Schrödinger picture

Remarks

For $\psi \in D(H)$, clause 1 gives the solution of the Schrödinger equation $i\hbar\, \psi'(t) = H\psi(t)$ with $\psi(0) = \psi$. For a density operator and bounded $H$, it gives the von Neumann equation $i\hbar\, \rho'(t) = [H, \rho(t)]$.

Example: on $\C^2$ let $H = \frac{\hbar\omega}{2}\sigma_z$ with $\omega > 0$, and $\psi = (1, 1)/\sqrt{2}$. Then $\psi(t) = (e^{-i\omega t/2}, e^{i\omega t/2})/\sqrt{2}$, and with $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, $\ip{\psi(t)}{\sigma_x\psi(t)} = \operatorname{Re}(e^{i\omega t}) = \cos \omega t$. The spin precesses about the vertical axis.

Non-example: moving both the state by $U(t)$ and the observable to $U(t)^* A U(t)$ is not a valid description. It gives $\ip{\psi}{U(t)^* U(t)^* A U(t) U(t)\psi} = \ip{\psi(2t)}{A\psi(2t)}$, the expectation value at time $2t$ instead of $t$.