Why can we use time-dependent perturbations when evaluating the S-matrix?

Suppose we have Hamiltonian $H_0 + V$. When working in the interaction picture we may derive the evolution operator of $|\psi_I(0)\rangle$ which is given by $$S(t,t_0) = T\left[\exp \left( -i \int_{t_0}^{t} V_I(t) dt \right) \right]$$ where $T$ is the time ordering symbol and $V_I$ is the interaction Hamiltonian in the interaction picture. This derivation seems to rely on the interaction part being constant in time in the Schrodinger picture, but it is immediately followed up by being used in a scenario where the interaction goes to zero at $t \to \pm \infty$ in the Schrodinger picture. I cannot reconcile this with the derivation. Can anyone help?

• It seems you can derive this result without assuming $V$ is time independent, although not all books do this. – user110503 Jun 29 '18 at 16:16
• can you provide references? Schrodinger operators are defined to be time-independent. – Kosm Jun 30 '18 at 3:00
• @Kosm The book I am using is Many Body Physics by Piers Coleman. On page 88 onwards he derives the S matrix equation assuming that $V$ is constant. Then later in the driven Harmonic oscillator section on page 92, the interaction Hamiltonian is time-dependent. – user110503 Jun 30 '18 at 10:12
• I don't have that book. But the interaction Hamiltonian $V_S$ in the Schrodinger picture is time independent, like any other operator. The relation between $V_S$ and $V_I$ is $V_I=e^{iH_0t}V_Se^{-iH_0t}$. What seems to be the problem? – Kosm Jun 30 '18 at 10:44
• i'm not sure how the Schrodinger picture is described there, but operators are defined to be time-independent is the S. picture, while states are time-dependent. In the Heisenberg picture, it's the opposite. – Kosm Jun 30 '18 at 13:51