Expectation value of time-independent operators

Question

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{equation} \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} \end{equation}

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?

Answer 1

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.