FY3464 Quantum Field Theory Exercise sheet 10

Transcription

1 Exercie heet 10 Scalar QED. a. Write down the Lagrangian of calar QED, i.e. a complex calar field coupled to the photon via D µ = µ + iqa µ. Derive the Noether current and the current to which the photon couple (defined by A µ = j µ. b. Find the vertice of thi theory. [Pay attention to the ign of the momentum of calar particle.] c. Write down the matrix element for calar Compton cattering φγ φγ and how that it i gauge invariant. d. Derive the cro ection for calar Compton cattering. a. Excluding a calar elf-interaction, the Lagrangian i or expanded with D µ = µ +iqa µ, L = (D µ φ D µ φ m 2 φ φ 1 4 F2. L = µ φ µ φ iqa µ φ µ φ+iqa µ ( µ φ φ+q 2 A µ A µ φ 1 φ }{{} 4 F2. L I We obtain the current on the RHS of the wave-equation for the photon from j µ = L [ ] [ ] I = i φ D µ φ (D µ φ φ = i φ µ φ ( µ φ φ 2q 2 A µ A µ φ φ. (1 A µ Note that j µ contain A µ, becaue L I i quadratic in A µ. A a reult, the plitting (we are ued from fermion into external current and the reulting electromagnetic field become ambiguou. The Noether current follow with δφ = iqφ, δφ = iqφ, and δa µ = 0, j µ = δl δl [ ] δφ+ δ µ φ δ µ φ δφ = iq φ D µ φ (D µ φ φ. (2 where we ued that A µ i invariant under global gauge tranformation. Thu the two current agree a expected. They are conerved and gauge invariant. Note that the uual j µ A µ coupling rule applie only for the linear term. For term quadratic in A, L I / A µ implie an additional factor 1/2. b. We ee from L I that there exit a φφa vertex iqa µ [φ µ φ ( µ φ φ] and a φφaa vertex given by q 2 η µν A µ A ν φ φ. In the latter cae we have to think about the ymmetry factor (2, becaue the two photon can be permutated. In the former cae about the ign of the momenta. Starting from the operator for the two real field, we obtain a ± (k = [a 1 (k±ia 2 (k]/ 2 and a ± (k = [a 1 (k±ia 2 (k]/ 2

2 with a ± (k 0 = 0 and a ± (k 0 = k,±. The field operator φ and φ are φ(x [a + (ke ikx +a (keikx ] and φ (x [a + (keikx +a (ke ikx ] Conider now e.g. an incoming particle with momentum k and an outgoing particle with momentum k. Then 0 φ(x k,+ = N k e ikx and thu k,+ ( µ φ φ k,+ ik e i(k kx and k,+ φ µ φ k,+ ike i(k kx Thu the vertex in thi cae iq(k + k µ. (A k come with an i, another one from exp(is int, another one from the prefactor +iq. Thi can be repeated for the other 3 type of vertice, with the reult that the momentum of anti-particle enter with a minu ign conitent with our interpretation of antiparticle a particle moving backward in time and croing ymmetry. c. At tree-level, three diagram contribute: Two are analogou to the one of fermionic QED, the third one i the o-called eagull diagram, ia = ( iq 2[ ε (2p +k i (p+k 2 m 2 ε (2p+k+ ε (2p k i ] (p k 2 m 2 ε (2p k 2iε ε. (3 Performing the gauge tranformation Λ(x = iλ exp( ikx, the polariation vector change a ε µ ε µ = ε µ +λk µ. Since λ i arbitrary, replacing ε by k (or ε k in A ha to give zero. Inerting ε k and uing k 2 = 0, (p+k 2 m 2 = 2pk and (p k 2 m 2 = 2pk give A ε (2p +k 1 2pk 2pk+ε (2p k 1 2pk 2p k 2ε k (4 = ε (2p +k 2p+k 2k = 2ε (p +k p k = 0. (5 d. Ue tranvere polarized photon in the ret frame of initial φ. Then ε p = ε p = 0. Since alo ε k = ε k = 0, only the ε ε term urvive. A 2 = 4e 4 (ε ε 2. (6 We recognie thi a one term in the normal Compton cattering amplitude, and copy the differential cro ection (lab frame from there, dσ dω = 1 4mω ω 2 16π 2 mω A 2 = α2 4m 2 ω 2 ω 2 [ 4(ε ε 2] (7 with α e 2 /(4π. Moreover, it i r,r (ε ε 2 = 1+co 2 ϑ and ω [ω(1 coϑ+m] = mω.

3 Vacuum polariation and the optical theorem. a. Derive the imaginary part of the photon polariation. Π on (q 2 = Π(q 2 Π(0 = 2α π 1 0 dxx(1 xln ] [1 q2 m 2x(1 x. b. Ue the optical theorem to connect I(Π on to the decay of a virtual photon γ into a fermion pair, γ ff. a. For ome background ee chapter 6.1 of the note, in particular example 6.1. We can contruct the imaginary part of the photon propagator from Π on (q 2 = Π(q 2 Π(0 = 2α 1 ] dxx(1 xln [1 q2 π m 2x(1 x, (8 0 uing variou way: Option 1: Find the x range for which the log i negative for a given q 2, and ue then Iln(x+iε = π. From 1 q2 m 2 x(1 x = 0, it follow x = 1 2 ± 1 2 β with β = 1 4m 2 /q 2. Then IΠ on (q 2 +iε = 2α π ( π β dxx(1 x = α β 3 β(1+2m2 /q 2. (9 Option 2: Do firt the x integral: change variable x = 1 2 (1+η and ue then dηln [ 1+x(1 η 2 ] = 4(1 ϑcotϑ (10 dηη 2 ln [ 1+x(1 η 2 ] = (1 ϑcotϑcot2 ϑ (11 with in 2 ϑ = q 2 /(4m 2. Ue then arccotz = iarccothiz to obtain the real part for q 2 > 4m 2 and arccothz = 1 z +1 ln 2 1 z to find the imaginary part via the dicontinuity. b. We apply the optical theorem to the cae of a decay , 2IT ii = TinT ni = dφ (2 A ina ni (12 n 1, 2 We want to compare the RHS to the vacuum polariation Π(q 2, where we factored out the external photon (and a tranvere polariation projector. Therefore we et A = ε µ A µ, and calculate with A µ = ū(p 1 (ieγ µ v(p 2 A µ in Aν ni = e 2 tr[(p/ 1 +mγ µ (p/ 2 mγ ν ] (13 1, 2 = 4e 2[ p µ 1 pν 2 +pµ 2 pν 1 2 ηµν] (14

4 where we introduced = k 2 = (p 1 +p 2 = 2p 1 p 2 +m 2. Then we recall and dφ (2 = 1 16π 2 p cm dω, p 2 cm = λ(,m2 1,m2 2 4 for m m 1 = m 2. Combining thi, we obtain 2IT ii = dφ (2 A ina ni = 1, 2 = 4 dω 1 16π 2 ] [1 4m2 2 ] 1/2 [1 4m2 4e 2 [ ] (15a ] 1/2 = 2 [1 4m2 e 2 dω 4π 4π [ ] (15b with = 2α [1 4m2 S µν = ] 1/2 S µν (15c dω 4π [ p µ 1 pν 2 +p µ 2 pν 1 2 ηµν]. (16 Since S µν can be only a function of k (p 1,2 are integrated, we ue a anatz We determine the coefficient, firt contracting with k µ k ν, S µν = Ak µ k ν +Bη µν. (17 k µ k ν S µν = (A+B (18 and econd contracting with η µν, = 2(kp 1 (kp = 0 (19 η µν S µν = A+4B (20 Thi reult in B = (+2m 2 /3 and A = (+2m 2 /3. Thu = 2(p 1 p 2 2 = (+2m 2 2. (21 S µν = 1 3 (1+2m2 /(k 2 η µν k µ k ν = S(k(k 2 η µν k µ k ν (22 and we ee that S µν i tranvere a required by gauge invariance. Now we factor out the tranvere projection operator, and conider only the calar part of Eq. (15. Comparing then with a. we find agreement. IT ii = α 3 ] 1/2 [1 4m2 (1+2m 2 / (23

5 Remark 1: We obtained IΠ(k 2 a follow: We cut the virtual line, then the diagram T ii decompoe into Tin and T ni. The virtual fermion line correpond now to two external on-hell particle, and therefore the integration i over d 3 k/(2π 3 2ω intead d 4 k/(2π 4. Thi how alo that the imaginary part of loop diagram are finite. In general, the et of uch recipe to obtain the imaginary part of a loop diagram are called Cutowky cutting rule. Remark 2: From (12 we ee that IT ii = ωγ( hold. Thu the imaginary part IT ii hould be poitive. Otherwie the choen vacuum i untable, and the intenity of a beam of photon would grow a I(t = I 0 exp( 2IT ii t/ω. Gauge boon propagator in axial gauge. The axial gauge condition i n µ A a µ = 0 where n i a fixed vector. a. Show that the Fadeev-Popov term i independent of A a µ and, thu, can be aborbed in the normaliation of the path integeral. b. Derive the gauge boon propagator uing a the correponding gauge-fixing term. L gf = 1 2ξ (nµ A µ 2 0. It may be ueful to recall the cae of the Maxwell equation. Chooing in µ F µν = µ ( µ A ν ν A µ = A ν µ ν A µ = j ν a covariant gauge a e.g. the Lorenz gauge µ A µ = 0 reduce the d.o.f. by one to three. We can till add to the potential A µ any µ χ atifying χ = 0, etting e.g. A 0 = 0, giving A = 0. Now we have only 2 d.o.f., but explicit Lorentz invariance i lot. Thi choice i with n µ = (1,0,0,0 a pecial cae of the axial gauge. a. We have to evaluate ( δg a det δϑ b δ(g a for the gauge condition g a = n µ A a µ = 0. An infiniteimal gauge tranformation lead to δg a = δ(n µ A a µ = n µ D ac µ ϑ c = n µ (δ ac µ gf abc A b µϑ c = n µ µ ϑ a, where we ued n µ A a µ = 0 in the lat tep. Thu the Fadeev-Popov Lagrangian i independent of the gauge field, L FP = c G ϑ c = cnµ µ c, and change only the normaliation of the generating functional. b. We have to conider only the quadratic term, thu the non-abelian term play role, and we uppre the group index a. L eff = L cl + L gf = 1 2 A µ(η µν µ ν A ν + 1 2ξ A µn µ n ν A ν = 1 2 A µp µν A ν. x,a

6 The propagator D µν ha to atify FT give P µν (kd νλ (k = δ λ µ with P µν D νλ (x y = δ λ µδ(x y. Plugging the tenor decompoition P µν (k = η µν k 2 +k µ k ν + 1 ξ nµ n ν. D µν (k = Aη µν +B(n µ k ν +n ν k µ +Ck µ k ν +Dn µ n ν into the anatz and comparing the coefficient lead to A = 1 k 2, B = 1 nk 1 k 2, C = ξk 2 n2 (nk 2, and D = 0. k2 Thu the gauge boon propagator in axial gauge i given by D µν (k = 1 [ k 2 η µν + 1 nk (nµ k ν +n ν k µ n2 ξk 2 ] (nk 2 k 2 kµ k ν.