Derivative of an Expectation value/Ehrenfest theorem

1. Sep 28, 2010

OGrowli

Show that (d/dt)<x^2>=(1/m)(<x(p_x)>+<(p_x)x>)

For a three dimensional wave packet

2. Relevant equations

1. <O>=Int_v(d^3r)(psi*Opsi), where O is some operator

Ehrenfest Theorem:

2. ihbar(d/dt)<O>=<[O,H]>+<(partial)(d/dt)O>, H is a hamiltonian.

3. The attempt at a solution

I evaluted (d/dt)<x^2> using both number one and 2. With each I come to the expression:

(-ihbar/2m)int_v(d^3r)[psi*[del^2(x^2psi)-(x^2)(del^2(psi))], let psi'=(partial)(d/dx)psi

=(-ihbar/2m)int_v(d^3r)[psi*[(partial)(d/dx)[2xpsi+x^2psi']-x^2psi'']],the y and z parts of the laplacian in the first term cancel with those of the second.

=(-ihbar/2m)int_v(d^3r)[psi*(2psi+2xpsi'+2xpsi'+x^2psi''-x^2psi'')]

take the 2 out, bring the -ihbar in and you have

=(1/m)int_v(d^3r)[psi*(-ihbar)psi+psi*x(-ihbar(partial)(d/dx))psi+
psi*(-ihbar(partial)(d/dx))xpsi

=(1/m)(<x(p_x)>+<(p_x)x>)-(ihbar/m)

My question is, what does this extra term mean? how do I get rid of it? Is is a result of error on my part? I assumed that <(partial)(d/dt)(x^2)>=0 since there was no term to account for this in the integral method. Thanks a lot for your help.