When is it appropriate to use $\hat U$, the unitary time evolution operator? For example, say I had a system in a certain potential that is changed to a different one at time $t = 0$. Would it be valid to operate $\hat U$ on the new wavefunction (at $t = 0$) to calculate the wavefunction at later time $t > 0$?
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2 Answers
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The answer to your question "Would it be valid to..." is "yes."
Whenever your Hamiltonian is time-independent over a time interval beginning at some reference time $t_0$, then Schrodinger's equation takes the form
$$|\psi(t)\rangle= \hat{U}\,|\psi(t_0)\rangle $$ where
$$\hat{U} = e^{-i H t/\hbar} $$
is the unitary time evolution operator. It doesn't matter how the wavefunction arrived at its state at $t = t_0$.
answered Dec 13, 2013 at 10:53
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There is some restrictions on previous answer. Hamiltonian must be time-independent to use $U = e^{-iHt}$ rule. For time-dependent hamiltonian, time-evolution in form $\psi(t) = U(t,t_0) \psi(t_0)$ takes U in more general form $U(t,t_0) = \mathcal{T}\,\exp(-i \int _{t_0} ^ t H(\tau) d\tau)$. Of course, if your potential is constant over $t$ to $t_0$ period, you can use simple form.
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$\begingroup$ Hi Vladimir, I just added the time-ordering operator to your expression as it is needed in general when $H(\tau)$ doesn't commute with itself at different values of $\tau$: please check carefully. PS I'm a bit scared of $\mathcal{T}$, I prefer to think of the full Peano-Baker series, but $\mathcal{T}$ is in general use and seems to be less fearsome to physicist in general than to me. $\endgroup$ Commented Dec 15, 2013 at 10:18
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$\begingroup$ Thanks, WetSavannaAnimal aka Rod Vance, I've just forgot to write it. But I fear that after that even more explaination is needed. You are right, I'ts crucial sometimes, that $H(t)$ doesn't commute with $H(t^\prime)$ in general. $\endgroup$– VladimirCommented Dec 15, 2013 at 10:32