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?

  • $\begingroup$ Related: physics.stackexchange.com/q/103503/2451 $\endgroup$
    – Qmechanic
    Commented Sep 22, 2017 at 12:18
  • $\begingroup$ 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. $\endgroup$
    – DWin
    Commented Sep 23, 2017 at 2:12

1 Answer 1


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?"?

  • 1
    $\begingroup$ But if the Hamiltonian $\hat{H}$ is dependent of $t$, is it anymore the Schrodinger picture? @EmilioPisanty $\endgroup$
    – SRS
    Commented Sep 22, 2017 at 11:49
  • $\begingroup$ Well, time-dependent Hamiltonians within the context of the Schroedinger equation mean that there's an external interaction of the system which manifests itself by a time-dependent potential term $\endgroup$
    – DanielC
    Commented Sep 22, 2017 at 11:54
  • $\begingroup$ @DanielC How does your response answer my question about the Schrodinger picture? $\endgroup$
    – SRS
    Commented Sep 22, 2017 at 11:56
  • $\begingroup$ @SRS Yes, of course it's still the Schrödinger picture. Why would you think otherwise? $\endgroup$ Commented Sep 22, 2017 at 11:56
  • 5
    $\begingroup$ @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 $\hat A =f(t) \mathbb I$. $\endgroup$ Commented Sep 22, 2017 at 12:57

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