Particle in a 1D box (momentum representation) I have this problem.
I want to find the wave function in the momentum space for the particle in a 1D box.
We know that the wave function in the position space is:
$$Y_n(x) = A\sin{(n\pi x/L)}$$
Well, if I write the Schrödinger equation in the moment space I have:
$$\frac{p^2}{2m}Y_n(p) = E_nY_n(p)$$
So, this equation doesn't give me any information about the wave function $Y_n(p)$
I know that I can solve this problem just using the Fourier tranformation, but I'm asking myself if there is another posibility to solve this problem.
PD. If I use the Fourier transformation, do I have to integrate just from $O$ to $L$?
 A: The solution will become that $Y_n(p)$ will give a $\delta$-distribution, namely:
$Y_n(p) = A\delta(p-\sqrt{2mE}) + B\delta(p+\sqrt{2mE})$,  Where A and B are  normalisation-constants. This is because only for $\frac{p^2}{2m} = E_n$ the equation is valid (hence the delta-distribution). This solution is the solution for the free particle, so for your problem without any boundry-conditions.
If we try to impose the boundry-conditions we will run into problems, this will give rise to integrals and hence integral-equations. A few examples in momentum-space are worked out in this article.
Since the 1D box only has wavefunctions defined within the box (outside yields zero) your integration for the Fourierseries will go from 0 to L.
A: You can rewrite the momentum space equation as
$$(p^2 - 2mE_n)Y_n(p) = 0.$$
This shows that if $p \neq \pm \sqrt{2m E_n}$, $Y_n(p) = 0$. But at $p = \pm \sqrt{2m E_n},$ $Y_n$ can take any value, so the most general solution is
$$Y_n(p) = a \, \delta(p - \sqrt{2m E_n}) + b \, \delta(p + \sqrt{2m E_n})$$
for some constants $a,b.$ These constants will depend on the boundary conditions, so you need to go back to real space first
$$Y_n(x) = \int Y_n(p) e^{i x p} \mathrm{d}p = a e^{i \sqrt{2mE_n} x} + b  e^{-i \sqrt{2mE_n} x}$$
and then use the boundary conditions $Y_n(0) = Y_n(L) = 0.$
