# Heisenberg representation: why position operator has no explicit time dependance?

In Heisenberg representation 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 representation. 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 representation?

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

• Related. Inspect the Schroedinger representation equivalent. – Cosmas Zachos Oct 8 '20 at 14:40

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