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:

  1. 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?
  2. 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?
  • 1
    $\begingroup$ I would be quite grateful if you can provide the specific reference within Merzbacher, and any other sources you've found. $\endgroup$ Sep 19, 2016 at 12:06
  • $\begingroup$ The sources I used are chapter 23 of marzbecher, the online resource that I gave the link for in the question, and Shiff's quantum mech, though find the last one less clear. You can also find the lecture notes of Ben Simons on radiative transitions, from his QM2 course $\endgroup$
    – Yair M
    Sep 19, 2016 at 16:07
  • $\begingroup$ Thanks. If you've got a specific page or section within Schiff with this manipulation, that'd be mighty appreciated. I was recently looking for examples of this precise manipulation on standard sources, but it's subtle and with a small footprint on the page so it's a bit tedious to look for. $\endgroup$ Sep 19, 2016 at 17:08
  • $\begingroup$ Which exact manipulation? As I said Schiff is rather unclear to me, I recommend marzbecher $\endgroup$
    – Yair M
    Sep 19, 2016 at 17:10
  • $\begingroup$ As in, using $\mathbf p\cdot \mathbf A$ and $\mathbf A\cdot \mathbf p$ interchangeably, with the radiation-gauge justification in my answer. Believe it or not I got flak from a referee saying this was wrong, and it'd be nice to have more examples of just how many textbooks do this, but again it's a tedious thing to hunt for. $\endgroup$ Sep 19, 2016 at 17:13

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.