I am studying the fine structure of the hydrogen atom. The non relativistic Hamiltonian of the Hydrogen atom is given by
$$\hat{H}_{\mathrm{non-rel}}=\frac{\hat{\textbf{P}}^{2}}{2m_{e}}-\frac{e^{2}}{4\pi\epsilon_{0}|\hat{\textbf{R}}|}$$
We can replace the non relativistic kinetic energy term with a first order expansion of the relativistic kinetic energy, doing so obtains the folowing the Hamiltonian
\begin{align*}
\hat{H}&=\frac{\hat{\textbf{P}}^{2}}{2m_{e}}-\frac{e^{2}}{4\pi\epsilon_{0}|\hat{\textbf{R}}|}-\frac{(\hat{\textbf{P}}^{2})^{2}}{8m_{e}^{3}c^{2}}\\&=\hat{H}_{\mathrm{non-rel}}+\hat{V}
\end{align*}
For
$$\hat{V}=-\frac{(\hat{\textbf{P}}^{2})^{2}}{8m_{e}^{3}c^{2}}$$
Since the perturbation, $\hat{V}$, is spherically symmetric, it is possible to apply non degenerate perturbation theory to calculate the perturbed energy. In order to evaluate this perturbation I expressed the perturbation as follows
\begin{equation}
\hat{V}=-\frac{1}{2m_{e}c^{2}}\left(\hat{H}_{\mathrm{non-rel}}+\frac{e^{2}}{4\pi\epsilon_{0}|\hat{\textbf{R}}|}\right)^{2}
\end{equation}
As seen in Sakurai Modern quantum mechanics. The result is that we have a first order perturbation of
$$\Delta_{nl}^{(1)}=\frac{(E_{n}^{(0)})^{2}}{2m_{e}c^{2}}\left(3-\frac{4n}{l+\frac{1}{2}}\right)$$
where
$$E_{n}^{(0)}=-\frac{m_{e}}{2\hbar^{2}}\left(\frac{e^{2}}{4\pi\epsilon_{0}}\right)^{2}\frac{1}{n^{2}}$$
is the unperturbed energy. I have now read an alternative source [1], which suggests this derivation is not correct for $l=0$, since the operator $(\hat{\textbf{P}}^{2})^{2}$ is not Hermitian for $l=0$, however coincidentally this derivation provides the correct result for $l=0$. Therefore, my questions are:
- Why would $(\hat{\textbf{P}}^{2})^{2}$ by non Hermitian for $l=0$?
2.Why does this mean that my expression $\hat{V}=-\frac{1}{2m_{e}c^{2}}\left(\hat{H}_{\mathrm{non-rel}}+\frac{e^{2}}{4\pi\epsilon_{0}|\hat{\textbf{R}}|}\right)^{2}
$ is invalid?
Best Answer
The problem is that some of the boundary terms at $r=0$ when trying to show self-adjointness $$ \langle \psi_{n00}, p^4 \psi_{m00}\rangle = \langle p^4\psi_{n00},\psi_{m00}\rangle$$ do not disappear for the $\ell = 0$ hydrogen states. See this answer by ZeroTheHero for a similar issue with the mere $p_r$, the problem for $p_r^4$ is similar but more severe since it contains higher powers of $\frac{1}{r}$.
Perturbation theory assumes the Hamiltonian is Hermitian. The results may be correct, but strictly speaking you cannot apply perturbation theory to something that's not Hermitian.