What I wanted to say is that one must not teach students "old quantum theory" as if it was still considered correct. The photoelectric effect, at the level of accuracy described in Einstein's paper, does not show that the electromagnetic field is quantized, as shown by the standard calculation provided in my Insights article (the only thing, I've never found is the argument given there, why one can omit the interference term between the two modes with ##\pm \omega## of the em. field, which are necessarily there, because the em. field is real).
I've not calculated the cross section to the end, because I thought that's an unnecessary complication not adding to the point at the level of the (in my opinion false) treatment in introductory parts of many QM1 textbooks. You can do this quite easily yourself, using as an example the analytically known hydrogen wavefunctions for the bound state and a plane-wave free momentum eigenstate for the continuum state. Then you integrate out the angles and rewrite everything in terms of energy instead of ##\vec{p}##. You can find the resul in many textbooks, e.g., Sakurai, where this example is nicely treated.
Of course, what I've calculated is the leading-order dipole approximation. Perhaps one should ad a paragraph showing this explicitly, but I don't know, whether one can add something to a puglished insight's article. There are also some typos :-(.
So here is the derivation. What we need is the right-hand side of Eq. (15), i.e., the matrix element in the Schrödinger picture (which coincides by assumption with the interaction picture at ##t=t_0##). First of all we note that in the interaction picture
$$\dot{\hat{\vec{x}}}=\frac{1}{\mathrm{i} \hbar} [\hat{\vec{x}},\hat{H}_0]=\frac{1}{m} \hat{\vec{p}}.$$
Thus we have
$$\langle E (t_0)|\hat{\vec{p}}(t_0)|E_n(t_0) \rangle=\frac{m}{\mathrm{i} \hbar} (E-E_n) \langle E(t_0)|\hat{\vec{x}}|E \rangle.$$
Now if you plug this into (20) then due to the energy-conserving ##\delta## distribution and making use of the fact that this piece relevant for the absorption (photoeffect) transition rate only comes from the positive-frequency piece ##\propto \exp(-\mathrm{i} \omega t)## in ##\vec{A}##, you find that what enters is in fact
$$\alpha^2 \propto |\vec{E}_0 \cdot \langle E(t_0)|\hat{\vec{x}}(t_0) E_n(t_0) \rangle|^2,$$
and this is nothing else than the electric-field amplitude times the dipole-matrix transition matrix element.
The whole calculation also shows that there's no absorption of frequency modes of the em. field if ##\hbar \omega## is smaller than the binding energy of the initial state of the electron and that the rate of absorption processes is proportional to the intensity of the external field (for small fields so that perturbuation theory is still applicable).
For those who like to print the article, I've put it on a new website, I've just created:
http://fias.uni-frankfurt.de/~hees/pf-faq/