Electron photon interaction - commutation of $\mathbf{A}$ and $\mathbf{p}$ I'm trying to figure out the radiative transition rates between electronic levels due to EM radiation using FGR as done by Merzbacher, this online source, and others.
I have two questions regarding the derivation process:


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*In all of the derivations I found the authors use: $\mathbf{A}\cdot\mathbf{p}=\mathbf{p}\cdot\mathbf{A}$. Is there some reason for which $[A_i,p_i]=0$ for all $i$, or is it just the extension of the Coulomb gauge (i.e. $\nabla\cdot\mathbf{A}=0$) to the QM case? Could it be that this is an approximation in the spirit of the dipole approximation?

*There're some references to the "radiation gauge" in multiple sources. According to my understanding this is just the Coulomb gauge with the addition of $\phi=0$. Is it a different gauge altogether, or is it just the Coulomb gauge with the additional requirement of no charge distribution. This question can actually be phrased in a more general way: Is it possible to gauge out $\phi$ without use of Lorenz gauge? 

 A: The Coulomb gauge and the radiation gauge are essentially the same thing. The radiation gauge is generally open to having a nonzero scalar potential $\phi$, but this is restricted only to the electrostatic fields of any particles present: in the typical case you will have a bunch of charged particles (like, say, electrons and nuclei in a molecule) interacting with each other through their usual electrostatic interactions (plus any relativistic effects you need to bring in), which are then subjected to an external radiation field. Working in the radiation gauge means that this external radiation field is described exclusively via a divergence-free vector potential.
The vector potential $\mathbf A(\mathbf r,t)$ and the momentum $\mathbf p$ commute, in their inner product, if and only if the vector potential is in this gauge - or, more specifically, if and only if it obeys $\nabla \!\cdot\! \mathbf A(\mathbf r,t)=0$. To see this, you just calculate:
\begin{align}
\mathbf A(\mathbf r,t)\cdot\mathbf p -\mathbf p\cdot\mathbf A(\mathbf r,t)
& =A_j(\mathbf r,t)p_j-p_jA_j(\mathbf r,t)
\\ & = i\hbar \frac{\partial A_j}{\partial x_j}(\mathbf r,t)
\\ & = i\hbar \nabla \!\cdot\! \mathbf A(\mathbf r,t).
\end{align}
In general, the commutator of $p_i$ and $A_j(\mathbf r,t)$ gives a nonzero derivative of the potential, but the contraction reduces it to the divergence. This is, of course, exact, and it does not rely on the dipole approximation.
