# Justification of dropping term in Hamiltonian and expectation Values

While reading Sakurai's Modern QM, I was stuck at the point where he explains the absorption and emission of light quanta in atoms. He proceeds with Hamiltonian:

$$H= p^2/2m + e\phi(x) -e/mc A\cdot p$$ where $$p$$ is momentum, $$m$$ is mass of an electron, $$A$$ is magnetic potential.

Next, he says to drop terms $$p\cdot A$$ and $$|A|^2$$ terms. He says that since $$\nabla \cdot A=0$$ the term can be dropped. However, I do not see any reason to drop $$|A|^2$$ term. Can anyone justify this?

My next question was about expectation value of $$A\cdot p$$. Sakurai writes:

$$\langle x|p\cdot A|y\rangle = -ih\nabla\cdot (A(x)\langle x| \rangle) = \langle x|A\cdot p| \rangle + \langle x| \rangle [-ih\nabla\cdot A(x)]$$

Can anyone explain how he manipulated the above equation to arrive at the result?

It would be justified to drop $$\left|A\right|^2$$ if $$A$$ were small, so that $$\left|A\right|^2$$ is negligible relative to $$A$$. This must be Sakurai's reason - the EM field is a small perturbation. Note, however, that one could also absorb the $$\left|A\right|^2$$ term into a redefinition of $$\phi$$, so it doesn't qualitatively change the problem unless we're given a specific form for the potentials.
The relevance of $$\nabla \cdot A=0$$ is that $$\left[\nabla ^\mu,A_\mu\right]=\nabla \cdot A$$ so these operators commute in the Coulomb gauge. This means that upon expanding the $$~\left(p-eA\right)^2$$ term in the Hamiltonian it takes the form (5.7.1).