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Home , Dirac spinor, Majorana equation, Transpose

An Introduction to Quantum Theory (Peskin and Schroeder) Solutions

Andrzej Pokraka

December 12, 2017

Contents

1TheDiracEquation 1 1.1 ! ...... 1 1.2 Gordon Identity ! ...... 7 1.3 products ! ...... 7 1.4 Majorana ! ...... 11 1.5 Supersymmetry ...... 21 1.6 Fierz transformations ! ...... 22 1.7 Discrete symmetries of the Dirac field ! ...... 27 1.8 Bound states ...... 32

1TheDiracEquation

1.1 Lorentz group !

The Lorentz commutation relations are

[J µν ,Jρσ ]=i (gνρJ µσ gµρJ νσ gνσJ µρ + gµσJ νρ) . − −

(a) Define the generators of rotations and boots as 1 Li = ϵijkJ jk and Ki = J 0i, 2 where ijk is a of (123).AninfinitesimalLorentztransformationcanthenbewritten

Φ (1 iθ L iβ K)Φ. → − · − · Write the commutation relations of these vector operators explicitly. Show that the combinations 1 1 J+ = (L + iK) and J = (L iK) 2 − 2 −

1 commute with one another and separately satisfy the combination of . Proof: The L operator commutation relations are

Li,Lj = LiLj Lj Li − 1 ! " = ϵiklJ kl,ϵjmnJ mn 4 1 ! " = ϵiklϵjmn J kl,Jmn 4 i ! " = ϵiklϵjmn glmJ kn gkmJ ln glnJ km + gknJ lm 4 − − i ! " = ϵiklϵjmnglmJ kn ϵiklϵjmngkmJ ln ϵiklϵjmnglnJ km + ϵiklϵjmngknJ lm 4 − − for! the second and last term k l " i ↔ = ϵiklϵjmnglmJ kn ϵilkϵjmnglmJ kn ϵiklϵjmnglnJ km + ϵilkϵjmnglnJ km 4 − − i ! " = ϵiklϵjmnglmJ kn + ϵiklϵjmnglmJ kn ϵiklϵjmnglnJ km ϵiklϵjmnglnJ km 4 − − for! the third and last term m n " i ↔ = ϵiklϵjmnglmJ kn + ϵiklϵjmnglmJ kn ϵiklϵjnmglmJ kn ϵiklϵjnmglmJ kn 4 − − i ! " = ϵiklϵjmnglmJ kn + ϵiklϵjmnglmJ kn + ϵiklϵjmnglmJ kn + ϵiklϵjmnglmJ kn 4 = iϵikl! ϵjmnglmJ kn " = iϵiklϵjmn δlm J kn − = iϵiklϵjlnJ kn − # $ = iϵiklϵjnlJ kn = i δij δkn δinδkj J kn − = i δij J kk J ji # − $ ij = iJ# $ = iϵijkLk where

J kk = i δk δk δkδk αβ α β − β α =0# $ and 1 ϵijkLk = ϵijkϵklmJ lm 2 1 = ϵkij ϵklmJ lm 2 1 = δilδjm δimδjl J lm 2 − 1 # $ = J ij J ji 2 − = J ij#. $

2 The K commutation relations are

Ki,Kj = KiKj Kj Ki − = J 0iJ 0j J 0j J 0i ! " − = i gi0J 0j g00J ij gij J 00 + g0j J i0 . − − This is simplified using properties of the metric gi0#=0, g00 = 1, gij = 1 and the$ generators J 00 =0 − − Ki,Kj = iJ ij − = iϵijkLk. ! " − Next, we need 1 Li,Kj = ϵikl J kl,J0j 2 ! " i ! " = ϵikl gljJ k0 + gkjJ l0 2 − i # $ = ϵikl δljJ k0 δkj J l0 2 − i # $ = ϵikj J k0 ϵijlJ l0 2 − i # $ = ϵijkJ 0k + ϵijkJ 0k 2 = iϵijl# Kl. $

Lastly we compute the angular momentum commutators 1 1 [J+, J ]= (L + iK) , (L iK) − 2 2 − % & 1 = [L, L] i [L, K]+i [K, L]+[K, K] 4 { − } i = [K, L] 2 i = Ki,Lj eˆ eˆ 2 i · j i ! " = Ki,Li 2 =0! "

1 1 J i ,Jj = Li iKi , Lj iKj ± ± 2 ± 2 ± % & ' ( 1 # $ # $ = Li,Lj i Li,Kj i Ki,Lj Ki,Kj 4 ± ± − 1 )! " ! " ! " ! "* = iϵijkLk i iϵijlKl i iϵjilKl + iϵijkLk 4 ± ∓ = i)ϵijkLk ϵijlK# l ϵijl$Kl +# iϵijkLk$ * ∓ ± 1 = ) iϵijkLk ϵijlKl * 2 ∓ i ) * = ϵijk Lk iKk 2 ± i ) * = ϵijkJ k 2 ±

3 (b) The finite-dimensional representations of the rotation group correspond precisely the to the allowed values for angular momentum: integers or half-integers. The result of part (a) implies that all finite-dimensional representations of the Lorentz group correspond to pairs of integers or half integers, (j+,j ),correspondingtopairsofrepresentationsofthe rotation group. Using the fact that J = σ/2 in the -1/2 representation− of angular momentum, write explicitly the 1 1 transformation laws of the 2-component objects transforming according to the 2 , 0 and 0, 2 representations of the Lorentz group. Show that these correspond precisely to the transformations of ψ and ψ giving in (3.37). L# $ R # $ Proof: The representations of the Lorentz group are denoted by (m, n) πm,n where m, n are either half-integers or integers. The irreducible representations are given by ≡

π Li = I J (n) + J (m) I m,n (2m+1) ⊗ i i ⊗ (2n+1) π #Ki$ = i I J (n) J (m) I . m,n (2m+1) ⊗ i − i ⊗ (2n+1) + , 1 # $ With J = σ/2 the 2 , 0 representation is found to be

# $ σi σi σi σi π Li = I I + I = I + I = I = m,n 2 ⊗ 1 2 ⊗ 1 2 2 ⊗ 1 2 ⊗ 1 2 - . # $ σi σi σi σi π Ki = i I I I = i I I = i I = i . m,n 2 ⊗ 1 − 2 ⊗ 1 2 − 2 ⊗ 1 − 2 ⊗ 1 − 2 - . - . # $ Thus, with Li = σi/2 and Ki = iσi/2 the transformation law becomes − i µν (ωµν J ) Φ 1 e− 2 Φ 1 ( 2 ,0) → ( 2 ,0) i µν = 1 ωµν J Φ 1 − 2 ( 2 ,0) - . i 0ν i iν = 1 ω0ν J ωiν J Φ 1 − 2 − 2 ( 2 ,0) - . i 00 i 0i i i0 i ij = 1 ω00J ω0iJ ωi0J ωij J Φ 1 − 2 − 2 − 2 − 2 ( 2 ,0) - . i 0i i 0i i ij = 1 ω0iJ ω0iJ ωij J Φ 1 − 2 − 2 − 2 ( 2 ,0) - . i i ijk k = 1 iω0iK ωij ϵ L Φ 1 − − 2 ( 2 ,0) - . =(1iβ K iθ L) Φ 1 − · − · ( 2 ,0) 1 i = 1 β σ θ σ Φ 1 − 2 · − 2 · ( 2 ,0) - . where we have defined βi = ω = ω and θk = ω ϵijk. 0i − i0 ij 1 Now for the 0, 2 representation we have # $ σi σi σi π Li = I + I I = I + I = m,n 1 ⊗ 2 1 ⊗ 2 1 ⊗ 2 2 2 - . # $ σi σi σi π Ki = i I I I = i I I = i I . m,n 1 ⊗ 2 − 1 ⊗ 2 2 − 2 ⊗ 1 2 ⊗ 1 - . - . # $

4 Thus, with Li = σi/2 and Ki = iσi/2 the transformation law becomes

i µν (ωµν J ) Φ 1 e− 2 Φ 1 (0, 2 ) → (0, 2 ) =(1iβ K iθ L) Φ 1 − · − · (0, 2 ) 1 i = 1+ β σ θ σ Φ 1 . 2 · − 2 · (0, 2 ) - . Upon comparison with equation (3.37) we identify

Φ 1 = ψL and Φ 1 = ψR. ( 2 ,0) (0, 2 )

Thus, the left- and right-handed spinor transform accordingtoseparaterepresentationsoftheLorentzgroup. "

(c) The identity σT = σ2σσ2 allows us to rewrite the ψ transformation in the unitarily equivalent form − L

ψ′ ψ′ (1 + iθ σ/2+β σ/2) , → · · T 2 1 1 where ψ′ = ψL σ .Usingthislaw,wecanrepresenttheobjectthattransformsas 2 , 2 as a 2 2 that has the ψR transformation law on the left and simultaneously, the transposed ψ transforms on the right.× Parametrize this matrix as L # $ V 0 + V 3 V 1 iV 2 . V 1 + iV 2 V 0 − V 3 - − . Show that the object V µ transforms as a 4-vector. Proof: Left-handed transform according to

1 i ψ 1 β σ θ σ ψ . L → − 2 · − 2 · L - . T 2 With ψ′ = ψL σ we verify the transformation law

T 1 i 2 ψ′ 1 β σ θ σ ψ σ → − 2 · − 2 · L -- . . 1 i T = ψT σ2σ2 1 β σ θ σ σ2 L − 2 · − 2 · - . 2 1 T i T 2 = ψ′σ 1 β σ θ σ σ − 2 · − 2 · - . 1 i = ψ′ 1+ β σ + θ σ . 2 · 2 · - . 1 1 1 1 Now we are interested in the transformation properties of a 2 , 2 object. We parameterize the 2 , 2 object as the matrix

0 3# $1 2 # $ V + V V iV µ Φ 1 1 = 1 2 0 − 3 = V σµ ( 2 , 2 ) V + iV V V - − .

5 µ where it will be shown that V is a 4-vector. Applying the transformation to Φ 1 1 we get ( 2 , 2 )

1 i 1 i Φ 1 1 1+ β σ θ σ Φ 1 1 1+ β σ + θ σ ( 2 , 2 ) → 2 · − 2 · ( 2 , 2 ) 2 · 2 · - . - . 1 1 = 1+ (β iθ) σ V µσ 1+ (β σ + iθ) σ 2 − · µ 2 · · - . - . 1 1 = V µσ + (β iθ) σV µσ + V µσ (β + iθ) σ + θ2,β2 µ 2 − · µ 2 µ · O V µ V µ # $ = V µσ + β (σσ + σ σ) iθ (σσ σ σ) µ 2 · µ µ − 2 · µ − µ V µ V µ = V µσ + βi σ ,σ iθi [σ ,σ ] µ 2 ·{ i µ}− 2 · i µ V 0 V j V 0 V j = V µσ + βi σ ,σ + βi σ ,σ iθi [σ ,σ ] iθi [σ ,σ ] µ 2 ·{ i 0} 2 ·{ i j }− 2 · i 0 − 2 · i j V 0 V j V 0 V j = V µσ + βi σ , I + βi σ ,σ iθi [σ , I] iθi [σ ,σ ] µ 2 ·{ i } 2 { i j }− 2 i − 2 i j V 0 V j V j = V µσ + βi (2σ )+ βi (2δ ) iθi 2iϵ σk µ 2 i 2 ij − 2 ijk = V µσ + V 0βiσ V iβ + V j θiϵ σk # $ µ i − i ijk i k i j i where σi, I =2σ , [σi, I]=0, σi,σj =2δij , [σi, I]=2iϵijkσ .Alsonotethatβ V δij = β Vi because gij = δij . { } { } k − − Recall that we have defined the anti-symmetric ω0i = βi and ωij = ϵijkθ .Insertingtheseexpressionsintothe above, we have

µ 0 i i j i k µ 0 i i j k V σµ + V β σi V βi + V θ ϵijkσ = V σµ + V ω0iσ V ω0i + V ωjkσ − µ 0 i − i 0 i j = V σµ + V ω0iσ + V ωi0σ + ωij V σ µ ν µ = V σµ + ωµν σ V ν ν µ = δµ + ω µ Vν σ .

We would like to show that this is identical to equation (3.19)inP&S.P&Sassertthata4-vector# $ V µ transforms as follows

i V α δα ω ( µν )α V β → β − 2 µν J β - . for ( µν ) = δµδν δν δµ.Withthisdefinitionof( µν )α the transformation condition becomes J αβ α β − α β J β i i δα ω ( µν )α = δα ω ( µν ) gγα β − 2 µν J β β − 2 µν J γβ i = δα ω δµδν δν δµ gγα β − 2 µν γ β − γ β i + , = δα (ω ω ) gγα β − 2 γβ − βγ = δα iω gγα β − γβ = δα iωα. β − β µ µ This is the identical result obtained from transforming V σµ.Thus,weseethatV is indeed a 4-vector. "

6 1.2 Gordon Identity !

Derive the Gordon identity, µ µ µν µ p′ + p iσ qν u¯ (p′) γ u (p)=¯u (p′) + u (p) , 2m 2m % & where q =(p′ p). − Proof: The computation is straightforward:

µν i µ ν σ q = [γ ,γ ](p′ p ) ν 2 ν − ν i µ µ i µ µ = (γ ✁p′ p✁′γ ) (γ ✁p ✁pγ ) 2 − − 2 − i µν µ µ i µ µν µ = (2g p′ ✁p′γ ✁p′γ ) (γ ✁p 2g pν + γ ✁p) 2 ν − − − 2 − µ µ µ µ = i (p′ ✁p′γ ) i (γ ✁p p ) − − − µ µ µν µ µ µ µ µ µ p′ + p iσ qν p′ + p (p′ ✁p′γ )+(γ ✁p p ) u¯ (p′) + u (p)=¯u (p′) − − − u (p) 2m 2m 2m % & % & µ µ ✁p′γ + γ ✁p =¯u (p′) u (p) 2m % & mγµ + γµm =¯u (p′) u (p) 2m % & µ =¯u (p′) γ u (p) where we have used the Fourier transformed andit’sadjointequation (✁p m) u (p)=0and u¯ (p)(✁p m)=0. − − "

1.3 Spinor products !

Together is Problems 5.3 and 5.6 we develop an efficient computational method for processes involving massless . Let kµ, kµ be fixed 4-vectors satisfying k2 =0, k2 = 1, k k =0.Definebasicspinorsinthefollowingway:Letu 0 1 0 1 − 0 · 1 L0 be the left-handed spinor for for a with momentum k0.LetuR0 = k✁1uL0.Then,foranyp such that p is lightlike p2 =0 define u (p)= 1 pu and u (p)= 1 pu L √2p k0 ✁ R0 R √2p k0 ✁ L0 . # $ · · This set of conventions defines the phases of spinors unambiguously (except when p is parallel to k0).

(a) Show that k✁0uR0 =0.Showthat,foranylightlikep, ✁puR (p)=0. Proof:

k✁0uR0 = k✁0k✁1uL0 µ ν = k0 k1 γµγν uL0 = kµkν (2g γ γ ) u 0 1 µν − ν µ L0 =2k k u k✁ k✁ u 0 · 1 L0 − 1 0 L0 =0

7 where we have used the Dirac equation for a massless k✁0uL0 =0and the k0 k1.Nowforanylightlike 4-momentum p the definition of the spinors above satisfy the massless Diracequation · 1 pu✁ R (p)= ✁p✁puL0 √2p k · 0 1 2 = p uL0 √2p k · 0 =0.

"

(b) For the choices k =(E,0, 0, E), k =(0, 1, 0, 0),constructu , u , u (p),andu (p) explicitly. 0 − 1 L0 R0 L R Proof: For this problem we will need the operator k✁ which in the Chiral or Weyl is given by

0 k0 k σ k✁ = k0 + k σ −0 · - · . 00k0 k3 k1 ik2 − − − 00k1 + ik2 k0 + k3 = # $ . ⎛ k0 + k3 k1 ik2 − 00⎞ − # $ ⎜ k1 + ik2 k0 k3 00⎟ ⎜ − ⎟ ⎝ ⎠ From equation (3.50) in P&S the Dirac spinors for are given by

√k0 σξ u (k0)= · √k σξ¯ - 0 · . E (σ0 + σ3)ξ = E (σ0 σ3)ξ - 5 − . 5 20 ξ 00 = √E ⎛ 6- . ⎞ ⎜ 00 ⎟ ⎜ ξ ⎟ ⎜ 6 02 ⎟ ⎜ - . ⎟ ⎝ 10 ⎠ ξ 00 = √2E ⎛ - 00. ⎞ ξ ⎜ 01 ⎟ ⎜ - . ⎟ ⎝ ⎠ T T T where ξ 10 , 01 and σ¯ = σ0 σ .Takingξ = 10 the left-handed spinor is ∈ − 7# $ # $ 8 # $ # $ 10 1 1 00 0 0 u = √2E = √2E L0 ⎛ - 00.- 1 . ⎞ ⎛ 0 ⎞ ⎜ 01 0 ⎟ ⎜ 0 ⎟ ⎜ - .- . ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ T while the right-handed spinor is given by taking ξ = 01

10# $ 0 0 00 1 0 u = √2E = √2E . R0 ⎛ - 00.- 0 . ⎞ ⎛ 0 ⎞ ⎜ 01 1 ⎟ ⎜ 1 ⎟ ⎜ - .- . ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

8 The chiral spinors for any momentum p are then given by 1 uL (p)= ✁puR0 √2p k · 0 00p0 p3 p1 ip2 0 √2E 00p1−+ ip2 − p0 +−p3 0 = ⎛ − # $ ⎞ ⎛ ⎞ 0 3 p0 + p3 p1 ip2 000 2E (p + p ) − # $ ⎜ p1 + ip2 p0 p3 00⎟ ⎜ 1 ⎟ 5 ⎜ − ⎟ ⎜ ⎟ p⎝1 + ip2 ⎠ ⎝ ⎠ 1 −p0 + p3 = ⎛ ⎞ p0 + p3 0 9 ⎜ 0 ⎟ ⎜ ⎟ ⎝ ⎠ 1 uR (p)= ✁puL0 √2p k · 0 00p0 p3 p1 ip2 1 − − − √2E 00p1 + ip2 p0 + p3 0 = ⎛ − # $ ⎞ ⎛ ⎞ 0 3 p0 + p3 p1 ip2 000 2E (p + p ) − # $ ⎜ p1 + ip2 p0 p3 00⎟ ⎜ 0 ⎟ 5 ⎜ − ⎟ ⎜ ⎟ ⎝0 ⎠ ⎝ ⎠ 1 0 = ⎛ ⎞ p0 + p3 p0 + p3 9 ⎜ p1 + ip2 ⎟ ⎜ ⎟ ⎝ ⎠ "

(c) Define the spinor products s (p1,p2) and t (p1p2),forp1,p2 lightlike, by

s (p1,p2)=¯uR (p1) uL (p2) and t (p1,p2)=¯uL (p1) uR (p2) .

Using the explicit forms for the uλ given in part (b), compute the spinor products explicitly andshowthatt (p1,p2)= s (p ,p )∗ and s (p ,p )= s (p ,p ).Inaddition,showthat 1 2 1 2 − 2 1 s (p ,p ) 2 =2p p . | 1 2 | 1 · 2 Thus the spinor products are the square roots of 4-vector dot products. Proof:

9 s (p1,p2)=¯uR (p1) uL (p2) 1 2 0010 p2 + ip2 − 0 3 1 0 3 1 2 0001 p2 + p2 = 0 3 0 3 00p1 + p1 p1 ip1 ⎛ ⎞ ⎛ ⎞ 6(p1 + p1)(p2 + p2) − 1000 0 # $ ⎜ 0100⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ ⎠ 1 0 = 00p0 + p3 p1 ip2 ⎛ ⎞ (p0 + p3)(p0 + p3) 1 1 1 − 1 p1 + ip2 6 1 1 2 2 − 2 2 # $ ⎜ p0 + p3 ⎟ ⎜ 2 2 ⎟ p0 + p3 p1 + ip2 + p1 ip2 p0 + p3 ⎝ ⎠ = 1 1 − 2 2 1 − 1 2 2 0 3 0 3 # $# (p1 + p$1)(p#2 + p2) $# $ p0p1 p3p1 + ip0p2 + ip3p2 + p1p0 ip2p0 + p1p3 ip2p3 = − 1 2 − 1 52 1 2 1 2 1 2 − 1 2 1 2 − 1 2 0 3 0 3 (p1 + p1)(p2 + p2) p1 p0 + p3 p0 p3 p1 + i p0 + p3 p2 p2 p0 + p3 = 1 2 2 − 1 −51 2 1 1 2 − 1 2 2 0 3 0 3 # $ # (p$1 + p1)(##p2 + p2)$ # $$ = s (p ,p ) − 2 1 5

1 0 3 0 3 1 0 3 2 2 0 3 2 2 p1 p2 + p2 p1 p1 p2 + i p1 + p1 p2 p1 p2 + p2 s (p1,p2) = − − 0 3 0 3 − | | : # $ # (p$1 + p1)(##p2 + p2) $ # $$: : 1 0 3 0 3 1 0 3 2 2 0 3 :2 p1 p2 + p2 p1 p1 p2 + i p1 + p1 p2 p1 p2 + p2 = − − 0 3 0 3 − : # $ # (p$1 + p1)(##p2 + p2) $ # $$: : 1 0 3 0 3 1 2 0 3 2 2 0 3 : 2 p1 p2 + p2 p1 p1 p2 + p1 + p1 p2 p1 p2 + p2 = − − 0 3 0 3 − # # $ # (p$1 +$p1)(p##2 + p2) $ # $$ 1 2 0 3 2 1 0 3 0 3 1 0 3 1 2 p1 p2 + p2 2p1 p2 + p2 p1 p1 p2 + p1 p1 p2 = − 0 3 0 − 3 − − # $ # $ #(p1 + p1$#)(p2 + p2)$ # $# $ 0 3 2 2 2 0 3 2 2 0 3 2 2 0 3 2 p1 + p1 p2 2 p1 + p1 p2p1 p2 + p2 + p1 p2 + p2 − 0 3 0 3 ×# $ # $ # (p1 + p$1)(p2#+ p2) $ # $ # $ 0 3 0 3 p2 + p2 1 2 2 2 p1 + p1 1 2 2 2 1 1 2 2 = 0 3 p1 + p1 + 0 3 p2 + p2 2 p1p2 + p2p1 #(p1 + p1$) #(p2 + p2)$ − 0 3 +# $ # $ , 0 3 +# $ # $ , # $ p2 + p2 1 2 2 2 p1 + p1 1 2 2 2 1 1 2 2 = 0 3 p1 + p1 + 0 3 p2 + p2 2 p1p2 + p2p1 #(p1 + p1$) #(p2 + p2)$ − +# $ # $ , +# $ # $ , # $ Since p1 and p2 are lightlike 2 2 2 2 p1 + p2 = p0 p3 = p0 + p3 p0 p3 i i i − i i i i − i we have # $ # $ # $ # $ # $# $ s (p ,p ) 2 = p0 + p3 p0 p3 + p0 + p3 p0 p3 2 p1p1 + p2p2 | 1 2 | 2 2 1 − 1 1 1 2 − 2 − 1 2 2 1 = p0p0 + p0p3 p3p0 p3p3 + p0p0 + p3p0 p0p3 p3p3 2 p1p1 + p2p2 #1 2 $#1 2 − 1 2$− #1 2 1$#2 1 2$− 1# 2 − 1 2 − $ 1 2 2 1 =2p0p0 2 p1p1 + p2p2 + p3p3 1 2 − 1 2 2 1 1 2 # $ =2p p . 1 · 2 # $ "

10 1.4 Majorana fermions !

Recall form Eq. (3.40) that one can write a relativistic equation for massless 2-component fermion field that transforms as the upper two components of a (ψL).Callsucha2-componentfieldχa (x), a =1, 2. (a) Show that it is possible to write an equation for χ (x) as a massive field in the following way:

2 iσ¯ ∂χ imσ χ∗ =0. · − That is, show, that this equation is relativistically invariant and, second, that it implies the Klein-Gordon equation, ∂2 + m2 χ =0.ThisformofthefermionmassiscalledMajoranamassterm. Proof:# $

1 The , Λ 2 ,whichLorentztransformsafermionfield(spinor)isgivenbyequation(3.30)

Λ 1 0 i µν ( 2 ,0) Λ 1 (θ, β)=exp ωµν S = 2 −2 0Λ1 - . ; (0, 2 ) < where i Sµν = [γµ,γν ] 4 i k ijk is the generator for spinor, β = ω0i = ωi0 and θ = ωij ϵ .Thistransformationmatrixis block diagonal. This is seen from the block diagonal form of the− generators of boost (3.26) and (3.27)

i σi 0 S0i = −2 0 σi - − . 1 Sij = ϵijkΣk 2 where the spin operator of Dirac theory is σi 0 Σk = . 0 σi - . The blocks Λ 1 and Λ 1 are the left and right handed representations of the Lorentz group. A spinor which ( 2 ,0) (0, 2 ) 1 transforms according to Λ 2 is called a Dirac spinor. Because of the block diagonal form left and right spinors transform in different representations of the Lorentz group

Λ 1 0 Λ 1 ψL ( 2 ,0) ψL ( 2 ,0) Λ 1 ψ = = . 2 0Λ1 ψR Λ 1 ψR ; (0, 2 ) < - . ; (0, 2 ) < Under infinitesimal rotations by θ and boosts β the transformation laws for the left and right handed spinorsare

1 i Λ 1 = 1 β σ θ σ ( 2 ,0) − 2 · − 2 · - . 1 i Λ 1 = 1 β σ + θ σ . (0, 2 ) − 2 · 2 · - . Now equation (3.40) tells us that we can write the relativistic field equation for a massless fermion which transforms under µ Λ 1 as iσ¯ ∂χ =0.Furthermore,weknowhowγ transforms. This is given by equation (3.29) ( 2 ,0) ·

1 µ µ ν Λ−1 γ Λ 1 =Λ γ . 2 2 ν

On the left the Lorentz transformations are acting on the spinor indices of γµ while on the right the Lorentz transformation

11 acts on the space time index µ.Wecanworkoutthetransformationforσ¯ because σ¯µ is just the off-diagonal blocks of γµ

µ ν 1 µ Λ γ =Λ−1 γ Λ 1 ν 2 2 1 µ µ ν 0Λ−1 σ Λ 0, 1 0Λν σ ( 2 ,0) ( 2 ) µ ν = 1 µ . Λ ν σ¯ 0 ⎛ Λ− 1 σ¯ Λ 1 ,0 0 ⎞ - . (0, 2 ) ( 2 ) ⎝ ⎠ µ 1 ν ν 1 From the above we conclude that σ¯ Λ− =Λ 0, 1 σ¯ Λ−1 .Wewillalsoneedthefactthatifχtransforms according to µ ( 2 ) ( 2 ,0) 2 2 Λ 1 then σ χ∗ transforms under Λ 1 (i.e., χ is left-handed and σ χ∗ is right-handed). We prove this here. Applying ( 2 ,0) #(0, 2 )$ 2 Λ 1 to σ χ∗ yields ( 2 ,0)

2 2 1 i ∗ σ χ∗ σ 1 β σ θ σ χ → − 2 · − 2 · -- . . 2 1 i = σ 1 β σ∗ + θ σ∗ χ∗ − 2 · 2 · - . 1 i = 1+ β σ θ σ χ∗ 2 · − 2 · - . =Λ1 χ∗. (0, 2 )

Lorentz transforms; review of a field χ (no spatial orientation).

1 1. χ:UnderLorentztransformationsthescalarχ transforms as χ (x) χ′ (x′)=χ Λ− x′ where x x′ =Λx. → → 2. ∂µχ:UnderLorentztransformations∂µχ transforms as # $

ν ∂x ∂ 1 β α ∂µχ (x) ∂µ′ χ′ (x′)= µ ν χ Λ− α x′ → ∂x′ ∂x 1 ν +ρ , ∂ Λ− x′ # $ ρ ∂ 1 β α = µ ν χ Λ− α x′ # ∂x$′ ∂x + , 1 ν ρ ∂ # 1 β $ α = Λ− g χ Λ− x′ ρ µ ∂xν α # 1$ν +1# $ , = Λ− µ (∂ν χ) Λ− x′

# $ # µ $ µ µ µ ν where x x′ =Λx.Noticethatthisisoppositetohowthevectorx transforms x x′ =Λν x .Thisshows → µ → that while x is a contravariant vector, ∂µ is a covariant vector. Applying the Lorentz transformation to the field equation we have

2 µ 1 ν 1 2 1 iσ¯ ∂χ(x) imσ χ∗ (x)=0 iσ¯ Λ− ∂ν Λ 1 χ Λ− x imΛ 1 σ χ∗ Λ− x =0 · − → µ ( 2 ,0) − (0, 2 )

µ 1 #ν $ ν 1 # $ # $ where we know how ∂µχ transforms. Using σ¯ Λ− =Λ 0, 1 σ¯ Λ−1 we conclude µ ( 2 ) ( 2 ,0) # $ 2 ν 1 1 2 1 iσ¯ ∂χ(x) imσ χ∗ (x)=0 iΛ 0, 1 σ¯ Λ−1 ∂ν Λ 1 ,0 χ Λ− x imΛ 0, 1 σ χ∗ Λ− x =0 · − → ( 2 ) ( 2 ,0) ( 2 ) − ( 2 ) ν 1 2 1 =Λ1 iσ¯ ∂ν χ Λ− x im# σ χ∗$ Λ− x =0. # $ (0, 2 ) −

)2 # $ # $* Thus, we see that the field equation iσ¯ ∂χ(x) imσ χ∗ (x)=0is relativistically invariant. · − 2 Next, we must show that iσ¯ ∂χ(x) imσ χ∗ (x)=0implies the Klein-Gordon equation. The equation iσ¯∗ ∂χ∗ (x)+ 2 · 1 −2 − 2· imσ ∗χ (x)=0implies χ∗ (x)= σ σ¯ ∂χ(x).Substitutionintothecomplexconjugateofiσ¯ ∂χ(x) imσ χ∗ (x)=0 m · · −

12 we have µ 2 iσ¯ ∗∂ χ∗ (x) imσ ∗χ (x)=0 µ − µ 1 2 ν 2 iσ¯ ∗∂ σ σ¯ ∂ χ (x) imσ ∗χ (x)=0 µ m ν − 2 2 µ 2 ν 2 2 σ σ σ¯ ∗σ σ¯ ∂µ∂ν χ (x)+m σ ∗χ (x)=0 2 µ ν 2 2 σ σ σ¯ ∂µ∂ν χ (x)+m σ ∗χ (x)=0 σµσ¯ν ∂ ∂ χ (x) m2χ (x)=0 µ ν − gµν ∂ ∂ χ (x) m2χ (x)=0 − µ ν − ∂2χ (x)+m2χ (x)=0 as required. "

2 (b) Does the follow from a Lagrangian? The mass term would seem to be the variation of σ ab χa∗χb∗; however, since σ2 is antisymmetric, this expression would vanish if χ (x) were an ordinary c-number field. When we go # $ to quantum field theory, we know that χ (x) will become an anti-commuting quantum field. Therefore, it makes sense to develop its classical theory by considering χ (x) as a classical anti-commuting field, that is, as a field that takes as values Grassmann numbers which satisfy αβ = βα for any α,β. − Note that this relation implies that α2 =0.AGrassmannfieldξ (x) can be expanded in a basis of functions as

ξ (x)= αnφn (x) , n = where the φn (x) are orthogonal c-number functions and the αn are a set of independent Grassmann numbers. Define the of a product of Grassmann numbers to reverse the order:

(αβ)∗ β∗α∗ = α∗β∗. ≡ − This rule imitates the Hermitian conjugation of quantum fields. Show that the classical action,

4 im T 2 2 S = d x χ†iσ¯ ∂χ + χ σ χ χ†σ χ∗ , ˆ · 2 − % & T # $ (where χ† =(χ∗) )isreal(S∗ = S),andthatvaryingthisS with respect to χ and χ∗ yields the Majorana equation. Proof:

The trick to this problem is realizing that S C and so S∗ = S†.Thecomplexconjugateoftheactionistherefore ∈

4 µ im T 2 2 † S∗ = d x χ†iσ¯ ∂ χ † + χ σ χ χ†σ χ∗ ˆ µ 2 − > - . ? # $ # $ 4 µ im 2 T 2 = d x (∂ χ)† iσ¯ †χ†† χ†σ †χ † χ∗†σ †χ†† ˆ − µ − 2 − % & # $ 4 µ im 2 T 2 = d x (∂ χ)† iσ¯ χ χ†σ χ∗ χ σ χ ˆ − µ − 2 − % & # $ 4 µ 4 im T 2 2 = d x (∂ χ)† iσ¯ χ + d x χ σ χ χ†σ χ∗ ˆ − µ ˆ 2 − % & ' ( # $ µ 4 µ 4 im T 2 2 = dS iχ†σ¯ χ d x χ†iσ¯ ∂ χ + d x χ σ χ χ†σ χ∗ IBP ˛ µ − − ˆ − µ ˆ 2 − % & ! " ! # $" # $ 4 µ 4 im T 2 2 = d x χ†iσ¯ ∂ χ + d x χ σ χ χ†σ χ∗ ˆ µ ˆ 2 − % & = S. !# $" # $

13 Notice that we have assumed that the field χ vanishes at the boundary of the integration region. Recall that the action is the integral of the Lagrange density .Inourcasewehave L

im T 2 2 = χ†iσ¯ ∂χ + χ σ χ χ†σ χ∗ L · 2 − µ im# 2 2 $ = χ∗iσ¯ ∂ χ + σ χ χ σ χ∗χ∗ a ab µ b 2 ab a b − ab a b # $ .TheEuler-Lagrangeequationsarederivedfromvariationoftheaction,requiringthattofirstorderδχ that δS =0. However, we cannot just use the Euler-Lagrange equations because χ is a Grassmann field. Varying the action with respect to χ∗ yields

δS S [χ∗ + δχ∗] S [χ∗] ≡ − 4 4 = d x (χ∗ + δχ∗,∂ χ∗ + ∂ δχ∗) d x (χ∗,∂ χ∗) ˆ L µ µ − ˆ L µ

4 µ im 2 = d x (χ∗ + δχ∗) iσ¯ ∂ χ + σ (χ χ (χ∗ + δχ∗)(χ∗ + δχ∗)) ˆ a a ab µ b 2 ab a b − a a b b % & 4 µ im 2 d x χ∗iσ¯ ∂ χ + σ (χ χ χ∗χ∗) − ˆ a ab µ b 2 ab a b − a b % & 4 µ im 2 = d x δχ∗iσ¯ ∂ χ + σ ( χ∗δχ∗ δχ∗χ∗) ˆ a ab µ b 2 ab − a b − a b % & 4 µ im 2 2 = d x δχ∗iσ¯ ∂ χ + σ δχ∗χ∗ σ δχ∗χ∗ ˆ a ab µ b 2 − ab a b − ab a b % & 4 µ #2 $ = d xδχ∗ iσ¯ ∂ χ imσ χ∗ ˆ a ab µ b − ab b ! " =0

Since this holds for all integration volumes µ 2 iσ¯ ∂ χ imσ χ∗ =0 µ − which is the Majorana equation. Similarly, with the action

4 µ im 2 T 2 S = S† = d x (∂ χ)† iσ¯ χ χ†σ χ∗ χ σ χ ˆ − µ − 2 − % & # $ and varying with respect to χ yields the complex conjugate equation

µ 2 iσ¯ ∂ χ∗ imσ χ =0. µ − "

(c) Let us write a 4-component Dirac field as ψ ψ (x)= L , ψ - R . and recall that the lower components of ψ transform in a way equivalent by a unitary transformation to the complex conjugate of the representation ψL.Inthiswaywecanrewritethe4-componentDiracfieldintermsoftwo2-component spinors: 2 ψL (x)=χ1 (x) and ψR (x)=iσ χ2∗ (x) .

Rewrite the Dirac Lagrangian in terms of χ1 and χ2 and note the form of the mass term.

14 µ T 2 T Proof: The Dirac Lagrange density is = ψ¯ (iγ ∂ m) ψ.TheDiracspinorisψ = ψL ψR = χ1 iσ χ∗ . LD µ − 2 = ψ¯ (iγµ∂ m) ψ # $ # $ LD µ − µ T 2 01 miσ∂µ χ1 = χ1† iχ2 σ −µ 2 − 10 iσ¯ ∂µ m iσ χ∗ - .- − .- 2 . # $ µ 2 T 2 mχ1 + iσ ∂µiσ χ2∗ = iχ2 σ χ1† − µ 2 − iσ¯ ∂µχ1 miσ χ2∗ - − . # T 2 $ µ 2 µ 2 = iχ σ mχ + iσ ∂ iσ χ∗ + χ† iσ¯ ∂ χ miσ χ∗ − 2 − 1 µ 2 1 µ 1 − 2 T 2 2 T 2 2 = iχ2 σ σ #∂ σ χ2∗ + iχ†σ¯ ∂χ1$ + imχ#2 σ χ1 miχ†σ χ2∗$ · 1 · − 1 T 2 2 T 2 2 = iχ†σ¯ ∂χ +# iχ σ$ σσ ∂χ∗ + im χ σ χ χ†σ χ∗ 1 · 1 2 · 2 2 1 − 1 2 T T+ 2 2 , = iχ†σ¯ ∂χ + iχ σ¯∗ ∂χ∗ + im χ σ χ χ†σ χ∗ 1 · 1 2 · 2 2 1 − 1 2 T + T 2 2 , = iχ†σ¯ ∂χ + iχ σ¯∗ ∂χ∗ + im χ σ χ χ†σ χ∗ 1 · 1 2 · 2 2 1 − 1 2 + , 2 µ 2 µ where we have used σ σ σ =¯σ ∗. T µ T µ Since the second term, iχ σ¯∗ ∂χ∗,isjustanumberwecansimplyitfurtherusingthefactthat(¯σ ) = σ 2 · 2 − T µ T µ T iχ2 σ¯ ∗∂µχ2∗ = iχ2 σ¯ ∗∂µχ2∗ µ = i#∂µχ2† σ¯ †χ2 $ µ = i∂µχ2† σ¯ χ2.

Thus, the Lagrange density is

T 2 2 = iχ†σ¯ ∂χ + iχ†σ¯ ∂χ + im χ σ χ χ†σ χ∗ . LD 1 · 1 2 · 2 2 1 − 1 2 + ,

Notice that if χ1 = χ2 we have twice the Majorana Lagrange density from part (b):

T 2 2 = iχ†σ¯ ∂χ + iχ†σ¯ ∂χ + im χ σ χ χ†σ χ∗ LM 1 · 1 1 · 1 1 1 − 1 1 + , im T 2 2 =2iχ†σ¯ ∂χ + χ σ χ χ†σ χ∗ . 1 · 1 2 1 1 − 1 1 @ + ,A "

(d) Show that the action of part (c) has a global symmetry. Compute the divergences of the currents

µ µ µ µ µ J = χ†σ¯ χ and J = χ†σ¯ χ χ†σ¯ χ , 1 1 − 2 2 for the theories of parts (b) and (c), respectively, and relate your results to the symmetries of these theories. Construct atheoryofNfreemassive2-componentfermionfieldswithO (N) symmetry (that is, the symmetry of rotations in an N-dimensional space). Proof: The action 4 T 2 2 S = d x iχ†σ¯ ∂χ + iχ†σ¯ ∂χ + im χ σ χ χ† σ χ∗ ˆ 1 · 1 2 · 2 2 1 − 1 2 ' + ,( is invariant under the U (1) symmetry ψ eiθψ. →

15 To verify this we write

ψ eiθψ → iθ χ1 e χ1 2 2 iθ . iσ χ∗ → iσ e− χ2 ∗ - 2 . - . # $ iθ iθ iθ Thus, in terms of the χ′s the transformation ψ e ψ is χ e χ and χ e− χ .Becauseineachtermofthe → 1 → 1 2 → 2 action χ1 is partnered with either its complex conjugate or χ2 and vice versa for χ2 the action is invariant under the U(1) iθ iθ symmetry χ e χ and χ e− χ . 1 → 1 2 → 2 We now compute the divergences of the currents,

µ µ J = χ†σ¯ χ µ µ µ J = χ†σ¯ χ χ†σ¯ χ , 1 1 − 2 2 for the theories of part (b) and (c). To derive these currents we apply equation (2.12). Part (c): Under the U(1) transformation the Lagrangian of part (c) is invariant. To compute the currents we need to know the equations of motion. The equation of motion is obtained from the Euler-Lagrange equations

∂ ∂ ∂ L L =0. µ ∂ (∂ χ ) − ∂χ - µ i . i

With χi = χ1 the EL equation yields

µ T 2 ∂ iχ†σ¯ imχ σ =0 µ 1 − 2 + ,µ T 2 ∂ χ† σ¯ mχ σ =0 µ 1 − 2 µ + , † 2 T σ¯ † ∂ χ† mσ †χ † =0 µ 1 − 2 µ 2 + σ¯ ∂, χ mσ χ∗ =0. µ 1 − 2

Similarly, for χi = χ2

µ ∂ T 2 ∂µ iχ2†σ¯ + im χ2 σ χ1 =0 ∂χ2 + , µ ∂ # T 2 $ ∂µχ2† σ¯ m χ1 σ χ2 =0 − ∂χ2 + , µ # T $2 ∂ χ† σ¯ mχ σ =0 µ 2 − 1 µ 2 +σ¯ (∂,χ ) mσ χ∗ =0. µ 2 − 1

Now the divergence of the current can be computed

µ µ µ ∂ J = ∂ χ†σ¯ χ χ† σ¯ χ µ µ 1 1 − 2 2 + µ µ , µ µ = ∂ χ†σ¯ χ ∂ χ† σ¯ χ + χ†σ¯ ∂ χ χ†σ¯ ∂ χ µ 1 1 − µ 2 2 1 µ 1 − 2 µ 2 µ µ µ µ =(¯σ ∂ χ )† χ (¯σ ∂ χ )† χ + χ† (¯σ ∂ χ ) χ† (¯σ ∂ χ ) µ 1 1 − µ 2 2 1 µ 1 − 2 µ 2 2 2 2 2 = mσ χ∗ † χ mσ χ∗ † χ + χ† mσ χ∗ χ† mσ χ∗ 2 1 − 1 2 1 2 − 2 1 T 2 T 2 2 2 = #mχ2 σ χ$1 mχ#1 σ χ2 +$ mχ† σ χ2∗# mχ†σ$ χ1∗ # $ − 1 − 2 =0.

16 Since the divergence of the current vanishes, the current is conserved under U(1) transformations. Part (c): Under the U(1) transformation the Lagrangian of part (b) is NOT invariant

im T 2 2 iφ iφ im iφ T 2 iφ iφ 2 iφ χ†iσ¯ ∂χ + χ σ χ χ†σ χ∗ χe † iσ¯ ∂ χe + χe σ χe χe † σ χe ∗ · 2 − → · 2 − # $ # $ im# T$ 2 2iφ+# $2 2iφ # $ # $ , = χ†iσ¯ ∂χ + χ σ χe χ†σ χ∗e− · 2 − = . # $ ̸ L However, if m =0then the Lagrangian is symmetric under U(1) phase rotations. The equations of motion are µ 2 σ¯ ∂µχ mσ χ∗ =0 µ − T 2 ∂ χ† σ¯ mχ σ =0. µ − Thus the divergence of the current is # $ µ µ µ ∂µ χ†σ¯ χ = ∂µχ†σ¯ χ + χ† (¯σ ∂µχ) T 2 2 # $ = #mχ σ $χ + χ† mσ χ∗ T 2 2 = #mχ σ χ$+ mχ†σ# χ∗. $ Like the Lagrangian the current is only conserved if m =0. For the last part of this question we construct a theory of N free massive 2-component fermion fields with O(N) symmetry (the symmetry of rotations in an N-dimensional space). Each free massive particle is described by the Lagrangian of part (b) im T 2 2 = χ† iσ¯ ∂χ + χ σ χ χ† σ χ∗ La a · a 2 a a − a a where a 1, 2, ..., N .ThetotalLagrangianisthesumoftheindividualLagrangia# $ ns ∈{ } im T 2 2 = χ† iσ¯ ∂χ + χ σ χ χ† σ χ∗ . L a · a 2 a a − a a a = # $ Each fermion field satisfies the equation of motion µ 2 σ¯ ∂µχa mσ χa∗ =0 µ − T 2 ∂ χ† σ¯ mχ σ =0. µ a − a To have O(N) symmetry the Lagrangian must# be invariant$ under rotations in N-dimensional space. Let the rotation th operator be denoted by Rab where a, b are the components of the operator. Applied to the b fermion field the rotation operator takes the bth fermion field to the ath fermion field

Rabχb = χa. 1 T Since R is a rotation, it is an (i.e., R− = R and R R). Application of R to the Lagrangian yields ab ∈ im T 2 2 (R χ )† iσ¯ ∂ (R χ )+ (R χ ) σ (R χ ) (R χ )† σ (R χ )∗ L→ ab b · ab b 2 ab b ab b − ab b ab b =ab + , im T T 2 2 = R† R χ†iσ¯ ∂χ + R R χ σ χ R† R∗ χ†σ χ∗ ab ab b · b 2 ab ab b b − ab ab b b =ab + , im T 2 2 = R R χ†iσ¯ ∂χ + R R χ σ χ R R χ†σ χ∗ ba ab b · b 2 ba ab b b − ba ab b b =ab + , im T 2 2 = χ†iσ¯ ∂χ + χ σ χ χ†σ χ∗ b · b 2 b b − b b =b + , = . L

17 Thus, we see that the Lagrangian for massive two-component fermions is invariant under O(N) transformations. " (e) Quantize the Majorana theory of parts (a) and (b). That is,promoteχ (x) to a quantum fields satisfying the canonical anti-commutation relation (3) χ (x) ,χ† (y) = δ δ (x y) , a b ab − construct a Hermitian Hamiltonian, and find7 a representatio8 nofthecanonicalcommutationrelationsthatdiagonalizes the Hamiltonian in terms of a set of creation and annihilationoperators.(Hint:Compareχ (x) to the top two components of the quantized Dirac field.) Proof: The book suggests comparing χ to the upper two component of the quantized Dirac field

3 d p 1 s s ip x s s ip x ψ = a u (p)e− · + b †v (p)e · ˆ (2π)3 p p 2Ep s 3 = # s $ s d p 5 1 s √p σξ ip x s √p ση ip x = a · e− · + b † · e · . ˆ (2π)3 p √p σξ¯ s p √p ση¯ s 2Ep s = - - · . - − · . . 5 2 Setting ψR = iσ ψL in the Dirac Lagrangian yields the Majorana Lagrangian. Thus, we expect the Dirac field to be a s 2 solution to the Majorana equation under certain restrictions (i.e., η = iσ ξ). Setting χ equal to ψL,thequantized Majorana field becomes − 3 d p p σ s s ip x s s ip x χ = 3 · apξ e− · + bp†η e · . ˆ (2π) 2Ep 9 s = # $ The condition of charge conjugation

s 2 s s 2 s u (p)= iγ (v (p))∗ and v (p)= iγ (u (p))∗ − − places a restriction on the spinors, namely ηs = iσ2ξ.Thus,theMajoranafieldbecomes − 3 d p p σ s 2 s ip x s s ip x χ = 3 · apiσ η ∗e− · + bp†η e · . ˆ (2π) 2Ep 9 s = # $ µ 2 We will now test our solution in the Majorana equation of motion iσ¯ ∂µχ = imσ χ∗.TheRHSbecomes

3 µ µ d p p σ s 2 s ip x s s ip x iσ¯ ∂µχ = iσ¯ ∂µ 3 · apiσ η ∗e− · + bp†η e · ˆ (2π) 2Ep 9 s 3 = # $ µ d p p σ s 2 s ip x s s ip x = iσ¯ 3 · apiσ η ∗ ( ipµ) e− · +(ipµ) bp†η e · ˆ (2π) 2Ep − 9 s = # $ 3 d p (¯σ p)(p σ) s 2 s ip x s s ip x = i 3 √σ¯ p · · apσ η ∗e− · + ibp†η e · ˆ (2π) · 2Ep 6 s = # $ 3 d p σ¯ p s 2 s ip x s s ip x = im 3 · apσ η ∗e− · + ibp†η e · ˆ (2π) 2Ep 6 s = # $ where we have used the fact that (¯σ p)(σ p)=m2.TheLHSis · · 3 ∗ 2 2 d p p σ s 2 s ip x s s ip x imσ χ∗ = imσ 3 · apiσ η ∗e− · + bp†η e · ˆ (2π) 2Ep ; 9 s < = # $ 3 2 d p p σ∗ s 2 s ip x s s ip x = imσ 3 · iap†σ ∗η e · + bpη ∗e− · . ˆ (2π) 2Ep − 6 s = # $ 18 To deal with the square root of σ∗ we note that the square root is defined as a taylor series. Inserting unity in the form 2 2 n 2 2 σ σ between each consecutive term of σ∗, σ∗ = σ σσ¯ and the taylor series becomes one in σ¯ instead of σ∗.Withthis in mind the LHS becomes

3 2 2 d p 2 p σ¯ 2 s 2 s ip x s s ip x imσ χ∗ = imσ 3 σ · σ iap†σ ∗η e · + bpη ∗e− · . ˆ (2π) 2Ep − 6 s = # $ 3 d p p σ¯ s s ip x s 2 s ip x = im · a †iη e · + b σ η ∗e− · . ˆ (2π)3 2E p p 6 p s = # $ Comparing the LHS and RHS we see that they are equal only if

s s bp = ap.

Thus, the quantized Majorana field is

3 d p p σ s 2 s ip x s s ip x χ = 3 · apiσ η ∗e− · + ap†η e · . ˆ (2π) 2Ep 9 s = # $ From this mode expansion of the Majorana field we can see that the Majorana particle is its own anti-particle. s To determine the commutation relations of the particle operator ap we simplify the commutator

(3) χ (x) ,χ† (y) = δ δ (x y) a b ab − 7 8 s in terms of ap commutators. Simplifying, we have

(3) δ δ (x y)= χ (x) ,χ† (y) ad − a d 7 3 3 8 1/2 1/2 d p d p′ p σ p′ σ = 3 3 · · ˆ (2π) ˆ (2π) 2Ep 2Ep′ sr % &ab % &ed = ′ ′ s 2 s ip x s s ip x r 2 r ip y r r ip y ia σ η ∗e− · + a †η e · , ia †′ σ η e · + a ′ η ∗e− · p bc c p b − p fe f p e 7 3 3 1/2 + 1/2 ,8 # d p d p′ p σ $ p σ = 3 3 · · ˆ (2π) ˆ (2π) 2Ep 2Ep sr % &ab % &ed = ′ ′ 2 s 2 r s r ip x+ip y 2 s r s r ip x ip y σbcηc∗σfeηf ap,ap†′ e− · · + iσbcηc∗ηe∗ ap,ap′ e− · − ·

′ ′ ' s 2 r 7s r 8ip x+ip y s r s r ip x ip y iη σ η a †,a †′ e · · + η η ∗ a †,a ′) e · −* · . − b fe f p p b e p p 7 8 ( ) s * This is very complicated and it is not obvious what commutation relations for ap yields the RHS. However, since we are s using a modified Dirac field we would expect that ap satisfies the usual relations

s r s r ap,ap′ =0= ap†,ap†′

s r 7 3 8 )a ,a †′ * =(2π) δ δ(p p′). p p rs − 7 8

19 Lets test if these commutation relations are indeed the ones we require

3 3 1/2 1/2 d p d p′ p σ p σ χ (x) ,χ† (y) = · · a d ˆ (2π)3 ˆ (2π)3 2E 2E sr % p &ab % p &ed 7 8 = ′ ′ 2 s 2 r s r ip x+ip y s r s r ip x ip y σ η ∗σ η a ,a †′ e− · · + η η ∗ a †,a ′ e · − · × bc c fe f p p b e p p ' 3 7 8 1/2 1/2 ) * ( d p 3 p σ p σ = d p′ · · ˆ (2π)3 ˆ 2E 2E sr % p &ab % p &ed = ′ ′ 2 s 2 r ip x+ip y s r ip x ip y δ δ(p p′) σ η ∗σ η e− · · + η η ∗e · − · × rs − bc c fe f b e ' 1/2 1/2 ( d3p p σ p σ = · · ˆ (2π)3 2E 2E s % p &ab % p &ed = 0 0 0 0 2 s 2 s i(Epx p x)+i(Epy p y) s s i(Epx p x) i(Epy p y) σ η ∗σ η e− − · − · + η η ∗e − · − − · × bc c fe f b e ' ( When taken at equal times the commutator becomes

3 1/2 1/2 d p p σ p σ 2 s 2 r ip x ip y s s ip x+ip y x † y χa ( ) ,χd ( ) = 3 · · σbcηc∗σfeηf e · − · + ηb ηe∗e− · · ˆ (2π) 2Ep 2Ep s % &ab % &ed 7 8 = ! " 3 1/2 1/2 1/2 1/2 d p p σ 2 s s 2 p σ ip (x y) p σ s s p σ ip (x y) = 3 · σbcηc∗ηf σfe · e · − + · ηb ηe∗ · e− · − ˆ (2π) 2Ep 2Ep 2Ep 2Ep s > ab ed ab ed ? = % & % & % & % & To simplify this further we sum over sand use the identity,

s s ηa∗ηb = δab, s = to get

3 1/2 1/2 1/2 1/2 d p p σ 2 2 p σ ip (x y) p σ p σ ip (x y) χ (x) ,χ† (y) = · σ δ σ · e · − + · δ · e− · − a d ˆ 3 bc cf fe be (2π) > 2Ep ab 2Ep ed 2Ep ab 2Ep ed ? 7 8 % & % & % & % & 3 d p p σ ip (x y) p σ ip (x y) = · e · − + · e− · − ˆ (2π)3 2E 2E %% p &ad % p &ad & 3 d p p σ ip (x y) p σ ip (x y) = · e · − + · e− · − . ˆ (2π)3 2E 2E %% p &ad % p &ad & To simplify further we need to know (p σ) .Here,weworkoutthematrixelementsofp σ, · ad ·

(p σ) =(Ep p σ) · ad − · ad i i = Epδ p σ . ad − · ad Substitution into the commutation relation yields

3 i i i i d p Epδad p σad ip (x y) Epδad p σad ip (x y) x † y χa ( ) ,χd ( ) = 3 − · e · − + − · e− · − ˆ (2π) 2Ep 2Ep 7 8 % & ✭✭ ✭✭✭✭ d3p δ δ d3p pi σi ✭p✭i ✭σi ad ip (x y) ad ip (x y) ad ✭ip✭(x✭y✭) ad ip (x y) = 3 e · − + e− · − 3 ·✭✭e · − + · e− · − ˆ (2π) 2 2 − ˆ (2π)✭✭✭2Ep 2Ep % & ✭✭✭ % & = δ δ(x y) ad −

20 where the second integral is odd in p. We now turn to calculating the Hamiltonian for Majorana fields. The Hamiltonian is the Legendre transform of the Lagrangian,

H = d3x (πχ˙ ) ˆ −L

3 im T 2 2 = d x iχ†χ˙ iχ†σ¯ ∂χ + χ σ χ χ†σ χ∗ ˆ − · 2 − - . # $ where

im T 2 2 = iχ†σ¯ ∂χ + χ σ χ χ†σ χ∗ , L · 2 − ∂ # $ π = L = iχ†. ∂ (˙χ) The simplification is straight forward and messy. We just note the result here d3p H = E as as . ˆ 3 p p† p (2π) s = Which is exactly 1/2 the Dirac Hamiltonian,

3 d p s s s s H = E a †a + b †b , D ˆ 3 p p p p p (2π) s = # $ s s with the restriction bp = ap. We simplify the Hamiltonian term by term. The first term is

3 3 ′ ′ d p d p′ s sT 2 ip x s s ip x p σ p′ σ r 2 r ip x r r ip x ′ †′ iχ†χ˙ = i 3 3 iapη σ e · + ap†η †e− · · ∂0 · ap iσ η ∗e− · + ap η e · ˆ (2π) ˆ (2π) − 2Ep 2Ep′ rs 9 6 = # $ + , 3 3 ′ ′ d p d p′ s sT 2 ip x s s ip x p σ p′ σ r 2 r ip x r r ip x ′ ′ †′ = i 3 3 Ep iapη σ e · + ap†η †e− · · · ap σ η ∗e− · + iap η e · ˆ (2π) ˆ (2π) − 2Ep 2Ep′ rs 9 6 = # $ + , "

1.5 Supersymmetry

It is possible to write field theories with continuous symmetries linking fermions and bosons; such transformations are called super-symmetries. (a) The simplest example of a supersymmetric field theory is the theory of a free complex boson and a free Weyl fermion, written in the form µ = ∂ φ∗∂ φ + χ†iσ¯ ∂χ + F ∗F. L µ · Here F is an auxiliary complex scalar field whose field equation is F =0.ShowthatthisLagrangianisinvariant(uptoa total divergence) under the infinitesimal transformation δφ = iϵT σ2χ, − 2 δχ = ϵF σ ∂φσ ϵ∗, − · δF = iϵ†σ¯ ∂χ, − ·

21 where the parameter ϵa is a 2-component spinor of Grassmann numbers. (b) Show that the term 1 ∆ = mφF + imχT σ2χ +(complex conjugate) L 2 % & is also left invariant by the transformation given in part (a). Eliminate F from the complete Lagrangian +∆ by solving its field equation, and show that the fermion and boson fields φ and χare given the same mass. L L (c) It is possible to write supersymmetric nonlinear field equations by adding cubic and higher-order terms to the La- grangian. Show that the following rather general field theory, containing the field (φi,χi), i =1, ..., n,issupersymmetric:

µ = ∂ φ∗∂ φ + χ†iσ¯ ∂χ + F ∗F L µ i i i · i i i ∂W [φ] i ∂2W [φ] + F + χT σ2χ + c.c , i ∂φ 2 ∂φ ∂φ i j - i i j . 3 where W [φ] is an arbitrary function of the φi,calledthesuper-potential.Forthesimplecasen =1and W = gφ /3, write out the field equations for φ and χ (after elimination of F).

1.6 Fierz transformations !

Let ui, i =1, ...4,befour4-componentDiracspinors.Inthetext,weprovedtheFierzrearrangementformulae(3.78)and (3.79). The first of these formulae can be written in 4-component notation as 1+γ5 1+γ5 1+γ5 1+γ5 u¯ γµ u u¯ γ u = u¯ γµ u u¯ γ u . 1 2 2 3 µ 2 4 − 1 2 4 3 µ 2 2 - . - . - . - . The above identity plays an important part in the weak interaction. Thus, we derive this equation from (3.78) and (3.79)

(¯u σµu )(¯u σ u )= (¯u σµu )(¯u σ u ) 1R 2R 3R µ 4R − 1R 4R 3R µ 2R (¯u σ¯µu )(¯u σ¯ u )= (¯u σ¯µu )(¯u σ¯ u ) . 1L 2L 3L µ 4L − 3L 4L 1L µ 2L 5 5 1+γ 1 γ Denoting the handedness operators by PR = 2 and PL = −2 we have the following:

µ µ (¯u1γ PRu2)(¯u3γµPRu4)=(¯u1Rγ u2R)(¯u3Rγµu4R) µ =(¯u1Rσ u2R)(¯u3Rσµu4R) = (¯u σµu )(¯u σ u ) − 1R 4R 3R µ 2R = (¯u γ P u )(¯u γ P u ) − 1 µ R 4 3 µ R 2 (¯u γµP u )(¯u γ P u )= (¯u γ P u )(¯u γ P u ) . 1 L 2 3 µ L 4 − 3L µ L 4L 1L µ L 2L In fact, there are similar rearrangement formulae for any product

A B u¯1Γ u2 u¯3Γ u4 , where ΓA, ΓB are any of the 16 combinations of Dirac# matrices$# listed$ in Section 3.4. (a) To begin, normalize the 16 matrices ΓA to the convention Tr ΓAΓB =4δAB.

This gives ΓA = 1,γ0,iγj, ... ;writeall16elementsofthisset.! " Proof: We already) know that* the 16 Dirac matrices are orthogonal subject to the inner product Tr [AB].Thus,weonly have to find the normalization constant.

22 1. Scalars (ΓA =1): Tr [1] = 4 thus, the identity is properly normalized. 2. Vectors (ΓA = γµ): 2 Tr γ0 = Tr [1] = 4 '# $ ( 2 2 σi 0 Tr γi = Tr − = Tr [1] 4 i 2 − − >; #0 $ σ

i 2 Tr (σµν )2 = Tr ([γµ,γν])2 2 - . ' ( 1 ' ( = Tr [γµγν γµγν γν γµγµγν γν γµγµγν + γν γµγν γµ] −4 − − 1 = 2Tr [γµγν γµγν ] 2Tr [γν γµγµγν ] −4 { − } 1 = 2Tr [γµγν γµγν ]+2Tr [γµγν γµγν ] 4gµν Tr [γµγν ] −4 { − } 1 = 4Tr [γµγν γµγν ] 16gµνgµν −4 { − } 1 = 16 [gµν gµν gµµgνν + gµν gµν ] 16gµνgµν −4 { − − } 1 = 16 [gµν gµν gµµgνν] −4 − = 4 gµν gµν gµµgνν − { − } 0 µ = ν = 4gµµgνν µ = ν B ̸ 0 µ = ν = 4 µ = ν and both µ, ν are spacelike or timelike ⎧ ̸ ⎨⎪ 4 µ = ν and µ is spacelike (timelike) and ν is timelike (spacelike) − ̸ Thus, the normalized tensor matrices⎩⎪ are: σ0i, σi0,σ00,σii. − − 4. Pseudo-vector (ΓA = γµγ5):

Tr γµγ5 = Tr γµγ5γµγ5 µ 5 5 µ !# $" = Tr !γ γ γ γ " µ µ = Tr ![γ γ ] " =4gµµ 4 µ =0 = − B4 µ = i

Thus, the normalized pseudo-vector matrices are: γ0γ5,γiγ5. − 5. Pseudo-scalar (ΓA = γ5): Tr γ5γ5 =1 Thus, γ5 is already properly normalized. ! "

23 " (b) Write the general Fierz identity as an equation

A B AB C D u¯1Γ u2 u¯3Γ u4 = C CD u¯1Γ u4 u¯3Γ u2 C,D # $# $ = # $# $ AB A with unknown coefficients C CD.Usingthecompletenessofthe16Γ matrices, show that 1 CAB = Tr ΓC ΓAΓDΓB . CD 16 ! " Proof: Using the completeness of the we write

A B AB C AB D AB C D ΓabΓcd = C C Γad C D Γbc = C CDΓadΓbc =C =D =CD from which it follows A B AB C D u¯1Γ u2 u¯3Γ u4 = C CD u¯1Γ u4 u¯3Γ u2 . C,D # $# ′ ′ $ = # $# $ AB C D We now calculate C C′D′ .MultiplyingbyΓdaΓbc we have ′ ′ C D A B AB C D C D ΓdaΓbc ΓabΓcd = C CDΓdaΓbcΓadΓbc C,D= which implies

′ ′ ′ ′ C A D B AB C C D D Tr Γ Γ Γ Γ = C CDTr Γ Γ Tr Γ Γ ' ( C,D ' ( ' ( = ′ ′ AB CC DD =16 C CDδ δ C,D= AB =16C C′D′ . AB 1 C A D B Thus, we see that C CD = 16 Tr Γ Γ Γ Γ . " ! " µ (c) Work out explicitly the Fierz transformation laws for theproducts(¯u1u2)(¯u3u4) and (¯u1γ u2)(¯u3γµu4). Proof: A B A,B First, the product (¯u1u2)(¯u3u4)= u¯1Γ u2 u¯3Γ u4 where Γ =1.Thus,weneedallnonzerotracesoftheform 1 # $#C11 $= Tr ΓC1ΓD1 CD 16 1 ! " = Tr ΓCΓD 16 1 ! " = δ . 4 CD Thus, the product rule for ΓA,B =1becomes

AB C D (¯u1u2)(¯u3u4)= C CD u¯1Γ u4 u¯3Γ u2 C,D = # $# $ 1 = δ u¯ ΓCu u¯ ΓDu 4 CD 1 4 3 2 C,D = # $# $ 1 = u¯ ΓC u u¯ ΓC u 4 1 4 3 2 C = # $# $ 24 where the index C cycles through all 16 ΓC. µ AB Next, we examine the product (¯u1γ u2)(¯u3γµu4).ThecoefficientsC CD are determined as usual 1 Cµµ = Tr ΓC γµΓDγ . CD 16 µ ! " Using the matrices from our normalized basis of part (a) we define Γµ = γ0,iγ .IntermsofΓµ we have # $ 1 3 Cµµ = Tr ΓCΓ0ΓDΓ0 Tr ΓC iΓi ΓD iΓi CD 16 − − − B i=1 G ! " = ! # $ # $" 1 3 = Tr ΓCΓ0ΓDΓ0 + Tr ΓC ΓiΓDΓi . 16 B i=1 G ! " = ! " Suppose that ΓC = γ0 then

3 1 Cµµ = Tr Γ0Γ0ΓDΓ0 + Tr Γ0ΓiΓDΓi 0D 16 B i=1 G = ! "3 ! " 1 = Tr ΓDΓ0 Tr Γ0ΓD 16 − B i=1 G ! " = ! " = δD0. − Similarly, if ΓC = γj then

3 1 Cµµ = Tr Γj Γ0ΓDΓ0 + Tr Γj ΓiΓDΓi jD 16 B i=1 G = ! " 3 ! " 1 = Tr Γj ΓD + Tr 2gij ΓiΓj ΓDΓi 16 − − − B i=1 G = ! "3 !# 3 $ " 1 = 4δDj 2 gij Tr ΓDΓi Tr ΓiΓj ΓDΓi 16 − − − B i=1 i=1 G = = ! 3 " ! " 1 = 4δDj 2Tr ΓDΓj Tr Γj ΓD 16 − − − B i=1 G = 1 ! " ! " = 4δDj 8δDj 12δDj 16 − − − 24) * = δDj −16 3 = δDj −2 Thus, we see that for ΓC = γν we must have ΓD = γν and vice versa. C,D µµ For Γ =1we have C CD =0. The remaining matrices are the anti-symmetric matrices formthesetΓ= σµν ,γµγ5,γ5 we have { }

25 for: ΓC = γ5

3 1 Cµµ = Tr Γ5Γ0ΓDΓ0 + Tr Γ5ΓiΓDΓi 5D 16 B i=1 G = ! " 3 ! " 1 = Tr Γ5ΓD Tr Γ5ΓD 16 − − B i=1 G ! " = ! " = δD5 − for: ΓC = γ0γ5

3 1 Cµµ = Tr Γ05Γ0ΓDΓ0 + Tr Γ05ΓiΓDΓi ν5,D 16 B i=1 G = ! " 3 ! " 1 = Tr γ0γ5γ0ΓDγ0 + Tr γ0γ5 iγi ΓDΓi 16 − − B i=1 G = ! "3 !# $# $ " 1 = Tr γ5γ0ΓD + Tr iγi γ0γ5 ΓDΓi 16 − − B i=1 G = ! " 3 !# $# $ " 1 = Tr Γ05ΓD + Tr Γ05ΓD 16 B i=1 G = 1 ! " ! " = 16δ05,D 16 = δ05,D for: ΓC = γiγ5

3 1 Cµµ = Tr Γj5Γ0ΓDΓ0 + Tr Γj5ΓiΓDΓi j5,D 16 B i=1 G = ! " 3 ! " 1 = Tr Γ0Γj5ΓDΓ0 + Tr iγjγ5 iγi ΓDΓi 16 B i=1 G = ! "3 !# $# $ " 1 = Tr Γj5ΓD Tr iγj iγi γ5 ΓDΓi 16 − B i=1 G = ! "3 !# # $ $ " 1 = 4δj5,D Tr i22gij iγi iγj γ5ΓDΓi 16 − − B i=1 G = 3 !# # $ 3$ " 1 = 4δj5,D +2 Tr gij γ5ΓDΓi + Tr iγi iγjγ5ΓDΓi 16 B i=1 i=1 G = = ! 3 " !# $ " 1 = 4δj5,D +2Tr γ5ΓDΓj + Tr Γj5ΓD 16 B i=1 G = ! " 3 ! " 1 = 4δj5,D +2Tr ΓDΓj5 + Tr Γj5ΓD 16 B i=1 G = 24 ! " ! " = δj5,D 16 3 = δj5,D 2

26 for: ΓC =Γµν = σ0i, σi0,σ00,σii − − i σµν ,γα = γµγν γν γµ,γα { } 2 { − } i i = γµγν ,γα γν γµ,γα 2 { }−2 { } i i i i = γµ,γα γν + γµ [γν ,γα] γν ,γα γµ γν [γµ,γα] 2 { } 2 − 2 { } − 2 = igµαγν + iγµ (γνγα gνα) igναγµ iγν (γµγα gµα) − − − − = igµαγν + iγµγν γα iγµgνα igναγµ iγν γµγα + iγν gµα − − − =2igµαγν 2igναγµ + iγµγν γα iγνγµγα − − =2igµαγν 2igναγµ + iγµγν γα iγνγµγα − −

1.7 Discrete symmetries of the Dirac field !

This problem concerns the discrete symmetries P, C,andT . (a) Compute the transformation properties under P, C, and T of the fermion bilinears, ψσ¯ µν ψ,with µν i µ ν σ = 2 [γ ,γ ].Thiscompletesthetableofthetransformationpropertiesof bilinears at the end of this chapter. Proof: Starting with the transformation we have i P ψσ¯ µν ψP = P ψ¯PP [γµ,γν ] PPψP 2 0 i µ ν 0 = η∗η ψ¯ (t, x) γ [γ ,γ ] γ ψ (t, x) a a − 2 − i = ψ¯ (t, x) γ0 (γµγν γν γµ) γ0ψ (t, x) . − 2 − − Notice that µ = ν otherwise the commutator vanishes. There are two cases: ̸ 1. µ = ν where µ, ν =1, 2, 3. ̸ i P ψσ¯ ij ψP =(1)2 ψ¯ (t, x) γiγj γjγi γ0γ0ψ (t, x) − − 2 − − = ψ¯ (t, x) σij ψ (t, #x) . $ − − 2. µ = ν and either µ =0or ν =0.Forν =0the parity transformation is ̸ i P ψσ¯ i0ψP = ψ¯ (t, x) γ0 γiγ0 γ0γi γ0ψ (t, x) − 2 − − #i $ =(1) ψ¯ (t, x) γiγ0 γ0γi γ0γ0ψ (t, x) − − 2 − − = ψ¯ (t, x) σi0ψ (#t, x) . $ − − − Similarly, for µ =0the parity transformation is

P ψσ¯ 0j ψP = ψ¯ (t, x) σ0j ψ (t, x) . − − −

27 The parity transformation of the bilinear tensor can be summarized by the following formula

P ψσ¯ µν ψP =( 1)µ ( 1)ν ψ¯ (t, x) σµν ψ (t, x) − − − − where 1 µ =0 ( 1)µ = − 1 µ =1, 2, 3. B− Next we look at the time reversal transformation i T ψσ¯ µν ψT = T ψ¯TT [γµ,γν ] TTψT 2 i ∗ = ψ¯ ( t, x) γ1γ3 [γµ,γν ] γ1γ3 ψ ( t, x) − − 2 − - . # 1 $3 i µ ν # ν µ$ 1 3 = ψ¯ ( t, x) γ γ (γ ∗γ ∗ γ ∗γ ∗) γ γ ψ ( t, x) − − − 2 − −

2 2 # $ # $ Since γ ∗ = γ , − 1 3 µ µ 1 3 γ γ γ µ =0 γ ∗ γ γ = 1 3 µ B# γ γ$ γ µ =1, 2, 3 # $ − and we can simplify the transformation law. Like the parity# transformation$ there are two cases:

1. µ = ν and µ, ν =1, 2, 3. ̸ i T ψσ¯ ij ψT = ( 1)2 ψ¯ ( t, x) γ1γ3 γ1γ3 γiγj γj γi ψ ( t, x) − − − − 2 − − = ψ¯ ( t, x) σij ψ ( #t, x) . $# $ # $ − − − 2. µ = ν and either µ =0or ν =0.Forν =0the time reversal transformation is ̸

i0 1 3 i i 0 0 i 1 3 T ψσ¯ ψT = ψ¯ ( t, x) γ γ γ ∗γ ∗ γ ∗γ ∗ γ γ ψ ( t, x) − − − 2 − − # $ # i $# $ = ( 1) ψ¯ ( t, x) γ1γ3 γ1γ3 γiγ0 γ0γi ψ ( t, x) − − − − 2 − − = ψ¯ ( t, x) σµ0ψ ( #t, x) . $# $ # $ − − Similarly, for µ =0the parity transformation is

T ψσ¯ 0j ψT = ψ¯ (t, x) σ0j ψ (t, x) . − −

The time reversal transformation of the bilinear tensor can be summarized by the following formula

T ψσ¯ µν ψT = ( 1)µ ( 1)ν ψ¯ (t, x) σµν ψ (t, x) . − − − − − Lastly, we work out how the bilinear tensor transforms under charge conjugation

Cψσ¯ µν ψC = Cψ¯Cσµν CψC T T = iγ0γ2ψ σµν iψγ¯ 0γ2 − − T = # ψγ¯ 0γ2 $(σµν )T #γ0γ2ψ $ − = ψγ'¯ 0γ2 (σµν )T γ0γ2ψ ( −

28 T T where in the last line we have used the fact that iγ0γ2ψ σµν iψγ¯ 0γ2 is just a number so the does not change anything. The transpose of the gamma matrices− are − # $ # $ T 0 σµ 0(¯σµ)T (γµ)T = = . σ¯ν 0 (σν )T 0 - . ; < Since,

T σ0 = σ0 T #σ1$ = σ1 T #σ2$ = σ2 − T #σ3$ = σ3 we see that # $ γµ µ =0, 2 (γµ)T = γµ µ =1, 3. B− Therefore, γ0γ2 (γµ)T = γµ γ0γ2 − and we can show that γ0γ2 (σµν )T = σµν γ0γ2#.Thus,thebilineartensorunderchargeconjugationbecome$ # $ s − Cψσ¯ µν ψC = ψσ¯ µν γ0γ2γ0γ2ψ = ψσ¯ µν ψ.

The charge conjugation transformation of the bilinear tensor can be summarized by the following formula

Cψσ¯ µν ψC = ψ¯ (t, x) σµν ψ (t, x) . − − − "

(b) Let φ (x) be a complex-valued Klein-Gordon field, such as we consideredinProblem2.2.Findunitaryoperators P, C and an anti-unitary operator T (all defined in terms of their action on the annihilation operators ap and bp for the Klein-Gordon particles and ) that give the following transformations of the Klein-Gordon fields:

Pφ(t, x) P = φ (t, x); − Tφ(t, x) T = φ ( t, x); − Cφ(t, x) C = φ∗ (t, x) .

Find the transformation properties of the components of the current

µ µ µ J = i (φ∗∂ φ ∂ φ∗φ) − under P, C and T . Proof: The quantized Klein-Gordon field is

3 d p 1 ip x ip x φ (t, x)= 3 ape− · + bp† e · ˆ (2π) 2Ep 3 # $ d p 5 1 ip x ip x φ∗ (t, x)= 3 ap† e · + bpe− · ˆ (2π) 2Ep # $ 5

29 The natural definition for the parity operator is

PapP = a p − PbpP = b p. − Let us test this definition on the KG field. Application of P yields

3 d p 1 ip x ip x † PφP = 3 PapPe− · +(PbpP ) e · ˆ (2π) 2Ep + , d3p 1 5 iEpt+ip x iEpt ip x = 3 a pe− · + b† pe − · . ˆ (2π) 2Ep − − + , Defining p˜ =(Ep, p˜) (Ep, p),theaboveequationmaybewrittenas5 ˜ ≡ − 3 d ( p) 1 iE−pt ip x iE−pt+ip x PφP =p p −3 a pe− − · + b† pe · ↔− ˆ (2π) 2E p − − − 3 + , d p˜ 51 ip˜ (t, x) ip˜ (t, x) = 3 ap˜ e− · − + bp†˜ e · − ˆ (2π) 2Ep˜ + , = φ(t, x). − 5 Thus, P takes φ(t, x) to φ(t, x) as required. We note here that − Pφ∗P = φ∗(t, x). − This will be used later to find the transformation properties of the current. Define the time reversal operator to be anti-linear so that when commuted past a c-number it conjugates the c-number. The time reversal operator acting on the creation operators is defined to be:

TapT = a p − TbpT = b p. − Let us test this definition on the KG field. Application of T yields

3 d p 1 ip x ip x † TφT = 3 TapTe · +(TbpT ) e− · ˆ (2π) 2Ep 3 + , d p 5 1 ip x ip x = 3 a pe · + b† pe− · ˆ (2π) 2Ep − − 3 + , d p˜ 5 1 ip˜ (t, x) ip˜ (t, x) = 3 ap˜ e · − + bp†˜ e− · − ˆ (2π) 2E˜p 3 + , d p˜ 5 1 ip˜ ( t,x) ip˜ ( t,x) = 3 ap˜ e− · − + bp†˜ e− · − ˆ (2π) 2E˜p + , = φ( t, x) − 5 as required. We note here that Tφ∗T = φ∗( t, x). − This will be used later to find the transformation properties of the current. Define the time reversal operator to be anti-linear so that when commuted past a c-number it conjugates the c-number. The time reversal operator acting on the creation operators is defined to be:

CapC = bp

CbpC = ap.

30 Let us test this definition on the KG field. Application of T yields

3 d p 1 ip x ip x † CφC = 3 CapCe− · +(CbpC) e · ˆ (2π) 2Ep 3 + , d p 5 1 ip x ip x = 3 bpe− · + ap† e · ˆ (2π) 2Ep 3 # $ d p 5 1 ip x ip x = 3 bpe− · + ap† e · ˆ (2π) 2Ep # $ = φ∗(t, x) 5 as required. We note here that Cφ∗C = φ(t, x). This will be used later to find the transformation properties of the current.

Now, we can find the transformation properties of the current.Inadditiontoknowinghowφ and φ∗ transform under P , T ,andC,weneedthetransformationrulesforthederivative∂µ.Paritytakesx x and t t and therefore, ∇ ∇ →− → →− and ∂0 ∂0.Timereversaltakesx x and t t and therefore, ∇ ∇ and ∂0 ∂0.Chargeconjugationdoesnot effect the→ derivative. With gµµ defined→ such that→− there is no sum we can→ write the transformati→− ons of the derivative in a compact way,

P∂µP = gµµ∂µ T∂µT = gµµ∂µ − C∂µC = ∂µ.

Thus, under P , T ,andC,thecurrentbecomes

µ µ µ PJ (t, x)P = i (Pφ∗(t, x)P∂ Pφ(t, x)P ∂ Pφ∗(t, x)PPφ(t, x)P ) µµ µ − µµ µ = i (φ∗(t, x)g ∂ φ(t, x) (g ∂ φ∗(t, x)) φ(t, x)) − − − − − =(1)µ J µ(t, x) − −

µ µ µ TJ (t, x)T = i (Tφ∗(t, x)T∂ Tφ(t, x)T T∂ Tφ∗(t, x)TTφ(t, x)T ) µµ µ − µµ µ = i (φ∗( t, x)( g ) ∂ φ( t, x) (( g ) ∂ φ∗( t, x)) φ( t, x)) − − − − − − − = ( 1)µ J µ( t, x) − − −

µ µ µ CJ (t, x)C = i (Cφ∗(t, x)C∂ Cφ(t, x)C ∂ Cφ∗(t, x)CCφ(t, x)C) µ µ − = i (φ(t, x)∂ φ∗(t, x) (∂ φ(t, x)) φ∗(t, x)) µ − µ = i (φ(t, x)∂ φ∗(t, x) (∂ φ(t, x)) φ∗(t, x)) − = J µ(t, x). − "

(c) Show that any Hermitian Lorentz-scalar local operator built from ψ (x), φ (x),andtheirconjugateshasCPT =+1.

Proof: A Hermitian, Lorentz invariant operator must have φ paired with φ∗ and ψ paired with ψ¯.Additionally,the Lorentz indices must be contracted to form a Lorentz scalar. Looking at table in P&S we can see that any combination of these bilinears which yields a Hermitian, Lorentz invariant will have CPT =1. "

31 1.8 Bound states

Two spin-1/2 particles can combine to a state of total spin either 0 or 1. The wavefunctions for these states are odd and even, respectively, under the interchange of the two spins. (a) Use this information to compute the quantum numbers under P and C of all -positron bound states with S, P, or D wavefunctions.

The Ps Schrodinger equation is the Schrodinger equation for aparticleofelectricchargee and reduced mass µ = me/2 in the Coulomb potential V ( x )= α/ x where e (< 0)istheelectroncharge,me is the electron mass, α 1/137 is the fine structure constant and| |x is− the separation| | distance between the electron and positroninPs.Insteadofdirectly≈ solving the Ps Schrodinger equation, the position space wavefunctionsofPscanbeobtainedbytakingadvantageofthe similarity between the Schrodinger equation for Ps and that for hydrogen. Specifically, the position space wave functions can be obtained by replacing the hydrogen Bohr radius with the Ps Bohr radius in the hydrogen position space wave functions. These wave functions are characterized by the principal quantum number (or energy quantum number), n,theorbitalangularmomentumquantumnumber,l,andtheorbitalangular momentum projection quantum number, ml.Thepositionspacewavefunctionsare

3 l 2 (n l 1)! x /na 2 x 2l+1 2 x m ψ (x)= − − e−| | | | L | | Y (θ,φ), (1.1) nlml na 3 na n l 1 na l 6- . 2n [(n + 1)!] - . − − - . m ml where a =2/meα is the Bohr radius of Ps, Ln m are associated Laguerre polynomials and Yl are spherical harmonics. − Now that we have the position space wave functions of Ps, we canconstructthePsboundstate.Inthecentreofmass frame, the Ps bound state can be expressed by d3k 1 1 Ψnlml;s,ms = √2mPs ψnlml (k) k; k; s, ms , (1.2) | ⟩ ˆ (2π)3 √2me √2me | − ⟩ where s is the total spin of the Ps bound state and ms is the spin projection along the z-axis. The momentum space wavefunction, 3 ik r ψ (k) d xe · ψ (x), (1.3) nlml ≡ ˆ nlml gives the amplitude for finding Ps in a given configuration where the electron has momentum k.Thefreestateinequation (1.2)is 1 1 1 1 2 2 2 2 k; k;0, 0 = ak †b−k † ak− †b †k 0 /√2, | − ⟩ − − − | ⟩ + , for p-Ps (s = ms =0)and

1 1 2 2 ak †b †k 0 for ms =1, 1 − |1 ⟩ 1 1 2 2 2 2 k; k;1,ms = ⎧ ak †b−k† + a−k †b †k 0 /√2 for ms =0, | − ⟩ ⎪ 1 − 1 − | ⟩ ⎪ 2 2 ⎨a+−k †b−k † 0 , for ms = 1, − | ⟩ − ⎪ ⎩⎪ for o-Ps (s =1)where 0 is the vacuum state. | ⟩ Applying the parity operator, we obtain 1 d3k P Ψnlml;00 = ψnlml (k)P k; k;00 | ⟩ √me ˆ (2π)3 | − ⟩ 1 d3k 1 1 1 1 k 2 † − 2 † − 2 † 2 † = 3 ψnlml ( )P ak b k ak b k P 0 √2me ˆ (2π) − − − | ⟩ + , 1 d3k 1 1 1 1 2 k 2 † − 2 † − 2 † 2 † = ηa 3 ψnlml ( )P ak b k ak b k P 0 −| | √2me ˆ (2π) − − − − | ⟩ + ,

32 where we have assumed that the vacuum state is even under parity. To simplify the above further, note from equation

(1.3), that the parity of ψnlml (k) is the same as ψlnml (x).Therefore,wecaninvertthespatialcoordinatein(1.1)to obtain the parity of the momentum space wavefunction. The only part of (1.1)thatissensitivetosuchaninversionis the spherical harmonic, Y (θ,φ) ( 1)lY (θ,φ).Thus,p-Pstransformsas lm → − lm P Ψ =( 1)l+1 Ψ | nlml;00⟩ − | nlml;00⟩ under parity. Similarity, o-Ps transforms as

P Ψ =( 1)l+1 Ψ | nlml;1ms ⟩ − | nlml;1ms ⟩ under parity and has an P eigenvalue of ( 1)l+1.Noticethattherelativephasedifferencebetweenthefermion and − anti-fermion inversion phases, ηb∗ = ηa,contributesafactorof 1 to the parity of Ps (p-Ps and o-Ps) in addition to the parity of the wave function; this extra− factor is called the intrinsic− parity of a fermion--anti-fermion system. The remaining C and T eigenvalues of Ps are obtained in a similar manner and are listed below in Table 1.

Discrete Transform Ps Eigenvalue P ( 1)l+1 − C ( 1)l+s − T ( 1)s+1 −

Table 1: P , C and T eigenvalues of the Ps state Ψnlml;sms .Here,l is the orbital angular momentum quantum number and s is the spin angular momentum quantum number.| ⟩

(b) Since the electron-photon coupling is given by the Hamiltonian

∆H = d3xeA jµ, ˆ µ where jµ is the electric current, electrodynamics is invariant to p and C if the components of the vector potential have the same P and C parity as the corresponding components of jµ.Showthatthisimpliesthefollowingsurprisingfact: The spin-0 ground state of positronium can decay to 2 photon, but the spin-1 state must decay to 3 photons. Find the selection rules for the annihilation of higher positronium states, and for 1-photon transitions between positronium levels.

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