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While reading about the interaction of matter with the quantised electromagnetic field I found that, after applying the minimal coupling $\hat{p_i}\rightarrow \hat{p_i}-\frac{e_i}{c}\hat{A}(\vec{r}_i,t)$ I arrive to terms involving $\hat{p_i}.\hat{A}$. It is suggested that we should apply these operators on a wavefunction $\psi$: $\hat{p}.\hat{A}\psi=(\hat{p}.\hat{A})\psi+\hat{A}.(\hat{p}\psi)$. How does one arrive to this expression? $\psi$: $\hat{p}.\hat{A}\psi$ is just $\hat{A}$ acting on $\psi$, followed by the application of $\hat{p}$, right? Afterwards I wrote the expression like this:$\hat{p}.\hat{A}\psi-\hat{A}.(\hat{p}\psi)=[\hat{p},\hat{A}]\psi$, isn't it how it should be? What is $(\hat{p}.\hat{A})?$. Thank you.

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  • $\begingroup$ Quick Hint : Coulomb gauge. $\endgroup$ – Sunyam Nov 24 '18 at 19:37
  • $\begingroup$ $\vec{\nabla} .A=0$? I would use it after writing the momentum as $-i\hbar \vec{\nabla}$, but I can't seem to understand I get how one arrives to the expression I mentioned. $\endgroup$ – RicardoP Nov 24 '18 at 20:23
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If you imagine $p$ behave like a derivative, $p\cdot (FG)=(p\cdot F)G+F(p\cdot G)$ is just the product rule. If we work in the position representation, and $A$ is just a function of $x$, then the answer follows.

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