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Use the state vector $(\psi ,{\psi }^{\prime }{)}^T$ immediately to the right of a delta well. Free propagation to the next well and the derivative jump there areThus the Floquet matrix in this convention isChoosing the state on the other side of each well conjugates the matrix and leaves its eigenvalues unchanged.

The Bloch theorem requires a real quasimomentum $q$ with a Floquet eigenvalue $e^{iqa}$. Since $\det M=1$, this occurs exactly whenDividing the two inequalities by the positive $\sinh \kappa a$ givesThe upper boundary in $y$ is the unique intersection of the increasing line with the strictly decreasing $\coth y$, which runs from infinity to $1$. Call it $y_{-}$. The concave function $\tanh y$ starts with slope $1$ and approaches $1$. If $\lambda a\le 2$, the line has slope at least $1$ and stays above it for every $y>0$, so allowed $y$ run from $0$ to $y_{-}$. If $\lambda a>2$, there is exactly one positive crossing $y_{+}$ with $\tanh y$; below it the line is inadmissible, and above it the line stays above $\tanh y$. Because $\coth y>\tanh y$, $y_{+}<y_{-}$. This establishes the negative-energy band of an attractive delta comb: there is exactly one negative-energy band:where $E_{\pm }=-{\hslash }^2(2y_{\pm }/a{)}^2/(2m)$, without needing to evaluate the roots.

As $a\to \infty$ with $\lambda$ fixed, both positive boundary values of $\kappa$ tend to $\lambda$. The band collapses to the isolated delta-well bound state $\boxed{E=-{\hslash }^2{\lambda }^2/(2m)}$, as tunnelling between widely separated wells disappears.

Choose any basis of two independent local solutions and collect their coefficients into a column $c_n$ immediately to the right of the $n$th cell boundary. The Floquet matrix is the one-period matrix $F(E)$ defined byEquivalently, it advances the Cauchy data $(\psi ,{\psi }^{\prime }{)}^T$ by one period. The Wronskian of two solutions of the stationary Schrodinger equation is constant between the delta functions, and the derivative jump at each delta function also has determinant one. ConsequentlyFor a real potential the Floquet matrix in the real Cauchy-data basis is real, so its trace is real in every basis. Its Floquet multipliers ${\rho }_{\pm }$ satisfyBy Bloch theorem, an energy is in the interior of an allowed energy band when ${\rho }_{\pm }=e^{\pm iqa}$ are distinct unit-modulus numbers. Since $\mathrm{tr}F$ is real, this is equivalent to

For the attractive delta-comb Kronig-Penney model, integrating the Schrodinger equation across a lattice point givesAt negative energy write $E=-{\hslash }^2{\mu }^2/(2m)$ and use the local basis $e^{\pm \mu (x-na)}$. Free propagation through one cell followed by the derivative jump givesThusThe band condition is thereforeUsing $(\cosh u-1)/\sinh u=\tanh (u/2)$ and $(\cosh u+1)/\sinh u=\coth (u/2)$ gives the requested inequality

At positive energy write $E={\hslash }^2{k}^2/(2m)$ and use $e^{\pm ik(x-na)}$. The corresponding matrix issoHence the positive-energy bands are exactly the values of $k>0$ for whichFinally, the zero-energy limit has $\frac{1}{2}\mathrm{tr}F=1-\lambda a$. If $\lambda a>2$, then $|1-\lambda a|>1$, and therefore zero energy lies in a forbidden gap.

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.

A Bravais lattice is the integer span of three linearly independent primitive vectors. Its reciprocal lattice consists of vectors $G$ satisfying $G\cdot r\in 2\pi \mathbb{Z}$ for every lattice vector $r$.

Translations by lattice vectors commute with the Hamiltonian. Their simultaneous eigenvalues have the form $e^{ik\cdot r}$, so an energy eigenfunction may be chosen withThus ${\psi }_k=e^{ik\cdot x}{u}_k$ with lattice-periodic $u_k$, which is Bloch theorem. Wavevectors differing by a reciprocal-lattice vector describe the same translation character, and the Brillouin zone is a fundamental Wigner-Seitz cell of the reciprocal lattice.

Every primitive basis of the Bravais lattice has the formwhere $p,q,r,s\in \mathbb{Z}$ andThe unimodular determinant condition is exactly what makes the integer change of basis invertible over $\mathbb{Z}$.

Let $b_i$ be the primitive vectors of the reciprocal lattice, withSolving these four equations givesThe six reciprocal-lattice points nearest the origin areall at distance $4\pi /\sqrt{3}$.

The Wigner-Seitz cell is bounded by the perpendicular bisectors of the segments joining the origin to those six points. It is the regular hexagon with verticesIts area equals the reciprocal primitive-cell area:This is the reciprocal lattice and Wigner-Seitz cell of the unit triangular lattice.

For a periodic potential, Bloch theorem identifies wavevectors differing by a reciprocal-lattice vector. The first Brillouin zone is the Wigner--Seitz cell just found. More generally, the $n$th Brillouin zone consists of wavevectors reached from the origin after crossing exactly $n-1$ reciprocal-lattice Bragg planes; its boundaries are the perpendicular bisectorsFor this triangular reciprocal lattice, the first zone is the central regular hexagon. The second is the sixfold-symmetric collection of regions immediately outside its six sides, bounded next by the bisectors associated with the next reciprocal points. These zones organize free-particle states into bands: Bragg coupling is strongest at their boundaries and opens energy gaps when the periodic potential is introduced.

For a potential of period $a$, define the Floquet matrix for a one-dimensional periodic potential byThe Schrodinger equation has no first-derivative term, so the Wronskian is constant and $\det M=1$. Its multipliers are therefore reciprocal. They lie on the unit circle precisely whenThose energies form continuous allowed bands of bounded Bloch theorem solutions. Values with $|\mathrm{tr}M|>2$ have a growing and a decaying multiplier and form forbidden gaps. The band edges have $\mathrm{tr}M=\pm 2$, as in Floquet discriminant and energy bands.

For the delta-comb Kronig-Penney model, start immediately to the right of one delta function. Free propagation to the next delta and the derivative jump there are represented byThusConsequently the Floquet discriminant of the delta-comb Kronig-Penney model isAll band edges are therefore determined byIn factorized form, the periodic edges $\sigma =1$ obeywhile the antiperiodic edges $\sigma =-1$ obey

To express the same edges through the single-barrier scattering data, the One-dimensional transfer matrix from scattering amplitudes givesAt an edge put $z=e^{-ika}$ and $\mathrm{tr}M=2\sigma$. Thenso the scattering-amplitude equations for one-dimensional band edges areExplicitly,For the delta barrier, $t-r=1$ andwhich reproduces the four factorized edge equations above.

Using one even and one odd site per unit cell, Bloch theorem reduces the Hamiltonian toIts eigenvalues are

In position space,Henceup to the common normalization factor $1/\sqrt{3}$.

Under $x\mapsto x+a$, reindexing $m=n-1$ gives the same three terms modulo three sites. The term shifted by three sites is unchanged because ${\psi }_{m+3}={\psi }_m$ and $e^{3ika}=1$. Therefore

The one-dimensional Bloch theorem states that an eigenfunction in a potential of period $a$ can be chosen in the formIt applies because the three-atom ring is invariant under translation by one lattice spacing and obeys periodic boundary conditions.

Bloch theorem follows because lattice translations commute with $H$ and with each other: simultaneous eigenstates obey $T_R\psi =e^{ik\cdot R}\psi$, so ${\psi }_k(r)=e^{ik\cdot r}{u}_k(r)$ with $u_k$ lattice-periodic. Conversely this form has the required translation character.

Bloch theorem $|k⟩\propto {\sum }_R{e}^{ik\cdot R}|R⟩$ are eigenstates with $E(k)=E_0-2\mu {\sum }_{i=1}^3\cos (k\cdot a_i)$.