI ε∂aJ + ... 4π I,J

Solution: For this problem and the next, we follow J. Fr¨ohlich and A. Zee, “Large Scale Physics of the Quantum Hall Fluid,” Nucl. Phys. B364 (1991) 517-540. For notational convenience we sometimes use differential forms and work with the Lagrangian volume-form L ≡ 4πL d3 x. The gauge potential aµ for the conserved electromagnetic current 1 ∗ da, where ∗ denotes the Hodge-star operator that takes J µ is defined by the equation J = 2π 1 µνλ p-forms to (3 − p)-forms (see IV.4.2). The Chern-Simons Lagrangian L = 4π ε aµ ∂ν aλ is written L = a da. (We set the coefficient of the Chern-Simons theory that results from the long-distance physics of the non-interacting Hall fluid to 1.) e Consider a 2D system of non-interacting electrons with m Landau levels filled: ν = 2πn = m. B The large gap between Landau levels implies that it is reasonable to take the levels as dynamically independent, so that the current of electrons belonging to each Landau level is separately conserved.

Let I = 1, 2, ..., m label the m levels. For each conserved electromagnetic current PmJI , we in1 ∗ troduce a gauge potential aI defined by JI = 2π daI . The new Lagrangian is L = I = 1 aI daI . Now turn on interactions between the electrons, which couple the different Landau levels. As discussed in the text, the resulting long-distance theory will be described by another 180

Chern-Simons Lagrangian. The interactions should only involve the total electromagnetic current ! m m X 1 ∗ X daI J= JI = 2π I =1 I =1 not the individual currents JI . As a result, electron-electron can only change P interactions P the Lagrangian by adding a Cern-Simons term of the form ( I aI )( J daJ ). Thus, the new Lagrangian is m m m X X X L= aI daI + p aI daJ ≡ aI KIJ daJ I =1

I,J = 1

I,J = 1

with p a real number. Here we have defined the m-by-m matrix p+1 p ... p .. p+1 . p K= . . . .. .. p p ... p p+1 Unpackaging the forms notation, we have arrived at the Lagrangian L0 =

m 1 X µνλ ε aIµ KIJ ∂ν aJλ . 4π I,J = 1

Denote the current of quasiparticles (the “vortex current”) in the I th Landau level by jIµ . This couples to the Chern-Simons gauge potential of each level as L1 =

m X

aIµ jIµ

I =1

by definition. (“Here we define the quasiparticles as the entities that couple to the gauge potential...” above equation (5) on p. 326.) Now subject the system to additional external electromagnetic fields described by the vector potential Aµ . This couples to the Chern-Simons gauge potentials through the electromagnetic current interaction for each level: m m m X X X 1 µνλ 1 µνλ µ L2 = JI Aµ = ε ∂ν aIλ Aµ = − ε aIλ ∂ν Aµ 2π 2π I =1 I =1 I =1 where we have dropped a total derivative. The full Lagrangian L = L0 + L1 + L2 is m m X 1 X µνλ 1 µνλ µ L= ε aIµ KIJ ∂ν aJλ + aIµ jI − ε ∂ν Aλ . 4π I,J = 1 2π I =1

181

1 µνλ ε ∂ν Aλ is called the “reduced vorticity current.” To The effective current ˜jIµ = jIµ − 2π understand the long-distance physics described by L, we integrate out the gauge fields aI to obtain a matrix-valued version of the Hopf Lagrangian of equation (6) on p. 326: m X −1 µνλ ˜jIµ (K )IJ ε ∂ν 1 ˜jJλ . Leff = π ∂2 I,J = 1

The inverse of the matrix K is K −1

p p p − 1+mp . . . − 1+mp 1 − 1+mp p − p 1+mp 1 − 1+mp = . . . . . . p p 1 − 1+mp − 1+mp

with 1 − p/(1 + mp) on the diagonals and −p/(1 + mp) everywhere else. This is the field theory whose consequences we will study in the next problem.

3. For the Lagrangian in (13), derive the analogs of (8), (9), and (11). (8)

µ Jem =

(9)

L=

(11)

1 µνλ ε ∂ ν Aλ 4πk

1 Aµ j µ k 1 θ = π k

Solution: In the previous problem, VI.2.2, we derived the effective Lagrangian m X 1 ˜ −1 µνλ ˜ jIµ (K )IJ ε ∂ν Leff = π jJλ ∂2 I,J = 1 with the matrix

p p p 1 − 1+mp − 1+mp . . . − 1+mp p − p 1+mp 1 − 1+mp = . .. .. . . p p − 1+mp 1 − 1+mp

K −1

This Lagrangian Leff contains three types of terms: AA, Aj and jj. The self-interaction of the electromagnetic field is given by ! m X 1 (AA) Leff = − (K −1 )IJ εµνλ Aµ ∂ν Aλ . 4π I,J = 1 182

Varying this with respect to A gives the expectation value of the total electromagnetic current: ! m X 1 µ (K −1 )IJ εµνλ ∂ν Aλ . hJEM i = 2π I,J = 1 As explained on p. 327 of the text, the µ = 0 component of this equation implies that the filling factor is ν=

m X

(K −1 )IJ =

I,J = 1

m 1 2 3 = , , , ... 1 + mp 1 + p 1 + 2p 1 + 3p

This is what replaces k −1 in equation (8). Thus for p = 2 we have derived the theory of quantum Hall fluids with filling factors 1 2 3 , , , ... 3 5 7 which realizes the claim of problem VI.2.2. ν=

µ From the spatial components of hJEM i ∝ εµνλ ∂ν Aλ , we also learn that an electric field in thePy-direction produces a current in the x-direction with a proportionality constant −1 )IJ = ν, known as the Hall conductance. σH = m I,J = 1 (K

Next we want to compute the electric charge of each quasiparticle. To do this, consider the coupling of the quasiparticle currents jI to the applied gauge field A: LAj eff

= Aµ

m X

µ (K −1 )IJ jJµ ≡ Aµ JEM

I,J = 1

where we have identified the total electromagnetic current induced by a quasiparticle currents jIµ . (All of these currents are actually current densities, but as is common we are being sloppy with the language.) Recall from earlier that the total electromagnetic current is comprised of a sum of m individually conserved electromagnetic currents for each Landau level: µ JEM

=

m X

JIµ

I =1

m X 1 µνλ ε ∂ν aIλ . = 2π I =1

Varying the total Lagrangian m m X 1 X µνλ 1 µνλ µ L= ε aIµ KIJ ∂ν aJλ + aIµ jI − ε ∂ν Aλ 4π I,J = 1 2π I =1 with respect to aIµ and recognizing

1 µνλ ε ∂ν aIλ 2π

jIµ =

m X

= JIµ , we obtain the matrix equation

KIJ JJµ .

J =1

183

Since we have already computed the matrix inverse of K, this can be inverted immediately to obtain m X JIµ = (K −1 )IJ jJµ . J =1

The µ = 0 component of this equation determines the charge density, which we can integrate to obtain the total electric charge induced in the I th Landau level: Z qI ≡

d2 xJI0 =

m X

(K −1 )IJ ΦJ

J =1

R where we have defined the “vorticities” ΦI = d2 x jI0 . In this language, qI is the electric charge bound to the vorticity ΦI in the I th Landau band. Using the explicit form of K −1 , the charge is qI =

p 1− 1 + mp

m X p ΦI − ΦJ 1 + mp J = 1 , J6=I

m X p = ΦI − ΦJ . 1 + mp J = 1

For ΦI = 1, I = 1, ..., m we find qI =

1 . 1 + mp

This is the replacement of q = 1/k from equation (9). Now let us turn to the jj interaction term: (jj) Leff

=π

m X

εµνλ jIµ (K −1 )IJ ∂ ν

I,J = 1

1 ∂2

jJλ .

As in the case of just one filled Landau level, this interaction describes a generalized version of the Aharonov-Bohm effect. An excitation of the system with vorticity vector (Φ1 , ..., Φm ) results in a statistics phase m X θ m = ΦI (K −1 )IJ ΦJ = π I,J = 1 1 + mp

where in the last line we have used ΦI = 1 for I = 1, ..., m. This is the generalization of equation (11).

184

VI.4

The σ Models as Effective Field Theories

1. Show that the vacuum expectation value of (σ, ~π ) can indeed point in any direction without changing the physics. At first sight, this statement seems strange since, by virtue of its γ5 coupling to the nucleon, the pion is a pseudoscalar field and cannot have a vacuum expectation value without breaking parity. But (σ, ~π ) are just Greek letters. Show that by a suitable transformation of the nucleon field parity is conserved, as it should be in the strong interaction. Solution: Before proceeding to the solution, it is useful to review the sigma model for nucleons and pions using slightly different notation from the text. The nucleon Lagrangian is ¯ L = g ψ(σI + i~π · ~τ γ5 )ψ = g ψ¯L (σI + i~π · ~τ )ψR + h.c. This problem is about transformation properties under the global SU (2)L ⊗SU (2)R symmetry whose diagonal piece SU (2)I ⊂ SU (2)L ⊗ SU (2)R is Heisenberg’s isospin. To remove possible sources of confusion about the Dirac spinor indices and to unclutter the notation as little as possible, let us write the Dirac field as two-component spinors: ψ = (χ, χ¯† )T . (Review Appendix E now if you are unfamiliar with this notation.) The above Lagrangian can be rewritten as L = −g χa εab Πbc˙ χ¯c˙ + h.c. where we have defined the matrix Πbc˙ ≡

4 X 1 µ=1

1 πµ (τ )bc˙ = 2 2 µ

σ + iπ3 i(π1 − iπ2 ) i(π1 + iπ2 ) σ − iπ3

. bc˙

Here {τ µ }3µ = 1 are the Pauli matrices times a factor of i:

0 1 τ ≡i 1 0 1

0 −i 1 0 3 , τ ≡i , τ ≡i i 0 0 −1 2

Also, τ 4 ≡ I is the 2 × 2 identity matrix, and π4 ≡ σ is the fourth meson field. (Review Appendix B for the local isomorphism SU (2) ⊗ SU (2) ' SO(4), which necessitates the inclusion of the fourth meson field.) The factors of 21 give the correctly normalized kinetic term L = tr(ΠΠ† ) = 12 [(∂σ)2 +(∂π3 )2 ]+ ∂π + ∂π − , where π ± ≡ √12 (π1 ∓ iπ2 ). Under SU (2)L , we have χa → (Lχ)a = Lab χb = χb Lab = χb (LT )b a and Πbc˙ → Lb c Πcc˙ , where L is a 2 × 2 unitary matrix with determinant 1. Since ~σ T ε = −ε~σ , we have LT ε = −εL† . So SU (2)L acts as χεΠ → χεL† LΠ = χεΠ, or in other words the Lagrangian is 185

invariant under SU (2)L . Under SU (2)R , we have χ¯c˙ → Rc˙ e˙ χ¯e˙ and Πbc˙ → Πbe˙ (R−1 )e˙ c˙ , so Πχ¯ → ΠR−1 Rχ¯ = Πχ, ¯ so the Lagrangian is invariant under SU (2)R . This is what is meant by stating that the Lagrangian is invariant under SU (2)L ⊗ SU (2)R . Now it is clear that the vacuum alignment may be chosen to point in an arbitrary direction in SO(4) ' SU (2)L ⊗ SU (2)R , since we can always perform SU (2)L ⊗ SU (2)R transformations on the nucleon fields to leave the physics invariant. In other words, the vacuum expectation values hπµ i just need to satisfy the constraint 3 X

hπµ i2 = hσi2 +

µ=0

3 X

hπi i2 ≡ v 2

i=1

with v some constant with dimensions of mass. The text chooses hσi = v and hπi i = 0, so that upon spontaneous symmetry breaking the Lagrangian contains the term L = gv χχ¯ + h.c. = ¯ which implies a mass M = gv for the nucleons. But any other gv ψ¯L ψR + h.c. = gv ψψ, choice of vacuum alignment can be transformed into this form using the SU (2)L ⊗ SU (2)R transformation hΠi → LhΠiR−1 , as long as we also redefine the nucleon fields by χ → Lχ and χ¯ → Rχ. ¯ For example, choose hπ2 i = v and hσi = hπ1 i = hπ3 i = 0. Then 0 v hΠi = . −v 0 Now choose the matrices L=

0 −1 −1 0

= −τx and R =

1 0 0 −1

= τz .

The transformation hΠi → LhΠiR† yields 0 v 0 −1 0 v 1 0 v 0 → = −v 0 −1 0 −v 0 0 −1 0 v So upon the field redefinitions χ → −τx χ and χ¯ → τz χ, ¯ we recover the mass term L = ¯ ¯ gv χχ¯ + h.c. = gv ψL ψR + h.c. = gv ψψ, as we must by SO(4) ' SU (2)L ⊗ SU (2)R invariance.

186

VI.5

Ferromagnets and Antiferromagnets

1. Work out the two branches of the spin wave spectrum in the ferromagnetic case, paying particular attention to the polarization. Solution: For this problem and the next we follow X. G. Wen and A. Zee, “Spin Waves and Topological Terms in the Mean-Field Theory of Two-Dimensional Ferromagnets and Antiferromagnets,” Phys. Rev. Lett., Vol. 61 No. 8, 22 Aug 1988. The equations of motion are14 2

− 12 iω − ωg2 + h(~k) 2 + 12 iω − ωg2 + h(~k)

!

! ~ ξ (k, ω) =0. ξ y (~k, ω) x

where h(~k) ≈ 2|J|a2 k 2 for small k ≡ |~k | (and J < 0). Setting the determinant of the matrix to zero gives the quadratic equation r g2 g2 16 g2 2 ~ ∓ 2 1 + 2 h(k) . ω ± ω − g h(k) = 0 =⇒ ω (k) = ∓ + 2 4 4 g We have chosen the + root in the quadratic equation to keep only positive-frequency solutions. For small k, we have: r g2 g2 16 g2 + ω =+ + 1 + 2 h(k) = + + O(k 2 ) 4 4 g 2 r 2 2 g g 16 ω− = − + 1 + 2 h(k) ≈ +2h(k) = +4|J|a2 k 2 4 4 g so that there is a high frequency branch ω + = g 2 /2 + O(k 2 ) and a low-frequency branch ω − ∝ k 2 , which is the typical dispersion relation for a nonrelativistic particle. Now let us work out the polarizations for each branch. Plugging in ω + ≈ 21 g 2 , dropping all terms of order k 2 and dividing through by −g 2 /4, the equations of motion become x 1 +i ξ =0. −i 1 ξy This is satisfied by (ξ x , ξ y ) ∝ (1, +i). Plugging in ω − ≈ 4|J|a2 k 2 and h(k) ≈ 2|J|a2 k 2 , dropping terms of higher order in k, and dividing through by 2|J|a2 k 2 , the equations of motion become x 1 −i ξ =0. +i 1 ξy This is satisfied by (ξ x , ξ y ) ∝ (1, −i). 14

The text uses the notation δni ≡ ξi . We choose to change notation in order to distinguish linearizing about the ground state from the arbitrary variation ~ni → ~ni + δ~ni used to obtain the equations of motion. But of course you are free to use whichever notation you like.

187

2. Verify that in the antiferromagnetic case the Berry’s phase term merely changes the spin wave velocity and does not affect the spectrum qualitatively as in the ferromagnetic case. As given on p. 346, linearizing around the ground state ~ni (t) = (−1)i eˆz + ξ~i (t) implies15 ! ! 2 x ~ − 21 iω − ωg2 + f (~k) ξ (k, ω) 1 ω2 y ~ ~ ~ ~ ω) = 0 + 2 iω − g2 + f (k + Q) ξ (k + Q, P where f (~k) = 4J[2 + 2µ = 1 cos(kµ a)] and J > 0. Note that since Qµ = πa , and cos(x + π) = ~ = 4J[2 + P cos(kµ a + π)] = 4J[2 − P cos(kµ a)]. For small kµ , − cos x, we have f (~k + Q) µ µ we have ~ ≈ +2Ja2 k 2 f (~k) ≈ 16J − 2Ja2 k 2 and f (~k + Q) where k ≡ |~k|. Returning to the equation of motion, setting the determinant of the matrix equal to zero results in a quadratic equation for ω 2 : ~ − 1 g4ω2 = 0 . [ω 2 − g 2 f (~k)][ω 2 − g 2 f (~k + Q)] 4 This has the solutions:

2 ω±

g2 ~ + 1 g2 = f (~k) + f (~k + Q) 2 4

v u ~ u 4f (~k)f (~k + Q) 1±u 1 − i2 h t 1 2 ~ ~ ~ f (k) + f (k + Q) + 4 g

.

~ = 16J + O(k 4 ) and f (~k)f (~k + Q) ~ = 32J 2 a2 k 2 + O(k 4 ). For small k, we find f (~k) + f (~k + Q) The solutions simplify to: 1 2 64J 2 a2 k 2 g2 2 ω± ≈ . 16J + g 1± 1− 2 4 [16J + 14 g 2 ]2 The spectrum splits into two branches: 1 2 ω+ ≈ g 2 (16J + g 2 ) + O(k 2 ) 4 2 2 2 32J a g 2 ≈ ω− k2 . 16J + 14 g 2 The low-frequency branch is a linear dispersion relation ω ∝ k with spin-wave velocity c= 15

32J 2 a2 g 2 . 16J + 14 g 2

There is a typo in equation (6) in the text. See the addendum at the end of this section.

188

The question now is to track the effect of the Berry’s phase term. Without this term (namely the ±iω terms in the equations of motion), setting the determinant of the matrix to zero in the equations of motion implies ~ =0 [ω 2 − g 2 f (~k)][ω 2 − g 2 f (~k + Q)] which is still a quadratic equation for ω 2 . The solutions are ω 2 = g 2 f (~k) ≈ g 2 (16J − 2Ja2 k 2 ) ~ ≈ g 2 (2Ja2 k 2 ). Again, there is a high-frequency branch and ω 2 = g 2 f (~k + Q) 2 ωhigh = 16g 2 J + O(k 2 )

and a low-frequency branch 2 ωlow = 2Ja2 g 2 k 2

which is a linear dispersion relation with spin-wave velocity c = 2Ja2 g 2 . We see that the spectrum is qualitatively the same, and the Berry’s phase only changes the particular values of the parameters in the dispersion relations.

Addendum: Deriving the Equations of Motion Here we fill in theR steps absent from the text in order to derive the equations of motion. The action is S = dt L with Lagrangian N X X 1 † L= i zi ∂t zi + 2 ∂t~ni · ∂t~ni − J ~ni · ~nj 2g i=1 hiji

where ~ni = zi†~σ zi is the Pauli-Hopf map introduced in the text, and the sum over nearest neighbors can be written explicitly as N

X hiji

N

d

1 XX X δi, j+ˆeµ + δi, j−ˆeµ . = 2 i=1 j =1 µ=1

The number of lattice sites is N , and the number of lattice spatial dimensions is d. (We specialize to d = 2.) The vector eˆµ is a unit vector pointing in the µth direction, and the kronecker deltas are to be interpreted according to the example N X

δi, j+ˆeµ f (~xi ) = f (~xj+ˆeµ ) ≡ f (~xj + a eˆµ )

i=1

where a is the lattice spacing. Using the variation Z Z i † dt δ zi ∂t zi = dt δ~ni · (~ni × ∂t~ni ) 2 189

given in the text, we have # " Z N X d N X X 1 2 1 δi, j+ˆeµ + δi, j−ˆeµ ~nj . δS = dt δ~ni · − ~ni × ∂t~ni − 2 ∂t ~ni − J 2 g j =1 µ=1 i=1 When deriving local equations of motion by setting δS = 0, we must assume that the variations δ~ni (t) are arbitrary functions of i and t – that is, that the sum over i and the integral over t are unconstrained. But here the sum is constrained by ~ni · ~ni = 1. We need to incorporate this constraint into the path integral: Z Z=

Dn Z Dn

=

N Y

δ(~ni · ~ni − 1) e i

i=1 N Z Y

Dν e i

R

R

dt L

dt (~ ni ·~ ni −1)ν

ei

R

dt L

i=1

Z =

DnDν ei

R

dt [

PN

]

ni ·~ ni −1)ν+L i = 1 (~

So the total unconstrained Lagrangian is Ltot = L +

N X

~ni · ~ni ν − N ν .

i=1

P ni · ~ni ν, so that the full varied To the previous variation we therefore add the term 2 N i = 1 δ~ action is ( " # ) Z N N d X X X 1 1 2 δStot = dt δ~ni · − ~ni × ∂t~ni − 2 ∂t ~ni − J δi, j+ˆeµ + δi, j−ˆeµ − 2νδij ~nj 2 g µ=1 i=1 j =1 Now the sum over i is totally unconstrained, so the equation of motion is the term in {...} set equal to zero: N X 1 1 ~ni × ∂t~ni + 2 ∂t2~ni + Jij ~nj = 0 2 g j =1 with Jij ≡ J

d X

δi, j+ˆeµ + δi, j−ˆeµ − 2νδij .

µ=1

First consider the ferromagnet (J < 0). The ground state of the ferromagnet is the state for which all spins point in the positive z-direction. Linearize about this state as ~ni (t) = eˆz + ξ~i (t) to obtain ~ni × ∂t~ni = (ˆ ez + ξ~i ) × ∂t (ˆ ez + ξ~i ) = eˆz × ∂t ξ~i + O(ξ 2 ) . As mentioned in the text, the constraint ~n2i = (ˆ ez + ξ~i )2 = 1 + 2 eˆz · ξ~i + O(ξ 2 ) = 1 implies eˆz · ξ~i = 0, so that the linearized deviation from the ground state occurs only in the (xy)-plane. 190

Since ξ~ lies purely in the (x, y)-plane, the equation of motion in the z-direction is: N X

Jij = 0 =⇒ ν = Jd .

j =1

This fixes the value of the Lagrange multiplier ν. In the (x, y)-plane, the equations of motion are: N X 1 1 2~ ~ eˆz × ∂t ξi (t) + 2 ∂t ξi (t) + Jij ξ~j (t) = 0 2 g j =1

where now

" Jij = J

#

d X

δi,j+ˆeµ + δi,j−ˆeµ − 2d δij

µ=1

and as usual d = 2. Now that the equation of motion is linear in ξ~i (t), we may Fourier transform using the convention N Z ∞ X ~ ~ dt e−i(ωt+k·~xi ) fi (t) . f (k, ω) ≡ −∞

i=1

We have

N Z X i=1

∞

~ ~ ~k, ω) dt e−i(ωt+k·~xi ) ∂t2 ξ~i (t) = −ω 2 ξ(

−∞

and N Z X i=1

∞

−i(ωt+~k·~ xi )

dt e

−∞

N X

N Z X

Jij ξ~j =

j =1

=

i=1

N Z ∞ X j =1

" =J

"

∞

−i(ωt+~k·~ xi )

dt e

J

−∞

dt e

J

−∞

d X

−i~k·(~ xj +aˆ eµ )

e

e

−i~k·ˆ eµ a

+e

+i~k·ˆ eµ a

# − 2d

µ=1

= 2J

d X

δi, j+ˆeµ + δi, j−ˆeµ − 2d δij ξ~j (t)

+e

−i~k·(~ xj −aˆ eµ )

# −i~k·~ xj

− 2d e

µ=1

d X

"

#

µ=1

" −iωt

d X

N Z X j =1

∞

~

dt e−i(ωt+k·~xj ) ξ~j (t)

−∞

# ~ ~k, ω) ξ(

cos(kµ a) − d

, kµ ≡ ~k · eˆµ .

µ=1

For the Berry’s phase term, we have: N Z ∞ N Z X X −i(ωt+~k·~ xi ) ~ dt e eˆz × ∂t ξi (t) = i=1

−∞

i=1

∞

~

dt (−1)(−iω)e−i(ωt+k·~xi ) eˆz × ξi (t) + O(ξ 2 )

−∞

~ ~k, ω) . = +iω eˆz × ξ( 191

ξ~j (t)

In the first step, the first minus sign is from moving the time derivative to act on the exponential using integration by parts. The Fourier transformed equation of motion now takes the simple form ( " d #) X ω2 ~ ~k, ω) . ~ ~k, ω) = − 1 iω eˆz × ξ( − 2 + 2J cos(kµ a) − d ξ( g 2 µ=1 Using eˆz × eˆx = +ˆ ey and eˆz × eˆy = −ˆ ex and moving everything to the left-hand side yields ! ! 2 − 21 iω − ωg2 + h(~k) ξ x (~k, ω) =0 2 + 12 iω − ωg2 + h(~k) ξ y (~k, ω) where " h(~k ) ≡ 2|J| d −

d X

# cos(kµ a)

µ=1

# 1 2 2 ≈ 2|J| d − 1 − kµ a 2 µ=1 1~ 2 2 = 2|J| d − d − |k | a 2 2 2 = +|J|a k "

d X

where k ≡ |~k |. On p. 346 of the book, the function h(k) is defined as h(k) ≡ 4J[2 − P2 2 2 µ = 1 cos(kµ a)] ≈ 2|J|a k , which is a factor of two larger than the one we have derived. This can be traced back to when we included a factor of 21 in the explicit form of the sum over nearest neighbors. Had we not included this 21 , we would be counting each nearest neighbor twice rather than once. In the reference, Wen and Zee have implicitly normalized the coupling J such that the nearest neighbor sum does not include the 21 . Thus we can take the results in the text and rescale J → 12 J in order to match our treatment. For the antiferromagnet (J > 0), the ground state is where the spins alternate in orien~ ≡ ( π , π ) be a tation, so we expand around the ground state as ~ni (t) = (−1)i eˆz + ξ~i (t). Let Q a a ~ xi µ i iQ·~ 2-dimensional vector. We can write (−1) = e , where ~xi = xi eˆµ is the coordinate of the

192

ith lattice site. Thus, for the antiferromagnet we have: " d # N N X X X ~ Jij (−1)j = J δi,j+ˆeµ + δi,j−ˆeµ − 2ν δij e iQ·~xj j =1

µ=1

j =1

=J

d X

~ xi −aˆ iQ·(~ eµ )

e

+e

~ xi +aˆ iQ·(~ eµ )

~

− 2ν e iQ·~xi

µ=1

" = 2e

~ xi iQ·~

J

d X

# cos(Qµ a) − ν

µ=1

" = 2e

~ xi iQ·~

J

d X

# cos π − ν

µ=1 ~

= 2 e iQ·~xi (−1) (Jd + ν) . Therefore the z-component of the equation of motion for the antiferromagnet fixes the Lagrange multiplier to be ν = −Jd rather than +Jd. The matrix Jij for the antiferromagnet is therefore: # " d X Jij = J δi,j+ˆeµ + δi,j−ˆeµ + 2d δij . µ=1

For the antiferromagnet, the Berry’s phase term becomes: N Z ∞ X ~ ~ ~k + Q, ~ ω) dt e−i(ωt+k·~xi ) ξ~i (t) × ∂t ξ~i (t) = +iω eˆz × ξ( i=1

−∞

again dropping terms of order ξ 2 . This results in the equations of motion ω2 1 ~ ~ ω) − 2 + f (k) ξ x (~k, ω) = + iω ξ y (~k + Q, g 2 ω2 1 ~ ~ ω) . − 2 + f (k) ξ y (~k, ω) = − iω ξ x (~k + Q, g 2 h i Pd ~ ~ where f (k ) ≡ 2J d + µ = 1 cos(kµ a) = h(~k + Q). ~ To put this into a matrix notation, we need to shift the second equation by ~k → ~k + Q ~ ~k + 2Q, ~ ~k, ω). We obtain ~ ω) = ξ( and use ξ( ! ! 2 − ωg2 + f (~k) − 21 iω ξ x (~k, ω) 2 ~ ~ ω) = 0 . + 12 iω − ωg2 + f (~k + Q) ξ y (~k + Q, ~ in the lower-right component of the above matrix. This is accidenNote the argument ~k + Q tally absent in the book, but present in equation (7) of the reference. 193

VI.6

Surface Growth and Field Theory

3. Field theory can often be cast into apparently rather different forms by a change of 1 variable. Show that by writing U = e 2 gh we can change the action (7) to 2 Z 2 −1 2 D −1 ∂ U −U ∇ U S = 2 d x dt U g ∂t a kind of nonlinear σ model. (7)

1 S= 2

Z

D

d x dt

2 g ∂ 2 2 − ∇ h − (∇h) ∂t 2

Solution: Let h = implies

2 g

ln U . Then ∇j ln U = U −1 ∇j U =⇒ ∇i ∇j ln U = −U −2 ∇i U ∇j U + U −1 ∇i ∇j U

g g 2 ∇ h + (∇h)2 = − −U −2 ∇i U ∇i U + U −1 ∇2 U + 2 g 2 2 = U −1 ∇2 U g 2

2 2 (U −1 ∇i U )(U −1 ∇i U ) g

Since ∂t ln U = U −1 ∂t U , we have 2 S= g

VI.7

Z

2 dD x dt U −1 ∂t − ∇2 U .

Disorder: Replicas and Grassmannian Symmetry

1. Work out the field theory that will allow you to study Anderson localization. [Hint: Consider the object 1 1 (x, y) (y, x) z−H w−H for two complex numbers z and w. You will have to introduce two sets of replica fields, − commonly denoted by ϕ+ a and ϕa .] Solution: Using the identities in the chapter we can write Z R D 0 † † 1 0 (x, y) = i lim Dϕ†+ Dϕ+ e i d x {∂ ϕ~ + ∂ ϕ~ + +[V (x )−z]~ϕ+ ϕ~ + } ϕ+1 (x)ϕ†+1 (y) n→0 z−H where we have denoted the n-dimensional vector of replica fields by ϕ+1 ϕ ~ + = ... . ϕ+n 194

Similarly, we can also write Z R D 0 † † 1 0 (y, x) = i lim Dϕ†− Dϕ− e i d x {∂ ϕ~ − ∂ ϕ~ − +[V (x )−w]~ϕ− ϕ~ − } ϕ−1 (y)ϕ†−1 (x) m→0 w−H where we have denoted the additional m-dimensional vector of replica fields by ϕ−1 ϕ ~ − = ... . ϕ−m Multiplying these together, we obtain 1 1 (x, y) (y, x) = z−H w−H Z R D 0 † † † † 2 Dϕ†+ Dϕ+ Dϕ†− Dϕ− e i d x {∂ ϕ~ + ∂ ϕ~ + +∂ ϕ~ − ∂ ϕ~ − −zϕ~ + ϕ~ + −wϕ~ − ϕ~ − } × i lim e

n,m→0 R i dD x0 V (x0 )(~ ϕ†+ ϕ ~ + +~ ϕ†− ϕ ~−)

ϕ+1 (x)ϕ†+1 (y)ϕ−1 (y)ϕ†−1 (x)

where we have separated out the terms linear in V (x) to prepare for averaging over disorder. R D R − 1 d x V (x)2 We now carry out the average hO(V )i = N DV e 2g2 O(V ) using the formula Z 1 1 −1 N DV e− 2 V ·M ·V +j·V = e 2 j·M ·j

where in our case j = i(~ ϕ†+ ϕ ~+ + ϕ ~ †− ϕ ~ − ) and M = 1/g 2 . We arrive at the expression 1 1 (x, y) (y, x) = z−H w−H Z R D 0 0 − lim Dϕ†+ Dϕ+ Dϕ†− Dϕ− e i d x L(x ) ϕ+1 (x)ϕ†+1 (y)ϕ−1 (y)ϕ†−1 (x) n,m→0

with the Lagrangian density 2 1 † L = ∂ϕ ~ +† ∂ ϕ ~ + + ∂ϕ ~ −† ∂ ϕ ~− − z ϕ ~ +† ϕ ~+ − w ϕ ~ −† ϕ ~ − + i g2 ϕ ~ +ϕ ~+ + ϕ ~ −† ϕ ~− . 2

195

2. As another example from the literature on disorder, consider the following problem. Place N points randomly in a D-dimensional Euclidean space of volume V . Denote the locations of the points by ~xi (i = 1, ..., N ). Let ~

dD k e ik·~x (2π)D ~k 2 + m2

Z f (~x ) = −

Consider the N by N matrix Hij = f (~xi − ~xj ). Calculate ρ(E), the density of eigenvalues of H as we average over the ensemble of matrices, in the limit N → ∞, V → ∞, with the density of points ρ0 ≡ N/V held fixed. Hint: Use the replica method and arrive at the field theory action # " n Z X Pn 2 S = dD x |∇ϕa |2 + m2 |ϕa |2 − ρ0 e−(1/z) a = 1 |ϕa | a=1

This problem is not entirely trivial; if you need help consult M. M`ezard et al., Nucl. Phy. B559: 689, 2000, cond-mat/9906135. Solution: The same steps leading to equation (3) on p. 352 result in " n N # N Z X Y Y Pn PN † 1 tr = i lim dϕ†ai dϕai e i a = 1 i,j = 1 ϕai (Hij −zδij )ϕaj ϕj1 ϕ†j1 . n→0 z−H a=1 i=1 j =1 Then to “average over disorder,” we average over the locations of the N points {~xi }N i = 1: # Z "Y N dD xi h(...)i = (...) . V i=1 Just as we find the propagator by taking derivatives of the partition function, to find the operator tr[1/(z − H)] we will instead compute the function #Z Z "Y N Pn PN † dD xi dnN ϕ† dnN ϕ e i a = 1 i,j = 1 ϕai (Hij −zδij )ϕaj ξN ≡ V i=1 where dnN ϕ† dnN ϕ ≡

Qn

a=1

QN

i=1

dϕ†ai dϕai . If we define the functions φa (~x ) ≡

N X

δ D (~x − ~xi )ϕai

i=1

196

and their hermitian conjugates, then Z dD x dD y φ†a (~x )f (~x − ~y )φa (~y ) = Z

dD x dD y

N X

! δ D (~x − ~xi )ϕ†ai f (~x − ~y )

i=1

=

N X

N X

! δ D (~y − ~xj )ϕaj

j =1

ϕ†ai f (~xi − ~xj )ϕaj =

i,j = 1

N X

ϕ†ai Hij ϕaj ,

i,j = 1

R P so we can replace ij ϕ†ai Hij ϕaj with dD x dD y φ†a (~x)f (~x − ~y )φa (~y ). To make this a useful substitution, we need to be able to treat the variables x ) and φ†a (~x ) as unconstrained R φa (~ field variables. In other words, we want to integrate Dφ† Dφ as if φa (~x ) and φ†a (~x ) were usual bosonic quantum fields that transform as a vector underPan O(n) symmetry, and we D can do so as long as we include the constraint that φa (~x) − N x − ~xi )ϕai = 0 and i = 1 δ (~ P N † D † φa (~x) − i = 1 δ (~x − ~xi )ϕai = 0. To implement this, insert the number 1 into the path integral as follows: # Z " Z Z N R D X PN (D) † ~ (D) (n) ~ φ(~x ) − ϕ ~ δ (~x − ~x ) = Dφ Dµ† e i d x µ~ (~x)·[φ(~x)− i = 1 ϕ~ i δ (~x−~xi )] 1 = Dφ δ i

i

i=1

and the same for the hermitian conjugate: " # Z Z Z N R D X PN † (D) ~† † (D) † (n) ~ † † 1 = Dφ δ φ (~x ) − ϕ ~ δ (~x − ~xi ) = Dφ Dµ e−i d x µ~ (~x)·[φ (~x)− i = 1 ϕ~ i δ (~x−~xi )] i

i=1

We chose the minus sign in this version of 1 for later convenience. If you are getting bogged down in functional integral notation, then orient yourself with the familiar example from ordinary 1-dimensional calculus: Z ∞ Z ∞ Z ∞ dµ ±iµ(x−c) e 1= dx δ(x − c) = dx −∞ −∞ −∞ 2π where c is some real number. In the functional integrals, the factors of 2π are swept into the definitions of the measures. Inserting both of these factors of 1 into the integral for ξN , we arrive at ξN = #Z "Z # Z Z "Y N n X dD xi Dφ† DφDµ† Dµ dnNϕ† dnNϕ exp i dD x dD y φ†a (~x )f (~x − ~y )φa (~y ) × V a=1 i=1 " ( )# Z n N n N XX † X X exp −i z ϕai ϕai + dD x i µ†a (~x )[φa (~x) − ϕai δ D (~x − ~xi )] + h.c. . a=1 i=1

a=1

i=1

197

perform the Gaussian integral over the ϕai and ϕ†ai variables. First note that RWeDcan now D d x µa (~x )δ (~x − ~xi ) = µa (~xi ), and then define the “sources” Ja (~x ) ≡ −iµa (~x ). Then the integral we have to perform is ( n N ) Z i XXh I ≡ dnN ϕ† dnN ϕ exp −iz ϕ†ai ϕai + i µ†a (~xi )ϕai − i ϕ†ai µa (~xi ) Z =

( nN

d

† nN

ϕd

ϕ exp

a=1 i=1 n X N h X

−iz

ϕ†ai ϕai

+

Ja† (~xi )ϕai

+

i

)

ϕ†ai Ja (~xi )

a=1 i=1

=

n Y N Z Y

o n dϕ†ai dϕai exp −(iz)ϕ†ai ϕai + Ja† (~xi )ϕai + ϕ†ai Ja (~xi ) .

a=1 i=1

Using Z

dϕ† dϕ e−α ϕ

† ϕ+J † ϕ+ϕ† J

=

2πi + 1 J † J e α α

with α = iz, we have n Y N Y 1 † 2π I= exp Ja (~xi )Ja (~xi ) z iz a=1 i=1 ( ) nN n N 2π i XX † = exp − J (~xi )Ja (~xi ) z z a=1 i=1 a ) ( nN n X N X i 2π exp − µ†a (~xi )µa (~xi ) . = z z a=1 i=1 Therefore the original integral ξN we wanted is now ξN = # #Z "Z nN Z " Y n N X 2π dD xi † † † D D φa (~x )f (~x − ~y )φa (~y ) × Dφ DφDµ Dµ exp i d x d y z V a=1 i=1 ( ) Z n N n X i XX † exp − µ (~xi )µa (~xi ) + i dD x µ†a (~x )φa (~x ) − φ†a (~x )µa (~x ) . z a=1 i=1 a a=1 Now we can perform the Gaussian integral over the fields φa (~x ) and φ†a (~x ): Z R D Pn R D D Pn † † † Dφ† Dφ e i d x d y a = 1 φa (~x)f (~x−~y)φa (~y)+ d x a = 1 [Ja (~x)φa (~x)+φa (~x)Ja (~x)] = (det f )−1 e+i

R

d D x dD y

Pn

a=1

Ja† (~ x)f −1 (~ x−~ y )Ja (~ y)

where again Ja (~x ) = −iµa (~x ). Since f (x) = − f

−1

Z (x) = −

R

,

dD k 1 (2π)D k2 +m2

e ikx , the inverse of f (x) is

dD k 2 (k + m2 ) e ikx . (2π)D 198

To convince yourself this is right, compute Z dD z f −1 (x − z)f (z − y) Z Z Z dD p 2 1 dD k 2 +ikx −ipy 2 (k + m ) 2 e e dD z e+i(p−k)z = (−1) (2π)D (2π)D p + m2 Z Z dD k dD p 2 1 = (k + m2 ) 2 e+ikx e−ipy (2π)D δ D (p − k) D D 2 (2π) (2π) p +m Z D d k +ik(x−y) = e = δ D (x − y) . X (2π)D To summarize our progress so far, we have #Z # " nN Z " Y N n X N D X 2π d xi i ξN = Dµ† Dµ exp − µ†a (~xi )µa (~xi ) × z V z a=1 i=1 i=1 " # Z n X exp − tr ln f + i dD x dD y µ†a (~x )f −1 (~x − ~y )µa (~y ) . a=1

Given the explicit form of f −1 (~x ), the last term simplifies (suppress the index a for clarity): Z Z Z dD k ~ 2 D D † −1 D D † 2 i~k·(~ x−~ y) d x d y µ (~x )f (~x − ~y )µ(~y ) = d x d y µ (~x ) − (|k | + m ) e µ(~y ) (2π)D Z Z dD k † ~ ~ D D µ (~x ) e+ik·~x (|~k |2 + m2 ) e−ik·~y µ(~y ) = − d xd y D (2π) Z Z h i dD k † +i~k·~ x 2 2 −i~k·~ y = − dD x dD y µ (~ x ) e (−∇ + m )e µ(~y ) y (2π)D Z Z dD k +i~k·(~x−~y ) 2 D D = −(−1) d xd y e µ† (~x )(−∇2y + m2 )µ(~y ) D (2π) Z = − dD x dD y δ D (~x − ~y )µ† (~x )(−∇2y + m2 )µ(~y ) Z = − dD x µ† (~x )(−∇2 + m2 )µ(~x ) Z h i ~ † (~x ) · ∇µ(~ ~ x ) + m2 µ† (~x )µ(~x ) . = − dD x +∇µ Now this is starting to approach the desired result. At this point we have the field theory Z † ξN = C Dµ† Dµ AN e−S0 [ µ , µ ] with the action †

S0 [µ , µ] = i

Z

dD x

n h X

i ~ †a (~x ) · ∇µ ~ a (~x ) + m2 µ†a (~x )µa (~x ) , ∇µ

a=1

199

the function

A=

2π z

n Z

dD x −(i/z) Pna = 1 µ†a (~x )µa (~x ) e , V

and the irrelevant overall constant C = e−tr ln f . We will now drop C since it will drop out of all correlation functions as do all overall constants in the path integral. We will also take n → 0 in the prefactor (2π/z)n in A. Passing to a grand canonical formulation of the disorder16 , we compute ! Z Z ∞ ∞ X X 1 1 † N † N −S0 [µ† , µ] ξ≡ ξN α = Dµ Dµ (Aα) e = Dµ† Dµ e Aα−S0 [µ , µ] . N! N! N =0 N =0 R We therefore arrive at the result ξ = Dµ† Dµ e−S , where the full action S = S0 − αA is " n !# Z n α X X i ~ †a (~x )· ∇µ ~ a (~x )+m2 µ†a (~x )µa (~x ) − exp − S[µ† , µ] = dD x i ∇µ µ†a (~x )µa (~x ) . V z a=1 a=1 The last thing to determine is the meaning of the parameter α. The average number of points is17 N = αhAi, so that in the limit n → 0 we have N = α and hence α/V = ρ0 , the density of points. We have arrived at the action !# " n Z n X X i ~ † (~x )· ∇µ ~ a (~x )+m2 µ† (~x )µa (~x ) −ρ0 exp − µ†a (~x )µa (~x ) . ∇µ S[µ† , µ] = dD x i a a z a=1 a=1 Perform the field redefinition µ → −iµ while keeping µ† unchanged to obtain " n !# Z n X X 1 ~ †a (~x )· ∇µ ~ a (~x )+m2 µ†a (~x )µa (~x ) −ρ0 exp − µ†a (~x )µa (~x ) S[µ† , µ] = dD x ∇µ z a=1 a=1 which is the desired result. This is just as in statistical mechanics. The partition function Z = P tr e−βH can be put into a grand ∞ canonical formulation by introducing a chemical potential µ via Z(µ) = tr N = 0 N1 ! e−β(H−µN ) . 17 The average particle number in a grand canonical ensemble described by the partition function Z(µ) ∂ is hN i = β1 ∂µ ln Z(µ). In our case, we have β = 1 and µ = ln α. The average value of the function A ∂ ∂ ∂ defined previously is hAi = ∂α ln ξ, and hence αhAi = α ∂α ln ξ = ∂ ln α ln ξ. Since ln α = µ, we have ∂ αhAi = ∂µ ln ξ = N . 16

200

VI.8

Renormalization Group Flow as a Natural Concept in High Energy and Condensed Matter Physics

1. Show that the solution of dg/dt = −bg 3 + ... is given by 1 1 = + 8πbt + ... α(t) α(0) where we defined α(t) = g(t)2 /4π. Solution: Z

g

g0

dg 0 = −b g 03

Z

t

0

1 dt =⇒ − 2 0

1 1 − 2 2 g g0

= −b t =⇒

1 1 = 2 + 2bt 2 g g0

Define g 2 ≡ 4πα and multiply by 4π to get 1 1 = + 8πb t . α α0

2. In our discussion of the renormalization group, in λϕ4 or in QED, for the sake of simplicity we assumed that the mass m of the particle is much smaller than µ and thus set m equal to zero. But nothing in the renormalization group idea tells us that we can’t flow to a mass scale below m. Indeed, in particle physics many orders of magnitude separate the top quark mass mt from the up quark mass mu . We might want to study how the strong interaction coupling flows from some mass scale far above mt down to some mass scale µ below mt but still large compared to mu . As a crude approximation, people often set any mass m below µ equal to zero and any m above µ to infinity (i.e., not contributing to the renormalization group flow). In reality, as µ approaches m from above the particle starts to contribute less and drops out as µ becomes much less than m. Taking either the λϕ4 theory or QED study this so-called threshold effect. Solution: First consider QED. Equation (6) on p. 358 gives the renormalization group flow equation for the QED coupling: µ

1 d d e(µ) = − e(µ)3 µ Π(µ2 ) + O(e5 ) dµ 2 dµ

The lowest order solution to this is 1 1 1 µ = 2 − 2 ln . 2 e (µ) e (µ0 ) 6π µ0 Suppose we have n + 1 electrons in this theory, one having mass m and the others massless, and take the initial condition to be at some superheavy mass scale M . 201

For m < µ < M , all n + 1 particles contribute to the vacuum polarization function, and we have 1 1 n+1 M = 2 + . ln 2 e (µ) e (M ) 6π 2 µ For 0 < µ < m, the lowest order approximation is to use 1 e2 (µ)

=

1 e2 (M )

+

n M . ln 2 6π µ

Including the threshold correction at the scale m amounts to replacing the lowest order expression with the following: 1 e2 (µ)

n m ln 2 6π µ n+1 M n m 1 + ln + ln = e2 (M ) 6π 2 m 6π 2 µ 1 n M 1 M = 2 + 2 ln + 2 ln . e (M ) 6π µ 6π m =

1

e2 (m)

+

We see that there is an extra term that depends explicitly on m. This result is usually written in terms of the µ-dependent coupling for µ > m, which we now denote by18 eG (µ). With −2 n+1 M e−2 (M ) = eG (µ) − 6π2 ln µ , we can rewrite the threshold-corrected coupling for µ < m as: 1 n+1 M n M 1 M 1 = 2 − ln + 2 ln + 2 ln 2 2 e (µ) eG (µ) 6π µ 6π µ 6π m 1 m 1 − 2 ln . = 2 eG (µ) 6π µ

The analysis for ϕ4 scalar field theory proceeds in the same way. Equation (5) on p. 357 along with appendix 1 in chapter III.1 (equation (14) on p. 168) gives the renormalization group flow equation for λϕ4 theory: µ

d 1 d λ(µ) = − λ(µ)2 µ Π(µ2 ) + O(λ3 ) dµ 16 dµ

where we have defined the function 1 Π(s) ≡ 2 2π

Z

1

dx ln

0

18

Λ2 m2 − x(1 − x)s

We use this notation to evoke the connection to threshold corrections in grand unified theories. See chapter VII.6 in the main text, as well as S. Weinberg, “Effective Gauge Theories,” Phys. Lett. 91B, No. 1 (1980) and L. Hall, “Grand Unification of Effective Gauge Theories,” Nucl. Phys. B178 (1981) 75-124.

202

using a notation intentionally evocative of the vacuum polarization function in QED. [Here Λ is an arbitrary upper cutoff on the momentum, and m is the mass of the quanta (“mesons”) of the scalar field ϕ.] The analysis proceeds in exactly the same fashion as for QED: take n + 1 mesons, one with mass m and the others massless, and assume an initial condition at M m. For m < µ < M , we use 1 n+1 M 1 = + . ln λ(µ) λ(M ) 16π 2 µ For 0 < µ < m without the threshold correction we use λ−1 (µ) = λ−1 (M ) + with the threshold correction we use instead

n 16π 2

ln M , but µ

1 1 n m = + ln 2 λ(µ) λ(m) 16π µ 1 n+1 M n m = + ln + ln 2 2 λ(M ) 16π m 16π µ n M 1 M 1 + ln + ln . = 2 2 λ(M ) 16π µ 16π m As before, we relabel the high-energy coupling as λG (µ) and rewrite the low-energy coupling M n+1 using λ−1 (M ) = λ−1 G (µ) − 16π 2 ln µ : 1 1 n+1 M n M 1 M = − ln + ln + ln 2 2 2 λ(µ) λG (µ) 16π µ 16π µ 16π m 1 m 1 − ln . = 2 λG (µ) 16π µ

203

˜ 4. In S(h) only derivatives of the field h can appear and not the field itself. (Since the transformation h(~x, t) → h(~x, t) + c with c a constant corresponds to a trivial shift of where we measure the surface height from, the physics must be invariant under this transformation.) Terms involving only one power of h cannot appear since they are all total divergences. Thus, ˜ ˜ S(h) must start with terms quadratic in h. Verify that the S(h) given in (17) is indeed the 2 most general. A term proportional to (∇h) is also allowed by symmetries and is in fact generated. However, such a term can be eliminated by transforming to a moving coordinate frame h → h + ct. Solution: The action is supposed to be the most general compatible with invariance under the Galilean transformation h(~x, t) → h0 (~x, t) = h(~x + g ~u t, t) + ~u · ~x + 21 gu2 t with ~u a constant velocity. In the solution for VI.8.3 it is shown that the combination g ~ 2 ∂t h − (∇h) 2 is invariant under the Galilean transformation, as is ∇2 h. Thus in general the action should be a linear combination of these terms: g ~ 2 X ≡ α ∂t h − (∇h) − β∇2 h . 2 As discussed in the problem, the action should also not involve terms linear in h since all such R D terms2 are total divergences. Thus to lowest order in h the action is proportional to d x dt X , which is the form given in (17). 6. Calculate the h propagator to one loop order. Extract the coefficients of the ω 2 and k 4 terms in a low frequency and wave number expansion of the inverse propagator and determine α and β. Solution: We need the free-field propagator and the cubic vertex (all momenta point into the vertex): 1 = 2

4

1 = 3

2 1

1

) k2

2 3

2

2

204

2

) k3

2 1

3

3

) k1

2

Fortunately the quartic vertex does not contribute to this order. The only diagram we need to compute is

Calculating this diagram results in a self-energy Z Λ d 4d2 − d − 6 4 d K 1 2 1 Σ(ω, k) = 2 ω + k . d 2 d(d + 2) µ (2π) K Here µ and Λ are arbitrary lower and upper cutoffs, respectively, on the integral. The idea is to integrate over an infinitesimal shell, so that the lower cutoff can be taken as µ = (1−δL)Λ. The integral is now Z Λ d Λd−2 d K 1 = δL ≡ f (Λ, d) δL . d 2 2d−1 π d/2 Γ( d2 ) µ (2π) K Adding the 1-loop self-energy to the zeroth-order propagator, we obtain the desired coefficients g2 f (Λ, d) δL 8 g 2 4d2 − d − 6 β =1− f (Λ, d) δL . 8 d(d + 2) α=1−

See M. Karder and A. Zee, “Matrix generalizations of some dynamic field theories,” Nucl. Phys. B464 (1996) 449-462 for further discussion.

205

VII

Grand Unification

VII.1

Quantizing Yang-Mills Theory and Lattice Gauge Theory

3. Consider a lattice gauge theory in (D + 1)-dimensional space with the lattice spacing a in D-dimensional space and b in the extra dimension. Obtain the continuum D-dimensional field theory in the limit a → 0 with b kept fixed. Solution: Consider a rectangular plaquette P with sides of unequal lengths a and b:

l

Uk l Ul i

i

k b

P Uj k Ui j

j

a We have S(P ) =Re trUij Ujk Uk` U`i , where 1 1 Uij = e+iaAµ (x) , µ ˆ = (xj − xi ) , x = (xi + xj ) a 2 1 a b 1 ˆ + νˆ Ujk = e+ibAν (y) , νˆ = (xk − xj ) , y = (xj + xk ) = x + µ b 2 2 2 −iaAµ (x0 ) 0 Uk` = e , x = x + bˆ ν a b 0 U`i = e−ibAν (y ) , y 0 = y − aˆ µ=x− µ ˆ + νˆ 2 2 Since Ujk and U`i are inverses when evaluated at the point x + 2b νˆ, let Ujk (x + 2b νˆ) ≡ U and U`i (x + 2b νˆ) = Ui`† (x + 2b νˆ) ≡ U † . Taylor expanding in powers of a, we have 1 • Uij = e+iaAµ (x) = I + iaAµ (x) − a2 Aµ (x)Aµ (x) + O(a3 ) 2 a b a 1 a 2 ∂µ ∂µ U + O(a3 ) • Ujk (x + µ ˆ + νˆ) = U + ∂µ U + 2 2 2 2 2 1 2 −iaAµ (x0 ) 0 • Uk` = e = I − iaAµ (x ) − a Aµ (x0 )Aµ (x0 ) + O(a3 ) 2 a b a 1 a 2 † † • U`i (x − µ ˆ + νˆ) = U − ∂µ U + − ∂µ ∂µ U † + O(a3 ) 2 2 2 2 2 206

For brevity suppress the direction µ and write Aµ (x) ≡ A and Aµ (x0 ) ≡ A0 . The trace of the product of these four matrices (ignoring terms of O(a3 ) and higher) is tr Uij Ujk Uk` U`i

1 = tr I + iaA − a2 A2 2

a a † a2 2 † a2 2 1 2 02 † 0 U + ∂U + ∂ U I − iaA − a A U − ∂U + ∂ U 2 8 2 2 8

1 1 2 1 2 2 ∂U + iAU + a ∂ U + iA∂U − A U × = tr U + a 2 2 4 1 2 1 2 † 1 † † 0 † 0 † 02 † U −a ∂U + iA U + a ∂ U + iA ∂U − A U 2 2 4

1 1 † † † 0 † = tr[ U U + a ∂U U + iAU U − U ∂U − iU A U ] 2 2 1 1 2 † 1 1 † 2 0 † 02 † 0 † + a tr[ U ∂ U + iA ∂U −A U − ∂U +iAU ∂U +iA U 2 4 2 2 1 1 2 2 + ∂ U +iA∂U −A U U † ] 2 4 †

Since U U † = I, the first term is a constant and can be dropped. Also, trAU U † = trA = 0. Using the cyclicity of the trace, we also have trU A0 U † = trA0 U † U = trA0 = 0. Since U = e ibAν (y) , we have ∂U = U ib ∂Aν (y), and so tr(∂U U † − U ∂U † ) = 2tr(∂U U † ) ∝ tr(U ∂AU † ) = tr(∂AU U † ) = tr(∂A) = 0 .

207

Therefore the O(a) terms in tr Uij Ujk Uk` U`i are zero. Dropping the O(a0 ) additive constant, we have: tr Uij Ujk Uk` U`i 1 = a2 tr U ∂ 2 U † + ∂ 2 U U † − 2∂U ∂U † 8 1 + i a2 tr U A0 ∂U † + A∂U U † − AU ∂U † − ∂U A0 U † 2 1 1 2 2 02 † 0 † † + a tr − U A U + AU A U − A U U 2 2 1 = − a2 tr ∂U ∂U † 2 1 2 + i a tr (∂U U † − U ∂U † )A + (∂U † U − U † ∂U )A0 2 1 2 2 − a tr A + A02 − 2 AU A0 U † 2 Consider a derivative operator defined as DU = ∂U + iAU − iU A0 . This is the covariant ¯ under SU (N ) gauge transforderivative appropriate for a field U transforming as ∼ N ⊗ N mations. To clarify this remark, return to ordinary continuum field theory for a moment and ¯ of SU (N ). This has one lower index a = 1, ..., N and one consider a matter field ψ ∼ N ⊗ N upper index a ¯ = 1, ..., N , meaning that the components of ψ are ψa a¯ . The gauge-covariant derivative of ψ is a ¯

a ¯

(Dµ ψ)a = ∂µ ψa + i

2 −1 N h X

¯

AIµ (TNI )ab ψb a¯ + AIµ (TNI¯ )a¯¯b ψa b

i

I =1

where TNI are the generators of the N -representation (“fundamental”) of SU (N ), and TNI¯ are ¯ -representation (“anti-fundamental”) of SU (N ). (As usual, the index the generators of the N 2 I = 1, ..., N − 1 counts the number of generators.) It is true in general that TNI¯ = −(TNI )∗ . Since the generators are hermitian, we have (TNI )∗ = (TNI )T , or in components [(TNI )ab ]∗ = (TNI )b a . Therefore: ¯

¯

¯

¯

(TNI¯ )a¯¯b ψa b = −[(TNI )a¯¯b ]∗ ψa b = −(TNI )¯b a¯ ψa b = −ψa b (TNI )¯b a¯ where the last equality is simply to bring the indices in matrix multiplication order. The covariant derivative is therefore a ¯

a ¯

(Dµ ψ)a = ∂µ ψa + i

2 −1 N X

i h ¯ AIµ (TNI )ab ψb a¯ − ψa b (TNI )¯b a¯ .

I =1

Suppressing the matrix indices in the usual way and defining the matrix-valued gauge field P Aµ ≡ I AIµ TNI , we have Dµ ψ = ∂µ ψ + iAµ ψ − iψAµ 208

which matches our lattice covariant derivative DU = ∂U + iAU − iU A0 . Now we return to the problem. Since (DU )† = ∂U † − iU † A + iA0 U † , we have tr(DU )(DU )† = tr ∂U ∂U † − i∂U (U † A − A0 U † ) + i(AU − U A0 )∂U † + (AU − U A0 )(U † A − A0 U † ) = tr ∂U ∂U † − i tr ∂U U † − U ∂U † A + ∂U † U − U † ∂U A0 + tr A2 − 2 AU A0 U † + A02 Recall our previous result trUij Ujk Uk` U`i = const + 1 − a2 tr ∂U ∂U † − i (∂U † U − U † ∂U )A0 + (∂U U † − U ∂U † )A + A02 − 2AU A0 U † + A2 . 2 Therefore tr Uij Ujk Uk` U`i = − 12 a2 tr[(DU )(DU )† ] for ijk` bounding a rectangular plaquette. The full lattice gauge theory action a2 S= Re tr Uij Ujk Uk` U`i + 2 b P square X

X

Re tr Uij Ujk Uk` U`i

P rectangle

in the continuum limit a → 0 with b held fixed becomes N Z X 1 1 † 4 D µν µ S=a d x tr − (Fi )µν (Fi ) − (Dµ U )i (D U )i 4 2 i=1 where bN is the length of the lattice in the extra dimension. This is the action of N copies of a D-dimensional continuous SU (N ) Yang-Mills theory with a Lorentz-scalar U transform¯ )-dimensional representation. For further discussion in the context of ing under the (N ⊗ N Kaluza-Klein theory, see http://arxiv.org/pdf/hep-th/0104005v1. 4. Study the alternative limit b → 0 with a kept fixed so that you obtain a theory on a spatial lattice but with continuous time. The result of the previous problem carries over immediately: N Z X 1 1 † µν µ 4 S=b dt tr − (Fi )µν (Fi ) − (Dµ U )i (D U )i 4 2 i=1 where aD N is the volume of the lattice. We now have N copies of a 1-dimensional SU (N ) ¯. gauge theory with a Lorentz-scalar U ∼ N ⊗ N 5. Show that for lattice gauge theory the Wilson area law holds in the limit of strong coupling. [Hint: Expand (20) in powers of f −2 .] Z hY i P 1 − 1 S(P ) hW (C)i = dU e 2f 2 P W (C) (20) Z 209

Solution: Consider an SU (N ) lattice gauge theory whose partition function is19 Z P 1 tr KP + KP† Z = DU e β P SP , SP = 2N where β ≡ 2N/g 2 , and the function KP is defined as KP ≡ Uij Ujk Uk` U`i for a plaquette P whose four corners are the lattice sites xi , xj , xk , x` as in the diagram below:

l

Uk l Ul i

i

k

P Uj k Ui j

j

Now consider a large closed rectangular curve C on the lattice whose sides have lengths R and T . Define the Wilson loop WC for the curve C as WC = tr Uij Ujk ...Umn Uni where xi → xj → xk → ... → xm → xn → xi are the links that comprise the curve C. We are interested in the expectation value of the Wilson loop: Z P 1 hWC i = DU WC e β P SP Z " ∞ # Z Y X 1 1 = DU WC (βSP )n Z n! n=0 P 19

We have written the theory in this form to suggest a connection to statistical mechanics with temperature T ≡ β −1 . Our “strong-coupling expansion” is analogous to the high-temperature expansion T → ∞ or β → 0.

210

To find the leading contribution in the limit of small β, we have to understand how to carry out integrals over the link variables Uij . The rules are as follows20 : Z DU UAB = 0 Z DU UAB UCD = 0 Z 1 D B δ δ DU UAB (U † )CD = N A C Z 1 1 2 N DU UAB UAB ... UAB εA A ...A εB1 B2 ...BN = 1 2 N N! 1 2 N for any N -by-N link variable U . (The indices A, B, ... run from 1 to N .) To get the leading nonzero contribution to hWC i, we need to pair up each Uij in WC with exactly one Uij† from Q P 1 n (P in Σ) n n! (βSP ) , where Σ is the surface of minimal area whose boundary is the curve C. Thus the leading contribution from the product over plaquettes in Σ is when n = 1 in the sum above, and the integral can be approximated to leading order as21 n∗ Z β hWC i ≈ DU WC (trK1† )...(trKn† ∗ ) 2N where n∗ ≡ RT /a2 is the number of plaquettes that fit inside the region Σ, and a2 is the area of a square plaquette with sides of length a:

C P

P

P

P

P

P

P

P

P

P

P

P

T

R To understand how to compute the integral, first consider the overly simplified but illustrative case for which the curve C encloses only 1 plaquette, bounded by the lattice sites (i, j, k, `) as in the diagram above. In this case, the Wilson loop is simply WC = tr Uij Ujk Uk` U`i = (Uij )AB (Ujk )BC (Uk` )CD (U`i )DA 20 See p. 90 and p. 222 of the text “Lattice Gauge Theories - An Introduction, 2nd Ed.” by Heinz. J. Rothe for more details. 21 The denominator Z contributes only a factor of 1 to this order. To the next order in β, we have to worry about nontrivial contributions from Z.

211

and the only K that contributes is K1 = tr Uij Ujk Uk` U`i , so that † † † † K1† = U`i† Uk` Ujk Uij† = (U`i† )EF (Uk` )FG (Ujk )GH (Uij† )HE .

The expectation value of the Wilson loop to leading order22 is Z β † † )GH (Uij† )HE )FG (Ujk hWC i ≈ DU (Uij )AB (Ujk )BC (Uk` )CD (U`i )DA (U`i† )EF (Uk` 2N Z Z β † E † B H C = DUij (Uij )A (Uij )H DUjk (Ujk )B (Ujk )G × 2N Z Z † † F D G A DUk` (Uk` )C (Uk` )F × DU`i (U`i )D (U`i )E 1 H C 1 G D 1 F A 1 E B β δ δ δ δ δ δ δ δ = 2N N A H N B G N C F N D E β 1 A1 B1 C1 D δ δ δ δ 2N N A N B N C N D β = . 2N If we now consider the second simplest class of example, for which the curve C contains β 2 exactly two plaquettes, we will find hWC i ≈ 2N , where again all of the factors of N1 will cancel. As soon as the curve C is large enough such that there can actually be an interior to the surface Σ – that is, for which some plaquettes will not be along the perimeter C – then the cancellation of N1 factors will be incomplete. After the integrations, there will be n∗ − 1 = RT /a2 − 1 factors of (1/N ) left over, so the leading contribution to the expectation value of the Wilson loop is RT /a2 RT /a2 −1 RT /a2 β β 1 =N . hWC i ≈ 2N N 2N 2 Recalling the definition β = 2N/g 2 , we obtain RT /a2 RT 2 1 = N e− a2 ln(g N ) hWC i ≈ N 2 g N The behavior hWC i ∼ e−(#)RT is called the area law. As explained on p. 377, the energy between a quark and an antiquark is 1 E(R) = − lnhWC i = σ R + O(1/T ) T where we have defined the string tension ln(g 2 N ) . a2 We can neglect the constant of order 1/T in the limit of taking the curve C arbitrarily large. σ=

22

R Remember integral DU is written in a condensed notation to denote integration over all link R that the R Q variables. So DU = [ hiji DUij ], where Uij connects the sites i and j on the lattice and hiji denotes that sites i and j are nearest neighbors.

212

VII.2

Electroweak Unification

1. Unfortunately, the mass of the elusive Higgs particle H depends on the parameters in the double well potential V = −µ2 ϕ† ϕ + λ(ϕ† ϕ)2 responsible for the spontaneous symmetry breaking. Assuming that H is massive enough to decay into W + W − and ZZ, determine the rates for H to decay into various modes. Solution: We compute the rates for H → W + W − , H → ZZ and H → `+ `− at tree level. We fix unitary gauge, which is the statement that we write the Higgs doublet ϕ as 0 ϕ = √1 (v + H) 2 where H is a real scalar field whose quantum is the spin-0 Higgs boson. As described on p. 382, after using the above parameterization for ϕ in the Lagrangian L = (Dµ ϕ)† (Dµ ϕ) and identifying the photon A and the Z-boson through the rotation 3 Z cos θ − sin θ W = B A sin θ cos θ you should eventually discover an interaction Lagrangian 2 1 2 L = − H(MW W +µ Wµ− + MZ2 Z µ Zµ ) v 2 Including the field H in addition to the constant v in equation (2) on p. 381 yields the Higgs-lepton interaction L=−

m` ¯ m` ¯ H `L `R + h.c. = − H `` v v

where ` is the 4-component Dirac field for either an electron, a muon or a tau. These three interactions yield the following vertices: 2 (HW W ) = −2iv −1 MW gµν , (HZZ) = −2iv −1 MZ2 gµν , (H``) = −im` v −1

Note the factor of 2 in the HZZ vertex due to the interchange symmetry of the Zs. The amplitude M for H → ZZ is M = −2iv −1 MZ2 gµν εµλ1 (k1 )ενλ2 (k2 ) which implies 2

|M| = 4

Mz2 v

2

εµλ1 (k1 )ενλ1 (k1 )∗ [ελ2 µ (k2 )ελ2 ν (k2 )∗ ]

213

Summing over outgoing polarizations using the rule X µ kµkν ελ (k)ενλ (k)∗ = g µν − MZ2 λ P gives the polarization-averaged amplitude-squared M2 ≡ λ1 ,λ2 |M|2 : 2 2 Mz k1µ k1ν k2µ k2ν µν 2 M =4 g − gµν − v MZ2 MZ2 " µ 2 # 2 2 k1 k2µ 1 Mz 4 − 2 (k12 + k22 ) + =4 v MZ MZ2 !2 2 2 0 0 ~ ~ Mz 2 + k1 k2 − k1 · k2 =4 v MZ2 where in the last line we have used the fact that the outgoing Z bosons are on-shell, meaning k12 = k22 = MZ2 . Now we use equation (38) on p. 141 for the differential decay rate in the center of mass frame: 1 d3 k1 d3 k2 dΓ = (2π)4 δ (4) (K − k1 − k2 ) M2 0 0 3 3 2mH (2π) 2k1 (2π) 2k2 √ where mH ≡ λ v is the mass of the Higgs boson and K is its 4-momentum. When integrating over the outgoing momenta, the fact that there are two identical outgoing Z bosons forces R 1 us to divide by 2, meaning that the total decay rate is Γ = 2 dΓ. Now we compute: Z 3 Z 3 1 1 1 d k2 d k1 Γ= δ(mH − k10 − k20 ) δ (3) (~k1 + ~k2 ) M2 0 2 2 2 2mH 2 (2π) k1 k20 Z q 1 4π dk k 2 = 6 2 δ(m − 2 k 2 + MZ2 ) M2 H 2 π mH k 2 + MZ2 p 1 δ(k − k∗ ), where k∗ is Define f (k) ≡ mH − 2 k 2 + MZ2 and use the formula δ(k) = |f 0 (k ∗ )| m2

defined by f (k∗ ) ≡ 0. f (k) = 0 =⇒ k 2 = 4H − MZ2 . The derivative of f (k) at this value of k is −2k 4k f 0 (k) = p =− 2 2 mH k + MZ where we wait until later to plug in the value for k. Using this, we have 1 k2 mH 2 M 2 4 2 2 πmH k + MZ 4k 1 k = 4 M2 2 2 π 4(k + Mz2 ) p m2H /4 − MZ2 1 M2 = 4 2π m2H s 2 1 2MZ = 5 1− M2 2 πmH mH

Γ=

214

If the Z bosons were spin-0 particles, p then there would be no momentum dependence in M and we would get the behavior Γ ∝ 1 − (2MZ /mH )2 we expect from the discussion of 1 → 2 meson decay on p. 142. The decay rate is only real if mH > 2MZ , which reflects the perfectly sensible fact that the Higgs can only decay into two Z bosons if its mass is larger than twice the mass of the Z boson. Now let’s see what the spin-1 polarizations do. Using the integrations over the delta functions, we have: 2 2 2 2MZ2 mH m2H m4H 2 0 0 2 2 2 2 2 ~ ~ + − MZ = 1− 2 (k1 k2 − k1 · k2 ) = [(k + MZ ) + k ] = 4 4 4 mH Therefore: s " 2 # 2 2MZ2 1 2MZ MZ4 m4H 1− 2 Γ= 5 1− 4 2 2+ 2 πmH mH v 4MZ4 mH s 2 2 2 " 2 # mH 2MZ m4H MZ 2MZ2 = 1− 1+ 1− 2 4π mH v mH 8MZ4 mH It is often convenient to display these decay rates as polynomials in the variable x ≡ 4MZ2 /m2H . Substituting MZ2 = xm2H /4 gives m4H 1+ 32MZ4

2M 2 1 − 2Z mH

2

2 m4H 2(xm2H /4) =1+ 1− 8(xm2H /4)2 m2H 2 x 2 2 x2 x 2 =1+ 2 1− = 2 + 1− x 2 x 2 2 2 2 1 2 x x = 2 +1−x+ = 2 (3x2 − 4x + 4) x 2 4 2x

Therefore the decay rate is Γ(H → ZZ) =

m3H √ 4MZ2 2 1 − x (3x − 4x + 4) , x ≡ 27 πv 2 m2H

√ When comparing to other sources23 substitute v for the Fermi constant GF ≡ √ 1/(v 2 2 ), although be aware that different definitions of GF exist, for example without the 2 . To get an idea for the size of the prefactor, take mH = v ≈ 246 GeV, for which m3H /(27 πv 2 ) ∼ 0.6 GeV and x ∼ 0.5 so that Γ ∼ 1.8 GeV. To calculate the decay rate for H → W + W − , note that almost everything would go through in exactly the same way as for H → ZZ with MZ replaced by MW . The only difference is 23 For example, B. W. Lee, C. Quigg and H. B. Thacker, “Weak Interactions at Very High Energies: the Role of the Higgs Boson Mass,” FERMILAB-Pub-77/30-THY March 1977.

215

that since W + and W − are not identical particles, there is no factor of 1/2 when integrating dΓ to get Γ. Therefore we can immediately write down Γ(H → W + W − ) =

2 4MW m3H p 2 1 − y (3y − 4y + 4) , y ≡ 26 πv 2 m2H

Note that Γ(H → W + W − )/Γ(H → ZZ) is not quite 2, since MW /MZ < 1. Finally, we compute the decay width of the Higgs boson into a lepton pair `+ `− , which becomes the dominant decay channel if mH < 2MW ∼ 190 GeV. The vertex is ¯ = imv −1 (H ``) so the amplitude is simplyP M = imv −1 u¯(p1 , s1 )v(p2 , s2 ), which implies that the spin-summed amplitude squared M2 ≡ s1 ,s2 |M|2 is 6 p1 + m 6 p2 − m 1 m2 2 = 2 ( pµ1 p2µ − m2 ) M = 2 tr v 2m 2m v The integrals over the momentum-conserving delta function will work out in exactly the same way as for the H → ZZ mode, which implies pµ1 p2µ − m2 = 2p2 , where p ≡ |~p1 | = |~p2 |. Therefore, as far as the decay rate is concerned, we have M2 =

4p2 v2

The decay rate formula is Z Z 1 d3 p1 d3 p2 (2π)4 δ(mH − p01 − p02 )δ (3) (~p1 + p~2 ) M2 Γ= 2mH (2π)3 (p01 /m) (2π)3 (p02 /m) We write the formula explicitly to remind you of the factor p0 /m for the fermions, although everything else will proceed in the same way as before, leaving the result 3/2 4m2` m2` + − Γ(H → ` ` ) = mH 1 − 2 16πv 2 mH √ Note that m` = f v/ 2 , where f is defined by the Lagrangian L = f ψ¯L ϕ `R + h.c. (using the notation of pp. 380-381), so the limit m` → 0 is perfectly fine and just leaves behind Γ(H → `+ `− ) →

f2 mH . 32π

2. Show that it is possible to stay with the SU (2) gauge group and to identify W 3 as the photon A, but at the cost of inventing some experimentally unobserved lepton fields. This theory does not describe our world: For one thing, it is essentially impossible to incorporate the quarks. Show this! [Hint: We have to put the leptons into a triplet of SU (2) 216

instead of a doublet.] Solution: We take the gauge group to be G = SU (2), and we will use 2-component spinor notation. Consider a lepton triplet written as a 2-by-2 symmetric matrix: ! √1 `0 `+ 2 . ` ∼ 2 ⊗S 2 =⇒ `ij = √1 0 − ` ` 2 The superscript labels indicate that we are anticipating Pa bit and denoting the electric charge of each component. With the gauge bosons (Wµ )i j ≡ 3a = 1 Wµa ( 12 σ a )i j , the covariant derivative acting on `ij with gauge coupling g is ! √1 (W + `− + W − `+ ) Wµ3 `+ + Wµ+ `0 µ µ 2 i(Dµ `)ij = i∂µ `ij + g √1 . + − − + 3 − − 0 (W ` + W ` ) −W ` + W µ µ µ µ ` 2 The currents defined by L = i`† σ ¯ µ Dµ ` = i`† σ ¯ µ ∂µ ` + Wµ3 J3µ + Wµ+ J −µ + Wµ− J +µ are therefore: J3µ = g[(`+ )† σ ¯ µ `+ − (`− )† σ ¯ µ `− ] J −µ = g[(`+ )† σ ¯ µ `0 + (`0 )† σ ¯ µ `− ] J +µ = g[(`0 )† σ ¯ µ `+ + (`− )† σ ¯ µ `0 ] If Wµ3 remains massless, then we see that J3µ is the correct definition of the electromagnetic current, with the electric coupling defined as e ≡ g. Consider a Higgs field φ ∼ 2 ⊗S 2. Its covariant derivative is the same as that for `, and its interactions with the gauge bosons comes from the kinetic term L = (Dµ φ)† Dµ φ. When φ0 obtains a vacuum expectation value, we find + 0 Wµ 0 + ... (Dµ φ)ij = −ighφ i 0 Wµ− and therefore (Dµ φ)† Dµ φ = 2g 2 |hφ0 i|2 Wµ+ W −µ + ... √ The charged bosons W ± get a mass mW = 2 g|hφ0 i| and the neutral boson W 3 remains massless. Given this and the lepton currents derived previously, we can indeed identify W 3 as the photon.

217

However, there is a problem in assigning masses to the charged leptons while keeping the neutral one massless (or at least approximately massless). From `ij ∼ 2 ⊗S 2, we can form the singlet 1 ` εik εj` `k` = − 12 `0 `0 + `− `+ . 2 ij If we couple an SU (2)-singlet Higgs ϕ with a nonvanishing vacuum expectation value hϕi to this term, then we find a Majorana mass for the neutral lepton with the same magnitude as the Dirac mass for the charged lepton. This is clearly incompatible with the near masslessness of the neutrino that participates in nuclear beta decay. For further discussion as well as for other models, we refer the reader to the literature.

Addendum: Covariant Derivative for the Triplet Here we derive the electroweak SU (2) ⊗ U (1) covariant derivatives for a field transforming as φ ∼ (2 ⊗S 2 = 3, y), for arbitrary values of the hypercharge y. (Here we will use the convention for which the hypercharge generator is denoted by Y rather than 12 Y .) For the generators of SU (2), we have: √ √ a σ 1 1 cZ + sA W3 2 W+ 2 W+ a √ √ W = = 2 W − −W 3 2 W − −(cZ + sA) 2 2 2 As usual, we define the neutral boson Z and the photon A as a linear combination of W 3 and B: 3 3 Z c −s W W c s Z cZ + sA ≡ =⇒ = = A s c B B −s c A −sZ + cA where s ≡ sin θ and c ≡ cos θ define the weak mixing angle s/c ≡ g 0 /g, where g is the SU (2) gauge coupling and g 0 is the U (1) gauge coupling. This implies s s B = s(− Z + A) c c which is a combination that appears in the covariant derivative. We have: 0 0 0 g s 0 0 0 (Dφ)ij = ∂φij − i [W a (σ a )i i δj j + W a δi i (σ a )j j + 2y Bδi i δj j ]φi0 j 0 ] . 2 c In detail: s g i[(D − ∂)φ]11 = + [W a (σ)1 i δ1 j + W a δ1 i (σ a )1 j + 2y Bδ1 i δ1 j ]φij 2 c g a a i s a a j = [W (σ )1 φi1 + W (σ )1 φ1j + 2ys(− Z + A)φ11 ] 2 c √ g s + = [2(cZ + sA)φ11 + 2 W (φ12 + φ21 ) + 2ys(− Z + A)φ11 ] 2n co √ s 2 + = g [c(1 − y( ) )Z + s(1 + y)A]φ11 + 2 W φ12 c 218

g s i[(D − ∂)φ]12 = [W a (σ a )1 i δ2 j + W a δ1 i (σ a )2 j + 2y Bδ1i δ2j ]φij 2 c s g a a i a a j = [W (σ )1 φi2 + W (σ )2 φ1j + 2y Bφ12 ] 2 c √ √ s g 3 3 + = [(W − W )φ12 + 2 W φ22 + 2 W − φ11 + 2ys(− Z + A)φ12 ] 2 c s 1 + − = g[ys(− Z + A)φ12 + √ (W φ22 + W φ11 )] c 2 s g i[(D − ∂)φ]22 = [W a (σ a )2 1 φ12 + W a (σ a )2 2 φ22 + W a (σ a )2 1 φ21 + W a (σ a )2 2 φ22 + 2y Bφ22 ] 2n c o √ s =g 2 W − φ12 + [−c(1 + y( )2 )Z + s(−1 + y)A] c To summarize: [i(D − ∂)φ]ij = 2

[c(1 − y sc2 )Z + s(1 + y)A]φ11 + g ×

√ 2 W + φ12

ys(− sc Z + A)φ12 + √12 (W + φ22 + W − φ11 ) √ 2 [−c(1 + y sc2 )Z + s(−1 + y)A]φ22 + 2 W − φ12

Let us check that the couplings to the photon come out as planned. The generator of electric 3 i0 0 3 j0 0 0 0 0 0 charge is Q = T 3 + Y , or in components Qiji j = σ2 δj j + δi i σ2 + yδii δjj . We find: i

j

(Qφ)11 = [(+ 12 ) + (+ 12 ) + y]φ11 = (1 + y)φ11 X (Qφ)12 = [(+ 21 ) + (− 12 ) + y]φ12 = yφ12 X (Qφ)22 = [(− 21 ) + (− 21 ) + y]φ22 = (−1 + y)φ22 X To fix the normalization of the field, notice that φ†ij φij = |φ11 |2 + 2|φ12 |2 + |φ22 |2 . So in general, for φ ∼ (2 ⊗S 2, y) of SU (2) ⊗ U (1), the components of φij can be identified as follows: ! φ(1+y) √12 φ(y) φij = √1 (y) (−1+y) φ φ 2 where the superscript denotes the electric charge. For example: y = −1 implies φij =

φ0 √1 φ− 2

√1 φ− 2 −−

!

φ

with covariant derivative: [i(D − ∂)φ]ij = 1 Zφ0 c

! + W + φ− −s(A − sc Z) √12 φ− + √12 (W + φ−− + W − φ0 ) 2 × [−c(1 − sc2 )Z − 2sA]φ−− + W − φ−

This Higgs triplet can couple to the left-handed lepton triplet `(i `j) to generate neutrino masses. 219

!

VII.3

Quantum Chromodynamics

1. Calculate C in (9). [Hint: If you need help, consult T. Appelquist and H. Georgi, Phys. Rev. D8: 4000, 1973; and A. Zee, Phys. Rev. D8: 4038, 1973.] ! X 2 σ(e+ e− → hadrons) 2 1+C + ... = 3 Qa (9) R(E) ≡ 2 σ(e+ e− → µ+ µ− ) n ) ln(E/µ) (11 − f 3 a Solution: The details of this two-loop calculation can be found on p. 415 (chapter 8-4-4) of the quantum field theory text by Itzykson and Zuber. If you compute the integrals, you can check your answer with the Particle Data Group24 : ! X αS (E) 2 2 R(E) = 3 Qa 1+ + O(αS ) π a Then equation (6) on p. 389, which is αS (E) =

1+

1 (11 2π

−

αS (µ) 2 n )αS (µ) ln(E/µ) 3 f

→

2π (11 −

2 n ) ln(E/µ) 3 f

for large E implies C = π. 2. Calculate (2). (2)

11 g3 dg = − T2 (G) dt 3 16π 2

Solution: We will use the background field gauge. The goal is to compute the 1-loop effective potential for pure Yang-Mills theory in the presence of a constant background gauge field.25 Including gauge-fixing and ghost terms (see p. 372), the full Yang-Mills Lagrangian with counterterms is L = LYM + Lgf + Lghost + Lct where 1 a aµν a F , Fµν = ∂µ Aaν − ∂ν Aaµ + gf abc Abµ Acν LYM = − Fµν 4 1 Lgf = − Ga Ga 2ξ Lghost = (Dµ c¯)a (Dµ c)a , (Dµ c)a = ∂µ ca + gf abc Abµ cc 1 a Lct = − (Zg − 1)Fµν F aµν + (Zc − 1)(Dµ c¯)a (Dµ c)a 4 24 25

http://pdg.ihep.su/2009/reviews/rpp2009-rev-qcd.pdf As suggested on p. 388 of the text, here we follow the calculation of S. Weinberg.

220

We will choose the gauge fixing function Ga shortly. a F aµν + ... to the bare Lagrangian Comparing the renormalized Lagrangian L = − 41 Zg Fµν 1 aµν a L = − 4 (FB )µν (FB ) + ..., we find the relation

gB = Zg1/2 g µ ˜ε/2 between the bare coupling gB and the renormalized coupling g. Here we have anticipated using dimensional regularization d = 4 − ε to regulate the forthcoming divergent integral, and therefore replaced g → g µ ˜ε/2 to keep g dimensionless in d 6= 0. The goal is now to compute the O(g 2 ) contribution to Zg and thereby obtain the 1-loop beta function for the gauge coupling g. We now split up the gauge field Aaµ into a constant background field A¯aµ and a fluctua0 ¯ tion A0a µ : A = A + A . (See p. 504 in chapter N.3 of the text.) The full (infinitesimal) gauge transformation δAaµ = ∂µ εa − f abc εb Acµ is split up into δ A¯aµ = ∂µ εa − f abc εb A¯cµ abc b 0c δA0a ε Aµ µ = −f

so that A¯ transforms as a gauge field while A0 transforms as a matter field in the adjoint representation. Putting A = A¯ + A0 into the Lagrangian and choosing the gauge-fixing function26 ¯ µ A0µ )a ≡ (δ ac ∂µ + gf abc A¯bµ )A0cµ Ga = (D we arrive at 1 ¯a ¯ µ A0 )a − (D ¯ ν A0 )a + gf abc A0b A0c 2 Fµν + (D ν µ µ ν 2 4g 1 ¯ 0µ a 2 Lgf = − [(D µA ) ] 2ξ ¯ µ c¯)a [(D ¯ µ c)a − gf abc cb A0cµ ] . Lghost = (D LYM = −

a Here F¯µν = gf abc A¯bµ A¯cν is the field strength of the constant background field (that is, ¯ µ is the covariant derivative with respect to the background field, as ∂µ A¯aν = 0), and D defined by the gauge fixing function Ga above.

Since A¯aµ is to be taken as a fixed classical background field, we do not integrate over it in the path integral. The theory is invariant under the formal transformation A¯aµ → A¯aµ + ∂µ εa − gf abc εb A¯cµ 26

Note that this differs from the usual Rξ gauge, in which the gauge fixing function is chosen as Ga = ∂µ Aaµ .

221

irrespective of whether the transformation is performed before or after integrating over the a a ¯ . Thus by this formal background gauge invariance, we can comfluctuation fields A0a µ,c ,c a ¯ aµν pute the 1-loop correction to the gauge coupling by finding the coefficient of − 41 F¯µν F ∼ 4 0 a (A¯µ ) in the 1-loop effective action, obtained by keeping only quadratic terms in A , c, c¯ and performing the resulting gaussian integrals. The quadratic action is 2 Z 1 1 ¯ 0µ a 2 1 ¯ 0 a a 0 4 ¯ ν Aµ ) ¯ ¯)a (D ¯ µ c)a − gf abc F¯ aµν A0bµ A0cν − [(D Squad = d x − [(Dµ Aν ) − (D µ A ) ] + ( Dµ c 4 2 2ξ Z Z 0bν 4 ¯a (x)N ab (x, y)cb (y) = d x d4 y − 12 A0aµ (x)Mab µν (x, y)A (y) + c where in the second line we have chosen the gauge ξ = 1 and defined the matrices ab ca ∂ cb ∂ cda ¯dα ceb ¯e Mµν (x, y) = ηµν δ δ − gf A − gf Aα δ 4 (x − y) ∂xα ∂y α ca ∂ cda ¯d ceb ¯e cb ∂ − δ − gf Aν δ − gf Aµ − (µ ↔ ν) δ 4 (x − y) ∂xν ∂y µ c 4 + gf cab F¯µν δ (x − y) and

∂ − f cda A¯dµ N (x, y) = δ µ ∂x ab

ca

cb ∂ ceb ¯eµ δ − f A δ 4 (x − y) . ∂yµ

The one-loop effective action can now be obtained by integrating over the fluctuations A0 and c, c¯: Z Z d4 p 1 4 ¯ iΓ1-loop (A) = d x − 2 tr ln M (p) + tr ln N (p) 4 (2π) ab (p) and N ab (p) are defined through the Fourier transforms of Mab where Mµν µν (x, y) and ab N (x, y) respectively: ab Mµν (p) = −ηµν ipα δ ca − gf cda A¯dα ipα δ cb + gf ceb A¯eα + ipν δ ca − gf cda A¯dν ipµ δ cb + gf ceb A¯eµ − (µ ↔ ν) + gf cab F¯ c µν

N ab (p) = − ipµ δ ca − gf cda A¯dµ

ipµ δ cb + gf ceb A¯eµ .

a ¯ aµν We are interested in extracting the coefficient of the term proportional to F¯µν F in the a ¯ one-loop effective action, so we need all the terms quartic in Aµ . Let On denote the part of an operator O that is of order (A¯aµ )n . Then the terms we are looking for are of the form: (tr ln O)4 = tr − 21 (O0−1 O2 )2 + (O0−1 O1 )2 O0−1 O2 − 41 (O0−1 O1 )4

for O = M, N . Each of these terms has a diagrammatic interpretation. The first term is 222

A

A O 0 −1 O2

O2 O 0 −1

A

A

In other words, reading from right to left, the term tr(O0−1 O2 O0−1 O2 ) starts with an operator with two external A¯ lines, turns into an internal propagator O0−1 , meets another operator with two external A¯ lines, then closes with another internal propagator O0−1 . The symmetry factors and minus signs are all automatically taken into account by the algebra at the level of the effective action. Similarly, the second term is represented graphically as A

A

O1 O 0 −1

O2

O 0 −1

O 0 −1 O1

A

A

and the third term is a box diagram: A

A O 0 −1 O1 O0

O1 O 0 −1

−1

O1

O1 O 0 −1 A

A

Applying this decomposition to the operators O = M and N and simplifying the momentum integrals using pµ pν → 41 p2 and pµ pν pα pβ → 4!1 (η µν η αβ +η µα η νβ +η µβ η να )(p2 )2 in the integrals,

223

we obtain d4 p 5 c ¯ dµν [tr ln M (p)]4 = − I g 2 f abc f abd F¯µν F 4 (2π) 3 Z d4 p 1 c ¯ dµν [tr ln N (p)]4 = + I g 2 f abc f abd F¯µν F . 4 (2π) 12

Z

Here we have defined a divergent integral Z d4 p 1 I≡ . 4 2 (2π) (p + i)2 The integral can be performed by computing in d = 4 − ε dimensions to regulate the UV divergence, and by imposing a lower cutoff m on the magnitude of the (Euclidean) momentum to regulate the IR divergence. After replacing g → g µ ˜ε/2 to take care of the mass dimension of the gauge coupling, we have 2 1 µ g2 2 2 + O(ε ) g I → i 2 1 + ε ln 8π ε 2 m2 R∞ where µ2 ≡ 4πe−(1−γ) µ ˜2 and γ = − 0 dt e−t ln t ≈ 0.58. For the color group SU (n), we have f abc f abd = n δ cd and therefore: Z 5 1 d4 p a ¯ aµν 1 [− tr ln M (p) + tr ln N (p)] = + + I g 2 n F¯µν F (2π)4 2 6 12 2 11 g 2 1 1 µ 1 ¯ a ¯ aµν + ln =i − n 2 − Fµν F 12 2π ε 2 m2 4 a ¯ aµν F in the 1-loop effective action. MulWe have therefore deduced the coefficient of − 14 F¯µν tiplying by −i, we find the renormalization factor Zg to O(g 2 ): µ 11 g 2 1 Zg = 1 − n 2 + ln . 12 2π ε m 1/2

The bare coupling gB is related to the running coupling g by gB = Zg g µ ˜ε/2 , so that ln gB = O(g, ε) + ln g + 12 ε ln µ ˜. where O(g, ε) ≡ 21 ln Zg . The bare coupling does not know about the parameter µ, so differentiating the above relation gives 0=

∂O(g, ε) dg + 12 ε . ∂g d ln µ

In general, the function O(g, ε) will contain finite terms as well as poles in powers of 1/ε: O(g, ε) = O0 (g) + 224

∞ X On (g) . n ε n=1

Plugging in this expansion, writing limε→0 β(g) =

1 2 0 g O1 (g) 2

dg d ln µ

= β(g) and matching powers of 1/ε we obtain

h i 2 0 0 1 − gO0 (g) + O (gO0 (g)) .

In our case, we have 11 g 2 1 µ 11 g 2 1 µ O(g, ε) = ln Zg = ln 1 − n 2 + ln = n 2 + ln 12 2π ε m 12 4π ε m 1 2

1 2

where we have dropped terms of O(g 4 ) and terms of higher powers in 1/ε, which are assumed to cancel order by order. Thus matching to the general expansion of O(g, ε), we have µ 11 g 2 n 2 ln 12 4π m 11 g 2 O1 (g) = − n 2 . 12 4π O0 (g) = −

Thus gO00 (g) is O(g 2 ) and can be dropped from the beta function at this order, since it multiplies g 2 O10 (g) = O(g 3 ). This is consistent with the general property of gauge theories that 1-loop beta functions are independent of the renormalization scheme. n Since O10 (g) = − 11 12

g 2π 2

= − 11 n 3

g , 8π 2

the 1-loop beta function β(g) = 12 g 2 O10 (g) is

β(g) = −

11 g3 n + O(g 5 ) . 3 16π 2

This is equation (3) on p. 388. In the non-abelian gauge theory of the strong interaction, we have n = 3.

225

VII.4

Large N

1. Since the number of gluons only differs by one, it is generally argued that it does not make any difference whether we choose to study the U (N ) theory or the SU (N ) theory. Discuss how the gluon propagator in a U (N ) theory differs from the gluon propagator in an SU (N ) theory and decide which one is easier. Solution: Pdim G a a j Let (Aµ )i j = a = 1 Aµ (T )i be the matrix-valued gluon field for G = U (N ) that acts on the defining representation, meaning that the indices i, j run from 1 to N . Then: h(Aµ )i j (x)(Aν )k` (0)i = hAaµ (x)Abν (0)i(T a )i j (T b )k` = ∆µν (x)δ ab (T a )i j (T b )k` = ∆µν (x)(T a )i j (T a )k` where

Z ∆µν (x) =

kµ kν d4 k e ik·x −ηµν + (1 − ξ) 2 (2π)4 k 2 + iε k + iε

is the usual abelian massless vector boson propagator in Rξ gauge. The new feature is just the group theory factor 2

N X

(T a )i j (T a )k` = C δi ` δk j

(1)

a=1

where C is some constant to be determined as follows. Suppose we normalize the generators as tr(T a T b ) = t δ ab where t is a number called the “index” of the defining representation of U (N ) to which the generators belong (we have been omitting the subscript N on the generators T a for typographical convenience.) Setting b = a and summing over a gives tr(T a T a ) = tδ aa = tdimU (N ) = tN 2 . In (1), setting k = j and summing over j, and setting ` = i and summing over i gives tr(T a T a ) = Cδi i δj j = CN 2 Therefore C = t, which from now on we take equal to 1. The propagator for the U (N ) gauge bosons is therefore h(Aµ )i j (x)(Aν )k` (0)iU (N ) = ∆µν (x) δi ` δk j . This provides a simple graphical interpretation, exactly of the form given in figure VII.4.4 on p. 398. Now suppose we insist on the gauge group G = SU (N ) rather than U (N ) = SU (N ) ⊗ U (1). This entails ensuring that the generators are traceless, or in other words insisting on not in2 cluding the U (1) generator (T a = N )i j = N 11/2 δi j . All we have to do to modify the propagator 226

is to take equation (1) (with C = 1 as per the previous discussion) and move the (N 2 )th generator from the left-hand side to the right-hand side. In other words, 2 −1 N X

(T a )i j (T a )k` = δi ` δk j −

a=1

1 j ` δ δ N i k

(10 )

The previous discussion for U (N ) carries through exactly, except with the group theory factor from equation (10 ) instead of (1). Therefore the SU (N ) gauge boson propagator is 1 j ` j ` ` j h(Aµ )i (x)(Aν )k (0)iSU (N ) = ∆µν (x) δi δk − δi δk . N This too has a straightforward graphical interpretation, where we subtract the trace from the U (N ) propagator: i j

l k

1 i N j

l k

2. As a challenge, solve large N QCD in (1 + 1)-dimensional spacetime. [Hint: The key is that in (1 + 1)-dimensional spacetime with a suitable gauge choice we can integrate out the gauge potential Aµ .] For help, see ’t Hooft, Under the Spell of the Gauge Principle, p. 443. Solution: 27 Define light cone coordinates p± = √12 (p0 ± p1 ), so that 2p+ p− = p20 − p21 = m2 , and fix light cone gauge: A− = 0. The Lagrangian for (1+1)-dimensional QCD is then L = − 12 tr(∂− A+ )2 + ψ¯i (i 6 ∂ − m)ψ i − g ψ¯i γ− (A+ )i j ψ j where (A+ )i j = Aa+ (T a )i j is the matrix-valued gauge field. Treating x+ as the temporal direction, we see that A+ has no dynamics and can be replaced by the 1D Coulomb potential. 0 1 The gamma matrices are defined by the Clifford algebra {γµ , γν } = 2 , so that 1 0 γ+2 = γ−2 = 0 and γ+ γ− +γ− γ+ = 2. The only interaction vertex in the theory is −igγ− (T a )i j , so the gamma matrices can be removed from the Feynman rules. For example, consider the 1-loop correction to the quark propagator due to single-gluon exchange: Z dk− dk+ γ− [γ− (6 k−6 p)+ + γ+ (6 k−6 p)− + m] γ− 2 a a i −g (T T ) j 2 (2π)2 [k− ][2(k − p)+ (k − p)− ] Since γ−2 = 0 and γ− γ+ γ− = γ− (−γ− γ+ + 2) = 0 + 2γ− , only the part of the fermion propagator proportional to γ+ contributes, and its contribution is simply a factor of 2. Thus the Feynman rules can be taken as: 27

We thank G. ’t Hooft for helpful discussion.

227

=

=

=

−i k −2 ik− 2 k + k −− m 2

−i2g

The large-N approximation is to neglect non-planar diagrams and to consider loops due to gluon exchange only. The quark self-energy Σ is related to the exact quark propagator S by iS(k) =

ik− 2k+ k− − m2 − k− Σ(k)

and satisfies the implicit equation

=

S

The above diagram reads Z i(k− + p− ) dk− dk+ −i 2 iΣ(p) = (−2g) 2 (2π)2 k− [2(k+ + p+ ) − Σ(k + p)](k− + p− ) − m2 Z k− + p− dk− dk+ 1 = 4g 2 2 2 (2π) k− [2k+ − Σ(k− + p− , k+ )](k− + p− ) − m2 where in the second line we have shifted k+ → k+ − p+ . The integral is independent of p+ , so Σ(p) is independent of p+ . We have Z Z 4g 2 k− + p− 1 iΣ(p− ) = dk− dk+ . 2 2 (2π) k− 2(k− + p− )k+ − (k− + p− )Σ(k− + p− ) − m2 R∞ RΛ The integral over k+ is UV divergent. Regularizing via −∞ dk+ → −Λ dk+ , the integral over k+ evaluates to 2(k−1+p− ) sgn(k− + p− )iπ. Therefore, we have Z g2 1 Σ(p− ) = dk− 2 sgn(k− + p− ) . 2π k− R∞ R∞ R −µ This integral diverges near k− → 0. To deal with this, replace −∞ dk− → µ dk− + −∞ dk− with µ > 0 a positive IR regulator. The self-energy is Z ∞ Z −µ 1 1 g2 dk− 2 sgn(k− + p− ) + dk− 2 sgn(k− + p− ) Σ(p− ) = 2π µ k− k− −∞ Z ∞ g2 1 = dk− 2 [sgn(p− + k− ) + sgn(p− − k− )] . 2π µ k− 228

d sgn(x) dx

= 2δ(x) to evaluate the integral by parts: Z g2 ∞ 1 d Σ(p− ) = − [sgn(p− + k− ) + sgn(p− − k− )] dk− 2π µ dk− k− " # ∞ Z ∞ 2 2 g 1 dk− =− (sgn(p− + k− ) + sgn(p− − k− )) − (δ(p− + k− ) − δ(p− − k− )) 2π k− k− µ k− = µ 1 1 g2 =− − (sgn(p− + µ) + sgn(p− − µ)) + 2sgn(p− ) 2π µ |p− | 2 1 g 1 sgn(p− ) − →+ π µ p−

Use

where in the last line we have taken µ → 0+ in the numerator. Thus we have solved for the quark self-energy. The denominator of the exact propagator is therefore g 2 |k− | 2k+ k− − − 1 − m2 . π µ In the limit µ → 0+ , the pole of the propagator is shifted to k+ → ∞. This indicates that there is no physical single-quark state. Next consider the implicit equation given by relating the nth and (n + 1)th ladder diagram: m1

p

p + k S

S

k

= S m2 p−q

p S S p−q

S p + k − q

Here the blob, which we denote ψ(p, q), stands for an arbitrary process in which a quarkantiquark pair emerges, the quark with mass m1 and momentum p, and the antiquark with mass m2 and momentum q − p. The source-free part of this equation is: i

iψ(p, q) = (−2g)2 (p− − q− )pi

2

2(p+ − q+ )(p− − q− ) − (m22 − gπ ) − Z i dk− dk+ −i iψ(p + k, q) . 2 2 2 g g (2π)2 k− 2p+ p− − (m21 − π ) − πµ |p− |

g2 |p πµ −

− q− |

×

After a series of manipulations, this can be put into the form of the eigenvalue equation28 Z 1 α1 α2 ϕ(y) 2 ρ ϕ(x) = + ϕ(x) − P dy x 1−x (y − x)2 0 28

The principal value integral is the average value of the integral across the pole: Z Z Z ϕ(x) 1 ϕ(x + i) ϕ(x − i) P dx 2 = dx + dx . x 2 (x + i)2 (x − i)2

229

R where we have defined ϕ(p− , q) ≡ dp+ ψ(p, r) and the dimensionless variables x ≡ p− /r− , αi ≡ π m2 − 1 and ρ2 ≡ gπ2 (2q+ q− ). Here ρ is the meson mass in units of g/π 1/2 . g2 i An approximate solution to the eigenvalue equation may be found by observing that the dominant contribution to the integral on the right-hand side comes from y ≈ x, at which point the denominator goes to zero. Using the trial function ϕ(x) = e iωx , we have Z 1 Z ∞ e iωy e iωy dy dy P ≈ P = −π|ω| e iωx 2 2 (y − x) (y − x) 0 −∞ so that e iωx is an approximate solution with eigenvalue ρ2 ≈ π|ω|, where we have further assumed that the quark and antiquark have equal masses (e.g., a meson made from a u¯ u 2 2 ¯ pair rather than a ud pair) and satisfy mq ≈ g /π, so that α1 = α2 ≈ 0. The appropriate boundary conditions are ϕ(0) = ϕ(1) = 0, so that we are to take the linear combination ϕω (x) ∝ e iωx − e−iωx ∝ sin(ωx) as the solution. Defining ω = πn, where n is any positive integer, we thereby arrive at the meson spectrum 2 m2π ≡ gπ ρ2 given by (m2π )n ≈ g 2 πn , n = 1, 2, 3, ... The approximation becomes better for larger values of n. In terms of the quark mass mq , we have (m2π )n ≈ π2n . m2q For corrections to this lowest order approximation as well as for mesons built of quarks with unequal masses, see the reference [G. ’t Hooft, “A two-dimensional model for mesons,” Nucl. Phys. B75 (1974) 461-470]. For a path integral treatment as well as for the inclusion of baryons in the spectrum, see E. Witten, “Baryons in the 1/N expansion,” Nucl. Phys. B160 (1979) 57-115.

3. Show that if we had chosen to calculate G(z) ≡ h(1/N )tr[1/(z − ϕ)]i, we would have to connect the two open ends of the quark propagator. We see that figures VII.4.5b and d lead to the same diagram. Complete the calculation of G(z) in this way. Solution (due to J. Feinberg): If we want to calculate G(z) directly, then we can’t have any free indices dangling on the random matrices ϕ. The trace takes the upper index on the incoming quark line and contracts it with the lower index on the outgoing quark (which behaves like an incoming antiquark, therefore in an antifundamental rep.) Therefore, we have really closed the loop by connecting the two quark lines in the diagram from the text. We will calculate G(z) by making use of a helpful recursion relation.

230

As we know from the text, we can write G(z) =

∞ X n=0

1 z 2n+1

1 tr ϕ2n N

where the matrix propagator gives a factor of 1/(N m2 ). To get the large N limit, the only surviving terms are the planar graphs, which means that the trace, when computed as Wick contractions, breaks into one trace per propagator (plus the original quark line trace). Thus, we can write ∞ X 1 G(z) = Cn 2n+1 z n=0 where Cn is the number of planar terms with n propagators. Let’s now derive a recursion relation for the Cn , starting with the zero propagator case, C0 = 1. Let the notation [ϕ...ϕ] denote the canonical Wick contraction, where the two fields at each end (ie, adjacent to each bracket) are contracted. Consider the case with n propagators, or 2n matrices. (To get planar graphs, there must be an even number of matrices between each contracted pair.) Thus, the total number of Wick contractions for 2n fields is given by (all for 2n) = [ϕ(all for 2n − 2)ϕ] + [ϕ(all for 2n − 4)ϕ][ϕϕ] + ...[ϕϕ](all for 2n − 2) Cn = C0 Cn−1 + C1 Cn−2 + ... + C0 Cn−1 =

n−1 X

Ck Cn−1−k .

k=0

Our expression for G(z) becomes n ∞ 1 1X Cn G(z) = z n=0 z 2 m2 " k n0 −k # ∞ n0 1 XX 1 1 1 1+ 2 2 Ck Cn0 −k = z z m n0 =0 k=0 z 2 m2 z 2 m2 n !2 ∞ X 1 1 1 Cn = 1 + 2 2 2 2 z z m z m n=0 1 1 2 = 1 + 2 G(z) . z m In the second line, we have used the recursion relation and set n0 = n − 1. In the third line, we noted that the second line was just the power series expansion of the product of two series. Now we have the same quadratic equation as in the text, leading to ! r m2 4 z − z2 − 2 G(z) = . 2 m

231

4. Suppose the random matrix ϕ is real symmetric rather than hermitean. Show that the Feynman rules are more complicated. Calculate the density of eigenvalues. [Hint: The double-line propagator can twist.] Solution: First, as clarification, this question is intended to follow the procedure on p. 397 with the quadratic (Gaussian) potential V (ϕ) = 21 m2 ϕ2 . In group theory language, the complication here is that there is no longer any distinction between up and down indices. For ϕ an N ×N hermitian matrix, meaning (ϕ† )i j ≡ (ϕj i )∗ = ϕi j , we know hϕi j ϕk` i ∝ δi ` δk j by matching upper and lower indices. But if ϕ is an N × N real symmetric matrix, meaning ϕij = ϕji with all entries being real numbers, then matching indices tells us hϕij ϕk` i = C1 δi` δjk + C2 δik δj` , with two a priori undetermined constants. The first term corresponds to the hermitian case (just lower the indices ` and j from before), while the second term involves a matrix transpose, which diagrammatically appears as a twist in the double-line propagator. The difference between the complex Hermitian case and the real symmetric case is in calculating the correlation function h(ϕ2 )ij i. The “propagator” hϕik ϕ`j i is given by the Gaussian integral Z 1 2 2 1 dϕ e−N tr 2 m ϕ ϕik ϕ`j = C1 δij δk` + C2 δik δj` Z where we will now fix the constants C1 and C2 by taking special cases of the above expression. First notice that trϕ2 = ϕij ϕji = ϕ211 + 2ϕ212 + ..., meaning that diagonal terms ϕ2ii (no sum) come in with a factor of 1, while off-diagonal terms ϕ2i,j6=i come in with a factor of 2. Let usR orient ourselvesp with a familiar example ordinary calculus. Define the integral Qfrom ∞ N − 21 αx2 = 2π/α . Then for Z ≡ i = 1 Z1N , we have Z1 ≡ −∞ dx e 1 Z

Z

N

x2 − 12 α~

d xe

∂ 1 ln Z1 δij = + δij . xi xj = −2 ∂α α

Now we will solve for the constants C1 and C2 in our matrix theory. First set i = k = j = ` = 1 to get Z 1 1 1 − 21 (N m2 )ϕ211 +... 2 dϕ e ϕ11 = = = C1 + C2 . Z α α = N m2 N m2 Next set i = j = 1, k = ` = 2. Paying special attention to the extra factor of 2 mentioned earlier, we obtain Z 1 1 1 1 2 2 dϕ e− 2 (2N m )ϕ12 +... ϕ212 = = = C1 . Z α α = 2N m2 2N m2 232

Therefore C2 =

1 N m2

−

1 2N m2

hϕik ϕ`j i ≡

1 Z

= Z

1 . 2N m2

We arrive at the somewhat intuitively clear result 1

dϕ e−N tr 2 m

2 ϕ2

ϕik ϕ`j =

1 (δij δk` + δik δj` ) 2N m2

so that the previous factor 1/(N m2 ) is split evenly among the two tensors dictated by group theory. Setting k = ` and summing over k gives h(ϕ2 )ij i =

1 1 1 (N δij + δik δjk ) = (N + 1)δij ≈ δij 2 2 2N m 2N m 2m2

where we have used the large N limit N + 1 ≈ N . Thus so far we have computed the n = 0 and the n = 1 terms in the expansion for Gij (z): Gij (z) =

∞ X n=0

1 z 2n+1

2n

h(ϕ )ij i =

1 1 1 + + ... δij . z z 3 2m2

As in the text, the next step is to study the planar diagrams contributing to the n = 2 term and thereby arrive at a quadratic equation for G(z). The first half of the algorithm (“repeat”) translates to the present case exactly as in the text, so that again we arrive at equation (7) on p. 399: 1 . G(z) = z − Σ(z) Next we look at Figure VII.4.6b on p. 400 and observe that in our case we have two terms, one which is depicted in the figure and another one for which the overarching propagator twists. In other words, we have 1 Σ(z) = G(z) 2m2 with the extra factor of 1/2 arising just as in the above calculation of h(ϕ2 )ij i. Together these result in the quadratic equation G(z) =

1 =⇒ G2 − 2m2 z G + 2m2 = 0 z − G(z)/(2m2 )

which has the solution r G(z) = m2 z −

z2 −

2 m2

! .

This matches equation (9) on p. 400 of the text, rescaled by m2 → 2m2 . We can understand this result in the following way: Let Φ = √12 (R + iS) be a complex matrix field. Then the action S ≡ m2 tr Φ† Φ = 21 m2 tr(R2 + S 2 ) is the action for two real matrix fields R and S. If Φ is Hermitian, then Φ† = Φ and therefore S = m2 tr Φ† Φ = m2 tr Φ2 . The action in the text is normalized as 21 m2 tr Φ2 , which is brought into the canonical normalization by the rescaling m2 → 2m2 . 233

5. For hermitean random matrices ϕ, calculate 1 1 1 1 1 1 1 1 Gc (z, w) ≡ tr tr − tr tr N z −ϕN w−ϕ N z−ϕ N w−ϕ for V (ϕ) = 12 m2 ϕ2 using Feynman diagrams. [Note that this is a much simpler object to study than the object we need to study in order to learn about localization (see exercise VI.6.1).] Show that by taking suitable imaginary parts we can extract the correlation of the density of eigenvalues with itself. For help, see E. Br´ezin and A. Zee, Phys. Rev. E51: p. 5442, 1995. Solution: We follow arxiv: cond-mat/9508135v2 9 Oct 1995. Let h...ic stand for the connected ensemble average, so that 1 1 1 1 tr tr Gc (z, w) = N z−ϕ N w−ϕ c ∞ 1 1 X D ϕ n ϕ m E tr tr = 2 N zw n,m=1 z w c ∞ D ϕ n ϕ m E X 1 tr = 2 ∂z ∂w tr . N z w c n,m = 1

In terms of diagrams, each tr(ϕn ) is a quark loop, and the average of tr(ϕn )tr(ϕm ) tells us that the two quark loops are to be attached to one another through gluon (double-line) propagators. Since the ensemble is assumed to be gaussian, V (ϕ) = 12 m2 ϕ2 , Wick’s theorem tells us that the average htr[(ϕ/z)n ]tr[(ϕ/w)m ]ic is equal to a product of free-field propagators, each consisting of one “z-type” ϕ and one “w-type” ϕ, that is one from each trace. Since hϕij i = 0, we conclude that only terms for which n = m give a nonzero result: ∞ ∞ D ϕ n ϕ m E X X 1 tr = tr htr(ϕn )tr(ϕn )ic . n z w (zw) c n=1 n,m = 1

The sum consists of two classes of diagrams, the first in which index contractions are made between the two traces only, and the second in which we allow contractions within the same trace. For the first class, consider an individual diagram for a fixed value of n, and draw the z-type quark loops inside the w-type quark loops. We have n different ways to attach the z-type loops to the w-type loop (using a double-line propagator connecting from the z-diagram to 234

one of the loops in the w-diagram). Once this choice is made, there is only one way to attach the rest of the z-loops to the w-loops without having the propagators cross. (Recall we are working in the large-N limit, in which planar diagrams dominate.) Thus each such diagram has a symmetry factor 1/n. Given the gluon propagator on p. 398, each attachment of a gluon propagator contributes a factor 1/(N m2 ), and each resulting closed loop contributes a factor of N . The factors of N cancel, and we obtain for the sum of these diagrams: n ∞ ∞ X X 1 1 1 1 n n htr(ϕ )tr(ϕ )ic = n n (zw) (zw) n m2 n=1 n=1 class 1 1 . = − ln 1 − zwm2 Now consider the second class of diagrams, in which we allow contractions within the same trace. In the large-N limit, these diagrams serve only to dress the bare propagator z1 to 1 (see p. 399). Therefore the full sum of diagrams can be the full propagator G(z) = z−Σ(z) achieved by taking the result from class 1 and simply replacing 1/z with G(z) and 1/w with G(w). We have 1 1 Gc (z, w) = − 2 ∂z ∂w ln 1 − 2 G(z)G(w) N m where (as on p. 400) m2 G(z) = 2

r z−

4 z2 − 2 m

! .

From this we may obtain the connected correlation between eigenvalues: ρc (E, E 0 ) = −

1 [Gc (+, +) + Gc (−, −) − Gc (+, −) − Gc (−, +)] 4π 2

where Gc (±, ±) ≡ limε,ε0 →0+ Gc (E ± iε, E 0 ± iε0 ). The result is ρc (E, E 0 ) = −

1 1 4 − m2 EE 0 p . 4π 2 N 2 (E − E 0 )2 (4 − m2 E 2 )(4 − m2 E 02 )

235

6. Use the Faddeev-Popov method to calculate J in the Dyson gas approach. Solution: We will follow footnote 31 on p. 84 of arXiv:hep-th/9304011v1. The Faddeev-Popov determinant (or Jacobian) J is defined by Z 1 ≡ J(ϕ) Dg δ (f (ϕg )) with ϕg = g † Λg. For the infinitesimal case we have g = e iθ = I + iθ + O(θ2 ), so ϕg = Λ − i[θ, Λ]. Moreover, we have for the commutator: [θ, Λ]ij = (θΛ)ij − (Λθ)ij =

N X

(θik λk δkj − λi δik θkj ) = (λj − λi )θij

k=1

(There are no sums on i and j in the last equality.) So if we choose our gauge transformation ϕg = g † Λg such that ϕg = Λ, or equivalently if we choose the gauge fixing function f (ϕ) = ϕ − Λ, then the integral over the delta function is: Z Z Dg δ (f (ϕg )) = Dθ δ(−i[θ, Λ]) YZ = d2 θij δ (2) ((λj − λi )θij )) ij

=

Y ij

1 (λj − λi )2

We wrote d2 θij for each matrix element θij because these generator matrices are complex, ∗ ). This and we integrate over their real and imaginary parts (or equivalently, over θij and θij generates the exponent 2 in the determinant. Therefore, the Faddeev-Popov determinant (Jacobian) is Y J= (λj − λi )2 ij

This is the result on p. 401 of the text.

236

7. For V (ϕ) = 12 m2 ϕ2 + gϕ4 , determine ρ(E). For m2 sufficiently negative (the double well potential again) we expect the density of eigenvalues to split into two pieces. This is evident from the Dyson gas picture. Find the critical value m2c . For m2 < m2c the assumption of G(z) having only one cut used in the text fails. Show how to calculate ρ(E) in this regime. Solution: p √ The function z 2 − a2 = (z − a)(z + a) has a single p branch cut of finite length along the real axis from −a to +a. More generally, the function (z − a)(z − b) with b > a has a single branchpcut of finite length along the real axis from a to b. From this it is clear that the function (z − a)(z − b)(z + c)(z + d) , with a, b, c, d real and positive and b > a, d > c, has two branch cuts of finite length along the real axis, one from a to b and one from −d to −c. In particular, −a, d = kc = −ka for which the function p consider the case b = ka, c = p (z − a)(z − ka)(z + a)(z + ka) = [z 2 − a2 ][z 2 − (ka)2 ] contains two disconf (z) ≡ nected branch cuts of length ka, one from a to ka and the other from −ka to −a. We know that the two cuts have the same length by the Z2 : z → −z symmetry of the potential V (z), hence the length ka for both of them. The parameter a fixes how far away from the origin the cuts begin. Together these constitute two unknown parameters. The function f (z) has the expansion 1 1 1 1 2 f (z) = z − C − D 2 + O , C = a2 (k 2 + 1) , D = a4 (k 2 − 1)2 4 z z 2 8 for z → ∞. Following the text, we are motivated to postulate a form G(z) =

1 0 [V (z) − P (z)f (z)] 2

where P (z) is a polynomial in z. Since V (z) = 21 m2 z 2 + gz 4 =⇒ V 0 (z) = m2 z + 4gz 3 , we must have P (z) at most linear in z to avoid the introduction of a z 4 term, which cannot be canceled by the cubic polynomial in V 0 (z). Moreover, P (z) should not have a constant term since that will result in a z 2 term, which also cannot be canceled by the odd polynomial V 0 (z). Thus we postulate the form P = Bz, which introduces a third unknown parameter. Requiring the cubic and linear terms in G(z → ∞) to vanish results in two equations. The requirement G(z → ∞) → (+1) z1 provides the third equation. The three equations for the three unknowns B, C, D imply B = 4g , C = −4g m2 , D =

237

1 . 2g

Notice that the definition C = 12 a2 (k 2 + 1) implies that C > 0, so that this solution is only valid for m2 < 0, as expected. Solving for a and k gives two solutions s s 1 ∓8g 3/2 m2 − (4g 3/2 m2 )2 − 1 a± = −4g m2 ± √ , k± = g 1 − (4g 3/2 m2 )2 where the ± signs are correlated, meaning that (a+ , k+ ) is one solution and (a− , k− ) is the second solution. The solution (a− , k− ) is only valid when |m2 | ≥ 1/(4g 3/2 ), which defines the critical value of m2 : 1 −m2c = 3/2 4g For |m2 | < |m2c |, the two-cut solution ceases to be valid. Let us make sure this is compatible with the solution for k− . Let q ≡ 4g 3/2 m2 < 0, so that 2 k− =

+2q − q 2 − 1 . 1 − q2

2 > 0 is satisfied if 1 − q 2 < 0, which again implies |m2 | > 1/(4g 3/2 ) with The condition k− 2 2 m = −|m | < 0. Thus we have two solutions for the function G(z): q 1 0 G+ (z) = V (z) − 4gz [z 2 − a2+ ][z 2 − (k+ a+ )2 2 q 1 0 2 2 2 2 V (z) − 4gz [z − a− ][z − (k− a− ) G− (z) = 2

where (a+ , k+ ) and (a− , k− ) are given above as functions of the parameters m2 < 0 and g > 0. The density of states ρ(E) = − π1 Im G(z) is given by p 2gE p[a2+ − E 2 ][(k+ a+ )2 − E 2 ] for a+ ≤ E ≤ k+ a+ ρ(E) = 2gE [a2− − E 2 ][(k− a− )2 − E 2 ] for a− ≤ E ≤ k− a− 0 otherwise We see that the density of states has split into two disconnected regions. For a treatment involving orthogonal polynomials, see N. Deo, “Multiple Minima in Glassy Random-Matrix Models,” J. Phys.: Condensed Matter, Vol. 12 No. 29, 24 July 2000.

238

8. Calculate the mass of the soliton (25). mS =

N mF π

(25)

Solution: After introducing the auxiliary field σ and integrating out the fermions the action is Z N S[σ] = − d2 x 2 [σ(x, t)]2 − i 12 N tr ln(−∂ µ ∂µ − ∂x σ − σ) 2g up to an overall additive constant. We know the shape of the time-independent soliton 2 configuration is given by σ(x) = m tanh(mx), where m = µe1−π/g , which satisfies the differential equation ∂x σ + σ = m2 . Thus the entire tr ln(...) term is merely an additive constant, and so the energy of the configuration is Z ∞ N M= dx 2 m2 C − tanh2 (mx) 2g −∞ where we have written explicitly the additive constant C, which can be thought of as a counterterm to set the vacuum energy to zero. Recall that the coupling constant g = g(µ) is a function of the renormalization point µ. Choosing µ = m implies g 2 = π, so that the energy is Z Nm ∞ du C − tanh2 u . M= 2π −∞ For a formal derivation of the constant C, see A. Klein, “Bound states and solitons in the Gross-Neveu model,” Phys. Rev. D, Vol. 14 No. 2, 15 Jul 1976. Here we will content ourselves with observing that limu→∞ tanh u = 1, so that to get a finite answer we must have C = 1. A physical way to see what is going on is to plot the function 1 − tanh2 u = sech2 u: Sech2 u

0.25

0.20

0.15

0.10

0.05

u -10

-5

5

10

We see that it is a peak localized atRthe origin, except that it asymptotes to zero only when ∞ C = 1. The value of the integral is −∞ du sech2 u = 2 tanh(∞) = 2, so we find M=

Nm . π

239

VII.5

Grand Unification

1. Write down the charge operator Q acting on 5, the defining representation ψ µ . Work out the charge content of the 10 = ψ µν and identify the various fields contained therein. Solution: The generator of electric charge is Q = T 3 + 21 Y , where T 3 is the third generator of SU (2). Acting on the ψ µ ∼ 5 of SU (5), we have 1 1 1 1 (− 3 )ψ −3 ψ 0 1 1 2 2 0 −3 ψ 3 (− 31 )ψ 3 µ 1 + ψ = (− )ψ 0 −3 [Qψ] = 4 3 4 1 1 ψ (+1)ψ +2 2 1 1 ψ5 −2 +2 (0)ψ 5 Recalling the definition in the text of the ψµ ∼ ¯5 in terms of the familiar fields at low energy d¯α ψµ = ν e you might worry about the charges of the lower two components of the ψ µ ∼ 5 worked out above. When working out the generator of electric charge for the ¯5, be careful to note that although the SU (2) generator T 3 is the same, the sign of the hypercharge generator flips sign: 1 1 0 +3 ψ1 (+ 3 )ψ1 0 ψ2 (+ 1 )ψ2 + 13 31 1 + ψ3 = (+ )ψ3 + 0 [Qψ]µ = 3 3 1 1 ψ4 (0)ψ4 − 2 2 − 12 − 12 (−1)ψ5 ψ5 So the electron field e = ψ5 indeed has charge −1, and the neutrino field ν = ψ4 is neutral. Now for the 10. Given a representation R and its generators TR , and given a second representation R0 and its generators TR0 , the generators of the product representation R ⊗ R0 are given by TR⊗R0 = TR ⊗ IR0 + IR ⊗ TR0 . Explicitly in terms of components, if µ, ν, ... denote the matrix indices of representation R and if a, b, ... denote the matrix indices of representation R0 , then (TR⊗R0 )µaνb = (TR )µ ν δ a b + δ µ ν (TR0 )a b . If the representations R0 and R are the same, then we can symmetrize and antisymmetrize the two upper indices (or the two lower indices). The generator of the symmetric product representation R ⊗S R is (TR⊗S R )µνρσ = (TR )(µρ δ ν)σ + δ (µρ (TR )ν)σ 240

where the parentheses mean symmetrization in those indices: 1 M (µν) ≡ (M µν + M νµ ). 2 Similarly, the generator of the antisymmetric product representation R ⊗A R is (TR⊗A R )µνρσ = (TR )[µρ δ ν]σ + δ [µρ (TR )ν]σ where the brackets mean antisymmetrization in those indices: 1 M [µν] ≡ (M µν − M νµ ). 2 The 10 of SU (5) is the antisymmetric product representation 5 ⊗A 5, so the generator of electric charge acting on the 10 representation is (Q10 )µνρσ = (Q5 )[µρ δ ν]σ + δ [µρ (Q5 )ν]σ where Q5 = diag(− 31 , − 13 , − 31 , +1, 0) is the generator of electric charge acting on the 5 of SU (5). Defining ψ 0µν = (Q10 )µνρσ ψ ρσ , we find 2 ψ 0αβ = (− )ψ αβ =⇒ ψ αβ = εαβγ u¯γ 3 045 ψ = (+1)ψ 45 =⇒ ψ 45 = e¯ . We also have ψ 0αi = (Q10 )αiβj ψ βj = (− 31 δ i j + δ i 4 δ 4j )ψ αj , so: 1 2 ψ 0α4 = (− + 1)ψ α4 = + ψ α4 =⇒ ψ α4 = uα 3 3 1 1 α5 0α5 ψ = (− + 0)ψ = − ψ α5 =⇒ ψ α5 = dα . 3 3 ν Actually there is a slight mistake here. The lepton doublet `i ≡ is defined correctly e c dα with a lower SU (2) index, since it is part of the ¯5 ψµ = . The usual Standard-Model `i u convention for the quark doublet is qi ≡ with a lower SU (2) index just like for the d d leptons. This implies q i = εij qj = , so that ψ α5 = dα as written, but ψ α4 = −uα with −u an extra minus sign arising from ε54 = −ε45 .

241

2. Show that for any grand unified theory, as long as it is based on a simple group, we have at the unification scale P 2 T 2 sin θ = P 32 Q where the sum is taken over all fermions. Solution: Consider the case for one fermion ψ that transforms under some representation of the simple group G. Its gauge-covariant derivative in the electroweak subgroup SU (2) ⊗ U (1) is √ 1 a ψ Dµ ψ = ∂µ − ig Aµ Ta + Bµ α Y 2 As discussed on p. 410 of the text for the particular case α = 3/5, we fix α by demanding that√the SU (2) and U (1) generators are normalized equally, or in other words tr[(T3 )2 ] = tr[( α 12 Y )2 ]. This implies tr[(T3 )2 ] α= tr[( 12 Y )2 ] √ √ We define the weak mixing angle by tan θ ≡ g1 /g2 = α , so since tan θ = s/c = s/ 1 − s2 , where c ≡ cos θ and s ≡ sin θ, we have s2 =

α . 1+α

The generator of electric charge is Q = T3 + 12 Y , so tr[( 12 Y )2 ] = tr(Q2 − 2QT3 + T32 ) and therefore tr(T32 ) s2 = . 2 tr[T3 (T3 − Q)] + tr(Q2 ) But T3 − Q = − 21 Y , and tr(T3 12 Y ) = 0. Therefore, for a single fermion, we have s2 =

tr(T32 ) tr(Q2 )

at the unification scale. In the case of multiple fermions, the currents in each direction of the Lie algebra get contributions from eachPfermion the group. In other words, we fix α P √ f that transforms (funder (f ) ) by f tr[(T3 )2 ] = f tr[( α 12 Y (f ) )2 ], where T3 and 12 Y (f ) are the generators appropriate for the representation to which fermion f belongs. Everything carries through as before, and we arrive at the result P (f ) 2 f tr[(T3 ) ] 2 s =P . (f ) )2 ] f tr[(Q

242

3. Check that the SU (3) ⊗ SU (2) ⊗ U (1) theory is anomaly-free. [Hint: The calculation is more involved than in SU (5) since there are more independent generators. First show that you only have to evaluate tr Y [Ta , Tb ] and tr Y 3 , with Ta and Y the generators of SU (2) and U (1), respectively.] Solution (due to J. Feinberg): Actually, we have to calculate slightly more than what is stated in the original problem. In addition to denoting SU (2) generators by T , denote normalized SU (3) generators as t. One thing we can note: since the charges are repeated in each generation of particles, we only need to calculate the anomaly cancellation for one of them. Note that when we say “tr” here, we mean to treat the gauge groups as tensor products (so the traces factorize) and to put an additional minus sign on the right-handed field traces. We start with anomalies involving SU (3). Since SU (3) couples equally to left and right handed particles, the one with 3 ts vanishes automatically. With only one t, one of the factors in the trace of generators is tr t = 0, so that always vanishes also. Now consider the traces with 2 ts. We have tr(ta tb )trT c = 0 and tr(ta tb )trY = 12 δ ab trq Y . The trace now runs only over the quarks because the leptons have t = 0. We have trq Y = 2 × ( 16 ) − 23 − (− 31 ) = 0. Note that the factor of 2 in the first term is because the left-handed quarks are in an SU (2) doublet, and that we have subtracted the right-handed ones. Now take anomalies involving SU (2) generators. Since the fermions are all in doublets or singlets, we can use the Pauli matrix anticommutator, which gives {T a , T b } = 12 δ ab . Thus, the 3T anomaly is tr({T a , T b }T c ) = 12 δ ab tr(T c ) = 0. Similarly, tr(Y 2 T a ) = trY 2 trT c = 0. The last case is with 2 T s, tr(Y T a T b ) = 12 δ ab trL Y , where the trace is now over left-handed fields. This gives trL Y = 3( 16 ) − 12 = 0. Note the factor of 3 because the quarks are in an SU (3) triplet. Finally, we have traces involving only hypercharge. This is trY 3 = 3 × 2 × ( 16 ) + 2(− 12 ) − 3( 32 ) − 3(− 31 ) − (−1) = 0. Technically, one should also calculate gravitational couplings, but we are not considering that in this model.

243

4. Construct grand unified theories based on SU (6), SU (7), SU (8), ... , until you get tired of the game. People used to get tenure doing this. [Hint: You would have to invent fermions yet to be experimentally discovered.] Solution: This is of course an open-ended problem with many solutions. We will simply point towards some of the literature on unified models based on SU (n) for n > 5: • SU (6): M. Singer and K. S. Viswanathan, “SU(6) Grand Unified Theory without Fundamental Scalars,” Phys. Rev. D Vol. 24 No. 11, 1 Dec 1981 • SU (7): H. Goldberg, T. W. Kephart and M. T. Vaughn, “Fractionally Charged ColorSinglet Fermions in a Grand Unified Theory,” Phys. Rev. Lett. Vol. 47, No. 20, 16 Nov. 1981 • SU (9): Y. Fujimoto and P. Sodano, “SU(9) Grand Unified Theory,” Phys. Rev. D Vol. 23 No. 7, 1 Apr. 1981 • SU (11): H. Georgi, “Towards a Grand Unified Theory of Flavor,” Nucl. Phys. B, 1979 • SU (15): P. Frampton and B. H. Lee, “SU(15) Grand Unification,” Phys. Rev. Lett. Vol. 64 No. 6, 5 Feb 1990

244

VII.6

Protons Are Not Forever

1. Suppose there are F 0 new families of quarks and leptons with masses of order M 0 . Adopting the crude approximation described in exercise IV.8.2 of ignoring these families for µ below M 0 and of treating M 0 as negligible for µ above M 0 , run the renormalization group flow and discuss how various predictions, such as proton lifetime, are changed. Solution: For energies below the scale M 0 we still have only F “massless” fermion families, so equations (1)-(3) on p. 414 become 0 1 1 M 1 = + (4F − 33) ln 0 αS (µ) αS (M ) 6π µ 0 2 2 0 sin θ(µ) sin θ(M ) 1 M = + (4F − 22) ln 0 α(µ) α(M ) 6π µ 0 2 2 0 cos θ(µ) cos θ(M ) 1 20 M = + F ln . 0 α(µ) α(M ) 6π 3 µ Above the scale M 0 we excite the new degrees of freedom, so for M 0 < µ < MGUT , these equations become exactly those on p. 414 except with F replaced by F + F 0 : 1 1 MGUT 1 0 = + [4(F + F ) − 33] ln αS (µ) αS (MGUT ) 6π µ sin2 θ(µ) sin2 θ(MGUT ) 1 MGUT 0 = + [4(F + F ) − 22] ln α(µ) α(MGUT ) 6π µ 2 2 cos θ(µ) cos θ(MGUT ) 1 20 MGUT = + (F + F 0 ) ln . α(µ) α(MGUT ) 6π 3 µ As discussed in the text, the couplings are assumed to unify at the scale of grand unification α(MGUT ) = αS (MGUT ) ≡ αGUT , and the angles are normalized by tan2 θ(MGUT ) = 3/5. The best data we have is provided at the mass of the Z boson, MZ = 91.188 GeV. We use the first set of equations to specify αS (M 0 ), α(M 0 ) and sin2 θ(M 0 ) in terms of the parameters evaluated at µ = MZ . We then use these values of the parameters at M 0 as inputs into the second set of equations, which run up to MGUT . This results in 1 1 1 MGUT 1 MGUT 0 = − (4F − 33) ln − 4F ln αS (MGUT ) αS (MZ ) 6π MZ 6π M0 sin2 θ(MGUT ) sin2 θ(MZ ) 1 MGUT 1 MGUT 0 = − (4F − 22) ln − 4F ln α(MGUT ) α(MZ ) 6π MZ 6π M0 cos2 θ(MGUT ) cos2 θ(MZ ) 1 20 MGUT 1 20 0 MGUT = − F ln − F ln . α(MGUT ) α(MZ ) 6π 3 MZ 6π 3 M0 You can see what is happening: if F 0 = 0, then we recover the equations on p. 414 with µ = MZ . The presence of the new families when F 0 6= 0 affects the running between the 245

scales M 0 and MGUT . See W. J. Marciano, “Weak mixing angle and grand unified gauge theories,” Phys. Rev. D, Vol. 20 No. 1, 1 July 1979 for running the renormalization group flow to compare SU (5) GUT predictions with experiment.

2. Work out proton decay in detail. Derive relations between the following decay rates: Γ(p → π 0 e+ ), Γ(p → π + ν¯), Γ(n → π − e+ ), and Γ(n → π 0 ν¯). Solution: We will use two-component spinor notation (see Appendix E). As explained on p. 407 in the text, we write all fields as left-handed with the following transformation properties under GSM ≡ SU (3)c ⊗ SU (2)W ⊗ U (1)Y : 2 1 1 1 ν u ¯ ∼ (1, 2, − ), e¯ ∼ (1, 1, 1) q≡ ∼ (3, 2, ), u¯ ∼ (¯3, 1, − ), d ∼ (¯3, 1, ), ` ≡ e d 6 3 3 2 The indices α, β, γ run from 1 to 3 and denote the 3-representation of SU (3)c , and the indices i, j run from 1 to 2 and denote the 2-representation of SU (2)W . The antisymmetric tensor εαβγ is invariant under SU (3)c and the antisymmetric tensor εij is invariant under SU (2)W . d¯α and (ψ10 )µν = −(ψ10 )νµ , whose components are The SU (5) fermion fields are (ψ¯5 )µ ≡ ` i α u and (ψ10 )ij = εij e¯. As pointed out in problem VII.5.1, (ψ10 )αβ = εαβγ u¯γ , (ψ10 )αi = qiα = dα we have (ψ10 )αi = εij (ψ10 )αj so that (ψ10 )α4 = ε45 dα = +dα and (ψ10 )α5 = ε54 uα = −uα . Let us now work out the new currents that arise from the SU (5) unified theory. The covariant derivative acting on a 5 of SU (5) is (Dψ5 )µ = ∂ψ5µ − ig X µν ψ5ν , X µν =

24 X

X a (T5a )µν .

a=1

At this point we should note an annoying circumstance: Since we have already used µ = 1, ..., 5 as an index for SU (5), α = 1, 2, 3 as an index for SU (3), i = 1, 2 as an index for SU (2) and a = 1, ..., 24 as an index for the adjoint representation of SU (5), we will denote Lorentz vector indices by M = 0, 1, 2, 3, and Lorentz spinor indices by m = 1, 2 and m ˙ = 1, 2 when we choose to display them explicitly. The covariant derivative acting on a ¯5 of SU (5) is ¯ µ ν ψ¯5 ν , X ¯µν = (Dψ¯5 )µ = ∂ψ¯5 µ − ig X

24 X a=1

246

X a (T¯5a )µν .

The adjoint components X a are the same for the 5 and for the ¯5. The generators of the ¯5 are T¯5a = −(T5a )∗ = −(T5a )T (since T5a are hermitian, we have (T5a )∗ = (T5a )T ). In components, this means (T¯5a )µν = −[(T5a )T ]µν = −(T5a )ν µ . Therefore the covariant derivative of a ¯5 is (Dψ¯5 )µ = ∂ψ¯5µ + ig

24 X

X a (T5a )ν µ ψ¯5ν

a=1

so the kinetic term in the Lagrangian for ψ¯5 is L = iψ¯5† Dψ¯5 = iψ¯5† ∂ψ¯5 − g

24 X

ψ¯5† µ X a (T5a )ν µ ψ¯5ν .

a=1

This form for the current term is awkward because the SU (5) indices are not in matrix multiplication order. To take care of this properly, we should display the Lorentz indices: a ˙ (ψ¯5† µ )m˙ σ ¯ M mm (T5a )ν µ (ψ¯5ν )m ψ¯5† µ X a (T5a )ν µ ψ¯5ν = XM †µ m a M a ν = − XM (ψ¯5ν )m σm ) m ˙ (T5 ) µ (ψ¯ 5

= − ψ¯5ν X a (T5a )ν µ ψ¯5† µ Now the SU (5) indices are in matrix multiplication order. The kinetic term in the Lagrangian for ψ¯5 is L = iψ¯5† ∂ψ¯5 + gψ¯5 Xψ¯5† P a where X = 24 a = 1 X , and now all indices can be suppressed without any confusion. The idea now is to see what this current implies for the low-energy fields d¯ and `. Since ψ¯5µ X µν ψ¯5† ν = ψ¯5α X αβ ψ¯5† β + ψ¯5α X αi ψ¯5† i + ψ¯5i X i α ψ¯5† α + ψ¯5i X i j ψ¯5† j and ψ¯5α = d¯α , ψ¯5i = `i , we can identify the new baryon-number-violating interactions from the ¯5 as =0 L∆B6 = g (d¯α X αi `†i + h.c.) ¯ 5 Now for the 10 of SU (5). The covariant derivative of ψ10 is µν (Dψ10 )µν = ∂ψ10 − ig

24 X

ρσ a µν X a (T10 ) ρσ ψ10

a=1

where as discussed in problem VII.5.1 the generators of the 10 = 5 ⊗A 5 are a µν (T10 ) ρσ = (T5a )[µρ δ ν]σ + δ [µρ (T5a )ν]σ

where we remind you of the notation f [µν] ≡ 21 (f µν − f νµ ). From the kinetic term L = † 1 iψ10 Dψ10 we therefore get the baryon-number-violating interactions 2 =0 † † L∆B6 = g[(ψ10 )αβ X αi ψ iβ + (ψ10 )αj X αi ψ ij + h.c.] 10

= g[εαβγ u¯†γ X αi (−εij qjβ ) + (εjk qα†k )X αi εij e¯ + h.c.] = g[−εij εαβγ u¯†γ X αi qjβ + qα†i X αi e¯ + h.c.] 247

Putting together the contributions from the ¯5 and the 10, we have ∆B6=0 α ¯ M †i ij †γ M β †i M L = g(XM )i dα σ ` − ε εαβγ u¯ σ ¯ qj + qα σ ¯ e¯ + h.c. ν Expanding out the SU (2) index with `i = , we arrive at the interactions e L∆B6=0 = g(XM )α4 d¯α σ M ν † − εαβγ u¯†γ σ ¯ M dβ + u†α σ ¯ M e¯ + h.c. ¯ M e¯ + h.c. ¯ M uβ + d† σ + g(XM )α d¯α σ M e† + εαβγ u¯†γ σ α

5

¯ † ) = −Q(d) + Q(ν) = Recall the electric charge of the down quark Q(d) = − 31 . Since Q(dν 1 + 3 + 0, invariance under electromagnetism tells us that the gauge boson (XM )α4 has electric ¯ † ) = −Q(d) − Q(e) = + 1 + 1 = + 4 , the gauge boson (XM )α has charge − 31 . Since Q(de 5 3 3 electric charge − 34 . Let us therefore write (X −1/3 )αM ≡ (XM )α4 and (X −4/3 )αM ≡ (XM )α5 and rewrite the above interactions as −4/3 α L∆B6=0 = g (X −1/3 )αM (J +1/3 )M )M (J +4/3 )M α + g (X α + h.c.

where we have defined the electrically charged currents ¯ M † (J +1/3 )M ¯†γ σ ¯ M dβ + u†α σ ¯ M e¯ α ≡ dα σ ν − εαβγ u (J +4/3 )M ≡ d¯α σ M e† + εαβγ u¯†γ σ ¯ M uβ + d† σ ¯ M e¯ . α

α

As a consistency check, we should verify the electric charges of all of the terms in the currents. We have Q(¯ u† d) = +Q(u)+Q(d) = + 32 − 31 = + 13 and Q(u† e¯) = −Q(u)−Q(e) = − 23 +1 = + 13 , ¯ † ) = + 1 . Also Q(¯ u† u) = 2Q(u) = + 34 and Q(d† e¯) = −Q(d) − Q(e) = which both match Q(dν 3 ¯ † ) = + 4 as they must. + 13 + 1 = + 43 , which match Q(de 3 Suppose the gauge bosons X −1/3 and X −4/3 have the same GUT-scale mass MX ∼ 1016 GeV so that they can be integrated out at low energies. Performing this integration gives the low-energy effective Lagrangian for baryon-number-violating processes such as proton decay: =0 L∆B6 = eff

g 2 +1/3 M −1/3 α −4/3 α (J )α (J )M + (J +4/3 )M )M . α (J 2 MX

The products of currents in this Lagrangian are J +1/3 J −1/3 = (d¯α σ M ν † − εαβγ u¯†γ σ ¯ M dβ + u†α σ ¯ M e¯)(νσM d¯†α − εαδ d†δ σ ¯M u¯ + e¯† σ ¯ M uα ) J +4/3 J −4/3 = (d¯α σ M e† + εαβγ u¯†γ σ ¯ M uβ + d† σ ¯ M e¯)(eσM d¯†α + εαδ u† σ ¯M u¯ + e¯† σ ¯ M dα ) α

δ

For this problem, we are interested in only those interactions that contribute to proton decay and neutron decay, so we are interested in operators of the schematic form ∼ qqq`. These are: J +1/3 J −1/3 = (d¯α σ M ν † )(−εαβγ d†β σ ¯M u¯γ ) + (−εαβγ u¯†γ σ ¯ M dβ )(¯ e† σ ¯M uα ) + h.c. and J +4/3 J −4/3 = (d¯α σ M e† )(+εαβγ u†β σ ¯M u¯γ ) + (+εαβγ u¯†γ σ ¯ M uβ )(¯ e† σ ¯M dα ) + h.c. 248

These can be simplified using the identities ˙ n n˙ ˙ ˙ (σ M )mm˙ (¯ σM )nn = 2 δm δm˙ and (¯ σ M )mm (¯ σM )nn = 2 εmn εm˙ n˙ ˙˙

where we use the metric η = (+, −, −, −) and the convention ε12 = ε12 = +1. Using these, we have (ξσ M ν † )(ψ † σ ¯M χ) = −2(ξχ)(ψ † ν † ) and (ξ † σ ¯ M ν)(ψ † σ ¯M χ) = +2(ξ † ψ † )(νχ) for any four Weyl spinors ξ, ν, ψ and χ. (Relations like these are called Fierz identities.) So the above products of currents simplify to u†γ e¯† )(dβ uα ) J +1/3 J −1/3 = +2 εαβγ (d¯α u¯γ )(d†β ν † ) − 2 εαβγ (¯ J +4/3 J −4/3 = −2 εαβγ (d¯α u¯γ )(u† e† ) + 2 εαβγ (¯ u†γ e¯† )(uβ dα ) . β

Adding these gives the low-energy effective Lagrangian i 2g 2 h ∆B6=0 u†γ e¯† )(dβ uα ) + h.c. . Leff = 2 εαβγ (d¯α u¯γ )(d†β ν † − u†β e† ) − 2 εαβγ (¯ MX We would now like to interpret this Lagrangian in terms of the low-energy hadron fields. As emphasized on p. 342, any two Lagrangians that exhibit identical symmetries must describe the same low-energy physics. This leads to the replacements29 2 i + π †α †β γ 3 0 1 ¯ εαβγ (d u¯ )(eu ) → µ e p + π n + √2 π p + O 2 f f 2 i π †γ † α β 3 † + † † 0 † 1 εαβγ (¯ u e¯ )(d u ) → µ e¯ p¯ − π n ¯ + √2 π p¯ + O 2 f f where f is the pion decay constant, µ is a strong-interaction parameter with dimensions of mass, π + is the charged pion, π 0 is the neutral pion, (p, p¯† ) is the proton and (n, n ¯ † ) is the neutron, and we drop terms of order π 2 /f 2 . Therefore, the part of the baryon-number violating effective Lagrangian relevant for proton decay is i + i + † ∆B6=0 0 † 0 † † 1 1 √ √ π n + 2 π p + 2¯ e p¯ − π n ¯ + 2 π p¯ Leff =λ e p+ + h.c. f f where we have defined the coupling µ3 λ ≡ 2g MX2 2

which has dimensions of mass. Since all calculations in the text are done with 4-component Dirac spinors instead of 2=0 component spinors, let us now rewrite L∆B6 in terms of Dirac spinors. Define eff e¯ p n c E ≡ † , P≡ , N ≡ e p¯† n ¯† 29

See the addendum to this chapter of solutions.

249

where E c is the Dirac field for the positron, or in other words the charge-conjugate of the e Dirac field E ≡ † for the electron. Since E c PL = ep and E c PR = e¯† p¯† , we have: e¯ i 0 ∆B6=0 c = λ E (PL + 2PR ) + π (PL − 2PR ) P + h.c. Leff 2fπ i +λ√ π + E c (PL − 2PR )N + h.c. 2 fπ where PL ≡ 12 (I − γ 5 ) and PR ≡ √ constant fπ ≡ f / 2 .

1 (I 2

+ γ 5 ), and we have defined a modified pion decay

We will first compute the rate for p → π 0 e+ . We see from the above Lagrangian that one contribution to the amplitude arises from the cubic π 0 e+ p vertex e+ = − 2f

p

P L − 2P R

0

There is also a contribution from the pion-nucleon interactions derived from the chiral Lagrangian: gA Lπ0 pn = ∂µ π 0 Pγ µ γ 5 P − N γ µ γ 5 N 2fπ where gA is the axial vector coupling. The contributing diagram to the proton decay amplitude is e+

p

0

=0 where the solid dot indicates the iλE c (PL + 2PR )P vertex from L∆B6 . Let k0 , k1 , k2 be eff the momenta of the proton, positron, and pion respectively. Then the second diagram contributes: gA 5 u¯1 [iλ (PL + 2PR )] [iSp (k1 )] − 6 k2 γ u0 2fπ λgA 6 k1 + mp =+ 6 k2 γ 5 u0 . u¯1 (PL + 2PR ) 2 2fπ k1 − m2p

Using γ 5 γ µ , we can move the 6 k1 to the left of PL + 2PR , picking up some minus signs. We will neglect the electron mass me mp , and so u¯1 6 k1 = u¯1 me ≈ 0. In the denominator, we also have k12 = m2e ≈ 0. Furthermore, we have k2 = k0 − k1 , so that again we can move 6 k1 to the left and get 250

zero when acted on by u¯1 . Finally, using PL γ 5 = −PL and PR γ 5 = +PR , we arrive at two powers of mp in the numerator to cancel out the 1/m2p from the propagator and obtain: −

λgA u¯1 (PL − 2PR )u0 2fπ

for the total contribution to the amplitude from the second diagram. This is exactly the same as the contribution from the π 0 e+ p vertex, up to a factor of gA . The amplitude is λ u¯1 (PL − 2PR )u0 . iM = −(1 + gA ) 2fπ Taking the magnitude squared, averaging over the proton spins and summing over the positron spins using the usual manipulations gives 1 2

X

|M|2 =

s0 ,s1

m2p − m2π 5 2 2 λ (1 + g ) A 32fπ2 mp me

where we have set me ≈ 0 in the numerator but kept the mass of the neutral pion. The decay rate is (p. 141) Z X d3 k2 d3 k1 4 4 1 (2π) δ (k − k − k ) |M|2 Γ= 0 1 2 2 3 3 (2π) (ω1 /me ) (2π) 2ω2 s ,s 0

1

p where ωi = |~pi |2 + m2i . The amplitude is a constant, and the 2-body phase space integral is given by equation (40) on p. 142 suitably adjusted to account for the fermion factor ω1 /me : 2 Z mp − m2π d3 k1 d3 k2 4 δ (k0 − k1 − k2 ) ≈ πme (ω1 /me ) 2ω2 m2p where again we have taken me ≈ 0 wherever possible. The decay rate is therefore Γ(p → e+ π 0 ) =

5 λ2 (1 + gA )2 (m2p − m2π )2 . 128π fπ2 m3p

Actually this is still not quite complete, since the coefficients of the hadronic operators in the effective Lagrangian are subject to renormalization group corrections. See F. Wilczek and A. Zee, “Operator analysis of nucleon decay,” Phys. Rev. Lett. Vol. 43 No. 21, 19 Nov 1979. Using the effective Lagrangian, the other decay rates p → ν¯π + , n → e+ π − and n → ν¯π 0 may be computed in exactly the same way. For general relations between the rates, see problems VIII.3.3 and VIII.3.4.

251

3. Show that SU (5) conserves the combination B −L. For a challenge, invent a grand unified theory that violates B − L. Solution: We follow F. Wilczek and A. Zee, “Conservation or Violation of B − L in Proton Decay,” Phys. Lett. Vol. 88B, No. 3,4 17 Dec 1979. The Higgs field ϕµ ∼ 5 of SU (5) couples to the fermion fields ψµ ∼ ¯5 and ψ µν ∼ 5 ⊗A 5 = 10 through the Yukawa interactions given in equations30 (11) and(12) on p. 417: L = −f1 ψµ Cψ µν ϕ†ν − f2 ψ µν Cψ λρ ϕσ εµνλρσ where f1 and f2 are coupling constants. These terms (and the rest of the Lagrangian) are invariant under an accidental U (1) symmetry. Let us call this symmetry U (1)X with generator X. Then to leave the Yukawa terms invariant, we require X(¯5) + X(10) − X(ϕ) = 0 and 2X(10) + X(ϕ) = 0, where X(¯5) is the U (1)X charge of the field ψµ ∼ ¯5, and so forth. So X(ϕ) = −2X(10) and thus X(¯5) = −3X(10). Thus choosing the normalization convention X(10) = 1, the Lagrangian is invariant under U (1)X if the fields have charges X(ψµ ) = −3, X(ψ µν ) = +1, X(ϕµ ) = −2. After ϕµ acquires a vacuum expectation value, the U (1)X symmetry gets broken, but a subgroup remains α unbroken. Recall that in terms of low-energy fields, the SU (5) fermions u d¯α α αβ αβγ α are (ψ¯5 )µ ≡ for the ¯5, and for the 10: (ψ10 ) = ε u¯γ , (ψ10 )i = qi = dα `i and (ψ10 )ij = εij e¯. Again recalling problem VII.5.1, we have (ψ10 )αi = εij (ψ10 )αj so that (ψ10 )α4 = ε45 dα = +dα and (ψ10 )α5 = ε54 uα = −uα . We can thereby evaluate the generator 21 Y of hypercharge on the ψµ and ψ µν to find that the combination X + 4(Y /2) generates a symmetry even after spontaneous symmetry breaking of SU (5). The other conserved generators are Q, the generator of electric charge, and {T a }8a = 1 , the generators of color SU (3). By explicitly evaluating the generator X + 4(Y /2) on the components of ψµ and ψ µν , we will find that X + 4(Y /2) is a multiple of B − L: [X [X [X [X [X

+ 4( 12 Y )]d¯ = [−3 + 4(+ 13 )]d¯ = 5(− 13 )d¯ (← B = − 13 , L = 0 X) + 4( 12 Y )]` = [−3 + 4(− 12 )]` = 5(−1)` (← B = 0 , L = +1 X) + 4( 12 Y )]¯ u = [+1 + 4(− 23 )]¯ u = 5(− 13 )¯ u (← B = − 13 , L = 0 X) + 4( 21 Y )]q = [+1 + 4(+ 16 )]q = 5(+ 13 )q (← B = + 13 , L = 0 X) + 4( 12 Y )]¯ e = [+1 + 4(+1)]¯ e = 5(+1)¯ e (← B = 0 , L = −1 X)

Therefore, B − L = 15 [X + 4( 12 Y )]. A deeper understanding of this seemingly accidental symmetry can be obtained by embedding the SU (5) theory into SO(10), as explained in the next chapter. One finds that B − L 30

We write ϕ†µ ≡ (ϕµ )† ∼ ¯ 5 to emphasize the complex conjugation.

252

is a generator of the SO(10) theory due to the presence of the right-handed gauge-singlet neutrino which can be assigned a lepton number of −1. One way to violate B − L in the SU (5) theory is to add another Higgs field, H, transforming as H µν ∼ 5 ⊗A 5 = 10. The field H couples to fermions through the Yukawa interaction L = fIJ ψIµ CψJν (H † )µν where the repeated indices I, J label the different families31 . At this stage H could simply be assigned a charge under X such that B − L remains conserved, but in the absence of further restrictions there is a cubic scalar interaction: L = µ H µν H ρσ ϕλ εµνρσλ where µ is a coupling with dimensions of mass. The clash between these two terms violates B − L by two units and thus implies (B − L)-violating processes such as n → µ− K + and p → µ− K + π + .

Addendum: Chiral Lagrangian for SU (3) ⊗ SU (3) Here we review the nonlinear sigma model used to parameterize low-energy QCD, also known as the chiral Lagrangian. For more details, see M. Claudson and M. Wise, “Chiral Lagrangian for deep mine physics,” Nucl. Phys. B195 (1982) 297-307 and O. Kaymakcalan, L. ChongHuah and K. C. Wali, “Chiral Lagrangian for proton decay,” Phys. Rev. D, Vol. 29 No. 9, 1 May 1984. The Lagrangian for three generations of massless quarks is L=i

3 X

qi† Dq 6 ¯ i + q¯i† D¯ 6 ¯ qi

i=1

¯ s¯) are the two-component spinors for the up, down and where q = (u, d, s) and q¯ = (¯ u, d, µ strange quarks, and D 6¯ ≡ σ ¯ Dµ is the gauge-covariant derivative for each quark field. This Lagrangian exhibits the global symmetry SU (3)L ⊗ SU (3)R , under which the quarks transform as q ∼ (3, 1) and q ∼ (1, ¯3). We will use the indices A = 1, 2, 3 and A0 = 1, 2, 3 to label the 3-dimensional representations of SU (3)L and SU (3)R respectively. We also choose upper indices for the fundamental, 3, and lower indices for the anti-fundamental, ¯3. There is also the global U (1)V symmetry (q, q¯† ) → e−iθ (q, q¯† ). The axial U (1)A : (q, q¯) → e−iθ (q, q¯) is anomalous. 31

Since H µν = −H νµ , the coupling f must be antisymmetric: fIJ = −fJI . Thus this interaction couples one generation of fermions to another. This is reminiscent of the interaction fab h+ εij `ai `bj present in the Zee model of neutrino masses. See problem VIII.3.2.

253

As discussed in the text, the QCD vacuum is supposed to break the global symmetry SU (3)L ⊗ SU (3)R down to the diagonal subgroup SU (3)V through the formation of a chiral condensate: hq A q¯A0 i = −v 3 δ AA0 where v is a positive parameter with dimensions of mass, and the minus sign arises from the fact that fermions contribute negative energy to the vacuum. On general principles, breaking a global symmetry group G down to a subgroup H results in massless excitations that parameterize the coset space G/H. In other words, breaking SU (3)L ⊗ SU (3)R → SU (3)V results in Nambu-Goldstone bosons that parameterize the “axial” SU (3) coset space SU (3)L ⊗ SU (3)R = SU (3)A . SU (3)V The broken group SU (3)A has 32 − 1 = 8 generators. For each broken generator there is one massless particle. These are the parity-odd spinless mesons: 1 + + √ π 0 + √1 η π K 2 π− 6 − √12 π 0 + √16 η K0 Π= . 2 0 − ¯ K − √6 η K Each field parameterizes a fluctuation along a particular direction in SU (3)A , generated by T a = 21 λa , where λa are the eight Gell-Mann matrices: i 0 0 1 0 0 −i σ 0 λi = , λ4 = 0 0 0 , λ5 = 0 0 0 0 0 1 0 0 i 0 0 1 0 0 0 0 0 0 0 0 1 λ6 = 0 0 1 , λ7 = 0 0 −i , λ8 = √ 0 1 0 . 3 0 0 −2 0 1 0 0 i 0 The σ i are the Pauli matrices: 0 1 0 −i 1 0 1 2 3 σ = , σ = , σ = . 1 0 i 0 0 −1 We may map a spacetime point xµ to any point on the group manifold of SU (3)A by treating the components of the meson octet as fields and exponentiating them to form a 3-by-3 special unitary spacetime-dependent matrix: Σ(x) ≡ e i2Π(x)/f where f is the pion decay constant, f ≈ 139 MeV. Under a general SU (3)L ⊗ SU (3)R transformation, the matrix Σ transforms as Σ → LΣR† , where L is an element of SU (3)L and R is an element of SU (3)R . The diagonal subgroup SU (3)V is given by transformations for 254

which L = R. It will prove convenient to define the “square root” of the matrix Σ as ξ(x) ≡ eiΠ(x)/f . Under a general SU (3)L ⊗ SU (3)R transformation, the matrix ξ is defined to transform as ξ → LξU † = U ξR† where U is a 3-by-3 unitary matrix that is defined by the above transformation law and thereby depends nonlinearly on the matrices L, R and Π. This ensures that Σ = ξ 2 transforms as stated. After seeing how the mesons arise from chiral symmetry breaking, we need the spectrum of baryons at low energy.32 The quark fields transform under SU (3)L ⊗ SU (3)R as u u¯ q ≡ d ∼ (3, 1) and q¯ ≡ d¯ ∼ (1, ¯3) . s s¯ Therefore q¯† transforms as (1, 3), and q¯† q¯† ∼ (1, 3) ⊗ (1, 3) = (1, ¯3A ) ⊕ (1, 6S ) . A baryon is a color-singlet bound state of three quarks, so the color indices of the above product of quarks are to be contracted with the 3-index antisymmetric tensor of SU (3)c . The Lorentz-singlet part33 of q¯† q¯† forms a color anti-triplet †β †γ d¯ s¯ †β †γ 1 Qα ≡ 2 εαβγ q¯ q¯ = s¯†β u¯†γ u¯†β d¯†γ which transforms as (1, ¯3) under global SU (3)L ⊗ SU (3)R transformations. (Here an upper α = 1, 2, 3 denotes the fundamental of color and a lower index α denotes the antifundamental.) Thus the color-singlet composite field α †β †γ u (d¯ s¯ ) uα (¯ s†β u¯†γ ) uα (¯ u†β d¯†γ ) ~ α )T = εαβγ dα (d¯†β s¯†γ ) dα (¯ s†β u¯†γ ) dα (¯ u†β d¯†γ ) ~q α (Q sα (d¯†β s¯†γ ) sα (¯ s†β u¯†γ ) sα (¯ u†β d¯†γ ) transforms as (3, ¯3) under global SU (3)L ⊗SU (3)R transformations. From the electric charges (+ 32 , − 13 , − 31 ) of (up, down, strange), the electric charges of the components of the above composite field are 0 +1 +1 −1 0 0 . −1 0 0 32

We thank Tim Tait and Mark Srednicki for helpful discussions on this. We are considering the 2-component Lorentz-spinor indices contracted as u ¯† d¯† ≡ u ¯†a˙ d¯†a˙ ≡ u ¯†a˙ εa˙ c˙ d¯†c˙ , so that u ¯† d¯† = d¯† u ¯† for Grassmann-valued fermion fields u ¯†a˙ and d¯†a˙ . 33

255

At low energy, the QCD vacuum breaks SU (3)L ⊗ SU (3)R → SU (3)V , under which (3, ¯3) → 3 ⊗ ¯3 = 1 ⊗ 8. We therefore find an octet of baryons34 1 + √ Σ0 + √1 Λ0 Σ p 2 Σ− 6 − √12 Σ0 + √16 Λ0 n B= Ξ− Ξ0 − √26 Λ0 ~ α )T − 1 Itr[~q α (Q ~ α )T ] → µ3 B, where µ is a parameter with given by the replacement ~q α (Q 3 dimensions of mass and I is the 3-by-3 identity matrix. For example, the proton and neutron are p ∼ u(¯ u† d¯† ) and n ∼ d(¯ u† d¯† ), as they should be. ¯ which can be derived analogously. The baryon There is also the octet of anti-baryons, B, B Dirac field is given by B = ¯ † . B Under a general SU (3)L ⊗ SU (3)R transformation, the baryon octet transforms as B → U BU † , where again the matrix U is defined by the transformation law ξ → LξU † = U ξR† . Therefore, we have ξBξ → L ξBξ R† so that ξBξ transforms as Σ = ξ 2 . Similarly, ξ † Bξ † transforms as Σ† . The next step is to write all possible interaction terms between baryons and mesons that respect SU (3)L ⊗ SU (3)R symmetry and parity. The result is ¯ ∂ − m)B] LπB = 18 f 2 tr(∂µ Σ∂ µ Σ† ) + tr[B(i6 ¯ µ (ξ∂µ ξ † + ξ † ∂µ ξ)B + B(∂µ ξ ξ † + ∂µ ξ † ξ) + 12 i tr Bγ µ 5 ¯ γ B(∂µ ξ ξ † − ∂µ ξ † ξ) − 21 i(D − F ) tr Bγ µ 5 ¯ γ (ξ∂µ ξ † − ξ † ∂µ ξ)B . + 1 i(D + F ) tr Bγ 2

The factor of 18 f 2 is so that the meson fields that comprise Σ = e i2Π/f have canonically normalized kinetic terms, e.g. 12 ∂µ π 0 ∂ µ π 0 and ∂µ π + ∂ µ π − . The parameters F = 0.44 and D = 0.81 are measured from semileptonic baryon decays. Expanding LπB using ξ = I + iΠ/f + ... and the components of the baryon octet, we find interactions between the protons, neutrons and pions: LπB = 21 ∂µ π 0 ∂ µ π 0 + ∂µ π + ∂ µ π − + P(i6 ∂ − m)P + N (i6 ∂ − m)N i D+F h 1 √ ∂µ π 0 Pγ µ γ 5 P − N γ µ γ 5 N + ∂µ π + Pγ µ γ 5 N + h.c. + + ... 2 f where the “...” stand for terms involving strange mesons and baryons, as well as terms of higher order in ∂µ π/f . It is often customary to define the axial vector coupling gA = D + F 34

We also find an SU (3)V -singlet, electrically-neutral baryon where we have suppressed the color indices.

256

√1 3

~ T) = tr(~q Q

√1 3

[u(d¯† s¯† )+d(¯ s† u ¯† )+s(¯ u† d¯† )],

√ and a modified pion decay constant fπ = f / 2 , so that the coupling of the neutral pion to the proton and neutron can be written as gA ∂µ π 0 Pγ µ γ 5 P − N γ µ γ 5 N 2fπ gA = ∂µ π 0 Ψγ µ γ 5 I3 Ψ , fπ 1 P +2 0 where Ψ ≡ is the nucleon doublet, and I3 ≡ is the third generator of N 0 − 12 isospin. Lπ0 pn =

Introducing nonzero quark masses results in nonzero masses for the meson octet and splits the degeneracy among the masses in the baryon octet. For further details, see the references.

VII.7

SO(10) Unification

1. Work out the Clifford algebra in d-dimensional space for d odd. Solution: The Clifford algebra for d = 2n + 1 is that of d = 2n with the addition of the γ FIVE from d = 2n. For example, consider d = 3. The gamma matrices for d = 2 are just γ 1 = σ 1 and γ 2 = σ 2 as explained in the text. The chirality matrix γ FIVE ≡ −iγ 1 γ 2 = −iσ 1 σ 2 = σ 3 anticommutes with γ 1 and γ 2 , and it squares to 1. Therefore the Clifford algebra for d = 3 is {γ i , γ j } = 2δ ij , γ i = σ i which you already encountered way back in problem II.1.12, where you discovered that the Dirac mass term in (2 + 1)-dimensional spacetime violates parity and time reversal. In general, the γ FIVE for d = 2n anticommutes with the {γ i }2n i = 1 from d = 2n (since 2n is always even) and squares to 1, and therefore will form a perfectly good (2n + 1)th gamma matrix γ 2n+1 ≡ γ FIVE ≡ (−i)n γ 1 γ 2 ...γ 2n for d = 2n + 1. The point is that while the spinor representation of SO(2n) is reducible into two chiral irreducible representations, the spinor representation of SO(2n + 1) is not reducible.

2. Work out the Clifford algebra in d-dimensional Minkowski space. Solution: The defining equation of the Clifford algebra in d-dimensional Minkowski space is {γ µ , γ ν } = 2η µν , η µν = diag(+1, −1, −1, −1, ..., −1)

257

where µ, ν = 0, 1, 2, ..., d − 1. Let i, j = 1, 2, ..., d − 1 denote purely spatial indices. Then the above can be written as (γ 0 )2 = +1, (γ i )2 = −1, {γ i , γ j } = −2δ ij All we have to do is to take the Clifford algebra for d-dimensional Euclidean space and throw in some factors of i to generate the appropriate minus signs in the metric. Immediately we find j j 0 0 γSO(d−1,1) = γSO(d) , γSO(d−1,1) = iγSO(d) for j = 1, ..., d − 1. Note that we are using slightly different notation for the SO(d) gamma matrices from the text. In the chapter, the vector indices run from 1 to d, whereas here our indices run from 0 to d − 1. If you want to stick to the notation in the book, you can write j j+1 0 1 = γSO(d) , γSO(d−1,1) = iγSO(d) γSO(d−1,1)

for j = 1, ..., d − 1. 3. Show that the Clifford algebra for d = 4k and for d = 4k + 2 have somewhat different properties. (If you need help with this and the two preceding exercises, look up F. Wilczek and A. Zee, Phys. Rev. D25: 553, 1982.) Solution: FIVE The = σ3 ⊗...⊗σ3 acts on a chiral spinor |ε1 , ..., εn i as γ FIVE |ε1 , ..., εn i = matrix γ Q chirality n j = 1 εj |ε1 , ..., εn i. The “left-handed” chiral spinor |ε1 , ..., εn iL is defined by

γ FIVE |ε1 , ..., εn i = (−1)|ε1 , ..., εn i, and the “right-handed” chiral Qspinor|ε1 , ..., εn iR is defined n determines whether by γ FIVE |ε1 , ..., εn iR = (+1)|ε1 , ..., εn iR . That is, the sign of j = 1 εj the spinor is left-handed or right-handed. Now, the charge conjugation matrix C = iσ2 ⊗ ... ⊗ iσ2 acts on the SO(2n) chiral spinor |ε1 , ..., εn i as C|ε1 , ..., εn i = |(−ε1 ), ..., (−εn )i, which means that the chirality matrix γ FIVE acts on a charge-conjugated spinor as γ FIVE C|ε1 , ..., εn i = γ FIVE |(−ε1 ), ..., (−εn )i =

n Y

! (−εj ) |(−ε1 ), ..., (−εn )i = (−1)n C|ε1 , ..., εn i .

j =1

Let’s process what this means. Consider the case for which n is even, so that (−1)n = +1. For that case, if γ FIVE |ε1 , ..., εn i = +γ FIVE C|ε1 , ..., εn i, then γ FIVE C|ε1 , ..., εn i = +C|ε1 , ..., εn i. If γ FIVE |ε1 , ..., εn i = −γ FIVE C|ε1 , ..., εn i, then γ FIVE C|ε1 , ..., εn i = −C|ε1 , ..., εn i. That is, |ε1 , ..., εn i and C|ε1 , ..., εn i have the same eigenvalue under γ FIVE ; the states |ε1 , ..., εn iL and |ε1 , ..., εn iR are therefore self-conjugate (that is, not conjugate to each other) if n is even.

258

Now consider the case for which n is odd, meaning (−1)n = −1. Under that assumption, we have the opposite situation from before. If γ FIVE |ε1 , ..., εn i = +γ FIVE C|ε1 , ..., εn i, then γ FIVE C|ε1 , ..., εn i = −C|ε1 , ..., εn i. If γ FIVE |ε1 , ..., εn i = −γ FIVE C|ε1 , ..., εn i, then we have γ FIVE C|ε1 , ..., εn i = +C|ε1 , ..., εn i. That is, |ε1 , ..., εn i and C|ε1 , ..., εn i have opposite eigenvalues under γ FIVE ; the states |ε1 , ..., εn iL and |ε1 , ..., εn iR are therefore conjugate to each other if n is odd. Rephrasing somewhat, let d = 2n. If n is even, then n = 2k for some integer k. We have shown that the chiral spinors are self-conjugate when n is even, or in other words when d = 4k = 0 mod 4. If n is odd, then n = 2k + 1 for some integer k. We have shown that the chiral spinors are conjugate to each other when n is odd, or in other words when d = 4k + 2 = 2 mod 4. As an important aside, note that the situation is reversed for SO(d − 1, 1), that is for d-dimensional Minkowski spacetime rather than d-dimensional Euclidean spacetime. For d = 4k (that is, d = 0 mod 4), the two chiral spinor representations are conjugate to each other. For d = 2 + 4k (d = 2 mod 4), each chiral spinor is its own conjugate. If you desire a second reference for this problem (and other properties of SO(2n) and SO(2n − 1, 1) spinors), consult Volume II, Appendix B of J. Polchinski’s textbook on string theory.

4. Discuss the Higgs sector of the SO(10). What do you need to give mass to the quarks and leptons? Solution: Let ψa ∼ 2n−1 (with a = 1, ..., 2n−1 ) be the SO(2n) spinor that contains a single family of matter fields. (We will specialize to n = 5 later.) The SO(2n) invariant tensors at our disposal are the charge conjugation matrix C ab and its inverse Cab , and the gamma matrices (γ µ )ab , where µ = 1, ..., 2n denotes the index for the vector (“defining”) representation of SO(2n). Note that once we pick the convention for which the spinor ψ has one index down, the rest of the index placements are fixed. For example, an SO(2n) transformation acts as µν ψ → e iωµν σ ψ = ψ + iωµν σ µν ψ + O(ω 2 ), so δψa = iωµν (σ µν )ab ψb . Since σ µν = 2i [γ µ , γ ν ], the index placements must be (γ µ )ab . We can raise and lower indices using C and C −1 , which is the whole point of defining such a matrix in the first place. The conventions we use are that indices are raised and lowered by contracting with C or C −1 always on the second index: ψ a ≡ C ab ψb and ψa ≡ Cab ψ b . This will introduce some minus signs, since in SO(2n) we have the property C T = (−1)n(n+1)/2 C (A11) For a thorough discussion of this point as well as for other aspects of SO(2n), consult the 259

paper “Families from Spinors,” Phys. Rev. D25: 553, 1982 by F. Wilczek and A. Zee, as referenced in problem VII.7.3. (We label the above equation as (A11) since that is the equation number in the reference.) To write a mass term for the fermions contained within the spinor ψa , we need a term quadratic in ψ that is invariant under SO(10) (and under the Lorentz group). For a single family of fermions, we are restricted to terms of the form L = −y ψa C ab (γ µ1 ...γ µK )b c ψc φµ1 ...µK + h.c. where φµ1 ...µK is a Lorentz-scalar field that is completely antisymmetric in its SO(2n) vector indices. (y is just a coupling constant.) We will suppress the Lorentz-spinor indices using the conventions in Appendix E. From the property (A11), raising and lowering a pair of up-and-down indices introduces the sign (−1)n(n+1)/2 , meaning: ψ a ψa = (−1)n(n+1)/2 ψa ψ a , where we remind you that ψ a ≡ C ab ψb and ψb ≡ Cbc ψ c . This means that raising and lowering two pairs of up-and-down indices can always be done without changing a sign regardless of the value of n. In other words, for any matrix Ma b , we have ψ a Ma b ψb = ψa M ab ψ b . So the above Lagrangian can equally well be written as L = −y ψ a Cab (γ µ1 ...γ µK )b c C cd ψd φµ1 ...µK + h.c. The matrix sandwiched between the ψs is C −1 (γ µ1 ...γ µK )T C. We can use the property C −1 (γ µ )T C = (−1)n γ µ

(A10)

to simplify this expression: C −1 (γ µ1 ...γ µK )T C = C −1 (γ µK )T ...(γ µ1 )T C = C −1 (γ µK )T CC −1 ...CC −1 (γ µ1 )T C = (−1)Kn γ µK ...γ µ1 = (−1)Kn (−1)K−1 γ µ1 γ µK γ µK−1 ...γ µ2 = (−1)Kn (−1)K−1 (−1)K−2 γ µ1 γ µ2 γ µK γ µK−1 ...γ µ3 = ...(keep on anticommuting)... PK−1

= (−1)Kn (−1)

j=1

j µ1

γ ...γ µK

= (−1)Kn (−1)K(K−1)/2 γ µ1 ...γ µK

Therefore: ψ a Cab (γ µ1 ...γ µK )b c C cd ψd = (−1)Kn+K(K−1)/2 ψ a (γ µ1 ...γ µK )ad ψd Now recall what we said about contracting indices with the second index of the charge conjugation matrix: ψ a ≡ C ab ψb = ψb C ab = ψb (−1)n(n+1)/2 C ba . 260

Finally, after relabeling some dummy indices, we have the result ψ a Cab (γ µ1 ...γ µK )b c C cd ψd = (−1)Kn+K(K−1)/2+n(n+1)/2 ψ a Cab (γ µ1 ...γ µK )b c C cd ψd In other words, the mass term for ψ can only be nonzero when the quantity Kn +

(n + K)(n + K + 1) K(K − 1) n(n + 1) + = −K 2 2 2

is an even number. (This is (A55) in the reference.) Since odd+odd = even and even+even = even while odd+even = odd, this implies that K and 12 (n + K)(n + K + 1) are both even or are both odd. Since odd×odd = odd, odd×even = even and even×even = even, we will always have (n + K)(n + K + 1) = even. So the question is where 12 (n + K) is even or odd. n+K = even ≡ 2p =⇒ K = 4p − n 2 n+K = odd ≡ 2q + 1 =⇒ K = 4q + 2 − n 2 where p and q are any integers, and of course we must have K > 0. The case at hand is SO(10), which means n = 5. This implies 5+K = even K = 4p − 5 = 3, 7, 11, ... 2 5+K K = 4p + 2 − 5 = 1, 5, 9, ... = odd 2 Moreover, recall that SO(10) has the invariant tensor εµ1 ...µ10 . That means any number of antisymmetrized indices greater than 5 is equivalent to a number less than 5. For example, the definition φµ1 = εµ1 µ2 ...µ10 φµ2 ...µ9 is a statement invariant under SO(10), so K = 9 is equivalent to K = 1. So the above is really 5+K K = 4p − 5 = 3 = even 2 5+K K = 4p + 2 − 5 = 1, 5 = odd 2 for SO(10). However, the first of these statements is a contradiction. Recall that we said we must have 21 (5 + K)(6 + K) and K either both even or both odd. For K = 3, we have 1 (5 + 3)(6 + 3) = 12 × 8 × 9 = 36, which is even. Finally we have deduced that only the terms 2 for which K = 1 or 5 are allowed in the Lagrangian. (This is the conclusion reached below (A56b) in the reference.) In other words, the masses for one family of fermions comes from the Lagrangian L = −y ψCγ µ ψ φµ − y 0 ψCγ µ1 ...γ µ5 ψ φ0µ1 ...µ5 + h.c. where φ and φ0 are Lorentz-scalar fields, and y and y 0 are coupling constants. From looking at the indices (and recalling that they are all implicitly antisymmetrized), we see the transformation properties φµ ∼ 10 and φ0µ1 ...µ5 ∼ 10 ⊗A 10 ⊗A 10 ⊗A 10 ⊗A 10 = 126 of SO(10). 261

Note that for multiple families of fermions, the analysis becomes more complicated. The mass terms have the form X L=− yF F 0 ψF Cγ µ1 ...γ µK ψF 0 φµ1 ...µK + h.c. F,F 0

where F and F 0 run from 1 to the number of fermion families. Lorentz-scalars φ satisfying the previous constraints would contribute terms for which yF F 0 = yF 0 F , and those not satisfying those constraints would contribute terms for which yF F 0 = −yF 0 F . 5. The group SO(6) has 6(6 − 1)/2 = 15 generators. Notice that the group SU (4) also has 42 − 1 = 15 generators. Substantiate your suspicion that SO(6) and SU (4) are isomorphic. Identify some low dimensional representations. Solution: The iterative construction of the spinor representation of SO(2n) is given on p. 422. In particular, for n = 3 we obtain the 23 -dimensional spinor representation of SO(6), which is generated by the 21 (6 × 5) = 15 hermitian matrices σ µν = i 12 [γ µ , γ ν ], with gamma matrices given by: µ γ(4) 0 µ µ = 0,1,2,3 γ = γ(4) ⊗ σ3 = µ 0 −γ(4) 0(4) I(4) 4 γ = I(4) ⊗ σ1 = I(4) 0(4) 0(4) −iI(4) 5 γ = I(4) ⊗ σ2 = . +iI(4) 0(4) This representation is reducible, since we can define the matrix γ FIVE ≡ −iγ 0 γ 1 γ 2 γ 3 γ 4 γ 5 FIVE = +i(γ(4) ⊗ σ34 )(I(4) ⊗ σ1 σ2 ) FIVE = +i(γ(4) ⊗ I)(I(4) ⊗ iσ3 ) FIVE = −γ(4) ⊗ σ3 FIVE −γ(4) 0(4) = FIVE 0(4) +γ(4)

which commutes with σ µν . Thus if Ψ transforms as a 23 -dimensional spinor, the chiral spinors ΨL ≡ 21 (1 − γ FIVE )Ψ and ΨR ≡ 12 (1 + γ FIVE )Ψ have 23−1 = 4 components and transform irreducibly under SO(6). Since the chiral spinors are 4-dimensional, the Lorentz generators σ µν acting on them should split up into two sets of fifteen 4-by-4 traceless hermitian matrices. The generators of SU (4) 262

are precisely the set of fifteen 4-by-4 traceless hermitian matrices, so that constructing the irreducible rotation generators acting on the chiral spinor would constitute a proof that SO(6) is locally isomorphic to SU (4). The rotation generators σ µν take the following forms in terms of 4-dimensional matrices. For µ, ν = 0, 1, 2, 3 only, we have: µ ν ⊗ σ3 , γ(4) ⊗ σ3 ] σ µν = i 21 [γ µ , γ ν ] = i 12 [γ(4) µ µ ν ν γ(4) ⊗ σ32 − γ(4) γ(4) ⊗ σ32 = i 21 γ(4) µ ν , γ(4) ]⊗I = i 12 [γ(4) µν σ(4) 0(4) µν = σ(4) ⊗ I = (µ, ν = 0, 1, 2, 3 only) . µν 0(4) σ(4)

For µ = 0, 1, 2, 3, we also have: σ

µ4

µ 4

= iγ γ = i =

µ iγ(4)

µ γ(4)

⊗ iσ2 =

⊗ σ3

0(4) µ −iγ(4)

I(4) ⊗ σ1 µ iγ(4) (µ = 0, 1, 2, 3 only) 0(4)

and σ

µ5

µ 5

µ γ(4)

⊗ σ3 I(4) ⊗ σ2 µ 0(4) γ(4) µ = iγ(4) ⊗ (−iσ1 ) = (µ = 0, 1, 2, 3 only) . µ γ(4) 0(4)

= iγ γ = i

Finally, we also have σ 45 = iγ 4 γ 5 = i I(4) ⊗ σ1 I(4) ⊗ σ2 −I(4) 0(4) = iI(4) ⊗ (iσ3 ) = . 0(4) +I(4) Let us now examine the chiral spinors in more detail. Consider the left-handed chiral spinor ΨL ≡ PL Ψ, where 1 FIVE (I + γ(4) ) 0(4) pR 0 FIVE 1 2 (4) )= ≡ PL ≡ 2 (I − γ . 1 FIVE 0(4) (I − γ(4) ) 0 pL 2 (4) The 8-dimensional spinor Ψ is therefore projected to a 4-dimensional as expected, spinor, ψ but the components are arranged slightly nontrivially. Write Ψ = , where ψ and ψ 0 ψ0 transform as (3 + 1)-dimensional Dirac spinors, which in turn reduce into two 2-dimensional chiral spinors: 0 χα χα 0 ψ= , ψ = . χ¯α˙ χ¯0α˙ 263

In terms of these, the SO(6) left-handed chiral spinor is 0 χ¯α˙ pR ψ ΨL = = 0 χ0β . pL ψ 0 So if we want to extract the parts of the rotation generators σ µν that act only on these components, then we may consider the matrices σLµν ≡ PL σ µν PL . It is now helpful to write the Dirac matrices for (3 + 1)-dimensional spacetime in terms of the Pauli matrices: 0 (σ µ )αβ˙ µ γ(4) = ˙ (¯ σ µ )αβ 0 ˙

˙ where numerically we have (σ µ )αβ˙ = (I, i~σ ) and (¯ σ µ )αβ ≡ εαβ εα˙ β (σ µ )β β˙ = (I, −i~σ ).

For µ, ν = 0, 1, 2, 3, we have: µν σ(4) 0(4) µν σ = µν 0(4) σ(4) 1 µ ν ¯ − σν σ ¯µ) 0(2) 0(2) 0(2) i 2 (σ σ σµσν − σ ¯ ν σµ) 0(2) 0(2) 0(2) i 12 (¯ . = 0(2) 0(2) i 21 (σ µ σ ¯ ν − σν σ ¯µ) 0(2) 1 µ ν ν µ 0(2) 0(2) 0(2) i 2 (¯ σ σ −σ ¯ σ ) Therefore, the middle 4-by-4 block of σLµν (for µ, ν = 0, 1, 2, 3 only) is given by the matrix 1 µ ν i 2 (¯ σ σ −σ ¯ ν σµ) 0(2) µν W = . 0(2) i 21 (σ µ σ ¯ ν − σν σ ¯µ) The middle 4-by-4 block of σLµ4 (for µ = 0, 1, 2, 3 only) is given by the matrix 0(2) +i¯ σµ µ X = −iσ µ 0(2) and that of σLµ5 is given by

0(2) σ ¯µ Y = . σ µ 0(2) µ

Finally, the middle 4-by-4 block of σL45 is given by −I(2) 0(2) Z= . 0(2) +I(2) The 12 (4 × 3) = 6 matrices W µν , the 4 matrices X µ , the 4 matrices Y µ and the matrix Z constitute 6+4+4+1 = 15 traceless hermitian 4-by-4 matrices. These therefore generate SU (4), 264

and we have proven that SO(6) and SU (4) are locally isomorphic. We have shown explicitly that the 4-dimensional representation of SU (4) is equivalent to the 6-dimensional representation of SO(6); in other words, the spinor of SO(6) is the fundamental of SU (4). Now we identify a few higher dimensional representations. = 6 of SU (4). Let Let ψA ∼ 4 and ψ A ∼ 4¯ of SU (4). Then ψ[AB] ∼ 4 ⊗A 4 ∼ 4×3 2 µ ψ ∼ 6 of SO(6). We have shown that the spinor of SO(6) is the fundamental of SU (4), and now we discover that the antisymmetric tensor of SU (4) is the fundamental of SO(6). Explicitly, we can construct a symbol Σµ[AB] out of the familiar Pauli matrices which is invariant under SO(6) ' SU (4) transformations on all of its indices simultaneously.35 = 10 of SU (4). Recall that SU (4) has the 4-index We also have ψ(AB) ∼ 4 ⊗S 4 ∼ 4×5 2 ABCD invariant tensor ε , so that 6 and ¯6 are equivalent representations (that is, two lower indices are group-equivalent to two upper indices): ψ AB = 21 εABCD ψCD . We also have ψAB ∼ 4 ⊗ ¯4 = 1 ⊕ 15, where the singlet is the trace tr ψ = δBA ψAB and the 15 is comprised of the traceless hermitian matrices displayed previously, or in other words the adjoint representation of SU (4). Also ψ [µν] ∼ 6 ⊗A 6 ∼ 6×5 = 15 of SO(6), which is the set 2 of traceless antisymmetric 6-by-6 matrices and therefore the adjoint of SO(6). Indeed, the adjoint of SO(6) is equivalent to the adjoint of SU (4), as we have just proven by showing that the two groups are locally isomorphic.

6. Show that (unfortunately) the number of families we get in SO(18) depends on which subgroup of SO(8) we take to be hypercolor. Solution: Let us use the notation SR to denote the “right-handed” spinor S, by which we mean the chiral spinor with eigenvalue +1 under the γ FIVE matrix, and SL to denote the chiral spinor with eigenvalue −1 under the γ FIVE matrix. As shown in the reference for problem VII.7.3 (equations A20a, b), the chiral spinors 2n+m−1 and 2n+m−1 of SO(2n + 2m) decompose under R R restriction to the subgroup SO(2n) ⊗ SO(2m) as m−1 m−1 2n+m−1 → (2n−1 ) ⊕ (2n−1 ) R R , 2R L , 2L n+m−1 n−1 m−1 n−1 m−1 2L → (2R , 2L ) ⊕ (2L , 2R )

For n = 5, m = 4 these become 256R → (16R , 8R ) ⊕ (16L , 8L ) 256L → (16R , 8L ) ⊕ (16L , 8R ) 35

See the appendix of http://arxiv.org/pdf/0902.0981.

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for the restriction of SO(18) to the subgroup SO(10) ⊗ SO(8). Let us choose the 256R as the spinor in which to place all of the fermions. We want a subgroup GHC of SO(8) such that 8L decomposes into a bunch of stuff that is confined, meaning non-singlets of GHC , while 8R decomposes into a bunch of free particles that do not participate in these strong “hypercolor” interactions, meaning we want 8R to yield singlets of GHC . Upon restriction to the special unitary subgroup SU (4) of SO(8), we have 8R → [0] ⊕ [2] ⊕ [4] = [0] ⊕ [2] ⊕ [0] 8L → [1] ⊕ [3] = [1] ⊕ [¯1] where [k] means the antisymmetric tensor with k indices of the 4-representation (the defining representation) of SU (4). Since SU (4) has the invariant tensor εABCD (where A, B, C, D are 4-rep indices, and therefore each runs from 1 to 4), the antisymmetrized 4-index tensor [4] is a singlet under SU (4). That is, ψ [ABCD] ∼ εABCD ψ ABCD ∼ 1 of SU (4). Similarly, ψ [ABC] ∼DABC ψ ABC ∼ ψD ∼ 4 of SU (4), or in other words [3] ∼ [¯1]. (We use the ∼ notation to mean “transforms as”.) In any case, we see that the restriction SO(8) → SU (4) exhibits the desired properties: 8R contains SU (4)-singlets, while 8L does not. At this stage, we see that the SO(10) ⊗ SU (4) spinors (16R , [0])1 and (16R , [0])2 (where “1” and “2” are merely labels) constitute two copies of the spinor S + of equation (14) on p. 425 of the text – that is, we have discovered exactly two families of fermions, which upon the decomposition SO(10) → SU (5) → SU (3)c ⊗ SU (2)W ⊗ U (1)Y become two families of the low-energy matter fields of the Standard Model. Since in fact we observe three families, the theory as it stands with GHC = SU (4) is wrong, or incomplete. We must break SU (4) down further. Let ψA transform as a 4 of SU (4). Suppose we decide to restrict SU (4) to the subgroup SU (2) ⊗ SU (2), meaning 2×2 special unitary matrix O 4×4 special unitary matrix → O 2×2 special unitary matrix The index A = 1, 2, 3, 4 thus breaks up into an index a = 1, 2 to denote transformation under the upper 2 × 2 block, and an index a˙ = 1, 2 to denote transformation under the lower 2 × 2 block. Thus under the restriction SU (4) → SU (2) ⊗ SU (2), we have ψA → ψa ⊕ ψa˙ . In other words, [1] → [1, 0] ⊕ [0, 1]. Now consider the antisymmetric two-index tensor ψ[AB] ∼ 4 ⊗A 4 of SU (4). If we restrict to SU (2) ⊗ SU (2) in exactly the same way, we see ψ[AB] → ψ[ab] ⊕ ψaa˙ ⊕ ψ[a˙ b]˙ , or in other words [2] → [2, 0] ⊕ [1, 1] ⊕ [0, 2]. However, recall that SU (2) has the invariant tensor εab , so actually ψ[ab] ∼ εab ψab ∼ 1 is invariant under SU (2). Thus [2] → [0, 0] ⊕ [1, 1] ⊕ [0, 0], and we have discovered two more singlets! Therefore, if we take GHC = SU (2) ⊗ SU (2), the spinors (16R , [0])1 , (16R , [0])2 , (16R , [0])3 , (16R , [0])4 266

where the old guys “1” and “2” appeared at SO(8) → SO(4), while the new guys “3” and “4” appeared at SO(4) → SU (2) ⊗ SU (2), comprise four families of SO(10) spinors. With GHC = SU (2) ⊗ SU (2) ⊂ SO(8), we predict the three observed fermion families plus an additional as of yet unobserved fourth generation. At this point we see how the game is played and simply quote from section III of the reference: “...the number of V − A families is ‘predicted’ to be two, three, four, or five, respectively, according to whether SO(8) is broken down to SO(6), SO(5), SO(4), or SO(3). [These orthogonal subgroups are not embedded in SO(8) in the obvious way, however.]” For further details, consult the literature.

7. If you want to grow up to be a string theorist, you need to be familiar with the Dirac equation in various dimensions but especially in 10. As a warm up, study the Dirac equation in 2-dimensional spacetime. Then proceed to study the Dirac equation in 10-dimensional spacetime. Solution: The salient feature will be whether the solutions to the Dirac equation can be classified as Weyl, Majorana, or both.36 In general, let Ψ denote a Dirac spinor in any dimension. Then a Weyl spinor is one that satisfies Ψ = ΨL or R , where L, R denote chiral projections, and a Majorana spinor is one that satisfies Ψ = Ψc , where c denotes the charge conjugation operation in whichever dimension is relevant. First consider d = 1 + 1. The gamma matrices can be taken as 0 1 0 1 0 1 γ = , γ = −1 0 1 0 which satisfy the Clifford algebra {γ µ , γ ν } = 2η µν with η = diag(−, +). The Dirac operator i6 ∂ ≡ iγ µ ∂µ is diagonal: ∂0 − ∂1 0 i6 ∂ = − 0 ∂0 + ∂1 ψ and therefore, writing Ψ = ˜ , we see that the two spinors ψ, ψ˜ do not mix under the ψ Dirac equation and therefore can be taken as independent degrees of freedom. 0 1 If we define the chiral projector γ FIVE ≡ γ γ = diag(+1, −1), then wecan define the 0 ψ , which satisfy chiral spinors ΨL ≡ 21 (I − γ FIVE )Ψ = ˜ and ΨR ≡ 12 (I + γ FIVE )Ψ = 0 ψ 36

We follow http://www.kitp.ucsb.edu/ joep/JLBS.pdf as well Appendix B of String Theory, Volume II by J. Polchinski.

267

γ FIVE ΨL = (−1)ΨL and γ FIVE ΨR = (+1)ΨR . The Lorentz generators in the spinor representation are σ µν ≡ −i 41 [γ µ , γ ν ], which in our case reduces to σ 01 = −i 12 γ 0 γ 1 ∝ γ FIVE , so that [γ FIVE , σ 01 ] = 0. Therefore the spinors ΨL,R transform irreducibly under the Lorentz group. Therefore, given a Dirac spinor Ψ, we may impose a Weyl condition Ψ = ΨL or Ψ = ΨR in two spacetime dimensions. Furthermore, since it is possible to choose a basis in which the Dirac operator is purely ˜ real (as we have done), it is consistent to impose that its solutions are real: ψ ∗ = ψ, ψ˜∗ = ψ, ∗ or in other words Ψ = Ψ. Thus, if we define the charge-conjugated Dirac field Ψc ≡ Ψ∗ as well as the left-chiral Dirac field ΨL ≡ 21 (I − γ FIVE )Ψ, then we find it is consistent to impose the condition Ψ = Ψc = ΨL . A spinor satisfying this condition is called a Majorana-Weyl spinor. (We can also use ΨR instead of ΨL .) Before jumping to d = 9 + 1, let us compare this briefly with what we know from d = 3 + 1. We learned in Chapter II (and as explained in Appendix E), a Dirac spinor Ψ in (3 + 1)dimensional spacetime transforms reducibly as 4Dirac = 2L ⊕ 2R : ψα Ψ= χ¯α˙ where ψ ∼ (2, 1) and χ¯ ∼ (1, 2) of SU (2) L ⊗SU (2)R ' SO(3,1). It is therefore consistent to ψα 0 impose the Weyl condition Ψ = ΨL = or Ψ = ΨR = using the gamma matrix 0 χ¯α˙ conventions given in Chapter II. However, we also learned that in (3 + 1)-dimensional spacetime, hermitian conjugation interchanges the two chiral representations. In other words, ψα ∼ (2, 1) =⇒ (ψα )† = (ψ † )α˙ ∼ (1, 2), and χ¯α˙ ∼ (1, 2) =⇒ (χ¯α˙ )† = (χ¯† )α ∼ (2, 1). (In other words, 2R = ¯2L .) Therefore it is not possible to impose a condition such as “ψ † = ψ” or “χ¯† = χ” ¯ in d = 3 + 1. However, it is possible to impose the condition χ¯† = ψ. In other words, if we define the conjugate spinor † (χ¯ )α c Ψ ≡ (ψ † )α˙ then it is possible to impose the Majorana condition Ψ = Ψc . The key point is that it is not possible to impose both the Weyl and Majorana conditions ψ 0 simultaneously. For example, if Ψ = ΨL = , then Ψc = 6= Ψ. Conversely, if 0 ψ† 268

ψ ψ Ψ=Ψ = 6= Ψ. In d = 3 + 1, a Weyl spinor cannot be Majorana, † , then ΨL = ψ 0 and a Majorana spinor cannot be Weyl. c

Now consider d = 9 + 1. As explained in the text, the gamma matrices can be constructed iteratively starting from d = 1 + 1: 0 1 0 1 0 1 γ(d=2) = , γ(d=2) = −1 0 1 0 −1 0 0 1 0 −i µ = 0,...,7 µ 8 9 γ(d=10) = γ(d=8) ⊗ , γ(d=10) = I16 ⊗ , γ(d=10) = I16 ⊗ 0 1 1 0 i 0 d/2 where I16 is the 16 = 24 -dimensional identity matrix. (Thegamma matrices have size 2 , 0 1 which for d = 10 is 25 = 32. For instance, the matrix I16 ⊗ is indeed 32-by-32.) 1 0

Define the chiral projection matrix γ FIVE ≡ γ 0 γ 1 γ 2 ...γ 9 . The chiral spinor ΨL ≡ 12 (1−γ FIVE )Ψ satisfies γ FIVE ΨL = (−1)ΨL , and the chiral spinor ΨR ≡ 12 (1 − γ FIVE )Ψ satisfies γ FIVE ΨR = (+1)ΨR . The question now is whether these irreducible chiral representations are self conjugate or conjugate to each other. Under Lorentz transformations, the Dirac spinor Ψ transformas as µν

Ψ → e iωµν σ Ψ = (I + iωµν σ µν + ...)Ψ where σ µν ≡ −i 41 [γ µ , γ ν ] generates Lorentz transformations in the 25 = 32-dimensional (reducible) spinor representation of SO(9, 1). If we can find a matrix B such that Bσ µν B −1 = −(σ µν )∗ , then the spinor B −1 Ψ∗ transforms under infinitesimal Lorentz transformations as: B −1 Ψ∗ → B −1 (I + iωµν σ µν )∗ Ψ∗ = B −1 (I − iωµν (σ µν )∗ )Ψ∗ = (I + iωµν σ µν )B −1 Ψ∗ which is identical to the transformation property of Ψ. Thus the (reducible) Dirac representation Ψ is self-conjugate. The question is whether its irreducible chiral components are also self-conjugate. By explicitly anticommuting the gamma matrices, we see that the matrix B = γ 3 γ 5 γ 7 γ 9 has the desired property: Bσ µν B −1 = −(σ µν )∗ . Now that we have an explicit form for B, we can compute its effect on the chiral projection matrix γ FIVE : Bγ FIVE B −1 = +γ FIVE . −1 ∗ Therefore, if Ψ = ΨL so that γ FIVE Ψ = (−1)Ψ, then ΨB L ≡ B ΨL is also left-handed: B γ FIVE ΨB L = (−1)ΨL . Thus the left-handed chiral spinor is self-conjugate. Similarly, the right-handed chiral spinor is also self-conjugate. With an extra factor of γ 0 , we define the charge conjugation matrix C ≡ Bγ 0 so that 269

¯ T = BΨ∗ denotes the charge conjugate spinor. We have therefore learned that, just Ψc ≡ C Ψ like d = 1 + 1 but unlike d = 3 + 1, it is consistent to impose the condition Ψ = ΨL = Ψc in d = 9 + 1 dimensions. Again, a spinor satisfying this condition is called a Majorana-Weyl spinor.

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VIII

Gravity and Beyond

VIII.1

Gravity as a Field Theory and the Kaluza-Klein Picture

1. Work out T µν for a scalar field. Draw the Feynman diagram for the contribution of one-graviton exchange to the scattering of two scalar mesons. Calculate the amplitude and extract the interaction energy between two mesons sitting at rest, thus deriving Newton’s law of gravity. Solution: For this and the next problem, we will need to vary the determinant of the metric. −1 δg)

det(g + δg) = det g det(1 + g −1 δg) = det g etr ln(1+g −1 = det g etr(g δg) = det g 1 + tr(g −1 δg)

So defining as usual δ det g = det(g + δg) − det g, we get δ det g = det g tr(g −1 δg). It will be convenient to vary with respect to g −1 rather than g, so to relate the two consider the definition gg −1 = 1. Varying this yields δgg −1 + gδg −1 = 0 =⇒ δg = −g δg −1 g. Therefore we have δ det g = − det g tr(δg −1 g) √ We actually need δ − det g , which is: δ

1 1p − det g = √ δ det g = − − det g tr(δg −1 g) 2 2 − det g

p

Just to clarify, tr(δg −1 g) = δg µν gµν . Now we are ready for the problem. The action for a real scalar field φ is Z p 1 S = d4 x − det g L , L = g µν ∂µ φ ∂ν φ − V (φ) . 2 Treating this as a function of g −1 , the variation δS ≡ S[g −1 + δg −1 ] − S[g −1 ] is Z h p i p 4 δS = d x δ − det g L + − det g δL Z p 1 1 4 = d x − det g − gµν L + ∂µ φ ∂ν φ δg µν . 2 2 So the stress tensor is Tµν ≡ 1 µν g ∂µ φ ∂ν φ − V (φ). 2

√

2 − det g

δS/δg µν = ∂µ φ ∂ν φ − gµν L with the Lagrangian L =

For a free field theory, V (φ) = 12 m2 φ2 , so the stress tensor is 1 Tµν = ∂µ φ ∂ν φ − gµν g ρσ ∂ρ φ ∂σ φ − m2 φ2 . 2 271

Now we need the weak field action for gravity coupling to matter. This is given in equation (10) on p. 437, which we repeat below: Z 1 µν λ µ λ ν µν 4 1 1 ∂λ h ∂ hµν − 2 ∂λ hµ ∂ hν − h Tµν S= dx 2 32πG √ √ Define MP ≡ 1/ 16πG and rescale the graviton as hµν → M1P 2 hµν to obtain Z 1 µν λ µ λ ν 4 µν 1 1 S = d x 2 ∂λ h ∂ hµν − 2 ∂λ hµ ∂ hν − √ h Tµν 2 MP We have the interaction term Sint = − √21M P space vertex represented pictorially as

R

d4 x T µν hµν , which leads to the momentum

k

k’ which is equal to

−i µ 0ν k k + k ν k 0µ − η µν (k · k 0 + m2 ) 2 MP where both momenta flow into the vertex. √

We also have the graviton propagator k

which, in the harmonic gauge, is equal to i∆µν,λσ (k) ≡

i (ηµλ ηνσ + ηµσ ηνλ − ηµν ηλσ ) . 2k 2

Consider the case of two different scalar fields φ1 and φ2 , each with its own stress tensor. We want to compute the gravitational potential between a φ1 particle and a φ2 particle both sitting at rest, or in other words in the extreme nonrelativistic limit. In this limit, only the t-channel diagram for φ1 φ2 → φ1 φ2 2-to-2 scattering contributes. Also, the 00-component of the stress tensor dominates, so the vertex becomes −i 0 00 ~ ~ 0 √ k k + k · k − m2 δ0µ δ0ν 2 MP The amplitude is iM = 272

p 1’

p1 q

p 2’

p2

−i 0 0 −i 0 0 =√ −p1 p10 − p~1 · p~10 − m21 [i∆00,00 (q)] √ −p2 p20 − p~2 · p~20 − m22 2 MP 2 MP −i = 2 2 2 (p01 p010 + p~1 · p~10 + m21 )(p02 p020 + p~2 · p~20 + m22 ) 2 MP q where q = p1 − p10 = p20 − p2 is the momentum transfer. The minus signs on the momenta in the second line come from the fact that in the diagram, one momentum flows into the vertex while the other flows out. In the definition of the vertex given above, both momenta flow into the vertex. If the two incoming particles are at rest, then pµ1 = (m1 , ~0) and pµ2 = (m2 , ~0), which implies (p01 p010 + p~1 · p~10 + m21 )(p02 p020 + p~2 · p~20 + m22 ) = (2m21 )(2m22 ) = 4m21 m22 . In the denominator we have q 2 = (p1 − p10 )2 = p21 + p210 − 2p1 · p10 q 2 2 = m1 + m1 − 2m1 |~p10 |2 + m21 |~p10 |2 |~p10 |4 2 2 = 2m1 − 2m1 1 + +O 2m21 m41 = −|~p10 |2 + (higher order)

The amplitude is M=

m21 m22 . MP2 |~p10 |2

Scattering from a potential in nonrelativistic quantum mechanics gives37 Z 1 d3 q i~q·~x V (~x ) = − e M(~q ) 4m1 m2 (2π)3 where ~q = p~1 − p~10 = −~p10 is the transfer of 3-momentum. Since Z d3 q i~q·~x 1 1 e = 3 2 (2π) |~q | 4π|~x | 37

See Section 4.5.7 of http://arxiv.org/abs/0812.1594 and the textbook M. Maggiore, A Modern Introduction to Quantum Field Theory (Oxford University Press, Oxford, UK, 2005) pp. 167-170 .

273

we have

1 V (~x ) = − 4m1 m2

m21 m22 MP2

1 4π|~x |

=−

m1 m2 . 16πMP2 |~x |

The Planck mass is defined in terms of Newton’s constant as MP2 = 1/(16πG), so we get V (~x ) = −

Gm1 m2 |~x |

which is Newton’s inverse square law. Note that various conventions exist for defining the Planck mass MP in terms of Newton’s constant G, and so various factors of 2 can appear between references. For a discussion of graviton scattering in various contexts, see K. Hinterbichler, “Theoretical aspects of massive gravity,” arXiv:1105.3735v1 [hep-th].

2. Work out T µν for the Yang-Mills field. Solution: The Lagrangian for the Yang-Mills field is 1 1 1 a a Fρσ L = − tr(Fµν F µν ) = − g µρ g νσ tr(Fµν Fρσ ) = − g µρ g νσ Fµν 2 2 4 R √ Varying the action S = d4 x − det g L gives with respect to the inverse metric gives Z p 1 1 4 a aµν νσ a a µρ a a δS = − d x − det g − gαβ Fµν F + g Fαν Fβσ + g Fµα Fρβ δg αβ 4 2 a a the last two terms are the same, so we get = −Fνµ From the antisymmetry Fµν

1 Yang-Mills a a a F aµν Tαβ = g µν Fαµ Fβν − gαβ Fµν 4 a a a Note that since γ αβ g µν Fαµ Fβν = Fµν F aµν and g αβ gαβ = 4, this stress tensor is in fact traceless.

274

3. Show that if hµν does not satisfy the harmonic gauge, we can always make a gauge transformation with εν determined by ∂ 2 εν = ∂µ hµν − 12 ∂ν hλλ so that it does. All of this should be conceptually familiar from your study of electromagnetism. Solution: Using the gauge transformation hµν = h0µν + ∂µ εν + ∂ν εµ , we have: 1 1 µ µ λ 0λ ∂µ hµν − ∂ν hλλ = ∂µ (h0µ ν + ∂ εν + ∂ν ε ) − ∂ν (hλ + 2∂λ ε ) 2 2 1 0µ 2 0λ = ∂µ hν − ∂ν hλ + ∂ εν + ∂µ ∂ν εµ − ∂ν ∂λ ελ 2 1 0µ 2 = ∂µ hν − ∂ν h0λ λ + ∂ εν 2 We can choose ∂ 2 εν = ∂µ hµν − 12 ∂ν hλλ to make h0 satisfy the harmonic gauge.

4. Count the number of degrees of polarization of a graviton. [Hint: Consider a plane wave hµν (x) = hµν (k)e ikx just because it is a bit easier to work in momentum space. A symmetric tensor has 10 components and the harmonic gauge kµ hµν = 12 kν hλλ imposes 4 conditions. Oops, we are left with 6 degrees of freedom. What is going on?] [Hint: You can make a further gauge transformation and still stay in the harmonic gauge. The graviton should have only 2 degrees of polarization.] Solution: Suppose we have already imposed the condition k µ hµν = 12 kν hλλ as indicated, so that we have 6 degrees of freedom. Consider the gauge transformation hµν = h0µν − kµ εν − kν εµ . Putting this into the above condition gives 1 k µ (h0µν − kµ εν − kν εµ ) = kν (h0λλ − 2k λ ελ ) 2 1 k µ h0µν − k 2 εν − kν k µ εµ = kν h0λλ − kν k λ ελ 2 1 k µ h0µν = kν h0λλ 2 since k 2 = 0. The function h0µν also satisfies the harmonic gauge for arbitrary εµ , so we have not completely used up our gauge freedom. Indeed, we still have 4 functions to choose, namely {εµ }3µ=0 , which cuts the number of physical degrees of freedom from 6 to 2. Thus the graviton has 2 physical degrees of freedom.

275

5. The Kaluza-Klein result that we argued by symmetry considerations can of course be derived explicitly. Let me sketch the calculation for you. Consider the metric ds2 = gµν dxµ dxν − a2 [dθ + Aµ (x)dxµ ]2 where θ denotes an angular variable 0 ≤ θ < 2π. With Aµ = 0, this is just the metric of a curved spacetime, which has a circle of radius a attached at every point. The transformation θ → θ + Λ(x) leaves ds invariant provided that we also transform Aµ (x) → Aµ (x) − ∂µ Λ(x). Calculate the 5-dimensional scalar curvature R5 and show that R5 = R4 − 41 a2 Fµν F µν . Except for the precise coefficient 14 this result follows entirely from symmetry considerations and from the fact that R5 involves two derivatives on the 5-dimensional metric, as explained in the text. After some suitable rescaling this is the usual action for gravity plus electromagnetism. Note that the 5-dimensional metric has the explicit form gµν − a2 Aµ Aν −a2 Aµ 5 gAB = . −a2 Aν −a2 Solution: We will compute the result using differential forms. First, find the veilbein 1-forms em = m µ m m M = −ω mn en , then = em em µ dx + eθ dθ, then find the connection 1-forms ω n from de M dx compute the curvature 2-form Rmn = dω mn + ω mp ω pn . From the curvature 2-form, extract the components via Rmn = 21 Rmnrs er es . Finally, compute the curvature scalar R ≡ η ns Rmnms . The conventions we use for the indices are: Flat: m, n, p, ... = 0, 1, 2, 3, 4 ; a, b, c, ... = 0, 1, 2, 3 Curved: M, N, P, ... = 0, 1, 2, 3, 4 ; µ, ν, ρ, ... = 0, 1, 2, 3 where x4 ≡ aθ is the fifth coordinate. The 5D metric is ds2 = GM N dxM dxN = (gµν − a2 Aµ Aν )dxµ dxν − a2 dθ2 − aAµ (dxµ dθ + dθdxµ ) Defining the veilbeins by µ ν θ θ n GM N = em M ηmn eN = eM ηµν eN − eM eN

we have: Gθθ = −a2 =⇒ eθθ = a, eaθ = 0 Gµθ = −aAµ =⇒ eθµ = aAµ Gµν = gµν − a2 Aµ Aν =⇒ eaµ = ea(4)µ M Now define the 1-forms em ≡ em M dx . This implies

ea = ea(4) , eθ = a(A + dθ) 276

where ea(4) are the 4D veilbeins defined by gµν = ea(4) ηab eb(4) , and A ≡ Aµ dxµ . Recall that the point of using forms notation is to be able to expand in any basis we choose, not necessarily the coordinate basis dxµ . Thus we define the component fields Aa via A ≡ Aa ea(4) , and for later convenience the partial derivatives ∂a via dxµ ∂µ ≡ ea(4) ∂a . Now we need to find the connection 1-forms ω mn , which are defined by the equation dem + ω mn en = 0. Since eθ = aA, we have38 deθ = d(dθ + aA) = 0 + aF = a 21 Fab ea(4) eb(4) . Since ω θθ = 0, the definition deθ = −ω θm em = −ω θa ea(4) implies 1 ω θa = + a Fab eb(4) 2 where we have been careful to keep track of the order of the 1-forms ea eb when extracting the components Fab = −Fba , hence the + sign in the above. a eb(4) . We have dea ≡ −ω am em = −ω ab eb − ω aθ eθ , and39 Next we need dea = dea(4) ≡ −ω(4)b ω aθ = ηθθ ω aθ = −ηθθ ω θa = −ηθθ η ab ω θb = −ηθθ η ab ( 12 a Fbc ec(4) ) = −ηθθ 12 a F ac ec(4) . Therefore a ec(4) = ω ac ec(4) − ηθθ 21 a F ac ec(4) eθ = ω ac ec(4) + ηθθ 12 a F ac eθ ec(4) , where we have used we have ω(4)c the Grassmann property ec eθ = −eθ ec . Moving the last term over to the other side of the equation, we can read off the coefficient of ec(4) :

1 a ω ac = ω(4)c − a ηθθ F ac eθ . 2 Now we are ready to define the curvature 2-form, Rmn = dω mn + (ω 2 )mn . First compute a − 12 a ηθθ d(F ab eθ ), and d(Fab eθ ) = ∂c Fab ec(4) eθ + Rab = dω ab + (ω 2 )ab . We have dω ab = dω(4)b Fab (aF ) = ∂c Fab ec(4) eθ + 12 aFab Fcd ec(4) ed(4) . To summarize: 1 1 a dω ab = dω(4)b − a ηθθ ∂c F ab ec(4) eθ − a2 ηθθ F ab Fcd ec(4) ed(4) . 2 4 Next we need (ω 2 )ab = ω am ω mb = ω ac ω c b + ω aθ ω θb . Multiplying the terms out gives 1 1 2 a c a (ω 2 )ab = (ω(4) ) b + a ηθθ (F ac ω(4)b − ω(4)c F cb )eθ − a2 F ac Fbd ec(4) ed(4) . 2 4 Therefore, the Rab part of the curvature 2-form is 1 1 a c a Rab = R(4)b − a2 (F ac Fbd + ηθθ F ab Fcd )ec(4) ed(4) + a ηθθ (F ac ω(4)b − ω(4)c F cb − ∂c F ab ec(4) )eθ . 4 2 38

To make sure the meaning of the last equality here is clear: In the coordinate basis, we have F = so that F is a 2-form and the components Fµν are just numbers. We use the definition of the veilbein 1-forms ea ≡ eaµ dxµ to define the c-number components Fab via Fµν ≡ eaµ ebν Fab . So while ω ab is a collection of 1-forms, F ab ≡ η ac Fcb is not a collection of forms but instead is just a collection of numbers. 39 We will keep the explicit factor of ηθθ present to show that it cancels out of the final result, and hence that the resulting action is independent of whether we use the metric convention η = (+, −, −, −, −) or the one more commonly used in the gravitational literature, η = (−, +, +, +, +). 1 µ ν 2 Fµν dx dx ,

277

A few comments are in order at this point. First, to compute the curvature scalar we will not need coefficients of the 2-form ec(4) eθ , so we will drop them from now on. Second, look at the terms in parentheses multiplying ec(4) ed(4) . Since Fcd = −Fdc , the second term in parentheses is manifestly antisymmetric in (cd), but the first term is not. Thus to extract the components Rabcd from the 2-form Rab = 21 Rabcd ec ed , we must write F ac Fbd ec ed = 12 (F ac Fbd − F ad Fbc )ec ed . Therefore, we have 1 1 Rabcd = (R(4) )abcd − a2 (F ac Fbd − F ad Fbc ) − a2 ηθθ F ab Fcd . 4 2 m Finally, let us make sure the dimensions are correct. The e have dimensions of length (recall µ m θ em = em µ dx with eµ dimensionless), so writing e = aA implies A is dimensionless. Since A = Aa ea , we find Aa has dimensions of inverse length. Since ea ∂a = dxµ ∂µ , the derivatives ∂a have dimensions of inverse length (just like ∂µ ), so Fab = ∂a Ab − ∂b Aa has dimensions of 1/length2 . Thus a2 F ac Fbd has dimensions of 1/length2 , as it should be to match Rabcd . Note that the operator d = ea ∂a is dimensionless, so that the 1-forms ω mn defined by dem = −ω mn en are also dimensionless. To compute the curvature scalar we still need the 2-form Rθ a = dω θa + (ω 2 )θ a . Recall that ω θa = 12 a Fab eb(4) , so we have dω θa = 12 a d(Fab eb(4) ) = 21 a (∂c Fab ec(4) eb(4) + Fab deb(4) ). Since b deb(4) = −ω(4)c ec(4) by definition, we have 1 1 b ec(4) . dω aθ = a ∂c Fab ec(4) eb(4) − a Fab ω(4)c 2 2 Again as a pedagogical reminder, we note that were we to extract the coefficient of ec(4) eb(4) in the above, we would have to antisymmetrize ∂c Fab in the indices (cb), but we will not need to do that here. Now to compute (ω 2 )θ a = ω θm ω ma = ω θb ω b a + 0, where we have used ω θθ = 0. When simplifying this expression, it is important to keep track of whether a particular term b is a 1-form, and ec(4) is a 1-form. is a form or a number. In particular, Fbc is a number, ω(4)a Therefore: 1 1 b (ω 2 )θ a = ( a Fbc ec(4) )(ω(4)a − a ηθθ F ba eθ ) 2 2 1 1 2 b c = − a Fbc ω(4)a e(4) − a ηθθ Fbc F ba ec(4) eθ 2 4 where in the first term, the minus sign was generated from moving the 1-form ec(4) to the b right of the 1-form ω(4)a . In contrast, no minus sign was generated in the second term when c moving the 1-form e(4) to the right of the number F ba . Putting it all together, we have40 1 1 1 1 b b ec(4) − a Fbc ω(4)a ec(4) − a2 ηθθ Fbc F ba ec(4) eθ . Rθ a = a ∂c Fab ec(4) eb(4) − a Fab ω(4)c 2 2 2 4 Let us again pause to check the dimensions. Fab has dimensions of 1/length2 , and ec(4) has dimensions of length. a has dimensions of length, and ∂c has dimensions of inverse length. So the first term is dimensionless. b Since ω(4)c is dimensionless, the second and third terms are also dimensionless. The fourth term has two powers of length from a2 , one power of length from each of ec(4) and eθ , which are indeed canceled by four powers of inverse length from the product Fbc F ba . 40

278

To get the curvature scalar, we need only the components Rmamc = Rb abc +Rθ aθc , so we need to extract Rθ aθb from the above expression for Rθ a . Since Rθ a = 12 Rθ amn em en = 12 Rθ abc eb(4) ec(4) + 2 × 12 Rθ aθc eθ ec(4) , we compare “Rθ a = − 41 ηθθ Fbc F ba ec(4) eθ + ... = + 14 ηθθ Fbc F ba eθ ec(4) + ...” with “Rθ a = Rθ aθc eθ ec(4) + ...” to get 1 Rθ aθc = ηθθ Fbc F ba . 4 Next, recall the previously obtained result 1 1 Rabcd = (R(4) )abcd − a2 (F ac Fbd − F ad Fbc ) − a2 ηθθ F ab Fcd 4 2 and contract with δ c a to get 1 1 Rabad = (R(4) )abad − a2 (0 − F ad Fba ) − a2 ηθθ F ab Fad 4 2 1 2 a 1 2 a = (R(4) ) bad + a F d (−Fab ) − a ηθθ F ab Fad 4 2 1 2 a = (R(4) ) bad − a (1 + 2ηθθ )Fab F ad . 4 Since we already found Rθ bθd = 14 ηθθ Fab F ad , we arrive at the following expression for Rbd ≡ Rmbmd = Rabad + Rθ bθd : 1 Rbd = (R(4) )bd − a2 (1 + ηθθ )Fab F ad . 4 We use the calligraphic font to denote Rbd ≡ Rmbmd to avoid potential confusion with the previously defined 2-form Rac , which can display a lowered index via Rac = ηab Rb c . To belabor the point, the object Rab is a collection of ordinary numbers, while the object Rab is a collection of 2-forms. To complete the calculation we need the quantity Rθθ ≡ Rmθmθ = Raθaθ = η ab Rbθaθ = η ab Rθbθa = η ab ηθθ Rθ bθa = η ab ηθθ ( 41 a2 ηθθ Fcb F ca ) = 41 a2 Fcb F cb . Finally, we can compute the scalar curvature: R ≡ η mn Rmn = η bd Rbd + η θθ Rθθ 1 1 = R(4) − a2 (1 + ηθθ )Fab F ab + η θθ a2 Fab F ab 4 4 1 2 = R(4) − a Fab F ab . 4 Notice that the terms with an explicit factor of η θθ = ηθθ drop out, reflecting the fact that the action should not depend on our convention for the signs in the metric. Switch from flat coordinates to spacetime coordinates by writing Fab = eµa eνb Fµν and using eµa eaν = δνµ to get Fab F ab = Fµν F ρσ eµa eνb eaρ ebσ = Fµν F ρσ δρµ δσν = Fµν F µν . We have therefore obtained the desired result 1 R = R(4) − a2 Fµν F µν . 4 279

6. Generalize the Kaluza-Klein construction by replacing the circles by higher dimensional spheres. Show that Yang-Mills fields emerge. Solution: Indices: M = 0, 1, 2, ..., D ← full spacetime, x¯M µ = 0, 1, 2, 3 ← non-compact part, xµ m = 4, 5, ..., D ← compact part, θm a = 1, 2, ..., K ← isometry group of compact part; generated by ξam (θ) ˆ = 0, 1, 2, ..., D ← locally flat frame of full spacetime M µ ˆ = 0, 1, 2, 3 ← locally flat frame of non-compact part m ˆ = 4, 5, ..., D ← locally flat frame of compact part ˆ

ˆ

ˆ

M M N Full metric GM N , full veilbein EM : GM N = EM EN ηMˆ Nˆ .

In terms of the compact and non-compact parts: µˆ eµ (x) −Bµmˆ (x, θ) ηµν 0 ˆ M ηMˆ Nˆ = , EM (x, θ) = ˆ 0 −δmˆ 0 em ˆn m (θ) T

GM N (x, θ) = (EηE )M N

=

ˆ m [Bµmˆ (x, θ) ≡ em m (θ)Bµ (x, θ)]

µˆ νˆ eµ (x) −Bµmˆ (x, θ) 0 ηµˆνˆ 0 eν (x) = ˆ −Bνnˆ (x, θ) ennˆ (θ) 0 −δmˆ 0 em ˆn m (θ)

n ˆ m ˆ n ˆ eµµˆ (x)ηµˆνˆ eννˆ (x) − Bµmˆ (x, θ)δmˆ ˆ n Bν (x, θ) Bµ (x, θ)δmˆ ˆ n en (θ) n ˆ ˆ n ˆ ˆ −em em ˆ n en (θ) ˆ n Bν (x, θ) m (θ)δmˆ m (θ)δmˆ

gµν (x) − Bµm (x, θ)gmn (θ)Bνn (x, θ) Bµm (x, θ)gmn (θ) = gmn (θ)Bνn (x, θ) −gmn (θ) ˆ

ˆ

M 0M Under the transformation x¯M → x¯0M , we have EM (¯ x) → EM (¯ x0 ) = the primes:

ˆ ∂x ¯N E M (¯ x). ∂x ¯0M N

∂ x¯0N 0Mˆ 0 ∂x0ν 0Mˆ 0 ∂θ0n 0Mˆ 0 E (¯ x ) = E (¯ x ) + E (¯ x). ∂ x¯M N ∂ x¯M ν ∂ x¯M n = −Bµmˆ , with x0µ = xµ and θ0m = θm + ξam (θ)εa (x): ˆ

M EM (¯ x) =

Look at Eµmˆ

∂x0ν 0mˆ 0 0 ∂θ0n 0mˆ 0 B (x , θ ) + e (θ ) ∂xµ ν ∂xµ n = −Bµ0mˆ (x, θ0 ) + ξan (θ)∂µ εa (x)en0mˆ (θ0 ) .

−Bµmˆ (x, θ) = −

280

Rearrange

Write Bµm (x, θ) ≡ ξam (θ)Aaµ (x) to get: 0 a 0m ˆ n ξamˆ (θ)Aaµ (x) = ξa0mˆ (θ0 )A0a µ (x) − ξa (θ)∂µ ε (x)en (θ )

Be mindful of the primes on ξa0mˆ (θ0 ) on the left-hand side: we have to ask how a Killing vector transforms under an infinitesimal transformation of the compact coordinates. For 0m = δnm + ∂n ξam (θ)εa (x), we have: θm → θ0m = (Λ−1 )mn θn , with (Λ−1 )mn = ∂θ ∂θn ξan (θ) → ξa0n (θ0 ) = Λnp ξap (Λ−1 θ) = δpn − ∂p ξbn (θ)εb (x) ξap (θ) + ξbq (θ)εb (x)∂q ξap (θ) + O(ε2 ) = ξan (θ) − [ξa (θ)·∂ξbn (θ) − ξb (θ)·∂ξap (θ)] εb (x) = ξan (θ) + gfabc ξcn (θ)εb (x) 0 0m ˆ ˆ where in the last equality we have used Killing’s equation. With this and with em m (θ) = em (θ ) x0ν m m ˆ m ˆ [since ∂θm = 0], we can strip off the overall contraction with ξa (θ) ≡ em (θ)ξa (θ) to obtain a 0b c a Aaµ (x) = A0a µ (x) + gfbc Aµ (x)ε (x) − ∂µ ε (x) .

For further discussion, see A. Salam, J. Strathdee, “On Kaluza-Klein theory,” Ann. Phys. 141, 316-352 (1982) and F. Cianfrani, G. Montani, “Non Abelian gauge symmetries induced by the unobservability of extra-dimensions in a Kaluza-Klein approach,” Mod.Phys.Lett. A21 (2006) 265-274 (arXiv:gr-qc/0511100v1).

281

8. The veilbeins for a spacetime with Minkowski metric is defined by gµν (x) = eaµ (x)ηab ebν (x), where the Minkowski metric ηab replaces the Euclidean metric δab . The indices a and b are to be contracted with ηab . For example, Rab = dω ab + ω ac ηcd ω db . Show that everything goes through as expected. Solution: We will follow the discussion beginning on p. 443 replacing δab with ηab . We begin with gµν (x) = eaµ (x)ηab ebν (x)

(16M )

For your convenience, we will label the equations as in the chapter, with the subscript M to denote “Minkowski.” Consider the “Minkowskian 2-sphere” defined by the line element ds2 = dt2 − sin2 t dϕ2 . From the metric (gtt = 1, gϕϕ = sin2 t) we can read off e1t = 1 and e2ϕ = sin t, with all other components zero. (Remember that ηab = diag(+1, −1) in this case!) We define the 1-forms ea = eaµ dxµ , so that e1 = dt , e2 = sin t dϕ. Now define the curvature 1-form dea = −ω ab eb

(17M )

where now we are careful to use upper and lower indices because we have a nontrivial norm on the spacetime. In our baby example, we have de1 = 0 and de2 = cos t dt dϕ, so as in the text the connection has only one nonvanishing component ω 12 = −ω 21 = − cos t dϕ. In the Euclidean case, the rotation eaµ (x) = Oab (x)e0bµ (x) leaves the metric invariant, that T 0b is gµν (x) = eaµ (x)δab ebν (x) = e0a µ (x)δab eν (x) if O O = 1. This time, under the same transfora 0b a mation eµ (x) = O b (x)eµ (x), we have (suppressing the spacetime arguments) gµν = eaµ ηab ebν = Oac ecµ ηab Ob d edν = ecµ (OT )c a ηab Ob d edν = ecµ ηcd edν only when OT η O = η. This is precisely the definition of the Lorentz transformation. That is, the rotation eaµ = Oab eb leaves the curved metric gµν invariant if O leaves the flat metric ηab invariant. Everything else up through equation (20) on p. 444 works out in a formally identical manner, including the curvature of +1 in our Minkowskian 2-sphere.

282

VIII.3

Effective Field Theory

1. Consider

1 (∂ϕ1 )2 + (∂ϕ2 )2 − λ(ϕ41 + ϕ42 ) − g ϕ21 ϕ22 2 We have taken the O(2) theory from chapter I.10 and broken the symmetry explicitly. Work out the renormalization group flow in the (λg)-plane and draw your own conclusions. L=

Solution: First let us make a few changes in notation. Normalize the couplings such that we won’t have to worry about unnecessary numerical factors in the Feynman rules, meaning λ → 4!1 λ and g → 212 g. To eliminate notational clutter, write ϕ1 ≡ ϕ and χ ≡ ϕ2 . For this problem we will use dimensional regularization and therefore separate the mass parameter µ ˜ from the ε ε couplings: λ → λ˜ µ and g → g µ ˜ . The Lagrangian is 1 1 1 µε (ϕ4 + χ4 ) − Zg g µ ˜ ε ϕ2 χ 2 L = Z (∂ϕ)2 + (∂χ)2 − Zλ λ˜ 2 24 4 1 1 1 (∂ϕ0 )2 + (∂χ0 )2 − λ0 (ϕ40 + χ40 ) − g0 ϕ20 χ20 = 2 24 4 In the first line we have included the renormalizing Z-factors, or equivalently the counterterms Ai ≡ Zi − 1. In the second line we have written the Lagrangian in terms of bare fields and parameters, denoted by the subscript 0. For future reference, we will need the relationships between the bare parameters and the renormalized parameters: λ0 = Z −2 Zλ λ˜ µε and g0 = Z −2 Zg g µ ˜ε The Feynman rules are as follows. We use a solid line for the ϕ propagator, and we use a dashed line for the χ propagator. Since we assume these fields have the same mass (namely m2 = 0), both propagators are equal, so that (dashed line) = (solid line) = i∆(k), where ∆(k) ≡ 1/(k 2 − m2 ) = 1/k 2 . A vertex connecting 4 solid lines equals a vertex connecting 4 dashed lines, both of which equal −iZλ λ˜ µε . A vertex connecting 2 solid lines and 2 dashed lines equals −iZg g µ ˜ε . At 1-loop order, the propagators do not get renormalized, so Z = 1 at 1-loop. The ϕ4 vertex has the following 1-loop corrections:

k1

k2

iV =

+ k4

(2)

(1)

1 2

+

(3)

+

k3

+

+ (4)

+

+ (higher order) (6)

(5)

283

We use conventions for which all momenta flow into the vertex. We also use the notational convention for which the diagram is just the picture as written, and the symmetry factor (in this case 21 ) is written explicitly multiplying the picture (as shown above). The first 1-loop diagram is: Z dd ` ε 2 (1) = (−iZλ λ˜ µ) i∆(`)i∆(` − k1 − k4 ) (2π)d Z dd ` 1 1 ε 2 = (λ˜ µ) d 2 2 (2π) (` − m ) [(` − k1 − k4 )2 − m2 ] Z Z 1 dd ` 1 ε 2 = (λ˜ µ) dx d 2 2 (2π) 0 {x(` − m ) + (1 − x)[(` − k1 − k4 )2 − m2 ]}2 The denominator simplifies to [` − (1 − x)(k1 + k4 )]2 + D, where D ≡ x(1 − x)(k1 + k4 )2 − m2 . Defining p ≡ ` − (1 − x)(k1 + k4 ), we have Z 1 Z 1 dd p ε 2 (1) = (λ˜ µ) dx d 2 (2π) (p + D)2 0 Z 1 i dx = (λ˜ µε )2 Γ(2 − d/2)D−(2−d/2) d/2 (4π) 0 ε Z 1 4π ε/2 i ε 2 = (λ˜ µ) Γ dx (4π)2 2 0 D The other two diagrams with only solid lines are the same as this one but with different dependence on the external momenta. Define Di ≡ x(1 − x)(k1 + ki )2 − m2 . Then the three diagrams are ε Z 1 4π ε/2 i ε 2 Γ dx (1) = (λ˜ µ) (4π)2 2 0 D4 ε/2 Z ε 1 4π i ε 2 Γ dx (2) = (λ˜ µ) (4π)2 2 0 D2 ε/2 Z ε 1 i 4π ε 2 (3) = (λ˜ µ) Γ dx (4π)2 2 0 D3 Moreover, the other three diagrams (containing dashed lines) are numerically identical to the three diagrams above, except with the replacement λ → g. Therefore, at 1-loop order we have ( Z 1 " #) 2 2 ε/2 2 ε/2 2 ε/2 g ε 4π µ ˜ λ 4π µ ˜ 4π µ ˜ 1+ 2 Γ dx Vλ = −λ˜ µε Zλ − + + 2(4π)2 λ 2 0 D2 D3 D4 Now we need some expansions in ε → 0+ , namely C ε/2 = 1 + 2ε ln C + O(ε2 ) for some number R∞ 2 C, and Γ(ε/2) = ε − γ + O(ε), where γ ≡ 0 dt e−t ln t ≈ 0.577. Together these imply ε 2 Γ C ε/2 = + ln Ce−γ + O(ε) . 2 ε 284

Now let µ2 ≡ 4πe−γ µ ˜2 . Using the above expansion and taking ε → 0+ wherever possible gives 2 2 Z 1 2 g2 6 µ µ µ λ 1+ 2 + dx ln + ln + ln Vλ = −λ Zλ − 2 2(4π) λ ε D2 D3 D4 0 We choose the “modified minimal subtraction” renormalization scheme, for which the counterterm Aλ = Zλ − 1 is chosen purely to cancel the 1/ε pole and nothing more. To this order, we therefore have 3 g2 1 Aλ = . λ 1+ 2 (4π)2 λ ε Now we need to renormalize the ϕ2 χ2 vertex. At 1-loop, we have

iV g =

+

+

1 2

+

+ (higher order)

+

Fortunately, we have already done all of the computational work. The two s-channel diagrams are numerically equal to the s-channel diagram from the ϕ4 vertex, except with the replacement λ2 → λg. The t-channel and u-channel diagrams are numerically equal to the t- and u- channel diagrams from the ϕ4 vertex, except with the replacement λ2 → g 2 . Immediately we get the result 1 2 Vg = −g Zg − (λ + 2g) + (finite) (4π)2 ε so that choosing the counterterm Ag ≡ Zg − 1 purely to cancel the divergent piece gives Ag =

2 1 (λ + 2g) . 2 (4π) ε

Now we compute the beta functions. Define H(λ, g) ≡ ln(Z −2 Zλ ) and G(λ, g) ≡ ln(Z −2 Zg ). Taking the logarithm of the relationships between the bare and renormalized couplings gives ln λ0 = H(λ, g) + ln λ + ε ln µ ˜ and

ln g0 = G(λ, g) + ln g + ε ln µ ˜.

The bare parameters are independent of the unphysical parameter µ, so differentiating the above with respect to ln µ gives 0=λ

dλ dG dg dH + + ελ and 0 = g + + εg . d ln µ d ln µ d ln µ d ln µ 285

Upon using the chain rule and rearranging a bit, these become dλ ∂H dg ∂H +1 +λ + ελ 0= λ ∂λ d ln µ ∂g d ln µ

(1)

and

∂G dg ∂G dλ 0= g +1 +g + εg (2) ∂g d ln µ ∂λ d ln µ The renormalizing Z-factors in the MS-bar renormalization scheme to this order are Z=0 3 g2 1 Zλ = 1 + λ 1+ 2 (4π)2 λ ε 2 1 Zg = 1 + (λ + 2g) 2 (4π) ε so the functions H and G are (after expanding ln(1 + x) = x + O(x2 )) g2 1 ∂H 3 ∂H 6g 1 g2 1 3 λ 1 + =⇒ = and = 1 − H(λ, g) = (4π)2 λ2 ε ∂λ (4π)2 λ2 ε ∂g (4π)2 λ ε 2 1 ∂G 2 1 ∂G 4 1 G(λ, g) = (λ + 2g) =⇒ = and = 2 2 (4π) ε ∂λ (4π) ε ∂g (2π)2 ε With these, the renormalization group equations (1) and (2) become g2 1 3 dλ 6 1 dg λ 1 − + 1 + g + ελ 0= (4π)2 λ2 ε d ln µ (4π)2 ε d ln µ 4 1 dg 2 1 dλ 0= g +1 + g + εg 2 2 (4π) ε d ln µ (4π) ε d ln µ

(10 ) (20 )

Now as per the usual procedure with dimensional regularization, we write dg dλ = −ελ + βλ and = −εg + βg d ln µ d ln µ and demand that βλ and βg be finite in the limit ε → 0+ . Putting these into equation (10 ) implies 3 (λ2 + g 2 ) + (things we insist sum to zero) 0 = βλ − 2 (4π) 0 while equation (2 ) implies 0 = βg −

2 (λ + 2g)g + (things we insist sum to zero) . (4π)2

Thus we have the 1-loop beta functions dλ 3 βλ = =+ (λ2 + g 2 ) 2 d ln µ (finite) (4π) dg 2 βg = =+ (λ + 2g)g d ln µ (finite) (4π)2 286

(100 ) (200 )

Before discussing the renormalization group dynamics in the (λ, g)-plane, let us perform a check. There should be a value of the parameter g for which we recover an O(2) symmetric ~ = (φ1 , φ2 )T . An O(2)-invariant theory would theory. Consider the 2-component vector φ have the interaction Lagrangian 1 1 ~ ~ 2 · φ ) = − λ0 φ41 + φ42 + 2φ21 φ22 L = − λ0 (φ 8 8 with some coupling λ0 . Compare this to our interaction Lagrangian L=−

1 1 1 λ(ϕ4 + χ4 ) − g ϕ2 χ2 = − λ(ϕ4 + χ4 ) + 3! g ϕ2 χ2 . 4! 4 4!

We recover the O(2) symmetric theory only for the precise value 3! g = 2λ =⇒ g = 31 λ. If we plug this particular value into the beta functions, we should recover a single beta function 1 10 2 2 for the coupling λ. If g = 13 λ, then g 2 = 91 λ2 , so that βλ = (4π) 2 3 λ . Also, βg = (4π)2 (1 + 2 1 1 10 2 1 )( )λ2 = (4π) 2 9 λ . Using the chain rule, we have dg/d ln µ = (dg/dλ)(dλ/d ln µ) = 3 βλ . So 3 3 1 10 2 βλ = (4π) 2 3 λ , which is exactly what we got before. Now let us study the dynamics implied by (100 ) and (200 ). Equation (100 ) implies that dλ/d ln µ > 0, so the coupling λ increases in magnitude as the parameter µ increases, irrespective of the signs of λ and g. Equation (200 ) is more interesting. Since d ln g/d ln µ ∝ λ+2g, the relative signs of λ and g change the running of g qualitatively: if λ + 2g < 0, then the coupling g decreases in magnitude as µ increases. At low energy there is an O(2) symmetric fixed point.

2. Assuming the nonexistence of the right handed neutrino field νR (i.e., assuming the minimal particle content of the standard model) write down all SU (2) ⊗ U (1) invariant terms that violate lepton number L by 2 and hence construct an effective field theory of the neutrino mass. Of course, by constructing a specific theory one can be much more predictive. Out of the product lL lL we can form a Lorentz scalar transforming as either a singlet or triplet under SU (2). Take the singlet case and construct a theory. [Hint: For help, see A. Zee, Phys. Lett. 93B: p. 389, 1980.] Solution: We use two-component spinor notation for the fermion fields. The neutrinos reside in SU (2) doublets ν (`a )i = a ∼ (2, − 12 ) of SU (2) ⊗ U (1) ea where i = 1, 2 labels the fundamental of SU (2), and a = e, µ, τ labels the flavor. From this field we can form the SU (2) singlet `a · `b ≡ εij (`a )i (`b )j = νa eb − ea νb 287

which has hypercharge 2(− 12 ) = −1. Note that this singlet is antisymmetric in the flavor indices. If we introduce a complex scalar field h+ with hypercharge +1 that is a singlet under SU (2), then we can form the interaction terms Lh`` = 12 fab h+ `a · `b + h.c. where the sums over a, b = e, µ, τ are implied, and due to fermi statistics41 only the antisymmetric part of f contributes. At this stage, we may simply assign two units of lepton number to the field h+ , and so this interaction does not generate Majorana neutrino masses on its own. Let us now augment the standard model with additional Higgs doublets: 0 ϕI (ϕI )i = ∼ (2, − 12 ) ϕ− I where again i = 1, 2 labels the fundamental of SU (2), while I = 1, 2, ..., nH labels the flavor of Higgs. This allows for the dimension-3 interaction Lhϕϕ = MIJ h+ ϕI · ϕJ + h.c. where MIJ is an antisymmetric matrix of size nH ×nH whose entries have dimensions of mass. I If the Higgs fields give charged lepton masses in the usual way, LYuk = −yab ϕI · `a e¯b + h.c., then they must be assigned lepton number zero. Thus, the clash between Lhϕϕ and Lh`` breaks lepton number by 2 and thereby generates neutrino masses at 1-loop.

For simplicity, let us assume that only ϕ1 gives mass to the charged leptons, and that nH = 2. Then neutrino masses are generated by the diagram +

h−

2

l −

h+ p

2

e fe

l+p

41

e

e

p

Recall that νe = eν for two Grassmann-valued spinors να and eα contracted with the antisymmetric tensor εαβ . See appendix E.

288

The diagram is42 Z d4 ` ∗ iAµe (p) = ve (p)(+iye )[iSe (` + p)](+ifµe )uµ (p)[i∆ϕ±2 (`)](+iM12 hϕ01 i∗ )[i∆h (`)] (2π)4 ∗ = fµe M12 m2e ve (p)uµ (p) I(p) where Z I(p) = −

1 1 1 d4 ` . 2 4 2 2 2 2 (2π) (` + p) − me ` − Mϕ± ` − Mh2 2

We want the mass term, so take p → 0. Also, me Mϕ±2 , Mh so take me → 0 inside the integral. Define I0 ≡ limme →0 I(p = 0). Let M> denote the greater of Mϕ±2 and Mh , and M< the lesser of the two. Z d4 ` 1 1 1 I0 = − 4 2 2 2 2 (2π) ` ` − M> ` − M2 ) + x3 (`2 − M2 + x3 M i 1 ≈ ln 2 2 (4π) M> M2 M2

M>2 M 1 then we have more indices to play with; if the left-hand side is symmetric under the exchange (α, I) ↔ (β, J), then the right-hand side can also be symmetric and nonzero if Z IJ = −Z JI . So Z is an antisymmetric matrix whose components are ordinary c-numbers.

3. From the fact that we do not know how to write consistent quantum field theories with fields having spin greater than 2 show that the N in the previous exercise cannot exceed 8. Theories with N = 8 supersymmetry are said to be maximally supersymmetric. Show that if we do not want to include gravity, N cannot be greater than 4. Supersymmetric N = 4 Yang-Mills theory has many remarkable properties. Solution: To set the notation, consider N = 1 supersymmetry. Let |si denote a state of spin-s. The generator Q raises the spin by +1/2, so that Q|si = |s + 21 i. Moreover, since Q is a 297

Grassmann generator we have Q2 = 0. Thus, for example, starting with the state s = 0 we find the state Q|0i = | 21 i, and similarly Q† |0i = | − 12 i. Thus we find the chiral supermultiplet containing the states s = 0 and s = ±1/2. Now generalize to the case N > 1. The supersymmetry generators QI obtain an index I = 1, 2, ..., N . Each QI increases the spin by 1/2. Starting with a state |si, the highest spin we can obtain using the QI is given by the action of all of the QI together: Q1 Q2 ...QN |si = |s + 12 N i. If we begin from s = −1 and we want to restrict to spin-0, spin-1/2 and spin-1 states only, then we require −1 + 21 N ≤ 1 =⇒ N ≤ 4. If we begin from s = −2 and demand that no states higher than spin-2 exist, then we require −2 + 12 N ≤ 2 =⇒ N ≤ 8. 4. Show that ∂θα /∂θβ = εαβ . Solution: ∂ ∂ θα = εαγ β θγ = εαγ δβγ = εαβ . β ∂θ ∂θ Alternatively, ∂ ∂ θα = εαγ β θγ = εαγ β ∂θ ∂θ

Z

dθβ θγ = εαγ δβγ = εαβ .

¯ α˙ )Φ. 5. Work out δϕ, δψ, and δF precisely by computing δΦ = i(ξ α Qα + ξ¯α˙ Q Solution: The chiral superfield is Φ=ϕ+

√

√ ν¯ ¯ µ ϕ − 1 (θσ µ θ)(θσ ¯ ¯ µψ 2 θψ + (θθ)F + i(θσ µ θ)∂ θ)∂µ ∂ν ϕ + i 2 θ(θσ µ θ)∂ 2

where as explained in the text, the field on the left depends on x and θ¯ only in the combination ¯ and it depends separately on θ, while the fields on the right depend only on x. y ≡ x + iθσ θ, The variation is therefore √ δΦ = δϕ + 2 θδϕ + (θθ)δF + ... So when computing the action of the supercharges on Φ, we need to extract the term without any θ or θ¯ to get δϕ, the term of order θ to get δψ, and the term of order (θθ) to get δF .

298

The supercharges are ∂ − iσαµα˙ θ¯α˙ ∂µ ∂θα ¯ α˙ = − ∂ + iθα σ µ ∂µ Q αα˙ ∂ θ¯α˙ Qα = +

Direct computation yields √ √ √ ¯ µ ψα + 2 (θ¯θ)θ ¯ α (θ∂ 2 ψ) − iσ µ θ¯α˙ (θθ)∂µ F − 2θα F Qα Φ = 2 ψα + i 2 (θσ µ θ)∂ αα˙ √ α˙ µαα ˙ µαα ˙ ¯ Q Φ = −¯ σ θα ∂µ ϕ − 2 2 i σ ¯ θα (θ∂µ ψ) ¯ α˙ )Φ gives so that the variation δΦ = i(ξ α Qα + ξ¯α˙ Q √ √ ˙ ˙ ¯ µαα θα θβ ∂µ ψβ + ... δΦ = i 2 ξ α ψα + ξ¯α˙ σ ¯ µαα θα ∂µ ϕ − 2iθα F + 2 2 ξ¯α˙ σ Since θα θβ = 12 (θθ)δαβ , we have √ √ ˙ ˙ ¯ µαα θα ∂µ ϕ − 2iθα F + 2 ξ¯α˙ σ δΦ = i 2 ξ α ψα + ξ¯α˙ σ ¯ µαα (θθ)∂µ ψα + ... Matching the terms order by order in θ as described above gives √ √ √ 1 ˙ ˙ ¯ µαα ∂µ ϕ + i 2 ξ α F , δF = 2 ξ¯α˙ σ ¯ µαα ∂µ ψα . δϕ = i 2 ξ α ψα , δψ α = − √ ξ¯α˙ σ 2 If instead you desire δψα = εαβ δψ β , then use ˙ µαβ ˙ ˙ µαβ ˙ ¯ ˙ = −ξ¯β σαµβ˙ ¯ ˙ = −εαβ εβ˙ α˙ ξ¯β σ εαβ ξ¯α˙ σ ¯ µαβ = εαβ εα˙ β˙ ξ¯β σ

√ to get δψα = + √12 ∂µ ϕσαµα˙ ξ¯α˙ + i 2 ξα F , which is the form given under equation (9) on p. 465.

6. For any polynomial W (Φ) show that [W (Φ)]F = F [dW (ϕ)/dϕ]+ terms not involving F . Show that for the theory (11) the potential energy is given by V (ϕ† , ϕ) = |∂W (ϕ)/∂ϕ|2 . Solution: This will follow from the binomial expansion n

(p + q) =

n X

n! p j q n−j . j!(n − j)! j =0

√ A chiral superfield has the expansion Φ = ϕ+ 2 θψ+(θθ)F +..., where the (...) are terms that will not contribute to the F term of W (Φ). From on we drop the (...). Any polynomial Pnow D W (Φ) of degree D can be written as W (Φ) = n = 0 Cn Φn , where Cn are constants whose 299

√ numerical values are unimportant for this problem. Let p = (θθ)F and q = ϕ + 2 θψ in the binomial expansion: n

Φ =

n X

√ n! [(θθ)F ] j (ϕ + 2 θψ)n−j j!(n − j)! j =0

Since (θθ)2 = 0, the series terminates at j = 1: √ √ Φn = (ϕ + 2 θψ)n + n(θθ)F (ϕ + 2 θψ)n−1 Moreover, since (θθ)θn−1 = 0 for n > 1, we have √ Φn>1 = (ϕ + 2 θψ)n + n(θθ)F ϕn−1 We could continue to simplify the first term, but we are asked to consider only the terms that involve F . So as far as this problem is concerned, we are free to write (Φn>1 )F = nF ϕn−1 + ... Since nϕn−1 =

d n ϕ , dϕ

W (Φ) =

D X n=0

we are just about done: n

Cn Φ =

D X n=0

Cn nF ϕ

n−1

D d X d + ... = F Cn ϕn = F W (ϕ) + ... dϕ n = 0 dϕ

where the (...) does not involve the field F . Now integrate out the auxiliary fields {Fa }3a = 1 using the Gaussian identity Z R 4 R 4 † † † † DF DF † e i d x(F F +J F +F J) = e−i d x J (x)J(x) d d where J † = dϕ W (ϕ) and J = (J † )† = [ dϕ W (ϕ)]∗ . After integrating out the F field, the Lagrangian gets the term ∗ dW (ϕ) 2 dW (ϕ) dW (ϕ) † LW = −J J = − = − dϕ dϕ dϕ

The potential energy is V = −LW = |dW (ϕ)/dϕ|2 .

7. Construct a field theory in which supersymmetry is spontaneously broken. [Hint: You need at least three chiral superfields.] Solution:

300

Let Ξ(x, θ) = ξ(x) + ... and {Φi (x, θ) = ϕi (x) + ...}ni= 1 be chiral superfields, and consider a superpotential of the form46 n X W (Ξ, Φ) = Φi fi (Ξ) i=1

with the fi (Ξ) being as-of-yet unspecified functions of Ξ. Supersymmetry is spontaneously broken at tree level if there exist values of the scalar components ξ(x) and ϕi (x) such that the potential is zero. From the previous problem, we know that this is equivalent to the conditions n

∂W (ξ, ϕ) X = ϕi fi0 (ξ) = 0 ∂ξ i=1

and

∂W (ξ, ϕ) = fi (ξ) = 0 . ∂ϕi

The first condition is satisfied by the field values ϕi = 0. The second equation, however, imposes n conditions on a function of just one variable, ξ. If n ≥ 2, then these conditions cannot be satisfied and supersymmetry is spontaneously broken. For example, take n = 2 and consider the functions f1 (ξ) = ξ − λ and f2 (ξ) = ξ 2 , with λ an arbitrary nonzero constant. Then ξ = λ sets f1 (ξ) = 0 but f2 (ξ) 6= 0, while ξ = 0 sets f2 (ξ) = 0 but f1 (ξ) 6= 0.

8. If we can construct supersymmetric quantum field theory, surely we can construct supersymmetric quantum mechanics. Indeed, consider Q1 ≡ 12 [σ1 P + σ2 W (x)] and Q2 ≡ 12 [σ2 P − σ1 W (x)], where the momentum operator P = −i(d/dx) as usual. Define Q ≡ Q1 + iQ2 . Study the properties of the Hamiltonian H defined by {Q, Q† } = 2H. Solution: We will briefly follow section 2 of arXiv:hep-th/9405029v2. Consult this reference for an extensive review. 2

d As described in the problem, it is possible to factorize a Hamiltonian H = − 21 dx 2 + V (x) as 1 d 1 d † † √ √ H = h h, where h ≡ 2 dx + W (x) and h ≡ − 2 dx + W (x). In terms of these variables, and df d using dx [W (x)f (x)] − W (x) dx = dW f (x), we find V (x) = W (x)2 − √12 W 0 (x). If the ground dx state has zero energy, then a state ψ0 that satisfies hψ = 0 automatically satisfies Hψ = 0 ψ 0 (x) and thus is the ground state. Solving Aψ0 = 0 gives W (x) = − √12 ψ00 (x) . Thus if we know ψ0 (x), then we know W (x) and V (x).

˜ ≡ hh† and put it into the form H ˜ = − 1 d22 + V˜ (x), If we define a new Hamiltonian H 2 dx 46

Here we follow Section 26.5 in Weinberg, Volume III. The original paper is L. O’Raifeartaigh, “Spontaneous symmetry breaking for chiral scalar superfields,” Nucl. Phys. B96 (1975) 331-352.

301

then by direct computation we find V˜ (x) = W (x)2 + new Hamiltonian is the following.

√1 2

W 0 (x). The reason for defining this

˜ Let ψ be an eigenstate of H with eigenvalue E, meaning Hψ = Eψ. Then H(hψ) = † hh hψ = hHψ = hEψ = E(hψ). Therefore, if ψ is an eigenstate of H with eigenvalue E, ˜ with eigenvalue E. Similarly, if ψ˜ is an eigenstate of H ˜ with then hψ is an eigenstate of H † ˜ then h ψ˜ is an eigenstate of H with eigenvalue E. ˜ eigenvalue E, ˜ or vice versa. ˜ ψ˜ = E˜ ψ, Therefore we can solve Hψ = Eψ by solving H The underlying reason for this algebra. The operators H H≡ 0

is that the two Hamiltonians belong to a supersymmetry 0 ˜ H

, Q≡

0 0 0 h† † , Q ≡ 0 0 h 0

satisfy the algebra [H, Q] = [H, Q† ] = {Q, Q} = {Q† , Q† } = 0 and {Q, Q† } = H. An important property of this Hamiltonian H is that the energy of any state is non-negative. A state |invi invariant under supersymmetry satisfies Qi |invi = 0. A state |noti that is not invariant under supersymmetry satisfies Qi |noti 6= 0. Since H is the sum of squares of Hermitian operators, its eigenvalues are non-negative, so Qi |noti > 0. Since all states besides |invi have energy greater than zero, the state |invi is the vacuum state. This tells us that if a supersymmetric state exists, it must be the vacuum state of the theory. This has implications for trying to break supersymmetry spontaneously. For further details, consult Witten, “Dynamical Breaking of Supersymmetry,” Nucl. Phys. B185 (1981) 513-554.

IX

Part N

IX.1

N.2 Gluon Scattering in Pure Yang-Mills Theory

1. Work out the two polarization vectors for general µ and µ ˜ for a gluon moving along the third direction. Solution: Let pµ = E(1, 0, 0, 1) be the momentum of a gluon moving along the third spatial direction. The 2-by-2 matrix representation of pµ is 0 0 µ 0 3 . pαα˙ = σαα˙ pµ = E(σαα˙ − σαα˙ ) = 2E 0 1 Therefore we have

√ λα =

√ 0 ˜ α˙ = 2E 0 2E , λ 1 1 302

as on p. 487. The polarization vectors (in 2-by-2 matrix notation) are given on p. 489: + αα˙ =

˜ α˙ λα µ ˜α˙ µα λ , − αα˙ = hµλi [λµ]

˙˜ ˜ we have where hµλi ≡ εαβ µα λβ and [λµ] ≡ εα˙ β λ ˜β˙ . Using the explicit forms for λ and λ, α˙ µ

hµλi = ε12 µ1 λ2 + 0 = and

˙˙

˜˙µ [λµ] = 0 + ε21 λ 2 ˜ 1˙ = The numerators are ˜ α˙ = µα λ and

√ 2E µ1

√ 2E µ ˜1˙ .

√ √ 0 µ1 µ1 (0, 2E ) = 2E 0 µ2 µ2

√ 0 0 0 λα µ ˜α˙ = √ . ˜2˙ ) = 2E (˜ µ1˙ , µ ˜2˙ µ ˜1˙ µ 2E

Therefore + αα˙

0 = 0

1 µ2 µ1

, − αα˙ =

0 1

0

!

µ ˜2˙ µ ˜1˙

.

3. Show that the result in (17) satisfies the reflection identity A(1− , 2− , 3+ , 4+ ) = A(4+ , 3+ , 2− , 1− ). A(1− , 2− , 3+ , 4+ ) =

h12i4 p1 · p2 = h12ih23ih34ih41i p2 · p3

(17)

Solution: The Parke-Taylor formula in equation (20) on p. 493 implies A(a+ , b+ , c− , d− ) =

hcdi4 . habihbcihcdihdai

Let a = 4, b = 3, c = 2, d = 1 to get A(4+ , 3+ , 2− , 1− ) =

h21i4 (−1)4 h12i4 = = A(1− , 2− , 3+ , 4+ ) X h43ih32ih21ih14i (−1)4 h12ih23ih34ih41i

4. Show that the “all plus” and “all minus” cubic Yang-Mills vertices (see appendix 2) vanish. [Hint: Choose the µ spinors wisely.] Solution: 303

The color-stripped all-plus cubic amplitude is (see equation (23) on p. 495) + + + + + + + + A(1+ , 2+ , 3+ ) = (+ 1 · 2 )(3 · p1 ) + (2 · 3 )(1 · p2 ) + (3 · 1 )(2 · p3 ) .

Referencing p. 490, we have + + i · j =

hµi µj i[λi λj ] hµi λj i[λi λj ] and + . i · pj = hµi λi ihµj λj i hµi λi i

+ + + + If µ1 = µ2 = µ3 , then all + i · j = 0. Therefore A(1 , 2 , 3 ) = 0. Flipping all three helicities − from + to − will make each term in the amplitude have a factor of − i · j ∝ [µi µj ]. If we choose all µ ˜i equal, then A(1− , 2− , 3− ) = 0.

5. Why doesn’t the argument in the text that A(− + +... + +) vanish apply to A(− + +)? Solution: For the 3-point amplitude, either all of the hiji = 0 or all of the [ij] = 0 (see equation (17) on p. 508). Therefore there are zeros in the denominator of the amplitude, so that choosing the µ spinors wisely yields an indeterminate zero over zero situation rather than just zero. Indeed, in appendix 2 on p. 496 we see the µ factors cancel out explicitly, leaving behind the vertex (25) or (26).

6. Insert the expansion for the cubic vertex into (21) and derive M (W1+ , W2+ , Z3− ). Z α ˜i) M (Wi ) = d2 λi e i˜µi λiα M (λi , λ (21a) Z ˜ i e iµαi˙ λ˜iα˙ M (λi , λ ˜i) M (Zi ) = d2 λ (21b) Solution: We follow a lecture given by A. Zee and documented by Oscar Castillo-Felisola. Recall that the amplitudes M are related to the amplitudes A by including the momentumconserving delta function: ! n X ˜i . λi λ M (1, ..., n) = A(1, ..., n) δ (4) i=1

The cubic vertex is A(1+ , 2+ , 3− ) =

304

[12]3 . [23][31]

We can write a fourier representation of the delta function ! Z n X P 1 ˜ ˙ (4) (i) ˜ (i) 4 ixαα˙ i λiα λiα δ λ λ = d x e 4 (2π) i=1 and perform the relevant twistor transforms, using W for + and Z for −. The (2π)4 will cancel from the resulting fourier transforms, so from now on we will simply drop the factors of 2π. We have: Z 3 Z P αα ˙ ˜ 3 ) [12] ˜ ˙ i(˜ µ1 λ1 +˜ µ2 λ2 +µ3 λ + − 2 2 2˜ + i λiα λiα d4 x e ix M (W1 , W2 , Z3 ) = d λ1 d λ2 d λ3 e [23][31] Z Z Z Z ˜3 i(µ3 +xλ3 )λ ˜ 1 )λ1 ˜ 2 )λ2 3 4 2 i(˜ µ1 +xλ 2 i(˜ µ2 +xλ ˜3 e = [12] dx d λ1 e d λ2 e d2 λ [23][31] Z Z ˜ i(µ3 +xλ3 )λ3 ˜ 1 ) δ 2 (˜ ˜2) ˜3 e . = [12]3 d4 x δ 2 (˜ µ 1 + xλ µ 2 + xλ d2 λ [23][31] Since two spinors span the 2-dimensional spinor space, we can write a third spinor as a linear ˜ 3 = a1 λ ˜ 1 +a2 λ ˜ 2 . The goal is to perform a change of integration combination of the first two: λ 2˜ variables from d λ3 to da1 da2 . The reader seeing this calculation for the first time may be distracted by the indices and tildes, but in fact the change of variables is completely straightforward. Let fα˙ ≡

˜ ˜ 3 )α˙ ∂(λ ˜ 1 )α˙ , gα˙ ≡ ∂(λ3 )α˙ = (λ ˜ 2 )α˙ . = (λ ∂a1 ∂a2

Then we have ˜ 3 )α˙ = fα˙ da1 + gα˙ da2 d(λ and therefore, just as in the usual calculus, ˜ 3 = |f1 g2 − g1 f2 |da1 da2 d2 λ ˙ ˜ ˜ = |εα˙ β (λ 1 )α˙ (λ2 ) ˙ da1 da2 β

= |[12]| da1 da2 . ˜ 3 = |[12]|da1 da2 = [12]sgn([12])da1 da2 . We also have [13] = a2 [12] and Thus we find d2 λ [32] = a1 [12] from writing the linear dependence equation as |3] = a1 |1]+a2 |2] and contracting with the appropriate spinors. Therefore, the amplitude is Z Z Z da1 i(µ3 +xλ3 )λ˜1 a1 da2 i(µ3 +xλ3 )λ˜2 a2 + + − 2 4 M (W1 , W2 , Z3 ) = [12] sgn([12]) d x e e a1 a2 ˜ 1 )δ 2 (˜ ˜2) × δ 2 (˜ µ 1 + xλ µ 2 + xλ 1 Using the property δ(ax) = |a| δ(x), the delta functions bring two powers of [12] in the denominator, which cancels the factor of [12]2 in the numerator. In the exponentials, the

305

˜ 1 = −˜ ˜ 2 = −˜ delta functions set xλ µ1 and xλ µ2 . We also recognize the integral representation R dy of the sign function as sgn(x) = y e ixy . We have: ˜1 − µ ˜2 − µ M (W1+ , W2+ , Z3− ) = sgn([12]) sgn(µ3 λ ˜1 λ3 ) sgn(µ3 λ ˜ 2 λ3 ) We recognize the Lorentz invariant products as W1 IW2 , W1 · Z3 and W2 · Z3 . Thus we obtain the result M (W1+ , W2+ , Z3− ) = sgn(W1 IW2 ) sgn(W1 · Z3 ) sgn(W2 · Z3 ) .

7. Show that M (W1+ , Z2− , W3+ , Z4− ) reproduces (19). A(1− , 2+ , 3− , 4+ ) =

h13i4 h12ih23ih34ih41i

(19)

Solution: To clarify, we hope to reproduce (19) but with the opposite helicity: A(1+ , 2− , 3+ , 4− ) =

[13]4 [12][23][34][41]

The amplitude is given on p. 495 as M (W1+ , Z2− , W3+ , Z4− ) = sgn(W1 · Z2 )sgn(Z2 · W3 )sgn(W3 · Z4 )sgn(Z4 · W1 ) ˜ and Z = (λ, µ) are and the inverse transforms for W = (˜ µ, λ) Z Z 2 dµ ˜i −i˜µαi λiα 2 i˜ µα λiα ˜ ˜ i M (Wi ) = d λi e M (λi , λi ) =⇒ M (λi , λi ) = e M (Wi ) (2π)2 Z Z 2 d µi −iµαi˙ λ˜iα˙ ˙λ ˜ iα˙ 2˜ iµα ˜ ˜ i e M (Zi ) M (Zi ) = d λi e M (λi , λi ) =⇒ M (λi , λi ) = (2π)2 R The integral representation of the sign function is sgn(x) = da e iax . The amplitude is a Z da1 da2 da3 da4 i(a1 W1 ·Z2 +a2 Z2 ·W3 +a3 W3 ·Z4 +a4 Z4 ·W1 ) − + − + M (W1 , Z2 , W3 , Z4 ) = e . a1 a2 a3 a4

306

Inverse twistor transforming gives Z 2 Z 2 Z 2 Z 2 dµ ˜3 −˜µ3 λ3 dµ ˜1 −i˜µ1 λ1 d µ2 −iµ2 λ˜2 d µ4 −iµ4 λ˜4 + − + − M (1 , 2 , 3 , 4 ) = e e e e (2π)2 (2π)2 (2π)2 (2π)2 Z da1 da2 da3 da4 i[a1 (˜µ1 λ2 −λ˜1 µ2 )+a2 (λ2 µ˜3 −µ2 λ˜3 )+a3 (˜µ3 λ4 −λ˜3 µ4 )+a4 (λ4 µ˜1 −µ4 λ˜1 )] e × a1 a2 a3 a4 Z 2 Z 2 da1 da4 dµ ˜1 i(−λ1 +a1 λ2 +a4 λ4 )˜µ1 d µ2 −i(λ˜2 +a1 λ˜1 +a2 λ˜3 )µ2 = ... e e 2 a1 a4 (2π) (2π)2 Z 2 Z 2 dµ ˜3 i(−λ3 +a2 λ2 +a3 λ4 )˜µ3 d µ4 −i(λ˜4 +a3 λ˜3 +a4 λ˜1 )µ4 × e e 2 (2π) (2π)2 Z

Z

da1 da4 2 ˜ 2 + a1 λ ˜ 1 + a2 λ ˜3) ... δ (−λ1 + a1 λ2 + a4 λ4 ) δ 2 (λ a1 a4 ˜1) . ˜ 3 + a4 λ ˜ 4 + a3 λ × δ 2 (−λ3 + a2 λ2 + a3 λ4 ) δ 2 (λ =

Now we Rneed the “inverse” R R 2 of the steps in the previous problem, namely to turn an integral dai /ai daj /aj into d λ. The first delta function sets λ1 = a1 λ2 + a4 λ4 , or in other words |1i = a1 |2i + a4 |4i. Multih21i plying on the left with h2| gives h21i = a4 h24i =⇒ a4 = h24i . Instead multiplying on the left with h4| gives h41i = a1 h42i =⇒ a1 = h41i . As in N.2.6, we have d2 λ1 = |h24i| da1 da4 h42i and therefore da1 da4 |h24i| 2 = d λ1 . a1 a4 h12ih41i We can therefore perform the integrals over a1 and a4 to get Z h41i ˜ da2 da3 |h24i| 2 ˜ + − + − ˜3) λ 1 + a2 λ δ (λ2 + M (1 , 2 , 3 , 4 ) = a2 a3 h12ih41i h42i ˜ 4 + a3 λ ˜ 3 + h21i λ ˜1) . × δ 2 (−λ3 + a2 λ2 + a3 λ4 ) δ 2 (λ h24i Now repeat the previous procedure to perform the integral over a2 and a3 . The middle delta function in the above expression sets λ3 = a2 λ2 + a3 λ4 , or in other words |3i = a2 |2i + a3 |4i. h23i Multiplying on the left with h2| implies h23i = a3 h24i =⇒ a3 = h24i . Multiplying with h4i implies h43i = a2 h42i =⇒ a2 =

h43i . h42i

Using d2 λ3 = |h24i| da2 da3 , we get

da2 da3 |h24i| 2 = d λ3 . a2 a3 h23ih34i Therefore, we have |h24i| |h24i| ˜ 2 + h41i λ ˜ 1 + h43i λ ˜ 3 ) δ 2 (λ ˜ 4 + h23i λ ˜ 3 + h21i λ ˜1) M (1 , 2 , 3 , 4 ) = δ 2 (λ h12ih41i h23ih34i h42i h42i h24i h24i h24i4 ˜ 2 −h41iλ ˜ 1 −h43iλ ˜ 3 ) δ 2 (h24iλ ˜ 4 +h23iλ ˜ 3 +h21iλ ˜1) = δ 2 (h24iλ h12ih23ih34ih41i +

−

+

−

307

P P The hope is that the two delta functions recollect into δ 4 ( 4i = 1 pi ). The equation 4i = 1 pi = 0 ˜ 1 + λ2 λ ˜ 2 + λ3 λ ˜ 3 + λ4 λ ˜ 4 = 0. Contracting with λ4 implies is written in terms of spinors as λ1 λ ˜ ˜ ˜ h41iλ1 + h42iλ2 + h43iλ3 = 0, which is the condition mandated by the first delta function ˜ 1 + h23iλ ˜ 3 + h24iλ ˜ 4 = 0, which is above (h42i = −h24i). Contracting with λ2 implies h21iλ the condition mandated by the second delta function. We therefore arrive at the result 4 X h24i4 4 δ ( pi ) . M (1 , 2 , 3 , 4 ) = h12ih23ih34ih41i i=1 +

−

+

−

Recall that we wanted to show that the prefactor A ≡ A(1+ , 2− , 3+ , 4− ) of the delta function is equal to [13]4 . [12][23][34][41] We will now show that the two forms are indeed equivalent. We will make repeated use of momentum conservation written in the form |1i[1| + |2i[2| + |3i[3| + |4i[4| = 0 .

(∗)

[13] . We have Multiplying by h2|...|3] implies h24i = h12i [43]

A=

h24i4 h12i4 [13]4 = . h12ih23ih34ih41i [43]4 h12ih23ih34ih41i

Multiply (∗) by h2|...|4] to get h23i = h12i [14] . Therefore [34] A=

h12i2 [13]4 h12i4 [13]4 [34] = . [43]4 h12i2 h34ih41i[14] [34]3 h34ih41i[14]

Multiply (∗) by h4|...|2] to get h41i = h34i [32] and therefore [12] h12i2 [13]4 [12] A= = [34]3 h34i2 [14][32]

h12i[12] h34i[34]

2

[13]4 . [12][34][14][32]

Since (p1 +p2 )2 = (p3 +p4 )2 , the term in parentheses equals 1. Since [14][32] = (−[41])(−[23]) = [41][23], we have [13]4 A= [12][23][34][41] which is the desired result. Note that had we chosen to integrate over the delta functions in a different order we could have arrived at this form directly. It is however worth showing the relationship between the forms with angle and square brackets.

308

8. Show that SL(4, R) is locally isomorphic to the conformal group. [Hint: Identify the 15 = 42 − 1 generators of the conformal group (3 rotations J i , 3 boosts K i , 1 dilation D, 4 translations P µ , and 4 conformal transformations K µ ) with the 15 traceless real 4 by 4 matrices.] Solution: Our strategy will be to write down all possible traceless real 4-by-4 matrices and identify them with the conformal generators based on their commutation relations. The 15 generators M µν , P µ , D, K µ of the conformal group47 satisfy the commutation relations: [D, K µ ] = −iK µ , [D, P µ ] = +iP µ , [K µ , P ν ] = +2i(η µν D + M µν ) [K µ , M νρ ] = −i(η µν K ρ − η µρ K ν ) , [P µ , M νρ ] = −i(η µν P ρ − η µρ P ν ) [M µν , M ρσ ] = −i(η νρ M µσ + η µσ M νρ − η µρ M νσ − η νσ M µρ ) with all others zero. The generators of rotations are J i = 21 εijk Mjk , and the generators of boosts are B i = M 0i . (We will use B i to denote the boosts so as not to confuse them with the spatial components of the special conformal generators.) For two real spinors λα and µα˙ , we have learned in the text that conformal transformations act naturally on the 4-dimensional column vector λα Z= µα˙ which transforms under the 4-dimensional representation of SL(4, R). The lie algebra of SL(4, R) consists of the 15 real traceless 4-by-4 matrices, which we now attempt to construct. From learning about spinor representations of the Lorentz group back in Chapter II, we know how the rotations J i and the boosts B i act on the vector Z: 1 i 1 i −i 2 σ σ 0 0 i i 2 , B = . J = 0 12 σ i 0 +i 12 σ i Here we are working in the signature η = (−, +, +, +) of SO(3, 1). These generators can be repackaged into the relativistic notation 1 µν β (σ )α 0 µν 2 M = 0 − 21 (¯ σ µν )α˙ β˙ 47

Acting on scalar fields, these generators take the form: Mµν = i(xµ ∂ν − xν ∂µ ) Pµ = −i∂µ , D = ixµ ∂µ Kµ = −i(x2 ∂µ − 2xµ xν ∂ν ) .

In addition to these, spinors and vectors also transform with the appropriate matrices. These matrices are what we are trying to identify.

309

¯ ν − σν σ ¯ µ ) and σ ¯ µν ≡ − 12 i(¯ σµσν − σ ¯ ν σ µ ), and we have defined the where σ µν ≡ + 12 i(σ µ σ 4-vector ˙

˙ (σ µ )αα˙ ≡ (I, σ 1 , σ 2 , σ 3 ) , (¯ σ µ )αα ≡ εαβ εα˙ β σβµβ˙ = (I, −σ 1 , −σ 2 , −σ 3 ) .

These are defined so that xαα˙ ≡

σαµα˙ xµ

=

x0 + x3 x1 − ix2 x1 + ix2 x0 − x3

satisfies det x = x20 − x21 − x22 − x23 = −η µν xµ xν with the SO(3, 1) metric η µν = diag(−1, +1, +1, +1). Note that the components of the matrices σ µν and σ ¯ µν are numerically equal to: σ ij = +εijk σ k , σ 0i = −iσ i σ ¯ ij = −εijk σ k , σ ¯ 0i = −iσ i with index placements as indicated by the form of M µν given above. From looking at the index structure of these tensors as well as that of the spinors in Z = (λ, µ), the only other traceless 4-by-4 matrices we can write down are of the forms 1 β 1 µ 0 σ δ 0 µ 2 αβ˙ 2 a X± ≡ and Y ≡ ˙ 0 − 12 δ α˙β˙ ±1σ ¯ µαβ 0 2

up to an overall constant for each. By direct computation, we find [Y, X±µ ] = X∓µ , and therefore: [Y, X±µ + X∓µ ] = +(X±µ + X∓µ ) [Y, X±µ − X∓µ ] = −(X±µ − X∓µ ) We also find [X±0 , M ij ] = 0 ,

[X±k , M ij ] = −i(δ ik X±j − δ kj X±i )

[X±i , M 0j ] = iδ ij X±0 , [X±0 , M 0i ] = iX±i . Finally, we also have [X+µ + X−µ , X+ν − X−ν ] = −2(η µν Y + iM µν ) . If we identify D = iY, P µ = −i(X+µ + X−µ ) and K µ = −i(X+µ − X−µ ), then we get the correct commutation relations for the conformal group. This completes the problem.

310

IX.2

N.3 Subterranean Connections in Gauge Theories

1. Show that the structure of Lie algebra (21) emerges naturally. fabe fcde + face fbde + fade fbce = 0

(21)

Solution: First note that equation (21) on p. 509 has a sign error on the second term. The Jacobi identity should read fabe fcde + fcae fbde + fade fbce = 0 . Recall that the recursion scheme is to separate the diagram into sets L and R, with r L and s R. If (r, s) = (1, 4), we get one separation with a pole in s = (p1 + p2 )2 = (p1 + p3 )2 , and one with a pole in t = (p1 + p3 )2 = (p2 + p4 )2 : −M(1a , 2b , 3c , 4d ) = fabe fecd A(1, 2, 3, 4) + face febd A(1, 3, 2, 4) If (r, s) = (1, 3), then we get one pole in s and one pole in u = (p1 + p4 )2 = (p2 + p3 )2 : −M(1a , 2b , 3c , 4d ) = fabe fedc A(1, 2, 4, 3) + fade febc A(1, 4, 2, 3) If (r, s) = (1, 2), then we get one pole in t and one pole in u: −M(1a , 2b , 3c , 4d ) = face fedb A(1, 3, 4, 2) + fade fecb A(1, 4, 3, 2) To make sense of these, we should relate the various color-stripped amplitudes to each other. Since the amplitudes with all +, all −, or 3 + or 3 − are zero, we can without loss of generality consider the amplitude with two + helicities and two − helicities. From equation (19) on p. 493 and the surrounding discussion, we know how to compute the amplitude for any ordering of the + and − helicities, so without loss of generality we choose (1− , 2− , 3+ , 4+ ). Thus consider the amplitude A(1− , 2− , 3+ , 4+ ) =

h12i4 h12ih23ih34ih41i

from equation (17) on p. 492. The goal P is to relate all of the other As to this one. We will repeatedly use momentum conservation 4i = 1 pi = 0 in the form |1i[1| + |2i[2| + |3i[3| + |4i[4| = 0 .

(∗)

Consider the amplitude A(1− , 3+ , 2− , 4+ ). From relabeling equation (19) on p. 493, we know h12i4 h12i4 −h12ih34i − + − + A(1 , 3 , 2 , 4 ) = = . h13ih32ih24ih41i h12ih23ih34ih41i h13ih24i Multiply the momentum conservation equation (∗) by h1| on the left and by |4] on the right to get h12i [34] h12i[24] + h13i[34] = 0 =⇒ =− . h13i [24] 311

Thus

+[34]h34i s −h12ih34i = = . h13ih24i [24]h24i t

Therefore, we have A(1− , 3+ , 2− , 4+ ) =

s A(1− , 2− , 3+ , 4+ ) . t

Next, consider h12i4 A(1 , 2 , 4 , 3 ) = = h12ih24ih43ih31i −

−

+

+

h12i4 h12ih23ih34ih41i

−h23ih41i h24ih31i

Multiply (∗) by h2| and |1] to get h23i[31] + h24i[41] = 0 =⇒ h23i/h24i = −[41]/[31] and thus −h23ih41i [41]h41i u = = . h24ih31i [31]h31i t We have A(1− , 2− , 4+ , 3+ ) =

u A(1− , 2− , 3+ , 4+ ) . t

Next, consider h12i4 A(1 , 4 , 2 , 3 ) = = h14ih42ih23ih31i −

+

−

+

h12i4 h12ih23ih34ih41i

−h12ih34i h42ih31i

.

Multiply (∗) by h2| and [3] to get h21i[13] + h24i[43] = 0 =⇒ h12i/h42i = −[34]/[31] and thus +[34]h34i s −h12ih34i = = . h42ih31i [31]h31i t We have A(1− , 4+ , 2− , 3+ ) =

s A(1− , 2− , 3+ , 4+ ) . t

Two more to go. The next one is h12i4 A(1 , 3 , 4 , 2 ) = = h13ih34ih42ih21i −

+

+

−

h12i4 h12ih23ih34ih41i

−h23ih41i h13ih42i

.

Multiply (∗) by h3| and |4] to get h31i[14] + h32i[24] = 0 =⇒ h23i/h13i = −[41]/[42] and thus +[41]h41i u −h23ih41i = = . h13ih42i [42]h42i t We have A(1− , 3+ , 4+ , 2− ) =

u A(1− , 2− , 3+ , 4+ ) . t

Fortunately, the last one is easy: A(1− , 4+ , 3+ , 2− ) =

h12i4 h12i4 = = A(1− , 2− , 3+ , 4+ ) . h14ih43ih32ih21i h12ih23ih34ih41i 312

We therefore arrive at the amplitude s − + + − − + + , 4 ) = f f + f f −M(1− , 2 , 3 ace ebd A(1 , 2 , 3 , 4 ) abe ecd a c b d t u s = fabe fedc + fade febc A(1− , 2− , 3+ , 4+ ) t ut = face fedb + fade fecb A(1− , 2− , 3+ , 4+ ) . t The equalities reflect the physical necessity for the amplitude to be the same regardless of which (r, s) we choose for the recursion formula. Subtracting and multiplying by t gives the consistency conditions tfabe fecd + s(face febd − fade febc ) − ufabe fedc = 0 t(fabe fecd − fade fecb ) + sface febd − uface fedb = 0 Subtract (2) from (1) to get tfade fecb − sfade febc + u(face fedb − fabe fedc ) = 0 . Since fecb = −febc and since s + t + u = 0 =⇒ t + s = −u, we get u(fade febc + face fedb − fabe fedc ) = 0 . u 6= 0 and fedc = −fecd implies fade febc + face fedb + fabe fecd = 0 . This is the Jacobi identity. 2. In appendix 1 we recursed by complexifying the momenta of two external lines with helicity + and −. In the derivation of the recursion relation (3) we could have picked any two external lines to complexify. Determine the amplitude calculated directly in chapter N.2, namely A(1− , 2− , 3+ , 4+ ), by complexifying lines 1 and 2. This is an example of the self-consistency argument sketched in the text. Particle physics experimentalists are fond of saying that yesterday’s spectacular discovery is today’s calibration and tomorrow’s annoying background. The canonical example is the Nobel-winning discovery of the CP-violating decay of the KL meson into two pions. In theoretical physics, yesterday’s discovery is today’s homework exercise and tomorrow’s trivium. M(0) = −

X RL L,h

zL

=−

X ML (zL )MR (zL ) PL (0)2

L,h

(3)

Solution: We are interested in the color-ordered amplitude A(1− , 2− , 3+ , 4+ ) with legs 1 and 2 complexified. There is only one way to split up the amplitude into 3-point vertices for which legs 1 and 2 do not belong to the same partition while maintaining the color ordering: 313

1

2 1

2 K

K

4 4

3

3

The color-ordered amplitude is: ˆ −h , ˆ2− , 3+ ) ˆ h , 4+ ) 1 A(−K A(ˆ1− , K 2 K h=± h i 1 + + ˆ− − ˆ− + − ˆ− + + + ˆ− ˆ ˆ ˆ ˆ = − 2 A(K , 4 , 1 )A(−K , 2 , 3 ) + A(1 , K , 4 )A(3 , −K , 2 ) K

A(1− , 2− , 3+ , 4+ ) = −

X

In the second line we have cyclically permuted the arguments of the 3-point amplitudes to put them into the canonical forms: A(1+ , 2+ , 3+ ) =

[12]3 h12i3 , A(1− , 2− , 3+ ) = . [23][31] h23ih31i

The internal momentum is K = −(p1 + p4 ) = +(p2 + p3 ), which in spinor notation reads |Ki[K| = −|1i[1| − |4i[4| = |2i[2| + |3i[3| .

(A)

When the internal momentum K ≡ |Ki[K| flows out of the vertex, we can effect the replacement K → −K by replacing its spinors as |Ki → i|Ki, |K] → i|K]. We have: " ! ! ! !# ˆ 3 ˆ ˆ2i)3 ˆ 3 ˆ 3 ˆ [ K4] (ih K h 1 Ki (i[3 K]) 1 + A(1− , 2− , 3+ , 4+ ) = − ˆ ˆ ˆ ˆ 2])[ ˆ 1ˆK] ˆ ˆ ˆ 23] ˆ 2h23i[23] [41][ h23i(ih3 Ki) hK4ih4 1i (i[K # " ˆ 3 [3K] ˆ ˆ2i3 ˆ 3 ˆ 3 hK 1 hˆ1Ki [K4] =+ + . ˆ h23ih3 ˆ ˆ ˆ 2][ ˆ 1ˆK] ˆ ˆ [K ˆ 23] ˆ 2h23i[23] [41][ Ki hK4ih4 1i Here we choose (r, s) = (1, 2) and thereby complexify the momenta as: |ˆ1] = |1] + zL |2] , |ˆ2i = |2i − zL |1i with |1i and |2] unchanged. This effects the shift pˆ1 = p1 +zL q , pˆ2 = p2 −zL q with q = |1i[2|. Here we have removed the hats from |1i and |2] since they are not complexified. Now use the hatted version of equation (A): ˆ ˆ K4] ˆ = −h21i[4ˆ1] hˆ2|(A)|4] =⇒ hˆ2Ki[ ˆ ˆ1] =⇒ h3Ki[ ˆ K ˆ ˆ1] = −h34i[4ˆ1] h3|(A)|

314

Therefore:

ˆ 3 hK ˆ ˆ2i3 (h21i)[4ˆ1])3 h21i3 [4ˆ1] [K4] = = . ˆ ˆ23ih3Ki ˆ ˆ ˆ1K]h [4ˆ1]hˆ23i(h34i[4ˆ1]) hˆ23ih34i [41][

Conservation of momentum implies |1i[1| + |2i[2| + |3i[3| + |4i[4| = 0 . Multiply the hatted version of this by hˆ2|...|4] to obtain [ˆ14] [34] =− ˆ h21i h23i and therefore

h21i3 [4ˆ1] h21i2 [34] h21i3 = = h34i h23ih34i hˆ23ih34i

[34] h21i[23]

.

Multiplying momentum conservation (unhatted) by h1|...|3] implies that the term in parentheses equals −h41i, so all together we have48 ˆ 3 hK ˆ ˆ2i3 h12i3 [K4] = . ˆ ˆ23ih3Ki ˆ ˆ ˆ1K]h h23ih34ih41i [41][ This is half of the correct answer. Now let us simplify the second term. Go back to equation (A) and obtain: ˆ ˆ K3] ˆ = +h12i[23] h1|(A)|3] =⇒ h1Ki[ ˆ ˆ K2] ˆ = −h41i[12] . h4|(A)|2] =⇒ h4Ki[ Therefore: ˆ 3 [3K] ˆ 3 1 hˆ1Ki 1 −h12i3 [23]3 1 = ˆ ˆ ˆ2][ˆ23] ˆ1i [K h23i[23] hK4ih4 h23i[23] h41i[23] +h41i[12] =

[23] −h12i3 . h23ih41i h41i[12]

Multiply momentum conservation (unhatted) by h4|...|2] to get h41i[12] + h43i[32] = 0 =⇒

h41i[12] = −h34i [23]

so that the above term equals h12i3 . h23ih34ih41i 48

We thank Eric Dzienkowski for catching an error here.

315

This is the second half of the correct answer. Adding the two terms cancels the overall factor of 2, and we arrive at the result A(1− , 2− , 3+ , 4+ ) =

h12i4 . h12ih23ih34ih41i

3. Using the explicit forms given for A(1− , 2− , 3+ , 4+ ) and A(1− , 2+ , 3− , 4+ ) in the preceding chapter, check the estimated large z behavior in (13-15). 1 z 1 −− M (z) → z M+− (z) → z 3 M−+ (z) →

(13) (14) (15)

Solution: The amplitudes are given on pages 492-493: A(1− , 2− , 3+ , 4+ ) =

h13i4 h12i4 , A(1− , 2+ , 3− , 4+ ) = h12ih23ih34ih41i h12ih23ih34ih41i

and the physical amplitude specified by helicity labels for particles r and s is Mhr hs (z) ≡ [hr r (z)]µ Mµν (z)[hs s (z)]ν . The spinorial version of the deformations pr (z) = pr + zq and ˜i: ps (z) = ps − zq are given on p. 501, with pi = λi λ ˜ r (z) = λ ˜r + zλ ˜s λr (z) = λr , λ ˜ s (z) = λ ˜s λs (z) = λs − zλr , λ ˜ s . Choose (r, s) = (1, 2). Then the only z-dependence in the amplitude and q = λr λ − + − + A(1 , 2 , 3 , 4 ) potentially comes from h12i and h23i. But h12i(z) = h12i − zh11i = h12i, so really the only dependence on z comes from h23i(z) = h23i − zh13i ∼ −zh13i. Therefore we find M−+ (z) ∼ 1/z. The case for A(1− , 2− , 3+ , 4+ ) is identical, with the only z-dependence coming from a factor of h23 in the denominator, leading immediately to M−− (z) ∼ 1/z. Now choose (r, s) = (2, 3). Then h13i(z) = h13i − zh12i ∼ z, leading to a factor of z 4 in the numerator. This is canceled partially by a factor h34i(z) = h34i − zh14i ∼ z in the denominator, leading to M+− (z) ∼ z 3 .

316

4. Worry about the sloppy handling of factors of 2 in appendix 1. [Hint: The final result is correct because the polarization vectors in (5-6) are normalized to |ε|2 = 2 for convenience.] Solution: ˙ Let us be pedantic about distinguishing among pµ , pαα˙ and pαα . Given the Lorentz 4-vector µ µ αα ˙ µαα ˙ µαα ˙ αβ α˙ β˙ µ p , define 6 pαα˙ ≡ pµ σαα˙ and 6 p¯ ≡ pµ σ ¯ , where σ ¯ ≡ ε ε σβ β˙ . Then for lightlike mo˙ ˜ α˙ = εαβ εα˙ β˙ λβ λ ˜ ˙ = εαβ εα˙ β˙ 6 p ˙ =6 p¯αα ˜ α˙ . Then λα λ . menta p2 ≡ pµ pµ = 0, we define 6 pαα˙ ≡ λα λ β ββ Therefore: ˙ ˜ α˙ λ ˜ 0α˙ =6 pαα˙ 6 p¯αα hλλ0 i[λλ0 ] = λα λ0α λ = pµ pν tr(σ µ σ ¯ ν ) = 2 p µ pµ .

So the top of page 511 should read PL (0)2 ≡ PL (0)µ PL (0)µ = 2 p1µ pµ2 = h12i[12]. Thus the result (25) is correct.

317

X

Appendix E: Dotted and Undotted Indices

1. Show that ησ µν ψ = −ψσ µν η and χ¯ ¯σ µ ψ = −ψσ µ χ. ¯ Solution: First consider the second object. We have: ˙ χ¯ ¯σ µ ψ = χ¯α˙ σ ¯ µαα ψα ˙ = −ψα σ ¯ µαα χ¯α˙ ˙

˙ = −εαβ ψ β σ ¯ µαα εα˙ β˙ χ¯β

β˙ ˙ ¯ µαα χ¯ = −ψ β (−1)2 εβα εβ˙ α˙ σ ˙

= −ψ β σβµβ˙ χ¯β = −ψσ µ χ¯ X For the first object, recall that σ µν = 41 (σ µ σ ¯ ν − σν σ ¯ µ ). Therefore: ˙

4 ησ µν ψ = ησ µ σ ¯ ν ψ − (µ ↔ ν) = η α σαµβ˙ σ ¯ ν ββ ψβ − (µ ↔ ν) ˙

= −ψβ σαµβ˙ σ ¯ ν ββ η α − (µ ↔ ν) ˙˙

˙ = −ψβ [(εαγ εβ˙ γ˙ σ ¯ µγγ )(εβ δ εβδ σδνδ˙ )]η α − (µ ↔ ν) ˙

˙˙

˙ = −ψβ [εαγ εβδ (−δγδ˙ )¯ σ µγγ σδνδ˙ ]η α − (µ ↔ ν)

˙˙

˙ = −ψβ (−εαγ εβδ σ ¯ µγγ σδνδ˙ )η α − (µ ↔ ν) ˙ = +(εβδ ψβ )(¯ σ µγγ σδνδ˙ )(εαγ η α ) − (µ ↔ ν) ˙ = (−ψ δ )(¯ σ µγγ σδνδ˙ )(−ηγ ) − (µ ↔ ν)

= +ψσ ν σ ¯ µ η − (µ ↔ ν) [ put in matrix multiplication order ] = −[ψσ µ σ ¯ ν η − (µ ↔ ν)] = −4 ψσ µν η =⇒ ησ µν ψ = −ψσ µν η X ¯ = − 1 (θσ µ ξ)( ¯ χ¯ 2. Show that (θϕ)(χ¯ξ) ¯σµ ϕ). 2 Solution: First we need

˙

˙

σαµα˙ σ ¯µββ = Cδαβ δα˙ β which we know by SU (2) ⊗ SU (2) invariance. Choose α = β = α˙ = β˙ = 1 to get µ 11 C = σ11 σ ¯µ = (σ 0 )11 (¯ σ 0 )11 − (σ 3 )11 (¯ σ 3 )11 = 1 − (−1) = 2

Therefore

˙

˙

σαµα˙ σ ¯µββ = 2 δαβ δα˙ β .

318

˙

[ note: εβ˙ γ˙ εβ δ = −εγ˙ β˙ εβ δ = −δγδ˙ ]

Now compute: ˙ ¯ χ¯ ¯µββ ϕβ (θσ µ ξ)( ¯σµ ϕ) = θα σαµα˙ ξ¯α˙ χ¯β˙ σ ˙

= 2 θα ξ¯α˙ δαβ δα˙ β χ¯β˙ ϕβ = 2 θα ξ¯α˙ χ¯α˙ ϕα = 2(−1) θα χ¯α˙ ξ¯α˙ ϕα = 2(−1)(−1)2 θα ϕα χ¯α˙ ξ¯α˙ ¯ . = −2 (θϕ)(χ¯ξ) Dividing by 2 proves the statement. 3. Show that θα θβ = 12 (θθ)δ αβ . [Hint: simply evaluate the two sides for all possible cases.] Solution: The indices have to match up by SU (2) invariance, and then we can fix the constant. θα θβ = C(θθ)δ αβ Contract both sides with δ βα and note that δ αα = 2 to get 1 . 2 Alternatively, we can follow the hint and just check. α = 1, β = 2 implies 0 for the right-hand side, and θ1 θ2 = ε12 θ2 θ2 = 0 (θθ) = C(θθ)δ αα = 2C(θθ) =⇒ C =

for the left-hand side, since θ22 = 0 since the square of any Grassmann number is zero. For α = β = 1, we have 21 (θθ) = 12 (θ1 θ1 + θ2 θ2 ), and θ2 θ2 = ε21 θ1 ε21 θ1 = −(ε21 )2 θ1 θ1 = −θ1 θ1 = +θ1 θ1 , so 21 (θθ) = θ1 θ1 , which matches the left-hand side θα θβ for α = β = 1. Addendum: More on Two-Component Spinors Here we collect a few relations between two-component spinors and four-component spinors for use in calculating Feynman diagrams using two-component notation. For an extensive review, consult arXiv:0812.1594v5 [hep-ph]. eα (x) Take the Dirac field for the electron for definiteness: E(x) ≡ . The mode exe¯†α˙ (x) pansions for the fields eα (x) and e¯α (x) are: XZ d3 p −ip·x † +ip·x eα (x) = b(~ p , s)u (~ p , s) e + d (~ p , s)v (~ p , s) e α α (2π)3 2ωp s XZ d3 p d(~p, s)uα (~p, s) e−ip·x + b† (~p, s)vα (~p, s) e+ip·x . e¯α (x) = 3 (2π) 2ωp s 319

The creation operators b† and d† create an electron e− and a positron e+ respectively. The wave functions u(~p, s) and v(~p, s) are related to the 4-component wave functions U (~p, s) and V (~p, s) as uα (~p, s) vα (~p, s) U (~p, s) = , V (~p, s) = v †α˙ (~p, s) u†α˙ (~p, s) P The usual spin sum relations (see p. 110 of the text) p, s)U¯ (~p, s) =6 p + m and s U (~ P ¯ p, s)V (~p, s) =6 p − m, along with the gamma matrices in the Weyl basis (see appendix s V (~ E) 0 σµ µ γ = σ ¯µ 0 imply the spin sum relations ! ! β X uα v β uα u†˙ 6 p m δ ˙ α αβ β = αβ ˙ †α˙ β †α˙ † ¯ 6 p m δ α˙β˙ v v v uβ˙ s ! ! β X vα uβ vα v †˙ 6 p −m δ ˙ α αβ β = αβ ˙ †α˙ β †α˙ † ¯ 6 p −m δ α˙β˙ u u u v˙ s

β

αβ ˙ ˙ where we have defined 6 pαβ˙ ≡ σαµβ˙ pµ and 6 ¯p ≡ σ ¯ µαβ pµ . Since tr(σ µ σ ¯ ν ) = 2η µν [with ˙ βα η = (+, −, −, −)] we have 6 pαβ˙ 6 p¯0 = 2p · p 0 .

An incoming electron state with momentum p~ and spin s is |e− (~p, s)i = (2π)3 2ωp b† (~p, s)|0i. This implies the following rules for the external state wave function for an incoming electron: h0|e(x)|e− (~p, s)i = u(~p, s) e−ip·x , h0|¯ e(x)|e− (~p, s)i = 0 h0|e† (x)|e− (~p, s)i = 0 , h0|¯ e† (x)|e− (~p, s)i = v † (~p, s) e−ip·x . An outgoing electron state with momentum p~ and spin s is he− (~p, s)| = h0|b(~p, s)(2π)3 2ωp . This implies the following rules for the external state wave function for an outgoing electron: he− (~p, s)|e(x)|0i = 0 , he− (~p, s)|¯ e(x)|0i = v(~p, s) e+ip·x he− (~p, s)|e† (x)|0i = u† (~p, s) e+ip·x , he− (~p, s)|¯ e† (x)|0i = 0 . An incoming positron state with momentum p~ and spin s is |e+ (~p, s)i = (2π)3 2ωp d† (~p, s)|0i. This implies the following rules for the external state wave function for an incoming positron: h0|e(x)|e+ (~p, s)i = 0 , h0|¯ e(x)|e+ (~p, s)i = u(~p, s) e−ip·x h0|e† (x)|e+ (~p, s)i = v † (~p, s) e−ip·x , h0|¯ e† (x)|e+ (~p, s)i = 0 . An outgoing positron state with momentum p~ and spin s is he+ (~p, s)| = h0|d(~p, s)(2π)3 2ωp . This implies the following rules for the external state wave function for an incoming positron: he+ (~p, s)|e(x)|0i = v(~p, s) e+ip·x , he+ (~p, s)|¯ e(x)|0i = 0 he+ (~p, s)|e† (x)|0i = 0 , he+ (~p, s)|¯ e† (x)|0i = u† (~p, s) e+ip·x .

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