Fourier Series
Sinusoidal Functions.
\(e_m(x) = \frac{1}{2\ell}\exp\left(\frac{im\pi x}{\ell}\right) = \frac{1}{\sqrt{\tau}}\exp(-i\omega_m x)\), \(\omega_m = m\omega_0, \omega_0 = \frac{2\pi}{\tau}, \tau = 2\ell\).
\(\{\sin(n\pi x):n\in\mathbb{N}\}\)
\(L_2^1(-\pi,\pi)\) (Generalizations exist for larger intervals) The functions \(|e_m\rangle := e_m(\Theta) = \frac{1}{2\pi}\exp(im\Theta)\). \(\langle e_m|e_n\rangle = \delta_n^m = \frac{1}{2\pi}\int d\Theta \exp(-i(m-n)\Theta)\).
\(f(x) = \sum_{m=-\infty}^\infty f^m e_m(x) = \frac{1}{\sqrt{2\pi}}\sum_m\frac{\sqrt{2\pi}}{\sqrt{2\ell}}\frac{\ell}{\pi}f^m\exp(i\omega_mx)\frac{\pi}{\ell}\), \(\sqrt{\frac{\ell}{\pi}}f^m = \tilde{f}(\omega_m)\). Then, \(f(x) = \frac{1}{\sqrt{2\pi}}\sum_m \tilde{f}(\omega_m)\exp(i\omega_m x)\delta\omega\). \(\delta\omega = \omega_{n+1}-\omega_n\). So, as \(\ell\) increases, \(\delta \omega\to d\omega\). So, \(f(x)=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}d\omega \exp(i\omega x)\tilde{f}(\omega)=\mathcal{F}^{-1}[\tilde{f}], \tilde{f}(\omega)=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}dx\exp(-i\omega x)f(x)=\mathcal{F}[f]\). These are the Fourier transform pair.
Theorem
On \(L_2^1(\mathbb{R})\), \(f(x)\) and \(\tilde{f}(\omega)\) represent the same \(|f\rangle\).
Parselval-Planchera Theorem
The Fourier transform is a unitary map on \(L_2^1(\mathbb{R})\), i.e. \(\langle f|g\rangle = \langle \tilde{f}|\tilde{g}\rangle\).
Consequence
What this is telling us is that a FT is a change of basis for \(L_2^1(\mathbb{R})\).
\(f(x) = \langle x|f\rangle, \tilde{f}(\omega) = \langle \omega|f\rangle, \mathbb{I}=\sum_\mathbb{R}dx|x\rangle\langle x|\).
\(f(\omega)=\int_\mathbb{R}dx\langle \omega|x\rangle\langle x|f\rangle = \int_\mathbb{R} dx\langle \omega|x\rangle f(x)=\mathcal{F}[f]\Rightarrow \langle\omega|x\rangle = \exp(-i\omega x)/\sqrt{2\pi}\).
Comparing to QM
\(\varphi_p(x) = \langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}}\exp\left(\frac{ipx}{\hbar}\right)\). I.e. position and momentum spaces are Fourier transforms of each other.
DeBrolie Relation
The physical consequence is: \(p=\frac{2\pi\hbar}{\lambda}\).