# 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: $$p=\frac{2\pi\hbar}{\lambda}$$.

Created: 2023-06-25 Sun 02:25

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