Question related to The equivalence between Heisenberg and Schroedinger pictures.

I understand what's explained in the link provided above. My textbook (Breuer and Petruccione's Theory of Open Quantum Systems - but that's similar to what's done in the related Wiki page) does things a little differently, and I'm struggling to understand what's going on.

What they say is:

Schrödinger picture and Heisenberg picture operators are related through the canonical transformation $A_H(t)=U^\dagger(t,t_0)A(t)U(t,t_0)$, where we allow the Schrödinger picture operator $A(t)$ to depend explicitly on time.

They then derive the equation of motion as

$\frac{d}{dt}A_H(t)=i[H_H(t),A_H(t)]+\frac{\partial A_H(t)}{\partial t}$

where $H_H(t)$ is the Hamiltonian in the Heisenberg picture. Explicitely,

$\frac{\partial A_H(t)}{\partial t}=U^\dagger(t,t_0)\frac{\partial A(t)}{\partial t}U(t,t_0)$

Now, what does it mean "we allow $A(t)$ to depend explicitly on time"? Isn't the point of Schrödinger's picture that operators do not depend on time, but states do? What does that "canonical transformation" mean? The operator $A$ on the RHS should not depend on time, how can we make it time-dependant and still be in Schrödinger's picture? Shouldn't the term $\frac{\partial A(t)}{\partial t}$ always be $0$?


1 Answer 1


Normally, in most usual problems, H and also operators DO NOT depend explicitly on time in the Schroedinger picture. In Heisenberg picture they do depend on time according to the equations you gave, plus indeed $\partial A/ \partial t = 0$. Its just that there exist also problems where the external force or magnetic field or something else is time-dependent (already in the Schroedinger picture) then you will need the more general formulas.

  • $\begingroup$ Aren't we exiting Schrödinger's picture when we are using those time-dependant formulas? $\endgroup$ Commented Oct 12, 2018 at 11:28
  • 2
    $\begingroup$ The answer is no, the Hamiltonian and any other operators can be time dependent already in the Schrödinger picture. An operator can even BECOME time independent in Heisenberg picture. The only thing for sure is that the wave function is time independent in Heisenberg picture! $\endgroup$
    – Kphysics
    Commented Oct 13, 2018 at 19:44
  • $\begingroup$ Actually, it is pretty common for the Hamiltonian to depend on time in the Schrodinger picture. A system with some kind of applied drive will have a time dependent part, and in fact we often use the Heisenberg (or "interaction") picture to remove the time dependence of the drive. $\endgroup$
    – DanielSank
    Commented Jan 28, 2020 at 17:16
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    $\begingroup$ See here for example. $\endgroup$
    – DanielSank
    Commented Jan 28, 2020 at 17:17
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    $\begingroup$ @user3461126 When you are talking about time dependency in context of the Heisenberg and Schrodinger pictures you are talking about the time dependency that is caused by time evolution. You are basically asking "I have this time evolution operator $U(t)$, do I give this to the states or the operators?". The time dependency in $\frac{\partial A}{\partial t}$ does not come from $U(t)$ but from some external source. $\endgroup$ Commented Oct 1, 2020 at 14:18

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