Physics 443 Homework 2 Solutions

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1 Phyic 3 Homework Solution Problem 1 a Under the tranformation φx φ x λ a φλx Lagrangian Lx 1 µφx ρφx tranform a Lx 1 µ λ a φλx ρλ a φλx 1 λ +a 1 µφ λx ρλ a φλx. Moreover, note that action S d xlx i invariant if L undergoe a tranformation Lx λ Lλx. It i eay to ee that we have preciely thi etting when a 1. b Firtly let u tudy infiniteimal tranformation. The Lagrangian tranform to linear order in ɛ where λ 1 + ɛ a where we define J µ x µ L. The field φ tranform a where we define φ φ + x µ µ φ. Noether theorem give a conerved current a In particular, for our cae it i Lx 1 + ɛ L 1 + ɛx 3 Lx + ɛx µ µ L + ɛl Lx + ɛ µ x µ L 5 Lx + ɛ µ J µ, 6 φx 1 + ɛφ 1 + ɛx 7 J µ D φx + ɛφ + ɛx µ µ φ 8 φx + ɛ φ, 9 L µ φ φ J µ. 10 J µ D µ φ φ + x ν ν φ x µ 1 αφ ρφ. 11 Thi lead to a conerved charge, D, by integrating JD 0 over an equal time urface. Since JD 0 0 φ φ + t 0 φ + x φ 1 t 0 φ φ + tρφ, 1 1

2 D t 1 d 3 x 0φ + 1 φ + ρφ + d 3 x 0 φ 1 + x φ Uing the fact that the momentum field conjugate to φ i given by πx, t 0 φx, t that the Hamiltonian i 1 H d 3 x πx, t + 1 φx, t + ρφx, t 1 we have D th + d 3 x πx, t 1 + x φx, t By Heienberg equation of motion we have d dt D D + i H, D. 16 t For conerved D the total time dependence the left h ide of the equation above hould be zero. Since D alo ha explicit time dependence from the th term we obtain or equivalently 0 H + i H, D, 17 H, D ih. 18 Our lat tak i to verify thi behavior for ρ 0 cae by computing H, D. Note that at ρ 0 the Hamiltonian reduce to 1 H d 3 x πx, t + 1 φx, t. 19 Then, H, D H, th + d 3 x πx, t 1 + x φx, t 0 d 3 y d 3 x 1 πy, t + 1 φy, t, πx, t 1 + x φx, t 1 πy, t, φx, t 1 d 3 y d x{ 3 πx, t 1 + x i i x + 1 y j φy, t y j φy, t, πx, t 1 + x i i x } φx, t. Note that we keep track whether a derivative i an x derivative or a y derivative uing a upercript. { H, D d 3 y d 3 x πx, t 1 + x i i x iδ 3 x yπy, t 3 } φx, t. + y j φy, t y j iδ 3 x y 1 + x i i x Uing integration by part firt canceling boundary term due to the delta function then performing the y integration we get H, D d 3 x iπ + i i x i π π i j j φ1 + x i i φ Then, H, D i i d 3 x π + 3π + x i i ππ j j φφ j j φx i i φ 5 { 1 d 3 x π + i xi ππ 3 π j j φφ j φ j φ j j φx i i φ + 1 } 1 jφ j φ i jφ j φx i 6

3 Note that at equal time π commute with j π φ commute with j φ. Thi fact i ued implicitly in the lat line above. Finally, canceling boundary term we get 1 H, D i 1 d 3 x π + 1 φ 7 ih. 8 Problem P&S Problem 3.1 a We tart with L, L commutation relation. L i, L j 1 ɛimr ɛ jkn J mr, J kn 9 i ɛimr ɛ jkn g rk J mn g mk J rn g rn J mk + g mn J rk 30 iɛ imr ɛ jkn g rk J mn 31 iɛ imr ɛ jnr J mn 3 i δ ij δ mn δ in δ mj J mn 33 ij ji iɛ kji L k, 3 finally giving L i, L j iɛ ijk L k. We move on with L i, K j 1 ɛimr J mr, J 0j 35 giving L i, K j iɛ ijk K k. Latly, giving K i, K j iɛ ijk L k. i ɛimr g r0 J mj g m0 J rj g rj J m0 + g mj J r0 36 iɛ imj J m0 37 K i, K j J 0i, J 0j 38 i g i0 J 0j g 00 J ij g ij J 00 + g 0j J i0 39 ij ij, 0 1 It may not be clear why we are allowed to cancel boundary term in an operator valued field. When acting on arbitrary field configuration thee boundary term can actually give nonzero contribution. However, retricting our attention to local field excitation looking at matrix element aociated with thoe configuration the boundary term have vanihing contribution. 3

4 Now, let u tudy J + J commutation relation. J+, i J j 1 L i + ik i, L j ik j 1 1 L i, L j + i K i, L j i L i, K j + K i, K j i ɛijk L k + ik k ik k L k 3 0. J+, i J j + 1 L i + ik i, L j + ik j 5 i ɛijk L k + ik k + ik k + L k 6 iɛ ijk J k +. 7 J, i J j 1 L i ik i, L j ik j 8 i ɛijk L k ik k ik k + L k 9 iɛ ijk J k. 50 b Since 1 iθ L iβ K 1 iθ + β J + iθ β J, 51 Lorentz tranformation for a field in j +, j repreentation are given a Φ 1 iθ + β J j+ + I iθ β I J j Φ, 5 where J j+ + J j are pin j + pin j repreentation matrice of u Lie algebra. Spin- 1 repreentation matrice are J 1/ σ/. Therefore, a field in 1, 0 repreentation tranform a Φ 1 iθ σ/ β σ/ Φ, 53 a filed in 0, 1 repreentation tranform a Φ 1 iθ σ/ + β σ/ Φ. 5

5 c The matrix given in the problem can be parametrized a V 0 + V i σ i. Then the tranformed object i to firt order in θ β V V 0 + V i σ i 1 iθ σ/ + β σ/ 0 + V i 1 + iθ σ/ + β σ/ 55 V β σ + V σ i i i θj σ j, σ i + βj {σj, σ i } 56 In other word, V β j σ j + V i σ i + θ j ɛ jik σ k + β i. 57 V 0 V 0 + β i V i 58 V i V i + β i V 0 + ɛ ijk θ j V k, 59 thi i preciely how a -vector tranform under infiniteimal Lorentz tranformation. Problem 3 P&S Problem 3.6 a For thi quetion a pretty traightforward method i to ue an explicit realization of the Clifford algebra for example by uing the chiral bai for γ µ compute the trace uing the algebra of Pauli matrice. One then would find a uitably normalized bai by { 1, γ 0, iγ i, iσ 0i, σ ij, iγ 0 γ 5, γ i γ 5, γ 5}. 60 One can alo perform thee computation uing only the algebra of γ matrice following the argument in chapter 5.1 of P&S. b The completene of the matrice mean that we can exp any matrix M a M A M A, 61 where the contant M A can be computed uing the trace normalization a of part a. M 1 TrM. 6 A Uing the ummation convention for matrix indice we can rewrite thi equation a M ab 1 ab cdm dc. 63 If we ue a et of matrice T ef with entrie T ef ab δ ae δ bf we get a completene identity A δ ae δ bf 1 ab fe. 6 5 A

6 Uing thi relation we get abγ B cd aeδ eb δ fc Γ B fd 65 1 aeγ B fdγ D ef Γ D cb 66 D 1 geγ D ef Γ B fhδ ag δ dh Γ D cb 67 D 1 16 geγ D ef Γ B fhγ C adγ C hgγ D cb 68 C,D 1 Tr Γ D Γ B Γ C Γ C 16 adγ D cb. 69 C,D Thi matrix identity then yield the Fierz identity given in the problem. If intead one tart by auming the identity abγ B cd C AB EF Γ E adγ F cb, 70 E,F by multiplying both ide with Γ D ad ΓD cb the reult for CAB CD follow. c For ū 1 u ū 3 u we have Γ B 1 in previou part notation. Then, Therefore, uing the lit of part a we get C AB CD 1 16 Tr Γ D Γ C 1 δcd. 71 ū 1 u ū 3 u 1 ū 1 u ū 3 u + ū 1 γ µ u ū 3 γ µ u + 1 ū 1σ µν u ū 3 σ µν u ū 1 γ µ γ 5 u ū 3 γ µ γ 5 u + ū 1 γ 5 u ū 3 γ 5 u. 7 For ū 1 γ µ u ū 3 γ µ u the term coming from the Fierz identity hould have all their Lorentz indice contracted a well o that both ide can have the ame tranformation under continuou Lorentz tranformation. Uing the trace expreion which can again be computed either uing an explicit realization or uing the technique of ection 5.1 Tr γ µ γ µ 16, 73 Tr γ ρ γ µ γ λ γ µ 8g ρλ, 7 Tr γ 5 γ µ γ 5 γ µ 16, 75 Tr γ ρ γ 5 γ µ γ λ γ 5 γ µ 8g ρλ, 76 Tr γ ρ γ µ γ λ γ 5 γ µ 0, 77 Tr γ µ γ 5 γ µ 0, 78 Tr σ αβ γ µ σ λρ γ µ 0, 79 6

7 we get ū 1 γ µ u ū 3 γ µ u ū 1 u ū 3 u 1 ū 1γ µ u ū 3 γ µ u 1 ū 1γ µ γ 5 u ū 3 γ µ γ 5 u ū 1 γ 5 u ū 3 γ 5 u. 80 Problem Let u firt verify that given matrice atify the Dirac algebra. {γ 0, γ } 1 0 1, 81 {γ 0, γ i 1 0 } 0 σ i 0 σ 0 1 σ i + i σ i σ i 0 σ σ i + i 0 σ i 0 0, 83 {γ i, γ j 0 σ } i 0 σ j 0 σ σ i 0 σ j + j 0 σ i 0 σ j 0 σ i 0 σ i σ j 0 σ 0 σ i σ j + j σ i 0 0 σ j σ i {σ i, σ j } 0 0 {σ i, σ j } δ ij 1 87 verifying that {γ µ, γ ν } g µν. Our next tak i to look for a imilarity tranformation S atifying Sγ µ S 1 γ µ ch, where γµ ch are γ matrice in the chiral bai. We will olve the linear ytem Sγ µ γ µ chs for µ 0, 1,, 3. For µ 0 we have A B C D A B C D where we ue block for S a well. Thi condition contrain S by requiring A C B D. 88 For µ i we have A D A D 0 σ i 0 σ σ i i A D 0 σ i 0 A D. 89 So we need Aσ i σ i D Dσ i σ i A for all i. Both can be eaily atified by taking A D 1. Setting a proper normalization we find a unitary tranformation matrix S by S

8 Finally, if we tart with poitive negative frequency olution of the Dirac equation in the chiral bai ee equation of P&S u p σξ chp p σξ vchp p ση p ση 91 we can contruct the olution in our bai imply by up S u ch p vp S v ch p. Let u quickly prove that tatement for poitive frequency olution. γ µ p µ mup Sγ µ p µ ms u ch p 9 S Sγ µ S p µ mu ch p 93 S γ µ ch p µ mu ch p Exactly the ame method work alo for negative energy olution which need to olve γ µ p µ +mvp 0. That give the olution a u p 1 p σ + p σ ξ p σ + p σ ξ 96 v p 1 p σ p σ η p σ p σ η. 97 We hould note that S tranformation doe not change the normalization ince imilarly for v p. ū pu p u pγ 0 u p 98 u ch psγ0 S u chp 99 u ch pγ0 chu chp 100 mδ 101 One can further implify the expreion above by noting that p σ + m p σ p0 + m p σ p σ + m p0 + m. 10 To verify thee we hould firt note that the right h ide have nonnegative eigenvalue they quare to p σ p σ repectively. Uing thee two expreion we get u p p 0 + m 103 ξ σ p p 0 +m ξ v p σ p p 0 + m p 0 +m ξ ξ. 10 We could alo find thee expreion by explicitly olving the equation. For example, for poitive frequency olution γ µ p p µ mup 0 m σ p u1 σ p p m u σ p The equation i olved for u p 0 +m u 1. From the normalization condition we get ūu u 1 u 1 u u u σ p 1 1 p 0 + m u 1 m p 0 + m u 1 u So we et u 1 u 1 p 0 + m up to an overall phae factor we retrieve the form of up above. A imilar dicuion alo work for vp. 8

Quantum Field Theory 2011 Solutions

Quantum Field Theory 2011 Solutions Quantum Field Theory 011 Solution Yichen Shi Eater 014 Note that we ue the metric convention + ++). 1. State and prove Noether theorem in the context of a claical Lagrangian field theory defined in Minkowki