# Does the Hamiltonian always commute with the Time Evolution Operator?

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?

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 $$U^\dagger U = 1$$