Heisenberg picture: why position operator has no explicit time dependence?

Question

In Heisenberg picture we have: $$\frac{d}{dt}A_H(t)=\frac{1}{i\hbar}\left[A_H(t),H_H(t)\right]+\frac{\partial A_H(t)}{\partial t} \ \ \ \ \ \ \ (1)$$ where I use the subscript $H$ to make as explicit as possible that we are working in Heisenberg picture. If we apply $(1)$ to the position operator $\hat{q}_H$ we get: $$\frac{d}{dt}\hat{q}_H=\frac{1}{i\hbar}\left[\hat{q}_H,H_H(t)\right]$$ or at least this is what my lecture notes said. This derivation uses the fact that the position operator has no explicit time dependence: $$\frac{\partial \hat{q}_H}{\partial t}=0$$ I do not understand why this has to be true. How do we know that the partial derivative of $\hat{q}_H$ has to be zero in Heisenberg picture?

Since in the Schrödinger picture the states can explicitly evolve with time, changing their "relation" with the position operator, I would expect the same in the Heisenberg picture but mirrored onto the operators.

Answer 1

It literally means that this operator doesn't have an explicit time dependence, as opposed, for example to something like $$ V(x, t) = \hat{x} \cos(\omega t). $$