Derivative of an Expectation value/Ehrenfest theorem

[OGrowli]

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

For a three dimensional wave packet

Homework 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.

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.

 

[mark726]

Your solution is correct up to the point where you have the term (ihbar/m). This term arises because you have not taken into account the time dependence of the wavefunction. The expectation value <x^2> is a function of time, so when you take the time derivative, you need to use the chain rule to account for the time dependence of the wavefunction.

To get rid of this extra term, you can use the Ehrenfest theorem (equation 2) to evaluate the time derivative of <x^2>. This theorem states that the time derivative of the expectation value of an operator is equal to the expectation value of the commutator of that operator with the Hamiltonian, plus the expectation value of the partial time derivative of the operator. In this case, the operator is x^2, so we have:

(d/dt)<x^2> = <[x^2,H]> + <(partial)(d/dt)x^2>

The commutator [x^2,H] can be evaluated using the commutation relations for position and momentum operators. After some algebra, you should get:

[x^2,H] = 2ihbarx(p_x) + 2ihbar(p_x)x

Substituting this into the above equation and using the fact that the expectation value of the Hamiltonian is <H> = <E>, we get:

(d/dt)<x^2> = 2ihbar<x(p_x)> + 2ihbar<(p_x)x> + (partial)(d/dt)<x^2>

Rearranging this equation, we get:

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

And this is the desired result. So, to summarize, the extra term (ihbar/m) in your solution arises because you have not taken into account the time dependence of the wavefunction. By using the Ehrenfest theorem, we can properly account for this time dependence and get the correct result.