Quantization of the E-M field

Transcription

1 Quantization of the E-M field 0.1 Classical E&M First we will wor in the transverse gauge where there are no sources. Then A = 0, nabla A = B, and E = 1 A and Maxwell s equations are B = 1 E E = 1 B E = 0 B = 0 Combining the above gives a wave equation for the vector potential A 1 c A t = 0 Start with periodic boundary conditions. Then there are plane wave solutions where e i( r ωt) = π L (n xˆx + n y ŷ n + n z ẑ) The plane waves form a complete orthonormal set. 1 dr e i( ) r = δ 1 e i (r r ) = δ(r r ) To go from periodi boundary conditions to a continuous spectrum 1 d (π) 3, (π) 3 δ δ( ) 1

2 The expansion of the vector potential as plane waves is A(r, t) = 1 [e i r A (t) + c.c.] A is real. The transverse gauge conditions A = 0 A = 0. We get the time dependence from the wave equation A (t) = A e iωt Then E = 1 A = i c ω[e i r A (t) c.c.] B = A = i [e i r A (t) c.c.] Polarization We need a mutually orthogonal pair of polarization vectors. We can use the real orthogonal unit vectors ɛ i, i = 1, with ɛ 1 ɛ = ˆ, ɛ α ɛ β = δ αβ Or we can use complex circular polarization vectors e ±1 = 1 (ɛ 1 ± ɛ ) The total energy is H = 1 d 3 r(e + B ) and using the expansions of E and B in terms of A we can write the energy in terms of A as H = A (t) We want to quantize the fields. The usual strategy is to identify canonically conjugate variables, assign them operator status that obey the canonical commutation rule. Woring bacwards towards what we learned when quantizing the harmonic oscillator. Indeed recognizing that plane waves that represent A are a solution to the harmonic oscillator equation, we define Q (t) = 1 c [A (t) + c.c.], P (t) = i[a (t) c.c.]

3 We can rewrite H = 1 (P + ω Q ). and hamilton s equations Ṗ = H, Q = H Q P would allow us to wor from Hamiltonian to equations of motion for A Quantization We raise the canonical variables to operator status and the commutation rules are [Q λ, P λ ] = i δ δ λλ, [Q λ, Q λ ] = 0, [P λ, P λ ] = 0 Next define annihilation and creation operators and a. We find a λ = 1 ω (ωq λ + ip λ ) [a λ, a λ ] = δ δ λλ Define the vector operator a = a 1 e 1 + a e or a 1 e + a +1 e +. Then H = ω [a a + 1 ] = λ ω [N λ + 1 ], In the heisenberg picture the equations of motion for a are i ȧ (t) = [a, H] = ω a (t) The operators a are the quantum analog of the fourier coefficients A. Getting the units right we assign A c/a (t). Finally write the field operators in terms of a. A = E = i B = i c [ei( r) ωt a + h.c.] c [ei r ωt a h.c.] c [ei r ωt a (t) h.c.] 3

4 Heisenberg equations of motion for the field operators give Maxwell s equations. Momentum operator is constructed from field operators. Hermitian version is P = 1 dr[e B B E] = N λ, c λ The momentum operator is originally constructed as the generator of translations. Let s chec it out. The translation operator is Consider translation of the efield. T (d) = e id P/ T (d)e(r, t)t (d) = E(r d, t) An infinitesmal translation would have the form (1 + id P/ )E(r, t)(1 id P/ ) = E(r) d E r Note that the expectaion value of E or B in a state with a definite number of photons is zero. A coherent superposition of single photon states that loos lie γ = e γ / e γa 0 n=0 has nonzero expectation value for electric and magnetic fields and represents a classical field. The angular momentum in the classical field is J cl = 1 dr[r (E B)] cl c which can be separated into intrinsic angular momentum (spin) that does not depend on the position of the origin J spin = 1 dr(e A A E), c and the part that does J orb = 1 c dr 3 [E i (r A i ) + (r A i )E i ]. i=1 The spin part in terms of operators becomes J spin = i a a = λ λˆn λ 4

5 The one photon state has helicity λ = ±1. The relation between single photon states and states with definite linear and angular momentum is given by jmλ = a () 0 = j + 1 4π j j + 1 4π j=1 m= 1 dnd j mλ (n) A λ () 0, Dj mλ (n) jmλ Note, the sum starts at 1. There is no zero helicty single photon state so there is no j = 0 contribution. angular momentum. 5