Quantum Mechanics - Ehrenfest's Theorem

1. Oct 9, 2008

dsr39

We have to apply Ehrenfest's theorem and I don't think it was ever explained well to us. I have read that expectation values of measurable quantities behave according to classical physics equations

ie.
$$M\frac{d\left<x(t)\right>}{dt} = \left<p(t)\right>$$

I think I must be applying this idea wrong because I don't see how this result works out for the calculation of <E(t)> for an infinite square well in a particular energy eigenstate.

For an infinite square well in a stationary state.
$$\left<x(t)\right>=const \to M\frac{d\left<x(t)\right>}{dt} = 0$$

That quantity is equal to momentum so <p> is zero, and does that also mean <E(t)> is zero since

$$\left<E(t)\right> = \frac{\left<p(t)\right>^2}{2m}$$

But <E(t)> shouldn't be zero because the energies in stationary states for the infinite square well are very clearly defined by the boundary conditions and which stationary state you're in. How am I wrongly applying Ehrenfest's theorem... your help is greatly appreciated. Thank you.

2. Oct 9, 2008

faen

E is the eigenvalue of the hamiltonian (the mechanical energy operator). So you forgot to count in the potential energy. The expectation value of the kinetical energy which you calculated, is ofcorse zero in a stationary state.

3. Oct 9, 2008

dsr39

The Potential term is zero, I think because it is in an infinite well where the potential at the walls (-L/2 and +L/2) is defined to be infinity and the potential at -L/2<x<L/2 is defined to be zero.

4. Oct 9, 2008

Gokul43201

Staff Emeritus
No, it is not.

That's not correct. The square lives inside the (expectation value) sum/integral, not outside it.

5. Oct 9, 2008

borgwal

The last 2 statements are wrong. Next time, leave out the phrase "of course" if you don't know what you're talking about :-) Gokul is right.

Last edited: Oct 9, 2008
6. Oct 10, 2008

dsr39

I see where I was going wrong, thanks for the help.