Problem 1(a): At equal times or in the Schrödinger picture the quantum scalar fields ˆΦ a (x) and ˆΠ a (x) satisfy commutation relations

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1 PHY 396 K. Solutions for homework set #7. Problem 1a: At equal times or in the Shrödinger iture the quantum salar fields ˆΦ a x and ˆΠ a x satisfy ommutation relations ˆΦa x, ˆΦ b y 0, ˆΠa x, ˆΠ b y 0, ˆΦa x, ˆΠ b y δ ab iδ 3 x y. S.1 By Leibniz rule, ˆΦa yˆπ b y, ˆΦ x ˆΦ a y ˆΠb y, ˆΦ x + ˆΦa y, ˆΦ x ˆΠb y ˆΦ a y iδ b δ 3 y x + 0 ˆΠ b y S. iδ b ˆΦa x δ 3 y x and likewise ˆΦb yˆπ a y, ˆΦ a x iδ a ˆΦb x δ 3 y x. S.3 Hene, for the ˆQ ab harge as in eq. 5, ˆQab, ˆΦ x d 3 y ˆΦa yˆπ b y ˆΦ b yˆπ a y, ˆΦ x d 3 y iδ b ˆΦa x + δ a ˆΦb x δ 3 y x iδ b ˆΦa x + δ a ˆΦb x. S.4 Similarly, ˆΦa yˆπ b y, ˆΠ x ˆΦ a y ˆΠb y, ˆΠ x + ˆΦa y, ˆΠ x ˆΠb y ˆΦ a x 0 + iδ a δ 3 x y ˆΠ b y S.5 +iδ a ˆΠb x δ 3 y x and likewise ˆΦb yˆπ a y, ˆΠ x +iδ b ˆΠa x δ 3 y x, S.6 1

2 hene ˆQab, ˆΠ x d 3 y ˆΦa yˆπ b y ˆΦ b yˆπ a y, ˆΠ x d 3 y +iδ a ˆΠb x iδ b ˆΠa x δ 3 y x S.7 iδ b ˆΠa x + δ a ˆΠb x. Problem 1b: As we saw in lass bak in Setember, quantizing lassial fields with Lagrangian 1 leads to Hamiltonian oerator Ĥ d 3 x 1 ˆΠ + 1 ˆΦ + m ˆΦ + λ 4 ˆΦ. S.8 Eah of the 4 terms on the big arentheses here is SON invariant, and that makes it ommute with the ˆQ ab harges. Indeed, suose some N oerators ˆV whih ould be ˆΦ x, or ˆΠ x, or whatever satisfy ommutation relations similar to eqs. 6, namely ˆQab, ˆV iδ b ˆVa + iδ a ˆVb, S.9 then the ˆV oerator ommutes with the harges ˆQ ab. Here is the roof: ˆQab, ˆV ˆQab, ˆV { ˆV, ˆQab, ˆV } { } ˆV, iδ b ˆVa + iδ a ˆVb { i ˆVb, ˆV } a 0. { + i ˆVa, ˆV } b S.10 In artiular, letting ˆV ˆΠ x, or ˆV ˆΦ x, or ˆV ˆΦ x whih also satisfy ˆQab, ˆΦ x ˆQab, ˆΦ x iδ b ˆΦ a x + iδ a ˆΦ b x S.11

3 we immediately obtain ˆQab, ˆΠ x 0, ˆQab, ˆΦ x 0, ˆQab, ˆΦ x 0, S.1 hene also Ĥ and therefore ˆQab, 0. ˆQ ab, ˆΦ x 0, S.13 Problem 1: ˆQab, ˆQ d d 3 x ˆQ ab, d 3 x ˆΦ xˆπ d x ˆΦ d xˆπ b x d 3 x ˆQab, ˆΦ xˆπ d x ˆΦ d xˆπ b x ˆΦ x ˆQab, ˆΠ d x + ˆΦ d x ˆQab, ˆΠ x d 3 x ˆΦ iδ bd ˆΠa + iδ ad ˆΠb ˆQab, ˆΦ x ˆΠd x + ˆQab, ˆΦ d x ˆΠ x iδ b ˆΦa + iδ a ˆΦb ˆΠd iδ bd ˆΦa + iδ ad ˆΦb ˆΠ ˆΦ d iδ b ˆΠa + iδ a ˆΠb iδ bd d 3 x ˆΦ ˆΠa ˆΦ a + iδ ad + iδ b d 3 x ˆΦd ˆΠa ˆΦ a iδ d 3 x ˆΦ ˆΠb ˆΦ b d 3 x ˆΦd ˆΠb ˆΦ b iδ bd ˆQ a + iδ ad ˆQ b + iδ b ˆQ da iδ a ˆQ db iδ b ˆQ ad + iδ a ˆQ bd + iδ bd ˆQ a iδ ad ˆQ b. S.14 3

4 Problem 1d: In the Shrödinger iture, d ˆΦ 3 1 a x π 3 e +i x â E,a + e i x â,a, d ˆΠ 3 ie b x π 3 e +i x â E,b e i x â,b. S.15 Using these exansions and d 3 x e ±i x e ±i x π 3 δ 3 ± ±, S.16 we obtain d 3 x ˆΦ a xˆπ b x d 3 1 d 3 ie π 3 E π 3 E d 3 π 3 +â,a â,b π3 δ 3 + â,a â,b π3 δ 3 +â,aâ,b π3 δ 3 + â,aâ,b π3 δ 3 i â 4E,a â,b â,a â,b + â,aâ,b â,aâ,b Likewise, exhanging a b everywhere and also in two of the four terms, we have d 3 x ˆΦ b xˆπ a x d 3 π 3 S.17 i â 4E,a â,b â,a â,b + â,aâ,b â,aâ,b. S.18 Consequently, ˆQ ab d 3 x ˆΦa xˆπ b x ˆΦ b xˆπ a x d 3 i â π 3 4E,a â,b â,a â,b + â,aâ,b â,aâ,b d 3 i π 3 â 4E,b â,a â,b â,a + â,bâ,a â,bâ,a d 3 i π 3 â 4E,a â,b â,b â,a â,aâ,b + â,bâ,a d 3 i + â π 3 4E,aâ,b + â,b â,a â,bâ,a + â,a â,b S.19 4

5 where on the last two lines â,a â,b â,b â,a â,aâ,b + â,bâ,a but â,a, â,b â,a, â,b 0 0 0, S.0 â,aâ,b + â,b â,a â,bâ,a + â,a â,b â,aâ,b + â,a, â,b â,bâ,a â,b, â,a S.1 â,aâ,b â,bâ,a beause â,a, â,b â,b, â,a S. although the two ommutators here are infinite, it s the same infinity in both ases! Therefore, lugging eqs. S.0 and S. into the integrals on the last two lines of eq. S.19, we obtain d ˆQ 3 ab 0 + π 3 i â 4E,aâ,b â,bâ,a d 3 π 3 1 E iâ,aâ,b + iâ,bâ,a 8 Problem 1e: Ation of the ˆQ ab harges on single-artile states, follows from eq. 8 we have just roved: ˆQ ab, iδ b, a + iδ a, b S.3 and hene i Θ ab ˆQ ab, 1 Θ a, b + 1 Θ b, b +Θ d, d. S.4 Alying this oerator several times, we obtain i Θ ab ˆQ ab, i Θ ab ˆQ ab Θ d, d Θ d i Θ ab ˆQ ab, d Θ d Θ de, e, S.5 5

6 likewise i Θ ab ˆQ ab 3, Θd Θ de Θ ef, f, S.6 et., et., or in matrix form i Θ ab ˆQ n ab, Θ n, d. d S.7 Hene, exanding the exonential in eq. 9 to a ower series, we obtain DR, n0 1 i n! Θ ab ˆQ n ab, 1 Θ n, d n! d n exθ, d d 10 R d, d. Problem 1f: Bak in homework set# we saw that Fok-sae oerators that are linear ombinations of â â terms at on multi-artile states as additive one-body-at-a-time oerators, Â α,β A αβ â αâ β N artiles N i1 Â 1 i th artile for Â1 αβ α A αβ β. S.8 In artiular, the harges ˆQ ab have form 8 of additive one-body-at-a-time oerators, thus N artiles ˆQ ab N i1 ˆQ 1 ab ith artile S.9 ˆQ 1 where ab ith ats on the seies index of the i th artile aording to eq. S.3. The oerators ating on different artiles ommute with eah other, so exonentiating the ˆQ 1 ab 6

7 N artile hermitian oerator S.9 rodues a rodut of unitary one-artile oerators ex i Θ ab ˆQ N artiles ab N i1 ex i Θ ab ˆQ 1 ab i th artile. S.30 Stritly seaking, this formula resumes N indeendent artiles whose quantum states do not have to be Bose-symmetri. However, when the oerator S.30 ats on a state whose wave funtion haens to be totally symmetri, the result is also totally symmetri beause eah one-artile fator ex i Θ ab ˆQ 1 ab i th has exatly same form. Consequently, eq. S.30 works as written also in the Hilbert sae of N idential bosons. Therefore, alying the Fok-sae oerator 9 to an N-boson state with definite momenta and seies of all artiles would give us DR 1, 1,,,..., N, N N ex i Θ ab ˆQ 1 ab i th 1, 1,,,..., N, N i1 Symmetrized N i1 ex i Θ ab ˆQ 1 ab i, i in light of eq. 10 roved in art e N Symmetrized R i d i i, d i i1 R 1 d 1 R d R N d N 1, d 1,, d,..., N, d N imliit sum over d 1, d,... d N. In other words, eah artile is indeendently rotated by the same SON symmetry R. Problem 1g: The harged fields Φx and φ x deomose into reation and annihilation oerators for artiles and antiartiles as d ˆΦx π 3 e ix â E + e +ix +E ˆb, d ˆΦ 3 1 S.31 0 x π 3 e ix ˆb + e +ix â +E. E 7

8 Comaring this deomosition to the real fields ˆΦ a x d 3 π 3 1 E e ix â,a + e +ix â,a 0 +E, a 1,, S.3 we find that Φ Φ 1 + iφ / and Φ Φ 1 iφ / alls for â â,1 + iâ,, ˆb â,1 iâ,, â â,1 iâ,, ˆb â,1 + iâ,, and onversely, S.33 â,1 â + ˆb, â, â ˆb i, â,1 â + ˆb, â, â ˆb i. S.34 Consequently, iâ,â,1 + iâ,1â, i â ˆb â + ˆb + i â + ˆb â ˆb i i â â ˆb â + â ˆb ˆb ˆb â â ˆb ˆb. â â + ˆb â â ˆb ˆb ˆb S.35 Substituting this formula into eq. 8 for the ˆQ 1 harge immediately gives us d ˆQ π 3 iâ,â,1 + iâ,1â, E d 3 π 3 1 E â â ˆb ˆb ˆN artiles ˆN antiartiles. 11 8

9 Problem a: Let T µν λ K λµν. Regardless of the seifi form of the K λµ,ν φ, φ tensor, its antisymmetry with reset to its first two indies λ and µ guarantees µ T µν µ λ K λµ,ν 0 S.36 and hene µ T µν µ T muν Noether hoefully 0 15 Furthermore, d 3 x T 0ν i K i 0,ν d Area i K i 0 ν 0 S.37 boundary of sae when the integral is taken over the infinite 3D sae, hene Pnet ν d 3 x T 0ν d 3 x T 0ν Noether. 16 More generally, any Noether urrent J µ Noether divergene of some antisymmetri tensor, an be re-defined by adding to it a total J µ J µ Noether + λi λµ, I λµ I µλ, S.38 without soiling the urrent onservation or hanging the net harge, µ J µ µ J µ Noether hoefully 0, Q net d 3 x J 0 d 3 x JNoether 0. S.39 The roof of these equations is omletely similar to the above roof of eqs. 15 and 16. 9

10 Problem b: In the Noether s formula 1 for the stress-energy tensor, φ a stand for the indeendent fields, however labeled. In the eletromagneti ase, the indeendent fields are omonents of the 4 vetor A λ x, hene T µν Noether EM µ A λ ν A λ g µν L F µλ ν A λ gµν F κλ F κλ. S.40 While the seond term here is learly both gauge invariant and symmetri in µ ν, the first term is neither. Problem : Clearly, one an easily restore both µ ν symmetry and gauge invariane of the eletromagneti stress-energy tensor by relaing ν A λ in eq. S.40 with F ν λ, hene eq. 18. The orretion amounts to T µν T µν true T µν Noether F µλ F ν λ ν A λ λ A ν λ F µλ A ν A ν λ F µλ. S.41 Moreover, for the free EM fields, the last term on the right hand side vanishes by equation of motion λ F µλ J µ 0. Consequently, T µν T µν Noether + λk λµν where K λµν F µλ A ν K µλν, S.4 in erfet agreement with eq. 14. Problem d: Let s start with the Lagrangian 17. In omonent form, F i0 F 0i E i, F ij ɛ ijk B k. S.43 Therefore, F i0 F i0 F 0i F 0i E i E i where the minus sign omes from raising one sae index. Likewise, F ij F ij +ɛ ijk B k ɛ ijl B l +B k B k where the lus sign omes from raising 10

11 two sae indies at one. Altogether, L 1 4 F µν F µν F i0 F i0 + F 0i F 0i + F ij F ij 1 E B. S.44 Consequently, eq. 18 for the energy density gives H T 00 F 0i F 0 i L +E 1 E B 1 E + B S.45 in agreement with the standard eletromagneti formulæ note the 1, rationalized units here. Likewise, the energy flux and the momentum density are S i T i0 T 0i F 0j F i j E j +ɛ ijk B k +ɛ ijk E j B k E B i, S.46 in agreement with the Poynting vetor S E B again, in the 1, rationalized units. Finally, the 3 dimensional stress tensor is T ij EM F iλ F j λ gij L F i0 F j 0 F ik F j k + δij L E i E j + ɛ ikl B l ɛ jkm B m + 1 δij E B E i E j B i B j + δ ij B + 1 δij E B S.47 E i E j B i B j + 1 δij E + B. Problem e: In a sense, eq. 0 follows from eq. S.41, but it is just as easy to derive it diretly from Maxwell equations. Starting with eq. 18, we immediately have µ T µν EM µ F µλ F ν λ F µλ µ F ν λ + 1 F κλ ν F κλ. S.48 Using the antisymmetry F µλ F λµ, we rewrite the seond term on the right hand side as F µλ µ F ν λ F µλ µ F νλ +F µλ µ F λν relabel λ µ F λµ λ F νµ +F µλ λ F νµ 1 F µλ λ F νµ + µ F λν 1F µλ ν F µλ S.49 11

12 where the last equality follows from the homogeneous Maxwell equation ɛ κλµν λ F µν 0 λ F νµ + µ F λν + ν F µλ 0. S.50 Consequently, the seond and the third terms on the right hand side of eq. S.48 anel eah other and we are left with the first term only. Thus, µ T µν EM µ F µλ F ν λ Jλ F ν λ S.51 where the seond equality omes from the in-homogeneous Maxwell equation µ F µλ J λ. This roves eq. 0, and eq. 1 follows from that and the onservation 19 of the net stress-energy tensor. Problem 3a: As disussed in lass, Euler Lagrange equations for harged fields an be written in a manifestly ovariant form as In artiularly, for φ Φ, we have D µ D µ φ φ 0. S.5 D µ Φ Dµ Φ, Φ m Φ, whih gives us D µ D µ Φ + m Φ 0. S.53 Likewise, for φ Φ we have D µ Φ Dµ Φ, Φ m Φ, and therefore D µ D µ Φ + m Φ 0. S.54 As for the vetor fields A ν, the Lagrangian deends on µ A ν only through F µν, whih 1

13 gives us the usual Maxwell equation µ F µν J ν where J ν A ν. S.55 To obtain the urrent J ν, we notie that the ovariant derivatives of the harged fields Φ and Φ deend on the gauge field: D µ Φ A ν iqδ ν µφ, D µ Φ A ν iqδ ν µφ. S.56 Consequently, J ν D ν Φ iqφ iq ΦD ν Φ Φ D ν Φ. D ν Φ iqφ S.57 Note that all derivatives on the last line here are gauge-ovariant, whih makes the urrent J ν gauge invariant. In a non-ovariant form, J ν iqφ ν Φ iqφ ν Φ q Φ Φ A ν. S.58 To rove the onservation of this urrent, we use the Leibniz rule for ovariant derivatives, D ν XY XD ν Y + Y D ν X. This gives us µ Φ D µ Φ D µ Φ D µ Φ D µ Φ D µ Φ + Φ D µ D µ Φ, µ ΦD µ Φ D µ ΦD µ Φ D µ ΦD µ Φ + Φ D µ D µ Φ, S.59 and hene in light of eq. S.57 for the urrent, νj ν iq D ν ΦD ν Φ + ΦD ν D ν Φ iqφ D Φ iqφ D Φ + iq D ν Φ D ν Φ + Φ D ν D ν Φ by equations of motion iqφ m Φ iqφ m Φ 0. S.60 13

14 Problem 3b: Aording to the Noether theorem, where T µν Noether µ A λ ν A λ + µ Φ ν Φ + T µν µν Noether EM + T Noether matter µ Φ ν Φ g µν L S.61 similar to the free EM fields, and T µν Noether EM F µλ ν A λ gµν F κλ F κλ S.6 T µν Noether matter Dµ Φ ν Φ + D µ Φ ν Φ g µν D λ Φ D λ Φ m Φ Φ. S.63 Both terms on the seond line of eq. S.61 lak µ ν symmetry and gauge invariane and thus need λ K λµν orretions for some K λµν K µλν. We would like to show that the same K λµν F µλ A ν we used to imrove the free eletromagneti stress-energy tensor will now imrove both the T µν µν EM and T mat at the same time! Indeed, to imrove the salar fields stress-energy tensor we need T µν matter T µν µν true matter T Noether matter D µ Φ D ν Φ ν Φ + D µ ΦD ν Φ ν Φ D µ Φ iqa ν Φ + D µ Φ iqa ν Φ S.64 A ν iqφ D µ Φ iqφd µ Φ A ν J µ, while the imrovement of the EM stress-energy tensor was selled out in eq. S.41: T µν EM F µλ F ν λ ν A λ +F µλ λ A ν λ F λµ A ν + A ν J µ. S.65 Altogether, to symmetrize the whole stress-energy tensor, we need T µν tot T µν µν true total T Noether total λ F µλ A ν K λµν. 14

15 Problem 3: Beause the fields Φx and Φ x have oosite eletri harges, their rodut is neutral and therefore µ Φ Φ D µ Φ Φ D µ Φ Φ + Φ D µ Φ. Similarly, µ D µ Φ D ν Φ D µ D µ Φ D ν Φ + D µ Φ D µ D ν Φ m Φ D ν Φ + D µ Φ D ν D µ Φ + iqf µν Φ S.66 where we have alied the field equation D µ D µ + m Φ x 0 to the first term on the right hand side and used D µ, D ν Φ iqf µν Φ to exand the seond term. Likewise, µ D µ Φ D ν Φ D µ D µ Φ D ν Φ + D µ Φ D µ D ν Φ m Φ D ν Φ + D µ Φ D ν D µ Φ iqf µν Φ S.67 and µ g µν D λ Φ D λ Φ m Φ Φ ν D λ Φ D λ Φ + m ν Φ Φ D ν D µ Φ D µ Φ D µ Φ D ν D µ Φ + m Φ D ν Φ + m Φ D ν Φ. S.68 Together, the left hand sides of eqs. S.66, S.67 and S.68 omrise µ T µν mat f. eq. 6. On the other hand, ombining the right hand sides of these three equations results in massive anellation of all terms exet those ontaining the gauge field strength tensor F µν. Seifially, µ T µν mat D µ Φ iqf µν Φ + D µ Φ iqf µν Φ F µν iqφ D µ Φ iqφ D µ Φ F µν J ν S.69 In erfet agreement in eq. 1. In light of eq. 0 roved in roblem e, this means that the total stress-energy tensor 5 is onserved. 15