Periodic Boundary Conditions

Periodic boundary conditions identify the two ends of a finite interval. Instead of putting impenetrable walls at $x=0$ and $x=L$, one declares those endpoints to be the same physical point:

$$\psi(x+L)=\psi(x).$$

For the ordinary kinetic-energy Hamiltonian, the derivative must match as well:

$$\psi'(x+L)=\psi'(x).$$

This turns the interval into a circle of circumference $L$. It is also the standard finite-volume regulator for free particles, scattering calculations, many-body systems, numerical diagonalization, and field-theory limits.

Geometry, Not A Hard Box

Periodic boundary conditions do not describe two hard walls. A particle crossing $x=L$ reappears at $x=0$ with the same wavefunction and derivative. There is no reflection from an endpoint, no node forced at the boundary, and no standing-wave requirement.

The Hilbert space may be represented as square-integrable functions on one period, with inner product

$$\langle\phi\vert\psi\rangle = \int_0^L \phi^*(x)\psi(x)\,dx.$$

The periodicity condition says which functions on the interval represent allowed states on the circle. For the free Hamiltonian

$$\hat H =-\frac{\hbar^2}{2m}\frac{d^2}{dx^2},$$

the matching of both $\psi$ and $\psi'$ makes the kinetic-energy operator compatible with probability conservation and self-adjointness.

Momentum Quantization

A plane wave has the form

$$\psi_k(x)=A e^{ikx}.$$

Periodic boundary conditions require

$$Ae^{ik(x+L)}=Ae^{ikx}.$$

For a nonzero state,

$$e^{ikL}=1.$$

Thus

$$kL=2\pi n, \qquad n\in\mathbb Z,$$

and the allowed wavenumbers are

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

The corresponding momenta are

$$p_n=\hbar k_n =\frac{2\pi\hbar n}{L}, \qquad n\in\mathbb Z.$$

The integer $n$ can be positive, negative, or zero. Positive and negative values represent opposite circulating directions around the ring.

Plane-Wave Basis

Choose normalized modes

$$u_n(x) =\frac{1}{\sqrt L}e^{ik_nx}, \qquad k_n=\frac{2\pi n}{L}.$$

They are orthonormal:

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

For the free particle, they diagonalize both momentum and energy:

$$\hat p\,u_n =-i\hbar\frac{d}{dx}u_n =\hbar k_n u_n,$$

and

$$\hat H u_n =\frac{\hbar^2k_n^2}{2m}u_n.$$

Thus

$$E_n =\frac{\hbar^2}{2m} \left( \frac{2\pi n}{L} \right)^2.$$

The $n=0$ state is the constant mode with zero kinetic energy. For $n\ne0$, the levels $n$ and $-n$ have the same energy but opposite momentum.

Fourier Series Viewpoint

Any sufficiently regular periodic wavefunction can be expanded as

$$\psi(x) =\sum_{n\in\mathbb Z} c_n u_n(x),$$

with coefficients

$$c_n = \int_0^L u_n^*(x)\psi(x)\,dx.$$

Normalization becomes

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

The completeness relation is

$$\sum_{n\in\mathbb Z} u_n(x)u_n^*(x') =\delta_L(x-x'),$$

where $\delta_L$ is the periodic delta distribution:

$$\delta_L(x-x') = \frac{1}{L} \sum_{n\in\mathbb Z} e^{i2\pi n(x-x')/L}.$$

It acts like a delta function on periodic test functions. The mathematical background is ordinary Fourier Series, with the quantum interpretation supplied by the inner product and Hamiltonian.

Box Normalization And The Continuum Limit

The spacing between adjacent allowed wavenumbers is

$$\Delta k=\frac{2\pi}{L}.$$

For a smooth function $F(k)$, a large periodic box gives the replacement

$$\sum_{n\in\mathbb Z} F(k_n) \longrightarrow \frac{L}{2\pi} \int_{-\infty}^{\infty} F(k)\,dk$$

as $L\to\infty$. In momentum variables, since $p=\hbar k$,

$$\sum_n F(p_n) \longrightarrow \frac{L}{2\pi\hbar} \int_{-\infty}^{\infty} F(p)\,dp.$$

This is the finite-volume origin of the one-dimensional density of states:

$$\frac{\text{number of states}}{\text{length in }k} =\frac{L}{2\pi}.$$

Equivalently, the density per unit length in $k$-space is $1/(2\pi)$.

Periodic boxes are often introduced only to regularize continuum expressions. One computes with discrete, normalized states at finite $L$, converts sums to integrals, and then removes the regulator. The factors of $L$ cancel in physical quantities when states, densities, and probabilities are normalized consistently.

Density Of States In Energy

For the free particle,

$$E=\frac{\hbar^2k^2}{2m}.$$

Counting both $k\gt0$ and $k\lt0$ states for $E\gt0$ gives

$$dN = 2\frac{L}{2\pi}\,dk = \frac{L}{\pi}\,dk.$$

Since

$$\frac{dE}{dk} =\frac{\hbar^2k}{m},$$

the energy density of states is

$$\rho(E) =\frac{dN}{dE} = \frac{L}{\pi\hbar} \sqrt{\frac{m}{2E}}, \qquad E\gt0.$$

The divergence as $E\to0^+$ is a one-dimensional feature. In higher dimensions, the power of $E$ changes because the number of states grows with the volume of a shell in momentum space.

Current And Self-Adjointness Check

A periodic plane wave can carry nonzero probability current:

$$j_n = \frac{\hbar k_n}{mL}.$$

This is allowed because there is no wall. The current circulates around the ring.

The boundary-condition reason is visible from integration by parts. For the kinetic operator, the boundary form is proportional to

$$\left[ \phi^*(x)\psi'(x) -\phi'^*(x)\psi(x) \right]_{0}^{L}.$$

If both $\phi$ and $\psi$ obey the same periodic boundary conditions, the value at $L$ equals the value at $0$, and this expression vanishes. The condition is therefore compatible with a self-adjoint kinetic Hamiltonian. The distinction between a vanishing boundary form and equality with the adjoint domain is developed in Hermitian vs Self-Adjoint Operators.

Comparison With The Infinite Square Well

The infinite square well on $0\lt x\lt L$ uses hard-wall conditions:

$$\psi(0)=0, \qquad \psi(L)=0.$$

Its eigenfunctions are standing waves,

$$\sqrt{\frac{2}{L}} \sin\frac{n\pi x}{L}, \qquad n=1,2,3,\ldots$$

not momentum eigenstates. The periodic box instead uses traveling waves,

$$\frac{1}{\sqrt L} e^{i2\pi nx/L}, \qquad n\in\mathbb Z.$$

Both systems have discrete spectra because the spatial domain is compact, but the physics and boundary conditions differ. The hard-wall box models confinement by impenetrable walls. The periodic box models a ring or a finite-volume regulator without boundaries.

Twisted Boundary Conditions

A useful nearby generalization is

$$\psi(x+L)=e^{i\theta}\psi(x),$$

with the same phase relation for $\psi'$. Then

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

This is not the standard periodic condition unless $\theta=0$. Twisted boundary conditions appear in rings threaded by magnetic flux, Bloch theory, and finite-size diagnostics. The present page uses the untwisted case as the canonical starting point.

Thermodynamic-Limit Preview

In many-particle physics and field theory, one often computes in a periodic box of length $L$ and later takes a large-volume limit. For a single particle, this limit makes the momentum spacing $\Delta k=2\pi/L$ approach zero. For many particles, one usually keeps intensive quantities, such as particle density, fixed while sending the volume to infinity.

The periodic boundary is a regulator, not a claim that the universe is a literal loop. It reduces edge effects and preserves translation symmetry, which is why it is so common in numerical physics and continuum-limit derivations.

Common Mistakes

• Confusing periodic boundary conditions with hard-wall boundary conditions.
• Forgetting that $n$ runs over all integers, including negative values and zero.
• Using $k_n=n\pi/L$ instead of $k_n=2\pi n/L$ for a periodic box.
• Treating $\sin(n\pi x/L)$ as a momentum eigenfunction in the periodic problem.
• Dropping the factor $L/(2\pi)$ when replacing sums by integrals.
• Forgetting that derivative matching is part of the kinetic-energy domain.
• Interpreting finite-volume levels as physical walls when the box was only a regulator.

Where This Is Used

• Boundary Conditions gives the general role of endpoint and matching conditions.
• Particle on a Ring: First Encounter interprets the same periodic condition as angular motion on a circle.
• Propagators and Boundary Conditions gives the momentum and winding representations of periodic and twisted kernels.
• Plane Waves and Delta Normalization uses the periodic box to motivate continuum normalization.
• Momentum Eigenstates explains why periodic boxes admit genuine momentum eigenfunctions.
• Infinite Square Well contrasts hard-wall and periodic spectra.
• Normalization Conventions compares box, delta, bound-state, and flux normalizations.
• Translations and Momentum explains why plane waves are natural when translation symmetry survives.
• Crystalline Symmetry Preview shows how periodic translations become Bloch phases and crystal momentum labels.
• Boundary Conditions Table summarizes this condition alongside finite, singular, and scattering boundary rules.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Holt, Rinehart and Winston, 1976.
• M. Le Bellac, Quantum and Statistical Field Theory, Oxford University Press, 1991.

Exercises

1. Derive the allowed wavenumbers for a periodic box.

Solution

Use $\psi(x)=Ae^{ikx}$ and impose $\psi(x+L)=\psi(x)$:

$$Ae^{ik(x+L)}=Ae^{ikx}.$$

For $A\ne0$,

$$e^{ikL}=1.$$

Therefore

$$kL=2\pi n, \qquad n\in\mathbb Z,$$

so

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

2. Verify the orthonormality of $u_n(x)=L^{-1/2}e^{i2\pi nx/L}$.

Solution

Compute

$$\int_0^L u_m^*(x)u_n(x)\,dx = \frac{1}{L} \int_0^L e^{i2\pi(n-m)x/L}\,dx.$$

If $n=m$, the integral gives $L$, so the answer is $1$. If $n\ne m$,

$$\int_0^L e^{i2\pi(n-m)x/L}\,dx = \left. \frac{L}{i2\pi(n-m)} e^{i2\pi(n-m)x/L} \right\rvert_0^L =0,$$

because $e^{i2\pi(n-m)}=1$. Hence the result is $\delta_{mn}$.

3. Convert a large-$L$ sum over periodic-box wavenumbers into an integral.

Solution

The spacing is

$$\Delta k=\frac{2\pi}{L}.$$

For a smooth function $F(k)$,

$$\sum_n F(k_n) \approx \frac{1}{\Delta k} \int_{-\infty}^{\infty} F(k)\,dk = \frac{L}{2\pi} \int_{-\infty}^{\infty} F(k)\,dk.$$

In momentum variables, $dp=\hbar\,dk$, so

$$\sum_n F(p_n) \approx \frac{L}{2\pi\hbar} \int_{-\infty}^{\infty} F(p)\,dp.$$

4. Why can a periodic-box eigenstate have nonzero current while an infinite-square-well energy eigenstate has zero average current?

Solution

A periodic-box eigenstate can be a traveling wave,

$$u_n(x)=\frac{1}{\sqrt L}e^{ik_nx},$$

with current

$$j_n=\frac{\hbar k_n}{mL}.$$

There is no wall for the current to hit; it circulates around the identified interval. An infinite-square-well energy eigenstate is a real standing wave, proportional to $\sin(n\pi x/L)$, formed from equal left-moving and right-moving components. The opposite currents cancel, and the hard walls forbid net flux through the endpoints.