Background + Question
In classical mechanics we all know:
$$ \dot U =\frac{dU}{dt} = \sum_i \frac{\partial U}{\partial x_i} \frac{dx_i}{dt} = \sum_i v_i \frac{\partial U}{\partial x_i} = v.\nabla U $$
Hence, $$ \dot U - v. \nabla U = 0 $$
But when would something similar hold in quantum mechanics?
Assuming $U$ is a function of $x$, i.e they both commute. And simplifying to a $1$-d case:
$$ \left\langle \dot U \right\rangle - \left\langle \frac{ \frac{\partial U}{\partial x}}{m} \right\rangle \left\langle p \right\rangle = 0 $$
(Note: In the Heisenberg picture the velocity operator is merely the momentum operator divided by mass)
My Attempt
Using the Heisenberg picture:
$$\implies \left\langle \,[ \frac{\frac{\hat p^2}{2m}, \hat U \,]}{-i \hbar} \right\rangle - \left\langle \frac{ U_x}{m} \right\rangle \left\langle \hat p \right\rangle = 0 $$
$$\implies \left\langle \,[ \frac{ \hat p^2, \hat U \,]}{-i \hbar} \right\rangle - 2 \left\langle U_x \right\rangle \left\langle \hat p \right\rangle = 0 $$
$$\implies \left\langle \hat p^2 \hat U - \hat U \hat p^2 \right\rangle + 2 i \hbar \left\langle U_x \right\rangle \left\langle p \right\rangle = 0 $$
Going to the explicit integrals:
$$\implies -\hbar^2 \int_{-\infty}^{\infty} \bar{\psi} ( \frac{\partial^2 }{\partial x^2} (\hat U \psi) - \hat U \frac{\partial^2 \psi }{\partial x^2} ) dx + 2 i \hbar \left\langle U_x \right\rangle \left\langle p \right\rangle = 0 $$
$$\implies -\hbar^2 \int_{-\infty}^{\infty} \bar{\psi} ( \frac{\partial^2 }{\partial x^2} (\hat U \psi) - \hat U \frac{\partial^2 \psi }{\partial x^2} )dx + 2 i \hbar \left\langle U_x \right\rangle \int_{-\infty}^{\infty} \bar{\psi} (-i \hbar)\frac{\partial }{\partial x} \psi = 0 $$
$$ \implies -\hbar^2 \int_{-\infty}^{\infty} \bar{\psi} ( \frac{\partial^2 }{\partial x^2} (\hat U \psi) - \hat U \frac{\partial^2 \psi }{\partial x^2} )dx + 2 \hbar^2 \left\langle U_x \right\rangle \int_{-\infty}^{\infty} \bar{\psi} \frac{\partial }{\partial x} \psi = 0 $$
$$ \implies \int_{-\infty}^{\infty} \bar{\psi} ( \frac{\partial^2 }{\partial x^2} (\hat U \psi) - \hat U \frac{\partial^2 \psi }{\partial x^2} )dx - 2 \left\langle U_x \right\rangle \int_{-\infty}^{\infty} \bar{\psi} \frac{\partial }{\partial x} \psi = 0 $$
$$ \implies \int_{-\infty}^{\infty} \bar{\psi}( U_{xx} + 2(U_{x} - \left\langle U_x \right\rangle ) \frac{\partial }{\partial x })\psi dx = 0 $$
And now I'm stuck ...