Intuition for why d<p>/dt = -dV(<x>)/dx

[EquationOfMotion]

Is there any good physical or graphical intuition for why $\frac{d \langle p \rangle}{dt} = -\frac{\partial V(\langle x \rangle)}{\partial x}$? Classically this is apparently true.

Thanks.

 

[PeroK]

Is there any good physical or graphical intuition for why $\frac{d \langle p \rangle}{dt} = -\frac{\partial V(\langle x \rangle)}{\partial x}$? Classically this is apparently true.

Thanks.

In classical physics potential is defined so that $F = -\frac{\partial V}{\partial x}$.

Your equation is, however, not correct. It should be:

$\frac{d \langle p \rangle}{dt} = -\langle \frac{\partial V(x)}{\partial x} \rangle$

 

[EquationOfMotion]

In classical physics potential is defined so that $F = -\frac{\partial V}{\partial x}$.

Your equation is, however, not correct. It should be:

$\frac{d \langle p \rangle}{dt} = -\langle \frac{\partial V(x)}{\partial x} \rangle$

I think $\frac{d \langle p \rangle}{dt} = -\langle \frac{\partial V(x)}{\partial x} \rangle$ is the Ehrenfest theorem. The Wikipedia page however notes that were quantum expectation values to be consistent with Newtonian mechanics, we'd have $F = -\frac{\partial V}{\partial x}$. Unless I'm misunderstanding something.