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Let adjacent equal masses $m$ be separated by $a$, with alternating strong and weak spring constants $K$ and $\kappa$, $K\gg \kappa >0$. The pattern repeats after $2a$, even though the masses are identical. Thus the first Brillouin zone isChoose the strong bond inside each two-atom primitive unit cell. With displacement amplitudes $(u,v)e^{i(2naq-\omega t)}$, the normal mode equations areThe alternating-spring chain dispersion relation is consequentlyThe lower acoustic phonon branch has in-phase motion within the strongly bound pair and ${\omega }_{-}(0)=0$. Its small-$q$ slope is the speed of sound,The upper optical phonon branch has opposite motion within each pair, with ${\omega }_{+}(0)=\sqrt{2(K+\kappa )/m}$. Here “optical” names the gapped branch; equal neutral masses need not couple to light through an electric dipole.

In the strong-bond limit of the alternating-spring chain, the pairs act almost as rigid dimers in the low-frequency branch, while their internal stretch gives a high, nearly flat branch:At the zone edge the two frequencies are $\sqrt{2\kappa /m}$ and $\sqrt{2K/m}$, separated by a large forbidden frequency interval. Neighbouring cells are out of phase there. Both branches have horizontal tangents at the edge.

A site in a primitive cubic lattice has six nearest neighbours, at $\pm a\hat{\mathbf{x}}$, $\pm a\hat{\mathbf{y}}$ and $\pm a\hat{\mathbf{z}}$. Repeating the Bloch state calculation with equal hopping integrals $t$ along all six bonds givesThe cosine sum ranges from $-3$ to $3$ within the first Brillouin zone. ThereforeFor $t<0$ the minimum is at the zone centre; for $t>0$ the maximum is there.

At nonzero band hybridization, the crossing becomes an avoided crossing. The separation at $k_c$ is $2\alpha$. If $\alpha \ll E_{\mathrm{fl}}$, appreciable mixing occurs only where $|A{k}^2-E_{\mathrm{fl}}|\lesssim \alpha$; away from that region each branch retains nearly its original orbital character. For $\alpha \sim E_{\mathrm{fl}}$, the avoided crossing is broad and both branches are substantially reshaped, as shown in the figure.

Use $E_{-}$ for the lower branch and $E_{+}$ for the upper one. For $\alpha >0$,The lower branch increases towards $E_{\mathrm{fl}}$ from below. Thus an indirect gap exists in the exactly flat-band model. In the unbounded-parabola limit its lower bound isA finite Brillouin zone gives a finite maximum of $E_{-}$ below $E_{\mathrm{fl}}$ and hence a slightly larger indirect gap. This hybridization gap differs from the local direct gap $2\alpha$.

The material can therefore be a band insulator when the lower band is full and the upper band empty. With two spin-degenerate orbitals per unit cell, the required filling isWith spin degeneracy removed, replace two by one for each independent spin sector. Other fillings are metallic in this noninteracting model.

Use dimensionless spin operators and nearest-neighbour exchange $J_H>0$ on a regular bipartite graph with N sites and coordination number of a lattice z. Count each bond once and define ${\gamma }_{\mathbf{k}}=z^{-1}{\sum }_{\delta }{e}^{i\mathbf{k}\cdot \delta }$, real for the inversion-symmetric examples considered here.

For the ferromagnetic Heisenberg model, $H_F=-J_H{\sum }_{⟨ij⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$. Expanding the Holstein–Primakoff transformation about all spins along +z givesup to interactions of higher order in boson density, with $E_{\mathrm{cl}}=-Nz{J}_H{S}^2/2$. The isotropic spin-wave vacuum energy comparison starts with this normal ordering. The linear spin-wave approximation then givesEach ferromagnetic magnon lowers total $S^z$ by one. The fully aligned state is an exact eigenstate and ground state; there is no harmonic vacuum depletion or added zero-point energy in this normal-ordered Hamiltonian. The global spin-rotation multiplet remains degenerate. On a hypercubic lattice of spacing a, $z=2d$ and ${\gamma }_{\mathbf{k}}=d^{-1}{\sum }_j\cos k_{j}a$, so $E_F\simeq J_{H}S{a}^2{k}^2$. The exact one-magnon sector agrees with this result even at finite S; interactions enter for multiple magnons.

For the Heisenberg antiferromagnet, take $H_A=+J_H{\sum }_{⟨ij⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$ and expand around a Néel state. On A, $S^z=S-a^{†}a$ and $S^{+}\simeq \sqrt{2S}a$; on B, $S^z=-S+b^{†}b$ and $S^{+}\simeq \sqrt{2S}{b}^{†}$. A bond becomesWith N/2 momenta in the magnetic Brillouin zone,The bosonic Bogoliubov transformation has $u_k^2-v_k^2=1$ and $\tanh 2{\theta }_k={\gamma }_k$. For example $a_k=u_k{\alpha }_k-v_k{\beta }_{-k}^{†}$ and $b_{-k}=u_k{\beta }_{-k}-v_k{\alpha }_k^{†}$. It gives two branches in this reduced zone,The reduced-zone counting of bipartite spin waves gives N oscillators in total, not 2N. The negative vacuum correction and quantum depletion of Néel order, $\delta S=(2/N){\sum }_{\mathrm{MBZ}}{v}_k^2$, distinguish the antiferromagnet from the ferromagnet. At nonzero temperature replace $v_k^2$ by $v_k^2+(u_k^2+v_k^2)n_B(E_k)$. Hypercubic small-momentum modes have $E_A\simeq 2J_{H}Sa\sqrt{d}|k|$, hence linear physical frequency $E_A/\hslash$.

Self-consistency requires depletion much smaller than S and respects the Holstein–Primakoff occupation constraint. In one dimension the antiferromagnetic vacuum depletion diverges logarithmically, invalidating ordered spin-wave theory; it does not alone prove an exact gap or its absence. At positive temperature, isotropic short-range models in d≤2 cannot sustain such magnetic order by the Mermin-Wagner theorem. In three dimensions the quadratic ferromagnetic modes give the low-temperature magnetization reduction proportional to $T^{3/2}$, whereas the antiferromagnetic thermal reduction scales as $T^2$. Anisotropy, fields, frustration and magnon interactions alter these conclusions. ^[ref]

A periodic harmonic chain with $N$ masses $m_0$, spacing $a$ and nearest-neighbour spring constant $K$ hasHamilton's equations give $m_0{\ddot{u}}_j=K(u_{j+1}+u_{j-1}-2u_j)$. The discrete Fourier transform $u_j=N^{-1/2}{\sum }_k{Q}_k{e}^{ikja}$, $p_j=N^{-1/2}{\sum }_k{P}_k{e}^{ikja}$, with $k=2\pi l/(Na)$ in one Brillouin zone, diagonalizes the quadratic form:The harmonic-chain normal-mode reality condition means $Q_k^{†}=Q_{-k}$, not that $Q_k$ and $Q_{-k}$ are independent real coordinates. For canonical quantization, impose $[u_j,p_l]=i\hslash {\delta }_{jl}$, equivalently $[Q_k,P_{k^{\prime }}]=i\hslash {\delta }_{k,-k^{\prime }}$. For $k\ne 0$ setThen the quantum harmonic chain is a set of independent phonon oscillators:The centre-of-mass zero mode of a harmonic chain is the free coordinate $X=N^{-1}{\sum }_j{u}_j$, with $[X,P_{\mathrm{tot}}]=i\hslash$; it cannot be normalized by dividing by $\sqrt{{\omega }_0}$. Retaining it giveswith the corresponding momentum expansion and uniform term $P_{\mathrm{tot}}/N$. The nonzero modes alone give the site commutator $i\hslash ({\delta }_{jl}-1/N)$; the uniform term supplies the missing $1/N$. Each oscillator occupation is a nonnegative integer, and adding one phonon costs $\hslash {\omega }_k$ and crystal momentum $\hslash k$ modulo a reciprocal-lattice vector. ^[ref]

In the continuum limit of a harmonic chain, $\rho =m_0/a$, $Y=Ka$ andThe wave equation and linear low-$k$ dispersion relation follow. The lattice gives an ultraviolet cutoff; the acoustic approximation does not reproduce the zone-edge flattening.

With the centre of mass fixed, the oscillator partition function and Bose-Einstein distribution areFor a long chain, above its finite-size level spacing and below the zone-edge scale, integrating the two acoustic propagation directions gives $C/(Na)=\pi k_{\mathrm{B}}^{2}T/(3\hslash c)$. At high temperature each retained oscillator instead contributes $k_{\mathrm{B}}$ to the heat capacity.

The Holstein–Primakoff transformation converts a spin algebra into bosonic operators while retaining the finite physical spin Hilbert space. With $[a_n,a_m^{†}]={\delta }_{nm}$, writeThe Holstein–Primakoff occupation constraint is $0\le n_n\le 2S$. Acting on $|n⟩$, the raising matrix element is $\sqrt{n(2S-n+1)}$ and the lowering element is $\sqrt{(n+1)(2S-n)}$. Their squared difference is $2(S-n)$, reproducing $[S^{+},S^{-}]=2S^z$. Operator ordering inside the square roots is essential. For low boson occupation relative to $S$, expand them to obtain $S^{+}\simeq \sqrt{2S}a$ and $S^{-}\simeq \sqrt{2S}{a}^{†}$. Keeping quadratic terms gives the linear spin-wave approximation; higher terms describe magnon interactions. Large $S$ helps control the expansion, but a low-dimensional infrared divergence can invalidate it even at large finite $S$.

For the ferromagnet choose $H_F=-J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, with unit lattice spacing and $\hslash =1$. Expand around the all-up state. A longitudinal bond contributes $-J{S}^2+JS(n_n+n_{n+1})$, and the transverse exchange contributes $-JS(a_n^{†}{a}_{n+1}+a_{n+1}^{†}{a}_n)$. After a Fourier transform,These are ferromagnetic magnons. The polarized vacuum is an exact ground state of the full ferromagnetic Hamiltonian: every bond is in its maximum total-spin sector, minimizing that bond energy. It therefore has no quantum zero-point reduction of its magnetization. The one-magnon dispersion is also exact, although interacting many-magnon states need not be free bosons. Rotating the polarization generates the maximum-total-spin ground-state multiplet. Selecting a definite direction in a thermodynamic symmetry-breaking description breaks $SU(2)$ to $U(1)$; the two transverse spin components form a canonical pair, producing a single quadratic type-B Goldstone boson. A small aligning field gaps it by the Zeeman energy.

At nonzero temperature the one-dimensional ferromagnet is different: its thermal magnon number contains $\int dk[e^{\beta JS{k}^2}-1{]}^{-1}$, with infrared behaviour $\int dkT/k^2$. The divergent magnetization reduction signals the absence of finite-temperature long-range order in an infinite short-range isotropic chain, consistent with the Mermin-Wagner theorem. It does not disallow the polarized zero-temperature ground states just constructed.^[ref]

For the antiferromagnet use $H_{AF}=+J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, still $J>0$, on an even bipartite chain. The classical reference state alternates up and down. Rotate one sublattice through $\pi$ so that both local quantization axes point along their reference spins. In unrotated physical components, to leading order $S_A^z=S-a^{†}a$, $S_B^z=-S+b^{†}b$, $S_A^{+}\simeq \sqrt{2S}a$ and $S_B^{+}\simeq \sqrt{2S}{b}^{†}$. Choosing Fourier phases using the actual site positions givesThe reduced Brillouin zone has $M=N/2$ momenta and length $\pi$. The pair creation and annihilation terms express the fact that an antiparallel reference product state is not an eigenstate. A bosonic Bogoliubov transformation with $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, $\tanh 2{\theta }_k=\cos k$ diagonalizes each two-mode block. With the sign convention $a_k=u_k{\alpha }_k-v_k{\beta }_{-k}^{†}$ and $b_{-k}=u_k{\beta }_{-k}-v_k{\alpha }_k^{†}$, the result isThus the harmonic antiferromagnetic modes have linear, rather than quadratic, long-wave dispersion. In an unfolded Brillouin-zone description the same result is represented by zeros at both zero and the antiferromagnetic wavevector; these are not extra independent degrees of freedom beyond the two-sublattice accounting.

The Bogoliubov vacuum has no $\alpha$ or $\beta$ excitations but contains original spin bosons, with $v_k^2=\frac{1}{2}(|\sin k{|}^{-1}-1)$. Its staggered magnetization is $m_s=S-M^{-1}{\sum }_{k\in \mathrm{RBZ}}{v}_k^2$. This quantum depletion of Néel order diverges logarithmically at small $k$ in one dimension. Linear spin-wave theory therefore does not establish a stable ordered antiferromagnetic ground state for this chain. Its nominal gapless dispersion is a result of expanding around an invalid ordered reference, not a general proof of the exact excitation spectrum. For the nearest-neighbour isotropic chain, integer-spin systems have a Haldane gap, whereas half-integer-spin systems are gapless; the spin-one-half chain in particular has fractional spin excitations rather than a simple dilute gas of ordered-state magnons. This distinction requires physics beyond the harmonic expansion, including the topological term in the continuum theory.^[ref]

For occupations small compared with $S$, use $S^z=S-n$, $S^{-}\simeq \sqrt{2S}{a}^{†}$ and $S^{+}\simeq \sqrt{2S}a$. The longitudinal bond contributes $-J\Delta S^2+JS\Delta (n_m+n_{m+1})+O(S^0)$; the transverse pair terms contribute $JS(a_m{a}_{m+1}+a_m^{†}{a}_{m+1}^{†})+O(S^0)$. Thus the direct real-space expansion isTake $a_m=N^{-1/2}{\sum }_k{e}^{ikm}{a}_k$, with $k=2\pi j/N$ over one full Brillouin zone and the lattice spacing set to one in $k$. Symmetrizing $k$ and $-k$ changes the phase in the pair sum to $\cos k$, givingnormal ordering the second Nambu spinor component then yieldsThe printed matrix has a factor-of-two error on its diagonal. With a full-zone sum its printed $2\Delta$ would produce $4\Delta a_k^{†}{a}_k$, and even its vacuum expectation would differ from the direct expansion by $J\Delta NS$. A half-zone convention cannot repair both coefficients: one must then group each $k,-k$ pair and double its off-diagonal pair coefficient as well. The formula above follows from the specified Hamiltonian and has the expected gapless isotropic limit; subsequent physical results use this consistent expansion. Terms quartic in bosons first enter at order $S^0$.

For $\alpha >0$, the dispersion is periodic and has minima at $k=0,\pi$ in the rotated full-zone convention, withThe maximum is $2JS(1+\alpha )$ at $k=\pm \pi /2$. The easy-axis state has no continuously broken symmetry, so a gap is allowed. Near a minimum the energy behaves as $2JS\sqrt{\alpha (2+\alpha )+(\delta k{)}^2}$. At $\alpha =0$ it becomes ${ϵ}_k=2JS|\sin k|$, linear near its zeros, with small-$k$ slope $2JS$ in the chosen units. Restoring lattice spacing $a$ and $\hslash$ gives spin-wave speed $2JSa/\hslash$. Zone folding reorganizes the same modes into the magnetic Brillouin zone.

These are linear spin-wave predictions, not a statement that the exact one-dimensional quantum chain always has long-range Néel order. At the isotropic point the zero-point spin reduction contains $N^{-1}{\sum }_k(|\sin k{|}^{-1}-1)/2$, which diverges logarithmically as the infrared cutoff vanishes. This failure of the assumed ordered reference must not be mistaken for a calculation of the exact interacting spectrum.

Take a uniform nearest-neighbour harmonic chain with displacement $u_j$, conjugate momentum $p_j$, mass $m_0$, spring stiffness $\kappa$, lattice spacing $a$ and periodic length $\ell =Na$. These specify a standard model for the requested notes, not additional numerical exam data. Its Lagrangian and Hamiltonian areVariation gives $m_0{\ddot{u}}_j=\kappa (u_{j+1}-2u_j+u_{j-1})$. For a mode $e^{i(kja-\omega t)}$, periodicity requires $k=2\pi n/\ell$ andThe lattice makes $k$ periodic modulo $2\pi /a$, so choose one Brillouin zone. At long wavelength ${\omega }_k\simeq v|k|$, where $v=a\sqrt{\kappa /m_0}$ is the sound speed. The continuum limit of a harmonic chain expands $u_{j+1}-u_j\simeq a{\partial }_{x}u$ and replaces ${\sum }_j$ by $a^{-1}\int dx$. Defining mass density $\rho =m_0/a$ and one-dimensional elastic coefficient $K=\kappa a$ givesThe momentum density is $\Pi =\rho {\partial }_{t}u$, and the Hamiltonian density is ${\Pi }^2/(2\rho )+K({\partial }_{x}u{)}^2/2$. Translational invariance of $u$ makes total momentum conserved. The continuum approximation describes acoustic wavelengths large compared with $a$; it is not a justification for extending the lattice dispersion and zero-point sums to arbitrarily large wavevector.^[ref]

canonical quantization promotes the coordinates and momenta to operators with $[u_j,p_l]=i\hslash {\delta }_{jl}$ and the other equal-time commutators zero. In the continuum, $[u(x),\Pi (x^{\prime })]=i\hslash {\delta }_{\mathrm{per}}(x-x^{\prime })$. The nonzero normal modes can be represented by independent phonon operators obeying $[b_k,b_q^{†}]={\delta }_{kq}$. A useful lattice expansion iswhere $X=N^{-1}{\sum }_j{u}_j$ is the centre-of-mass displacement. The corresponding momentum isIncluding the centre-of-mass term restores the full site commutator: the nonzero-mode sum alone gives $i\hslash ({\delta }_{jl}-1/N)$. Substitution diagonalizes the Hamiltonian,A phonon is one quantum of a collective normal mode, carrying energy $\hslash {\omega }_k$ and lattice momentum $\hslash k$ modulo a reciprocal-lattice vector. Reality of $u_j$ relates its Fourier coordinates by $u_k^{†}=u_{-k}$; it does not remove the independent oppositely travelling phonon modes. Using one complete Brillouin zone avoids double-counting real standing-wave coordinates.^[ref]

The free centre of mass has no restoring force and must not be treated as an oscillator of zero frequency. Fixing it, weakly pinning it, or specifying a finite centre-of-mass volume defines the corresponding zero-mode factor. With a fixed centre of mass, the internal thermal partition function and occupations areThe internal energy is ${\sum }_{k\ne 0}\hslash {\omega }_k(⟨n_k⟩+1/2)$; its high-temperature heat capacity approaches $(N-1)k_B$. For a large one-dimensional chain at temperatures below the lattice cutoff but above its finite-size mode spacing,This one-dimensional acoustic phonon heat capacity shows that quantization turns classical normal waves into bosonic excitations with a linear low-temperature heat capacity in one dimension. The Euclidean field action $S_E=\frac{1}{2}{\int }_0^{\beta \hslash }d\tau dx[\rho ({\partial }_{\tau }u{)}^2+K({\partial }_{x}u{)}^2]$ is Gaussian, with periodic imaginary-time boundary conditions; its Fourier modes reproduce these oscillators. Absolute displacement correlations require the zero-mode convention, and the lattice regulates short-distance fluctuations.

The periodic boundary condition requires $e^{ikM}=1$, so there are exactly $M$ independent wavevectors in a Brillouin zone:The inverse discrete Fourier transform, with the sign convention of the paper, isFourier orthogonality gives the tight binding dispersion relation ${ϵ}_k=-2t\cos k$. For the interaction, first use the canonical anticommutation relations at distinct sites to write $n_{\sigma ,j}{n}_{\sigma ,j+1}=c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}^{†}{c}_{\sigma ,j+1}{c}_{\sigma ,j}$. Summing its discrete Fourier transform over $j$ enforces crystal momentum conservation, givingEvery wavevector is reduced modulo $2\pi$. The phase $e^{iq}$ comes from the displacement of the second site by one lattice spacing. Equivalently replace $e^{iq}$ by $\cos q$: exchanging the two incoming and the two outgoing fermions changes $q$ to $-q$ with two cancelling fermionic signs. Thus the usual interaction vertex is $2V\cos q$ accompanied by the factor $1/(2M)$.

Use the unitary discrete Fourier transformThe transformed operators obey $[a_k,a_q^{†}]={\delta }_{kq}$. The hopping terms contribute $2A\cos k$ to the number coefficient. In the pairing sum, the odd sine part cancels between $k$ and $-k$. ThuswhereThe factor one half in the full momentum sum avoids counting an anomalous pair twice. For $H_f>0$, ${\alpha }_k>|{\Delta }_k|$ for every allowed momentum. SetReal even angles ensure consistency between opposite momenta, and ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$ preserves the canonical commutation relations. The anomalous coefficient becomes ${\Delta }_k\cosh 2{\theta }_k-{\alpha }_k\sinh 2{\theta }_k=0$. Normal ordering then giveswithThe field-polarized planar ferromagnetic spin wave has energy gap $ϵ(0)=S\sqrt{H_f(H_f+2J)}$. The momenta run over the whole Brillouin zone once, including the self-inverse momenta $0$ and, for even $L$, $\pi$; these use single-mode squeezing rather than introducing an extra independent partner.

The zero-field case has a zero-mode obstruction to bosonic Bogoliubov diagonalization. Its nonzero-momentum modes have $ϵ(k)=2\sqrt{2}JS|\sin (k/2)|$, but ${\alpha }_0={\Delta }_0=JS$, so no finite squeezing angle exists for the uniform mode. With $Q_0=(a_0+a_0^{†})/\sqrt{2}$, its Hamiltonian is $JS{Q}_0^2-JS/2$, not an identically zero oscillator. For example, its matrix element between the vacuum and the two-boson number state is $JS/\sqrt{2}$, whereas $ϵ(0)b_0^{†}{b}_0$ would vanish. Consequently the requested oscillator diagonalization is literal for nonzero field; at zero field treat the collective orientation separately or introduce a field regulator before taking a limit.

Phonon scattering changes the occupation or propagation of lattice-vibration normal modes through interactions with boundaries, defects or other phonons. Anharmonic terms permit phonon-phonon scattering. Different processes need not relax heat flow equally: normal processes conserve total crystal momentum within the first Brillouin zone, whereas Umklapp scattering changes its representative by a nonzero reciprocal lattice vector.

A direct band gap compares two energy bands at the same crystal momentum. A global gap instead requires the lowest energy of the upper band to exceed the highest energy of the lower band, allowing the two extrema to occur at different momenta. For example, the periodic toy bands $E_1(k)=A\cos k$ and $E_2(k)=A\cos k+\delta$, with $A>0$ and $0<\delta <2A$, have positive direct separation $\delta$ everywhere but global overlap $2A-\delta$. A positive direct splitting at each Brillouin zone boundary therefore does not alone establish an insulating filled band. At a suitable electron count the overlap can produce both electron and hole pockets.

In a semiconductor, a conduction electron fills a vacant valence state and emits a photon. In a simple nondegenerate model the recombination rate is $Bnp$, while the equilibrium reverse generation rate is $B{n}_i^2$; their difference gives the displayed net pair-loss rate. This subtraction enforces detailed balance at thermal equilibrium. Photon momentum is small on the scale of a Brillouin zone, so a direct band gap permits a direct transition while an indirect band gap generally requires a phonon or another source of momentum.

Consider a spin-degenerate two-dimensional free electron model with rectangular periods $a>b$ and two electrons per unit cell. Write $K=\pi /a$, $L=\pi /b$, and $k_F^2=4\pi /(ab)$. If $K<k_F<L$, the circular Fermi contour crosses only the vertical boundaries of the first Brillouin zone. Folding translates its protruding caps back by a reciprocal lattice vector $(2K,0)$. A weak periodic potential mixes the intersecting plane waves, reconnecting the contours into two open first-band sheets and a closed second-band pocket. The pocket is centred on a zone boundary: its halves on opposite sides of a reduced-zone plot are identified periodically and belong to one closed sheet.

Two sublattices give two bosonic modes for each of N/2 wavevectors in a magnetic Brillouin zone. For a symmetric quadratic pairing block with number coefficient A and anomalous coefficient B, a bosonic Bogoliubov transformation gives two equal positive energies $E_k=\sqrt{A^2-B_k^2}$ and the displayed vacuum shift. There are N modes overall. The quantum depletion of Néel order per site is $(2/N){\sum }_{\mathrm{MBZ}}{v}_k^2$, where $v_k^2=(A/E_k-1)/2$. Equivalently one can use one branch over a full zone with the corresponding factors of one half. Mixing these conventions doubles the vacuum energy or depletion.

A crystal with $N$ primitive unit cells has $N$ distinct wavevectors in one Brillouin zone. For a rectangular periodic sample, $k_i=2\pi l_i/L_i$, and the product of the numbers of allowed values along the three directions equals the number of cells. Each orbital-derived energy band therefore holds $N$ states per spin orientation, or $2N$ with twofold spin degeneracy. This means two states per primitive cell per spin-degenerate band, independently of how many atoms the cell contains.

Take equal masses $m$, atomic spacing $a$, and alternating springs $K\gg \kappa >0$. The primitive unit cell contains a strong-bonded pair, so its length is $2a$ and the Brillouin zone ends at $\pm \pi /(2a)$. Expanding the exact alternating-spring chain dispersion relation givesThe low-frequency acoustic phonon branch moves almost rigid dimers of mass $2m$ coupled by weak springs; its speed of sound tends to $a\sqrt{2\kappa /m}$. The high-frequency optical phonon branch describes internal dimer stretch. At the zone edge the exact frequencies are $\sqrt{2\kappa /m}$ and $\sqrt{2K/m}$.

A superlattice is a spatial periodic structure with a repeat length larger than an underlying atomic lattice period. Its additional reciprocal lattice vectors fold electronic bands into a smaller Brillouin zone, and allowed couplings can open small gaps at the folded crossings. The gap sizes depend on potential Fourier components and symmetry, not on the geometric period alone.