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For equivalent minima with nearest-neighbour hopping magnitude $\Delta$, paths with $r$ right steps and $s$ left steps have $r-s=n-m$. Summing $(\Delta \tau /\hslash {)}^{r+s}/(r!s!)$ and applying the Fourier representation of a Kronecker delta gives the displayed modified Bessel function kernel in a normalized localized-site basis. Its Fourier exponent gives the tight-binding model energy $E(\theta )=E_{\mathrm{well}}-2\Delta \cos \theta$. Position-endpoint kernels can have an additional common local-wavefunction prefactor.

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.

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 minima occur at $x=(2j+1)\pi$, where $V=-2V_0$. Writing $x=(2j+1)\pi +\xi$ givesThe harmonic approximation near a potential minimum therefore has frequencybecause $\frac{1}{2}m{\omega }_0^2{\xi }^2=\frac{5}{2}{V}_0{\xi }^2$. The two lowest localized energies are

In the tight-binding model, tunnelling between neighbouring wells broadens each localized level into a band,For large $V_0$, the hopping amplitudes $J_n$ are exponentially small compared with the oscillator spacing. Hence the gap between the first two bands isup to exponentially small tight-binding bandwidth corrections.