It is the root of quantum mechanics that Heisenberg picture and Schrödinger picture are equivalent?
In most textbooks and wikipedia, the equivalence is proved with a time-independent Hamiltonian. However, some literature uses Heisenberg equation with time-dependent Hamiltonian.
$$i\hbar \frac{dA}{dt}~=~[A(t),H(t)]+i\hbar \frac{\partial A}{\partial t}.$$
So, does Heisenberg equation work with time-dependent Hamiltonian? If so, any proof?
The short answer is that the time evolution operator for a time-dependent Hamiltonian has two times, the initial and final $U (t,s) $.
Therefore defining $A (t,s)=U (s,t)A U (t,s) $, the Heisenberg equation is obtained differentiating with respect to $t $. Schroedinger equation is obtained differentiating $U (t,s) \psi (s)$ instead. The two are equivalent in the usual sense, i.e. they both give the same time-evolved transition amplitudes.
OK, I figured it out myself. Heisenberg equation still holds for time-dependent Hamiltonians:
$$i\hbar \frac{dA}{dt}~=~[A(t),H(t)]+i\hbar \frac{\partial A}{\partial t}.$$
However, now $H(t)$ is defined as $U^\dagger(t)H_S(t)U(t)$.
