Commuting the time evolution operator Given the time evolution operator $U(t, t_0)$, I don't understand why it is true that for a time-independent operator Q,
$$[Q, U(t, t_{0})] = 0 \Leftrightarrow [Q, H(t)] = 0 $$
where H is the Hamiltonian.
 A: The unitary time evolution operator is by construction
$$
U(t, t_0) = \text{e}^{-\frac{i}{\hbar} H (t - t_0)}
$$
which is understood as the series expansion of the exponential function!
Therefore one observes that
$$
\text{e}^{-\frac{i}{\hbar} H (t - t_0)} = \sum_{k = 0}^{\infty} \frac{(-\frac{i}{\hbar} H (t - t_0))^k}{k!} = 1 + -\frac{i}{\hbar} H (t - t_0) + ... (\text{powers of } H)
$$
So if you look at the series expansion you can clearly see that if $H$ commutes with any operator then $U$ commutes with this operator too, and vice versa! Therefore
$$[Q, U(t, t_{0})] = 0 \Leftrightarrow [Q, H] = 0 $$
A: You can derive the Heisenberg equation from the Schrödinger equation.
$$\frac{dQ_H}{dt}= \frac{d(U^\dagger Q_S U)}{dt}$$
Then after you write it complete and define $H_H=U^\dagger H_S U$ yo see the following:
$$\frac{dQ_H}{dt}=-i[Q_H,H_H]$$
And finally you can say $U$ and $H$ always commute for time dependent $H$ and at the end you can write your result.
A: One could use the expression for the evolution operator:
$$U(t,t_0)= e^{-H(t-t_0)}.$$
