INTRODUCTION TO QUANTUM ELECTRODYNAMICS by Lawrence R. Mead, Prof. Physics, USM

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1 INTRODUCTION TO QUANTUM ELECTRODYNAMICS by Lawrence R. Mead, Prof. Physics, USM I. The interaction of electromagnetic fields with matter. The Lagrangian for the charge q in electromagnetic potentials V and A is, L = 1 m v qv + q c A v, where the interaction is through the scalar potential (qv ) and the vector potential ( J A = q v A); the fields in terms of the potentials are, Canonical momenta are defined by, B = A, E = V 1 A c t. (1) p i = L ẋ i = mẋ i + q c A i, or p = m v + q c A. The Hamiltonian can thus be written as H = i p iẋ i L, or H = 1 m ( p q A/c) + qv. () To go over to quantum mechanics, one turns the canonical momentum into a quantum operator by the transcription p h/i. II. Maxwell s equations in free space and their solutions. The Maxwell equations in free space (Gaussian units) are, E = 0 B = 0, B = 1 c E t E = 1 c The two equations on the left imply that in free space the (time-dependent) fields can be written in terms of the potentials, as in (1), but where the scalar potential is zero and 1 B t.

2 (Coulomb Gauge) A = 0. The Hamiltonian for the fields only can therefore be written in terms of the vector potential alone as, H fields = 1 d 3 r ( E 8π + B ) = 1 d 3 r [( 1 (3) A 8π c t ) + ( A) ]. The second pair of the Maxwell equations imply that the vector potential satisfies the wave equation, 1 A c t A = 0. (4) Solutions of (4) are plane waves with ω = ck: A( r, t) = A 0 e ±i( k r ωt). The general solution in free space is therefore, [ ] C k,λ ɛ k,λ e i( k r ωt) + Ck,λ ɛ k,λ e i( k r ωt) / V, A( r, t) = k,λ where the wavevector and polarization vectors form an orthogonal triplet, k ɛk,λ = 0, ɛ k,1 ɛ k, = 0, ɛ k,1 ɛ k, = k/k. For convenience, define new dimensionless coefficents a, a from, π h C k,λ = c ω a k,λ. Then the general solution of the wave equation becomes, A( r, t) = π h [ ] c a k,λ ɛ k,λ e i( k r ωt) + c.c., (5) ωv k,λ together with the associated orthonormality of the plane waves, d 3 r e i k r e i k r = V δ k, k. Substitution of (5) into (3) results in a very simplified form of the Hamiltonian, H fields = (a k,λa k,λ + 1 ) hω. (6) k,λ

3 III. Photons To quantize the theory defined by (5) and (6) we promote the a-coefficients to creation and destruction operators, equipped with Bose commutation relations, a k,λ â k,λ a k,λ â k,λ, [â k,λ, â k,λ ] = δ k,k δ λ,λ. The Hamiltonian then becomes a sum of number operators counting photon numbers of differing wavevector and polarization. The infinite zero-point energy which is the sum over 1/ is discarded as unphysical. States of an indefinite number of photons ( Fock states ) are defined from the action of the creation and destruction operators: general state: n k1,λ 1, n k,λ,, â k,λ n k,λ = n k,λ n k,λ 1 â k,λ n k,λ = n k,λ + 1 n k,λ + 1. Other physical observables may also of course be written in terms of the new operators; for example, the electromagnetic momentum operator, P = 1 d 3 r ( E 4πc B) = h k â k,λâk,λ. k,λ The complete Hamiltonian, which treats the atom only nonrelativistically, is then, where Ĥ = 1 ( p e A) m e c e r + 1 [ d 3 r ( 1 A 8π c t ) + ( A) ] (7) Ĥ0 + Ĥint, Ĥ 0 = ( h m e e r ) + Ĥfield, whose eigenstates (without electron spin) are, and the interaction Hamiltonian is, n, l, m photon numbers, Ĥ int = e m e c A h i + 3 e m e c A. (8)

4 Equation (8) follows because of the identity ( Aψ) = ( A)ψ + ( A )ψ and that we are working in a gauge where A = 0. Also, the above neglects the interaction of the electron and proton fields which is down by a factor of m e /m p. IV. Perturbation Theory The interactions given in (8) are sufficient to calculate quantum electrodynamic effects. Here, we will be content with first-order, time-dependent theory which results in the Fermi Golden Rule approximation for the transition rate between two quantum states labelled i and f, Γ = π h f Ĥi i ρ dɛ δ(ɛ f ɛ i ± hω). (9) λ The density of states ρ for photons required is, ρ(ɛ)dɛ = V ω (π) 3 dω dω. (10) c3 V. Spontaneous Emission The states of the unperturbed hydrogen atom n, l, m are stationary; if left alone the states will remain forever. However, in the quantum theory of fields governed by (6) and (8) there are always photons present even in a vacuum; their presence may cause spontaneous emission from an excited initial state. In particular, the first order term A in Ĥi contains a linear combination of creation and destruction operators and thus creates or destroys a single photon. The A term contains quadratic combinations of the photon operators and can thus contribute to two photon processes. Note that the exact time-development operator in the interaction picture is the time-ordered exponential, { Û(t) = T exp ī t Ĥ i (t ) dt }, h 0 which has all possible numbers of photons created or destroyed. Let us now consider a spontaneous transition between a P state and the 1S ground state of the hydrogen atom. Then i =, 1, m, and f = 1, 0, 0. The essential matrix element needed for first order theory is, f Ĥi i = 1 k,λ 1, 0, 0 Ĥi, 1, m 0 = e m e π h ωv d 3 r Ψ 100 (r)e i k r ɛ k,λ h Ψ i 10 (r) 1 k,λ â k,λ 0. (11) 4

5 Note that if there were one photon initially present, the photon part for stimulated emission would be k,λ â k,λ 1 k,λ =, and thus the stimulated emission rate (to first order) would be larger by a factor of from the spontaneous emission rate that we obtain below. VI. Spontaneous emission rate in the electric dipole approximation For optical transitions in the range of Å, such as the P 1S transition in hydrogen, k r << 1, so that one may make the approximation exp{ i k r} 1 (r a 0 1Å). Thus, we may write the transition rate per unit solid angle, for polarization λ, as, d Γ λ π h = (c dω dω π h ωv ) d 3 r Ψ 100 ɛ k,λ p Ψ 1m ( e mc ) ρ. The matrix element of the momentum operator may be conveniently computed as follows: p = im h [ p /m e /r, r ], and thus 100 p 1m = im h [H 0, r] = im h (E E 1 ) r. = iωm r We now need to compute, dγ λ dω = π h π h dω (c ωv ) e ω c d 3 r Ψ 100 ɛ k,λ r Ψ 1m ( V ω ) π hc δ(ɛ 3 f ɛ i hω). (1) For unpolarized photons, and random values of m l = 0, ±1, we average over all possible initial states: we take 1 3 m=0,±1. Also, r ɛ = = r q=0,±1 4π 3 r q ɛ q ( 1) q (ɛ x + iɛ y Y 1, 1 ɛ x iɛ y ) Y 1,1 + ɛ z Y 1,0, and Y 1,1 = Y 1, 1 Y 1,0 = 1 4π dω Y 1,mY 1,m = δ mm. 5

6 The integral above is therefore, = 1/3 ( ɛ x + iɛ y d 3 r 4π r R 1,0 4π 3 (same as above) R,1Y 1,m δ m,1 + ɛ x iɛ y ) δ m, 1 + ɛ z δ m,0 dr r 3 R 1,0 R,1. 0 The last radial integral has the value 3/( 8 /3 5 )a 0. So taking 1 3 expression yields a (ɛ x + ɛ y + ɛ z ) = a 0. m in the last Putting together all of the above pieces into (1) and restoring the sum over polarizations produces, Γ p 1s = dω αω3 15 πc 3 11 a 0 λ = αω3 c a 0, where α = e / hc 1/137 is the fine structure constant. The final result for the transition probability is, Γ p 1s = ( 3 )8 α 5 ( mc h ) Hz. The mean transition time for spontaneous emission from P to 1S is, τ p 1s = Γ 1 p 1s = sec. VII. Q.E.D. processes: Feynman Diagrams Electron Self Energy: + + e + e - Scattering: