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Regard $V(x)$ as an electron potential energy, and take $G=2\pi /a$. Each cosine supplies half its coefficient to each of its two Fourier coefficients. The nonzero coefficients areThe Fourier coefficient connecting the two plane waves at $k=\pm \pi /a$ is $V_G$, so the two-wave band gap isAt $\delta =0$ the actual periodic potential has period $a/2$ and no Fourier harmonic at $G$. The crossing at the artificially halved Brillouin zone boundary then has zero band gap. Higher-order weak-potential corrections can change the proportionality coefficient at nonzero $\delta$; the displayed coefficient belongs to the given two-wave approximation.

There is also a small-displacement qualification in the source: the claimed positions of the potential energy minima are correct to first order in $\delta$, not exactly. Put $t=2\pi x/a$. ThenAt $\delta =0.01$ and $x/a=1/4-\delta =0.24$, this equals approximately $-2.802\times 10^{-3}$, rather than zero. For $V_0>0$, expanding about $x/a=1/4$ and $3/4$ gives minima at $1/4-\delta +O({\delta }^3)$ and $3/4+\delta +O({\delta }^3)$. This does not alter the leading Fourier coefficient or the linear opening of the band gap.

Construct the first Brillouin zone by the perpendicular bisectors between the origin and reciprocal lattice points. For a test point $(h{a}^{\ast },0,l{c}^{\ast })$, the nearest relevant reciprocal points are $(0,0,\pm 2c^{\ast })$ and $(\pm a^{\ast },0,\pm c^{\ast })$. Their inequalities areSet $\rho =(c^{\ast }/a^{\ast }{)}^2=(a/c{)}^2=0.113024$. The six vertices in conventional $(h,l)$ coordinates areNumerically these are $(\pm 0.55651,0)$ and $(\pm 0.44349,\pm 1)$. The $k\ne 0$ reciprocal lattice vectors must in principle be checked because the requested object is a section of a three-dimensional zone. For example $(\pm 1,\pm 1,0)$ gives the weaker bound $|h|\le 1$, and $(0,\pm 1,\pm 1)$ gives a weaker vertical bound; neither clips this hexagon. Direct comparison with surrounding three-dimensional reciprocal lattice vectors confirms the displayed section. It is not the rectangular reciprocal conventional cell.

The primitive unit cell contains two carbon atoms and four edge-shared oxygen atoms, each contributing one half: $n=2+4/2=4$ atoms. Equivalently, one molecular repeat contributes its two carbon atoms but only half of each of its four shared oxygen atoms. There are therefore $3n=12$ phonon branches.

At the Brillouin zone centre, three rigid translations give three acoustic phonon branches; the remaining $3n-3$ are optical phonon branches:A free layer has two in-plane acoustic modes and an out-of-plane flexural acoustic mode. A two-dimensional lattice does not imply that its atoms can move only in two directions. Artificially suppressing all out-of-plane motion would instead define a different model with two acoustic and six optical branches.

For ordinary one-phonon infrared optical absorption, both an energy and a momentum selection rule must hold. The optical photon has negligible crystal momentum on the scale of the Brillouin zone, so crystal momentum conservation selects $\mathbf{q}\simeq 0$ modulo a genuine reciprocal lattice vector. In addition, the mode must change the macroscopic electric polarization, with a component along the incident polarization of an electromagnetic wave.

In terms of atomic displacements ${\mathbf{u}}_j$ and Born effective charge tensors $Z_j^{\ast }$, the relevant derivative is proportional to ${\sum }_j{Z}_j^{\ast }{\mathbf{u}}_j$ per primitive unit cell. Thus an optical phonon at $\mathbf{q}=0$ is not automatically infrared active: its dipole contributions may cancel. Disorder-assisted and multiphonon processes relax these simple first-order selection rules.

Apply the same change of reciprocal lattice coordinates to the phonon wave vectors:Although $(1,0{)}_r$ is an integer coordinate of the rectangular reciprocal grid, it is not a reciprocal lattice vector of the actual primitive crystal. Thus $P_1$ is not equivalent to the Brillouin zone centre.

With atomic displacement $u_j$, the harmonic potential energy is $\frac{1}{2}K{\sum }_j(u_{j+1}-u_j{)}^2$. Newton's second law givesFor a periodic chain of $N$ atoms, its normal modes are $u_j=U{e}^{i(kja-\omega t)}$, with $k=2\pi s/(Na)$ and one representative per Brillouin zone. Substitution yields the monatomic-chain dispersion relationA real general displacement is a superposition of these normal modes and their complex conjugates. The $k=0$ mode is a rigid translation. At small $|k|$, the acoustic phonon branch is linear, with speed of soundFor $0<k<\pi /a$, the group velocity is $a\sqrt{K/m}\cos (ka/2)$ and vanishes at the zone boundary. The dashed initial tangent in the sketch has slope $v$.

For a periodic potential, the Hamiltonian commutes with every lattice translation. The three primitive translations are commuting unitary operators, so within each energy eigenspace one can choose simultaneous translation eigenstates. Their eigenvalues have modulus one and can be written $e^{i\mathbf{k}\cdot {\mathbf{a}}_i}$. Translation by any lattice vector then givesThis proves Bloch's theorem in its eigenbasis form; arbitrary superpositions inside a degenerate eigenspace need not themselves have a single Bloch wavevector.

The reciprocal lattice consists of vectors $\mathbf{g}$ with $\mathbf{g}\cdot {\mathbf{a}}_i\in 2\pi \mathbb{Z}$. Changing $\mathbf{k}$ to $\mathbf{k}+\mathbf{g}$ and $u$ to $e^{-i\mathbf{g}\cdot \mathbf{r}}u$ gives the same physical wavefunction with a periodic new $u$. The eigenvalue problem on one cell has a discrete set of levels, giving bands indexed by $n$, and this equivalence yields $E_n(\mathbf{k}+\mathbf{g})=E_n(\mathbf{k})$. A reciprocal-lattice fundamental domain, conventionally a Brillouin zone, therefore labels inequivalent wavevectors.

Periodic boundary conditions in physical volume $\mathcal{V}$ give one allowed wavevector per reciprocal-space volume $(2\pi {)}^3/\mathcal{V}$. Including the two spin states, there are $2\mathcal{V}{d}^{3}k/(2\pi {)}^3$ states; per unit physical volume the count is the printed $2d^{3}k/(2\pi {)}^3$. Each occupied state carries charge $-e$ and velocity ${\hslash }^{-1}{\nabla }_k{E}_n$, soFor a full band, the integral of ${\nabla }_k{E}_n$ is a boundary integral. Opposite faces of a reciprocal fundamental cell have equal energies by periodicity and opposite normals, so their contributions cancel. Hence a full band carries zero electric current.

Label the two displacements in cell $n$ by $u_n,v_n$, with cell length $2a$. Choose the $K$ spring within a cell and the $G$ spring between cells. The equations areA Bloch wave $(u_n,v_n)=(u,v)e^{i(2aqn-\omega t)}$ gives a two-by-two eigenvalue problem, whose determinant isThus the alternating-spring chain dispersion relation isPeriodic boundaries require $e^{2iaqN}=1$, so $q=\pi j/(Na)$ modulo $\pi /a$. There are $N$ distinct wavenumbers in the Brillouin zone $-\pi /(2a)\le q<\pi /(2a)$ and two branches, accounting for $2N$ modes. The minus branch is acoustic, with zero frequency and in-phase motion at the centre; the plus branch is optical, with opposite cell displacements.

Expanding the square root near zero givesAt the zone boundary the square root is $K-G$, so the acoustic maximum is $\sqrt{2G/m}$ and the optical minimum is $\sqrt{2K/m}$. The frequency gap is thereforeQuantizing each harmonic mode gives phonons, bosonic excitations of energy $\hslash {\omega }_{\pm }(q)$ and crystal momentum $\hslash q$ defined modulo a reciprocal lattice vector. Their group velocity is $d\omega /dq$; acoustic phonons carry long-wavelength sound, while optical phonons have a nonzero centre frequency. There is one longitudinal polarization per branch here, and phonon number is not conserved in thermal equilibrium.

A periodic lattice Hamiltonian commutes with lattice translations. The Bloch theorem therefore classifies its states by characters of the translation group:where $u$ is lattice periodic. For each crystal momentum $\mathbf{k}$, solving on a primitive cell with the corresponding quasiperiodic boundary condition gives discrete eigenvalues indexed by $n$. Their continuous dependence on $\mathbf{k}$ produces the energy bands $E_n(\mathbf{k})$ as $\mathbf{k}$ ranges through the Brillouin zone. Bragg mixing opens gaps at suitable degeneracies, although bands can overlap in three dimensions and gaps are not guaranteed between every pair.

The quantum numbers are band index, wavevector modulo reciprocal lattice vectors, and spin where spin is conserved. In a finite crystal the allowed wavevectors are discrete; in the thermodynamic limit they become dense and the band description emerges. Spin-orbit coupling can replace a separately conserved spin label by an appropriate internal band label.

Dilute randomly placed impurities break exact translation symmetry, so crystal momentum is no longer an exact quantum number. Most extended states retain approximately the host band structure but acquire impurity scattering and disorder broadening. An impurity potential can also produce impurity levels in a crystal, often in a band gap: levels associated with a donor impurity lie near the conduction-band edge and can supply electrons, while those associated with an acceptor impurity lie near the valence-band edge and can create holes. The impurity's species and potential determine whether such levels exist and how deep they are; not every impurity necessarily makes a gap state. At low density these levels are nearly isolated, while overlap at increasing density can broaden them into an impurity band. Added or removed carriers shift the chemical potential as well.

For potential period $a$, a Bloch state has the form ${\psi }_k(x)=e^{ikx}{u}_k(x)$, with $u_k(x+a)=u_k(x)$, or equivalently ${\psi }_k(x+a)=e^{ika}{\psi }_k(x)$. Quasimomentum is defined modulo the reciprocal-lattice vector $2\pi /a$. A Brillouin zone is one fundamental interval for that identification, usually $[-\pi /a,\pi /a)$. Here $a=2\pi$, so it is $[-1/2,1/2)$.

For small $V_0>0$, the first free-particle degeneracy occurs at the zone edge $k=1/2$, between plane waves $e^{ix/2}$ and $e^{-ix/2}$ of energy $1/4$. In this two-dimensional subspace, the diagonal matrix elements of the potential vanish. Its Fourier coefficient at wave number one is $-V_0/2$; the $\cos 2x$ term cannot connect these two waves at first order. Thus degenerate perturbation theory gives the matrixwith energies $1/4\pm V_0/2+O(V_0^2)$. Near the edge the two uncoupled energies have opposite slopes, and diagonalization gives the avoided crossing, so the lowest band gap is

For large $V_0$, expand about each deepest minimum $x=2\pi j$:The local quantum harmonic oscillator $-d^2/d{x}^2+(5V_0/2)x^2$ has levels $(2n+1)\sqrt{5V_0/2}$. Its localization width is $O(V_0^{-1/4})$, so the quartic term changes these low levels by $O(1)$. Tunnelling between successive deep wells gives exponentially narrow energy bands. The secondary minima at $x=(2j+1)\pi$ have potential zero, and their levels lie far above the first two deep-well levels, whose energies are $-2V_0+O(\sqrt{V_0})$. Therefore the gap between the first and second bands is the first oscillator spacing to leading order:The $O(1)$ anharmonic shifts and exponentially small bandwidths do not change this leading estimate.

The Bloch theorem says that for a translation-invariant lattice Hamiltonian, eigenstates can be chosen as ${\psi }_k(x+a)=e^{ika}{\psi }_k(x)$, or ${\psi }_k(x)=e^{ikx}{u}_k(x)$ with $u_k$ periodic of period $a$. The localized site orbitals are translates, ${\varphi }_n(x)={\varphi }_0(x-na)$ and ${\chi }_n(x)={\chi }_0(x-na)$. In the tight-binding model, neglect intersite overlap and useProjecting the Schrödinger equation onto each localized orbital gives its onsite energy times its own coefficient, plus the four stated neighbor matrix elements. ThusFor the Bloch amplitudes, set $z=\cos ka$. The coefficient eigenproblem isso the two energy bands areA first Brillouin zone is $-\pi /a\le k<\pi /a$; wavevectors differing by $2\pi /a$ are equivalent. Put $\Sigma =E_0+E_1$, $D=\sqrt{(E_1-E_0{)}^2+16A^2}$. Each band is nonincreasing as a function of $z\in [-1,1]$, because $d{E}_{\pm }/dz=-2A\pm 8A^{2}z/\sqrt{(E_1-E_0{)}^2+16A^2{z}^2}\le 0$. HenceThe extrema occur at $k=0$ and at the zone edge; they may be nonunique when $E_0=E_1$, in which case the gap vanishes. In the trial wavefunction, replacing $x$ by $x+a$ and reindexing $n$ produces exactly the factor $e^{ika}$, verifying Bloch theorem directly.

An electrical conductor has available nearby unoccupied states in a partially filled band or overlapping bands. A band insulator has filled lower bands separated from empty higher bands by a large gap. A semiconductor has the same filled-band structure with a smaller gap, so thermal excitation or doping supplies mobile electrons and holes. Band filling and carrier availability, not just the existence of two bands, determine which description applies.

Bloch theorem states that energy eigenfunctions in a Bravais lattice potential may be chosen asThe wavevector is defined modulo a reciprocal lattice vector, so one can restrict it to the first Brillouin zone. This follows because lattice translations commute with the Hamiltonian and have simultaneous eigenvalues $e^{i\mathbf{k}\cdot \mathbf{l}}$.

In one dimension write $V(x)={\sum }_G{V}_G{e}^{iGx}$ with $G=2\pi j/a$ and expand a Bloch state as ${\psi }_k={\sum }_G{c}_G{e}^{i(k+G)x}$. The Schrödinger equation becomesFor zero potential, the parabolas $E_G^0(k)={\hslash }^2(k+G{)}^2/(2m)$ fold into the Brillouin zone. A weak potential shifts a nondegenerate level by $V_0$ at first order, with second-order correction ${\sum }_{G\ne 0}|V_G{|}^2/(E_0^0-E_G^0)$ where the denominators stay away from zero. This expansion fails near crossings, where degenerate perturbation theory is required.

At the first zone boundary, put $G=2\pi /a$ and $k=G/2+q$. The nearly degenerate waves with momenta $\hslash k$ and $\hslash (k-G)$ are coupled by $V_G$. Retaining their two-dimensional subspace givesDiagonalizing producesAt $q=0$ there is a gap $\boxed{\Delta E=2|V_G|}$: the two standing-wave combinations sample the periodic potential differently. Similar Bragg degeneracies occur at $k=j\pi /a$, with gaps controlled to first order by the corresponding Fourier components. If that component vanishes, a gap need not open at first order. Between the gaps the continuous energy eigenvalues form energy bands, periodically represented in the reduced-zone scheme. For the first zone edge, the curvature of the upper and lower branches gives a minimum and a maximum respectively for a sufficiently weak nonzero potential, so the avoided crossing is a genuine local forbidden energy interval.

At zero temperature electrons occupy states up to the Fermi energy. The Fermi surface is the set $E_n(\mathbf{k})=E_F$ separating occupied from unoccupied states within partially filled bands (Fermi points in one dimension). A partially filled band supplies arbitrarily nearby empty states, allowing occupations to respond to a weak electric field and conduct. Completely filled bands give no net current from such a redistribution; if the next empty band is separated by a positive band gap, the system is a band insulator. Band overlap can instead yield a metal even at an integer band filling. At finite temperature an insulator can conduct through thermally excited electrons and holes.

Use $a_n=N^{-1/2}{\sum }_k{e}^{ikn}{a}_k$ with $k=2\pi j/N$, choosing $N$ distinct values modulo $2\pi$ in the Brillouin zone. The lattice spacing is one; for spacing $a$ the dimensionless variable here would be $ka$. Orthogonality of the discrete Fourier modes diagonalizes the bilinear Hamiltonian:This is the nearest-neighbour ferromagnetic magnon dispersion. The printed expression is missing a factor of four for the stated exchange Hamiltonian. If ${\omega }_k$ denotes a physical frequency, $\hslash {\omega }_k={\epsilon }_k$; setting $\hslash =1$ does not remove that factor. The printed coefficient would require an exchange constant $J/4$ in the original Hamiltonian.

An independent check uses one spin lowering at site $n$. The exact one-magnon Hamiltonian relative to $E_0$ has diagonal element $2JS$ and off-diagonal elements $-JS$ at $n\pm 1$, giving the same Fourier eigenvalue. At small $k$, ${\epsilon }_k\simeq JS{k}^2$, and the $k=0$ state belongs to the degenerate ground multiplet. The quadratic low-energy branch is the ferromagnetic type-B Goldstone boson.

Apply a discrete Fourier transform, with lattice spacing one:The wavevectors lie in a Brillouin zone, which we choose as $[-\pi ,\pi )$. Pairing $k$ with $-k$ makes the quadratic HamiltonianA bosonic Bogoliubov diagonalization uses $a_k=\cosh {\theta }_k{\alpha }_k-\sinh {\theta }_k{\alpha }_{-k}^{†}$. The canonical commutation relations hold because ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$. The anomalous terms vanish when $\tanh 2{\theta }_k=B_k/A=\cos k$, givingNear $k=0$ and $k=\pi$, the dispersion relation is linear: ${\omega }_k\sim 2JS|k-k_0|$. These are the low-energy antiferromagnetic spin waves; the two zeroes are related by the two-sublattice description. The transformation is singular at the exact zero modes, which require an infrared regulator or separate treatment of the collective rotation.

The normalized Bloch states $|k⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{ikna}|n⟩$ satisfy periodicity when $k=2\pi j/(Na)$. Acting with the hopping terms shifts the coefficients by one site, givingThe first Brillouin zone is a fundamental cell of reciprocal space, for example $[-\pi /a,\pi /a)$; wavevectors differing by $2\pi /a$ label the same state. It contains exactly $N$ distinct allowed wavevectors and hence $N$ orbital states. The written Hamiltonian has no spin index; if electron spin degeneracy is included there are $2N$ states.

For a narrow wave packet, group velocity and band-curvature effective mass areFor the usual $J>0$, the mass is negative in the outer half of the zone, $\pi /(2a)<|k|\le \pi /a$, taking one endpoint representative. At $|k|=\pi /(2a)$ the inverse mass vanishes and the mass diverges. For arbitrary nonzero $J$, the sign criterion is $J\cos ka<0$; $J=0$ is a flat band.

For electron charge $-e$ with $e>0$ in a constant electric field $\mathcal{E}$, the single-band semiclassical equations are $\hslash \dot{k}=-e\mathcal{E}$, $\dot{x}=v(k)$. Thus $k(t)=k_0-e\mathcal{E}t/\hslash$, and for $\mathcal{E}\ne 0$,These are Bloch oscillations of period $2\pi \hslash /(|e\mathcal{E}|a)$. The zero-field limit is $x-x_0=v(k_0)t$. This assumes a narrow packet and neglects scattering and interband transitions.

For many electrons obeying the Pauli exclusion principle, a partially filled band has available nearby states and can carry current under a displaced occupation distribution: it describes a metal within band theory. A completely filled band has cancelling velocities over the zone, so its current is zero; if separated from empty higher bands by a nonzero gap, it describes an insulator at sufficiently low temperature and weak electric field. With spin degeneracy this isolated band is filled at two electrons per site. The single-electron Hamiltonian alone does not specify the filling or a gap to a next band, so those extra physical assumptions are required for classifying a material.

Label the two atoms in primitive cell $j$ by $u_{2j}$ and $u_{2j+1}$, with $j$ understood modulo $N$ because of the periodic boundary condition. Choose the spring between them to have constant $\lambda$. Newton's second law gives

For a longitudinal normal mode, putPeriodicity requires $e^{2iNka}=1$, so $k=\pi r/(Na)$. Since the primitive-cell length is $2a$, wavevectors differing by $\pi /a$ describe the same mode, and the first Brillouin zone may be chosen asThe amplitudes obeySetting its determinant to zero gives the alternating-spring chain dispersion relationThe lower sign is the acoustic phonon branch and the upper sign is the optical phonon branch.

At the edge of the Brillouin zone,so the frequency band gap isIt vanishes at $\alpha =1$, when the apparent two-atom primitive cell can be reduced to one atom.

Near the centre of the Brillouin zone, a Taylor expansion givesandThe group velocity of the acoustic branch at long wavelength is the speed of sound,

A three-dimensional Bravais lattice is the set of all integer combinations of three linearly independent primitive vectors. Its reciprocal lattice isand the first Brillouin zone is the Wigner-Seitz cell of the origin in reciprocal space.

For the stated FCC lattice, the primitive-cell volume isThe reciprocal-basis formula givesA reciprocal primitive cell, and hence the Brillouin zone, has volume

In the first Born approximation, with momentum transfer $q=k-k^{\prime }$, the scattering amplitude isFor the spherical shell, align the polar axis with $q$ and use its radial Dirac delta function:Therefore

Translation of each atomic potential multiplies its transform by a phase, so the crystal lattice structure factor givesFor an infinite lattice, the lattice Poisson summation formula turns the sum intoThus the amplitude vanishes away from reciprocal-lattice momentum transfers.

In elastic scattering, $|k|=|k^{\prime }|$ and henceThe two shortest relevant reciprocal-vector lengths of this FCC lattice areBoth are kinematically available when $k>2\pi /a$. Assigning their scattering angles ${\theta }_2$ and ${\theta }_1$, respectively, gives

Bloch theorem states that the energy eigenstates of a Hamiltonian invariant under translations by a Bravais lattice can be chosen as Bloch statesIndeed, the unitary operators $T_r$ commute with one another because $T_r{T}_s=T_{r+s}=T_s{T}_r$, and they commute with $H$ by hypothesis. The simultaneous diagonalization theorem therefore lets us diagonalize all translations within each energy eigenspace. Their eigenvalues form a unitary character $\chi$ of the additive lattice:Writing $\chi (a_j)=e^{-ik\cdot a_j}$ on a primitive basis gives $\chi (r)=e^{-ik\cdot r}$. In position space, with $(T_r\psi )(x)=\psi (x-r)$, this implies ${\psi }_k(x+r)=e^{ik\cdot r}{\psi }_k(x)$. Hence $u_k(x)=e^{-ik\cdot x}{\psi }_k(x)$ is lattice-periodic. Adding a reciprocal lattice vector to $k$ leaves the character unchanged, so the crystal momentum lies in a Brillouin zone.

Forthe equations $a_i\cdot b_j=2\pi {\delta }_{ij}$ give the reciprocal basisThe reciprocal lattice is triangular, so the first Brillouin zone is its Wigner-Seitz cell, a regular hexagon. Its six corners areReciprocal-lattice translations identify these corners in two classes of three. Representatives areFor example, $(b_1-b_2)/3=K^{\prime }-b_2$, while the other equivalences follow by symmetry and reciprocal translations. This gives the requested sketch: a regular hexagon with the vertical edge from $K^{\prime }$ to $K$ at $k_x=2\pi /(\sqrt{3}a)$ and alternating $K,K^{\prime }$ corner classes.

For the tight-binding model, introduce the normalized Bloch sumEach hop by $\pm a_j$ multiplies this state by $e^{\mp ik\cdot a_j}$. Thus the two-direction nearest-neighbour tight-binding dispersion isAlong the boundary edge from $K^{\prime }$ to $K$, $k_x=2\pi /(\sqrt{3}a)$ and $|k_y|\le 2\pi /(3a)$, soFor $\lambda >0$, it rises from $E_0+2\lambda$ at either corner to $E_0+4\lambda$ at the midpoint; for $\lambda <0$ the ordering reverses. Globally the two cosines in the first expression can simultaneously equal $1$ or $-1$, and therefore

Each orbital band contains two one-electron states per lattice site because an electron has two spin states. By band filling, a valency of one leaves this band half-filled and the material conducts, whereas a valency of two fills it. A filled band can be insulating only if it is separated from every empty band by a positive band gap.

The first band's maximum is $E_0+4|\lambda |$. If the second band's minimum is $E_0+\Delta$, a gap exists precisely when $\Delta >4|\lambda |$. Consequently, among the nontrivial fillings described here,For valency one the first band is partially filled. If $\Delta \le 4|\lambda |$, the bands overlap or touch, so even at valency two the overlapping energy bands prevent a band insulator.

For a lattice of period $a$, wavevectors differing by the reciprocal-lattice vector $2\pi /a$ describe the same translation eigenvalue. The first Brillouin zone may therefore be chosen as

Bloch theorem states that every energy eigenstate in a periodic potential may be chosen in the formTo prove it, let $T_a$ be translation by one period:Periodicity gives $[T_a,H]=0$. The translation operator is unitary, so it can be diagonalized within each energy eigenspace. Choose a common eigenstate,Unitarity gives $|\lambda |=1$, so $\lambda =e^{ika}$ for some real $k$. Thus $\psi (x+a)=e^{ika}\psi (x)$. Defining $u_k(x)=e^{-ikx}\psi (x)$ giveswhich proves the theorem.