Bloch’s Theorem

Bloch’s theorem classifies one-particle states by the characters of a crystal’s discrete translation group. For a Hamiltonian with exact Bravais-lattice symmetry, its stationary states can be chosen so that translation by any lattice vector changes the state only by a phase. In position space,

$$\psi_{n\mathbf k}(\mathbf r+\mathbf R) = e^{i\mathbf k\cdot\mathbf R} \psi_{n\mathbf k}(\mathbf r),$$

or, equivalently,

$$\psi_{n\mathbf k}(\mathbf r) = e^{i\mathbf k\cdot\mathbf r} u_{n\mathbf k}(\mathbf r), \qquad u_{n\mathbf k}(\mathbf r+\mathbf R) = u_{n\mathbf k}(\mathbf r).$$

The theorem is exact whenever its symmetry assumptions hold. It does not assume a weak potential, nearly free electrons, inversion symmetry, time-reversal symmetry, or one atom per primitive cell. It supplies the kinematic organization into crystal-momentum sectors; it does not by itself determine the dispersion, filling, lifetime, or experimental spectral weight.

Crystals and Lattices defines the translation group, Reciprocal Lattice defines its dual characters, and Brillouin Zones chooses representatives of those characters. This page is the canonical home for the theorem, its proof, its finite-size normalization, and its limitations.

Required background. Crystals and Lattices supplies the discrete translation group and finite crystal; Reciprocal Lattice supplies its translation characters; Brillouin Zones supplies one representative set for the resulting crystal-momentum classes.

Theorem

Let $\Lambda$ be a Bravais lattice,

$$\Lambda = \left\{ \mathbf R = \sum_{i=1}^{d}n_i\mathbf a_i : n_i\in\mathbb Z \right\},$$

and use the active translation convention

$$\left( T_{\mathbf R}\psi \right)(\mathbf r) = \psi(\mathbf r-\mathbf R).$$

Suppose a self-adjoint one-particle Hamiltonian $H$ and its operator domain are invariant under every $T_{\mathbf R}$:

$$T_{\mathbf R}HT_{\mathbf R}^{-1} = H, \qquad \mathbf R\in\Lambda.$$

Then the Hilbert space decomposes into translation-character sectors labeled by a wavevector $\mathbf k$, and stationary states may be chosen to satisfy

$$H|\psi_{n\mathbf k}\rangle = E_n(\mathbf k) |\psi_{n\mathbf k}\rangle,$$ $$T_{\mathbf R} |\psi_{n\mathbf k}\rangle = e^{-i\mathbf k\cdot\mathbf R} |\psi_{n\mathbf k}\rangle.$$

The label $\mathbf k$ is defined only modulo a reciprocal-lattice vector:

$$\mathbf k \sim \mathbf k+\mathbf G, \qquad \mathbf G\in\Lambda^\ast.$$

On a finite Born–von Karman crystal, these are ordinary normalizable eigenvectors at a discrete set of $\mathbf k$ values. On an infinite crystal, the precise statement is a Bloch–Floquet direct-integral decomposition; the individual Bloch waves are generally generalized eigenfunctions rather than square-integrable vectors on all of $\mathbb R^d$.

Assumptions and Scope

The theorem needs three structural ingredients.

1. A full-rank translation lattice. The Hamiltonian repeats under the same primitive translations throughout the bulk.
2. A symmetry-compatible operator problem. The differential expression, internal matrix structure, nonlocal kernels, operator domain, and boundary conditions must all respect the lattice translations being used.
3. A stationary linear one-particle problem. This includes exact one-particle Hamiltonians and effective linear problems such as mean-field, Kohn–Sham, photonic, phononic, and Bogoliubov–de Gennes eigenproblems, with the qualifications appropriate to each.

For the familiar scalar Schrödinger operator,

$$H = \frac{\mathbf p^2}{2m} + V(\mathbf r), \qquad V(\mathbf r+\mathbf R) = V(\mathbf r),$$

the kinetic term is invariant under all continuous translations and the potential reduces that symmetry to $\Lambda$. The same argument works for periodic matrix potentials, spinors with spin–orbit coupling, periodic effective masses, and suitable periodic nonlocal operators because the proof uses commutation with translations rather than the detailed form of $H$.

External surfaces and arbitrary finite boundaries do not obey the premise. Born–von Karman boundary conditions are a controlled finite-volume construction that preserves translation symmetry; an actual open crystal must instead be treated with its reduced bulk, surface, or slab symmetries.

Finite-Crystal Proof

The cleanest proof starts with a finite periodic crystal. Take $N_i$ primitive cells along $\mathbf a_i$ and identify

$$\mathbf r \sim \mathbf r+N_i\mathbf a_i.$$

The primitive translation operators

$$T_i \equiv T_{\mathbf a_i}$$

are unitary, commute with one another, and satisfy

$$T_i^{N_i} = I.$$

They therefore represent the finite abelian group

$$\mathbb Z_{N_1} \times \cdots \times \mathbb Z_{N_d}.$$

Every unitary is normal, and commuting normal operators admit a simultaneous spectral decomposition. The Hilbert space can consequently be split into common eigenspaces of all $T_i$. Because $[H,T_i]=0$, the Hamiltonian preserves each such eigenspace and can be diagonalized within it.

Let a common translation eigenvector obey

$$T_i|\psi\rangle = \lambda_i|\psi\rangle.$$

Unitarity gives $|\lambda_i|=1$, while $T_i^{N_i}=I$ gives

$$\lambda_i^{N_i} = 1.$$

Write the phases as

$$\lambda_i = e^{-i\mathbf k\cdot\mathbf a_i}.$$

For a general lattice vector $\mathbf R=\sum_i n_i\mathbf a_i$,

$$\begin{aligned} T_{\mathbf R}|\psi\rangle &= \prod_i T_i^{n_i}|\psi\rangle \\ &= \prod_i e^{-in_i\mathbf k\cdot\mathbf a_i} |\psi\rangle \\ &= e^{-i\mathbf k\cdot\mathbf R} |\psi\rangle. \end{aligned}$$

This proves the translation-eigenvalue form. Since $H$ preserves each character sector, its eigenstates can be chosen with definite $\mathbf k$ and an additional label $n$ that distinguishes eigenstates within that sector.

What degeneracy changes

The statement “an energy eigenstate is a translation eigenstate” needs care. If an energy is degenerate across different crystal momenta, an arbitrary linear combination of those energy eigenvectors need not have definite $\mathbf k$. The correct statement is:

A complete energy eigenbasis can be chosen to diagonalize the commuting lattice translations simultaneously.

Within a degenerate energy eigenspace, diagonalize the restricted translation operators. Degeneracy does not invalidate the theorem; it makes the choice of basis nonunique.

Position-Space Form

Project the translation equation onto position:

$$\psi_{n\mathbf k}(\mathbf r-\mathbf R) = e^{-i\mathbf k\cdot\mathbf R} \psi_{n\mathbf k}(\mathbf r).$$

Replacing $\mathbf r$ by $\mathbf r+\mathbf R$ gives the quasiperiodic boundary condition

$$\psi_{n\mathbf k}(\mathbf r+\mathbf R) = e^{i\mathbf k\cdot\mathbf R} \psi_{n\mathbf k}(\mathbf r).$$

Now define

$$u_{n\mathbf k}(\mathbf r) \equiv e^{-i\mathbf k\cdot\mathbf r} \psi_{n\mathbf k}(\mathbf r).$$

Then

$$\begin{aligned} u_{n\mathbf k}(\mathbf r+\mathbf R) &= e^{-i\mathbf k\cdot(\mathbf r+\mathbf R)} \psi_{n\mathbf k}(\mathbf r+\mathbf R) \\ &= e^{-i\mathbf k\cdot\mathbf r} \psi_{n\mathbf k}(\mathbf r) \\ &= u_{n\mathbf k}(\mathbf r). \end{aligned}$$

Thus $u_{n\mathbf k}$ is lattice periodic and the wavefunction has Bloch form,

$$\psi_{n\mathbf k}(\mathbf r) = e^{i\mathbf k\cdot\mathbf r} u_{n\mathbf k}(\mathbf r).$$

This equivalence also works for multicomponent wavefunctions. The periodic object may be a spinor, an orbital vector, a Nambu spinor, or a field profile; translation symmetry does not require it to be a scalar.

Crystal Momentum and Reciprocal Equivalence

Two wavevectors label the same translation character when

$$e^{-i(\mathbf k'-\mathbf k)\cdot\mathbf R} = 1$$

for every $\mathbf R\in\Lambda$. By definition this holds exactly when

$$\mathbf k'-\mathbf k = \mathbf G \in \Lambda^\ast.$$

Crystal momentum therefore lives on the quotient

$$\mathbb R^d/\Lambda^\ast,$$

which is a $d$-dimensional torus. A Brillouin zone is a choice of one representative from each equivalence class, not an additional physical boundary in momentum space.

Reciprocal sewing

The same physical Bloch wave can be relabeled by $\mathbf k+\mathbf G$:

$$e^{i(\mathbf k+\mathbf G)\cdot\mathbf r} u_{n,\mathbf k+\mathbf G}(\mathbf r) = e^{i\mathbf k\cdot\mathbf r} u_{n\mathbf k}(\mathbf r).$$

One compatible choice is

$$u_{n,\mathbf k+\mathbf G}(\mathbf r) = e^{-i\mathbf G\cdot\mathbf r} u_{n\mathbf k}(\mathbf r).$$

The factor $e^{-i\mathbf G\cdot\mathbf r}$ is itself lattice periodic. Consequently, the spectrum is reciprocal-periodic as a set:

$$\operatorname{Spec}H(\mathbf k+\mathbf G) = \operatorname{Spec}H(\mathbf k).$$

At crossings, a fixed numerical band index can be permuted by continuation, so setwise periodicity is more fundamental than a globally fixed ordering of individual branches.

What crystal momentum is not

$\hbar\mathbf k$ is the generator label for discrete lattice translations. It is not generally the mechanical momentum $m\mathbf v$, nor is it necessarily the expectation value of the canonical momentum operator. A Bloch wave contains ordinary momentum components $\hbar(\mathbf k+\mathbf G)$, and a periodic potential exchanges reciprocal-lattice momentum among those components.

The Cell Hamiltonian

Substituting $\psi_{n\mathbf k}=e^{i\mathbf k\cdot\mathbf r}u_{n\mathbf k}$ into

$$H\psi_{n\mathbf k} = E_n(\mathbf k)\psi_{n\mathbf k}$$

gives a family of cell problems,

$$H(\mathbf k) u_{n\mathbf k} = E_n(\mathbf k) u_{n\mathbf k},$$

where

$$H(\mathbf k) \equiv e^{-i\mathbf k\cdot\mathbf r} H e^{i\mathbf k\cdot\mathbf r}.$$

For a scalar periodic potential,

$$H(\mathbf k) = \frac{1}{2m} \left( -i\hbar\nabla + \hbar\mathbf k \right)^2 + V(\mathbf r).$$

The operator acts on cell-periodic functions. Instead of solving one differential equation over the entire crystal, one solves a parameterized eigenproblem on one primitive cell for each $\mathbf k$ in a Brillouin zone.

Reciprocal shifts are implemented by unitary conjugation:

$$H(\mathbf k+\mathbf G) = e^{-i\mathbf G\cdot\mathbf r} H(\mathbf k) e^{i\mathbf G\cdot\mathbf r}.$$

This relation explains both spectral periodicity and the change of the cell-periodic eigenvector across an identified zone boundary.

Infinite Crystals and Bloch–Floquet Theory

In an infinite crystal, a function with nonzero periodic density cannot be normalized over all space. The literal finite-dimensional simultaneous-diagonalization proof must therefore be replaced by a spectral decomposition.

Under standard hypotheses for a periodic self-adjoint operator, the Bloch–Floquet transform gives

$$L^2(\mathbb R^d) \cong \int_{\mathcal B}^{\oplus} \mathcal H_{\mathbf k}\, \frac{\Omega_c\,d^dk}{(2\pi)^d},$$

and

$$H \cong \int_{\mathcal B}^{\oplus} H(\mathbf k)\, \frac{\Omega_c\,d^dk}{(2\pi)^d},$$

where $\mathcal B$ is any Brillouin-zone fundamental domain and $\mathcal H_{\mathbf k}$ is a cell Hilbert space with $\mathbf k$-quasiperiodic boundary conditions. Each fiber problem has discrete eigenvalues under the usual elliptic conditions, while their union over continuous $\mathbf k$ forms energy bands.

The symbols $\psi_{n\mathbf k}$ in an infinite perfect crystal should therefore be read like plane waves: they are generalized eigenfunctions used to resolve normalizable wavepackets,

$$\Psi(\mathbf r) = \sum_n \int_{\mathcal B} \frac{\Omega_c\,d^dk}{(2\pi)^d} \, a_n(\mathbf k) \psi_{n\mathbf k}(\mathbf r).$$

A physical bulk state has an integrable amplitude $a_n(\mathbf k)$, or is first defined in a finite periodic volume and then taken to the thermodynamic limit.

Finite-Size Quantization and State Counting

Born–von Karman periodicity imposes

$$\psi(\mathbf r+N_i\mathbf a_i) = \psi(\mathbf r).$$

Combining this with Bloch quasiperiodicity gives

$$e^{i\mathbf k\cdot N_i\mathbf a_i} = 1.$$

If reciprocal primitive vectors obey

$$\mathbf a_i\cdot\mathbf b_j = 2\pi\delta_{ij},$$

the allowed momenta can be represented as

$$\mathbf k = \sum_{i=1}^{d} \frac{m_i}{N_i} \mathbf b_i, \qquad m_i=0,1,\ldots,N_i-1.$$

There are exactly

$$N_c = \prod_{i=1}^{d}N_i$$

distinct $\mathbf k$ values in one reciprocal primitive cell. Thus every isolated band contributes $N_c$ one-particle states before spin or other internal multiplicities. This counting is the momentum-space version of the fact that a lattice with $N_c$ cells has $N_c$ independent translates of each localized orbital.

Normalization

Use the cell-average inner product

$$\langle u|v\rangle_c = \frac{1}{\Omega_c} \int_{\mathrm{cell}} d^dr\, u^\dagger(\mathbf r)v(\mathbf r).$$

If

$$\langle u_{n\mathbf k} | u_{m\mathbf k} \rangle_c = \delta_{nm},$$

then the finite-volume normalized state is

$$\psi_{n\mathbf k}^{\mathrm{norm}}(\mathbf r) = \frac{1}{\sqrt{N_c\Omega_c}} e^{i\mathbf k\cdot\mathbf r} u_{n\mathbf k}(\mathbf r).$$

The frequently written form $\psi=e^{i\mathbf k\cdot\mathbf r}u$ suppresses this overall normalization. For two allowed momenta, the sum over cells produces

$$\sum_{\mathbf R} e^{i(\mathbf k'-\mathbf k)\cdot\mathbf R} = N_c\, \delta_{\mathbf k,\mathbf k'},$$

so states with different translation characters are orthogonal.

Plane-Wave Content

Because $u_{n\mathbf k}$ is lattice periodic, it has a reciprocal-lattice Fourier series:

$$u_{n\mathbf k}(\mathbf r) = \sum_{\mathbf G} c_{n\mathbf k}(\mathbf G) e^{i\mathbf G\cdot\mathbf r}.$$

Therefore

$$\psi_{n\mathbf k}(\mathbf r) = \sum_{\mathbf G} c_{n\mathbf k}(\mathbf G) e^{i(\mathbf k+\mathbf G)\cdot\mathbf r}.$$

A Bloch state has one crystal momentum $\mathbf k$ but generally many canonical plane-wave momenta $\hbar(\mathbf k+\mathbf G)$. The periodic Hamiltonian couples precisely those plane-wave components whose wavevectors differ by a reciprocal-lattice vector.

Translation changes a Bloch amplitude by a cell-dependent phase while leaving its probability density periodic. The same state is a superposition of reciprocal components $\mathbf q=\mathbf k+\mathbf G$; crystal momentum labels the common equivalence class rather than a single canonical momentum.

The density of a single Bloch state is periodic:

$$|\psi_{n\mathbf k}(\mathbf r+\mathbf R)|^2 = |\psi_{n\mathbf k}(\mathbf r)|^2.$$

This does not imply that the state carries no current. A phase gradient and the internal structure of $u_{n\mathbf k}$ can support a nonzero expectation value of the velocity.

Worked Examples

Free particle in a chosen lattice

The free-particle Hamiltonian is invariant under every translation, so it is also invariant under any chosen lattice. Restrict $\mathbf k$ to one Brillouin zone and label plane waves by reciprocal vectors:

$$\psi_{\mathbf G,\mathbf k}(\mathbf r) = e^{i(\mathbf k+\mathbf G)\cdot\mathbf r}.$$

Its periodic factor and energy are

$$u_{\mathbf G,\mathbf k}(\mathbf r) = e^{i\mathbf G\cdot\mathbf r},$$ $$E_{\mathbf G}(\mathbf k) = \frac{\hbar^2}{2m} \left| \mathbf k+\mathbf G \right|^2.$$

Folding the free parabola into one zone produces many crossing branches. A periodic potential can couple branches with the same reduced $\mathbf k$, which is the starting point of Nearly Free Electrons. Bloch’s theorem itself is already exact before that approximation is made.

One-dimensional periodic potential

For $V(x+a)=V(x)$,

$$H(k) = \frac{1}{2m} \left( -i\hbar\frac{d}{dx} + \hbar k \right)^2 + V(x)$$

acts on periodic functions over $0\leq x<a$. For a regular scalar problem, one may solve with

$$u(a) = u(0), \qquad u'(a) = u'(0).$$

Equivalently, solve the original Hamiltonian on one cell with quasiperiodic conditions

$$\psi(a) = e^{ika}\psi(0), \qquad \psi'(a) = e^{ika}\psi'(0).$$

Scanning $k$ through $[-\pi/a,\pi/a)$ yields the complete bulk spectrum.

Nearest-neighbor tight-binding chain

Let $|R\rangle$ be one localized orbital in each cell and

$$H = -t \sum_R \left( |R+a\rangle\langle R| + |R\rangle\langle R+a| \right).$$

The normalized Bloch sum is

$$|k\rangle = \frac{1}{\sqrt N} \sum_R e^{ikR} |R\rangle.$$

Since $T_a|R\rangle=|R+a\rangle$,

$$T_a|k\rangle = e^{-ika}|k\rangle,$$

and direct application of $H$ gives

$$H|k\rangle = -2t\cos(ka)|k\rangle.$$

The form of the Bloch sum follows from translation symmetry; the cosine dispersion follows from the additional nearest-neighbor hopping model.

Two-site motif

A honeycomb lattice has a triangular Bravais lattice with two sites, conventionally labeled $A$ and $B$, in each primitive cell. Bloch’s theorem uses the triangular translations, not a fictitious one-site honeycomb Bravais lattice. At each $\mathbf k$, the cell-periodic object is a two-component orbital vector,

$$|u_{n\mathbf k}\rangle = \begin{pmatrix} u_{nA}(\mathbf k) \\ u_{nB}(\mathbf k) \end{pmatrix},$$

and the band index comes from diagonalizing a $2\times2$ Bloch Hamiltonian. More orbitals, spin, layers, and Nambu doubling enlarge this internal vector without changing the translation character.

Physical Interpretation

Bloch’s theorem separates two kinds of spatial structure:

• $e^{i\mathbf k\cdot\mathbf r}$ records how the state transforms between equivalent cells;
• $u_{n\mathbf k}(\mathbf r)$ records the detailed amplitude, phase, orbital, and spin structure within a cell.

The separation is basis dependent, but the translation character is physical. A local probe can resolve the intracell structure; a bulk momentum-resolved probe organizes spectral weight by crystal momentum modulo reciprocal vectors.

The band label $n$ is not supplied by the translation group. It enumerates the eigenstates left after fixing $\mathbf k$. Different microscopic Hamiltonians with the same lattice have the same allowed translation characters but different $E_n(\mathbf k)$, eigenvectors, gaps, and matrix elements.

Extensions and Limitations

Internal degrees of freedom and nonlocal operators

If a spinor or orbital Hamiltonian commutes with the lattice translations, each component carries the same overall translation character. Spin–orbit coupling may remove spin degeneracy and entangle spin with orbital motion, but it does not invalidate Bloch form. A periodic nonlocal kernel $K$ is compatible when

$$K(\mathbf r+\mathbf R,\mathbf r'+\mathbf R) = K(\mathbf r,\mathbf r').$$

Interacting many-body systems

For an interacting Hamiltonian invariant under simultaneous translation of all particles, many-body eigenstates can be labeled by a total crystal momentum modulo $\mathbf G$. This is a representation statement about the many-body translation group. It does not imply that the exact state is a Slater determinant of one-particle Bloch orbitals or that its excitations are sharp independent-electron bands.

Translation-invariant one-particle Green functions remain block diagonal in $\mathbf k$, but poles can broaden, split, lose weight, or be replaced by continua. The one-particle Bloch theorem and many-body crystal momentum must not be conflated.

Magnetic fields

In a uniform magnetic field, a vector potential is not generally invariant under ordinary translations. The Hamiltonian may commute instead with gauge-covariant magnetic translations, whose operators can fail to commute with one another. For rational magnetic flux through a cell, an enlarged magnetic unit cell and magnetic Brillouin zone may restore a Bloch-like classification. See Magnetic Translations.

Disorder, boundaries, and incommensurability

Static disorder breaks exact lattice translations in a given sample, even if an ensemble average restores them statistically. An open surface breaks translations normal to the boundary but can preserve those parallel to it, giving a surface crystal momentum. Quasicrystals and incommensurate structures lack a finite ordinary primitive cell; higher-dimensional embeddings or approximants may be useful, but the elementary theorem above does not apply unchanged.

Other periodic eigenproblems

The same representation logic organizes phonon polarization vectors, photonic modes, magnons, and other linear periodic problems. Their inner products, constraints, and generalized eigenvalue structures can differ from the electronic Schrödinger problem. Spatial Bloch theory should also be distinguished from time-domain Floquet theory: both exploit periodicity, but their spectra and physical interpretations are different.

Common Mistakes

Reversing the translation phase

With

$$(T_{\mathbf R}\psi)(\mathbf r) = \psi(\mathbf r-\mathbf R),$$

the translation eigenvalue is $e^{-i\mathbf k\cdot\mathbf R}$, while the shifted-coordinate relation is

$$\psi(\mathbf r+\mathbf R) = e^{i\mathbf k\cdot\mathbf R}\psi(\mathbf r).$$

Both signs are correct in their respective equations. Changing the definition of the active translation operator changes the bookkeeping.

Calling the Bloch wave periodic

$u_{n\mathbf k}$ and $|\psi_{n\mathbf k}|^2$ are periodic. The wavefunction $\psi_{n\mathbf k}$ is generally quasiperiodic and becomes cell periodic only when its translation character is trivial.

Assuming every energy eigenvector already has definite crystal momentum

An arbitrary vector in a degenerate energy subspace may mix translation sectors. Choose a simultaneous eigenbasis of $H$ and the translations.

Treating the theorem as a weak-potential approximation

Weak-potential and tight-binding methods approximate particular band dispersions. Bloch’s theorem precedes both and remains exact for a strongly varying periodic one-particle Hamiltonian.

Equating crystal and mechanical momentum

Crystal momentum is conserved modulo reciprocal vectors under lattice-symmetric dynamics. Mechanical momentum and velocity depend on the Hamiltonian and on the state’s full periodic structure.

Forgetting normalization volume

If the periodic factor has unit cell-average norm, $e^{i\mathbf k\cdot\mathbf r}u_{n\mathbf k}$ has norm squared $N_c\Omega_c$ in a finite crystal. Include $1/\sqrt{N_c\Omega_c}$ when an actual unit-normalized wavefunction is required.

Ignoring basis sewing at a zone boundary

$\mathbf k$ and $\mathbf k+\mathbf G$ label the same translation character, but their cell-periodic representatives differ by a periodic unitary. Raw eigenvector components and Berry connections need not be numerically identical across that boundary.

Problem-Solving Checklist

Before invoking Bloch’s theorem, check:

1. What translations leave the full operator problem invariant?
2. Is the translation convention active or passive?
3. What primitive direct and reciprocal vectors are being used?
4. Is $\mathbf k$ discrete in a finite periodic sample or continuous in an infinite-crystal limit?
5. Which normalization is used for $u_{n\mathbf k}$ and $\psi_{n\mathbf k}$?
6. Which internal basis and orbital embedding define the cell Hamiltonian?
7. Are degeneracies, reciprocal-boundary sewing, magnetic phases, disorder, or surfaces relevant?

Exercises

Exercise 1: derive Bloch form

Starting from

$$T_{\mathbf R}|\psi_{\mathbf k}\rangle = e^{-i\mathbf k\cdot\mathbf R} |\psi_{\mathbf k}\rangle,$$

derive the quasiperiodic position-space condition and prove that $u_{\mathbf k}(\mathbf r)=e^{-i\mathbf k\cdot\mathbf r}\psi_{\mathbf k}(\mathbf r)$ is lattice periodic.

Solution

The active convention gives

$$\psi_{\mathbf k}(\mathbf r-\mathbf R) = e^{-i\mathbf k\cdot\mathbf R} \psi_{\mathbf k}(\mathbf r).$$

Replace $\mathbf r$ by $\mathbf r+\mathbf R$:

$$\psi_{\mathbf k}(\mathbf r+\mathbf R) = e^{i\mathbf k\cdot\mathbf R} \psi_{\mathbf k}(\mathbf r).$$

Then

$$\begin{aligned} u_{\mathbf k}(\mathbf r+\mathbf R) &= e^{-i\mathbf k\cdot(\mathbf r+\mathbf R)} \psi_{\mathbf k}(\mathbf r+\mathbf R) \\ &= e^{-i\mathbf k\cdot\mathbf r} \psi_{\mathbf k}(\mathbf r) \\ &= u_{\mathbf k}(\mathbf r). \end{aligned}$$

Thus $\psi_{\mathbf k}=e^{i\mathbf k\cdot\mathbf r}u_{\mathbf k}$ with periodic $u_{\mathbf k}$.

Exercise 2: reciprocal equivalence

Prove that two wavevectors define the same character of every lattice translation if and only if their difference is a reciprocal-lattice vector.

Solution

The characters agree when

$$e^{-i\mathbf k'\cdot\mathbf R} = e^{-i\mathbf k\cdot\mathbf R}$$

for all $\mathbf R\in\Lambda$, or

$$e^{-i(\mathbf k'-\mathbf k)\cdot\mathbf R} = 1.$$

The reciprocal lattice is precisely

$$\Lambda^\ast = \left\{ \mathbf G: e^{i\mathbf G\cdot\mathbf R}=1 \ \text{for every}\ \mathbf R\in\Lambda \right\}.$$

Therefore $\mathbf k'-\mathbf k\in\Lambda^\ast$. The converse follows immediately from the same definition.

Exercise 3: finite momentum mesh

A rectangular two-dimensional crystal contains $N_x\times N_y$ primitive cells with

$$\mathbf a_x = a_x\hat{\mathbf x}, \qquad \mathbf a_y = a_y\hat{\mathbf y}.$$

Find the allowed crystal momenta under Born–von Karman boundary conditions and count them in the first Brillouin zone.

Solution

Global periodicity requires

$$e^{ik_xN_xa_x} = 1, \qquad e^{ik_yN_ya_y} = 1.$$

Thus

$$k_x = \frac{2\pi m_x}{N_xa_x}, \qquad k_y = \frac{2\pi m_y}{N_ya_y},$$

with $m_x$ and $m_y$ chosen from any $N_x$ and $N_y$ consecutive integers. There are $N_xN_y=N_c$ inequivalent pairs in one Brillouin zone, hence $N_c$ states per band before internal multiplicities.

Exercise 4: tight-binding translation

For the nearest-neighbor chain in the worked example, verify both the translation eigenvalue and the energy dispersion of $|k\rangle$.

Solution

Translate the Bloch sum and relabel $R'=R+a$:

$$\begin{aligned} T_a|k\rangle &= \frac{1}{\sqrt N} \sum_R e^{ikR}|R+a\rangle \\ &= \frac{1}{\sqrt N} \sum_{R'}e^{ik(R'-a)}|R'\rangle \\ &= e^{-ika}|k\rangle. \end{aligned}$$

For the Hamiltonian,

$$\begin{aligned} H|k\rangle &= -\frac{t}{\sqrt N} \sum_R e^{ikR} \left( |R+a\rangle+|R-a\rangle \right) \\ &= -t \left( e^{-ika}+e^{ika} \right) |k\rangle \\ &= -2t\cos(ka)|k\rangle. \end{aligned}$$

The sign of the translation eigenvalue follows from the active convention, while the energy is even in $k$ for this real, inversion-symmetric model.

Exercise 5: reciprocal plane-wave support

Show that a Bloch function’s Fourier transform can have support only at wavevectors $\mathbf q=\mathbf k+\mathbf G$. Then explain why this does not mean that every coefficient is nonzero.

Solution

Expand the periodic factor in its reciprocal Fourier series:

$$u_{\mathbf k}(\mathbf r) = \sum_{\mathbf G} c_{\mathbf k}(\mathbf G) e^{i\mathbf G\cdot\mathbf r}.$$

Multiplication by $e^{i\mathbf k\cdot\mathbf r}$ gives

$$\psi_{\mathbf k}(\mathbf r) = \sum_{\mathbf G} c_{\mathbf k}(\mathbf G) e^{i(\mathbf k+\mathbf G)\cdot\mathbf r}.$$

Thus no other wavevectors occur. Symmetry, the form of the periodic potential, orbital selection rules, or a special eigenstate can force particular coefficients $c_{\mathbf k}(\mathbf G)$ to vanish.

Exercise 6: degeneracy audit

Suppose $|\psi_{\mathbf k}\rangle$ and $|\psi_{\mathbf k'}\rangle$ have the same energy but distinct translation characters. Is

$$|\phi\rangle = \frac{1}{\sqrt2} \left( |\psi_{\mathbf k}\rangle + |\psi_{\mathbf k'}\rangle \right)$$

an energy eigenstate? Is it a translation eigenstate?

Solution

It is an energy eigenstate because both terms have the same energy:

$$H|\phi\rangle = E|\phi\rangle.$$

Under a lattice translation,

$$T_{\mathbf R}|\phi\rangle = \frac{1}{\sqrt2} \left( e^{-i\mathbf k\cdot\mathbf R} |\psi_{\mathbf k}\rangle + e^{-i\mathbf k'\cdot\mathbf R} |\psi_{\mathbf k'}\rangle \right).$$

This is not proportional to $|\phi\rangle$ unless the two characters coincide. The example shows why the theorem promises a simultaneous eigenbasis, not that every vector chosen inside a degenerate energy subspace has definite crystal momentum.

Exercise 7: diagnose a failed premise

For each system, state whether the elementary Bloch theorem applies without modification: (a) a perfect periodic scalar potential, (b) one impurity in an otherwise periodic crystal, (c) a slab periodic only in two directions, and (d) a charged particle in a periodic potential plus a uniform magnetic field.

Solution

(a) Yes, provided the operator domain or finite boundary conditions are translation compatible.

(b) No for the exact impurity Hamiltonian. Bulk Bloch states can still be used as a basis for scattering theory, but they are not exact eigenstates of the disordered system.

(c) Partly. Two-dimensional Bloch classification survives for translations parallel to the slab, while momentum normal to the surface is not a conserved crystal label.

(d) Not generally with ordinary translations because the vector potential changes under translation. Magnetic translations supply the appropriate gauge-covariant symmetry; their algebra depends on the flux through a cell.

Connections

• The chapter gateway places this exact symmetry theorem in the local dependency graph and separates it from later model approximations and outputs.
• Band Theory Overview carries the translation-sector result into fermionic filling, material classification, and the hierarchy of electronic-structure approximations.
• Brillouin Zones constructs a fundamental domain for the crystal-momentum labels proved here.
• Nearly Free Electrons applies degenerate perturbation theory to weakly coupled free branches in each Bloch sector.
• Tight-Binding Models constructs the same Bloch sectors from localized crystalline orbitals.
• Wannier Functions chooses and Fourier-transforms a Bloch frame into localized orbitals and owns the resulting gauge, localization, and obstruction questions; the theorem here supplies the translation fibers but does not choose that frame.
• Symmetry of Bloch States begins after the translation-sector theorem and owns the little-group representations, nontranslation band labels, compatibility relations, and crystalline degeneracy taxonomy within those fibers.
• Chern Numbers in Band Theory turns the cell-periodic Bloch fibers into an occupied bundle and a gauge-invariant Hall invariant.
• Translations and Momentum develops continuous and discrete translation operators abstractly.
• Translation-Invariant Hamiltonians explains symmetry sectors and conserved quantum numbers beyond the crystalline application.
• Periodic Boundary Conditions develops finite-volume momentum quantization.
• Tight-Binding Model applies Bloch sums to localized lattice orbitals.
• Bloch Theorem formula card is the compact lookup version of the result.
• Conventions for Quantum Matter fixes the translation sign, cell inner product, reciprocal sewing, and basis conventions used here.

References

• N. W. Ashcroft and N. D. Mermin, Solid State Physics (Holt, Rinehart and Winston, 1976), Chapters 8–9.
• F. Bloch, “Über die Quantenmechanik der Elektronen in Kristallgittern,” Zeitschrift für Physik 52, 555–600 (1929), doi:10.1007/BF01339455.
• M. S. P. Eastham, The Spectral Theory of Periodic Differential Equations (Scottish Academic Press, 1973).
• C. Kittel, Introduction to Solid State Physics, 8th ed. (Wiley, 2004), Chapter 7.
• P. Kuchment, “An overview of periodic elliptic operators,” Bulletin of the American Mathematical Society 53, 343–414 (2016), doi:10.1090/bull/1528.
• P. Kuchment, “Floquet theory for partial differential equations,” Russian Mathematical Surveys 37(4), 1–60 (1982), doi:10.1070/RM1982v037n04ABEH003965.
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