Periodic Functions and Fourier Series

A Fourier series expands a periodic function as a sum of discrete sinusoidal modes. On an interval of length $L$ with periodic identification, the normalized complex modes are

$$Finite sampled versions of these periodic expansions are computed with the [Fast Fourier Transform](/math-toolkit/fast-fourier-transform/). e_n(x) = \frac{1}{\sqrt L} \exp \left( \frac{2\pi i n x}{L} \right), \qquad n\in\mathbb Z.$$

For the phase notation behind these modes, see Complex Exponentials.

For suitable functions, and for every $L^2$ function in the norm-convergence sense,

$$f(x) \sim \sum_{n\in\mathbb Z} c_n e_n(x), \qquad c_n = \int_0^L e_n(x)^*f(x)\,dx.$$

The symbol $\sim$ is a reminder that convergence needs interpretation. In quantum mechanics, Hilbert-space norm convergence is often the right first meaning.

Why Quantum Mechanics Needs Fourier Series

Fourier series are the finite-volume version of Fourier transforms. They appear whenever boundary conditions make momenta discrete:

• a particle on a ring has periodic momentum modes;
• a particle in a box has sine or cosine standing-wave modes;
• numerical simulations often put systems in a finite periodic box;
• crystals and lattices use reciprocal-space ideas built from periodic modes;
• finite-temperature and imaginary-time methods use discrete frequency sums.

The full-line transform lives at Fourier Transform. This page explains the interval and periodic case, where sums replace integrals.

The reciprocal-lattice identity connecting periodic real-space sums to Fourier-space sums is Poisson Summation Formula.

Prerequisites

You should know:

• inner products and orthogonality from Inner Products;
• Hilbert-space expansions from Completeness and Orthonormal Bases;
• endpoint constraints from Boundary Conditions.

Orthogonality of Modes

The periodic modes are orthonormal on $[0,L]$:

$$\int_0^L e_m(x)^*e_n(x)\,dx = \delta_{mn}.$$

Indeed,

$$\int_0^L e_m(x)^*e_n(x)\,dx = \frac{1}{L} \int_0^L \exp \left[ \frac{2\pi i(n-m)x}{L} \right]dx.$$

If $n=m$, the integrand is $1$ and the result is $1$. If $n\ne m$, the exponential completes an integer number of periods and the result is $0$.

This is the same algebra as an orthonormal basis of vectors, but with an integral replacing a dot product.

Coefficients and Reconstruction

For a periodic function $f$, the coefficient

$$c_n = \langle e_n\vert f\rangle = \int_0^L e_n(x)^*f(x)\,dx$$

measures the component of $f$ in mode $n$. The partial sum

$$S_N(x) = \sum_{n=-N}^{N} c_n e_n(x)$$

is the projection of $f$ onto the modes with $\lvert n\rvert\le N$.

For $f\in L^2([0,L])$, the Fourier series converges to $f$ in $L^2$ norm:

$$\lim_{N\to\infty} \int_0^L \lvert f(x)-S_N(x)\rvert^2\,dx = 0.$$

Pointwise convergence requires additional hypotheses. Smooth periodic functions have rapidly decaying coefficients; functions with jumps have slower decay and may display Gibbs oscillations near discontinuities.

Sine and Cosine Forms

The complex exponential form is compact, but real functions are often expanded as

$$f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n\cos \left( \frac{2\pi n x}{L} \right) + b_n\sin \left( \frac{2\pi n x}{L} \right) \right].$$

The sine and cosine coefficients are equivalent to the complex coefficients. The best choice depends on symmetry and boundary conditions:

• even periodic functions use only cosines;
• odd periodic functions use only sines;
• hard-wall box eigenfunctions use a sine basis on $0\lt x\lt L$;
• Neumann endpoint conditions naturally lead to cosine modes.

Boundary Conditions Choose the Modes

The differential expression

$$-\frac{d^2}{dx^2}$$

has different eigenfunctions depending on the boundary conditions.

For periodic boundary conditions,

$$f(0)=f(L), \qquad f'(0)=f'(L),$$

the eigenfunctions are the complex modes $e_n(x)$ with wave numbers

$$k_n = \frac{2\pi n}{L}.$$

For Dirichlet boundary conditions,

$$f(0)=f(L)=0,$$

the normalized sine modes are

$$u_n(x) = \sqrt{\frac{2}{L}} \sin \left( \frac{n\pi x}{L} \right), \qquad n=1,2,3,\ldots .$$

This is the basis that appears in the Infinite Square Well. The mathematical reason these modes organize the problem is Sturm–Liouville Theory.

Particle on a Ring

For a particle on a ring of circumference $L$, wavefunctions satisfy periodic boundary conditions. The momentum eigenfunctions are

$$\psi_n(x) = \frac{1}{\sqrt L} e^{2\pi i n x/L}.$$

Applying the momentum operator gives

$$-i\hbar\frac{d}{dx}\psi_n(x) = p_n\psi_n(x), \qquad p_n = \hbar k_n = \frac{2\pi\hbar n}{L}.$$

The finite circumference makes the allowed momenta discrete. For a free particle on the ring, the energies are

$$E_n = \frac{p_n^2}{2m} = \frac{(2\pi\hbar n)^2}{2mL^2}.$$

Except for $n=0$, the states $n$ and $-n$ have the same energy but opposite momentum.

Relation to the Fourier Transform

The Fourier series of a periodic box becomes the Fourier transform in a large-volume limit. For periodic length $L$,

$$p_n = \frac{2\pi\hbar n}{L}, \qquad \Delta p = \frac{2\pi\hbar}{L}.$$

As $L\to\infty$, the spacing $\Delta p$ tends to zero and sums over modes become integrals:

$$\sum_n \Delta p\,F(p_n) \longrightarrow \int_{-\infty}^{\infty}F(p)\,dp.$$

This is the bridge between discrete finite-volume normalization and continuum momentum normalization. It is also the source of many factors of $2\pi$ and $\hbar$ in box-to-continuum calculations.

Parseval Identity

For a complete orthonormal Fourier basis,

$$\int_0^L \lvert f(x)\rvert^2\,dx = \sum_{n\in\mathbb Z} \lvert c_n\rvert^2.$$

In quantum mechanics, this identity says that the probability norm can be computed either from the wavefunction in position space or from its discrete mode coefficients. If the state is normalized, then

$$\sum_{n\in\mathbb Z} \lvert c_n\rvert^2 = 1.$$

The continuous analogue is the Plancherel property of the Fourier transform.

The theorem-level treatment, including inner products and derivative norms, is Plancherel and Parseval Theorems.

Common Mistakes

• Forgetting that the boundary conditions determine which Fourier modes are allowed.
• Treating a Fourier series as pointwise convergent without checking hypotheses.
• Mixing periodic modes $2\pi n/L$ with box sine modes $n\pi/L$.
• Dropping normalization factors such as $1/\sqrt L$.
• Confusing a finite-volume discrete momentum sum with a continuum momentum integral.
• Differentiating a Fourier series term by term when the function lacks enough regularity.
• Treating Gibbs oscillations near a discontinuity as a numerical error rather than a convergence phenomenon.

Cross-Links

• Fourier Transform
• Poisson Summation Formula
• Plancherel and Parseval Theorems
• Fourier Transform Conventions
• Fast Fourier Transform
• Completeness and Orthonormal Bases
• Boundary Conditions
• Sturm–Liouville Theory
• Infinite Square Well
• Position and Momentum Representations

References

• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992.
• G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2013.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.

Exercises

1. Prove the orthogonality relation for $e_n(x)=L^{-1/2}e^{2\pi i n x/L}$.

Solution

Compute

$$\int_0^L e_m(x)^*e_n(x)\,dx = \frac{1}{L} \int_0^L \exp \left[ \frac{2\pi i(n-m)x}{L} \right]dx.$$

If $n=m$, the integral is $L$, so the normalized result is $1$. If $n\ne m$, the antiderivative is proportional to

$$e^{2\pi i(n-m)x/L},$$

which has the same value at $x=0$ and $x=L$, so the integral vanishes. Thus the result is $\delta_{mn}$.

2. For $f(x)=1$ on $[0,L]$ with periodic boundary conditions, find the normalized Fourier coefficients $c_n$.

Solution

The coefficients are

$$c_n = \int_0^L \frac{1}{\sqrt L} e^{-2\pi i n x/L}\,dx.$$

For $n=0$,

$$c_0=\sqrt L.$$

For $n\ne0$, the exponential integrates to zero over a full period. Thus only the zero mode is present.

3. Why does an infinite square well use $n\pi/L$ rather than $2\pi n/L$?

Solution

The infinite square well imposes Dirichlet conditions $\psi(0)=\psi(L)=0$, not periodic conditions. The allowed standing waves are

$$\sin \left( \frac{n\pi x}{L} \right), \qquad n=1,2,3,\ldots .$$

Periodic modes instead satisfy equality of the values and derivatives at $0$ and $L$, leading to wave numbers $2\pi n/L$.

4. Show how the momentum spacing in a periodic box tends to a continuum.

Solution

The allowed momenta are

$$p_n = \frac{2\pi\hbar n}{L}.$$

Adjacent values differ by

$$\Delta p = \frac{2\pi\hbar}{L}.$$

As $L\to\infty$, $\Delta p\to0$, so sums over momentum labels can approach integrals when the summand is smooth enough and normalized with the appropriate factor of $\Delta p$.