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The time evolution operator $U(t, t_0)$ is given as the solution of the equation

$$ i\hbar \dfrac{\text{d}}{\text{d}t} U(t, t_0) = HU(t, t_0)$$

whether or not the system is conservative. When the system is conservative, the time evolution operator is given by

$$ U(t, t_0) = \text{e}^{-i(t-t_0)H/\hbar} \implies [U(t, t_0), H] = 0.$$

What can be said about the general case, in which $H$ depends on time explicitly?

Specifically:

  1. Do $U(t, t_0)$ and $H$ still commute?
  2. Is $U(t, t_0)$ still unitary?
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1 Answer 1

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What can be said about the general case, in which $H$ depends on time explicitly?

Specifically:

  1. Do $U(t, t_0)$ and $H$ still commute?

No, not in general.

In general, you can write $U$ as a time-ordered series : $$ U(t,t_0) = \mathcal{T}(e^{-i\int_{t_0}^t dt' H(t')}) $$ $$ \equiv 1 -i\int_{t_0}^t dt' H(t') -\int_{t_0}^t dt'\int_{t_0}^{t'} dt'' H(t')H(t'') + \ldots\;, $$ where I'm setting $\hbar =1$.

So, even at first order the commutator is seen to not necessarily be zero: $$ [H(t),U(t,t_0)] \approx -i\int_{t_0}^t dt' [H(t), H(t')]\;, $$ since $H(t)$ and $H(t')$ don't necessarily commute when $H$ depends on time.

  1. Is $U(t, t_0)$ still unitary?

Yes. We still demand that the wavefunction remains normalized throughout time. So, we still have $U$ being unitary. This is because $$ 1 = \langle\psi(t)|\psi(t)\rangle = \langle\psi(t_0)|U^\dagger U|\psi(t_0)\rangle \;, $$ but $t_0$ is arbitrary, so $$ U^\dagger U = 1 $$

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