Physics 582, Problem Set 6 Solutions

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1 Physics 582, Problem Set 6 Solutions TAs: Hart Goldman and Ramanjit Sohal Fall PATH ITEGRAL QUATIZATIO OF THE FREE ELECTROMAGETIC FIELD.B. The ordering of the sub-uestions is a little confusing. We will accet solutions that answer the sub-uestions in a different order. 1. In momentum sace, the roagator is G µν = iɛ [g µν + α 1 µ ν 2, 1.1 as will be derived in a subseuent art of the uestion. We have chosen the Feynman iɛ rescrition which yields a time-ordered roagator. 2. Inverting the above matrix euation, we find [ g µν 2 + α 1 α µ ν G νλ = δ µ λ. 1.2 One way to arrive at this exression without exlicitly inverting the matrix is to note that the inverse of G µν must be symmetric in µ and ν and so can only contain terms of the form g µν 2 and µ ν. The coefficients of these terms are then fixed by checking that the above exression is satisfied. ow, inverting the Fourier transform, we find that the Green function satisfies the PDE [ g µν 2 α 1 α µ ν G νλ x y = δ µ λδ 4 x y We can read off from art 1 that D F = iɛ = D F = which is the roagator for a massless, real scalar field. d 4 e i x x 2π iɛ 1.4 As calculated in a revious roblem set and in lecture, in Minkowski sace a massive, real scalar field has roagator Gx x = i m 4π K 1m s s 2

2 2 In the m 0 limit we find that using the asymtotic behavior of K 1 z discussed in chater 5 of the lecture notes i sace-like searation, x x 2 = s 2 < 0 D F x x 4π = 2 s i time-like searation, x x 2 = s 2 > 0 4π 2 s 2 4. a The Fadeev-Poov determinant is given by δg g [A µ = Det δu g=0 1.7 where the generalized gauge fixing condition is Under a U1 gauge transformation U, we have that Hence, ga µ = µ A µ cx. 1.8 ga U µ = µ A µ + µ φ cx = µ A µ + 2 φ cx. 1.9 δgx g [A µ = Det = Det δφy g= is indeendent of the gauge field configuration. b Under a gauge transformation, the field strength F µν is invariant, as was shown in a revious roblem set. The couling to the current transforms as follows: δa µ J µ = µ φj µ φ µ J µ 1.11 where we have droed total derivative terms. Hence the action is gauge invariant rovided the currents are conserved. However, in order to ensure that the full ath integral is invariant, we must check that 1 g [A µ DUδgA U µ 1.12 is also gauge invariant. Using the roerty of the Haar measure that DU = DUU for fixed U, we have that 1 g [A U µ = DUδgA U U µ = DU UδgA U U µ = DU δga U µ = 1 g [A µ 1.13

3 3 where we have set U = U U. Altogether, the ath integral Z[J = DA µ g [A µ DUδgA U µ e is[a,j 1.14 is gauge invariant. 5. We first massage the above ath integral by changing variables A µ A V µ which is related to A µ by the gauge transformation V. If we choose V = U 1 then we obtain [ Z[J = DU DA µ g [A µ δga µ e is[a,j 1.15 where DU is an infinite constant. ow, we consider a generalized gauge-fixing condition of gx = cx, where cx is some function, instead of gx = 0 nothing in the above derivation changes. Since hysical uantities and hence the ath integral do not deend on the gauge-fixing condition, we can average over all cx s, weighted by a Gaussian: Z[J = Det[ 2 = Det[ 2 where DA µ Dce i d 4 x cx2 2α δgaµ cxe is[a,j 1.16 DA µ e i d 4 xl α[a,j 1.17 L α = 1 4 F µν 2 J µ A µ 1 2α µa µ = 1 [ 2 A µ g µν 2 α 1 α µ ν A ν J µ A µ 1.19 we integrated by arts and skied some algebra to get to the second line. Performing the usual shift of variables and comleting the suare, we obtain [ Z[J = Det[ 2 Det g µν 2 α 1 α µ ν e i 2 d 4 x d 4 x J µxd µν x x J ν x 1.20 where D µν is the same D µν from art 1. This exression is indeendent of α since we started with a gauge-indeendent ath integral. The arameter α was introduced when we averaged over different gauge-fixing conditions with a Gaussian; since the ath integral was indeendent of the choice of gauge-fixing condition, it follows that it must also be indeendent of the choice of α. Indeed, we can note that the exonential is exlicitly indeendent of α since µ J µ x = 0 and so the Fourier transform satisfies µ J µ = 0. Hence the term in the roagator deendent on α vanishes when

4 4 contracted with J µ. However, we cannot conclude that the ath integral is gauge invariant by comuting correlators of the gauge field, T A µ1 x 1... A µn x n as these exression necessarily deend on the gauge. 2. PROPAGATORS AD CORRELATORS FOR THE OE-DIMESIOAL QUATUM HEISEBERG ATIFERROMAGET We first recall the results of the sin-wave aroximation from the revious roblem set. We write the sin oerators as: j even Ŝ + j = 2Sâj Ŝ j = 2Sâ j Ŝ 3 j = S ˆnj We can Fourier exand: j odd Ŝ + j = 2Sˆb j Ŝ j = 2Sˆbj Ŝ 3 j = S + ˆnj. 2.1 âj = 2 e ij â, ˆbj = 2 e ijˆbj 2.2 where the momentum sums are over the reduced Brillouin zone: [ π/2, π/2. In the sinwave aroximation, the Hamiltonian can be diagonalized via a Bogoliubov transformation a = cosh θ sinh θ c 2.3 b sinh θ cosh θ d where cosh 2θ = sin The ground state satisfies ĉ G = ˆd G = The single article excitations satisfy Ĥĉ G = ε ĉ G, Ĥ ˆd G = ε ˆd G 2.6

5 5 where ε = 2JS sin. 2.7 In the sin-wave aroximation we dro uantities of order less than 1/S. In articular, this means we can dro terms uadratic in ˆnj when comuting the roagators below as, roughly seaking, nn 1/ε 2 1/S In the sin-wave aroximation the unit cell is enlarged to two sites since we have assumed a éel i.e. antiferromagnetic ordering. As such, when comuting the roagators, we must consider different cases according to whether the sites are even or odd. a Exlicitly, D 33 nt, n t = i G T Ŝ3n, tŝ3n, t G = i G Ŝ3n, tŝ3n, t G θt t i G Ŝ3n, t Ŝ3n, t G θt t. 2.8 ow, G Ŝ3n, tŝ3n, t G = G [S ˆnn, t[s ˆnn, t G 2.9 S 2 S G ˆnn, t G S G ˆnn, t G + G ˆnn, tˆnn, t G Let us first consider the case where n and n are both even. We then have that G ˆnn, t G = G ˆnn G 2.11 = 2 G e inâ e i nâ G 2.12, = 2 e in G [cosh θ ĉ sinh θ ˆd [cosh θ ĉ sinh θ ˆd G, 2.13 = 2 e in sinh θ sinh θ δ, 2.14, = 2 sinh 2 θ. 2.15

6 6 We also find G ˆnn, tˆnn, t G = 4 e in e in G [cosh θ 2 ĉ sinh θ ˆd [cosh θ ĉ sinh θ ˆd,,, = 4 2 e Ĥt t [cosh θ ĉ sinh θ ˆd [cosh θ ĉ sinh θ ˆd G,,, e in e in [sinh θ sinh θ δ, sinh θ sinh θ δ, + sinh θ cosh θ e iε+ε t t cosh θ sinh θ δ,δ, = 4 sinh 2 θ 2 sinh 2 θ + 4 e in e in e iε+εt t sinh 2 θ 2 cosh 2 θ,, 2 2 = sinh 2 2 θ + e i n+n e iεt t sinh 2 θ 2 e i n+n e iεt t cosh 2 θ 2.16 where, in the last line, we slit the second double sum into a roduct of two sums over and, and then relabelled, using the fact that ε and

7 7 sinh θ, cosh θ are even under. So, D 33 nt, n t 2.17 = iθt t S2 4 S 2 sinh 2 2 θ + sinh 2 θ } 2 + e i n+n e iεt t sinh 2 2 θ e i n+n e iεt t cosh 2 θ + t t 2.18 = is 2 + 4i S 2 sinh 2 2 θ + sinh 2 θ 2 + e i n+n e iεt t sinh 2 2 θ e i n+n e iεt t cosh 2 θ θt t 2 + e i n+n e iεt t sinh 2 2 θ e i n+n e iεt t cosh 2 θ θt t The first three terms are just constants and so their Fourier transforms are trivial to write down. As for the last two terms, we note that they are each the roduct of two functions written as inverse momentum-sace Fourier transforms. the convolution theorem, their momentum-sace Fourier transforms will be the convolutions of the Fourier transforms of the factors in the roducts. obtain, id 33, T = S 2 4 S 2 sinh 2 2 θ + sinh 2 θ δ,0 + 2 sinh 2 θ e iɛ T cosh 2 θ e iεt θt + 2 sinh 2 θ e iɛ T cosh 2 θ e iεt θ T. By So, we 2.20 where we have set T = t t. By erforming a Fourier transform in time or, euivalently, using the reresentation of the Heaviside ste function as a contour

8 8 integral, θt = 1 eiωt dω, we can write 2πi ω iɛ D 33, ω = D + ; T e iωt dt 2.21 = is 2 + 4i S 2 sinh 2 2 θ i sinh 2 θ δ,0 δω sinh 2 θ cosh 2 1 θ ω + ε + ε iɛ 1 ω ε ε + iɛ = S S sinh 2 θ sinh 2 θ 1 δ,0 ω iɛ 1 ω + iɛ sinh 2 θ cosh 2 1 θ ω + ε + ε iɛ 1. ω ε ε + iɛ 2.25 The euality of the first and second line is checked by inverting the Fourier transform and treating the ω integral as a contour integral. When T > 0 we must close the contour in the uer half-lane and so we ick u the oles at ω = iɛ and ω = ε ε + iɛ. When T < 0 we must close the contour in the lower half lane and so we ick u the oles at ω = iɛ and ω = ε + ε iɛ. By identical calculations, one finds that for n and n both even, the roagator takes the same form. In contrast, when one of n, n is odd and the other even, the roagator takes the form D 33, ω = S 2 4 S 2 sinh 2 2 θ + sinh 2 θ 1 δ,0 ω iɛ 1 ω + iɛ sinh2θ sinh2θ ω + ε + ε iɛ 1 ω ε ε + iɛ 2.27.

9 9 b Exlicitly, D + nt, n t = i G T Ŝ+ n, tŝ n, t G 2.28 = i G Ŝ+ n, tŝ n, t G θt t i G Ŝ n, t Ŝ+ n, t G θt t. Let us first consider the case where both n and n are even. We then have 2.29 G Ŝ+ n, tŝ n, t G = G Ŝ+ ne iĥt t Ŝ n G S G âne iĥt t â n G 2.31 = 2S 2 G e inâe iĥt t â e i n G 2.32, = 2S 2 e in+i n G [cosh θ ĉ sinh θ ˆd, [ e iĥt t cosh θ ĉ sinh θ ˆd G = 2S 2 = 2S 2, e in+i n e iε t t cosh θ cosh θ δ, e i n+n e iεt t cosh 2 θ, 2.35 where we used the fact that Ĥĉ G = ε ĉ G. By a similar calculation, we obtain G Ŝ n, t Ŝ+ n, t G = 2S 2 e i n+n e iεt t sinh 2 θ Altogether, D + nt, n t = 2iS 2 e i n+n e iεt t cosh 2 θ θt t iS 2 e i n+n e iεt t sinh 2 θ θt t Setting T t t, we can read off that D + ; T = 2iS [ e iεt cosh 2 θ θt + e iεt sinh 2 θ θ T. 2.39

10 10 By erforming a Fourier transform in time or, euivalently, using the reresentation of the Heaviside ste function as a contour integral, θt = 1 eiωt dω, 2πi ω iɛ we can write D + ; ω = D + ; T e iωt dt = 2S [ cosh 2 θ ω + ε iɛ sinh2 θ. ω ε + iɛ 2.40 One can check that this iɛ rescrition satisfies the time-ordered boundary conditions. Indeed, when inverting the Fourier transform, we write the integral over freuency as a contour integral. When T > 0 we must close the contour in the uer half-lane and so we ick u the ole in the first term. When T < 0 we must close the contour in the lower half lane and so we ick u the ole in the second term. Calculation of the roagator for n and n both odd roceeds in much the same way. One finds D + ; ω = 2S [ sinh 2 θ ω + ε iɛ cosh2 θ ω ε + iɛ Lastly, when one of n and n is odd and the other even, we find [ 1 D + ; ω = S sinh 2θ ω + ε iɛ ω ε + iɛ 2. In the interaction icture, for an oerator Âx, t, the ground-state exectation value at time t is given by G U tâx, tut G 2.43 where Ut = T e i t dt H extt and Âx, t is time evolved using the unerturbed Hamiltonian i.e. without H ext. On exanding the above exectation value to first order in H ext, we find that the change in the exectation value of  is given by δ Âx, t i t dt G [H ext t, Âx, t G. 2.45

11 11 In our case, the erturbation is H ext = n Bn Sn = n B + ns + n + B ns n + S z nb z n 2.46 where B x = B + + B ; B y = ib + B So, δ Âx, t = i t dt [B z n, t G [Ŝzn, t, Âx, t G n + B + n, t G [Ŝ+ n, t, Âx, t G +B n, t G [Ŝ n, t, Âx, t G ow, let us take Ân, t = Ŝ3n, t. In the sin-wave aroximation, for n and n both even, G Ŝ+ n, tŝ3n, t G 2S S G ân e iĥt t G G ân e iĥt t â nân G It is immediate to see that this exression consists of exectation values of an odd number of ĉ and ˆd oerators and so must vanish. The same holds for n and n both odd or with one even and the other odd, as well as with Ŝ+ relaced with Ŝ. Hence, So, δ Ŝ3n, t = i G [Ŝ+ n, t, Ŝ3n, t G = G [Ŝ n, t, Ŝ3n, t G = = i t dt n B z n, t G [Ŝzn, t, Ŝzn, t G 2.51 dt n B z n, t G [Ŝzn, t, Ŝzn, t G θt t 2.52 = B z n, t χ 33 nt, n t 2.53 n where we have defined the suscetibility χ 33 nt, n t = i G [Ŝzn, t, Ŝzn, t G θt t DR 33nt, n t. 2.55

12 12 Here D R 33nt, n t is a retarded roagator. Let us now take Ân, t = Ŝ+ n, t. By logic similar to that above, we see that in the sin-wave aroximation, Hence, δ Ŝ+n, t = i = i t G [Ŝ+ n, t, Ŝ+n, t G = dt n B n, t G [Ŝ n, t, Ŝ+ n, t G 2.57 dt n B n, t G [Ŝ n, t, Ŝ+ n, t G θt t 2.58 = B n, t χ + nt, n t 2.59 n where we have defined the suscetibility χ + nt, n t = i G [Ŝ+ n, t, Ŝ n, t G θt t 2.60 Here D R + nt, n t is a retarded roagator. 1 DR + nt, n t ote that one must handle the cases of even and odd sites searately for the suscetibilities as well. Adding together the even-even, even-odd, and odd-odd roagators in momentum sace does not yield a hysically meaningful uantity. 3. We have already comuted the time-ordered correlators in the sin-wave aroximation in art We can deduce D R + from D + by changing the integration contour, which amounts to changing the sign of one of the iɛ s. Looking first at the even n even n roagator, we have that cosh2θ = sin 1 and ε = 2JS sin χ +, ω = 1 DR +, ω 2.62 = 2S [ cosh 2 θ ω + ε + iɛ sinh2 θ 2.63 ω ε + iɛ = 2S [ ω + iɛ ε cosh 2θ 2.64 ω + iɛ 2 ε 2 = 2S [ ω + iɛ 2JS ω + iɛ 2 2JS 2 sin

13 13 where we used some hyerbolic function identities in the third line. If we were to set ω = 0, it is clear that we would get a ole at = π. However, this would be a ole of order two and so would have vanishing residue. In order to extract a finite, non-zero uantity, we must kee ω finite which leads to two first-order oles near = π. Indeed, Taylor exanding the denominator about = π, we find χ +, ω = 2S [ ω + iɛ 2JS ω + iɛ 2 2JS 2 π 2 = 2S [ 1 ω + iɛ 2JS 2JS 2 ω + iɛ 2 /2JS 2 π 2 = 2S [ 1 ω + iɛ 2JS 2JS 2 π ω + iɛ/2js π + ω + iɛ/2js which has oles at = π ± ω + iɛ/2js. Focusing on one of these oles, we find the residue lim Res =π ω+iɛ/2jsωχ +, ω = lim ω 2S ω 0 ω 0 ote that added in a factor of ω to make the limit convergent. 1 ω 2JS 2JS 2 ω/js = S The odd n odd n suscetibility is given by χ +, ω = 2S [ sinh 2 θ ω + ε + iɛ cosh2 θ 2.70 ω + ε + iɛ = 2S [ ω + iɛ 2JS ω + iɛ 2 2JS 2 sin So, by similar maniulations to those above, we find lim Res =π ω+iɛ/2jsωχ +, ω = lim ω 2S ω 0 ω 0 1 ω 2JS 2JS 2 ω/js = S This exression also has a second order ole at = π with residue zero. Lastly, for one of n, n odd and the other even, the suscetibility is given by χ +, ω = S [ sinh 2θ 1 ω + ε + iɛ ω ε + iɛ = S [ sinh 2θ 2ɛ 2.74 ω + iɛ 2 ε 2 = S 4JS 2.75 ω + iɛ 2 ε 2

14 14 where we used the fact that sinh 2θ = cosh 2θ tanh 2θ = cos / sin. Exanding about = π, χ +, ω = S 4JS ω + iɛ 2 2JS 2 sin = S 1 4JS 2JS 2 π ω + iɛ/2js π + ω + iɛ/2js This has the same oles as the other roagators. Again, focusing on one of them, we find lim Res =π ω+iɛ/2jsωχ +, ω = S ω SPECTRAL FUCTIO FOR THE DIRAC PROPAGATOR 1. By definition, we have that 0 T ψ α x ψ β y 0 = θt 0 ψ α x ψ β y 0 θ t 0 ψ β yψ α x where t x 0 y 0. Let us consider the first of the two oint functions i.e. take x 0 > y 0. Inserting a comlete basis of states with definite momenta, we find 0 ψ α x ψ β y 0 = 0 ψ α x n n ψ 1 d 3 β y 0 2E 2π n Using the fact that P µ is the generator of translations and choosing the n to be eigenstates of momentum, we find that 0 T ψ α x ψ β y 0 = n 0 ψ α 0 n n ψ β 0 0 e i x x 1 2E = A αβ e i x x d 3 2π 3 n d 3 2π where we have set A αβ = 0 ψ α 0 n n ψ β 0 0. We can rewrite this as a contour integral over 0 : 0 ψ α x ψ A αβ β y 0 = i e i x x d 4 2 m 2 n + iɛ 2π n Finally, on inserting 1 = dµ 2 δµ 2 m 2 0 n, we obtain 0 ψ α x ψ d 4 β y 0 = i dµ 2 n δµ2 m 2 na αβ e i x x, x 2π 4 2 µ 2 0 > y 0 + iɛ 0 3.5

15 15 If we reeat the same analysis for the y 0 > x 0 term, 0 ψ β yψ α x 0, it would seem that we would get a different set of matrix elements, 0 ψ β 0 n n ψ α 0 0 and hence a different sectral function. However, if one were to reeat the same calculation for the anti-commutator, instead of the time-ordered roagator, one would find that it vanishes at sace-like searations only if these two sectral densities are the same. So, it must be the case that the y 0 > x 0 term, 0 ψ β yψ α x 0, rovides the same contribution as the the x 0 > y 0 term. 1 Hence, we can read off that in momentum sace, S F αα = i 0 dµ 2 ρ αα 2 µ 2 + iɛ 3.6 where ρ αα = n δµ 2 m 2 n 0 ψ α 0 n n ψ β The sectral function, ρ, is a 4 4 matrix and must be Lorentz invariant. Moreover, it can only be a function of the four-momentum and must have the same transformation roerties as the Dirac fields under boosts. Given these constraints, it follows that ρ 2 must be a linear combination of γ matrices and the identity of the following form ρ = ρ 1 2 / + ρ ρ 1 2 γ 5/ + ρ 2 2 γ ow, we recall that under a arity transformation, ψx 0, x γ 0 ψx 0, x and ψx 0, x ψx 0, xγ 0. So, the sectral function will transform as P ρp 1 = γ 0 ρ γ If the vacuum is invariant under arity, then the sectral function must be as well. Insection of the most general, Lorentz invariant form given above reveals that the only non-invariant terms are those containing γ 5. So, we obtain ρ = ρ 1 2 / + ρ 2 2 1, This may be interreted as being a conseuence of CP T invariance. See, for instance, the discussion in Weinberg, Vol. I on the Lehmann reresentation.

16 16 as reuired. ow, we have that S F = i But we know that in the free Dirac theory, So, we can read off 0 dm ρ 1m 2 / + ρ 2 2 m m 2 + iɛ / + m1 S F = i 2 m 2 + iɛ ρ 1 µ 2 = δµ 2 m 2, ρ 2 µ 2 = mδµ 2 m WICK S THEOREM 1. Due to the global O3 symmetry and the fact that the theory is free, the ground state is invariant under O3 rotations and so any n-oint function which is not O3 invariant must necessarily vanish. Consider the following correlation functions: a 0 T φ a xφ a x b 0 T φ a xφ b x φ b x c 0 T φ a xφ a x φ b x φ b x It is immediate to see that a and c are invariant under O3 rotations since all the flavor indices are contracted. Exlicitly, 0 T φ a xφ a x 0 0 T O ab φ b xo ad φ d x 0 = 0 T φ a xφ a x where we used the fact that for O O3, O ab O ad = Oba T O ad = δ bd. In contrast, b is not invariant under O3 rotations since it has one free index; indeed, it transforms as a vector: 0 T φ a xφ b x φ b x 0 0 T O ac φ c xo bd φ d x O be φ e x = O ac 0 T φ c xφ b x φ b x So, if b were to be non-zero, this would mean there is a referred direction in flavorsace, contradicting the fact that the ground state is invariant under O3 rotations. Hence b must vanish.

17 17 2. Since b vanishes, we only need to look at c. Using Wick s theorem, we obtain 0 T φ a xφ a x φ b x φ b x 0 = 0 T φ a xφ b x 0 0 T φ b x φ a x T φ a xφ b x 0 0 T φ b x φ a x Due to the global O3 symmetry, we have that + 0 T φ a xφ a x 0 0 T φ b x φ b x 0. 0 T φ 1 xφ 1 x 0 = 0 T φ 2 xφ 2 x 0 = 0 T φ 3 xφ 3 x 0 = T φ axφ a x Conversely, terms of the form 0 T φ a xφ b x 0, a b must vanish as they are not invariant under O3 rotations. We can thus write 0 T φ a xφ a x φ b x φ b x 0 = T φ axφ a x 0 0 T φ a x φ a x T φ axφ a x 0 0 T φ a x φ a x T φ a xφ a x 0 0 T φ a x φ a x 0 5. REDUCTIO FORMULAS First, we must construct the in state creation and annihilation oerators. A free, comlex scalar field has the mode exansion d 3 k φx = ake ik x + b ke ik x 5.1 2π 3 2ωk where k µ = ωk, k. We also have Π φ x = 0 φ d 3 k x = 2π 3 2ωk iωk a ke ik x + bke ik x. 5.2 In order to solve for ak and bk, we comute d 3 xe ik x φx = 1 2ωk ak + 1 2ωk b ke2iωkx d 3 xe ik x 0 φx = i 2 ak b ke2iωkx We can invert these exressions to find ak = d 3 x[ωkφx + i 0 φxe ik x. A similar exression for bk with φ φ can be found by considering the Fourier transforms of φ

18 18 and 0 φ instead. So we have a in k = i b in k = i d 3 xe ik x 0 φ in x 5.5 d 3 xe ik x 0 φ in x, 5.6 where we have used the notation f g = f g fg. As for the gauge field, we recall that in the A 0 = 0 gauge the transverse comonents have the mode exansion d 3 k Ax = εk[a 2π 3 α ke ik x + a 2 k αke ik x 5.7 α=1,2 where ε α are the olarization vectors. In analogy with the comutation for a real scalar field and making use of the orthonormality of the olarization vectors, ε α ε β = δ αβ, we find for the gauge field a α,in k = i e ik x 0 A µ in xε α µ kd 3 x. 5.8 Here we have defined ε α 0 = 0 to make use of four-vector notation. We now roceed to the comutation of the S-matrix element: +, ; out i, α; in = + S i, α 5.9 = +, ; out a α,in i 0; in 5.10 = ilim t Z 1/2 A d 3 xe i i x 0 +, ; out A µ x 0; in ε α µ i t 5.11 where Z A is the wavefunction renormalization for A µ which takes into account the fact that A µ,in x creates single article states whereas A µ x also creates multi-article states. Following the stes in the notes, we recall that for some function F x lim lim d 3 xf x = d 4 x 0 F x, 5.12 t f t i and so taking F x = e i i x 0 +, ; out A µ x 0; in ε α µ i, we can write +, ; out i, α; in = + ; out a α,out i 0; in + i d 4 x Z 1/2 0 [e i i x. 0 +, ; out A µ x 0; in ε α µ i A 5.13

19 19 where we have used the fact that 0 A µ x 1 = Z 1/2 A 0 A µ,inx 1 = Z 1/2 A 0 A µ,outx The first term clearly vanishes since it is an overla of states with different article secies. To simlify the second term, we note that the lane wave e i x must satisfy the gauge field EOM: 2 e i i x = So, we have that d 4 x 0 [e i i x 0 +, ; out A µ x 0; in ε α µ i 5.16 = d 4 x [ e ii x 0 2 +, ; out A µ x 0; in ε α µ i 0e 2 ii x +, ; out A µ x 0; in ε α µ i = = 5.17 d 4 x [ e ii x 0 2 +, ; out A µ x 0; in ε α µ i 2 e ii x +, ; out A µ x 0; in ε α µ i 5.18 d 4 x [ e ii x 2 +, ; out A µ x 0; in ε α µ i 5.19 where in the enultimate line we used the above EOM and in the final line we integrated by arts. Reeating the same calculation for the outgoing ± comlex scalar states, we obtain +, ; out i, α; in = i3 d 4 y Z φ Z 1/2 + d 4 y d 4 x [ e i i x+ + y + + y 5.20 A 2 y + + m 2 2 y + m 2 2 x 0; out T φy + φ y A µ x 0; in ε α µ i ote that the correlator is time-ordered because the in and out states are already arranged in a time-ordered fashion to begin with.