I'm working on time-independent degenerate perturbation theory for the Hydrogen first excited state. I have the following perturbation $H$:

$H = \lambda V_0 \sin^2 \theta \sin 2\phi = \lambda V$.

We can see that $P V P^{-1} = V$ so this means that $\langle n l m | V | n' l' m'\rangle$ might be $\neq 0$ if and only if $l+l'$ is an even integer.

We can also see that $\int_0^{2 \pi} \sin 2\phi \ \mathrm{d}\phi = \int_0^{2 \pi} \sin 2\phi \ e^{\pm i \phi} \mathrm{d}\phi = 0$, so we have to add that $\langle n l m | V | n' l' m'\rangle$ might be $\neq 0$ if and only if $m = -m' = \pm 1$.

So I arrive to the conclussion that $\langle n l m|V|n'l'm'\rangle = 0$ except from $\langle 211|V|21-1\rangle$ and $\langle21-1|V|211\rangle$.

But we can see that $\langle 211 |V| 21-1\rangle = -\langle21-1|V|211\rangle = V_0 \frac{(2a)^{-3}}{8 \pi a^2} \int_0^{\infty} r^4 e^{-r/a} \mathrm{d}r \int_0^{\pi} \sin^5 \theta \ \mathrm{d}\theta \int_0^{2\pi} e^{-2i\phi} \sin 2\phi \ \mathrm{d}\phi = \frac{2 V_0}{5} i$ which is a complex number.

What am I doing wrong or what is the interpretation for these results?

When I construct the matrix $\langle nlm |V| n'l'm'\rangle$ and diagonalize it I get 3 eigenvalues $0, \frac{2}{5}, \mathrm{and}\ -\frac{2}{5}$ in terms of $\lambda V_0$, so everything's fine in this way.

Thanks in advance, please help a confined quantum mechanics student!!!


1 Answer 1


You are not doing anything wrong. What you calculated is not the energy but the matrix element of the perturbation. It doesn't have an interpretation as an observable quantity. The eigenvalues that you calculated are real, and they are the numbers associated with the energy of the perturbation (I assume that they are measured in units of $\lambda V_0$?)

Generally speaking, an Hermitian operator $V$ which we associate with a physical observable must have real eigenvalues, which will lead to real expectation values, but doesn't have to have real matrix element. For example, the spin-$y$ operator for spin-$1/2$ particles has $i$ and $-i$ entries when represented in the basis of spin-$z$ eigenvectors, but it still has real eigenvalues and will generate real observable when measured.

If you are struggling with what to do with these matrix elements in degenerate perturbation theory, maybe the answer here might help.

  • $\begingroup$ You assume well, I forgot the $\lambda V_0$ term. Thank you very much for your explanation!!! $\endgroup$
    – Pablo
    Apr 14, 2020 at 11:51

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