Why doesn't Ehrenfest's theorem work for a particle in a an infinite square well? I'm reading Griffith's Intro to Quantum Mechanics, and he mentions that in an infinite potential well, a classical particle would simply bounce back and forth between the two walls indefinitely. He then proceeds to calculate the situation for a quantum particle, and gets a whole bunch of stationary state solutions. 
My question is: Using Ehrenfest's theorem, can't we immediately say that the expectation values would mimic the classical particle and bounce back and forth as well? I know that this would give us the wrong answer(i.e. a non-stationary solution), but I'd like to know why this reasoning is wrong. 
 A: Ehrenfest's theorem holds in the well but it doesn't say that the particle moves from one wall to another and back. Instead, it says that the average change of the momentum is the expectation value of the force:
$$\frac{d}{dt} \langle p\rangle = \langle F \rangle = \langle -V'(x)\rangle $$
Forces derived from a potential are equal to the minus derivative (gradient) of the potential. For a square well, $V'(x)$ is only nonzero at the boundaries of the well where it behaves like $c\delta'(x)$, a multiple of the derivative of the delta-function (with the opposite coefficients on both sides).
The expectation value of such $\delta'(x)$ operators is proportional to the squared absolute value of the derivative of the wave function over there. This expectation value (or the contribution from one end of the well) is nonzero as soon as the wave function is non-vanishing at this end of the well, so this is enough for the force to be nonzero.
Consequently, $\langle x\rangle$ will oscillate in an interval that is a proper subset of the well i.e. shorter than the well. There won't be any single, sharply defined moment when the particle suddenly hits the wall. If the well is infinitely high, the particle is strictly prohibited to visit places outside the well. But in such a case, its average position can't reach the end of the well, either, because that would mean $\Delta x =0$ and by the uncertainty principle, $\Delta p$ would have to be infinite i.e. the particle would have to carry an infinite kinetic energy.
