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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.

For $V(x)=\alpha {\sum }_j\delta (x-ja)$ with $\alpha >0$, put $P=m\alpha a/{\hslash }^2$ and $x=ka$. The Floquet discriminant of the delta-comb Kronig-Penney model is $f(x)=\cos x+P\sin x/x$. Extended Bloch states require $|f(x)|\le 1$. Since $f(0)=1+P>1$, there is a positive lower band edge; a repulsive potential does not support negative-energy states in this energy convention.

The $j$th positive-energy band ends at $x=j\pi$. Its lower edge lies in $((j-1)\pi ,j\pi )$ and solves $f(x)=(-1{)}^{j-1}$. A finite periodic ring selects only the discrete allowed values of crystal momentum within these bands. Nonzero band gaps do not by themselves imply an insulator: the band filling must also place the Fermi energy in a gap.

The Quantum Hamiltonian acting on the cell-periodic part of a Bloch state at fixed wave vector.

Crystal momentum is the translation quantum number $\hslash k$ of a Bloch state. Since $k$ and $k+G$ give the same translation character whenever $G$ belongs to the reciprocal lattice, crystal momentum is defined modulo reciprocal-lattice vectors.

An electron hole is an unoccupied electron quantum state in an otherwise nearly full energy band, often the valence band. “Filled” refers to occupation of the allowed quantum states, not to filling all space with a continuous fluid of electrons. Each state, including its spin label, can hold at most one electron by the Pauli exclusion principle. The almost full energy band and its few missing occupants are therefore simpler to describe as a background plus a few electron holes. The electron hole is a quasiparticle, not a Positron inserted into the crystal.

Removing an electron of electric charge $-e$ changes the background's electric charge by $+e$. Thus an electron hole carries positive electric charge relative to that background. Promoting an electron into the conduction band creates an electron hole without changing the whole crystal's electric charge: $(-e)+(+e)=0$. An unbound pair still consists of two separate charge carriers; only a bound state of their relative motion is an exciton.

The local cartoon explains why a vacancy can move: when a neighboring electron fills it, another vacancy appears where that electron was. The vacancy has moved opposite to that particular transfer of an electron. In an extended Bloch state, use crystal momentum instead of imagining a permanently empty atomic site. For a completely filled energy band the summed electric current vanishes. Omitting one electron with group velocity ${\mathbf{v}}_v$ removes its contribution $-e{\mathbf{v}}_v$, leaving a contribution $+e{\mathbf{v}}_v$. This fixes both the positive electric charge and the velocity convention of the electron hole.

More precisely, near a parabolic valence band maximum, set its energy $E_v(0)$ as reference and define the electron hole wave vector by ${\mathbf{k}}_h=-{\mathbf{k}}_v$. ThenThe occupied electron branch has negative curvature there, while the electron hole branch has positive effective mass $m_h^{\ast }$. The velocity equality compares the electron hole with the missing Bloch state's group velocity; it does not contradict the opposite vacancy motion in a local transfer cartoon. The sign conventions are listed in Phys 460 Lecture 17.

F. Duncan M. Haldane's 1988 tight-binding model adds complex second-neighbour hopping and opposite sublattice offsets to a honeycomb lattice. Directed phases break time-reversal symmetry while each primitive cell has zero net magnetic flux. Smaller loops can enclose magnetic flux; ordinary Bloch states remain available.^[ref]

Extend the nearest-neighbour model of graphene, ignoring physical spin and interactions. With bond length $a$, $t_1>0$, $t_2\ge 0$ and ordered sublattice basis $(A,B)$, chooseThe following Bloch Hamiltonian defines $\varphi$; reversing directed second-neighbour hopping sends $\varphi \mapsto -\varphi$:The Pauli matrices act on sublattice amplitudes, as in sublattice pseudospin in graphene. The identity matrix term $d_{0}I$ shifts both energy eigenvalues equally at each momentum; it changes neither the eigenvectors nor the direct separation $E_{+}-E_{-}$. A genuine band insulator additionally needs ${\min }_{\mathbf{k}}{E}_{+}>{\max }_{\mathbf{k}}{E}_{-}$. We use $t_2/t_1=0.08$ in the figures, within the original paper's sufficient nonoverlap restriction $|t_2/t_1|<1/3$.^[ref]

Take a planar honeycomb lattice with two sites $A,B$ per primitive cell and one localized atomic orbital per site. In this explicitly idealized tight-binding model, all three nearest-neighbour hopping matrix elements are $-t$, with $t>0$; there is no on-site offset or further-neighbour hopping. Ignore electron interactions and treat physical spin as a separate degeneracy. Let the bond length be $b$, and choose the three vectors from an $A$ site to its neighbours asThe Bloch state amplitudes on $A,B$ form a two-component vector. A consistent phase convention gives the HamiltonianIts characteristic equation is $E^2=t^2|f{|}^2$, giving the two displayed energy bands. The two-component structure comes from the two sites, rather than two values of physical spin. This is a particularly small matrix model in which a crystal's geometry creates an unexpected low-energy law.

Historically, Philip Russell Wallace used a tight-binding model to investigate the electronic band structure of graphite in 1947. The material in that paper was layered graphite; the single-sheet model here isolates one simpler part of the problem.^[ref]

For one localized orbital per site in a chain, take on-site matrix elements $H_{jj}=E_0$, nearest-neighbour matrix elements $H_{j,j\pm 1}=h$, overlaps $S_{jj}=1$ and $S_{j,j\pm 1}=s$, with other terms neglected and $h,s$ real. A Bloch state reduces the generalized eigenvalue problem $Hc=ESc$ toThe overlap metric must be positive for the retained model; $|s|<1/2$ is sufficient for all $k$. Dropping it yields the cosine band of the orthogonal tight-binding approximation.

An optical lattice is a periodic optical trap produced by interfering light fields. Its periodic potential gives atomic Bloch states and energy bands. For the displayed one-dimensional lattice, the recoil energy is $E_R={\hslash }^2{\pi }^2/(2m{d}^2)$ and $s$ measures the depth in recoil units.

electron confinement raises the conduction edge and hole confinement lowers the corresponding valence edge, so their positive confinement energies add. With $L=120\mathring{\mathrm{A}}=12$ nm, the common bare-mass scale isWriting ${\eta }_e=0.067$ and ${\eta }_j=m_{h,j}^{\ast }/m_e$, the quantum-well interband thresholds areUse $E_{g,\mathrm{hh}}=E_{g,\mathrm{lh}}=1.52$ eV and $E_{g,\mathrm{so}}=1.86$ eV. The thresholds in the requested window, using the printed masses without fitting, areThe next heavy-hole and split-off thresholds are about $2.2224$ and $2.3674$ eV, outside the interval. Each threshold marks a continuum onset or shoulder, not the stronger bound-exciton peak below it. The six heavy- and light-hole onsets follow the six paired peak groups in the supplied diagram. Their first members lie roughly 20 and 12 meV below their respective onsets, as the common solution above explains.

A qualification is necessary for the split-off branch. Its p-like $j=1/2$ bloch states have nonzero $p_x$ matrix elements to the s-like conduction states, so in-plane polarization does not generally forbid them. For example, a $j=1/2$ state has a nonzero $p_x\pm i{p}_y$ orbital component with the appropriate spin, which couples to an s-like state through $p_x$. The same-envelope rule therefore allows the two split-off thresholds in the table. The first nearly coincides with the third heavy-hole edge, while the second would give an additional onset near 2.086 eV. The supplied figure and twelve peak energies appear to retain only the heavy- and light-hole series; they do not specify a split-off-series oscillator strength or identify its extra features. Omitting the split-off transitions would require an additional approximation beyond the four-band information and polarization given. Their expected ideal two-dimensional first-state peaks would be near 1.9010 and 2.0701 eV; these are predictions of the stated model, not additional listed data.

Start from electrons moving in the periodic equilibrium ionic potential. Displace ion $l$ by ${\mathbf{u}}_l$ and expand its contribution as $V(\mathbf{r}-{\mathbf{R}}_l-{\mathbf{u}}_l)\simeq V(\mathbf{r}-{\mathbf{R}}_l)-{\mathbf{u}}_l\cdot \nabla V(\mathbf{r}-{\mathbf{R}}_l)$. Quantizing the harmonic normal modes expresses ${\mathbf{u}}_l$ as a sum of creation and annihilation operators with displacement scale $\sqrt{\hslash /(2{\rho }_{\mathrm{ion}}V{\omega }_{\mathbf{q}})}$ for a single simple branch. Expanding the electronic field in bloch states then gives a band-projected Fröhlich HamiltonianHere $c,c^{†}$ obey canonical anticommutation relations and remove/create an electron with crystal momentum $\mathbf{k}$ and spin $\sigma$; $b,b^{†}$ are boson operators for lattice modes. The matrix element $M_{\mathbf{q}}$ contains the derivative of the ionic potential, phonon polarization, displacement normalization and electronic wavefunction overlaps. With phases chosen appropriately, $M_{-\mathbf{q}}=M_{\mathbf{q}}^{\ast }$ ensures a Hermitian Hamiltonian. The displayed one-band, one-branch notation suppresses additional branch and band labels. “Fröhlich Hamiltonian” here means the linear electron-phonon model used in the question; it is not a requirement to use specifically the polar-optical-phonon coupling of a polar semiconductor. A harmonic lattice, small displacements and a weak enough coupling to organize perturbation theory are the underlying approximations.

The Fröhlich canonical transformation removes real linear phonon operators from the electron-only effective problem to first order. Choose an anti-Hermitian $S$ with $[S,H_0]=-V_{e\text{-}ph}$. For $A_{\mathbf{kq}\sigma }=c_{\mathbf{k}+\mathbf{q},\sigma }^{†}{c}_{\mathbf{k}\sigma }$ and ${\Delta }_{\mathbf{kq}}={ϵ}_{\mathbf{k}+\mathbf{q}}-{ϵ}_{\mathbf{k}}$, one possible generator isIts Hermitian conjugate is its negative after relabelling momenta. The Baker--Campbell--Hausdorff formula givesThe commutator describes virtual emission and absorption of a phonon. normal ordering yields an electron-electron term as well as one-electron energy renormalizations and phonon terms. Retain the former, absorb single-particle shifts into ${ϵ}_{\mathbf{k}}$ and eliminate the explicit phonon sector. For elastic pair scattering the two virtual time orderings have the combined amplitudeUsing the usual factor one-half to count unordered pairs gives the coefficient in the printed Hamiltonian, with $\Delta ={ϵ}_{\mathbf{k}}-{ϵ}_{\mathbf{k}-\mathbf{q}}$. The two outgoing momenta are $\mathbf{k}-\mathbf{q}$ and ${\mathbf{k}}^{\prime }+\mathbf{q}$, so total momentum and the spins are conserved. A fully off-shell effective operator can instead use symmetrized denominators for both external legs; the displayed expression is the usual elastic-scattering reduction. Eliminating nearly resonant real phonons requires more care than this second-order construction.

For $|\Delta |<{\Omega }_{\mathbf{q}}$ the denominator is negative: virtual phonon exchange is attractive in this low-energy channel. Physically, an electron changes the ionic density and a second electron can gain energy by responding to the delayed distortion. This is a retarded interaction, not an unscreened instantaneous attraction at every energy transfer. At large transfer the sign changes, and screened Coulomb repulsion must also be included when deciding the net low-energy pairing strength. The connection between lattice dynamics and superconductivity is supported by the isotope effect, but the mathematical attraction alone does not determine a measured transition temperature.^[ref]

For the pair calculation, make the model restrictions explicit. The filled Fermi sea is inert, and $|0⟩$ denotes that sea with no added particles, not an empty electronic vacuum. Project the interaction onto the previously empty shell $0\le {\xi }_{\mathbf{k}}\le D$, where ${\xi }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)-E_F$ and $D=\hslash {\omega }_D$. Measure the added-pair energy relative to $2E_F$; this is the negative $E$ in the final formula. If absolute one-electron energies are retained in the preceding eigenvalue equation, its eigenvalue must instead be $E_{\mathrm{abs}}=2E_F+E$. A shell-restricted state would not in general be an exact eigenstate of the completely unprojected, all-momentum contact Hamiltonian.

Write $g={\gamma }^2>0$ and $|\mathbf{k}⟩=c_{-\mathbf{k}\downarrow }^{†}{c}_{\mathbf{k}\uparrow }^{†}|0⟩$. Within the zero-total-momentum pair subspace the two opposite-spin terms cancel the interaction's factor one-half, while same-spin s-wave terms cancel by fermionic antisymmetry. Thus the projected Hamiltonian acts asFor $|\Psi ⟩={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}|\mathbf{k}⟩$ and $C={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}$, matching each basis coefficient givesThis is the stated coefficient formula in the consistent Fermi-level energy convention. Its even dependence on $\mathbf{k}$ produces an even orbital state with a spin singlet. The constant $C$ is fixed up to an overall phase by normalization; for a bound state, $|C{|}^{-2}=g^2{\sum }_{⟨\mathbf{k}⟩}(E-2{\xi }_{\mathbf{k}}{)}^{-2}$. It cannot be set to zero for the nontrivial bound solution.

Put $E=-B$ with $B>0$ and impose the sum defining $C$:Here $\rho (E_F)$ counts orbital states for one spin species, as appropriate to the unrestricted $\mathbf{k}$ sum in this pair basis, and is approximately constant over the thin shell. If a two-spin total density of states is used, it must be divided by two in this integral. State-counting and the volume convention in $g$ and $\rho$ must be consistent. With $\xi =g\rho (E_F)$ as the paper's dimensionless coupling, inversion yieldsThe first approximation is the constant-density-of-states, restricted-shell model; the final exponential additionally assumes weak attraction. The binding energy is nonanalytic at zero coupling, so a finite-order power expansion in $g$ cannot produce it.

Cooper's problem asks whether two electrons added above an inert Fermi sea bind despite Pauli exclusion of lower states. An arbitrarily weak net attraction binds a zero-momentum singlet below the free-pair threshold, because the near-Fermi-surface integral diverges logarithmically as $B\downarrow 0$. Its wavefunction draws on many time-reversed momentum states and becomes dominated by a narrow region near the Fermi surface at weak coupling; the corresponding pair is spatially extended. This establishes a Cooper instability of the unpaired reference state.^[ref] The full superconducting ground state needs a coherent many-electron treatment in which overlapping pairs and the reorganized sea are included. BCS theory supplies that step; the two-electron binding scale alone is neither a calculation of the many-body gap nor a derivation of zero resistance.^[ref]

Expand a single-electron wavefunction as $|\psi ⟩={\sum }_n{\psi }_n|n⟩$. Project the Schrödinger equation onto $⟨n|$ to obtainTranslation symmetry suggests the Bloch state ${\psi }_n=N^{-1/2}{e}^{ikna}$ on a periodic chain of $N$ sites. Substitution givesTake the conventional positive hopping $t>0$. The first Brillouin zone is $-\pi /a\le k<\pi /a$; the band has minimum ${ϵ}_0-2t$ at $k=0$ and maximum ${ϵ}_0+2t$ at the zone edges. Its width is $4t$. A negative hopping interchanges the extrema but does not change the density of states after replacing $t$ by $|t|$.

There are $N$ allowed wavevectors and two spin states per wavevector. For the total spin-summed density of states $D(E)$, each interior band energy has two roots, $\pm k$. Using the spacing $2\pi /(Na)$ givesThus the density of states per atom isIt is symmetric about ${ϵ}_0$ and has integrable inverse-square-root Van Hove singularities at both band edges. Its integral over the band is two states per atom. The plot labels the dispersion extrema and both singular edges.

Choose orthonormal localized orbitals $|\mathbf{R}⟩$. The on-site energy is $E_0=⟨\mathbf{R}|H|\mathbf{R}⟩$, and the sign convention in the question defines $t(\mathbf{T})=-⟨\mathbf{R}|H|\mathbf{R}+\mathbf{T}⟩$. Bloch state diagonalizes this translationally invariant hopping problem. For real, inversion-symmetric nearest-neighbor amplitudes, $t(\pm a\hat{x})=t_x$ and $t(\pm b\hat{y})=t_y$, so the exponential pairs yieldThe first zone is the rectangle $|k_x|\le \pi /a$, $|k_y|\le \pi /b$, wider in $k_x$ because $a<b$. Use $\Gamma =(0,0)$, $X=(\pi /a,0)$, $Y=(0,\pi /b)$ and $M=(\pi /a,\pi /b)$.

For positive $t_y=t$ and $t_x=2t$, the four corner energies and the requested path areAlong a path parametrized in reduced coordinates $x=k_{x}a$, $y=k_{y}b$, the full curve is simply $E-E_0=-4t\cos x-2t\cos y$. In particular the two appearances of $\Gamma$ and the two diagonal traversals must have consistent endpoint energies.^[ref] A reversed hopping sign changes which point is the band minimum; the usual positive-hopping convention is stated explicitly here.

The nearly-free electron model expands the Bloch state in plane waves,All basis functions have the same translation eigenvalues because $e^{i\mathbf{G}\cdot \mathbf{R}}=1$. A weak periodic potential mixes them most strongly near Bragg degeneracies and opens band gaps. Delocalized $s,p$ conduction states of simple metals, such as sodium and aluminium, are typical applications; the approximation need not describe tightly bound core states.

The single-particle Hamiltonian $H_0=-{\hslash }^2{\partial }_x^2/(2m)+V(x)$ has a periodic potential $V(x+d)=V(x)$. Define the unitary spatial translation operator $T_{d}f(x)=f(x+d)$. Its derivatives commute with translation, and the potential's periodicity gives $H_0{T}_d=T_d{H}_0$.

Each energy eigenspace is therefore preserved by $T_d$, so within it one may choose simultaneous energy and translation eigenstates. On a large periodic interval of $M$ cells, $T_d^M=1$ and its eigenvalues are phases $\lambda =e^{ikd}$, with $k=2\pi j/(Md)$; the infinite-volume limit gives continuous quasimomentum modulo $2\pi /d$. For a simultaneous state in band $j$, this provesThen $u_{jk}(x)=e^{-ikx}{\psi }_{jk}(x)$ obeys $u_{jk}(x+d)=u_{jk}(x)$, which constructs the usual Bloch state rather than assuming its form. The label $j$ distinguishes energy bands at the same crystal momentum $\hslash k$. In a degenerate energy eigenspace an arbitrary superposition need not itself have one translation phase; the result refers to the simultaneous Bloch states.

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 low-energy localized-well basis gives the tight-binding modelA Bloch state with coefficients $e^{ipna}$ is an eigenstate because the two neighbouring coefficients add to $2\cos (pa)e^{ipna}$. ThusHere $p$ is a Bloch wavenumber; physical momentum is $\hslash p$, and writing $p$ for physical momentum would require $pa/\hslash$. Spectral evolution of these eigenstates is exactly the Fourier integral in the preceding solution. The negative hopping lowers the symmetric $p=0$ state, and coherent quantum tunnelling spreads a formerly degenerate family of localized levels into a band of width $4\Delta ϵ$. Using the printed positive Euclidean prefactor instead would give on-site energy $-\hslash \omega /2$, showing its contradiction directly.

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.

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.

In a periodic crystal potential, atomic levels broaden into bands of allowed bloch states. Coupling between states of the same symmetry opens avoided crossings, including Bragg-coupled states in the nearly free-electron picture. Bonding and antibonding combinations provide a complementary tight-binding picture. A semiconductor has a filled valence band and an empty conduction band at zero temperature separated by a forbidden energy interval; thermal excitation, light and doping provide mobile electrons and holes. The gap depends on the crystal's periodic potential, bonding and symmetry, not on a classical collision barrier. In a direct gap, the relevant extrema coincide in wavevector.

For the stated parabolic edges, take $E_v(k)=-{\hslash }^2{k}^2/(2m_h^{\ast })$ and $E_c(k)=E_g+{\hslash }^2{k}^2/(2m_e^{\ast })$. Negligible photon momentum and no phonon make the transition nearly vertical in $E$ versus $k$. parabolic-band optical energy sharing divides the excess photon energy $0.32\mathrm{eV}$ in inverse proportion to the effective masses:Thus the initial and final electron energies, relative to the valence-band maximum, areTheir difference is $0.5000\mathrm{eV}$. The selected $|k|$ is about $9.33\times 10^8{\mathrm{m}}^{-1}$; either sign gives the same energy in this symmetric approximation.

A shallow donor supplies an electron in a screened impurity potential, so its localized level lies just below the conduction edge. A shallow acceptor binds a hole and has an electron-occupation level just above the valence edge; thermal ionization supplies mobile holes. The level positions are measured relative to the relevant band edge, and deep impurities need not obey this hydrogenic approximation.

For the supplied bound-carrier mass and unit effective ionic charge, dielectric screening and effective mass change the hydrogen binding energy toThis corresponds to $E_b/k_B\simeq 110\mathrm{K}$ and an effective Bohr radius about $6.35\mathrm{nm}$. The binding is shallow compared with the band gap and typical atomic binding. Trivalent denotes chemical valence, not an unscreened hydrogenic charge of three. An acceptor physically binds a hole; the question's stated $0.1m_e$ is used as its stipulated bound-carrier mass rather than silently replacing it by the separate band-edge $0.4m_e$.