# Is it possible to study a time-dependent Hamiltonian in Schrödinger picture?

Operators in Heisenberg picture are time-dependent while those in Schrödinger picture are time-independent, and they are related by $$A_H(t)=U^\dagger(t,t_0)A_S(t_0)U(t,t_0)$$ where $U(t,t_0)$ is the unitary evolution operator.

Does it mean it is not possible to work with the Schrödinger picture for time-dependent Hamiltonians? If yes, what does it even mean, in this case, to work in the Schrodinger picture because the operators are time-dependent?

• – Qmechanic Sep 22 '17 at 12:18
• The treatments I have seen start with a time-dependent wave equation and then just omit the parts that average to zero to form the more simple time-independent problem. – DWin Sep 23 '17 at 2:12

Yes, it's perfectly possible. You just pose the Schrödinger equation, $$i\hbar\partial_t |\psi(t)\rangle = \hat H(t)|\psi(t)\rangle,$$ and you solve it. Or what do you mean by "how does that work?"?
• But if the Hamiltonian $\hat{H}$ is dependent of $t$, is it anymore the Schrodinger picture? @EmilioPisanty – SRS Sep 22 '17 at 11:49
• @SRS No, that's an oversimplified view. This is what a picture is: it tells you how operators/states evolve, but it never rules out explicit time dependence of operators. If it helps, choose any hamiltonian you like, and look at how the different pictures treat $\hat A =f(t) \mathbb I$. – Emilio Pisanty Sep 22 '17 at 12:57