0
$\begingroup$

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?

enter image description here

enter image description here

$\endgroup$
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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