The scattering process clearly requires terms in that annihilate one photon and create another. The order does not matter. The is the square of the Fourier decomposition of the radiation field so it contains terms like and which are just what we want. The term has both creation and annihilation operators in it but not products of them. It changes the number of photons by plus or minus one, not by zero as required for the scattering process. Nevertheless this part of the interaction could contribute in second order perturbation theory, by absorbing one photon in a transition from the initial atomic state to an intermediate state, then emitting another photon and making a transition to the final atomic state. While this is higher order in perturbation theory, it is the same order in the electromagnetic coupling constant , which is what really counts when expanding in powers of . Therefore, we will need to consider the term in first order and the term in second order perturbation theory to get an order calculation of the matrix element.
Start with the first order perturbation theory term. All the terms in the sum that do not annihilate the initial state photon and create the final state photon give zero. We will assume that the wavelength of the photon's is long compared to the size of the atom so that .
This is the matrix element . The amplitude to be in the final state is given by first order time dependent perturbation theory.
Now we very carefully put the interaction term into the formula for second order time dependent perturbation theory, again using . Our notation is that the intermediate state of atom and field is called where represents the state of the atom and we may have zero or two photons, as indicated in the diagram.
The space-time diagram below shows the three terms in Time is assumed to run upward in the diagrams.
Looking again at the formula for the second order scattering amplitude, note that we integrate over the times and and that . For diagram (a), the annihilation operator is active at time and the creation operator is active at time . For diagram (b) its just the opposite. The second order formula above contains four terms as written. The and terms are the ones described by the diagram. The and terms will clearly give zero. Note that we are just picking the terms that will survive the calculation, not changing any formulas.
Now, reduce to the two nonzero terms. The operators just give a factor of and make the photon states work out. If is the intermediate atomic state, the second order term reduces to.
We have calculated all the amplitudes. The first order and second order amplitudes should be combined, then squared.
The final step to a differential cross section is to divide the transition rate by the
incident flux of particles.
This is a surprisingly easy step because we are using plane waves of photons.
The initial state is one particle in the volume moving with a velocity of ,
so the flux is simply
Also note that the formula yields an infinite result if . This is not a physical result. In fact the cross section will be large but not infinite when energy is conserved in the intermediate state. This condition is often refereed to as ``the intermediate state being on the mass shell'' because of the relation between energy and mass in four dimensions.