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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:

  1. 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?

doublefelix
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Adrien Amour
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  • https://physicaeducator.files.wordpress.com/2018/01/solutions-of-quantum-mechanics-by-griffith.pdf Page 165 (answer to Q6.15) has a good answer for this if anyone in the future has the same question :) – Adrien Amour Dec 04 '22 at 11:47

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  1. 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 $\frac{1}{r}$.

  2. 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.

ACuriousMind
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  • In the 2nd edition of Griffiths, he says that $\hat{p}^4$ is not Hermitian for the $l=0$ states. However, in the 3rd edition, he corrects this and says that in fact he was wrong; it is Hermitian, and to show this, we are meant to use the identity$$\nabla^4(e^{-kr}) = \left(-\frac{4k^3}{r}+k^4\right)e^{-kr}+8\pi k\delta^{(3)}(\vec{r})$$. He seems to imply that it's the extra delta-function that fixes the issue, but I'm not sure. What do you think? – march Jan 17 '23 at 22:49
  • @march The problem is probably more subtle and related to domain issues: Since $p$ is unbounded it can only be densely defined and you need to start with the correct domain of definition for $p$ to make it self-adjoint. Then you have to apply functional calculus carefully to figure out what the domain of $p^4$ is, and I think it indeed turns out that the $\ell = 0$ states will not lie inside its domain (but would be self-adjoint). This does not mean that you can't find some way to apply $p^4$ to states outside this domain, only that you mustn't do that if you want to keep it self-adjoint. – ACuriousMind Jan 18 '23 at 00:27
  • In physics treatments that don't care about these details of functional analysis such issues show up variously as non-real eigenvalues, unexpected boundary terms or other apparent paradoxes (e.g. this one). Note that Griffiths' cure is worse than the problem: If $p^4$ produces a $\delta$-function - which is not an element of the Hilbert space $L^2$! - then $p^4$ isn't even an operator from the Hilbert space to itself anymore, let alone self-adjoint. – ACuriousMind Jan 18 '23 at 00:27