1D Infinite Square Box Discrete Energy levels but Continous Momenta? In the 1d particle in the box the energy of the particle should be completely determined by the momentum of the particle that you observe correct? So how can you have discrete energy levels and a continuous momentum spectrum at the same time?
 A: The momentum operator $P$ in the infinite well can be defined as a self-adjoint operator by infinitely many ways with respect to the boundary conditions by:
$$P_\theta=-i\hbar\frac{d}{dx}\\
\mathcal{D}(P_\theta)=\left\{\psi\in \mathcal{H}^1[0,a]:\psi(a)=e^{i\theta}\psi(0)\right\},$$
where $\mathcal{H}^1[0,a]$ is the Sobolev space, on the interval $[0,a]$. In any case, the spectrum is purely discrete, which can be seen by solving the eigenvalue equation
$$P_\theta\psi_n=\lambda_{\theta,n}\psi_n$$
The equation yields the eigenvalues
$$\lambda_{\theta,n}=\frac{2\pi\hbar}{a}\left(n+\frac{\theta}{2\pi}\right), \qquad n\in\Bbb{Z},$$
with eigenfunctions
$$\psi_n(x)=\frac{1}{\sqrt a}\exp\left(\frac{i\lambda_{\theta,n}x}{\hbar}\right),$$
which constitutes a basis for the Hilbert space $\mathcal{H}=L^2[0,a]$.
Edit: concerning the comments, viewing the infinite well as a limiting procedure of finite well leads to boundary conditions on the Hamiltonian to be given by:
$$H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\\
\mathcal{D}(H)=\left\{\psi \in \mathcal{H}^2[0,a]:\psi(0)=\psi(a)=0\right\},$$
so, although the action of $H$ seems to give the idea that the momentum eigenfunctions are also energy eigenfunctions, this is false, because the action is not defined, since the functions are not in $\mathcal{D}(H)$.
Using boundary conditions such that $\psi(0)=\psi(a)=0$ for the moimentum operator does not give a self-adjoint operator, since the adjoint $P^*$ has a bigger domain, in this case $\mathcal{D}(P^*)=\mathcal{H}^1[0,a]$. As observables must be represented by self-adjoint and not just symmetric operators, this boundary condition is not valid. Check this article for a better discussion.
A: 
In the 1d particle in the box the energy of the particle should be
  completely determined by the momentum of the particle that you observe
  correct?

The Hamiltonian in the position basis is
$$\hat H = \begin{cases}-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}, & 0\lt x \lt a\\\infty, & \text{otherwise} \end{cases}$$
and the energy eigenfunctions are of the form
$$\psi_n(x,t) = \sqrt{\frac{2}{a}}\sin \left(\frac{n\pi}{a}x \right)\left[\theta(x) -  \theta(x - a)\right]e^{-i\omega_nt}$$
which are clearly not momentum eigenfunctions.  So, no, the quoted statement is not correct.

So how can you have discrete energy levels and a continuous momentum
  spectrum at the same time?

This potential doesn't allow a continuous momentum spectrum.
It is true that we can find the continuous $\phi_n(p,t)$ and expand $\psi_n(x,t)$ over the momentum eigenfunctions.
However, thinking clearly about this, the energy eigenfunctions are only complete on the interval $[0,a]$.  That is to say, we cannot expand a momentum eigenfunction over the energy eigenfunctions.
Put another way, the operator $i\hbar\frac{\partial}{\partial x}$ has no eigenfunctions for this potential.
