Fourier Series

Sinusoidal Functions.

$e_{m} (x) = \frac{1}{2 ℓ} \exp (\frac{i m π x}{ℓ}) = \frac{1}{\sqrt{τ}} \exp (- i ω_{m} x)$ , $ω_{m} = m ω_{0}, ω_{0} = \frac{2 π}{τ}, τ = 2 ℓ$ .

${\sin (n π x) : n \in N}$

$L_{2}^{1} (- π, π)$ (Generalizations exist for larger intervals) The functions $| e_{m} ⟩ := e_{m} (Θ) = \frac{1}{2 π} \exp (i m Θ)$ . $⟨ e_{m} | e_{n} ⟩ = δ_{n}^{m} = \frac{1}{2 π} \int d Θ \exp (- i (m - n) Θ)$ .

$f (x) = \sum_{m = - \infty}^{\infty} f^{m} e_{m} (x) = \frac{1}{\sqrt{2 π}} \sum_{m} \frac{\sqrt{2 π}}{\sqrt{2 ℓ}} \frac{ℓ}{π} f^{m} \exp (i ω_{m} x) \frac{π}{ℓ}$ , $\sqrt{\frac{ℓ}{π}} f^{m} = \tilde{f} (ω_{m})$ . Then, $f (x) = \frac{1}{\sqrt{2 π}} \sum_{m} \tilde{f} (ω_{m}) \exp (i ω_{m} x) δ ω$ . $δ ω = ω_{n + 1} - ω_{n}$ . So, as $ℓ$ increases, $δ ω \to d ω$ . So, $f (x) = \frac{1}{\sqrt{2 π}} \int_{R} d ω \exp (i ω x) \tilde{f} (ω) = F^{- 1} [\tilde{f}], \tilde{f} (ω) = \frac{1}{\sqrt{2 π}} \int_{R} d x \exp (- i ω x) f (x) = F [f]$ . These are the Fourier transform pair.

Theorem

On $L_{2}^{1} (R)$ , $f (x)$ and $\tilde{f} (ω)$ represent the same $| f ⟩$ .

Parselval-Planchera Theorem

The Fourier transform is a unitary map on $L_{2}^{1} (R)$ , i.e. $⟨ f | g ⟩ = ⟨ \tilde{f} | \tilde{g} ⟩$ .

Consequence

What this is telling us is that a FT is a change of basis for $L_{2}^{1} (R)$ .

$f (x) = ⟨ x | f ⟩, \tilde{f} (ω) = ⟨ ω | f ⟩, I = \sum_{R} d x | x ⟩ ⟨ x |$ .

$f (ω) = \int_{R} d x ⟨ ω | x ⟩ ⟨ x | f ⟩ = \int_{R} d x ⟨ ω | x ⟩ f (x) = F [f] \Rightarrow ⟨ ω | x ⟩ = \exp (- i ω x) / \sqrt{2 π}$ .

Comparing to QM

$φ_{p} (x) = ⟨ x | p ⟩ = \frac{1}{\sqrt{2 π ℏ}} \exp (\frac{i p x}{ℏ})$ . I.e. position and momentum spaces are Fourier transforms of each other.

DeBrolie Relation

The physical consequence is: $p = \frac{2 π ℏ}{λ}$ .