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By Ehrenfest’s theorem one finds that the quantum operators $\hat{q}$ and $\hat{p}$ satisfy $$\frac{d \langle\hat{q}\rangle}{dt} = \frac{\langle\hat{p}\rangle}{m}$$ and $$\frac{d \langle\hat{p}\rangle}{dt} = \langle F(\hat{q})\rangle$$ Classically, by applying Liouvilles theorem the canonical coordinates $p$ and $q$ satisfy $$\frac{d \langle q \rangle}{dt} = \frac{\langle p \rangle}{m}$$ and $$\frac{d \langle p \rangle}{dt} = \langle F(q)\rangle$$

It seems to me that the relations yield identical predictions for the values of $q$ and $p$ if the potential is the harmonic oscillator, does that seem correct? Is this always the case?

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  • $\begingroup$ The expectation value is in general, not the same as the classical value, however. $\endgroup$
    – UKH
    Commented Jan 27, 2017 at 2:19

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This is only true for potentials up to order two in the variables!

Thus it is correct for the harmonic oscillator.

better explained here, unfortunately only in german ;-)

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  • $\begingroup$ How would it fail for say the potential $V(x)=x^3$? $\endgroup$
    – A15234B
    Commented Jan 27, 2017 at 16:03
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    $\begingroup$ if you have a quadratic potential the force $F$ is a linear function of position. by linearity $ F(\langle x \rangle) = \langle F(x) \rangle$ which yields the classical analogon of EOM with $ x \mapsto \langle x \rangle$. $\endgroup$ Commented Jan 27, 2017 at 17:44

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