PHY 396 K. Solutions for problem set #7.

Transcription

1 PHY 396 K. Solution for problem et #7. Problem 1a: γ µ γ ν ±γ ν γ µ where the ign i + for µ ν and otherwie. Hence for any product Γ of the γ matrice, γ µ Γ 1 nµ Γγ µ where n µ i the number of γ ν µ factor of Γ. For Γ γ 5 iγ 0 γ 1 γ 2 γ 3, n µ 3 for any µ 0, 1, 2, 3; thu γ µ γ 5 γ 5 γ µ. Problem 1b: Firt, γ 5 iγ 0 γ 1 γ 2 γ 3 iγ 3 γ 2 γ 1 γ 0 +iγ 3 γ 2 γ 1 γ 0 +iγ 3 γ 2 γ 1 γ i γ 0 γ 3 γ 2 γ i γ 0 γ 1 γ 3 γ i γ 0 γ 1 γ 2 γ 3 S.1 +iγ 0 γ 1 γ 2 γ 3 +γ 5. Second, γ 5 2 γ 5 γ 5 iγ 0 γ 1 γ 2 γ 3 iγ 3 γ 2 γ 1 γ 0 γ 0 γ 1 γ 2 γ 3 γ 3 γ 2 γ 1 γ 0 +γ 0 γ 1 γ 2 γ 2 γ 1 γ 0 γ 0 γ 1 γ 1 γ 0 +γ 0 γ S.2 Problem 1c: Any four ditinct γ κ, γ λ, γ µ, γ ν are γ 0, γ 1, γ 2, γ 3 in ome order. They all anticommute with each other, hence γ κ γ λ γ µ γ ν ɛ κλµν γ 0 γ 1 γ 2 γ 3 iɛ κλµν γ 5. The ret i obviou. Problem 1d: iɛ κλµν γ κ γ 5 γ κ γ [κ γ λ γ µ γ ν] 4 1 γ κ γ κ γ [λ γ µ γ ν] γ [λ γ κ γ µ γ ν] + γ [λ γ µ γ κ γ ν] γ [λ γ µ γ ν] γ κ 4γ [λ γ µ γ ν] + 2γ [λ γ µ γ ν] + 4g [λµ γ ν] + 2γ [ν γ µ γ λ] 1 4 S γ[λ γ µ γ ν] γ [λ γ µ γ ν]. 1

2 Problem 1e: Proof by inpection: In the Weyl bai, the 16 matrice are σ i 1, γ 0, γ i σ i, 0 σ k iγ [i γ j] ɛ ijk 0 iσ i 0 σ k, iγ [0 γ i] 0 0 +iσ i, S σ i 1 0 γ 5 γ 0, γ 5 γ σ i, γ 5, and their linear independence i elf-evident. Since there are only 16 independent 4 4 matrice altogether, any uch matrix Γ i a linear combination of the matrice S.4. Q.E.D. Algebraic Proof: Without making any aumption about the matrix form of the γ µ operator, let u conider the Clifford algebra γ µ γ ν + γ ν γ µ 2g µν. Becaue of thee anticommutation relation, one may re-order any product of the γ a ±γ 0 γ 0 γ 1 γ 1 γ 2 γ 2 γ 3 γ 3 and then further implify it to ±γ 0 or 1 γ 1 or 1 γ 2 or 1 γ 3 or 1. The net reult i up to a ign or ±i factor one of the 16 operator 1, γ µ, iγ [µ γ ν], iγ [λ γ µ γ ν] ɛ λµνρ γ 5 γ ρ cf. d or iγ [κ γ λ γ µ γ ν] ɛ κλµν γ 5 cf. c. Conequently, any operator Γ algebraically contructed of the γ µ i a linear combination of thee 16 operator. Incidentally, the algebraic argument explain why the γ µ and hence all their product hould be realized a 4 4 matrice ince any leer matrix ize would not accommodate 16 independent product. That i, the γ are 4 4 matrice in four pacetime dimenion; different dimenion call for different matrix ize. Specifically, in pacetime of even dimenion d, there are 2 d independent product of the γ operator, o we need matrice of ize 2 d/2 2 d/2 : 2 2 in two dimenion, 4 4 in four, 8 8 in ix, in eight, in ten, etc.,etc.. In odd dimenion, there are only 2 d 1 independent operator becaue γ d+1 iγ 0 γ 1 γ d 1 the analogue of the γ 5 operator in 4d commute rather than anticommute with all the γ µ and hence with the whole algebra. Conequently, one ha two ditinct repreentation of the Clifford algebra one with γ d+1 +1 and one with γ d+1 1 but in each repreentation there are only 2 d 1 independent operator product, which call for the matrix ize of 2 d 1/2 2 d 1/2. For example, in three pacetime dimenion two pace, one time, can take γ 0, γ 1, γ 2 σ 3, iσ 1, iσ 2 for γ 4 iγ 0 γ 1 γ 2 +1 or γ 0, γ 1, γ 2 σ 3, iσ 1, iσ 2 for γ 4 1, 2

3 but in both cae we have 2 2 matrice. Likewie, we have 4 4 matrice in five dimenion, 8 8 in 7D, in 9D, in 11D, etc., etc. Problem 2a: Depite anticommutativity of the fermionic field, the Hermitian conjugation of an operator product revere the order of operator without any extra ign factor, thu Ψ αψ β +Ψ β Ψ α. Conequently, for any 4 4 matrix Γ, Ψ ΓΨ +Ψ Γ Ψ, and hence ΨΓΨ ΨΓΨ where Γ γ 0 Γ γ 0 i the Dirac conjugate of Γ. Now conider the 16 matrice which appear in the bilinear 1. Obviouly 1 +1 and thi give u S +S. We aw in cla that γ µ +γ µ, and thi give u V µ +V µ. We alo aw that iγ [µ γ ν] iγ [ν γ µ] +iγ [µ γ ν], and thi give u T µν +T µν. A to the γ 5 matrix, it i Hermitian cf. 1.b and anticommute with γ 0, hence γ 5 γ 0 γ 5 γ 0 +γ 0 γ 5 γ 0 γ 5 and therefore iγ 5 +iγ 5, which give u P +P. Finally, γ 5 γ µ γ µ γ 5 γ µ γ 5 +γ 5 γ µ, which give u A µ +A µ. Thu, by inpection, all the bilinear 1 are Hermitian. Q.E.D. Problem 2b: Under a continuou Lorentz ymmetry x x Lx, the Dirac pinor field and it conjugate tranform according to Ψ x MLΨx L 1 x, Ψ x Ψx L 1 x M 1 L, S.5 hence any bilinear ΨΓΨ tranform according to Ψ x ΓΨx ΨxΓ Ψx S.6 where Γ M 1 LΓML. S.7 Obviouly, for Γ 1, Γ M 1 M 1. According to homework et #5 problem 3d, for Γ γ µ, Γ M 1 γ µ M L µ νγ ν. Similarly, M 1 γ µ γ ν M M 1 γ µ MM 1 γ ν M L µ κγ κ 3

4 L ν λ γλ and hence for Γ γ [µ γ ν], Γ L µ κl ν λ γ[κ γ λ]. Conequently, S x Sx, V µ x L µ νv ν x, T µν x L µ κl ν λ T κλ x, S.8 which make S a Lorentz calar, V µ a Lorentz vector and T µν antiymmetric indice. a Lorentz tenor with two The γ 5 matrix commute with even product of the γ µ matrice uch a γ µ γ ν, hence it commute with all S µν and therefore with ML exp 2 i θ µνs µν. Conequently, for Γ γ 5, Γ M 1 γ 5 M γ 5 while for Γ γ 5 γ µ, Γ M 1 γ 5 γ µ M γ 5 M 1 γ µ M γ 5 L µ νγ ν L µ νγ 5 γ ν. Therefore, P x P x, A µ x L µ νa ν x, S.9 which make P a Lorentz calar and A a Lorentz vector. Q.E.D. Problem 2c: Under the parity ymmetry P, x, t x, +t and the Dirac field tranform according to Ψ x ±γ 0 Ψx, Ψ x ±Ψxγ 0 S.10 cf. problem 2 of the previou et. Hence, parity propertie of the Dirac bilinear 1 follow from the commutation relation of the 16 matrice 1.e with the γ 0 matrix. It i eay to verify that 1, γ 0, γ [i γ j] and γ 5 γ i commute with the γ 0 while γ i, γ 0 γ i, γ 5 γ 0 and γ 5 anticommute with the γ 0. Conequently, the S, V 0, T ij and A i remain invariant under parity, while the V i, T 0i, A 0 and P change their ign. In three-dimenional term, thi mean that S and V 0 are true calar, P and A 0 are peudocalar, V i a true or polar vector, A i a peudovector or axial vector, and the tenor T contain one true vector T 0i and one axial vector 1 2 ɛijk T jk. In pace-time term, we call S a Lorentz true calar, P a Lorentz peudocalar, V µ a Lorentz true vector and A µ an Lorentz axial vector. Pedantically peaking, T µν i a Lorentz true tenor while T κλ 1 2 ɛκλµν T µν i a Lorentz peudotenor, but few people are that pedantic. 4

5 Problem 2d: In the Weyl convention for the Dirac matrice, the charge conjugation ymmetry C act on the Dirac field according to Ψ x ±γ 2 Ψ x. Conequently Ψ x Ψ xγ 2 Ψ x Ψ xγ 0 Ψ xγ 2 γ 0, S.11 and therefore for any Dirac bilinear, Ψ ΓΨ Ψ γ 2 γ 0 Γγ 2 Ψ +Ψ γ 2 γ 0 Γγ 2 Ψ +Ψγ 0 γ 2 Γ γ 0 γ 2 Ψ ΨΓ c Ψ. S.12 The econd equality of thi formula follow by tranpoition of the Dirac andwich Ψ Ψ, which carrie an extra minu ign becaue the fermionic field Ψ and Ψ anticommute with each other. Problem 2e: By inpection, 1 c γ 0 γ 2 γ 0 γ The γ 5 matrix i ymmetric and commute with the γ 0 γ 2, hence γ5 c +γ 5. Among the four γ µ matrice, the γ 1 and γ 3 are anti-ymmetric and commute with the γ 0 γ 2 while the γ 0 and γ 2 are ymmetric but anti-commute with the γ 0 γ 2 ; hence, for all four γ µ, γµ c γ µ. Finally, becaue of the tranpoition involved, γ µ γ ν c γνγ c µ c +γ ν γ µ, hence γ [µ γ ν] c +γ [ν γ µ] γ [µ γ ν]. Likewie, γ 5 γ µ c γ µ c γ 5 c γ µ γ 6 +γ 5 γ µ. Therefore, according to eq. S.12, the calar S, the peudocalar P and the axial vector A µ are C even while the vector V µ and the tenor T µν are C odd. Problem 3a: Given the anticommutation relation 2, we have â αâ β â δ â αâ δ â β δ α,δ â δ â αâ β +â δ â αâ β δ α,δ â β S.13 and therefore [â αâ β, â δ ] δ α,δ â β. Likewie, â αâ β â γ â αδ β,γ â γâ β δ β,γ â α â αâ γâ β δ β,γ â α + â γâ αâ β S.14 and therefore [â αâ β, â γ] +δ β,γ â α. 5

6 Finally, by Leibniz rule [â αâ β, â γâ δ ] [â αâ β, â γ]â δ + â γ[â αâ β, â δ ] +δ β,γ â αâ δ δ α,δ â γâ β. S.15 Problem 3b: According to eq. S.15, the commutator [â αâ β, â γâ δ ] ha exactly the ame form a it boonic counterpart. Hence, the proof of [Â, ˆB] Ĉ proceed exactly a in the boonic cae, cf. homework et #3 problem 2b. Problem 3c: Uing the Leibniz rule and eq. S.13 and S.14, [â µâ ν, â αâ βâγâ δ ] δ να â µâ βâγâ δ + δ νβ â µâ αâ γ â δ δ µγ â αâ βâνâ δ δ µδ â αâ βâγâ ν. S.16 Problem 3d: Again, w have a fermionic analogue to the boonic econd-quantized operator we tudied in homework et #3 problem 2d. Given eq. 4 and S.16 in which we exchange γ δ, we have [Â, â α â βâδâγ] µ,ν µ Â1 ν [ â µâ ν, â αâ ] βâδâγ µ µ Â1 α â µâ βâδâγ + µ µ Â1 β â αâ µâ δ â γ S.17 ν δ Â1 ν â αâ βâνâ γ ν γ Â1 ν â αâ βâδâν 6

7 and conequently, in light of eq. 8, [Â, [ ˆB] α β ˆB 2 γ δ µ Â1 α â µâ βâδâγ + µ µ α,β,γ,δ µ Â1 β â αâ µâ δ â γ ] ν δ Â1 ν â αâ βâνâ γ ν γ Â1 ν â αâ βâδâν µ β Â11 t ˆB 2 γ δ â µâ βâδâγ µ,β,γ,δ + α µ Â12 nd ˆB 2 γ δ â αâ µâ δ â γ α,µ,γ,δ α β ˆB 2  1 2 nd γ ν â αâ βâνâ γ α,β,γ,ν α β ˆB 2  1 1 t ν δ â αâ βâδâν α,β,ν,δ renaming indice α,β,γ,δ α,β,γ,δ [ A1 α β 1 t + A 1 2 nd, ˆB ] 2 γ δ â αâ βâδâγ α β Ĉ2 γ δ â αâ βâδâγ Ĉ. Q. E. D. Problem 4.a: The implet form of the Dirac Lagrangian i L Ψiγ µ µ mψ, S.18 which involve pacetime derivative of the Ψ field but not of the Ψ. Conequently, by Noether theorem T µν N L µ Ψ ν Ψ g µν L Ψiγ µ ν Ψ g µν Ψiγ λ λ mψ. S.19 A uual for field of non-zero pin, the Noether tre-energy tenor i not ymmetric and the 7

8 true tre-energy tenor i T µν true T µν N + λk [λµ]ν S.20 for ome three-index tenor K [λµ]ν antiymmetric in it firt two indice cf. problem 1 of the et 1. Fortunately, the correction S.20 doe not affect the net energy and momentum of the Dirac field, thu E net d 3 x T 00 N x d 3 x Ψx i γ + mψx S.21 cf. eq. 5, and P net d 3 x T 0 ı N x d 3 x Ψx iγ 0 Ψx. S.22 Hence, in term of the quantum Dirac field ˆΨx and ˆΨ x, the net mechanical momentum operator i ˆP mech d 3 x ˆΨ i ˆΨ. 9a At thi point, let u expand the Dirac field in term of creation and annihilation operator. In the Schrödinger picture eq. 5 become ˆΨx ˆΨx d 3 p e +ipx 2π 3 2E p d 3 p e ipx 2π 3 2E p up, â p, + v p, ˆb p,, u p, â p, + v p, ˆb p,. S.23 Subtituting thee formulæ into eq. 9a for the momentum operator give ˆP mech d 3 p 2π 3 d 3 p 2π 3 p 2E p 2 p 2E p, u p, â p, + v p, ˆb p, â p,â p, where the econd equality follow from + ˆb p,ˆb p, up, â p, + v p, ˆb p, S.24 u p, up, v p, v p, 2E p δ, S.25 8

9 while u p, v p, v p, up, 0. S.26 Finally, we re-write the ˆbˆb part of the lat line of eq. S.24 a d 3 p p d ˆb p,ˆb 3 p p 2π 3 p, ˆb+p,ˆb 2E p 2π 3 2E p where the lat equality follow from by reaon of rotational ymmetry. arrive at Q.E.D. Problem 4b: ˆP mech d 3 p 2π 3 d 3 p 2π 3 p 2E p +p, d 3 p 1 +p 2π 3 2E ˆb p,ˆb p, p d 3 p p δ 3 0 ˆb p,ˆb p, + 2E p 2π 3 δ 3 0 S.27 0 S.28 Conequently, ubtituting eq. S.27 into eq. S.24, we p 2E p p â p,â p, + p ˆb p,ˆb p,. 9b Electron have charge q e, hence the gauge-covariant derivative of the electron field i D µ Ψx µ Ψx iea µ xψx. S.29 Conequently, the gauge invariant Lagrangian for the electron field Ψ coupled to the EM field A µ x i L L EM and hence the electric current + Ψiγ µ D µ mψ 1 4 F µνf µν + Ψiγ µ µ mψ + ea µ Ψγ µ Ψ, J µ S.30 L A µ eψγ µ Ψ. S.31 Note that thu current i a true Lorentz vector and i odd under the charge conjugation ymmetry C cf. problem 2 of thi et. 9

10 To be precie, eq. S.31 preume claical fermionic field which anticommute with each other, thu the charge denity J 0 can be written a either J 0 eψγ 0 Ψ eψ Ψ e α Ψ αψ α or J 0 +e α Ψ α Ψ α. S.32 Ala, in the quantum theory ˆΨ α ˆΨ β i not equal to ˆΨ β ˆΨ α, and thi give rie to the operator ordering ambiguity in defining the quantum electric charge. Fortunately, thi ambiguity amount to a contant. Indeed, the quantum Dirac field ˆΨ αx and ˆΨ α x are linear combination of the fermionic creation and annihilation operator cf. eq. 5, and the latter either anticommute with each other or have c-number anticommutator cf. eq. 6 and 7. Therefore, the anticommutator { ˆΨ αx, ˆΨ β y} i a c-number function of x y. We hall calculate thi function later in cla, but for the moment all we need to know it a c-number, and therefore e α ˆΨ αx ˆΨ α x +e α ˆΨ α x ˆΨ αx + a c-number contant. S.33 Conequently, however we order the creation and annihilation operator in the quantized electric charge operator, it will give u the ame reult up to a c-number contant which may be infinite. Hence, we may jut a well take the implet ordering and allow for an extra contant term, thu ˆQ e d 3 x ˆΨ x ˆΨx + a c-number contant. 10a Next, let u expand the field ˆΨx and ˆΨ x into creation and annihilation operator according to eq. S.23 and plug into the pace integral 10a. Proceeding a in the previou 10

11 quetion, we have d 3 x ˆΨ ˆΨ d 3 p 1 2π 3 2E p 2, d 3 p 1 2π 3 2E p u p, â p, + v p, ˆb p, â p,â p, d 3 p 1 â 2π 3 2E p,â p, + ˆb p,ˆb p, p d 3 p 1 2π 3 2E p up, â p, + v p, ˆb p, + ˆb p,ˆb p, â p,â p, ˆb p,ˆb p, + 2E p 2π 3 δ 3 0 d 3 p 1 â 2π 3 2E p,â p, ˆb p,ˆb p, p + infinite c-number contant, S.34 and therefore ˆQ d 3 p 1 e 2π 3 â 2E p,â p, + e ˆb p,ˆb p, p + C S.35 where C i the um of c-number contant from eq. 10.a and S.34. To determine the value of C, note that the vacuum tate of the theory i invariant under the charge conjugation ymmetry and therefore mut have zero electric charge, ˆQ 0 0. On the other hand, the vacuum tate 0 i annihilated by all the â p, and ˆb p, operator and hence by the term on the right hand ide of eq. S.35 except for the contant C. Conequently, ˆQ 0 C 0 and the electric charge of the vacuum i C. And ince thi charge mut vanih, we mut have C 0 i.e., omehow the contant term in eq. 10a and S.34 mut cancel each other and therefore Q.E.D. ˆQ d 3 p 1 e 2π 3 â 2E p,â p, + e ˆb p,ˆb p, p. 10b Problem 4c: In thi quetion, we tart with eq. 11 for the net pin operator a a pace integral of a Dirac 11

12 bilinear, o our firt tep i to expand the field into momentum mode S.23 and plug the expanion into eq. 11. Thi give u Ŝ net d 3 p 2π 3 1 2E p S p S.36 where S p 1 u p, â p, + v p, ˆb 2E p, S up, â p, + v p, ˆb p,. S.37 p, Next, we need to evaluate the Dirac andwiche u Su, v Sv, etc., where S 12 σ σ. S.38 For the non-relativitic mode p m we approximate up, u0, m ξ ξ, v p, v0, m η η, S.39 which give u u p, Sup, m ξ σξ, v p, Sv p, m η ση, u p, Sv p, O p 0, S.40 v p, Sup, O p 0, and conequently S p ξ σ 2 ξ â p,â p, + η σ 2 η ˆb p,ˆb bp,, + O p /m. S.41 At thi point, let u eparate the â â term from the ˆbˆb term in the momentum integral S.36, and then in the ˆbˆb part change the ign of the integration variable p. Putting the two part 12

13 back together now give u Ŝ net d 3 p 1 2π 3 Ŝ p 12 2E p where for the non-relativitic momenta Ŝ ξ σ 2 ξ â p,â p, + η σ 2 η ˆb p,ˆb bp,, + O p /m. S.42 To re-write thi formula in the form of eq. 13, we note that the two-component pinor ξ and η have oppoite pin. Specifically, η ξ σ 2 ξ and therefore At the ame time, η σ 2 η σ ξ σ 2 2 σ 2ξ σ ξ σ 2 2 σ 2 ξ ξ σ 2 ξ. S.43 ˆbp,ˆb p, ˆb p, ˆb p, + 2E p 2π 3 δ 3 0 δ,. S.44 Conequently, we may re-write the ˆbˆb part of eq. S.42 a, η σ 2 η ˆb p,ˆb bp, +, ξ σ 2 ξ ˆb p, ˆb p, 2E p 2π 3 δ 3 0 ξ σ 2 ξ S.45 where the econd term on the right hand ide vanihe becaue ξ σ 2 ξ tr σ 2 0. S.46 Hence, interchanging the ummation pin indice and, we have, η σ 2 η ˆb p,ˆb bp, + ξ σ 2 ξ ˆb p,ˆb p,, S.47, and plugging thi formula into eq. S.45 finally give u S p, ξ σ 2 ξ â p,â p, + ˆb p,ˆb p, + O p /m 13 for the non-relativitic mode p. Q.E.D. 13