I am studying quantum mechanics and I encountered the famous Ehrenfest Theorem, which states that given an observable $A$, its expectation value time evolution is governed by $\partial_t\langle A\rangle=\frac{1}{ih}\langle[A,H]\rangle$. This is famously analogous through canonical quantization of Poisson Brackets to the fact that in Hamiltonian mechanics one has $\frac{dF}{dt}=\{F,H\}$.

One has that the Hamilton equations have the following analogous form:$$\partial_t\langle x\rangle=\frac{1}{ih}\langle[x,H]\rangle=\frac{1}{m}\langle p_j\rangle$$ $$\partial_t\langle p\rangle=\frac{1}{ih}\langle[p,H]\rangle=-\langle\partial_x U\rangle$$ Whereas the first is completely analogous to the first Hamilton eq., to have the complete analogy one needs to have $$\langle\partial_x U\rangle=\partial_x\langle U\rangle$$ and this happens for at most quadratic potentials. But this must mean that at the classical level $\langle x^2\rangle=\langle x\rangle\langle x\rangle$, whereas $\langle x^2\rangle\neq\langle x\rangle\langle x\rangle$ at the quantum level. In some sense then at quantum level we have a kind of correlation.

What justifies this asymmetry in the two equations? Is there anything to this idea of correlation?

  • $\begingroup$ How are you defining $\partial_x \langle U\rangle$? $\langle U \rangle = \int dx \;\psi^*(x) U(x) \psi(x)$, so $x$ is an internal integration varaible. There is no external $x$ variable to differentiate with respect to. $\endgroup$ – By Symmetry Jun 12 '19 at 11:32
  • $\begingroup$ @BySymmetry you're correct, I mean $\partial_{<x>} U(<x>)$ $\endgroup$ – Francesco Bilotta Jun 12 '19 at 13:53

FWIW, the asymmetry is apparently directly tied to the assumption that the Hamiltonian $\hat{H}$ is quadratic in momenta $\hat{p}_i$ but not necessarily quadratic in positions $\hat{x}^i$. Generically, the rule $\langle f(\hat{A})\rangle= f(\langle \hat{A}\rangle)$ is only expected to hold for affine functions $f$.

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  • $\begingroup$ Ok, this is indeed true, but affinity is only sufficient condition for the rule you stated. What I mean is, classically we do have that non affine functions' expectation factorizes. $\endgroup$ – Francesco Bilotta Jun 11 '19 at 20:55

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