Quantum effects vanish when $\hbar \rightarrow 0$. The analysis of the $\hbar$ powers in the vertices and propagators results in a simple rule asserting that the contribution of a diagram containing $N-$ loops to the amplitudes is proportional to $\hbar^{N}$. Thus, we should expect that the classical amplitudes to be given exactly by the tree level diagrams.

However, there is one exception in the correspondence rule between the tree level diagrams and classical amplitudes. This exception is explained in the following work by: Holstein and Donoghue. Please see also, previous works of the same authors cited in the article, where more cases were analyzed.

The exception of the correspondence rule occurs when the loop diagram contains two or more massless propagators. In this case, it was observed by Holstein and Donoghue that contributions to the classical amplitudes occur at the one loop level due a certain non-analytical term in the momentum space. This tem can be recognized to contain $\hbar$ to the zeroth power when the loop diagram is expressed in terms of the momenta rather than the wave numbers as usually implicitly done when loop diagrams are solved. Holstein and Donoghue show that this term does not exist in the case of massive propagators where there is no contribution of the loop diagrams to the classical amplitudes.

The example given in the question electron-photon scattering does not suffer from the above problem. The expression given in the question is valid in the particular case of an electron in nonrelativistic motion as emphasized in Jackson's book "Classical electrodynamics" section 14.7. The fully relativistic classical cross section should be exactly equal to the tree level quantum (Klein-Nishina ) cross section.

Update

Answer to the first follow up question

In order to perform correctly the $\hbar$ expansion, the fields should be scaled by appropriate powers $\hbar$ in order to make their commutation relations proportional to $\hbar$, so they will commute in the $\hbar \rightarrow 0$ limit. The coupling constants should be scaled accordingly. For example the Dirac Lagrangian:
$$\mathcal{L}_D = c \bar{\psi}\bigg(\gamma^{\mu} (i \hbar \partial_{\mu} – e A_{\mu}) – m c \bigg)\psi$$
We need to absorb the $\hbar$ coming from the momentum operator into the field, thus by redefining
$$\mathcal{L}_D = c \bar{\tilde{\psi}}\bigg(\gamma^{\mu} (i \partial_{\mu} – \frac{e }{\hbar}A_{\mu}) – \frac{ m}{\hbar} c \bigg)\tilde{\psi}$$
Thus we need to take
$$ \tilde{e} = \frac{e}{\hbar}, \quad \tilde{m} = \frac{m}{\hbar} $$
To be fixed as $\hbar \rightarrow 0$.

Using the scaled electric charge, the fine structure constant takes the form:
$$\alpha = \frac{e^2}{4 \pi \epsilon_0 \hbar c} = \frac{\tilde{e} \hbar}{4 \pi \epsilon_0 c}$$
Now, $\alpha$ is linear in $\hbar$, therefore vanishes in the classical limit.
Please see the following work by Brodsky and Hoyer, for the scaling of the various fields and coupling constant.

Answer to the second follow up question:

In their original paper from 1929 Klein and Nishina computed the Compton cross section using Dirac's equation in the background of a classical radiation field. They did not use a quantized electromagnetic field. Therefore, they did not use QED. Please see the derivation also in Yazaki's article(There is a pdf version inside).

In my opinion, the only reason that they had to use the Dirac equation is to take into account the spin. I am quite sure that using more modern classical models for spin such as the Bargmann-Michel-Telegdi theory can be used to provide a fully classical derivation of the Klein-Nishina result. I couldn't find a reference that anyone has performed this work. I am sure it could be a nice project to do.

This post imported from StackExchange Physics at 2017-10-10 21:04 (UTC), posted by SE-user David Bar Moshe