Interaction picture

Remarks

Since $U_0(t)\psi_I(t) = U(t)\psi$, one has $\ip{\psi_I(t)}{A_I(t)\psi_I(t)} = \ip{U(t)\psi}{A U(t)\psi}$, so the expectation values are those of the Schrödinger picture. For $\psi \in D(H) = D(H_0)$, the state vector solves $$i\hbar\, \frac{d}{dt}\psi_I(t) = V_I(t)\, \psi_I(t), \qquad V_I(t) = U_0(t)^* V\, U_0(t).$$ Indeed, differentiating $U_0(t)^* U(t)\psi$ gives $\frac{i}{\hbar} U_0(t)^*(H_0 - H)U(t)\psi = -\frac{i}{\hbar} V_I(t) \psi_I(t)$. Iterating this equation in integral form gives the Dyson series.

Example: if $V = 0$, then $\psi_I(t) = \psi$ is constant and $A_I(t)$ is the observable of the Heisenberg picture. If $H_0 = 0$, then $\psi_I(t) = U(t)\psi$ and $A_I(t) = A$, the Schrödinger picture.

Non-example: the rule $\psi_I(t) = U_0(t) U(t)\psi$, with $U_0(t)$ in place of $U_0(t)^*$, is not an interaction picture. For $V = 0$ it gives $\psi_I(t) = U(2t)\psi$, and then $\ip{\psi_I(t)}{A_I(t)\psi_I(t)} = \ip{U(3t)\psi}{A\, U(3t)\psi}$, the expectation value at time $3t$ instead of $t$.