For my quantum homework, I was asked to prove if $f(x)$ is an eigenvector of $F(\hat{A})$ where $F$ is given as an "arbitrary differential function" and $f(x)$ is a known eigenfunction of $\hat{A}$ with an eigenvalue of $\lambda$.

I know that $F(\hat{A})$ can be expanded into a Taylor series to be represented as:

$e^{(\hat{A})} = \sum\limits_{n=0}^{\infty} \frac{1}{n!}\hat{A}^n$

I'm currently thinking of solving this problem by doing something along the lines of :

$e^{(\hat{A})} f(x) = f(x) + \hat{A}f(x) + \frac{1}{2!}\hat{A}^2f(x) + ...+ \frac{1}{n!}\hat{A}^nf(x)$

$ = f(x) + \lambda f(x) + \frac{1}{2!}\hat{A} \lambda f(x) + ...+ \frac{1}{n!}\hat{A}^{n-1} \lambda f(x)$

$ = f(x) + \lambda f(x) + \frac{1}{2!}\lambda^2 f(x) + ...+ \frac{1}{n!}\lambda^n f(x)$

This is really my first experience working with a series expansion, so I am not really sure if this expansion alone is sufficient to prove if $f(x)$ is an eigenfunction of $F(\hat{A})$ nor am I confident in how to pull an eigenvalue out of the expansion. Any guidance is appreciated.

  • $\begingroup$ Now the eigenvalue is $e^{\lambda}$ in the way you calculate it, if you let $n \rightarrow \infty$, so indeed $f(x)$ is an eigenfunction $\endgroup$ – Dani Sep 30 at 9:16
  • $\begingroup$ @Dani thank you! Much appreciated! $\endgroup$ – c2v_reactsonly Sep 30 at 13:23

Is $F(\hat{A}) $ an "arbitrary differential function" (I am not entirely sure what you mean with this) as you stated in the first sentence or is it explicitly given as the exponential of $\hat{A}$?

Lets begin with the second case. You can proceed with your calculation by noting \begin{align} e^{\hat{A}} f(x) &= f(x) + \lambda f(x) + \frac{\lambda^2}{2!} f(x) \, + \, ... \, + \frac{\lambda^n}{n!}f(x) \,+ \, ... \, \\ &= (1 + \lambda + \frac{\lambda^2}{2!} \, + \, ... \, + \frac{\lambda^n}{n!} \,+ \, ... )\, f(x) . \end{align} From here on you should be able to figure out the rest.

Now for the sake of completeness to the first case, if $F$ is an "arbitrary differential function" you would need to proof the spectral theorem, which I think is unlikely in a regular QM course.

  • $\begingroup$ I think $F$ is the arbitrary differential function, not $F\hat{A}$. Hopefully, I'm not expected to prove the spectral theorem, it does seem a little overkill for this course. $\endgroup$ – c2v_reactsonly Sep 30 at 13:14
  • $\begingroup$ Could this not be extended to the general case by writing $F(\hat{A})=\sum_{n=0}^{\infty}a_n\hat{A}^n$ where $a_n$ are some undetermined coefficients? $\endgroup$ – Tyberius Oct 1 at 1:10
  • $\begingroup$ @Tyberius I think this should be possible, although there are a few non trivial technicalities (convergence, domain, ...) you would need to take care of. If you are interested in these I would recomend looking at a functional analysis books. And as a side note, a power series is not the most general case for which the spectral theorem holds. $\endgroup$ – hof_a Oct 1 at 8:19

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