I'm struggling to write the Fourier expansion of the wave-function associated to a particle confined in an infinite potential well. Suppose that $V(x) = \left\{\begin{array}{c c} \infty & \text{ if } x\notin(0,a)\\ 0 & \text{ if } x\in(0,a)\\ \end{array}\right.\,.$
Then, the solution to the time-independent Schrödinger equation is:
$\psi_n(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)$.
In this case, the momentum is a discrete variable, so I would use a discrete Fourier expansion, that is, I use the following formulas for any function $f(x)$
$f(x) = \sum\limits_{m=-\infty}^{\infty}f_m\,e^{i\frac{2m\pi}{a}x}\;\;;\;\;f_m = \frac{1}{a}\int\limits_0^af(x)\,e^{-i\frac{2m\pi}{a}x}\,dx$.
If I use the formula for$f_m$ to obtain the expression in the momentum's domain of $\psi_n(x)$ and using $\sin(x) = \frac{1}{2i}\left(e^{ix}-e^{-ix}\right)$ I got to the following equation:
$f_m = \frac{1}{2ia}\int\limits_0^a\left[e^{i\frac{x\pi}{a}(n-2m)}-e^{-i\frac{x\pi}{a}(n+2m)}\right]\,dx$
which does not seem to have a simple orthogonality relation, because $n = \pm 2m$ does the trick to the zero-argument for the exponential, but if $n\neq \pm 2m$ the condition for the integrals to be zero would the that $\frac{n}{2}\pm m\in\mathbb{Z}$ which forces me to discard all the non-pair $n$'s!
If I think of a Fourier transformation with the same "argument-type" as the wave-function, that is, instead of $\frac{2\pi m}{a}x$ using $\frac{\pi m}{a}x$ so that the expansion is:
$f(x) = \sum\limits_{m=-\infty}^{\infty}f_m\,e^{i\frac{m\pi}{a}x}$,
I got cornered much in the same way. Does anyone have an idea what I'm doing wrong?
