# Ehrenfest theorem and correlation among observables at the quantum scale

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?

• 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. – By Symmetry Jun 12 '19 at 11:32
• @BySymmetry you're correct, I mean $\partial_{<x>} U(<x>)$ – 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 $$f$$.