# Heisenberg and Schrödinger pictures - clarification

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 $$0$$?

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.