Γ =
∫
dΩ∣ ∣ ∣ ∣ ∣ ∣
δniǫˆ·ˆǫ′−1
m ̄h∑
j[
〈n|ˆǫ′·~p|j〉〈j|ˆǫ·~p|i〉
ωji−ω+
〈n|ǫˆ·~p|j〉〈j|ˆǫ′·~p|i〉
ω′+ωji]
∣ ∣ ∣ ∣ ∣ ∣
2×
V ω′^2
(2πc)^3e^4
4 V^2 m^2 ω′ω
2 πdΓ
dΩ=
e^4 ω′
(4π)^2 V m^2 c^3 ω∣ ∣ ∣ ∣ ∣ ∣
δniˆǫ·ˆǫ′−1
m ̄h∑
j[
〈n|ˆǫ′·~p|j〉〈j|ˆǫ·~p|i〉
ωji−ω+
〈n|ˆǫ·~p|j〉〈j|ǫˆ′·~p|i〉
ω′+ωji]
∣ ∣ ∣ ∣ ∣ ∣
2Note that the delta function has enforced energy conservation requiring thatω′=ω−ωni, butwe
have leftω′in the formula for convenience.
The final step to adifferential cross sectionis to divide the transition rate by theincident flux
of particles. This is a surprisingly easy step because we are using plane waves of photons. The
initial state isone particle in the volumeV moving with a velocity ofc, so the flux is simply
c
V.
dσ
dΩ
=
e^4 ω′
(4π)^2 m^2 c^4 ω∣
∣
∣
∣
∣∣δniˆǫ·ˆǫ′−^1
m ̄h∑
j[
〈n|ˆǫ′·~p|j〉〈j|ˆǫ·~p|i〉
ωji−ω+
〈n|ǫˆ·~p|j〉〈j|ˆǫ′·~p|i〉
ω′+ωji]
∣
∣
∣
∣
∣∣
2Theclassical radius of the electronis defined to ber 0 = e
2
4 πmc^2 in our units. We will factor the
square of this out but leave the answer in terms of fundamental constants.
dσ
dΩ=
(
e^2
4 πmc^2) 2 (
ω′
ω)
∣ ∣ ∣ ∣ ∣ ∣
δniˆǫ·ˆǫ′−1
m ̄h∑
j[
〈n|ˆǫ′·~p|j〉〈j|ˆǫ·~p|i〉
ωji−ω+
〈n|ˆǫ·~p|j〉〈j|ˆǫ′·~p|i〉
ωji+ω′]
∣ ∣ ∣ ∣ ∣ ∣
2This is called theKramers-Heisenberg Formula. Even now, the three (space-time)Feynman
diagramsare visible as separate terms in the formula.
(They show up like
∣
∣
∣
∣
∣
c+∑
j(a+b)∣
∣
∣
∣
∣
2
.) Note that, for the very short time that the system is in anintermediate state,energy conservation is not strictly enforced. The energy denominators in
the formula suppress larger energy non-conservation. The formula can be applied to several physical
situations as discussed below.