Bloch Theorem

This card is a compact lookup. Bloch’s Theorem is the canonical article for the proof, finite-size normalization, Bloch–Floquet formulation, examples, and limitations.

Formula

For a single-particle Hamiltonian invariant under lattice translations $\mathbf R$,

$$T_{\mathbf R}H = HT_{\mathbf R},$$

energy eigenstates can be chosen as simultaneous translation eigenstates:

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

In position space this gives Bloch form:

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

where the cell-periodic part satisfies

$$u_{n\mathbf k}(\mathbf r+\mathbf R) = u_{n\mathbf k}(\mathbf r).$$

Crystal momentum is defined modulo reciprocal lattice vectors:

$$\mathbf k\sim\mathbf k+\mathbf G, \qquad e^{i\mathbf G\cdot\mathbf R}=1.$$

Assumptions

• The Hamiltonian has exact lattice-translation symmetry.
• A single-particle or mean-field one-particle Hamiltonian is being diagonalized.
• Boundary conditions are compatible with the lattice in the large-system or Born-von Karman sense.
• Disorder, surfaces, magnetic translations, interactions beyond an effective one-particle description, and time-dependent driving require extra structure.

Validity

Bloch theorem organizes eigenstates into bands labeled by $n$ and crystal momentum $\mathbf k$ in a Brillouin zone. It is a symmetry statement, not a statement that the electrons are free. The periodic factor $u_{n\mathbf k}$ contains the effect of the lattice potential.

In interacting many-body systems, translation symmetry still exists, but the elementary excitations and quantum numbers need not be simple single-particle Bloch waves.

Common Mistakes

• Treating crystal momentum as ordinary mechanical momentum without the modulo-reciprocal-lattice caveat.
• Forgetting that $u_{n\mathbf k}$ is periodic while $\psi_{n\mathbf k}$ generally is not periodic in a single unit cell.
• Applying Bloch theorem to disordered or surface-dominated systems without stating the approximation.
• Ignoring degeneracies when choosing a smooth band basis across the Brillouin zone.

Quick Check

If $\mathbf G$ is a reciprocal lattice vector, why do $\mathbf k$ and $\mathbf k+\mathbf G$ give the same translation eigenvalue?

Solution

For every lattice vector $\mathbf R$,

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

Thus translations cannot distinguish $\mathbf k$ from $\mathbf k+\mathbf G$.

Canonical Links

• Bloch’s Theorem
• Tight-Binding Model
• Math Needed for Quantum Matter
• Translations and Momentum
• Fourier Series
• Condensed Matter Roadmap

References

• F. Bloch, “Über die Quantenmechanik der Elektronen in Kristallgittern,” Zeitschrift für Physik 52, 555–600 (1929), doi:10.1007/BF01339455.
• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Holt, Rinehart and Winston, 1976.
• S. H. Simon, The Oxford Solid State Basics, Oxford University Press, 2013.