Free Particle: Time dependence of expectation value of position Paradox

Question

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?

Answer 1

I will only be giving an outline of the idea described in the paper "Ehrenfest’s theorem and the particle‐in‐a‐box" by D. S. Rokhsar in American Journal of Physics 64, 1416 (1996).

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 $V_0 \to \infty$.