An Introduction to Quantum Field Theory (Peskin and Schroeder) Solutions Andrzej Pokraka December 12, 2017 Contents 1TheDiracEquation 1 1.1 Lorentz group ! ................................................... 1 1.2 Gordon Identity ! .................................................. 7 1.3 Spinor products ! .................................................. 7 1.4 Majorana fermions ! ................................................ 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 permutation 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 relation of angular momentum. 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 spin-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 matrix 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 spinors 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 tensor 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 - .