Heisenberg picture

Remarks

Since $\ip{\psi}{U(t)^* A U(t)\psi} = \ip{U(t)\psi}{A U(t)\psi}$, the expectation values agree with those of the Schrödinger picture. Each $A_H(t)$ is again self-adjoint, with the same spectrum as $A$, and $H_H(t) = H$. If $\mathcal{H}$ is finite-dimensional, $A_H$ solves the Heisenberg equation $\frac{d}{dt}A_H(t) = \frac{i}{\hbar}[H, A_H(t)]$, with the commutator $[H, B] = HB - BH$.

Example: on $\C^2$ let $H = \frac{\hbar\omega}{2}\sigma_z$, so $U(t) = \mathrm{diag}(e^{-i\omega t/2}, e^{i\omega t/2})$. A direct computation gives $(\sigma_x)_H(t) = \cos(\omega t)\, \sigma_x - \sin(\omega t)\, \sigma_y$. For $\psi = (1, 1)/\sqrt{2}$ one has $\ip{\psi}{\sigma_x\psi} = 1$ and $\ip{\psi}{\sigma_y\psi} = 0$, so the expectation value is $\cos \omega t$, as in the Schrödinger picture.

Non-example: the rule $A \mapsto U(t) A\, U(t)^*$, with the factors in the other order, gives $\ip{U(-t)\psi}{A U(-t)\psi}$, the expectation value at time $-t$.