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An allowed energy band is an interval of energies supporting extended Bloch states. Adjacent bands may be separated by a band gap.

A band insulator has an integer number of completely filled energy bands, with the Fermi energy inside a positive band gap. At zero temperature, no arbitrarily low-energy charged excitation is then available.

A Bragg point is a crystal momentum at which two plane waves differing by a reciprocal lattice vector are degenerate. A periodic potential mixes the waves there, producing an avoided crossing and opening a band gap.

Mercury cadmium telluride is a semiconductor whose composition tunes its band gap and hence its long-wavelength absorption cutoff. It is widely used as the absorbing material in infrared detector arrays.

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,

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.

The potential has period $2\pi$, so the first Brillouin zone is $-1/2\le k<1/2$. At its boundary the free-particle statesare degenerate, with energyThe relevant Fourier coefficient is supplied by $V_0\cos x$:The term $-V_0\cos 2x$ changes wavevector by two and has no matrix element within this degenerate pair. Thus degenerate perturbation theory gives the matrixand the two band-edge energiesThe nearly-free electron model therefore predicts the lowest band gap

A semiconductor has an electronic band gap and a conductivity strongly affected by temperature, impurities and carrier injection. Optical absorption can create mobile electrons and holes, enabling a photodiode.