$$ V(x) = \left\{ \begin{split} 0&\quad \operatorname{in}\ [0, a] \\ +\infty&\quad \operatorname{elsewhere} \end{split} \qquad a>0, \right. $$
The Schrödinger equation for stationary states on the interval $[0, a]$, $$ -{\hbar}^2\frac{{\mathrm{d}}^2}{{\mathrm{d}}x^2} \psi(x) = E\psi(x), $$ with boundary conditions $\psi(0)=\psi(a)=0$ has solutions $$ \psi_n(x) = \sqrt{\frac{2}{a}}\sin\left(k_n x\right) \qquad \text{with eigenvalues} \qquad E_n=\frac{\hbar^2 k_n^2}{2m} $$ and $\displaystyle k_n=\frac{n\pi}{a}$ for $n\in\mathbb{N}$.
Here's the paradox: in what follows 1. and 2. should not hold together, but they do.
1.The energy eigenstates are not momentum eigenstates: $$ -i\hbar\frac{{\rm{d}}}{{\rm{d}}x}\psi_n(x)=-i\hbar k_n\sqrt{\frac{2}{a}}\cos(k_nx) \; \not\propto\; \psi_n(x) $$ 2. In this case, energy and momentum commute. Here's a quick proof by user115625:
\begin{equation} [\hat H, \hat p] = [\hat T+ V, \hat p] =[\dfrac{\hat p^2}{2m}+ V, \hat p] =[\dfrac{\hat p^2}{2m}, \hat p] + [V, \hat p] \end{equation} Note that here, in general, the potential is a function of $\hat x$, i.e. $V(x)$. Using the property of commutators $$[AB, C] =A[B,C]+[A,C]B$$ and the result that for any function $f$ $$[f, \hat p]=i \hbar \dfrac{\partial f}{\partial x}$$ we get $$ [\hat H, \hat p]=\dfrac{1}{2m}(\hat p[\hat p,\hat p]+[\hat p,\hat p]\hat p)+i \hbar \dfrac{\partial V}{\partial x}= i \hbar \dfrac{\partial V}{\partial x} $$ Therefore \begin{equation} \boxed{\dfrac{\partial V}{\partial x}=0 \implies [\hat H, \hat p] = 0} \end{equation}
