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The Quantum EM Fields and the Photon Propagator

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The Quantum EM Fields and the Photon Propagator The Quantum EM Fields and the Photon Propagator Quantizing the free electromagnetic tension fields E and B is fairly straightforward. The time-independent Maxwell equations Bˆ (x) = 0, Eˆ(x) = Jˆ0(x) 0 (forthefreefields) (1) ∇· ∇· → are imposed as operatorial constraints in the Hilbert space, while the time-dependent Maxwell equations ∂ ∂ Bˆ = Eˆ, Eˆ = + Bˆ (2) ∂t −∇ × ∂t ∇ × follow from the free Hamiltonian ˆ 3 1 ˆ 2 1 ˆ 2 H = d x 2 E (x) + 2 B (x) (3) Z and the equal-time commutation relations Eˆi(x, t), Eˆj(y, t) = 0, i j Bˆ (x, t), Bˆ (y, t) = 0, (4) ∂ Eˆi(x, t), Bˆj(y, t) = iǫijk δ(3)(x y). − ∂xk − In light of eqs. (1), we may expand the vector fields into momentum/polarization modes using only the transverse polarizations, for example the helicity modes λ = 1 only, thus ± k ek,λ = iλ k ek,λ, ek,λ k for λ = 1, × | | ⊥ ± 3 d k ixk † Eˆ(x) = e ek Eˆk , Eˆ = Eˆ , (2π)3 ,λ × ,λ k,λ − −k,λ Z λ=±X1 only 3 d k ixk † Bˆ (x) = e ek Bˆk , Bˆ = Bˆ , (2π)3 ,λ × ,λ k,λ − −k,λ Z λ=±X1 only 3 (3) ′ Eˆ k , Eˆ k′ ′ = 0, Bˆ k , Bˆ k′ ′ = 0, Eˆ k , Bˆ k′ ′ = iλ k δ ′ (2π) δ (k + k ). ,λ ,λ ,λ ,λ ,λ ,λ | | λλ × (5) Consequently, the photonic creation and annihilation operators are defined for the transverse polarizations only, aˆ = λBˆ + iEˆ , aˆ† = λBˆ† iEˆ† , λ = 1 only, (6) k,λ k,λ k,λ k,λ k,λ − k,λ ± 1 † † † 3 (3) ′ aˆ , aˆ ′ ′ = 0, aˆ , aˆ ′ ′ = 0, aˆ , aˆ ′ ′ = δλλ′ 2 k (2π) δ (k k ). (7) k,λ k ,λ k,λ k ,λ k,λ k ,λ × | | − In terms of these operators, the free EM Hamiltonian becomes d3k 1 ˆ † k H 3 ωkaˆk,λaˆk,λ + const for ωk = + , (8) (2π) 2ωk | | Z λ=±X1 only which makes for the usual time-dependence in the Heisenberg picture: aˆ aˆ e−iωkt, aˆ† aˆ† e+iωkt, (9) k,λ → k,λ × k,λ → k,λ × and hence (after a bit of algebra) d3k 1 Eˆ(x, t) = e−iωt+ikx( iωe )ˆa + e+iωt−ikx(+iωe∗ )ˆa† , (2π)3 2ω − k,λ k,λ k,λ k,λ Z λ=±X1 only d3k 1 Bˆ (x, t) = e−iωt+ikx( ik e )ˆa + e+iωt−ikx(+ik e∗ )ˆa† . (2π)3 2ω − × k,λ k,λ × k,λ k,λ Z λ=±X1 only (10) Up to now I have focused on the EM tension fields Fˆµν(x), but to couple the EM to come charged fields — like the electron field Ψ(ˆ x) — I would also need the quantum potential µ fields Aˆµ(x) to spell out the interactions Jµ A . Classically, the potential fields are L ⊃ − × subject to gauge symmetries Aµ (x) Aµ (x) = Aµ (x) ∂µΛ(x), (11) old → new old − but implementing such symmetries in the quantum theory is highly non-trivial since a trans- form changing the time-dependence of quantum fields must change the Hamiltonian oper- ator. There are more problems with quantum gauge theories, and I’ll deal with them next semester. For the moment, let me simply say that the canonical quantization of the potential fields Aµ(x) requires fixing a gauge. That is, we should remove the gauge-redundancy of the potential fields by imposing a local linear constraint at each x, for example A(x) 0 ∇· ≡ µ 3 (the Coulomb gauge), or ∂µA (x) 0 (the Landau gauge), or A (x) 0 (the axial gauge). ≡ ≡ 2 Once we impose such a constraint at the Lagrangian level, we find the canonical conjugates of the remaining independent fields, build the classical Hamiltonian, and then quantize the theory in the usual way to build the quantum fields, the Hilbert space where they act, and the Hamiltonian operator. In general, different gauge-fixing constraints give rise to different quantum theories, each having its own Hilbert space and the Hamiltonian operator. However, all such theories are physically equivalent to each other! For the free electromagnetic fields, I do not have to re-quantize the theory starting from the Lagrangian and some gauge-fixing constraint for the Aµ(x). Since I already have the quantum Eˆ(x) and Bˆ (x) fields as in eqs. (10), I can can obtain the free potential fields Aˆµ(x) by simply solving the equations ∂µAˆν(x) ∂ν Aˆµ(x)= Fˆµν(x) combined with the gauge-fixing − constraint. By linearity and translation invariance, the result has form d3k 1 Aˆµ (x) = e−ikx µ(k,λ) aˆ + e+ikx µ∗(k,λ) aˆ† (12) free (2π)3 2ω E × k,λ E × k,λ Z λ=±X1 only where the plane waves Aµ(x) = e−ikx µ(k,λ) and Aµ(x) = e+ikx µ∗(k,λ) for k0 =+ k (13) ×E ×E | | obey the gauge-fixing conditions as well as the Maxwell equations. For example, in the Coulomb gauge Aˆ = 0, the polarization vectors µ have simple form ∇· E µ(k,λ) = 0, e(k,λ) (14) E while in the axial gauge n A = 0 (where n is a fixed unit 3-vector) we have much messier · formulae n e(k,λ) 0 n e(k,λ) ~(k,λ) = e(k,λ) · k, (k,λ) = · ωk . (15) E − n k E − n k · · 3 Photon Propagator The photon propagator Gµν(x y) = 0 TAˆµ(x)Aˆν(y) 0 (16) F − h | | i depends on the gauge-fixing condition for the quantum potential fields Aˆµ(x). So let me first calculate it for the Coulomb gauge Aˆ 0, and then I’ll deal with the other gauges. ∇· ≡ Instead of calculating the propagator directly from eqs. (16), (12), and (14), let me use the fact that Gµν(x y) is a Green’s function of the Maxwell equations for the Aµ fields, − 2 ν ν µ ν µ 4 µν ∂ A (x) ∂ (∂µA (x)) = J (x) = A (x) = d y ( i)G (x y) Jν(y). (17) − ⇒ − − × Z Let’s start with the scalar potential A0(x, t). In the Coulomb gauge we have µ 0 0 ∂µA = ∂ A + A = ∂ A + 0, (18) 0 ∇· 0 which makes the equation for the A0(x) time-independent. Indeed, 0 2 0 0 µ 2 2 0 0 0 2 0 J = ∂ A (x) ∂ (∂µA (x)) = (∂ )A ∂ (∂ A ) = A (x) (19) − 0 −∇ − 0 −∇ 0 while the terms containing the time derivatives ∂0 cancel out. Consequently, the A (x, t) is the instantaneous Coulomb potential for the electric charge density J0(y, t), J0(y, same t) A0(x, t) = d3y (20) 4π x y Z | − | — which is why this gauge is called the Coulomb gauge. In terms of the Green’s function, this means iδ(x0 y0) G0i 0, G00(x y) = − , (21) ≡ − 4π x y | − | of after Fourier transform to the momentum space, i G00(k) = independent of k0. (22) k2 4 As to the vector potential A, it obeys 2 µ ∂ A + (∂µA ) = J (23) ∇ where in the Coulomb gauge µ 0 1 0 1 ∂µA = ∂ A = ∂ J = + ( J) (24) 0 − 2 0 2 ∇· ∇ ∇ where the last equality follows from the electric current conservation, ∂ J0 = J. Con- 0 −∇ · sequently, 1 (∂2 2)A = J ( J), (25) 0 −∇ − ∇ 2 ∇· ∇ or in momentum basis kikj (k2 k2) Ai(k) = Ji Jj. (26) − 0 − × − k2 × In terms of the Green’s function, this means i kikj Gi0 = 0, Gij(k) = δij . (27) k2 k2 × − k2 0 − Altogether, all the components of the Gµν(k) can be summarized as i Gµν(k) = Cµν(k) (28) k2 k2 × 0 − where kikj k2 k2 k2 Cij = δij , Ci0 = C0i = 0, C00 = 0 − = 0 1. (29) − k2 k2 k2 − A convenient way to summarize this Cµν(k) tensor is (k0, k) Cµν(k) = gµν + kµcν(k) + kνcµ(k) for cµ(k) = − . (30) − 2k2 In the coordinate space d4k ie−ik(x−y) Gµν(x y) = Cµν(k) (31) − (2π)4 k2 × Z where the denominator has poles at k0 = k . As usual, these poles should be regularized ±| | by moving them off the real axis, and different regularizations lead to different Green’s 5 functions. We are interested in the specific Green’s function — the Feynman propagator — so we should move the poles to k0 =+ k i0 and k0 = k + i0, thus | | − −| | d4k ie−ik(x−y) Gµν(x y) = Cµν(k) = gµν + kµcν(k) + kνcµ(k) . (32) F − (2π)4 k2 + i0 × − Z Now consider photon propagators for different gauge conditions for the EM potential fields Aˆµ(x). When we change a gauge condition, the potential fields change as Aˆµ (x) = Aˆµ (x) ∂µΛ(ˆ x) (33) new old − for some Λ(ˆ x) that depends on the old Aˆµ(x) and the new gauge condition. For linear gauge conditions, 4 µ Λ(ˆ x) = d y Lµ(x y)Aˆ (y) (34) − old Z for some kernel Lµ(x y), or in momentum space − ˆ ˆµ Λ(k) = Lµ(k)Aold(k). (35) Consequently, in the new gauge the Feynman propagator becomes Gµν (x y) = 0 TAˆµ (x)Aˆν (y) 0 new − h | new new | i ˆµ ˆν ∂ ˆ ˆν = 0 TAold(x)Aold(y) 0 0 TΛ(x)Aold(y) 0 h | | i − ∂xµ h | | i 2 ∂ ˆµ ˆ ∂ ˆ ˆ 0 TAold(x)Λ(y) 0 + 0 TΛ(x)Λ(y) 0 − ∂yν h | | i ∂xµ∂yν h | | i 2 µν ∂ ν ∂ µ ∂ = Gold(x y) + f (x y) + f (y x) + h(x y) − ∂xµ − ∂yν − ∂xµ∂yν − for some functions f µ(x y) and h(x y).
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