Hey does anybody have an idea of how to prove that(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\frac{d}{dt}\left\langle{XP}\right\rangle= 2\left\langle{T}\right\rangle-\left\langle{x\frac{dV}{dx}}\right\rangle[/tex] for a hamiltonian of form

[tex]H=\frac{P^2}{2M}+V(x)[/tex]

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

[tex]\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}[/tex]

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 [tex]\frac{\partial\psi^*}{\partial t}[/tex] and [tex]\frac{\partial\psi}{\partial t}[/tex] from shcrodingers. Can anybode help?

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

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