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?
