# Ehrenfest's Theorem

1. Jul 18, 2009

### roeb

1. The problem statement, all variables and given/known data
Show that $$\frac{d<p>}{dt} = < - \frac{\partial V}{\partial x}>$$

2. Relevant equations

3. The attempt at a solution

I am trying to repeat the derivation that griffiths gives for deriving <p>, but it doesn't seem to give me anything that would indicate this proof is correct.

<p> = $$\int \psi^* (\hbar \frac{\partial}{\partial x} ) \psi dx$$

d<p>/dt = $$\int -\hbar i \frac{\partial }{\partial x} \frac{\partial }{\partial t} | \psi |^2 dx$$

= $$\int -\hbar i \frac{\partial }{\partial x} ( \psi^* \frac{\partial \psi}{\partial x} - \psi \frac{ \partial \psi^*}{\partial x} )dx = -2 \hbar i \int \frac{\partial}{\partial x} ( \psi^* \frac{\partial \psi }{\partial x} ) dx$$

I was thinking about going about it the other way and calculating < -dV/dx >, but unfortunately I realized I can't really think of a way to do that directly with the schrodinger equation.

Last edited: Jul 18, 2009
2. Jul 18, 2009

### guguma

On your second step, you cannot carry $$\psi^*$$ over from the operators and multiply it with $$\psi$$

What you really need to do is replace $$\frac{\partial}{\partial t}\psi$$ with the hamiltonian according to the schrodinger equation. You can carry the operators over eachother that is correct. But do not forget that when you are taking the the time derivative inside, there is a product rule out there.

Then you will need to do several integrations by parts to be left with $$< - \frac{\partial V}{\partial x}>$$ take into account the fact that the boundary terms when integrating by parts will be zero due to our functions being normalizable.

P.S.

I attached my solution as a pdf if you want to look at it

Also when you are done with this problem, you may want to show that

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

with pretty much the same method you used for your original question. The equality above is the actual form of Ehrenfest's Theorem which is more general. I am not being an *** by saying this, I believe that it is good practice and this is merely a suggestion. I am telling this to avoid future misunderstandings.

Good Luck.

Last edited by a moderator: Jul 19, 2009
3. Jul 19, 2009

### roeb

here is what I have so far, I feel like I am getting close:

$$<p> = -i \hbar \int \Psi^* \frac{\partial \Psi}{\partial x}dx$$
$$\frac{d<p>}{dt} = -i \hbar \int \frac{\partial \Psi^*}{\partial t}\frac{\partial \Psi}{\partial x} + \Psi^* \frac{\partial^2\Psi}{\partial x \partial t} dx$$
$$= -i \hbar \int \frac{\partial \Psi^*}{\partial t}\frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \frac{\partial \Psi}{\partial t} dx$$
Substituting in $$\frac{\partial \Psi}{\partial t}$$ and $$\frac{\partial \Psi^*}{\partial t}$$ from Schrodinger's equation:

$$= -i \hbar \int \frac{i}{\hbar } V \frac{\partial}{\partial x} ( \Psi \Psi^* ) - \frac{i \hbar}{2 m } \frac{\partial}{\partial x} ( \frac{\partial \Psi}{\partial x} \frac{\partial \Psi^*}{\partial x} ) dx$$

So with the exception of a sign problem... I can see $$\int \frac{\partial V}{\partial x} \Psi \Psi^*dx$$ = <dV/dx>

But unfortunately it doesn't appear like the second term is going to disappear, is there anything I am missing?

(Note I used: $$\frac{\partial \Psi}{\partial t} = \frac{ i \hbar}{2 m }\frac{\partial^2\Psi}{\partial x^2} - \frac{i}{\hbar} V \Psi$$ and $$\frac{\partial \Psi^*}{\partial t} = -\frac{ i \hbar}{2 m }\frac{\partial^2\Psi^*}{\partial x^2} + \frac{i}{\hbar} V \Psi^*$$ as given by Griffiths)

4. Jul 21, 2009

### gabbagabbahey

Hmmm....

$$\Psi^* \frac{\partial^2\Psi}{\partial x \partial t}= \Psi^* \frac{\partial}{\partial x} \left(\frac{\partial\Psi}{\partial t}\right)\neq - \frac{\partial \Psi^*}{\partial x} \frac{\partial \Psi}{\partial t}$$

Why did you think you could transfer the partial derivative from $\frac{\partial \Psi}{\partial t}$ to $\Psi^*$ like this?

5. Jul 22, 2009

### cipher42

Underneath the integral, that's perfectly fine. Since the wavefunction is normalizable, it (and it's derivatives) will go to 0 at infinity and -infinity (the implied limits of integration), so integration by parts lets you transfer the derivative at the cost of a sign without having to worry about the boundary term.

As for roeb, you're so close! That term you mentioned will cancel out if you keep track of your signs closely. Remember to change the signs when you integrate by parts. And don't bother trying to write it as the derivative of something, it will cancel out before that if you keep all of your signs in the right places.

Also, good call by guguma on learning/proving the more general form of Ehernfest's Theorem; it can get you a lot of these special cases much more quickly (though doing this one by brute force is still a good exercise).