Take a unit box, the energy eigenfunctions are $\sin(n\pi x)$ (ignoring normalization constant) inside the box and 0 outside. I have read that there is no momentum operator for a particle in a box, since $\frac{\hbar}{i}\frac{d}{dx}\sin(n\pi x)=\frac{\hbar}{i}n\pi\cos(n\pi x)$ and this isn't 0 at the end points. Nonetheless, we can write $\sin(n\pi x)=\frac{e^{in\pi x}-e^{-in\pi x}}{2i}$, which seems to imply that there are two possible values of momentum: $n\pi$ and $-n\pi$, each with 50% probability.. Is this wrong? If you measured one of these momenta and the wavefunction collapsed to one of the eigenstates then it wouldn't solve the boundary conditions. So, what values of momentum could you obtain if you measured the momentum of a particle in a box?
Edit: I know that you can't measure the momentum of a particle exactly, but normally after a measurement of momentum, or such a continuous observable, the wavefunction collapses to a continuous superposition of momentum eigenstates corresponding to the precision of your measurement. But in this case since the wavefunction seems to just be a superposition of two momentum eigenstates, the wavefunction must have to collapse to one of them exactly, or so it seems.
Best Answer
There are two different issues. One of them is the sign of the momentum; the other one is whether the momentum is spread (it's not because of the unnatural boundary conditions).
Concerning the first point, the standing wave (sine) is a real function and every real wave function has the same probability to carry momentum $+p$ and $-p$. So indeed, both of them are equally likely.
But even if you write the sine as a difference of two complex exponentials, it's still true that these exponentials aren't equal to the wave function everywhere – just inside the box – so it's still untrue that the momentum is sharply confined to two values $p$ and $-p$.
To get the probabilities of different momenta, you need to Fourier transform the standing wave – a few waves of the sine. One has $$\int_0^1 dx\,\sin(n\pi x)\exp(ipx) = \frac{n\pi[-1 +e^{ip}(-1)^n]}{p^2-n^2\pi^2} $$ Square the absolute value to get the probability density that the momentum is $p$. The momentum $p$ should have the natural prefactors $\hbar/L$ etc. and the overall wave function should get another normalization factor for the overall probability to equal one. This changes nothing about the shape of the probability distribution: almost all values of $p$ have a nonzero probability.