Expectation value of time-independent operators

I'm reading from a lecture note on introductory quantum mechanics (here), which says on P.3 that "The expectation value of any time-independent operator $$\hat{Q}$$ on a stationary state is time-independent". It proceeds to prove the statement in the following way: $$$$\begin{split} \langle\hat{Q}\rangle_{\Psi}& \:=\int{dx\:\Psi^*(x,t)\hat{Q}\Psi(x,t)} \\ & = \int{dx\:e^{iEt/\hbar}\psi^*(x)\hat{Q}\psi(x)e^{-iEt/\hbar}}\\ & =\int{dx\:e^{iEt/\hbar}e^{-iEt/\hbar}\psi^*(x)\hat{Q}\psi(x)}\\ & = \langle\hat{Q}\rangle_{\psi} \end{split}$$$$

My Question is: why does the time-independence (i.e. no explicit time dependence) of the operator $$\hat{Q}$$ warrant the moving of the factor $$e^{-iEt/\hbar}$$ across the operator $$\hat{Q}$$? Why won't $$\hat{Q}$$ act on the factor even though itself is time-independent?

• Note that you have assumed $\Psi(x,t)=e^{-iE t/\hbar} \psi(x)$, i.e. your state is a state of definite energy. In general $\langle Q\rangle_\psi$ will depend on $t$ if $\Psi(x,t)=\sum_n e^{-iE_n t/\hbar} \psi_n(x)$ with different $E_n$ in the sum. – ZeroTheHero Sep 9 '20 at 13:40

The factor $$e^{-iEt/\hbar}$$ is just a ($$t$$-dependent) complex number, so it passes through $$\hat Q$$ because $$\hat Q$$ is linear. This would be true whether $$\hat Q$$ was time-independent or not. Remember that in this context, operators essentially act on functions of the position coordinate $$x$$, so a function of $$t$$ passes right through them just like a pure number.
The key ingredient here is that $$\hat Q$$ is the same operator at every moment in time. If $$\hat Q(t)\neq \hat Q(0)$$, then you'd need to plug in $$\hat Q(t)$$ in your second line, and your expected value would be different.