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

  1. Apr 29, 2008 #1
    Hey does anybody have an idea of how to prove that

    [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?
     
  2. jcsd
  3. Apr 30, 2008 #2
    Use Ehrenfest's theorem:

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

    If you don't know how to prove it, see any QM text.
     
  4. Apr 30, 2008 #3

    malawi_glenn

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    I would also advice you to use Ehrenfest's theorem.

    Look for a derivation on wikipedia or QM books like Sakurai etc.
     
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