thanks for your condescendence Emilio, but you missed the point. I edited with bold letters so you can read. The question was about the terminology that is vague. I could tell you that it's not because we have " amplification by spontaneous emission" that we have a laser at the end. I was hoping for more discussion and not lame comparison with a octopus

(if you only count positive k). For the B field with BC you get different phase shift. And it should explain why you have $[q_{i},m_{i} \dot{q}_{i}]$ non zero above. I can work that out later..

This is subtle. When you had boundary conditions (BC) you decrease the degrees of freedom (the number of entries in the matrices) so you have less chance for the fields to commute. An other way to see it is to see the standing waves in BC as a mix of returning waves. So even by taking the commutator at the same z you get a "mix of different z' ". I was trying to explain intuitively. The mathematical way is to do the BC case with $b_{k}=(a_{k}e^{i\alpha}+a_{-k}e^{i\beta})/\sqrt{2})$ as the new E field operator that satisfies the BC. The number of degrees of freedom have been divided by 2.

@Steven I realized I had answered a little too fast without reading the comments. Your original question and your last comment are interesting. So doing the computations for the 2nd-quantized expansion you saw that $[E(r,t),B(r',t)]$ is null for r=r' but not necessarily null otherwise if I am right. Because in this case there's a phase shift that applies differently to $a_{k}$ and $a^{+}_{k}$. I never saw the computations done so I am not 100% sure. But all this doesn't answer your original question. You can see from your equations with the boundary conditions you have $[E(z,t),B(z,t)]\not=0$