course, in cases where the orbital term gives vanishing contribution, the electron magnetic moment
term may become important, and will have to be included.) Hence, the sole remaining contribution
to the matrix element is given by,
〈f|H 1 |i〉 =(
〈ψf|m⊗〈∅|EMaα(k))
e
2 m(
A·p+p·A)(
|ψi〉m⊗|∅〉EM)
(20.8)We begin by evaluating the matrix element inHEM, with the help of the free electro-magnetic field
A, in transverse gauge,
A(t,x) =∫
d^3 k′
(2π)^31
√
2 |k′|∑β(
εβ(k′)aβ(k′)e−ik
′·x
+ε∗β(k)a†β(k′)eik
′·x
)
(20.9)wherek′·x=|k′|t−k′·x. Using the canonical commutation relations of the radiation oscillators,
we readily find that,
〈∅|aα(k)A(t,x)|∅〉=1
√
2 |k|ε∗a(k)e−ik·x (20.10)Note that the ordering betweenAandpis immaterial here, since the commutator of the two terms
is proportional to∇·A= 0 in transverse gauge. As a result, the full matrix element takes the
form,
〈f|H 1 |i〉=
e
m√
2 ωeiωt〈ψf|ε∗α(k)·pe−ik·x|ψi〉 (20.11)Upon taking the norm square, as needed to compute the rate, wehave,
|〈f|H 1 |i〉|^2 =
e^2
2 mω∣∣
∣∣〈ψf|ε∗α(k)·pe−ik·x|ψi〉∣∣
∣∣2
(20.12)Finally, the summation over all final states, required to derive the total rate, amounts to an inte-
gration over all photon momenta and summation over both photon polarizations. This gives,
Γ = 2π∫
d^3 k
(2π)^3e^2
2 mω∑
α∣∣
∣∣〈ψf|ε∗α(k)·pe−ik·x|ψi〉∣∣
∣∣2
δ(
E(0)i −Ef(0)−ωk)
(20.13)The radial part of thek-integration may be easily carried out in view of the presence of theδ-
function onωk=|k|, and we haved^3 k=ω^2 dω dΩ, wheredΩ is the solid angle volume element for
the directionk/|k|. Carrying out theωintegration gives,
Γ =
e^2
8 π^2 m(
Ei(0)−Ef(0))∑α∫
dΩ∣∣
∣∣〈ψf|ε∗α(k)·pe−ik·x|ψi〉∣∣
∣∣2
(20.14)We simplify this formula one more step. If the rate is to correspond to the decay of an excited
atomic state to a lower energy state or the ground state, thenthe energy differences are of order
mc^2 α^2 ∼c|k|. The typical size of the atom is the Bohr radius, which is given by|x|∼a 0 = ̄h/(mcα).