# Heisenberg equation with time-dependent Hamiltonian

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?

• – Qmechanic Nov 17 '16 at 6:49
• A downvote, why? I think it's a proper question. – Chong Wang Nov 19 '16 at 8:53

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.
$$i\hbar \frac{dA}{dt}~=~[A(t),H(t)]+i\hbar \frac{\partial A}{\partial t}.$$
However, now $H(t)$ is defined as $U^\dagger(t)H_S(t)U(t)$.