# Eigenvalues and Eigenfunctions for a function of an operator?

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.

• 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 – Dani Sep 30 at 9:16
• @Dani thank you! Much appreciated! – 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.
• 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. – c2v_reactsonly Sep 30 at 13:14
• 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? – Tyberius Oct 1 at 1:10