Ehrenfest theorem and Hamiltonian operator

[cleggy]

Homework Statement

Use the generalized Ehrenfest theorem to show that any free particle with the one-dimensional Hamiltonian operator

H= p^2/2m obeys

d^2<x^2> / dt^2 = (2/m)<p^2>,

Homework Equations

The commutation relation xp - px = ih(bar)

The Attempt at a Solution

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


H = p^2/2m + V(x)

then [x^2,H] = x^2[(p^2/2m) + V(x)) - ((p^2/2m) + V(x))x^2 = (1/2m)[x^2,p^2]


[x^2,p^2] = xxpp - ppxx = 2ih(bar)(xp + px)

I'm stuck at this point?

am i on the right track?

the < > brackets represent the expectation value

 

[Cyosis]

You're not using the generalized Ehrenfest theorem from the looks of it. T

The theorem you want to use is:
<br /> \frac{d \langle A \rangle}{dt}=\frac{1}{i \hbar}\langle [A,H] \rangle+\left\langle \frac{dA}{dt} \right\rangle<br />

With A an operator in the Schrodinger picture. Now use A=x and A=p.

 

[cleggy]

According to my textbook the generalized Ehrenfest theorem is

what you have put but without the derivative term of A at the end

 

[Cyosis]

Don't forget the tex bracket. Well you could write it like that, because the last term in my expression is usually zero for operators in the Schrodinger picture.

It wasn't clear from your original post that you were actually using that one however.

That said can you evaluate the equation for x and p?

 

[cleggy]

Which x and p ?

the term (xp + px) ?

I really thought i was on the right track.

I'm not sure where to go now. Pointers?

 

[cleggy]

If

[x^2,p^2] = xxpp - ppxx = 2ih(bar)(xp + px)

Then how do I find the term (xp + px) ?

 

[Cyosis]

Write the commutator as [x^2,pp] and use the commutator identity [A,BC]=[A,B]C+B[A,C] then do the same for the x^2.

 

[cleggy]

I'm still not following

I'm given the commutation relations in my text as


[x^2,p^2] = xxpp - ppxx = 2ih(bar)(xp + px)

[xp,p^2] = xxpp - ppxx = 2ih(bar)(p^2)

[px,p^2] = xxpp - ppxx = 2ih(bar)(p^2)

 

[Cyosis]

You can let a test function work on it. Evaluate (xp+px)f(x).

 

[cleggy]

This makes no sense to me. I think I'm going to call it a day. Thanks for all your help Cyosis

 

[Cyosis]

Do you know how you can show that [x,p]=i \hbar in the first place?