$$ H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2 +V_0 x^3 $$ Find $\frac{d\langle x\rangle}{dt}$, $\frac{d\langle p\rangle}{dt}$ and $\frac{d\langle H\rangle}{dt}$.

If this is the hamiltonian given then this should be impossible, as there is no time variable. Am i right? I have trouble evaluating all three.

  • $\begingroup$ Hint: $<x>$ is the average of the operator $x$ on a wave function. In this case, probably it is meant the wave function which solves the time-dependent Schrodinger equation. This is where the time-dependence may come from. $\endgroup$
    – fra_pero
    Commented May 6, 2020 at 9:52
  • $\begingroup$ so i have to include e^(iwt/hbar)? $\endgroup$
    – Hames
    Commented May 6, 2020 at 10:04
  • $\begingroup$ Can you clarify if you have trouble doing all three of the calculations asked or just the last one (from the question title it sounds like just the last one?) $\endgroup$
    – jacob1729
    Commented May 6, 2020 at 10:06
  • $\begingroup$ Hi. Welcome to the website. In the future you should use MathJax for for formulas and not attach them as a picture. $\endgroup$
    – RedGiant
    Commented May 6, 2020 at 10:57
  • 1
    $\begingroup$ are you familiar with the Heisenberg picture? $\endgroup$
    – user245141
    Commented May 6, 2020 at 11:35

1 Answer 1


Assuming that we work in the Schrödinger picture: the time evolution of the system is encoded in its wave function, as determined by the time-dependent Schrödinger equation: $$i\hbar\partial_t\Psi(x, t) = \hat{H}\Psi(x,t).$$ The time-dependent expectation values are then given by $$\langle x(t)\rangle = \langle \Psi(t)|\hat{x}|\Psi(t) = \int dx\Psi(x,t)^*\hat{x}\Psi(x,t), etc.$$ Generally speaking, these are time-dependent and have non-zero time derivatives.

However, if the system is in a stationary state, i.e. $$\hat{H}\psi(x) = E\psi(x),$$ then the time dependence of the wave function is trivial $$\Psi(x,t)=\psi(x)e^{-iEt/\hbar},$$ and the averages in question are time-independent.

Finally, let us note that $\frac{d\langle x(t)\rangle}{dt}$ is not the same as $\left\langle\frac{d x(t)}{dt}\right\rangle$ - the former is the derivative of an expectation value, whereas the latter is the expectation of the derivative, where the operator of the derivative is defined by $$\hat{\dot{x}} = \frac{d \hat{x}(t)}{dt} = \frac{1}{i\hbar}\left[\hat{x}, \hat{H}\right].$$ Note that, as we are still in the Schrödinger picture, this is definition of the operator of the derivative rather than a time derivative of the operator. In the Heisenberg and interaction pictures the latter is given by the same equation, but with time-dependent operators.


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