3b.1 — Bloch's Theorem¶

"In a crystal, the wavefunction does not have to be periodic; it has to be quasiperiodic." — Felix Bloch's 1928 thesis, paraphrased

Why does Bloch's theorem exist?

The problem before Bloch (1928): physicists could solve the Schrödinger equation for a single atom (the hydrogen atom problem, done by 1926) and for free electrons in a box. But a real crystal is neither. It contains $\sim 10^{23}$ atoms, each with electrons interacting with all the others. By 1928 nobody knew how to even start writing down the answer for "what does an electron do in a piece of copper?". The Hilbert space was infinite-dimensional; the Hamiltonian had no obvious special structure to exploit.

What Bloch noticed: the crystal looks the same if you slide it by one lattice vector. This is an exact symmetry. Quantum mechanics says that whenever a Hamiltonian has a symmetry, the eigenstates can be labelled by the eigenvalues of the symmetry operator. So instead of asking "what does the electron's wavefunction look like everywhere in the crystal at once?" — an infinite-dimensional question — Bloch could ask "what does the wavefunction look like in one unit cell, with a fixed phase relation to the next cell?" — a finite-dimensional question, parametrised by a single vector $\mathbf k$.

The picture to keep: imagine driving down a long road with bumps spaced every $a$. Your car's suspension vibrates as you go. The amplitude of the vibration repeats with the bumps (it has lattice periodicity), but you can also be at any phase in your bouncing cycle as you pass each bump. The two pieces — a lattice-periodic envelope $u(\mathbf r)$ and a global phase $e^{i\mathbf k\cdot\mathbf r}$ — completely describe the state of the system. That is Bloch's theorem.

What it buys us: the infinite eigenvalue problem becomes a one-parameter family of finite problems, one per $\mathbf k$ in the BZ. Every modern DFT code in existence — VASP, Quantum ESPRESSO, ABINIT, CASTEP — is implementation of this single insight.

The single most important theorem in solid state physics is Bloch's theorem. It is the reason the electronic structure of a crystal — an object containing $\sim 10^{23}$ electrons — is computable on a laptop. It reduces an infinite-dimensional eigenvalue problem to a one-parameter family of finite-dimensional eigenvalue problems indexed by a wavevector $\mathbf k$ that lives in a small, bounded region of reciprocal space called the first Brillouin zone. Every plane-wave DFT code in existence — VASP, Quantum ESPRESSO, ABINIT, CASTEP, GPAW, CP2K — is a careful, brutally optimised implementation of Bloch's theorem. So we had better understand it.

Intuition: a wave longer than the bumps

Before any equations, hold this picture in mind. Imagine a snake gliding over a corrugated sheet — the corrugation has period $a$, and the snake's body is a sinusoidal wave of wavelength $\lambda \gg a$. As the snake slithers forwards by one corrugation $a$, every bit of its body moves to a position previously occupied by a slightly different bit; the shape of the snake at that next instant looks almost the same as before, but rigidly translated by $a$ and additionally multiplied by a phase. That phase advance is the heart of Bloch's theorem: an electron wavefunction in a periodic potential need not be itself periodic — it is allowed to pick up a complex phase $e^{i\mathbf k\cdot\mathbf R}$ when translated by a lattice vector $\mathbf R$. The vector $\mathbf k$ labels how fast the phase winds, and is the only surviving label of translational symmetry in the eigenstate.

The picture also tells you why $\mathbf k$ lives in a bounded region of reciprocal space. If $\mathbf k$ is so large that the phase advance $\mathbf k\cdot\mathbf a$ exceeds $2\pi$, you cannot distinguish it from a smaller $\mathbf k'$ that gives the same phase modulo $2\pi$. The set of inequivalent $\mathbf k$ — the Brillouin zone — is a single primitive cell of the reciprocal lattice. Everything beyond that just folds back inside.

3b.1.1 The setting: a periodic Hamiltonian¶

Consider a single electron in a crystalline solid. The other electrons we will average over later — for the moment, regard their effect as part of an external potential, much as one does in Hartree–Fock or Kohn–Sham DFT. The crystal is described by a Bravais lattice with primitive vectors $\mathbf a_1, \mathbf a_2, \mathbf a_3$, and a lattice translation is

\[\mathbf R = n_1 \mathbf a_1 + n_2 \mathbf a_2 + n_3 \mathbf a_3, \qquad n_i \in \mathbb Z. \tag{3b.1.1}\]

Inside the crystal the electron experiences a potential $V(\mathbf r)$ generated by the nuclei (positioned at $\mathbf R + \boldsymbol\tau_\alpha$ for basis vectors $\boldsymbol\tau_\alpha$) and the mean field of the other electrons. The crucial property of $V$ is that it inherits the lattice periodicity:

\[V(\mathbf r + \mathbf R) = V(\mathbf r) \quad \text{for every lattice vector } \mathbf R. \tag{3b.1.2}\]

The single-particle Hamiltonian is therefore

\[\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf r), \qquad V(\mathbf r + \mathbf R) = V(\mathbf r). \tag{3b.1.3}\]

We want the eigenstates of $\hat{H}$. Naively this is hopeless: $V(\mathbf r)$ is defined on an infinite domain. Bloch saw that the symmetry of $V$ does almost all of the work for us.

3b.1.2 The translation operator¶

Define a translation operator $\hat T_\mathbf R$ that shifts a wavefunction by a lattice vector $\mathbf R$:

\[(\hat T_\mathbf R \psi)(\mathbf r) := \psi(\mathbf r + \mathbf R). \tag{3b.1.4}\]

This is a unitary operator (it preserves the norm: $\int |\psi(\mathbf r+\mathbf R)|^2 d^3 r = \int |\psi(\mathbf r)|^2 d^3 r$ by change of variables) and, because all lattice translations commute amongst themselves,

\[\hat T_\mathbf R \hat T_{\mathbf R'} = \hat T_{\mathbf R + \mathbf R'} = \hat T_{\mathbf R'} \hat T_\mathbf R. \tag{3b.1.5}\]

So the set of lattice translations $\{\hat T_\mathbf R\}_{\mathbf R}$ forms an abelian group.

Why this step? — translations commute

The commutativity in (3b.1.5) is not arbitrary; it follows from the fact that lattice translations live in $\mathbb R^3$ and addition of vectors is commutative. Compare this with rotations in 3D: rotating by $90^\circ$ about $\hat x$ and then $90^\circ$ about $\hat y$ is not the same as the reverse order. Rotations form a non-abelian group (SO(3)). For non-abelian symmetry groups the irreducible representations are higher-dimensional, and the labels of common eigenstates carry additional indices (think of orbital angular momentum $\ell$ and the magnetic quantum number $m$). The fact that lattice translations commute is precisely what makes the Bloch labels one-dimensional: a single complex phase per translation, parametrised by a single vector $\mathbf k$. If translations did not commute the labels would be matrices instead.

3b.1.3 $\hat T_\mathbf R$ commutes with $\hat{H}$¶

Here is the key calculation. Apply $\hat T_\mathbf R \hat{H}$ to an arbitrary state $\psi$:

\[(\hat T_\mathbf R \hat{H} \psi)(\mathbf r) = (\hat{H} \psi)(\mathbf r + \mathbf R) = -\frac{\hbar^2}{2m}\nabla^2 \psi(\mathbf r + \mathbf R) + V(\mathbf r + \mathbf R)\, \psi(\mathbf r + \mathbf R). \tag{3b.1.6}\]

The Laplacian is translation-invariant: $\nabla^2_{\mathbf r}\, \psi(\mathbf r + \mathbf R) = (\nabla^2 \psi)(\mathbf r + \mathbf R)$, because a constant shift does not change the second derivatives. And by (3b.1.2), $V(\mathbf r + \mathbf R) = V(\mathbf r)$. Hence

\[(\hat T_\mathbf R \hat{H} \psi)(\mathbf r) = -\frac{\hbar^2}{2m}\nabla^2 \psi(\mathbf r + \mathbf R) + V(\mathbf r)\, \psi(\mathbf r + \mathbf R) = (\hat{H} \hat T_\mathbf R \psi)(\mathbf r). \tag{3b.1.7}\]

Since this holds for any $\psi$,

\[\boxed{\; [\hat{H}, \hat T_\mathbf R] = 0 \quad \text{for every lattice vector } \mathbf R. \;} \tag{3b.1.8}\]

This is the cleanest statement of translational symmetry. We exploited only two facts: the kinetic operator is translation-invariant, and the potential is lattice-periodic.

Why we wanted commutation

In Chapter 4 you learnt that two commuting Hermitian operators can be simultaneously diagonalised: there exists an orthonormal basis in which both operators are diagonal. Here $\hat T_\mathbf R$ is not Hermitian (it is unitary), but the same logic works for any pair of commuting normal operators. The plan is therefore: diagonalise $\hat T_\mathbf R$ first — which is trivial, since it acts on functions by translation — and then within each eigenspace diagonalise $\hat{H}$. The first step is symmetry; the second step is dynamics.

3b.1.4 Eigenvalues of $\hat T_\mathbf R$¶

Because all $\hat T_\mathbf R$ commute amongst themselves (3b.1.5), they have a common eigenbasis. Suppose $\psi$ is a simultaneous eigenstate of every $\hat T_\mathbf R$:

\[\hat T_\mathbf R \psi = c(\mathbf R)\, \psi. \tag{3b.1.9}\]

What can the eigenvalue $c(\mathbf R)$ look like? Three constraints.

(i) Composition. Applying $\hat T_{\mathbf R'}$ after $\hat T_\mathbf R$ and using (3b.1.5):

\[\hat T_{\mathbf R + \mathbf R'} \psi = c(\mathbf R + \mathbf R')\, \psi, \qquad \hat T_{\mathbf R'} \hat T_\mathbf R \psi = c(\mathbf R')\, c(\mathbf R)\, \psi. \tag{3b.1.10}\]

Equating the two,

\[c(\mathbf R + \mathbf R') = c(\mathbf R)\, c(\mathbf R'). \tag{3b.1.11}\]

The only continuous functions of $\mathbf R$ that satisfy this functional equation are exponentials.

(ii) Unitarity. Because $\hat T_\mathbf R$ is unitary, $|c(\mathbf R)| = 1$: it must be a phase.

Why this step? — eigenvalues of unitary operators

Recall that if $\hat U$ is unitary, $\hat U^\dagger \hat U = \hat I$. For a normalised eigenstate $\hat U\psi = c\psi$ we therefore have $\langle\psi|\hat U^\dagger \hat U|\psi\rangle = |c|^2 \langle\psi|\psi\rangle = |c|^2 = 1$. The eigenvalue must lie on the unit circle. Combining with (3b.1.11), which forces an exponential, we are squeezed into precisely the family $c(\mathbf R) = e^{i\mathbf k\cdot\mathbf R}$ with $\mathbf k$ a real vector (any imaginary part of $\mathbf k$ would make $|c| \ne 1$). This is exactly the constraint we want: the label of irreducible representations of the lattice translation group is a real wavevector.

Combining (i) and (ii) we may write

\[c(\mathbf R) = e^{i\mathbf k \cdot \mathbf R} \tag{3b.1.12}\]

for some real vector $\mathbf k$. The set of simultaneous eigenstates of $\{\hat T_\mathbf R\}$ is therefore labelled by $\mathbf k$.

Why must $\mathbf k$ be real?

Strictly, (3b.1.11) and continuity together force $c(\mathbf R) = e^{i\mathbf k\cdot\mathbf R}$ with $\mathbf k$ possibly complex. The unitarity constraint $|c(\mathbf R)|=1$ then forces $\text{Im}(\mathbf k) = 0$ on an infinite, translationally invariant lattice. Allowing complex $\mathbf k$ — evanescent Bloch waves with $\text{Im}(\mathbf k) \ne 0$ — is, however, not nonsense; it is essential for describing electronic states inside a band gap (e.g. tunnelling through a barrier or surface states), where the wavefunction must decay rather than propagate. Real-$\mathbf k$ Bloch states are the bulk propagating states; complex-$\mathbf k$ Bloch states exist mathematically but are not normalisable on the infinite crystal. We restrict to real $\mathbf k$ throughout this chapter and return to evanescent waves in Ch 6 (surface states) and Ch 7 (tunnelling).

(iii) Born–von Kármán periodic boundary conditions. A real crystal is finite, but to avoid awkward surface effects one imagines an infinite crystal and instead imposes periodic boundary conditions: pretend the crystal is a torus that closes after $N_1$ unit cells along $\mathbf a_1$, $N_2$ along $\mathbf a_2$, $N_3$ along $\mathbf a_3$. The total number of unit cells is $N = N_1 N_2 N_3$. Periodicity demands

\[\psi(\mathbf r + N_i \mathbf a_i) = \psi(\mathbf r) \quad \Longrightarrow \quad e^{i N_i \mathbf k \cdot \mathbf a_i} = 1 \quad \Longrightarrow \quad \mathbf k \cdot \mathbf a_i = \frac{2\pi m_i}{N_i}, \quad m_i \in \mathbb Z. \tag{3b.1.13}\]

So $\mathbf k$ is discrete, with a spacing $\sim 1/N$ that becomes infinitesimal in the thermodynamic limit $N\to\infty$. Equivalently, writing $\mathbf k$ in the basis of reciprocal primitive vectors $\mathbf b_j$ (defined by $\mathbf a_i \cdot \mathbf b_j = 2\pi \delta_{ij}$),

\[\mathbf k = \sum_j \frac{m_j}{N_j}\, \mathbf b_j, \qquad m_j \in \mathbb Z. \tag{3b.1.14}\]

There are exactly $N$ inequivalent values of $\mathbf k$ in one reciprocal unit cell — this matters in Chapter 6 when you choose a Monkhorst–Pack mesh.

3b.1.5 Bloch's theorem¶

We are now ready to state and prove the theorem in its standard form.

Key idea — Bloch's theorem in one line

Periodic potential $\Rightarrow$ eigenstates are plane wave $\times$ lattice-periodic function $\Rightarrow$ band structure $E_n(\mathbf k)$ on the BZ. The infinite crystal problem reduces to a unit-cell problem at each $\mathbf k$. Everything else in this section unpacks this single sentence.

Theorem 3b.1.1 (Bloch)

Let $\hat{H}$ be a single-particle Hamiltonian with a lattice-periodic potential $V(\mathbf r + \mathbf R) = V(\mathbf r)$ and Born–von Kármán boundary conditions. Then there is an orthonormal basis of eigenstates of the form $$\boxed{\; \psi_{n\mathbf k}(\mathbf r) = e^{i\mathbf k\cdot\mathbf r}\, u_{n\mathbf k}(\mathbf r), \qquad u_{n\mathbf k}(\mathbf r + \mathbf R) = u_{n\mathbf k}(\mathbf r). \;} \tag{3b.1.15}$$ The label $\mathbf k$ runs over the first Brillouin zone; $n$ is a discrete band index.

Proof, step by step. Let us walk through this slowly so that no part of the logic is obscured.

Translations commute amongst themselves. Equation (3b.1.5) — derived from the vector-addition structure of the Bravais lattice.
Translations commute with the Hamiltonian. Equation (3b.1.8) — a consequence of the kinetic operator being translation-invariant and the potential being lattice-periodic.
Common eigenbasis exists. A set of mutually commuting normal operators admits a common eigenbasis. (For a single pair $[A,B]=0$ with both Hermitian, this is the standard simultaneous diagonalisation theorem; for the more general case of an abelian group of normal operators, the same conclusion holds and is sometimes called the spectral theorem for abelian C*-algebras.) Apply this to the family $\{\hat H\} \cup \{\hat T_\mathbf R\}_\mathbf R$ to conclude: there exists an orthonormal basis $\{\psi_\alpha\}$ in which every $\hat T_\mathbf R$ and $\hat H$ is diagonal.
Translation eigenvalues are phases parametrised by $\mathbf k$. Steps (i)–(iii) above forced $c(\mathbf R) = e^{i\mathbf k\cdot\mathbf R}$ for some real $\mathbf k$. The basis element $\psi_\alpha$ therefore satisfies $\hat T_\mathbf R \psi_\alpha = e^{i\mathbf k_\alpha\cdot\mathbf R}\psi_\alpha$; we re-label $\alpha$ by the pair $(n,\mathbf k)$ where $\mathbf k$ records the translation phase and $n$ enumerates eigenstates inside the same translation-eigenvalue sector.
Translation eigenstates have the Bloch form. This is the calculation (3b.1.16)–(3b.1.19), repeated here for clarity. We have shown that the simultaneous eigenstates of $\{\hat T_\mathbf R\}$ are labelled by a vector $\mathbf k$ such that

\[\psi(\mathbf r + \mathbf R) = e^{i\mathbf k \cdot \mathbf R}\, \psi(\mathbf r). \tag{3b.1.16}\]

Because $[\hat{H}, \hat T_\mathbf R]=0$, we may choose the eigenbasis of $\hat{H}$ to also be an eigenbasis of every $\hat T_\mathbf R$. Hence the eigenstates of $\hat{H}$ satisfy (3b.1.16).

Now define

\[u(\mathbf r) := e^{-i\mathbf k \cdot \mathbf r}\, \psi(\mathbf r). \tag{3b.1.17}\]

Compute $u(\mathbf r + \mathbf R)$:

\[u(\mathbf r + \mathbf R) = e^{-i\mathbf k \cdot (\mathbf r + \mathbf R)}\, \psi(\mathbf r + \mathbf R) = e^{-i\mathbf k \cdot \mathbf r}\, e^{-i\mathbf k \cdot \mathbf R}\, e^{i\mathbf k \cdot \mathbf R}\, \psi(\mathbf r) = e^{-i\mathbf k \cdot \mathbf r}\, \psi(\mathbf r) = u(\mathbf r). \tag{3b.1.18}\]

So $u$ is lattice-periodic. Solving (3b.1.17) for $\psi$,

\[\psi(\mathbf r) = e^{i\mathbf k \cdot \mathbf r}\, u(\mathbf r), \tag{3b.1.19}\]

which is Bloch's theorem in the form (3b.1.15). The discrete index $n$ arises because, within a fixed $\mathbf k$ sector, the Schrödinger equation reduces to a problem on a single unit cell (see below), and that problem has infinitely many discrete eigenvalues. $\blacksquare$

Pause and recall

Before reading on, try to answer these from memory:

Which two properties of the Hamiltonian were used to show $[\hat H, \hat T_\mathbf{R}] = 0$, and why does that commutation matter?
Why must the eigenvalue $c(\mathbf{R})$ of the translation operator have the form $e^{i\mathbf{k}\cdot\mathbf{R}}$ with $\mathbf{k}$ real — what role do the composition rule and unitarity each play?
Bloch's theorem writes $\psi_{n\mathbf{k}} = e^{i\mathbf{k}\cdot\mathbf{r}}\,u_{n\mathbf{k}}(\mathbf{r})$ — which factor is lattice-periodic, and why does this not mean the wavefunction itself is periodic?

If any of these is shaky, re-read the preceding section before continuing.

3b.1.6 The Bloch equation on the unit cell¶

Bloch's theorem reduces the problem to: find $u_{n\mathbf k}(\mathbf r)$ that is periodic on the unit cell. Plug (3b.1.15) into the Schrödinger equation $\hat{H} \psi = E \psi$:

\[-\frac{\hbar^2}{2m}\nabla^2 \left[e^{i\mathbf k\cdot\mathbf r} u\right] + V(\mathbf r)\, e^{i\mathbf k\cdot\mathbf r} u = E\, e^{i\mathbf k\cdot\mathbf r} u. \tag{3b.1.20}\]

Use $\nabla(e^{i\mathbf k\cdot\mathbf r} u) = e^{i\mathbf k\cdot\mathbf r}(i\mathbf k u + \nabla u)$ and $\nabla^2(e^{i\mathbf k\cdot\mathbf r} u) = e^{i\mathbf k\cdot\mathbf r}(\nabla + i\mathbf k)^2 u$. The exponential cancels on both sides:

\[\left[ -\frac{\hbar^2}{2m}(\nabla + i\mathbf k)^2 + V(\mathbf r)\right] u_{n\mathbf k}(\mathbf r) = E_{n\mathbf k}\, u_{n\mathbf k}(\mathbf r). \tag{3b.1.21}\]

This is the cell-periodic Bloch equation: a Hermitian eigenvalue problem for $u_{n\mathbf k}$ defined on a single unit cell, with periodic boundary conditions. The Hamiltonian on the unit cell is

\[\hat{H}_\mathbf k = -\frac{\hbar^2}{2m}(\nabla + i\mathbf k)^2 + V(\mathbf r). \tag{3b.1.22}\]

For each $\mathbf k$ there are countably many eigenvalues $E_{1\mathbf k} \le E_{2\mathbf k} \le \cdots$. The function $\mathbf k \mapsto E_{n\mathbf k}$ is the $n$-th band and the whole collection $\{E_{n\mathbf k}\}$ is the band structure. This is precisely the object you will plot in Chapter 6.

Computational consequence

The infinite-domain problem has been reduced to a unit-cell problem, parametrised by $\mathbf k$. In a plane-wave DFT code, $u_{n\mathbf k}(\mathbf r)$ is expanded in plane waves with the periodicity of the unit cell: $$u_{n\mathbf k}(\mathbf r) = \sum_\mathbf G c_{n\mathbf k}(\mathbf G)\, e^{i\mathbf G\cdot\mathbf r},$$ where $\mathbf G$ runs over reciprocal lattice vectors. The cutoff $|\mathbf G|^2 \le E_\text{cut}\cdot 2m/\hbar^2$ controls the basis-set size — this is the ENCUT parameter in VASP and ecutwfc in Quantum ESPRESSO. Convergence with respect to it is the second universal DFT chore, after $\mathbf k$-point convergence.

Sanity check: estimating basis-set size

Suppose you set ENCUT = 400 eV. Converting to joules, $400\,\mathrm{eV} = 6.41\times 10^{-17}\,\mathrm{J}$. The maximum $|\mathbf G|^2$ included is $$|\mathbf G|_\text{max}^2 = \frac{2 m_e\cdot 400 \text{ eV}}{\hbar^2} = \frac{2\cdot 9.109\times 10^{-31}\cdot 6.41\times 10^{-17}}{(1.055\times 10^{-34})^2} \approx 1.05\times 10^{22} \text{ m}^{-2},$$ so $|\mathbf G|_\text{max} \approx 1.02\times 10^{11}$ m$^{-1} = 10.2$ Å$^{-1}$. For a cubic unit cell of side $a = 4$ Å, the reciprocal lattice spacing is $2\pi/a \approx 1.57$ Å$^{-1}$, so the cutoff encloses roughly a sphere of radius $\sim 6.5$ in units of $2\pi/a$. The number of reciprocal lattice vectors inside is $\sim (4/3)\pi\cdot 6.5^3 \approx 1\,150$ plane waves per cubic unit cell. Doubling ENCUT to 800 eV roughly doubles $|\mathbf G|^2_{\max}$ and so multiplies the plane-wave count by $\sqrt{8} \approx 2.8$ — into the few-thousand range. For a transition-metal oxide with pseudopotentials that need higher cutoff (often 500–700 eV), the basis can reach tens of thousands of plane waves, and that is the typical regime of production DFT calculations.

3b.1.6a Crystal momentum vs real momentum¶

A persistent source of confusion: the wavevector $\mathbf k$ in Bloch's theorem is not the momentum eigenvalue of $\hat{\mathbf p} = -i\hbar\nabla$. We call $\hbar\mathbf k$ the crystal momentum to flag this distinction. Let us see why explicitly.

Apply the real-space momentum operator to a Bloch state:

\[\hat{\mathbf p}\, \psi_{n\mathbf k}(\mathbf r) = -i\hbar\nabla \left[ e^{i\mathbf k\cdot\mathbf r} u_{n\mathbf k}(\mathbf r) \right] = \hbar\mathbf k\, \psi_{n\mathbf k}(\mathbf r) - i\hbar\, e^{i\mathbf k\cdot\mathbf r}\, \nabla u_{n\mathbf k}(\mathbf r). \tag{3b.1.\text{$p$1}}\]

The first term is what you would naively expect — but the second term is not in general proportional to $\psi_{n\mathbf k}$, because the cell-periodic factor $u_{n\mathbf k}(\mathbf r)$ has its own spatial dependence. Hence

\[\hat{\mathbf p}\,\psi_{n\mathbf k} \ne \hbar\mathbf k\, \psi_{n\mathbf k} \quad\text{in general.} \tag{3b.1.\text{$p$2}}\]

A Bloch state is not an eigenstate of $\hat{\mathbf p}$, only of the translation operator $\hat T_\mathbf R$. The two coincide only in the empty-lattice limit $V=0$, where $u_{n\mathbf k}$ is constant.

So what is $\hbar\mathbf k$? Three useful interpretations:

Quantum number for translation symmetry. $\hbar\mathbf k$ labels how the wavefunction transforms under a lattice translation; nothing more, nothing less. It plays exactly the role that ordinary momentum plays for a continuous translation symmetry.
Conserved quantity modulo $\hbar\mathbf G$. In a perfect crystal, scattering processes conserve crystal momentum up to a reciprocal lattice vector: $\mathbf k_\text{in} + \mathbf k_\text{phonon} = \mathbf k_\text{out} + \mathbf G$, where $\mathbf G$ is a reciprocal lattice vector ("Umklapp"). This is the lattice analogue of momentum conservation.
Group velocity. The expectation value $\langle\hat{\mathbf p}\rangle$ in a Bloch state, divided by mass, gives the group velocity of a wavepacket centred on $\mathbf k$: $$\mathbf v_{n\mathbf k} = \frac{1}{\hbar} \nabla_\mathbf k E_{n\mathbf k}. \tag{3b.1.\text{$p$3}}$$ It is the velocity at which a localised electron actually moves through the crystal. This is the quantity that enters the Boltzmann transport equation and the Drude conductivity.

Common student trap

A pure Bloch state $\psi_{n\mathbf k}$ is a standing-wave-like extended state with definite crystal momentum $\hbar\mathbf k$ but indefinite real momentum. It does not describe a moving electron at definite velocity. To describe a propagating electron one must construct a wavepacket — a superposition of Bloch states with $\mathbf k$ values in a small window around $\mathbf k_0$. The wavepacket then has group velocity (3b.1.$p$3) and finite spatial extent. This is the standard semiclassical picture used in transport theory.

3b.1.7 Crystal momentum and the first Brillouin zone¶

The wavevector $\mathbf k$ entered through (3b.1.12) as a label for translation eigenvalues, not as a momentum. The quantity $\hbar\mathbf k$ is called crystal momentum; it is conserved (modulo a reciprocal lattice vector) in scattering processes, but it is not the expectation value of $-i\hbar\nabla$. To see why $\mathbf k$ is only defined modulo a reciprocal lattice vector $\mathbf G$, examine the translation eigenvalue:

\[e^{i(\mathbf k + \mathbf G)\cdot \mathbf R} = e^{i\mathbf k\cdot\mathbf R}\, e^{i\mathbf G\cdot\mathbf R} = e^{i\mathbf k\cdot\mathbf R} \cdot 1 \tag{3b.1.23}\]

because by definition of the reciprocal lattice $\mathbf G\cdot\mathbf R \in 2\pi\mathbb Z$. So $\mathbf k$ and $\mathbf k + \mathbf G$ give the same eigenvalue, hence the same translation symmetry sector. We can restrict $\mathbf k$ to lie in any single primitive cell of the reciprocal lattice. The most symmetric choice — the Wigner–Seitz cell of the reciprocal lattice — is the first Brillouin zone (BZ). For the simple cubic lattice it is a cube of side $2\pi/a$; for the face-centred cubic lattice it is the famous truncated octahedron with high-symmetry points $\Gamma, X, L, K, W$; for the hexagonal lattice it is a hexagonal prism with $\Gamma, M, K, A, H$.

When we plot a band structure $E_n(\mathbf k)$ we trace a path inside the BZ that connects these high-symmetry points. The total number of allowed $\mathbf k$ in the BZ is exactly $N$ (the number of unit cells), distributed uniformly with density $V_\text{crystal}/(2\pi)^3$.

Reduced, extended, and repeated zone schemes¶

Three ways to draw a band structure are in common use, and they often confuse students. They contain exactly the same information.

Reduced zone scheme. Plot $E_{n\mathbf k}$ with $\mathbf k$ restricted to the first BZ. Different bands $n$ are different curves. This is what every textbook and DFT plot uses. The advantage is that the plot is compact; the disadvantage is that you lose a sense of which "free-electron parabola" each band fragment came from.
Extended zone scheme. Allow $\mathbf k$ to range over all of reciprocal space; each band lives in a different region (the $n$-th band in the $n$-th BZ). The advantage is that, for an empty lattice, each band is a single smooth parabola; the disadvantage is that the plot extends infinitely.
Repeated zone scheme. Periodically repeat the band structure over reciprocal space: $E_{n,\mathbf k+\mathbf G} = E_{n,\mathbf k}$. This is useful when discussing Fermi surfaces that straddle BZ boundaries — they look smooth in the repeated scheme.

The choice is purely cosmetic. Modern DFT codes output the reduced zone; you should be comfortable mentally translating between them.

3b.1.8 The band index and the meaning of "bands"¶

For a fixed $\mathbf k$ the cell-periodic equation (3b.1.21) has discrete eigenvalues $E_{1\mathbf k} \le E_{2\mathbf k} \le \cdots$. The label $n = 1,2,\ldots$ is the band index. The pair $(n,\mathbf k)$ uniquely indexes every electronic single-particle state of the crystal. Two important consequences:

Total state count. The Hilbert space dimension per band is $N$ (one $\mathbf k$ per unit cell). The total number of electronic states up to band $n$ is $nN$. Including spin gives $2nN$. If the crystal has $Z$ valence electrons per unit cell, the lowest $\lceil Z/2 \rceil$ bands are filled at zero temperature.
Metals vs insulators. If $Z$ is even and the highest filled band lies entirely below the lowest empty band — i.e. there is a band gap — the material is an insulator (or semiconductor). If $Z$ is odd, or if bands overlap, some band is partially filled and the material is a metal. This is the most powerful single prediction of band theory, and it is the criterion by which you will classify materials in Chapter 6.

What band theory does not tell you

Single-particle band theory ignores electron–electron interactions beyond a mean field. Strongly correlated materials — Mott insulators, high-$T_c$ cuprates, Kondo lattices — have ground states that no single-determinant band theory can describe. Standard DFT (LDA, GGA) gets band gaps wrong by 30–100% for this reason, and gets some materials (NiO, MnO) qualitatively wrong, predicting a metal where experiment finds an insulator. Hybrid functionals (Ch 5), DFT+U, and many-body methods (GW, DMFT) exist to plug these gaps.

3b.1.9 Worked example: the empty lattice¶

Suppose $V(\mathbf r) = 0$ (free electrons in a box with imposed lattice periodicity — the "empty lattice"). The exact eigenstates are plane waves $e^{i\mathbf q\cdot\mathbf r}$ with energy $\hbar^2 q^2/2m$. To cast them in Bloch form, write $\mathbf q = \mathbf k + \mathbf G$ with $\mathbf k$ in the BZ:

\[\psi_{n\mathbf k}(\mathbf r) = e^{i(\mathbf k + \mathbf G_n)\cdot \mathbf r} = e^{i\mathbf k\cdot\mathbf r}\, e^{i\mathbf G_n\cdot\mathbf r}, \tag{3b.1.24}\]

so $u_{n\mathbf k}(\mathbf r) = e^{i\mathbf G_n\cdot\mathbf r}$ and the band energies are

\[E_n(\mathbf k) = \frac{\hbar^2}{2m}\, |\mathbf k + \mathbf G_n|^2. \tag{3b.1.25}\]

Each reciprocal lattice vector $\mathbf G_n$ generates a band. Plotting $E_n(\mathbf k)$ along a BZ path produces a system of folded parabolas. We will use this picture in §3b.2 as the starting point for the nearly-free-electron model — the band gaps open exactly where two of these parabolas cross.

Sanity check: degeneracies at zone boundaries

At a BZ boundary $\mathbf k$ satisfying $|\mathbf k|^2 = |\mathbf k + \mathbf G|^2$ (the Bragg condition) two empty-lattice bands cross. Adding any nonzero $V_\mathbf G$ lifts that degeneracy — that is the origin of the band gap, and it is the cleanest illustration of why translational symmetry alone is not enough; the strength of $V$ shows up only as gap sizes, not as the qualitative structure of bands.

Worked example: two explicit Bloch states in 1D

Let the lattice constant be $a = 4.0$ Å and pick the Bloch wavevector $k = \pi/(2a)$ (the midpoint of the BZ from $\Gamma$ to the zone boundary). Suppose the unit-cell potential is approximated as a single attractive Gaussian well centred at the origin, so the lowest cell-periodic function $u_1(x)$ is well-localised on the ion. Then two band states near this $k$ are:

Lowest band ($n=1$): $$\psi_{1,k}(x) = e^{i\pi x/(2a)}\, u_1(x), \qquad u_1(x+a) = u_1(x).$$ Translating by one lattice constant: $\psi_{1,k}(x+a) = e^{i\pi/2}\psi_{1,k}(x) = i\,\psi_{1,k}(x)$. The wavefunction picks up a phase of $i$ — a quarter rotation — as we step from one cell to the next. After four steps it has rotated by $e^{i 2\pi} = 1$, recovering the original value: the "wavelength of the envelope" is $4a$.
Second band ($n=2$): $$\psi_{2,k}(x) = e^{i\pi x/(2a)}\, u_2(x),$$ where $u_2$ is a higher cell-periodic eigenfunction (with one node inside the cell). Same translation phase $i$ — what distinguishes the two states is the interior of the unit cell, not the inter-cell phase.

The number $k = \pi/(2a)$ corresponds to a crystal momentum of $\hbar k \approx 0.082\, \hbar/\text{Å} \approx 8.7 \times 10^{-26}$ kg m/s. This is not the momentum expectation value of either state — that would be obtained by computing $\langle\psi|\hat{\mathbf p}|\psi\rangle$ and depends on the detailed shape of $u_n(x)$.

3b.1.9a Time-reversal symmetry and $E(\mathbf k) = E(-\mathbf k)$¶

A subtle but important consequence of time-reversal symmetry: in a non-magnetic crystal with no external magnetic field, the band energies satisfy

\[\boxed{\; E_n(\mathbf k) = E_n(-\mathbf k). \;} \tag{3b.1.tr}\]

This identity is responsible for many practical conveniences — most importantly, you only need to compute the band structure on half the Brillouin zone, since the other half is determined by symmetry.

Argument. The time-reversal operator $\hat\Theta$ acts on a spinless wavefunction by complex conjugation: $\hat\Theta\psi(\mathbf r) = \psi^*(\mathbf r)$. Because the Hamiltonian $\hat H = -\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf r)$ with real $V(\mathbf r)$ contains only real-coefficient operators (the kinetic term has only real coefficients when written in real space; the Laplacian is real), $\hat\Theta$ commutes with $\hat H$:

\[[\hat H, \hat\Theta] = 0. \tag{3b.1.tr2}\]

Now apply $\hat\Theta$ to a Bloch eigenstate $\psi_{n\mathbf k}(\mathbf r) = e^{i\mathbf k\cdot\mathbf r}u_{n\mathbf k}(\mathbf r)$ with energy $E_{n\mathbf k}$:

\[\hat\Theta \psi_{n\mathbf k}(\mathbf r) = \psi_{n\mathbf k}^*(\mathbf r) = e^{-i\mathbf k\cdot\mathbf r}\, u_{n\mathbf k}^*(\mathbf r). \tag{3b.1.tr3}\]

The right-hand side is itself a Bloch state, with cell-periodic part $u_{n\mathbf k}^*$ and wavevector $-\mathbf k$. Since $\hat\Theta$ commutes with $\hat H$, the result is also an eigenstate of $\hat H$ with the same energy $E_{n\mathbf k}$. But this same energy is, by construction, an energy of a Bloch state with wavevector $-\mathbf k$, hence equal to $E_n(-\mathbf k)$. Therefore $E_n(\mathbf k) = E_n(-\mathbf k)$.

When this fails

Time-reversal symmetry is broken by (i) an external magnetic field, which adds an imaginary vector potential term to $\hat H$; (ii) ferromagnetic order, which introduces a spontaneous magnetisation; and (iii) intrinsic angular momentum (spin) coupled to a magnetic background. In each case $E_n(\mathbf k) \ne E_n(-\mathbf k)$ is allowed, and is the microscopic origin of the anomalous Hall effect and Berry curvature dipoles. Topological insulators with spin–orbit coupling have $E_n(\mathbf k) = E_n(-\mathbf k)$ only after combining time reversal with a spin flip (Kramers degeneracy). All of these refinements are addressed in advanced solid-state courses; for our purposes (3b.1.tr) holds throughout this chapter.

3b.1.10 Why this matters for the rest of the book¶

You will encounter Bloch's theorem in every electronic structure calculation in the remainder of the book. A short forward catalogue:

Chapter 5 (DFT theory). The Kohn–Sham equations are exactly the Bloch equation (3b.1.21) with $V$ replaced by the self-consistent Kohn–Sham potential $V_\text{KS}[\rho]$. Self-consistency means $\rho(\mathbf r) = \sum_{n\mathbf k} f_{n\mathbf k} |\psi_{n\mathbf k}(\mathbf r)|^2$, with the sum running over the BZ.
Chapter 6 (running DFT). You will choose a $\mathbf k$-mesh, an energy cutoff, and a smearing scheme. Convergence with respect to mesh density is your first job; the equation behind it is (3b.1.21). You will plot band structures along $\Gamma$–X–L–$\Gamma$ paths — those are exactly slices through $E_{n\mathbf k}$.
Chapter 7 (MD). Even classical MD on a periodic supercell uses Bloch-like reasoning: the interatomic force is computed from the periodic image convention, which is a real-space analogue of the BZ.
Chapter 9 (MLIPs). The descriptors used by SchNet, NequIP, and MACE are built from local atomic environments. Their effectiveness rests on the fact that, by Bloch's theorem, the electronic structure of an infinite crystal is determined by the unit cell — local descriptors capture the local part exactly. Equivariance with respect to lattice translations is the analogue of crystal-momentum conservation.
Chapter 10 (GNNs). Message-passing on a crystal graph respects translational symmetry because the graph is defined modulo the lattice. The Brillouin zone never explicitly appears, but its existence is what makes the descriptor space finite-dimensional.

3b.1.11 Common pitfalls and how to avoid them¶

A short list of misunderstandings that appear in every cohort of students and that you should bullet-proof yourself against:

"$\mathbf k$ is the momentum of the electron." No. $\hbar\mathbf k$ is the crystal momentum, a label for translation-symmetry eigenvalues. The true momentum expectation value differs by terms involving $\nabla u_{n\mathbf k}$ — see §3b.1.6a.
"Bloch's theorem says the wavefunction is periodic." No. Bloch's theorem says the wavefunction has the form $e^{i\mathbf k\cdot\mathbf r}\, u_{n\mathbf k}(\mathbf r)$ where only the cell-periodic part $u_{n\mathbf k}$ is periodic. The wavefunction itself picks up a phase under lattice translation.
"$\mathbf k$ values are continuous." Only in the thermodynamic limit. For a finite crystal of $N$ unit cells, exactly $N$ inequivalent $\mathbf k$ values are allowed, spaced by $\sim 1/N$ across the BZ. This discreteness matters when choosing $\mathbf k$-meshes in DFT.
"Band gaps require a strong potential." No, even an arbitrarily weak periodic potential opens gaps at Bragg planes (§3b.2). What requires a strong potential is large gaps. The qualitative existence of gaps is generic.
"There is only one band." There are infinitely many — one for each eigenvalue of the cell-periodic problem. In practice we plot the few near the Fermi level, but every plane-wave DFT code computes hundreds of them under the hood.
"The BZ is unique." Any primitive cell of the reciprocal lattice is a valid choice. The Wigner-Seitz cell of the reciprocal lattice (the "first Brillouin zone") is the most symmetric, but parallelepipeds spanned by reciprocal lattice vectors are also valid and sometimes computationally convenient.
"Different bands at the same $\mathbf k$ are degenerate." Only if forced by some additional symmetry (e.g. spatial group symmetry, or a hidden gauge symmetry like the chiral symmetry of graphene at $K$). Generically, bands are non-degenerate at a generic $\mathbf k$.

Keep these in mind as you read the remainder of Chapter 3b. They will recur.

3b.1.12 What to remember in 3 months¶

If you forget every equation in this section, hold on to these:

The wavefunction in a crystal has the form $\psi = e^{i\mathbf k\cdot\mathbf r}\, u(\mathbf r)$ with $u$ lattice-periodic. The $e^{i\mathbf k\cdot\mathbf r}$ is the envelope phase; the $u$ is the cell pattern.
$\mathbf k$ lives in the first Brillouin zone, a bounded region of reciprocal space. There are exactly $N$ allowed $\mathbf k$ values (one per unit cell of a finite crystal of $N$ cells).
Each $\mathbf k$ gives a discrete ladder of energies $E_{n\mathbf k}$, called bands. Plotting $E_n$ along a BZ path is the band structure.
Crystal momentum $\hbar\mathbf k$ is not the same as the momentum expectation value. It is a label for translation-symmetry quantum numbers, conserved modulo $\hbar\mathbf G$ in scattering.
The infinite crystal problem reduces to a unit-cell problem at each $\mathbf k$. This is why DFT calculations on a laptop are even possible.
Time reversal forces $E_n(\mathbf k) = E_n(-\mathbf k)$ in non-magnetic crystals — half the BZ is enough.

Three months from now, if someone asks "what is Bloch's theorem?", say: "The electron wavefunction in a crystal can be written as a plane wave times a lattice-periodic function. This is a consequence of the lattice translation symmetry and reduces an infinite-domain problem to a unit-cell problem labelled by $\mathbf k$."

Where this is used later¶

Bloch's theorem is the conceptual backbone of every electronic structure chapter. Specifically:

Tier 1. §5.3 (Kohn–Sham equations in a periodic system), §6.2 ($\mathbf k$-point grids and Monkhorst–Pack meshes), §6.4 (band structure plotting), §6.5 (defect supercells and Brillouin-zone folding).
Tier 2. §9.4 (why local descriptors work), §10.2 (crystal graphs and periodic neighbour lists), §11.3 (active learning over $\mathbf k$-point sampled phonon spectra).
Capstone Project 1. Screening dopants in a semiconductor: you will compute $E_{n\mathbf k}$ near the band gap, identify the band-edge characters, and predict doping behaviour from the conduction band effective mass.

Read §3b.2 (nearly free electrons) next, where the abstract structure derived here is given flesh by an explicit perturbation calculation at a zone boundary.