Why is the momentum wave function of a particle in a box equal to this expression? https://en.wikipedia.org/wiki/Particle_in_a_box#Momentum_wave_function
Surely the Fourier transform of a sine wave amounts to two dirac deltas either side of zero. What is the cause of this additional complexity?
The energy eigenfunction in the one-dimensional box looks like $$\phi_n(x)=\sqrt{\frac{2}{L}}\sin\left( \frac{n\pi x}{L}\right)$$ which is a square integrable as in should be so that $\phi_n(x)\in l_2$ that suggest that the $\tilde{\phi}_n(p)$ must be square integrable too which is fourier transform of $\phi_n(x)$.
OP's suggestion that $\tilde{\phi}_n(p) $ should be dirac delta function knock down as dirac delta in no in $l_2$.
As suggested in the comment $\phi_n(x)$ is not a sin wave. The only way to settle this is to actually find out the Fourier transform of the $\phi_n(x)$.
