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.

  • 1
    $\begingroup$ Related. Inspect the Schroedinger representation equivalent. $\endgroup$ – 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 $$ V(x, t) = \hat{x} \cos(\omega t). $$

  • $\begingroup$ I get this; maybe I am just confused, but how do we check that this is true for the position operator in Heisenberg representation? $\endgroup$ – Noumeno Oct 8 '20 at 14:34
  • $\begingroup$ I mean: for your operator V we can clearly see, because we have the explicit form of it, that it has an explicit time dependence in the cosine. But how do we check for the position operator? $\endgroup$ – Noumeno Oct 8 '20 at 14:37
  • 1
    $\begingroup$ Position operator doesn't have time dependence in its explicit form. We know that when we convert to the Heisenberg picture, all its time dependence comes from the Hamiltonian. $\endgroup$ – Roger Vadim Oct 8 '20 at 14:39
  • $\begingroup$ Ok, and we know that it has no time dependence from the Schrödinger representation of it right? And so when we switch to the Heisenberg representation we know that all the time dependence comes from the Hamiltonian operator. I think I get it know, It was really obvious now that I think about it. Do you mind mathematically showing what you just said in your answer? Your current answer does not answer my question precisely I think. $\endgroup$ – Noumeno Oct 8 '20 at 14:47
  • $\begingroup$ Note that this is not a quantum mechanical feature - the Hamiltonian equations of motion in classical mechanics make the same distinction between the explicit time dependence and the one due to the Hamiltonian. $\endgroup$ – Roger Vadim Oct 8 '20 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.