Problem 1(a): Since the Hamiltonian (1) is a function of the particles momentum, the evolution operator has a simple form in momentum space,

Transcription

1 PHY 396 K. Solutions for problem set #5. Problem 1a: Since the Hamiltonian 1 is a function of the particles momentum, the evolution operator has a simple form in momentum space, exp iĥt d 3 2π 3 e itω S.1 where ω + M In the coordinate basis, the definite-momentum states have wave functions x e ix, therefore Ux y; t x exp iĥt x d 3 exp ix y itω. 2π 3 S.2 To simplify this 3D integral, we use spherical coordinates, θφ where, θ is the angle between and x y, d 3 d 2 d cos θ dφ, x y r cos θ, S.3 and d cos θ dφ e ix 4π sinr r 2π e +ir e ir. ir S.4 Consequently, Ux y; t 1 4π 2 i r 1 4π 2 i r 0 d e itω e +ir e ir d e +ir itω e ir itω π 2 d e +ir itω. i r S.5 Q.E.D. 1

2 Problem 1b: As explained in the mathematical notes lined to the homewor page, integrals of the form I dz fz e Agz Γ S.6 in the large A limit evaluate to I e Agz0 πη fz0 η 2 Ag z OA 1. S.7 Generally, f and g are complex analytic functions of a complex variable z which is integrated some contour Γ; quite often Γ is the real axis, but one should allow for its deformation in the complex plane. In eq. S.7, z 0 is a saddle point of gz where the derivative g z 0 0; if several saddle points are present near the contour Γ, the point with the largest Rg dominates the integral. Note that the g 0 does not have to lie on the original integration contour Γ; if it does not, we should deform Γ Γ so that Γ does go through the saddle point z 0. Finally, η is the direction dz of the Γ at the saddle point z 0 ; it should be chosen such that R η 2 g z 0, which assures that Γ crosses z 0 as a mountain path, from a valley to the lowest crossing point to another valley. For the integral 3 at hand, we identify A t, g i r t iω, f 4π 2 i r. S.8 The saddle point in the plane is where dg d ir t idω d ir t i ω 0, S.9 which for r < t happens at real 0 M r t 2 r, ω t 0 M 2 t 2 r. 2 S.10 2

3 At this point, Ag 0 ir 0 itω 0 im t 2 2, f 0 im 1 4π 2 t 2 r, 2 Ag 0 it M 2 ω 3 0 t2 r 2 3/2 imt 2, S.11 and the direction of the integration contour at 0 should be in the fourth quadrant of the complex plane, argη between 0 and π/2; the real-axis contour is marginally OK. Substituting all these data into eq. S.7 gives us πη fz0 η 2 Ag z 0 im3/2 t 4π 3/2 t 2 r 2 5/4 S.12 and therefore Ux y; t exp im t 2 r 2 im3/2 4π 3/2 Q.E.D. t t 2 r 2 5/4 1 + O 1 M t 2 r 2. 4 Problem 1c: Again we use the saddle point method to evaluate the integral 3, identify A, g, and f according to eq. S.8, and solve eq. S.9 to find the saddle point. But for r > t, the saddle point is imaginary 0 imr r 2 t 2, ω 0 imt r 2 t 2, S.13 and the integration contour must be deformed away from the real axis. At the saddle point S.13, Ag 0 ir 0 itω 0 M r 2 t 2, f 0 M 4π 2 1 r 2 t 2, Ag 0 it M 2 ω 3 0 r2 t 2 3/2 Mt 2, S.14 all real, and the deformed contour should cross 0 in the imaginary direction, argη π 2 ± π 4. 3

4 Consequently, in eq. S.7 πη fz0 η 2 Ag z 0 +im 3/2 t 4π 3/2 r 2 t 2, S.15 5/4 and therefore Ux y; t exp M r 2 t 2 im 3/2 4π 3/2 t r 2 r t 5/4 1 + O 1 M r 2 t 2. 5 The exponential factor here rapidly decays outside the light cone, which maes the probability of superluminal motion of relativistic particle very small. But tiny as it is, it does not vanish, which violates the relativistic causality. Problem 2: First, let s clarify eqs. 6. Fourier-transforming the EM fields Êx and ˆBx into Ê and ˆB, and then decomposing the latter into vector components according to the helicity basis e,, we get eqs. 6, while the reverse transform is given by Ê, d 3 e ix e, Êx, ˆB, d 3 e ix e, ˆBx. S.16 In general, the in eqs. 6 and S.16 should include all 3 polarizations, but the transversality Ê ˆB 0 translates to Ê ˆB 0, and hence in the helicity basis Ê, ˆB, 0 for 0. So from now on, we shall limit the polarization sums in both eqs. 6 and S.16 to ±1 only. Also, the EM fields Êx and ˆBx are Hermitian, but the mode operators Ê, and ˆB, are not. Instead, conjugating eqs. S.16 and taing the phase conditions 9 into account gives us Ê, Ê,+ and ˆB, ˆB,+. S.17 4

5 Problem 2a: According to eqs. 11 of homewor set#2, at equal times [Êi Êjy] [ x, ˆBi x, ˆB j y] 0, S.18 [Êi x, ˆB j y] iɛ ijl δ 3 x y. S.19 x l The first line here immediately implies [Ê,, Ê, ] [ ˆB,, ˆB, ] 0, S.20 but the mixed commutators tae more wor: [Ê,, ˆB ], d 3 x d 3 y e +ix i y e, i e, [Êi j x, ˆB ] j y d 3 x d 3 y e +ix i y e, i e, j iɛ ijl δ 3 x y x l pulling constants outside and integrating by parts +iɛ ijl e, i e, j d 3 x d 3 y δ 3 x y e +ix i y x l +iɛ ijl e, i e, j i l 2π 3 δ 3 2π 3 δ 3 e, e, S.21 i δ, 2π 3 δ 3, where the last equality follows from eqs. 7 and 8. Equivalently, [Ê,, ˆB ], i δ, 2π 3 δ 3 + or [Ê,, ˆB ], i δ, 2π 3 δ 3. S.22 As to the Hamiltonian of the free EM fields, in eq. 13 of homewor#2 we had Ĥ d 3 x 12 Ê 2 x + 1 ˆB 2 2 x. S.23 Fourier transforming the fields and expanding in components according to eqs. 6 leads to d d 3 3 d 3 x Ê2 x Ê 2π 3 Ê 2π 3 Ê,Ê,, S.24 and liewise for the magnetic fields. Consequently, the Hamiltonian S.23 taes form 10. 5

6 Problem 2b: The commutation relations of the operators 11 follow from eqs. S.20 and S.22. In particular, [â,, â, ] [ ] i ˆB,, Ê, + i [ Ê,, ˆB, ] i +i δ, 2π 3 δ i i δ, 2π 3 δ 3 + 2π 3 δ 3 + δ, 0 because when + 0, and liewise [â,, â, ] S.25 0 S.26 Finally, [â,, â, ] [ ] i ˆB,, Ê, + i [ Ê,, ˆB, ] i +i δ, 2π 3 δ 3 + i i δ, 2π 3 δ 3 2π 3 δ 3 + δ, + 2ω 2π 3 δ 3 + δ, S.27 where ω is the photon s energy; the last equality here follows from 2 +1 because the photons have the ±1 polarizations only. As to the Hamiltonian, reversing eqs. 11 and taing the Hermiticity relations S.17 into account, we have Ê, i 2 â, + i 2â,, ˆB, 2 â, + 2 â,. S.28 Consequently, Ê,Ê, + ˆB, ˆB, 1 4 â, â, â, â, â, + â, â, + â, 1 2â,â, + 1 2â,â, 1 2â,â, + 1 2â,â, + const, S.29 6

7 and therefore Ĥ d π 3 Ê,Ê, + 1 ˆB ˆB 2,, 1 4â,â, + 1 4â,â, + const d 3 2π 3 d 3 1 2π 3 2ω 1 2 â,â, + zero point energy ω â,â, + zero point energy. 12 Q.E.D. Problem 2c: Let us start with the Schrödinger-picture fields. Combining eqs. 6 and S.28, we have Êx d 3 2π 3 e ix e l i 2 â, + 2â i, d 3 1 S.30 2π 3 iω e +ix e 2ω, â, + iω e ix e, â, where the second equality follows from reversing the sign of in the second term in the and using the phase convention 9. Similarly, for the magnetic field, we have ˆBx d 3 1 2π 3 ω e +ix e 2ω, â, + ω e ix e, â,. S.31 In the Heisenberg picture, the EM fields are time-dependent according to Êx, t ˆBx, t d 3 1 iω 2π 3 e +ix e 2ω, â, t + iω e ix e, â, t, d 3 1 ω 2π 3 e +ix e 2ω, â, t + ω e ix e, â, t, S.32 where the creation and annihilation operators depend on time according to â, t e +iĥt â S, e iĥt e iω t â S,, â, t e+iĥt â S, e iĥt e +iω t â S, ; S.33 7

8 the second equality on each line here follows from [ ] â,, Ĥ +ω â,, [ ] â,, Ĥ ω â,. S.34 Combining eqs. S.32 and S.33 and switching to relativistic notations for the x µ and µ, we arrive at Êx ˆBx d 3 1 Schroedinger â,â 2π 3 iω e ix e 2ω, â, + iω e +ix e, â,, S ω d 3 1 Schroedinger â,â 2π 3 ω e ix e 2ω, â, + ω e +ix e, â,. S ω Finally, to mae our notations fully relativistic, we combine Êx and ˆBx into ˆF µν x. Accordingly, eqs. S.35 and S.36 combine into eq. 13 where f i0, f 0i, iω e, i and f ij, ω ɛ ijl e, l ɛ ijl i e, l. S.37 Problem 2d: Let us start with the Coulomb gauge Ax, t 0 and consequently 2 A 0 x, t ρx, t, which means that A 0 x, t is the instantaneous Coulomb potential due to the electric charge density ρy, t, hence the name Coulomb gauge. For the free EM fields i.e., no charges or currents, A 0 x, t 0. Applying this gauge to the quantum field Schrödinger picture Âx and expanding into modes, we have in the Âx d 3 2π 3 Â,, ±1 modes only, S.38 and thans to eq. 7 Â, 1 ˆB, 1 â 2, + â,. S.39 8

9 Consequently, similar to the free Ê and ˆB fields, Âx d 3 1 e +ix 2π 3 e 2ω, â, + e ix e, â,. S.40 In the Heisenberg picture, the time-dependent field becomes Âx d 3 2π 3 1 2ω ±1 Schroedinger â,â e ix e, â, + e +ix e, â,, S ω and together with Â0 x 0, this formula can be written in relativistic form as eq. 14 where e µ, 0, e,. In other gauges, we have  µ x µ Coulomb x + µ ˆΛx S.42 for some quantum field ˆΛx. In principle, any such field generates a valid gauge transform, even if ˆΛx involves some non-photonic degrees of freedom. But the problem at hand assumes no such extra-photonic degrees of freedom, and also a local linear gauge condition such as some differential operatorâµ 0. Consequently, ˆΛx has to be linearly comprised of the same plane waves and operators as the ˆF µν x fields, namely e ix â, and e +ix â, for 0 +ω. Thus, ˆΛx d 3 2π 3 1 2ω ±1 ic, e ix â, ic, e +ix â, 0 +ω S.43 for some coefficients C, ; their actual values depend on a particular gauge. Substituting eq. S.43into S.42 and combining with eq. S.41 gives us eq. 14 with e µ, 0, e, + C, µ. Q.E.D. 9

10 Problem 3a: The left hand side of eq. 15 is separately antisymmetric in µ ν and in α β, and its easy to see that the right hand side has the same antisymmetries. This gives 36 distinct choices of indices µ, ν, α, β, and when we arrange them according to time versus 3D space, we end up with four 3D equations to chec: ±1 ±1 ±1 f, i0 f, j0 ω 2 δ ij i j, S.44 f, il f jm, and the complex conjugate of S.46. i j δ lm + l m δ ij i m δ lj l j δ im ɛ iln ɛ jmp ω 2 δnp n p, S.45 f, i0 f, jm ω m δ ij ω j δ im, S.46 The F µν, are spelled out in eq. S.37 in terms of the transverse e, vectors. Those vectors form a complete basis of the transverse plane, hence i e, e j, i e, e j, i e,0 e,0 δ ij ±1 0,±1 i j 2. S.47 Consequently, ±1 f, i0 f j0, ω 2 i e, e j, ω 2 δij i j, S.48 ±1 which verifies eq. S.44 note ω 2 2. Liewise, ±1 f, il f jm, ω 2 ɛ iln ɛ jmp ±1 which verifies eq. S.45. Finally, using e, n e, p ɛ iln ɛ jmp ω 2 δnp n p, S.49 f jm, ɛ jmn i e, n iɛ jmn ɛ npq p e, q i j e, m + i m e, j, S.50 10

11 we obtain ±1 f, i0 f jm, +ω m ω m δ ij ±1 i j ω 2 e, i e, j j m ω m δ ij ω j δ im, j m S.51 which verifies eq. S.46. Q.E.D. Problem 3b: Choose any gauge where µ x is given by eq. 14 for some e µ,. Since ˆF µν µ  ν ν  µ should agree with eq. 13, the e µ, must satisfy i µ e ν, iν e µ, f µν,. S.52 Now, let Π µα ±1 e µ, e α, S.53 for that gauge. Applying eq. S.52 twice, we get ±1 f µν, f, αβ µ α Π νβ + ν β Π µα µ β Π να ν α Π µβ, S.54 and comparing this formula to eq. 15 tells us that µ α Π νβ + g νβ α β µ ν 0. S.55 Now, in 3D we now that if cross products of some two vectors is zero, then the vectors are parallel. Generalizing this rule to Lorentz vectors and hence to tensors, we find that if a nonzero vector µ and a tensor T ν α,...,β satisfy µ T ν α,...,β ν T µ α,...,β 0, then T ν α,...,β ν Q α,...,β 11

12 for some tensor Q α,...,β with one less index then T. Applying this generalized rule to eq. S.55, we immediately see that α Π νβ + g νβ β Π να + g να ν Q αβ S.56 for some antisymmetric tensor Q αβ. Moreover, multiplying both sides of eq. S.56 by some vector R ν such that R ν ν 0, we find Q αβ α q β β q α where g α R ν Π να + g να R ν ν. S.57 Substituting this formula bac into eq. S.56 and regrouping terms produces α Π νβ + g νβ ν q β α β 0, S.58 which in turn implies Π νβ + g νβ ν q β β q ν S.59 for some vector q ν. In other words, Π νβ g µν + ν q β + β q ν. S.60 Finally, the definition S.53 implies that Π νβ is an Hermitian matrix, Π νβ Π βν. Consequently, the vectors q ν and q ν in eq. S.60 must be complex conjugates of each other, q ν q ν, which reduces eq. S.60 to eq. 16. Q.E.D. Problem 3b, alternative solution: Let us start with the Coulomb gauge where e µ, 0, e,. In this gauge, eq. S.47 gives Π µα Coulomb δ µα µ α ω 2 0 otherwise, for µ, α 1, 2, 3, S.61 12

13 where Π µα is defined as in eq. S.53. Now let us define q c µ 1 2ω 2 ω,. S.62 A bit of algebra tells us that δ µα µ α g µα + µ qc α + q c µ α ω 2 for µ, α 1, 2, 3, 0 otherwise; S.63 comparing this formula to eq. S.62 verifies eq. 16 for the Coulomb gauge note q µ c is real. Now consider any other gauge consistent with eq. 14. In any such gauge, for some coefficients C,. Consequently, ±1 where e µ, Now let e µ, eµ, Coulomb + C, µ S.64 e α, e µ, e α Coulomb, Coulomb ±1 + µ C, e α, q µ ±1 + µ α ±1 C, 2 Coulomb + α ±1 C, e µ, Coulomb g µα + µ q α c + q µ c α + µ q α + α q µ + µ ν N ±1 C, e µ, Coulomb and N ±1 C, 2. S.65 S.66 q µ q µ c + q µ N µ. S.67 Using this q µ, we can rewrite the last line of eq. S.65 as g µα + µ q α + q α µ, S.68 which proves eq. 16 for the gauge at hand. Q.E.D. 13

14 Problem 3c: Let us expand the µ x and Âν y operators according to eq. 14. The field product comprises ââ, â â, â â, and â â terms, of which the first three types have zero vacuum expectation values while the fourth type has Consequently, 0 µ xâν y 0 0 â, â, 0 2ω 2π 3 δ 3 δ,. S.69 d 3 1 2π 3 2ω d 3 2π 3 1 d 3 1 2π 3 2ω e ix y 2ω e µ, e ν, d 3 1 2π 3 2ω e ix y g µα + µ q α + q α µ e ix e +i y e µ, e ν, 0 â, â, 0 where 0 +ω and q µ depends on and on the gauge as in part b. Problem 3d: Let us start with the un-modified time-ordered product of the two µ fields: Then Q.E.D. S.70 0 Tµ xâα y 0 { 0  µ xâα y 0 D µα x y for x 0 > y 0, 0 Âα yâµ x 0 D αµ y x for x 0 < y 0, S.71 where D µα x y is the right hand side of eq. 17. Now consider the right hand side of eq. 18; let s denote it G µα x y. Separating space and time integrals, we have G µα x, t d 3 d0 2π 3 eix 2π ie i0t 2 0 ω2 + iɛ gµα + µ q α + q α µ. S.72 Note that to define this integral we must analytically continue the q α functions to the offshell momenta with 0 ±ω. Let us assume for the moment that this continuation does not introduce any un-physical poles or other singularities into the integrand of S.72. In particular, in the d 0 integral, the only singularities are the poles at 0 ±ω iɛ and the asymptotic behavior of the integral for 0 ±i. 14

15 Consequently, we may evaluate the d 0 integral exactly as in class. For t > 0, we close the integration contour with a large-radius semicircle in the lower half-plane. Inside this contour, there is a pole at 0 +ω iɛ, hence d0 2π ie i0t [ 0 2 ω2 + iɛ gµα + µ q α + q µ α e i0t ] Residue 0+ω 0 2 ω2 + iɛ gµα + µ q α + q µ α g µα + µ q α + q µ α 2ω e iω t q +ω, S.73 where the bottom line is precisely the integrand of the d 3 integral in eq. 17. Consequently, integrating over d 3 in eq. S.72 yields G µα x D µα x for x 0 > 0. S.74 For x 0 < 0, we close the contour in the upper half of the complex 0 plane. The pole inside that contour is at 0 ω + iɛ, hence d0 2π ie i0t [ 0 2 ω2 + iɛ gµα + µ q α + q µ α e i0t ] Residue 0 ω 0 2 ω2 + iɛ gµα + µ q α + q µ α and consequently g µα + µ q α + q µ α 2ω e+iω t q ω, S.75 G µα x D µα x 0, +x D αµ x for x 0 < 0. S.76 Altogether, G µα x y { } D µα x y for x 0 > y 0 0 D αµ y x for x 0 < y 0 Tµ xâα y 0. S.77 The above analysis is over-simplified and overloos potential troubles. For starters, we haven t considered the x 0 y 0 case, which may hide a δ 4 x y singularity; if it s there, 15

16 we would need to modify the time-ordered product at x y to cancel it. But more importantly, we assume that q µ 0, has no singularities, and that can t be true unless q µ is a constant 4 vector. Otherwise, it would either have unphysical poles or worse singularities at some finite complex 0, or else it would blow up for 0 and obstruct closing of the integration contour. To see what s really going on with the EM Feynman propagator, let us switch from the gauge-dependent µ fields to the gauge invariant ˆF µν. Let us define D µν,αβ def 0 ˆF µν x ˆF αβ y 0. S.78 Then proceeding exactly as in part c of this problem, we evaluate D µν,αβ x y where d 3 2π 3 1 2ω d 3 2π 3 1 d 3 1 2π 3 2ω e ix y 2ω f µν, e αβ, d 3 1 [ 2π 3 e ix y Π ] 0 µν,αβ +ω, 2ω e ix e +i y f µν, e αβ, 0 â, â, 0 S.79 Π µν,αβ µ α g νβ ν β g µα + µ β g να ν α g µβ, S.80 cf. eq. 15. Note that Π µν,αβ is a quadratic polynomial in, which Fourier transforms into a second-order differential operator. Hence, D µν,αβ x y Π µν,αβ i d 3 1 2π 3 2ω e ix y 0 +ω Π µν,αβ i Dx y, S.81 where Dx y was defined in class in the scalar context. Although in the present case we should set M 2 0. Now let s put the ˆF µν x and ˆF αβ y into time order. Eq. S.81 gives us 0 T ˆF µν x ˆF αβ y 0 { Π µν,αβ i Dx y for x 0 > y 0, Π αβ,µν i Dy x for x 0 < y 0, S.82 16

17 and since Π αβ,µν Π µν,αβ, it loos lie the right hand side can be simplifies as { } Dx y for x Π µν,αβ 0 > y 0 i Π µν,αβ i G Dy x for x 0 < y 0 F x y where G F x y d 4 ie ix y 2π iɛ { Dx y for x 0 > y 0 Dy x for x 0 < y 0 S.83 is the Feynman propagator of the massless scalar field. But we saw in class that time derivatives do not commute with time ordering, and consequently { } 2 0G 2 0 Dx y for x 0 > y 0 F x y 0 2Dy x for iδ 4 x y. S.84 x0 < y 0 Therefore, for the EM fields we have 0 T ˆF µν x ˆF αβ y 0 Π µν,αβ i G F x y + iδ 4 x y δ µ0 δ α0 g νβ α β µ ν. S.85 To remedy the delta-singularity on the right hand side, we modify the time product of the EM fields according to T ˆF µν x ˆF αβ y T ˆF µν x ˆF αβ y iδ 4 x y Then the Feynman propagator for the ˆF µν fields becomes δ µ0 δ α0 g νβ α β µ ν. S.86 G µν,αβ F x y def 0 T ˆF µν x ˆF αβ y 0 Π µν,αβ i G F x y d 4 ie ix y Π µν,αβ 2π iɛ S.87 What we need is to translate this formula into the propagator for the Âν fields, G µα def F x y 0 T Â µ xâν y 0 S.88 17

18 for a suitably modified time-ordered product such that µ x α y G νβ F x y α β µ ν G µν,αβ F x y. S.89 Fourier transforming both propagators in all four dimensions converts this condition into µ α Gνβ F α β µ ν µν,αβ G F iπµν,αβ 2. S.90 + iɛ Proceeding exactly as in the first solution to part b of this problem, we solve this equation as G νβ F i gµα + µ q α + q µ α 2. S.91 + iɛ Finally, transforming this propagator bac to coordinate space, we arrive at eq. 18. The only question we haven t answered here is how exactly should we modify the timeordered product of the µ fields to mae eq. S.89 wor. Alas, eq. 20 is wrong, mea culpa. Itzyson and Zuber give a similar formula T  µ xâα y Tµ xâα y + i M 2 δµ0 δ α0 δ 4 x y S.92 for the massive µ x cf. homewor#1, problem 1, but it does not apply to the massless EM fields. Since these solutions are already waaay too long, I decided to sip any details of the EM modification of the T  µ xâα y. Suffices to say that it depends on the gauge and it s generally rather messy. Finally, I would lie to mae a comment about gauge fixing in QED Quantum ElectroDymamics, i.e. theory of the quantum EM fields coupled to quantum electron fields. For practical purposes, one usually fixes a gauge in terms of the photon propagator 18 rather than in terms of an explicit gauge condition on the µ x fields. In particular, the most commonly used gauge is the Feynman gauge where q µ is set to zero by fiat, even though this does not correspond to any condition on the µ x at all. But whenever one calculates some gauge-invariant properties, the actual gauge condition does not matter, and it s easier to use a simple propagator where G µα F d 4 ig µα e ix y 2π iɛ g µα G scalar F x y