# Hamiltonian operating on a function of time

I've seen a few people claiming:

$$\hat{H(t)}[\psi(x)T(t)] = \hat{H(t)}[\psi(x)]T(t)\tag{1}$$

i.e. an explicit function of t is not acted upon by H, even if H itself may be dependent on t.

A more specific example, Griffiths between equation 9.7 and 9.8 (implicitly):

$$\hat{H(t)}[\psi e^{iEt/\hbar}] = \hat{H(t)}[\psi] e^{iEt/\hbar}$$

Is this because t is within an exponential, or is the general statement (1) true? And why?

I feel like it has something to do with time being a parameter not a variable (although I don't fully get this concept either)

• $\hat H=\frac{-1}{2m}\partial_x^2+V(x)$, which has no time-derivatives. Therefore, time-dependent functions pass right through $\hat H$ -- nothing acts on them. – AccidentalFourierTransform Aug 3 '18 at 19:57
• – AccidentalFourierTransform Aug 3 '18 at 19:58

• Hamiltonian(including the potential $V$) can be time-dependent. In this case these time-dependent functions will just multiply the eigenvectors or the wavefunctions. But to change the form of eigenvectors or wavefunctions, you should have a operator(such as a time-derivative or time-integral) in Hamiltonian which does so. I can think of one case in which Hamiltonian affects the time-dependent wavefunctions which is only possible if your potential term contains a derivative or integral with respect to time. But now your Hamiltonian won't be unitary. So this is also not a physical case. – Jitendra Aug 7 '18 at 6:27