# Time derivate of the Expectation value for the product of position and momentum ops

bspoka
Hey does anybody have an idea of how to prove that

$$\frac{d}{dt}\left\langle{XP}\right\rangle= 2\left\langle{T}\right\rangle-\left\langle{x\frac{dV}{dx}}\right\rangle$$ for a hamiltonian of form
$$H=\frac{P^2}{2M}+V(x)$$
where X is the position operator, P is the momentum and T is the kinetic energy. I got through the first chain rule and got
$$\frac{d}{dt}\left\langle{XP}\right\rangle=2\frac{\hbar}{i}\int{x\psi^*\frac{\partial^2\psi}{\partial t\partial x}}-\frac{\hbar}{i}\int{x\frac{\partial\psi^*}{\partial t}\frac{\partial\psi}{\partial x}$$
with negative infinity to positive infinity bounds on the integrals. I tried integrating the first one by parts and it just gets even more messy.

I don't know where to go from here it's kind of starting to look like it should but I have no luck when I start substituting $$\frac{\partial\psi^*}{\partial t}$$ and $$\frac{\partial\psi}{\partial t}$$ from shcrodingers. Can anybode help?

lbrits
Use Ehrenfest's theorem:

$$\frac{d}{dt}\langle A\rangle = \frac{1}{i\hbar}\langle [A,H] \rangle + \left\langle \frac{\partial A}{\partial t}\right\rangle$$

If you don't know how to prove it, see any QM text.