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.

The proof below starts with a finite periodic crystal, where translation phases are ordinary eigenvalues. The infinite-crystal version then explains how these sectors combine into wavepackets.

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.

Bloch eigenstates under lattice translations

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:

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

For the usual Schrödinger operator on a finite Born–von Karman crystal, these are ordinary normalizable eigenvectors at a discrete set of $\mathbf k$ values. Translation symmetry alone guarantees the sector decomposition; a discrete eigenbasis within each sector additionally uses the spectral properties of the cell problem, such as compact resolvent for a regular elliptic operator on a compact cell. 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:

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

The eigenvalues sorted by energy can be chosen periodic, including their multiplicities. Following an eigenvector through a degeneracy may instead exchange branch labels; this is why a continuation label need not return to itself.

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 represented by cell-periodic functions, with periodic boundary conditions in the domain of the already conjugated operator $H(\mathbf k)$. Equivalently, one may use the untransformed differential operator $H$ on cell functions with $\mathbf k$-quasiperiodic boundary conditions. These are two representations of the same fiber problem; the momentum shift and the quasiperiodic boundary phase must not be applied together. Each fiber problem has discrete eigenvalues under the usual elliptic conditions, while their union over continuous $\mathbf k$ forms energy bands.

To make the normalization explicit, choose $u_{n\mathbf k}$ with unit cell-average norm as below and define $\phi_{n\mathbf k}=e^{i\mathbf k\cdot\mathbf r}u_{n\mathbf k}/\sqrt{\Omega_c}$. These are generalized eigenfunctions, normalized to a delta function in the measure $d\mu(\mathbf k)=\Omega_c d^dk/(2\pi)^d$. They resolve normalizable wavepackets,

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

With this convention,

$$\|\Psi\|^2=\sum_n\int_{\mathcal B}d\mu(\mathbf k)\,|a_n(\mathbf k)|^2.$$

The finite-volume normalized wave is $\phi_{n\mathbf k}/\sqrt{N_c}$. Thus the two normalizations differ by the number of cells, without changing the cell eigenproblem. One can also construct a wavepacket at finite volume and then take 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.

An explicit Bloch-form test function: $\psi_k(x)=e^{ikx}[1+\tfrac12\cos(Gx)]$, with $G=2\pi/a$ and $ka=\pi/3$. Its density is periodic; amplitudes at corresponding positions advance by $\pi/3$ per cell. Its only Fourier amplitudes are $c_0=1$ and $c_{\pm1}=1/4$. An overall normalization is omitted. This illustrates translation symmetry; it is not an energy eigenstate of the free Hamiltonian.

The density of a single Bloch state is periodic:

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

Bloch form alone also does not imply an energy eigenstate: an arbitrary periodic $u$ lies in the chosen translation sector but must still solve $H(\mathbf k)u=Eu$. Nor does a periodic density 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 and layers enlarge this internal vector without changing the translation character. A Nambu description also enlarges the vector, but introduces particle–hole redundancy rather than doubling the number of physical one-particle states.

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. In particular, a bosonic Bogoliubov dynamical matrix is not generally Hermitian in the ordinary positive inner product. Translation blocks still exist, but the self-adjoint spectral argument above does not establish bosonic stability or mode completeness. Spatial Bloch theory should also be distinguished from time-domain Floquet theory: both exploit periodicity, but their spectra and physical interpretations are different.

Exercises

Exercise 1: energy degeneracy and translation

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 2: identify the translation symmetry

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.

Exercise 3: a twisted boundary condition

A chain of $N$ cells satisfies $\psi(x+Na)=e^{i\theta}\psi(x)$ instead of strict periodicity. Find its allowed $k$ values. What does the active translation $T_{Na}$ do?

Solution

Bloch quasiperiodicity gives $e^{ikNa}=e^{i\theta}$, hence

$$k_m=\frac{2\pi m+\theta}{Na},\qquad m=0,\ldots,N-1$$

modulo $2\pi/a$. The active operator shifts the argument backward, so $T_{Na}\psi=e^{-i\theta}\psi$. There are still $N$ distinct momenta, but the mesh is shifted. Applying both this boundary twist and an equivalent uniform vector-potential shift would count the same twist twice.

Exercise 4: normalize the illustrated function

For the figure’s $u(x)=1+\tfrac12\cos(Gx)$, calculate the cell-average norm and the probabilities of the three canonical momenta. Is this a free-particle energy eigenstate?

Solution

Orthogonality of reciprocal Fourier modes gives

$$\frac1a\int_0^a|u(x)|^2dx=1+\frac18=\frac98.$$

Divide $u$ by $\sqrt{9/8}$ to obtain unit cell-average norm. The normalized momentum probabilities are $8/9$ at $\hbar k$ and $1/18$ at each of $\hbar(k\pm G)$. Their sum is one, and their symmetric offsets give $\langle p\rangle=\hbar k$. Thus a periodic density can carry current.

For the free Hamiltonian, however, these three components have different energies at $ka=\pi/3$. The function has definite crystal momentum but not definite energy. Free time evolution changes its intracell profile while preserving its translation character and periodic density.

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.
• P. Kuchment, “An overview of periodic elliptic operators,” Bulletin of the American Mathematical Society 53, 343–414 (2016), doi:10.1090/bull/1528.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. IV: Analysis of Operators (Academic Press, 1978), Section XIII.16.
• J. Zak, “Magnetic Translation Group,” Physical Review 134, A1602–A1606 (1964), doi:10.1103/PhysRev.134.A1602.