Bloch's Theorem

Learning Objectives

Sections 3.1–3.4 built the geometry of reciprocal space: why we need it, how to construct it, what the Brillouin zone looks like, and how to label its high-symmetry points. This section turns the geometry into physics. We ask the most consequential question in solid-state theory:

What does the periodicity of the crystal do to the wavefunctions of electrons inside it?

The answer — Bloch's theorem — is the single fact from which every band-structure picture, every density of states, every metal/semiconductor distinction, and every plane-wave DFT code (VASP included) descends. By the end of this section you should be able to:

1. State Bloch's theorem in both equivalent forms and explain why they are saying the same thing in two languages.
2. Derive the theorem from the single observation that lattice translations commute with the Hamiltonian.
3. Distinguish the three actors — $\psi_{\mathbf{k}}(\mathbf{r})$ (the Bloch state), $e^{i\mathbf{k}\cdot\mathbf{r}}$ (the plane-wave envelope), and $u_{\mathbf{k}}(\mathbf{r})$ (the cell-periodic part) — and say in one sentence what each does.
4. Explain the difference between true momentum and crystal momentum, and why $\mathbf{k}$ can always be reduced to the first Brillouin zone.
5. Recognise a Bloch wavefunction in the wild, including its instantly recognisable signature: the probability density $|\psi_{\mathbf{k}}|^2$ has the full periodicity of the lattice, but $\psi_{\mathbf{k}}$ itself only repeats up to a phase.
6. Connect the Bloch form directly to the plane-wave expansion VASP uses internally — and predict from that connection how VASP's ENCUT and KPOINTS tags relate to the mathematics.

The Question Bloch's Theorem Answers

Imagine an electron inside an infinite, perfect crystal. Wherever it sits, it sees the same potential it would have seen one lattice vector $\mathbf{R}$ away — the next cell looks identical. Two questions immediately arise:

1. Is the electron a plane wave $e^{i\mathbf{k}\cdot\mathbf{r}}$, like a free particle? It can't be — a plane wave has constant $|\psi|^2$, but a real crystal has nuclei that attract electrons more strongly at some places than others, so the density must wobble.
2. Is the electron localised on a single atom? It can't be either — by translation symmetry, the cell next door is just as attractive, so why would the electron be stuck on one particular nucleus and not the others?

Bloch's theorem (Felix Bloch, 1928, in his Leipzig PhD thesis under Heisenberg) is the elegant resolution: every stationary electron state in a periodic potential is a plane wave times a periodic function — delocalised globally, but modulated locally to feel the atoms. One formula does both jobs. Let us derive it from first principles.

The Setup — Periodic Potentials

Inside a crystal, the (effective single-particle) potential satisfies

That is the entire physics: identical unit cells, repeated forever. The single-particle Hamiltonian is then

The Translation Operator

Define an operator $\hat{T}_{\mathbf{R}}$ that translates a wavefunction by the lattice vector $\mathbf{R}$:

(The minus sign is conventional — it makes $\hat{T}_{\mathbf{R}}$ shift the wavefunction forward by $\mathbf{R}$, the way you would expect.) Two facts follow immediately from the definition:

1. Composition. $\hat{T}_{\mathbf{R}}\hat{T}_{\mathbf{R}'} = \hat{T}_{\mathbf{R}+\mathbf{R}'}$. Translating by $\mathbf{R}'$ and then by $\mathbf{R}$ is the same as translating by their sum.
2. Unitarity. $\hat{T}_{\mathbf{R}}^{\dagger}\hat{T}_{\mathbf{R}} = \mathbb{1}$ — translation preserves inner products and norms.

These two facts make the set $\{\hat{T}_{\mathbf{R}}\}$ a unitary representation of the abelian translation group of the lattice. Group theory will now do the heavy lifting.

Why T̂_R Commutes with the Hamiltonian

Apply $\hat{T}_{\mathbf{R}}\hat{H}$ to a test function $\psi(\mathbf{r})$:

The Laplacian is translation-invariant — $\nabla^2$ doesn't care where the origin is — and crucially $V(\mathbf{r}-\mathbf{R}) = V(\mathbf{r})$ because $\mathbf{R}$ is a lattice vector. So

Since this holds for any test function:

Eigenvalues of T̂_R Are Pure Phases

Let $\psi$ be a simultaneous eigenstate. Call its eigenvalue under $\hat{T}_{\mathbf{R}}$ the number $c(\mathbf{R})$:

Two constraints fix $c(\mathbf{R})$ almost completely:

1. Unitarity says $|c(\mathbf{R})| = 1$. Translation does not change the norm, so the eigenvalue of a unitary operator must lie on the unit circle. Write $c(\mathbf{R}) = e^{i\phi(\mathbf{R})}$.
2. Composition demands $c(\mathbf{R})\,c(\mathbf{R}') = c(\mathbf{R}+\mathbf{R}')$, i.e. $\phi(\mathbf{R}) + \phi(\mathbf{R}') = \phi(\mathbf{R}+\mathbf{R}')$. So $\phi$ is a linear function of $\mathbf{R}$.

A linear scalar function of a 3-vector is a dot product with some other vector. Call that vector $\mathbf{k}$:

Group theory has just delivered, for free, the central object: every joint eigenstate of $\hat{H}$ and $\hat{T}_{\mathbf{R}}$ is labeled by a wavevector $\mathbf{k}$, and that wavevector determines the phase factor $e^{i\mathbf{k}\cdot\mathbf{R}}$ the state picks up under translation by $\mathbf{R}$.

Bloch's Theorem — Two Equivalent Forms

Combining the two facts — eigenstates of $\hat{H}$ can be chosen as eigenstates of $\hat{T}_{\mathbf{R}}$, and those eigenvalues are exactly $e^{i\mathbf{k}\cdot\mathbf{R}}$ — gives the theorem in its first form:

Translating the wavefunction by one lattice vector multiplies it by a phase. Now define a new function by stripping out the plane-wave envelope:

Under translation by $\mathbf{R}$:

The two phase factors cancel exactly. Therefore:

Forms I and II are mathematically equivalent — given one, you can derive the other by the calculation above. They emphasise different aspects:

One sentence summary: a Bloch state is a plane wave whose amplitude has been "painted" with the periodicity of the crystal — an unbounded carrier of crystal momentum, modulated locally by the atomic landscape.

Interactive — A 1D Bloch State

The cleanest place to feel Bloch's theorem is 1D. The diagram below plots a model 1D Bloch state $\psi_k(x) = e^{ikx}\,u_k(x)$ with $u_k(x) = 1 + \alpha\cos(2\pi x/a)$ — a constant plus a single Fourier component at the smallest reciprocal lattice vector $G = 2\pi/a$. By construction $u_k(x+a) = u_k(x)$, so this is a legitimate Bloch state.

Three things to play with:

• Slide $k$. Watch the cyan $\mathrm{Re}\,\psi$ and purple $\mathrm{Im}\,\psi$ oscillate at different rates as the plane-wave envelope twists in the complex plane. At $k=0$ (Γ) the imaginary part vanishes; at the zone edge $k = \pi/a$ (X) the wavefunction flips sign every cell.
• Slide $\alpha$. At $\alpha = 0$ the wavefunction is a pure plane wave (free electron) — its $|\psi|^2$ is constant. As you crank $\alpha$ up, the pink $|\psi|^2$ develops bumps locked to the atoms — the lattice is "squeezing" electron density into the cells.
• Read the table. The phase column shows $\arg\psi(na)$ at successive lattice sites. The increment per row is exactly $ka$, independent of $\alpha$ — the visual fingerprint of Form I of the theorem.

The Cell-Periodic Part u_k(r)

Form II isolates the "atomic" ingredient of the Bloch state. Because $u_{\mathbf{k}}$ has the periodicity of the lattice, it is fully described by its values inside one unit cell — a finite chunk of space. And any periodic function can be expanded in the natural basis on a lattice: a Fourier series whose wavevectors are the reciprocal lattice vectors $\mathbf{G}$.

Substituting back into Form II:

That is one of the most consequential identities in solid-state physics. It says the Bloch state at wavevector $\mathbf{k}$ is a superposition of plane waves whose wavevectors form a single set: $\{\mathbf{k}+\mathbf{G}\}$. No other plane waves contribute. The full Hilbert space, infinite in principle, is block-diagonal in $\mathbf{k}$, with each block spanned only by plane waves shifted by some reciprocal lattice vector.

Crystal Momentum and Why k Lives in the BZ

The label $\mathbf{k}$ looks like a momentum, but there is a subtle redundancy. Replace $\mathbf{k}$ by $\mathbf{k}' = \mathbf{k} + \mathbf{G}$ for any reciprocal lattice vector $\mathbf{G}$. The phase factor under translation becomes

because $e^{i\mathbf{G}\cdot\mathbf{R}} = 1$ for every reciprocal-lattice / direct-lattice pair (this was the entire point of Section 3.2). So Bloch states at $\mathbf{k}$ and $\mathbf{k}+\mathbf{G}$ transform identically under every translation — $\mathbf{k}$ is only defined modulo $\mathbf{G}$.

The natural fundamental domain is the first Brillouin zone (Section 3.3). Every physically distinct Bloch wavevector can be reduced into the first BZ by adding or subtracting some $\mathbf{G}$. This is the famous reduced-zone scheme: instead of plotting bands over all of $\mathbb{R}^3$, plot them only over the BZ — the rest is redundant.

Worked Example — The Free-Electron Limit

Suppose $V(\mathbf{r}) = 0$ everywhere. Then the eigenstates are plane waves $\psi(\mathbf{r}) \propto e^{i\mathbf{q}\cdot\mathbf{r}}$ with energies $E = \hbar^2 q^2/2m$. Are these Bloch states? Yes — set $u(\mathbf{r}) = \text{constant}$ and $\mathbf{k} = \mathbf{q}$. The cell-periodic part is just a constant (the trivially periodic function) and the entire structure is in the plane-wave envelope.

But $\mathbf{q}$ can be anywhere in $\mathbb{R}^3$, not just the first BZ. The reduced-zone scheme handles this gracefully: write $\mathbf{q} = \mathbf{k} + \mathbf{G}$ with $\mathbf{k}$ in the first BZ. Then

and indeed $e^{i\mathbf{G}\cdot\mathbf{r}}$ is cell-periodic because $e^{i\mathbf{G}\cdot(\mathbf{r}+\mathbf{R})} = e^{i\mathbf{G}\cdot\mathbf{r}}$ (again, $e^{i\mathbf{G}\cdot\mathbf{R}}=1$). Each free-electron plane wave is a Bloch state at the BZ-folded $\mathbf{k}$, with cell-periodic part equal to the leftover plane wave indexed by $\mathbf{G}$.

Bloch's Theorem in VASP — The Plane-Wave Expansion

Putting Bloch's theorem and Fourier expansion together gives the equation that VASP, Quantum ESPRESSO, ABINIT, and every other plane-wave DFT code is built around:

where $n$ is the band index and the sum runs over all reciprocal lattice vectors $\mathbf{G}$ within a chosen energy cutoff:

This is the ENCUT tag in your VASP INCAR file. The variational unknowns in the calculation are the complex coefficients $c_{n,\mathbf{k},\mathbf{G}}$. For each band $n$ and each k-point sampled in the BZ, VASP solves a generalised eigenvalue problem in a basis of order $N_{\text{PW}} \sim V \cdot E_{\text{cut}}^{3/2}$ plane waves.

Code — Building and Verifying a Bloch State

Theorems are easier to trust once you have built one with your hands. The snippet below constructs a Bloch state by definition — $\psi_k(x) = e^{ikx} u_k(x)$ with a cell-periodic $u_k$ — and then verifies the defining identity $\psi_k(x+a) = e^{ika}\psi_k(x)$ numerically. Every line carries an explanation card; click any line in the right panel to see it.

1import numpy as np
2
3# Lattice setup: a single Bloch state in 1D
4a = 1.0           # lattice constant (arbitrary units)
5N_cells = 6
6N_per_cell = 200
7x = np.linspace(0, N_cells * a, N_cells * N_per_cell)
8
9# Cell-periodic part u(x) — must satisfy u(x + a) = u(x)
10def u_periodic(x, a, alpha=0.45):
11    G = 2 * np.pi / a
12    return 1.0 + alpha * np.cos(G * x)
13
14# Pick a wavevector inside the first Brillouin zone
15k = 0.25 * (2 * np.pi / a)   # k = pi / (2a), midway between Gamma and X
16
17# Construct the Bloch wavefunction
18psi = np.exp(1j * k * x) * u_periodic(x, a)
19
20# Verify the Bloch property: psi(x + a) = exp(ika) * psi(x)
21phase_factor = np.exp(1j * k * a)
22i0 = 0
23i_a = N_per_cell  # index corresponding to x = a
24
25print(f"psi(0)            = {psi[i0]: .4f}")
26print(f"psi(a)            = {psi[i_a]: .4f}")
27print(f"e^(ika) * psi(0)  = {phase_factor * psi[i0]: .4f}")
28print(f"|psi(0)|^2        = {abs(psi[i0])**2: .4f}")
29print(f"|psi(a)|^2        = {abs(psi[i_a])**2: .4f}")

Running this prints (rounded to 4 decimals):

Three observations to lock in:

1. The middle two lines match exactly. Translating $\psi$ by $a$ is identical to multiplying the on-site value by $e^{ika}$. This is Form I of the theorem, numerically.
2. The last two lines are equal. The probability density $|\psi|^2$ is lattice-periodic — Form I forces this because $|e^{ika}|=1$.
3. We never solved a Schrödinger equation. We chose a $u_k$ and got Bloch form for free. In a real DFT calculation, $u_k$ is determined by the self-consistent Kohn–Sham equation — but it is forced into this same Bloch shape by the symmetry argument we made earlier. The code is instructive precisely because it isolates the kinematics (Bloch form) from the dynamics (which $u_k$ is the right one).

Common Pitfalls

Summary

• A periodic potential makes lattice translations $\hat{T}_{\mathbf{R}}$ commute with the Hamiltonian. Eigenstates can therefore be chosen as joint eigenstates of $\hat{H}$ and every $\hat{T}_{\mathbf{R}}$.
• The eigenvalues of $\hat{T}_{\mathbf{R}}$ are pure phases $e^{i\mathbf{k}\cdot\mathbf{R}}$; the continuous label $\mathbf{k}$ is the crystal momentum.
• Bloch's theorem (Form I): $\psi_{\mathbf{k}}(\mathbf{r}+\mathbf{R}) = e^{i\mathbf{k}\cdot\mathbf{R}}\psi_{\mathbf{k}}(\mathbf{r})$.
• Bloch's theorem (Form II): $\psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}}\,u_{\mathbf{k}}(\mathbf{r})$ with $u_{\mathbf{k}}$ cell-periodic.
• The probability density $|\psi_{\mathbf{k}}|^2$ has the full periodicity of the lattice; the wavefunction itself only repeats up to the phase $e^{i\mathbf{k}\cdot\mathbf{R}}$.
• $\mathbf{k}$ is defined modulo any reciprocal lattice vector $\mathbf{G}$; the natural fundamental domain is the first Brillouin zone.
• Fourier-expanding $u_{\mathbf{k}}$ on the reciprocal lattice gives $\psi_{\mathbf{k}} = \sum_{\mathbf{G}} c_{\mathbf{k},\mathbf{G}}\,e^{i(\mathbf{k}+\mathbf{G})\cdot\mathbf{r}}$ — the plane-wave expansion VASP uses internally. ENCUT picks the number of $\mathbf{G}$; KPOINTS picks the set of $\mathbf{k}$.

Coming next: Section 3.6 — Structure Factor and Diffraction — where we use the Fourier-on-the-reciprocal- lattice machinery developed here to compute X-ray diffraction intensities, derive the Laue condition, and predict which Bragg peaks are systematically absent for a given crystal structure.