Bloch's Theorem from Translational Symmetry

Derivation

Bloch's Theorem from Translational Symmetry

D-249 Home PU-303 Threads symmetry · waves · matter Depends on Reciprocal Lattice from the Bravais Lattice, Degenerate Perturbation Theory

Statement

For a single particle in a potential with the periodicity of a Bravais lattice, \(V(\vec{r}+\vec{R})=V(\vec{r})\) for every lattice vector \(\vec{R}\), the energy eigenstates of the Hamiltonian may be chosen so that each has the form \(\psi_{n\vec{k}}(\vec{r})=e^{i\vec{k}\cdot\vec{r}}\,u_{n\vec{k}}(\vec{r})\) with \(u_{n\vec{k}}(\vec{r}+\vec{R})=u_{n\vec{k}}(\vec{r})\), equivalently \(\psi_{n\vec{k}}(\vec{r}+\vec{R})=e^{i\vec{k}\cdot\vec{R}}\psi_{n\vec{k}}(\vec{r})\), where the crystal momentum \(\vec{k}\) can be restricted to the first Brillouin zone and \(n\) is a discrete band index.

Why it matters

Bloch's theorem is the organizing principle of solid-state physics. It reduces the intractable problem of an electron moving in a potential produced by \(\sim 10^{23}\) atoms to a family of problems on a single primitive cell, each labelled by a conserved quantum number \(\vec{k}\). The continuous energy spectrum of free space fractures into bands \(E_n(\vec{k})\), and everything from conduction, semiconductor gaps, and effective masses to optical selection rules follows from the geometry of those bands.

The result is purely a statement about symmetry: it does not require the potential to be weak, and it holds for any Bravais lattice. That generality is what lets a two-line group-theoretic argument replace a many-body calculation, and it is why the same theorem reappears for phonons, photonic crystals, and any wave equation with discrete translational invariance.

Assumptions

The potential is exactly lattice-periodic: \(V(\vec{r}+\vec{R})=V(\vec{r})\) for all \(\vec{R}\) in the Bravais lattice. Drop it and the translation operators no longer commute with \(H\); \(\vec{k}\) ceases to be a good quantum number and eigenstates need not factor into a plane wave times a periodic part.

The Hamiltonian is a single-particle operator: \(H=-\frac{\hbar^2}{2m}\nabla^2+V(\vec{r})\), so translations act as ordinary c-number shifts. Drop it (genuine two-body interactions that depend on relative coordinates) and lattice translation of one particle is not a symmetry, invalidating the construction.

Born–von Kármán boundary conditions on a finite crystal of \(N=N_1N_2N_3\) cells: \(\psi(\vec{r}+N_i\vec{a}_i)=\psi(\vec{r})\). Drop it and the spectrum of allowed \(\vec{k}\) is not discrete; the counting of \(N\) states per band and the reduction to the first Brillouin zone lose their clean form, though the factorized form survives in the infinite-volume limit.

No magnetic field (or only a field commensurate with the lattice): The canonical momentum enters through \(-i\hbar\nabla\) alone. Drop it: a uniform \(\vec{B}\) makes the magnetic translation operators fail to commute among themselves except at rational flux, giving the Hofstadter problem rather than ordinary Bloch bands.

Derivation

\[ (T_{\vec{R}}\,\psi)(\vec{r})\equiv\psi(\vec{r}+\vec{R}),\qquad \vec{R}=\sum_i n_i\vec{a}_i,\ n_i\in\mathbb{Z} \]

Define the lattice-translation operator for each Bravais vector \(\vec{R}\) built from the primitive vectors \(\vec{a}_i\). A pure relabelling of the wavefunction's argument. A

\[ (T_{\vec{R}}\,H\psi)(\vec{r})=-\frac{\hbar^2}{2m}\nabla^2\psi(\vec{r}+\vec{R})+V(\vec{r}+\vec{R})\,\psi(\vec{r}+\vec{R}) \]

Apply \(T_{\vec{R}}\) to \(H\psi\). The Laplacian is invariant under a constant shift of coordinates, and by periodicity \(V(\vec{r}+\vec{R})=V(\vec{r})\). A

\[ T_{\vec{R}}\,H=H\,T_{\vec{R}}\quad\Longrightarrow\quad [H,T_{\vec{R}}]=0\ \ \forall\vec{R} \]

Comparing step 2 with \(H(T_{\vec{R}}\psi)\) shows the two orderings agree for every \(\psi\); the operators commute. This is the whole physical content: periodicity is a symmetry of \(H\). A

\[ T_{\vec{R}}\,T_{\vec{R}'}=T_{\vec{R}+\vec{R}'}=T_{\vec{R}'}\,T_{\vec{R}},\qquad [T_{\vec{R}},T_{\vec{R}'}]=0 \]

Successive shifts add, and vector addition is commutative, so the translation operators form an abelian group. A set of mutually commuting operators that also commute with \(H\) can be simultaneously diagonalized with \(H\). B

\[ H\psi=E\psi,\qquad T_{\vec{R}}\,\psi=c(\vec{R})\,\psi\quad\text{for all }\vec{R} \]

Choose common eigenstates. Because the \(T_{\vec{R}}\) commute with each other and with \(H\), a simultaneous eigenbasis exists; degenerate energy eigenspaces are decomposed by the translation eigenvalues (this is where degenerate-perturbation-theory-style block diagonalization is invoked). C

\[ c(\vec{R})\,c(\vec{R}')=c(\vec{R}+\vec{R}') \]

Act with \(T_{\vec{R}}T_{\vec{R}'}=T_{\vec{R}+\vec{R}'}\) on \(\psi\). The eigenvalues inherit the group multiplication: they form a one-dimensional representation of the translation group. B

\[ c(\vec{a}_i)=e^{i\theta_i}\quad(\theta_i\in\mathbb{R}) \]

Each \(T_{\vec{R}}\) is unitary (it preserves the norm under the Born–von Kármán conditions), so its eigenvalues have unit modulus. Write the eigenvalue for each primitive translation \(\vec{a}_i\) as a pure phase. B

\[ c(\vec{R})=c(\vec{a}_1)^{n_1}c(\vec{a}_2)^{n_2}c(\vec{a}_3)^{n_3}=e^{i(n_1\theta_1+n_2\theta_2+n_3\theta_3)} \]

Use the multiplicative law of step 6 on \(\vec{R}=\sum_i n_i\vec{a}_i\). The eigenvalue for any lattice vector is fixed by the three primitive phases. B

\[ \vec{k}\equiv\frac{1}{2\pi}\sum_i\theta_i\,\vec{b}_i,\qquad \vec{b}_i\cdot\vec{a}_j=2\pi\,\delta_{ij} \]

Introduce the reciprocal-lattice basis \(\vec{b}_i\) (from reciprocal-lattice-from-bravais) and define the crystal wavevector \(\vec{k}\) so that \(\vec{k}\cdot\vec{a}_i=\theta_i\). This is a definition; it re-expresses the three phases as one vector. C

\[ n_1\theta_1+n_2\theta_2+n_3\theta_3=\vec{k}\cdot\!\sum_i n_i\vec{a}_i=\vec{k}\cdot\vec{R}\ \Rightarrow\ c(\vec{R})=e^{i\vec{k}\cdot\vec{R}} \]

Substitute the reciprocal-basis identity \(\vec{b}_i\cdot\vec{a}_j=2\pi\delta_{ij}\) into step 8. The translation eigenvalue is now a single plane-wave phase in \(\vec{k}\). B

\[ \psi(\vec{r}+\vec{R})=(T_{\vec{R}}\psi)(\vec{r})=e^{i\vec{k}\cdot\vec{R}}\,\psi(\vec{r}) \]

Combine the eigenvalue equation of step 5 with \(c(\vec{R})=e^{i\vec{k}\cdot\vec{R}}\). This is Bloch's theorem in its first (phase) form. A

\[ u_{\vec{k}}(\vec{r})\equiv e^{-i\vec{k}\cdot\vec{r}}\,\psi(\vec{r}) \]

Define \(u_{\vec{k}}\) by stripping the plane-wave factor. A definition, motivated by wanting to isolate a periodic remainder. A

\[ u_{\vec{k}}(\vec{r}+\vec{R})=e^{-i\vec{k}\cdot(\vec{r}+\vec{R})}\psi(\vec{r}+\vec{R})=e^{-i\vec{k}\cdot\vec{r}}e^{-i\vec{k}\cdot\vec{R}}e^{i\vec{k}\cdot\vec{R}}\psi(\vec{r})=u_{\vec{k}}(\vec{r}) \]

Shift the argument of \(u_{\vec{k}}\) and insert step 11. The two exponential phases cancel exactly, so \(u_{\vec{k}}\) is lattice-periodic. B

\[ \psi_{\vec{k}}(\vec{r})=e^{i\vec{k}\cdot\vec{r}}\,u_{\vec{k}}(\vec{r}),\qquad u_{\vec{k}}(\vec{r}+\vec{R})=u_{\vec{k}}(\vec{r}) \]

Invert the definition of step 12. This is Bloch's theorem in its second (envelope) form; the two forms are algebraically identical. A

\[ e^{i(\vec{k}+\vec{G})\cdot\vec{R}}=e^{i\vec{k}\cdot\vec{R}}e^{i\vec{G}\cdot\vec{R}}=e^{i\vec{k}\cdot\vec{R}},\qquad \vec{G}\cdot\vec{R}=2\pi\times\text{integer} \]

For any reciprocal-lattice vector \(\vec{G}\), \(\vec{G}\cdot\vec{R}\in 2\pi\mathbb{Z}\), so \(\vec{k}\) and \(\vec{k}+\vec{G}\) give identical translation eigenvalues and label the same physics. Hence \(\vec{k}\) is defined only modulo \(\vec{G}\) and may be folded into the first Brillouin zone; the residual discrete label at fixed \(\vec{k}\) is the band index \(n\). C

Result

\[ \boxed{\ \psi_{n\vec{k}}(\vec{r})=e^{i\vec{k}\cdot\vec{r}}\,u_{n\vec{k}}(\vec{r}),\qquad u_{n\vec{k}}(\vec{r}+\vec{R})=u_{n\vec{k}}(\vec{r}),\qquad \vec{k}\in\text{1st BZ}\ } \]

Reading. Every eigenstate of a periodic Hamiltonian is a running plane wave \(e^{i\vec{k}\cdot\vec{r}}\) whose amplitude is modulated by a function \(u_{n\vec{k}}\) that repeats with the lattice. The plane wave carries the long-range coherence — it is the same at equivalent points of every cell up to a fixed phase \(e^{i\vec{k}\cdot\vec{R}}\) — while \(u_{n\vec{k}}\) carries the atomic-scale structure inside a cell. The wavevector \(\vec{k}\) is a conserved crystal momentum, unique within the first Brillouin zone; the discrete index \(n\) counts the bands, i.e. the tower of solutions of the cell problem at fixed \(\vec{k}\).

Units check. \(\vec{k}\) has dimensions of inverse length, so \(\vec{k}\cdot\vec{r}\) and \(\vec{k}\cdot\vec{R}\) are dimensionless and the exponentials are pure phases, as required. \(u_{n\vec{k}}\) carries the same dimension as \(\psi\) (length\(^{-3/2}\) in 3D for a normalized state), so the product \(e^{i\vec{k}\cdot\vec{r}}u_{n\vec{k}}\) is correctly normalizable. The reciprocal vectors satisfy \([\vec{b}_i]=[\vec{a}_i]^{-1}\), consistent with \(\vec{b}_i\cdot\vec{a}_j=2\pi\delta_{ij}\) being dimensionless.

Limiting cases

Empty lattice, \(V\to 0\): \(u_{n\vec{k}}\) becomes a constant (or a single reciprocal-lattice plane wave \(e^{i\vec{G}\cdot\vec{r}}\)), recovering free-particle states \(\psi\propto e^{i(\vec{k}+\vec{G})\cdot\vec{r}}\) with parabolic bands folded into the zone.
\(\vec{k}=0\) (zone centre, \(\Gamma\) point): \(\psi=u_{n0}\) is itself lattice-periodic; the state has the full translational symmetry of the crystal.
Tight-binding / isolated-atom limit: \(u_{n\vec{k}}\) concentrates on atomic orbitals and \(\psi_{n\vec{k}}\) reduces to a lattice sum \(\sum_{\vec{R}}e^{i\vec{k}\cdot\vec{R}}\phi(\vec{r}-\vec{R})\), the Bloch sum of Wannier/atomic functions.
1D chain of \(N\) sites: allowed \(k=2\pi m/(Na)\), \(m=0,\dots,N-1\), giving exactly \(N\) states per band in the zone \(-\pi/a<k\le\pi/a\).

Breaks when

The periodicity is broken. Any defect, impurity, surface, or disorder makes \([H,T_{\vec{R}}]\neq 0\); \(\vec{k}\) is no longer conserved. In strong disorder the extended Bloch states give way to Anderson-localized states with no plane-wave factor at all.
A uniform magnetic field is present. The gauge potential spoils ordinary translational symmetry: magnetic translation operators commute only when the flux per plaquette is a rational multiple of the flux quantum, producing the fractal Hofstadter spectrum rather than smooth bands \(E_n(\vec{k})\).
Genuine many-body interactions dominate. When electron–electron correlations cannot be absorbed into a single-particle periodic potential (e.g. a Mott insulator), the single-particle Bloch label loses meaning even though the crystal is periodic.
Quasicrystals / incommensurate potentials. With no Bravais lattice there is no discrete translation group; states are labelled by higher-dimensional or fractal schemes, not a Brillouin-zone \(\vec{k}\).

Failure modes

Confusing \(\vec{k}\) with mechanical momentum. Crystal momentum \(\hbar\vec{k}\) is conserved only modulo \(\hbar\vec{G}\); \(-i\hbar\nabla\) does not have \(\psi_{n\vec{k}}\) as an eigenstate because \(u_{n\vec{k}}\) is not constant. Umklapp processes exploit exactly this.
Claiming \(\psi_{n\vec{k}}\) is itself periodic. Only \(u_{n\vec{k}}\) is lattice-periodic; \(\psi\) picks up the phase \(e^{i\vec{k}\cdot\vec{R}}\) under a lattice shift and is periodic only for \(\vec{k}\) at a reciprocal-lattice point.
Assuming the theorem needs weak \(V\). Bloch's theorem is exact for any strength of periodic potential; it is a symmetry statement, not a perturbative one. Nearly-free-electron expansion is a separate approximation.
Forgetting the band index. Fixing \(\vec{k}\) does not fix the state: the cell problem has a discrete tower of solutions \(u_{n\vec{k}}\). Omitting \(n\) undercounts states and misidentifies gaps.
Using the wrong zone / double counting. Treating \(\vec{k}\) and \(\vec{k}+\vec{G}\) as distinct states double-counts; states must be reduced to one Brillouin zone.

Discussion

The deepest reading of Bloch's theorem is representation-theoretic. The lattice translations form an abelian group, and the irreducible representations of any abelian group are one-dimensional and labelled by a phase — a character \(\chi_{\vec{k}}(\vec{R})=e^{i\vec{k}\cdot\vec{R}}\). Bloch's theorem is nothing more than the statement that energy eigenstates can be organized by these characters. The Brillouin zone is the space of inequivalent characters (the group of characters of the lattice, itself a torus), and the band index \(n\) enumerates how many times each irreducible representation appears in the physical Hilbert space. Seen this way, the theorem is not special to the Schrödinger equation at all: it governs phonons, electromagnetic modes in photonic crystals, and any linear wave problem on a periodic medium.

The factorization has an immediate dynamical payoff. Substituting \(\psi_{n\vec{k}}=e^{i\vec{k}\cdot\vec{r}}u_{n\vec{k}}\) into \(H\psi=E\psi\) gives a \(\vec{k}\)-dependent Hamiltonian \(H(\vec{k})=\frac{1}{2m}(-i\hbar\nabla+\hbar\vec{k})^2+V\) acting on periodic functions on a single cell. The spectrum of that cell operator is the band structure \(E_n(\vec{k})\), and its \(\vec{k}\)-dependence encodes group velocity \(\vec{v}_n=\frac{1}{\hbar}\nabla_{\vec{k}}E_n\) and inverse effective mass \((m^*)^{-1}_{ij}=\frac{1}{\hbar^2}\partial^2 E_n/\partial k_i\partial k_j\). An electron in a perfect lattice therefore propagates without scattering — resistance comes from the breaking of periodicity, not from the ions themselves, resolving the classical puzzle of why metals conduct so well.

A subtler layer is the geometry of the periodic functions \(u_{n\vec{k}}\) as \(\vec{k}\) sweeps the zone. Because each \(u_{n\vec{k}}\) is defined only up to a \(\vec{k}\)-dependent phase, the bundle of Bloch states over the Brillouin-zone torus can be topologically nontrivial. The Berry connection \(\vec{A}_n(\vec{k})=i\langle u_{n\vec{k}}|\nabla_{\vec{k}}|u_{n\vec{k}}\rangle\) and its curvature integrate to Chern numbers that classify integer quantum Hall phases and, with additional symmetries, topological insulators. Bloch's theorem thus supplies not just a label but a fiber bundle whose global structure is physically observable — a fact invisible in the original 1928 derivation and central to modern condensed-matter physics.

Common misconceptions. Bloch's theorem does not say electrons are free plane waves — the modulation \(u_{n\vec{k}}\) can be strongly peaked at atoms. It does not require a weak or nearly-free potential. And \(\hbar\vec{k}\) is crystal momentum, a bookkeeping label for translation eigenvalues, not the expectation value of the momentum operator; only its conservation modulo \(\hbar\vec{G}\) is guaranteed.

Worked examples

\[ \textbf{1D Kronig–Penney chain: count states and locate zone edge} \] \[ V(x+a)=V(x),\quad a=0.30\ \text{nm},\quad N=1000\ \text{cells (BvK)} \]

Setup: a finite periodic chain of lattice constant \(a\) with Born–von Kármán periodicity over \(N\) cells. Symbols first. A

\[ e^{ikNa}=1\ \Rightarrow\ k=\frac{2\pi m}{Na},\quad m=0,\dots,N-1 \]

BvK quantizes \(k\); there are exactly \(N\) allowed values, hence \(N\) states per band. A

\[ \Delta k=\frac{2\pi}{Na}=\frac{2\pi}{1000\times 0.30\ \text{nm}}=2.09\times 10^{7}\ \text{m}^{-1} \]

Level spacing in \(k\). Numbers now inserted: \(Na=3.0\times10^{-7}\ \text{m}\). B

\[ k_{\text{BZ edge}}=\frac{\pi}{a}=\frac{\pi}{0.30\times10^{-9}\ \text{m}}=1.05\times 10^{10}\ \text{m}^{-1} \]

Zone boundary. The full zone \((-\pi/a,\pi/a]\) holds \(2\pi/a\div\Delta k=N=1000\) states, confirming the count. B

\[ N=1000\ \text{states/band},\quad k_{\text{edge}}=1.05\times10^{10}\ \text{m}^{-1},\quad \Delta k=2.09\times10^{7}\ \text{m}^{-1} \]

Reading. Each band holds one \(k\)-state per unit cell; the zone edge sits at \(\pi/a\), and the discrete grid spacing scales as \(1/(Na)\), becoming a continuum as \(N\to\infty\). Units: all wavevectors in \(\text{m}^{-1}\), consistent.

\[ \textbf{Bloch phase and group velocity for a nearly-free electron} \] \[ E_1(k)=\frac{\hbar^2 k^2}{2m},\quad k=0.5\,\frac{\pi}{a},\quad a=0.40\ \text{nm} \]

Setup: lowest band approximated as free near zone centre; evaluate the Bloch phase across one cell and the group velocity. Symbols first. A

\[ \psi(x+a)=e^{ika}\psi(x),\qquad ka=0.5\pi=1.571\ \text{rad} \]

Bloch phase per lattice step from the theorem. Numbers: \(k=\pi/(2a)\Rightarrow ka=\pi/2\). A

\[ v_g=\frac{1}{\hbar}\frac{dE_1}{dk}=\frac{\hbar k}{m} \]

Group velocity from the band slope (symbolic). B

\[ k=\frac{\pi}{2(0.40\times10^{-9})}=3.93\times10^{9}\ \text{m}^{-1} \] \[ v_g=\frac{(1.055\times10^{-34})(3.93\times10^{9})}{9.11\times10^{-31}}=4.55\times10^{5}\ \text{m/s} \]

Insert \(\hbar=1.055\times10^{-34}\,\text{J·s}\), \(m=9.11\times10^{-31}\,\text{kg}\). B

\[ \psi(x+a)=e^{i\pi/2}\psi(x)=i\,\psi(x),\qquad v_g=4.6\times10^{5}\ \text{m/s} \]

Reading. Advancing one lattice constant multiplies the wavefunction by \(i\); after four cells the phase completes a full turn (\(e^{i2\pi}\)), matching wavelength \(2\pi/k=4a\). The group velocity \(\sim 5\times10^5\ \text{m/s}\) is a typical band electron speed. Units: \(v_g\) in \(\text{J·s·m}^{-1}/\text{kg}=\text{m/s}\), correct.

Problems

Show that if \(\psi_{\vec{k}}\) is a Bloch state with wavevector \(\vec{k}\), then \(\psi_{\vec{k}}\) and \(\psi_{\vec{k}+\vec{G}}\) describe the same physical state for any reciprocal-lattice vector \(\vec{G}\).
Solution
Under any lattice translation, \(T_{\vec{R}}\psi_{\vec{k}}=e^{i\vec{k}\cdot\vec{R}}\psi_{\vec{k}}\). For \(\vec{k}\to\vec{k}+\vec{G}\) the eigenvalue becomes \(e^{i(\vec{k}+\vec{G})\cdot\vec{R}}=e^{i\vec{k}\cdot\vec{R}}e^{i\vec{G}\cdot\vec{R}}\). By definition of the reciprocal lattice \(\vec{G}\cdot\vec{R}=2\pi\times\text{integer}\), so \(e^{i\vec{G}\cdot\vec{R}}=1\) and both wavevectors give identical translation eigenvalues for every \(\vec{R}\). Since the states are distinguished only by these eigenvalues, they carry the same label and reduce to the same first-BZ state (differing at most by relabelling of \(u\)). Hence \(\vec{k}\) is defined modulo \(\vec{G}\).
For a 2D square lattice with constant \(a\), find the primitive reciprocal vectors and the area of the first Brillouin zone. If the crystal has \(N=N_1N_2\) cells with area \(A=N a^2\), how many allowed \(\vec{k}\) lie in the zone?
Solution
Real primitive vectors \(\vec{a}_1=a\hat{x},\ \vec{a}_2=a\hat{y}\). Reciprocal vectors satisfy \(\vec{b}_i\cdot\vec{a}_j=2\pi\delta_{ij}\), giving \(\vec{b}_1=\frac{2\pi}{a}\hat{x},\ \vec{b}_2=\frac{2\pi}{a}\hat{y}\). The first BZ is the square \(-\pi/a<k_{x,y}\le\pi/a\), area \(A_{\text{BZ}}=(2\pi/a)^2=4\pi^2/a^2\). Each allowed \(\vec{k}\) occupies area \((2\pi)^2/A=(2\pi)^2/(Na^2)\). Count \(=A_{\text{BZ}}\big/\big[(2\pi)^2/(Na^2)\big]=\frac{4\pi^2/a^2}{4\pi^2/(Na^2)}=N\). Exactly \(N\) states per band, as required.
A 1D lattice has \(a=0.25\ \text{nm}\) and \(N=2000\) cells. Compute the zone-edge wavevector, the spacing \(\Delta k\), and the total number of distinct \(\vec{k}\) in the lowest band.
Solution
Zone edge \(k=\pi/a=\pi/(0.25\times10^{-9})=1.257\times10^{10}\ \text{m}^{-1}\). BvK spacing \(\Delta k=2\pi/(Na)=2\pi/(2000\times0.25\times10^{-9})=2\pi/(5.0\times10^{-7})=1.257\times10^{7}\ \text{m}^{-1}\). Number of states \(=\) zone width \(/\Delta k=(2\pi/a)/(2\pi/(Na))=N=2000\). So 2000 distinct \(k\)-states per band.
Verify explicitly that the trial function \(\psi(x)=e^{ikx}\cos(2\pi x/a)\) is a valid 1D Bloch state and identify its \(u_k(x)\) and its crystal momentum. Is it also a momentum eigenstate?
Solution
Write \(u_k(x)=e^{-ikx}\psi(x)=\cos(2\pi x/a)\). Then \(u_k(x+a)=\cos(2\pi(x+a)/a)=\cos(2\pi x/a+2\pi)=\cos(2\pi x/a)=u_k(x)\): lattice-periodic. So \(\psi=e^{ikx}u_k\) is Bloch with crystal momentum \(\hbar k\). It is NOT a momentum eigenstate: \(\cos(2\pi x/a)=\tfrac12(e^{i2\pi x/a}+e^{-i2\pi x/a})\), so \(\psi=\tfrac12(e^{i(k+2\pi/a)x}+e^{i(k-2\pi/a)x})\) is a superposition of two momenta \(\hbar(k\pm2\pi/a)\), differing by \(\hbar G\) with \(G=2\pi/a\). Applying \(-i\hbar\partial_x\) does not return a single eigenvalue, illustrating that crystal momentum \(\ne\) mechanical momentum.
Near a band minimum an electron has \(E(k)=E_0+\frac{\hbar^2(k-k_0)^2}{2m^*}\) with effective mass \(m^*=0.20\,m_e\). Compute the group velocity at \(k-k_0=1.0\times10^{9}\ \text{m}^{-1}\) and the inverse curvature check.
Solution
Group velocity \(v_g=\frac{1}{\hbar}\frac{dE}{dk}=\frac{\hbar(k-k_0)}{m^*}\). With \(m^*=0.20\times9.11\times10^{-31}=1.82\times10^{-31}\ \text{kg}\): \(v_g=\frac{(1.055\times10^{-34})(1.0\times10^{9})}{1.82\times10^{-31}}=5.8\times10^{5}\ \text{m/s}\). Curvature check: \(\frac{d^2E}{dk^2}=\frac{\hbar^2}{m^*}\Rightarrow m^*=\hbar^2/(d^2E/dk^2)\); with \(d^2E/dk^2=\hbar^2/m^*=(1.055\times10^{-34})^2/(1.82\times10^{-31})=6.1\times10^{-38}\ \text{J·m}^2\), inverting returns \(m^*=1.82\times10^{-31}\ \text{kg}\), consistent. The small effective mass gives a large velocity for a given \(k-k_0\), as expected for a light band electron.