- Development from Fourier Series to Fourier Transform
Introduction
This document explains the mathematical development from Fourier series to Fourier transform, which is fundamental in quantum mechanics and signal processing.
Vector Basis Representation of Fourier Series
The Fourier series representation in vector basis form can be written as:
$$|\psi\rangle = \sum_{n=-\infty}^{\infty} \psi(n)|n\rangle$$
To give this summation a meaning of area, we can view $n$ as a value $x’$ on the x-axis:
$$|\psi\rangle = \sum_{n=-\infty}^{\infty} \Delta x’ \psi(n \Delta x’) |n \Delta x’\rangle$$
where $\Delta x’ = 1$.
As $\Delta x’ \to 0$ and $n \Delta x’ = x’$, we can express this as an integral:
$$|\psi\rangle = \int_{-\infty}^{\infty} dx’ \psi(x’) |x’\rangle$$
Expansion Using Completeness Relation
Using the completeness relation $1 = \sum_{n=-\infty}^{\infty} |n\rangle\langle n|$, we have:
$$\begin{align} \langle x|\psi\rangle &= \int_{-\infty}^{\infty} dx' \psi(x')\langle x|x'\rangle \\ &= \sum_{n=-\infty}^{\infty}\int_{-\infty}^{\infty} dx' \psi(x')\langle x|n\rangle\langle n|x'\rangle\\ \end{align}$$$$\tag{1}$$
Basis Functions in Fourier Series
In Fourier series, the basis function with period $L/n$ is:
$$\langle x|n\rangle = \frac{1}{\sqrt{L}} e^{i\frac{2\pi}{L/n} x}$$
If we define the wave number $k_n = \frac{2\pi}{L/n}$, then:
$$\Delta k = k_{n+1} - k_n = \frac{2\pi}{L}, \quad k_n = n \Delta k$$
Since $p=\hbar k$, we have:
$$p_n = \frac{2\pi\hbar}{L/n}, \quad \Delta p = p_{n+1} - p_n = \frac{2\pi\hbar}{L},$$
$$p_n = n \Delta p$$
Therefore:
$$\langle x|n\rangle = \frac{1}{\sqrt{L}} e^{i\frac{2\pi}{L/n} x} = \frac{\sqrt{\Delta p}}{\sqrt{2\pi\hbar}} e^{i \frac{p_n}{\hbar} x}$$
$$\langle n|x\rangle = \frac{\sqrt{\Delta p}}{\sqrt{2\pi\hbar}} e^{-i \frac{p_n}{\hbar} x}$$
Derivation of the Fourier Transform
Substituting these expressions into equation (1):
$$\begin{align} &\langle x|\psi\rangle \\ &= \sum_{n=-\infty}^{\infty}\int_{-\infty}^{\infty} dx' \psi(x')\langle x|n\rangle\langle n|x'\rangle \\ &= \sum_{n=-\infty}^{\infty}\int_{-\infty}^{\infty} dx'\\ &\quad \quad \quad \left ( \psi(x')\frac{\sqrt{\Delta p}}{\sqrt{2\pi\hbar}} e^{i \frac{p_n}{\hbar} x}\frac{\sqrt{\Delta p}}{\sqrt{2\pi\hbar}} e^{-i \frac{p_n}{\hbar} x'} \right )\\ &= \sum_{n=-\infty}^{\infty}\Delta p\frac{1}{\sqrt{2\pi\hbar}}e^{i \frac{p_n}{\hbar} x}\\ &\quad \quad \quad \left(\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx' \psi(x') e^{-i \frac{p_n}{\hbar} x'}\right) \end{align}$$As $\Delta p \to 0$ (which occurs when $L \to \infty$, and $p_n \to p$):
$$\begin{align} &\langle x|\psi\rangle \\ &= \lim_{\Delta p \to 0}\sum_{n=-\infty}^{\infty}\Delta p\frac{1}{\sqrt{2\pi\hbar}} e^{i \frac{p}{\hbar} x}\\ &\quad \quad \quad \quad \quad \left(\int_{-\infty}^{\infty} dx' \psi(x')\frac{1}{\sqrt{2\pi\hbar}}e^{-i \frac{p}{\hbar} x'}\right) \\ &= \int_{-\infty}^{\infty} dp \frac{1}{\sqrt{2\pi\hbar}} e^{i \frac{p}{\hbar} x}\\ &\quad \quad \quad\left(\int_{-\infty}^{\infty} dx' \psi(x')\frac{1}{\sqrt{2\pi\hbar}}e^{-i \frac{p}{\hbar} x'}\right) \end{align}$$The Fourier Transform Pair
This gives us the Fourier transform pair:
$$\langle p|\psi\rangle = \tilde{\psi}(p) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx \psi(x)e^{-i \frac{p}{\hbar} x}$$
$$\langle x|\psi\rangle = \psi (x) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dp \tilde{\psi}(p) e^{i \frac{p}{\hbar} x}$$
These equations represent the Fourier transform and its inverse, which are fundamental in quantum mechanics for transforming between position and momentum representations.