# Operator algebra for momentum and potential vector in second quantisation

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.

• Quick Hint : Coulomb gauge. – Sunyam Nov 24 '18 at 19:37
• $\vec{\nabla} .A=0$? 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. – RicardoP Nov 24 '18 at 20:23

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.