Suppose we have a periodic function $f_K(\vec x)$.
We want to show that $\int {f_K^*(\vec x) f_{K'}(\vec x) d\vec x} = \delta_{KK'}$, where the integration is over the period of $f(x)$.
I know this has to do with Fourier series, one form of which is:
$$f(x) = \frac {a_0}{2} + \sum_{n=1}^N [a_n \cos(nx)+ b_n \sin(nx)] ,$$
where
$$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx)dx, $$
$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx)dx .$$
I am unable to proceed because it looks to me as a circular logic. Cosines and Sines are themselves orthogonal functions, I do not understand how to use them to prove the orthogonality of another function which should depend – via Fourier Transform – on the orthogonality of the sines and cosines! I know, I am misunderstanding something here.
Best Answer
You are correct that cosines and sines are orthogonal. Without knowing something about $f_K(\vec x)$ and $f_{K'}(\vec x)$they are not automatically orthogonal. One way to prove it is to expand $f_K$ as a Fourier series, but that will only work if the relation between $f_K$ and $f_{K'}$ is right. It is not circular, you are reducing the problem to one you have already solved.