quantum mechanics - Is the momentum operator well-defined in the basis of standing waves?

Suppose I want to describe an arbitrary state of a quantum particle in a box of side $L$. The relevant eigenmodes are those of standing waves, namely $$ \left=\sqrt{\frac{2}{L}}\cdot \sin \left(\frac{n\pi x}{L}\right)$$ In this basis, the operate $\hat{p}^2$ is diagonal by construction, so every state has a definite energy.

But suppose I want to construct this operator from the matrix elements of $\hat{p}$? These are ($\hbar=1$) $$\left=-\frac{2imn((-1)^{m+n}-1)}{L\cdot(m^2-n^2)}$$ which vanishes for $m$ and $n$ with the same evenness, making this an off-diagonal "checkered" matrix.

I tried taking a finite portion of this matrix and squaring it in order to obtain $\hat{p}^2$, but from what I could tell, this yields an on-diagonal checkered matrix, contrary to the construction that it would be strictly diagonal.

I assume this has something to do with the fact that the wavefunctions obtained by applying $\hat{p}$ do not satisfy the same boundary conditions as the basis functions (do not vanish). Would this problem be solved I had taken the the infinite matrix in its entirety before squaring? Or is this entire problem (decomposing $\hat{p}$ in this basis) ill-defined?

Answer

The Hilbert space for a particle in a box has no associated momentum operator (as a momentum operator implies that the state space is invariant under space translations). So your attempt do define one leads to artifacts.

[Edit] In general, the Hamiltonian must be a densely defined operator on the physical Hilbert space. But a square well potential is too singular as an operator on the space of all square integrable functions on $R$ (i.e., no densely defined subspace is mapped by it into the Hilbert space), hence its Hilbert space cannot be the standard Hilbert space.

[Edit2] Note that the usual formula for the momentum is not a canonical momentum on the restricted space of wave functions defined only in the box, as it fails to satisfy either the commutation relation or does not preserve the boundary conditions.

The appropriate replacement for the momentum operator is the mode counting operator $\hat n$ with $\hat n|n\rangle=n|n\rangle$; see also the derivation of Qmechanic.