Phase in time evolution operator for time-dependent Hamiltonian [duplicate]

Question

In Quantum Mechanics, a state vector $|\psi\rangle$ will evolve in time according to $$|\psi(t)\rangle=e^{-\frac{i}{\hbar}\hat H t}|\psi(0)\rangle$$ Imagine we have a system such that, for a short period of time $T$, the Hamiltonian increases by a constant and then returns to normal, such that $$\hat H=\hat H_0+ \begin{cases} 0 & \text{($t\lt0,\, t\gt T$)}\\ A & \text{($0\leq t\leq T$)}\\ \end{cases} $$ At $t=T$ we will have $$|\psi(T)\rangle=e^{-\frac{i}{\hbar}AT}e^{-\frac{i}{\hbar}\hat H_0T}|\psi(0)\rangle$$ Now, following the first equation, since after $t=T$ there is no $A$, it should just become $$|\psi(t)\rangle=e^{-\frac{i}{\hbar}\hat H_0 t}|\psi(0)\rangle$$ But this seems strange, it's as if that period of interaction with whatever caused the extra energy had no effect on the particle whatsoever. I think it makes more sense to apply the time evolution operator separately and obtain $$|\psi(t)\rangle=\hat U(t-T)|\psi(T)\rangle=e^{-\frac{i}{\hbar}\hat H_0(t-T)}e^{-\frac{i}{\hbar}AT}e^{-\frac{i}{\hbar}\hat H_0T}|\psi(0)\rangle=e^{-\frac{i}{\hbar}\hat H_0t}e^{-\frac{i}{\hbar}AT}|\psi(0)\rangle$$ Is my idea wrong or is the time evolutikon operator different in this case? If so, the what would be the case for a time-dependent Hamiltonian?

Answer 1

Your first equation does not hold for a time dependent Hamiltonian.