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How do I prove this equation for any function $f$ (can this even be proven for an arbitrary function?), where $\hat a$ and $\hat a^{\dagger}$ are the creation and annihilation operators for the 1D harmonic oscillator?

$$\hat af(\hat a^{\dagger} \hat a) = f(\hat a^{\dagger} \hat a + 1) \hat a$$

I have tried expanding $f$ as a power series and using the commutation relation of $\hat a$ and $\hat a^{\dagger}$.

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I think the equation really only makes sense if it's acting on some state, say $|n\rangle$. Then it \begin{align*} a f(a^\dagger a) |n\rangle = a f(n) |n\rangle = f(n) \sqrt{n}|n-1\rangle \end{align*} and \begin{align*} f(a^\dagger a + 1) a |n\rangle = f(a^\dagger a + 1) \sqrt{n}|n-1\rangle = f(n-1+1) \sqrt{n}|n-1\rangle \end{align*} and the expressions are equal. Informally I'd argue that you can first measure how many particles exist in that state and then destroy a particle or you can first destroy one but then you have to add 1 to the number of particles that you measured afterwards.

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    $\begingroup$ To add: Since $\{|n\rangle\}_n$ forms a basis, the action of $f(a^\dagger a+1)\, a$ and $a\,f(a^\dagger a)$ are the same on every state $|\psi\rangle$: $f(a^\dagger a+1)\, a |\psi\rangle =a\,f(a^\dagger a) |\psi\rangle $. Thus, as operators, they are equal. $\endgroup$
    – Tobias Fünke
    Commented Jan 25, 2022 at 11:39
  • $\begingroup$ Am I wrong or are you assuming, in your first expression, that $a$ and $f(n)$ commute? Why so? $\endgroup$
    – Schiele
    Commented Apr 28, 2023 at 18:45
  • $\begingroup$ It would be clearer if I would have differentiated between the operator and $c$-numbers, i.e. $\hat{a} f(\hat{n})|n\rangle = \hat{a} f(n)|n\rangle = f(n) \hat{a}|n\rangle$. $\endgroup$
    – Wihtedeka
    Commented Apr 28, 2023 at 19:06

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