In the Heisenberg picture, I can define the velocity Operator $\hat{V}$ as the operator which satisfies $\hat{V}(t) = \frac{\partial \hat{x}}{\partial t}(t)$ for all $t$. The Heisenberg equation then tells me that $ \hat{V}(t) = \frac{i}{\hbar}[\hat{H}(t), \hat{x}(t)]$. If I want to calculate the time evolution of the velocity operator, a straight forward approach yields:

$$\dot{\hat{V}} = \frac{i}{ \hbar}(\dot{\hat{H}}\hat{x}-\hat{x}\dot{\hat{H}} + \hat{H}\dot{\hat{x}} -\dot{\hat{x}}\hat{H} ) = \frac{i}{ \hbar}(\dot{\hat{H}}\hat{x}-\hat{x}\dot{\hat{H}} +[\hat{H}(t), \hat{V}(t)]) $$

What I find strange about that is that the time evolution of $V$ is no longer described by the Heisenberg-equation, but instead by some more complicated form, at least if $\dot{\hat{H}}$ doesn't commute with $\hat{x}$. I see problems arising, since I now can't in general make statements on the time evolution of functions $\hat{f}(\hat{x}, \hat{V})$, like for example the lagrangian $\hat{L}(\hat{x}, \hat{V}, t)$. Since the time evolution of $\hat{V}$ isn't given by the Heisenberg equation, the time evolution of for example $\hat{p}(t) = \frac{\partial L}{\partial t}(\hat{x}(t), \hat{v}(t), t)$ also isn't necessarily given anymore by the Heisenberg equation.

I know that this is a unusual approach to the topic, since I assume observables to be functions of $\hat{x}$ and $\hat{v}$, and not functions of $\hat{x}$ and $\hat{p}$, but maybe somebody can still tell me where the mistakes are that I do in this derivations, or if it is an additional assumption that $\dot{\hat{H}}$ and $\hat{x}$ do always commute. If I did a mistake, what should I do different to still approach the topic from the lagrangian point of view?

  • 1
    $\begingroup$ Wouldnt the commutator in the first line be $[\hat x(t), \hat H(t)]$? $\endgroup$
    – sbp
    Sep 24, 2017 at 19:29
  • $\begingroup$ It would. The $i$ in the coefficient is misplaced: $\frac{d}{dt} \hat{A}(t) = \frac{i}{\hbar} [ \hat{H}(t), \hat{A}(t) ]$ $\endgroup$
    – Darkseid
    Sep 24, 2017 at 19:38
  • $\begingroup$ I see. I will edit the question accordingly $\endgroup$ Sep 24, 2017 at 20:13

1 Answer 1


It all comes down to the time derivative of $\hat{H}$. Regardless of being time dependent or independent in Schroedinger picture, one would have to use Heisenberg picture hamiltonian in the equation of motion. Heisenberg picture hamiltonian has vanishing time derivative, so the time evolution of $\hat{V}$ won't have $\dot{\hat{H}}$ in it.

See this thread for details on the time-dependent Hamiltonian and Heisenberg picture.

Updated answer

In Heisenberg picture equation of motion is:

$$\frac{d}{dt} \hat{A}_H = \frac{i}{\hbar} [ \hat{H}_H, \hat{A}_H ] + \left( \frac{\partial{\hat{A_S}}}{\partial{t}} \right)_H$$

While $\hat{x}$ is a time-independent operator in Schroedinger picture, the velocity may have non-zero partial derivative $\frac{\partial}{\partial{t}}$. The extra term that you get by doing things manually is going to be the same as the last term in the full equation of motion.

  • $\begingroup$ I don't see why the Heisenberg Hamiltonian should lose it's time dependence, to be honest. $\endgroup$ Sep 24, 2017 at 20:45
  • $\begingroup$ You're right. I'll correct the answer in a few minutes. $\endgroup$
    – Darkseid
    Sep 24, 2017 at 21:10
  • $\begingroup$ @Quantumwhisp answer is updated now $\endgroup$
    – Darkseid
    Sep 24, 2017 at 21:19
  • $\begingroup$ is it true that in those cases, the time evolution of another operator which is a function of $\hat{x}$ and $\hat{V}$, is not necesarilly given anymore by the Heisenberg equation? What about the special case of $\hat{p}$? $\endgroup$ Sep 24, 2017 at 21:50
  • $\begingroup$ Heisenberg equation holds for any observable, because it follows from the application of the product rule ($(ab)' = a'b + ab'$). See Wikipedia for derivation. $\endgroup$
    – Darkseid
    Sep 24, 2017 at 22:17

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