11  Fourier Series

Consider functions defined on the real interval \(x\in\left[-L,L\right)\). The aim is to decompose them into the following components:

\[f\left(x\right)=\frac{a_{0}}{2}+\sum^{\infty}_{n=1}a_{n}\cos\left(\frac{n\pi x}{L}\right)+\sum^{\infty}_{n=1}b_{n}\sin\left(\frac{n\pi x}{L}\right). \tag{11.1}\]

Recall from first year that

\[\begin{aligned} a_{0} & =\frac{1}{L}\int^{L}_{-L}f\left(x\right)\text{d}x\\ a_{n>0} & =\frac{1}{L}\int^{L}_{-L}f\left(x\right)\cos\left(\frac{n\pi x}{L}\right)\text{d}x\\ b_{n} & =\frac{1}{L}\int^{L}_{-L}f\left(x\right)\sin\left(\frac{n\pi x}{L}\right)\text{d}x. \end{aligned}\]

11.1 Fourier Series in Dirac notation

Define some abstract states in an \(\aleph_{0}\)-dimensional Hilbert space \(|0\rangle,|c_{n}\rangle,|s_{n}\rangle\in\mathcal{H}\) by their projections into the position basis:

\[\begin{aligned} \langle x|0\rangle & \triangleq\frac{1}{\sqrt{2L}}\\ \langle x|c_{n}\rangle & \triangleq\frac{1}{\sqrt{L}}\cos\left(\frac{n\pi x}{L}\right)\\ \langle x|s_{n}\rangle & \triangleq\frac{1}{\sqrt{L}}\sin\left(\frac{n\pi x}{L}\right). \end{aligned}\]

In this abstract Hilbert space, Equation 11.1 reads

\[|f\rangle=\frac{a_{0}}{2}|0\rangle+\sum^{\infty}_{n=1}a_{n}|c_{n}\rangle+\sum^{\infty}_{n=1}b_{n}|s_{n}\rangle\]

where

\[\begin{aligned} a_{0}/2 & =\langle0|f\rangle\\ a_{n>0} & =\langle c_{n}|f\rangle\\ b_{n} & =\langle s_{n}|f\rangle. \end{aligned}\]

Implicit in this construction is the fact that \(\left\{ |0\rangle,|c_{n}\rangle,|s_{n}\rangle\right\}\) form a complete orthonormal basis for \(\mathcal{H}\). That is,

\[\begin{aligned} \langle0|c_{n}\rangle & =\langle0|s_{n}\rangle=\langle c_{n}|s_{n}\rangle=0\\ \langle0|0\rangle & =1\\ \langle c_{n}|c_{m}\rangle & =\langle s_{n}|s_{m}\rangle=\delta_{nm}. \end{aligned}\]

Each of these can be shown using a resolution of the identity into the position basis. E.g.

\[\begin{aligned} \langle0|0\rangle & =\langle0|\mathbb{I}|0\rangle\\ & =\langle0|\left(\int^{L}_{-L}\text{d}x|x\rangle\langle x|\right)|0\rangle\\ & =\int^{L}_{-L}\text{d}x\langle0|x\rangle\langle x|0\rangle\\ & =\int^{L}_{-L}\text{d}x\frac{1}{\sqrt{2L}}\frac{1}{\sqrt{2L}}\\ & 1\,\,\checkmark \end{aligned}\]

Completeness is (I’m told) equivalent to the statement that all basis vectors other than \(|0\rangle\) are orthogonal to \(|0\rangle\).

11.2 QM example: the infinite potential well

Consider a quantum particle in a potential

\[V\left(x\right)=\begin{cases} \begin{array}{c} 0,\\ \infty, \end{array} & \begin{array}{c} 0\le x\le L\\ \text{otherwise} \end{array}\end{cases}\]

(the infinite potential well, or particle in a box). The task is to find the energy eigenstates defined by

\[\hat{H}|\phi_{n}\rangle=E_{n}|\phi_{n}\rangle\]

where

\[\begin{aligned} \hat{H} & =\frac{\hat{p}^{2}}{2m}+\hat{V}\\ \hat{V} & =\int^{\infty}_{-\infty}\text{d}xV\left(x\right)|x\rangle\langle x|. \end{aligned}\]

In the position basis this reads

\[\frac{-\hbar^{2}}{2m}\frac{\partial^{2}\phi_{n}\left(x\right)}{\partial x^{2}}+V\left(x\right)\phi_{n}\left(x\right)=E_{n}\phi_{n}\left(x\right)\]

where

\[\phi_{n}\left(x\right)\triangleq\langle x|\phi_{n}\rangle.\]

The solutions are

\[\phi_{n}\left(x\right)=\frac{1}{\sqrt{L}}\sin\left(\frac{n\pi x}{L}\right)\]

with eigenenergies

\[E_{n}=\frac{\hbar^{2}n^{2}\pi^{2}}{2mL^{2}}.\]

Note that

\[|\phi_{n}\rangle=|s_{n}\rangle\]

as defined above. With the boundary conditions \(x\in\left[0,L\right]\) it follows that \(a_{0}=b_{n}=0\). Since \(|s_{n}\rangle\) are the normalised eigenstates of the Hermitian operator \(\hat{H}\), with non-degenerate eigenvalues \(E_{n}\), they form a complete orthonormal basis for \(\mathcal{H}\). That is, any function \(f\left(x\right)\) on the interval \(x\in\left[0,L\right]\) can be written as a sum of \(\langle x|s_{n}\rangle\).