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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)

RelativisticDolphin
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1 Answers1

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Hamiltonian operator only have derivatives of space and that too they are partial derivatives, so they don't affect any time-dependent function. Hence your exponential acts as a constant for the Hamiltonian operator.

Jitendra
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  • Hi, but is it not possible that V is dependent on t such that it does affect time-dependent functions? – RelativisticDolphin Aug 06 '18 at 21:04
  • 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 07 '18 at 06:27