Is De Broglie's formula $\textbf{p}=\hbar \textbf{k}$ applicable to a discrete wave number system? I don't know if my question has sense at all but while doing my homework there appeared in my mind this question. 
Say, for a particle in a box, the confinement makes that the wave number k is discrete, depending on integer numbers $\textbf{n}=(n_{x}, n_{y},n_{z})$.
If De Broglie's formula holds, it doesn't mean that momentum $\textbf{p}=\hbar \textbf{k}$ is also discrete? 
 A: No. You can't use the De Broglie formula here because De Broglie's formula is true only for a free particle $(V(x)=0$ everywhere$)$.
For the infinite square well, we can write down the energy eigenvalue equation as:
$$\frac{-\hbar^2}{2m}\frac{\partial^{2}\psi(x)}{\partial x^2}+V(x)\psi(x)=E\psi(x)$$ such that $V(x)=0 $ inside the box and $V(x)\rightarrow\infty$ outside. Now, clearly we can see that the Energy operator $\hat{E}$ doesn't commute with $\hat p$ (although they do commute for a free particle potential.) If you try to use the definition of momentum operator $ \hat p_x=(\hbar/i)\partial_x $ on the stationary solutions of infinite well, you'll see that none of them is an eigenfunction of $\hat p$. 
The probable reason for your confusion here is naming the constant $2mE/\hbar^2$ as $\textbf k$. Here, $\textbf k$ is not the wavenumber as we had for a free particle. Just name it as something else, and logically try to argue if it still makes sense to call it the same wavenumber. The definition $\textbf p =\hbar\textbf k$ doesn't extend here.
Also, if you are thinking about three dimensions, you can extend the argument very similarly to higher dimensions by symmetry.
A: Yes. But remember, you can think of the wavefunction in quantum physics like a wave in the configuration space of the system, that is isomporphic to the physical space only for a single particle case. Momentum is discrete. Actually, 1 of the 1st quantum observed effects was the discreteness of the energy spectrum, that depends on momentum and electrostatic pontential  $ V\vec(r) $ for the electron in Hydrogen atom in an approximated electrostatic potencial within its interactions with the positive charge in nucleus.
A: Yes it is applicable. Note that the expectation value of momentum is zero for any stationary solution in a box, so $\langle \psi | \frac{\hbar}{i} \vec \nabla |\psi\rangle = 0$. Still $ \psi | \vec p |\psi = 0$, without the integration over space is the Noether momentum distribution belonging to the Schrödinger lagrangian ${\cal L} = i\hbar \psi^* \dot \psi = -\hbar^2 \vec \nabla \psi^* \cdot \vec \nabla \psi +qV\psi^* \psi$ .
