# Ehrenfest theorem

1. Jun 11, 2009

### cleggy

1. The problem statement, all variables and given/known data

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>,

2. Relevant equations

The commutation relation xp - px = ih(bar)

3. 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

2. Jun 11, 2009

### Cyosis

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

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

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

3. Jun 12, 2009

### cleggy

According to my text book the generalized Ehrenfest theorem is

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

4. Jun 12, 2009

### 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?

5. Jun 12, 2009

### 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?

6. Jun 12, 2009

### cleggy

If

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

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

7. Jun 12, 2009

### 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.

8. Jun 12, 2009

### 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)

9. Jun 12, 2009

### Cyosis

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

10. Jun 12, 2009

### cleggy

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

11. Jun 12, 2009

### Cyosis

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