# Free Particle: Time dependence of expectation value of position Paradox

It would be really appreciated if somebody could clarify something for me: I know that stationary states are states of definite energy. But are all states of definite energy also stationary state?

This question occurred to me when I considered the free particle (plane wave, not a Gaussian packet). The time derivative of the mean value of the position is not constant in time but equal to $\frac{\hbar k}{m}$ (via Ehrenfest's theorem because the expectation value of the momentum is equal to ${\hbar k}$).
But, plane waves are energy (and momentum) eigenstates for the free particle, that means that they are states of definite energy. The paradox comes up because the Ehrenfest theorem gives a non-zero value for the time derivative of the expectation value of the position but we know that for stationary states, expectation values do not change in time(for reference: Griffiths p.26)

So, am I getting something wrong here?

EDIT: After some responses in the comments that explain that all the above are due to the non-normalizability of those states, I have come across another similar case:
The same thing happens for the plane wave energy eigenstate in the case of a particle confined in a box with periodic boundary conditions. Now, these states are normalizable. So, what happens in this case?

• ...the plane wave doesn't have a position expectation value, as it is not a state (because it is not normalizable). I'm not sure what you mean. – ACuriousMind Jun 14 '16 at 14:02
• @ACuriousMind I edited. Hope it helps – TheQuantumMan Jun 14 '16 at 14:04
• Yeah, you can't apply Ehrenfest's theorem to things that are not states. In general, every weirdness that you encounter with plane waves is caused by them not actually being states. In this case, the issue is that there is no position expectation value for them. It just doesn't exist, so you can't apply any theorems to it. – ACuriousMind Jun 14 '16 at 14:12
• @TheQuantumMan I don't have access right now, but this: scitation.aip.org/content/aapt/journal/ajp/64/11/10.1119/… sounds interesting – Sanya Aug 7 '16 at 9:41
• What do you mean by stationary? Consider this definition: a state is stationary if it won't evolve in time. So if the Hamiltonian is time-independent, then all states with definite energy, means they are the Hamiltonian eigenvector, are indeed stationary; as the Hamiltonian in the finite box is. However, it does not mean the expected value of position is time-independent too. If the position operator commutes with the Hamiltonian, then its expected value is also time-independent. – heaven-of-intensity Aug 7 '16 at 11:48

Ehrenfest's theorem states that: $$\frac{d}{dt} \langle \Phi\lvert \hat{X} \rvert\Phi\rangle= \frac{1}{M}\langle \Phi\lvert \hat{P} \rvert\Phi\rangle$$ and $$\frac{d}{dt} \langle \Phi\lvert \hat{P} \rvert\Phi\rangle= \langle \Phi\lvert \frac{\partial V}{\partial X} \rvert\Phi\rangle$$ The problem, according to the paper, of applying this theorem to the particle in a box is the infinite value of $V$ outside the box. Of course, the probability density there is $0$ too, but putting $0 \cdot \infty = 0$ can result in funny outcomes, as apparently here. The author of the paper thus calculates the expectation value for a potential of finite height $V_0$ with $$\frac{\partial V}{\partial X} = V_0 \cdot \left( \delta(x-x_{right \phantom{0} end \phantom{0} of \phantom{0} box}) - \delta(x-x_{left \phantom{0} end \phantom{0} of \phantom{0} box}) \right)$$ and the wave functions being the wave functions for a step potential of finite size and seems to regain something reasonable in the limit of $V_0 \to \infty$.