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A Bravais lattice is the set of all integer combinations of linearly independent primitive translation vectors. Its reciprocal lattice consists of $g$ satisfying $g\cdot l\in 2\pi \mathbb{Z}$ for every translation $l$; equivalently $e^{ig\cdot l}=1$.

The original PDF's third family has a bare $j$, not $n_{2}j$. Taken literally, the listed set is not the claimed Bravais lattice: the claimed primitive vector $a(i+k)/2$ does not occur. Its half-integer first and third coordinates force it into that third family, whose second coordinate is always $a$, not zero. The intended repair is $j\mapsto n_{2}j$ in the third family. The following conclusions apply to that corrected set.

The four corrected families are exactlyThe allowed parity patterns are all even, or exactly two odd coordinates. An integer combination of the proposed primitive vectors has coordinates $(a/2)(m_2+m_3,m_1+m_3,m_1+m_2)$, so lies in this set. Conversely, solveAll three are integers precisely when $h+k+l$ is even. This proves that the corrected set is the face-centered cubic lattice with the proposed primitive basis.

Direct dot products give $a_i\cdot b_j=2\pi {\delta }_{ij}$ forThese vectors generate the entire reciprocal lattice, since the coefficients of an arbitrary reciprocal vector in this basis are its integer pairings with the $a_i$. Its vectors are $(2\pi /a)(h,k,l)$ with all three coordinates of the same parity, the body-centered cubic lattice.

In Bragg scattering, translating a scattering atom by $l$ changes the amplitude by $e^{i(k-k^{\prime })\cdot l}$. Constructive interference from every lattice translate requires this factor to be one, so $k^{\prime }-k=g$ is a reciprocal lattice vector. For elastic scattering, $|k^{\prime }|=|k|$; the chord between these two vectors has length $2|k|\sin (\theta /2)$. ThereforeThe shortest nonzero reciprocal vectors are the $(111)$ family of length $(2\pi /a)\sqrt{3}$, followed by the $(200)$ family of length $(2\pi /a)2$. Identical atoms on this Bravais lattice introduce no additional basis extinction. If both families are accessible at the incident wavelength, their scattering angles obey $\boxed{\sin ({\theta }_1/2)/\sin ({\theta }_2/2)=\sqrt{3}/2}$.

Universality means that systems with different microscopic interactions can have the same asymptotic critical behaviour. Under renormalization-group flow, irrelevant microscopic differences shrink, and systems whose critical flows approach the same renormalization-group fixed point belong to one universality class. Important determinants are spatial dimension, order-parameter symmetry and number of components, the range of interactions and any additional relevant constraints. Microscopic Bravais lattice shape alone does not normally change the short-range scalar universality class.

Critical exponents, properly normalized scaling functions and certain dimensionless amplitude ratios are universal within a universality class. For example, the universal specific-heat amplitude ratio compares the two singular amplitudes above and below the transition, canceling their common normalization factors. Scaling functions require choices of thermal, field and length metric factors before comparisons between microscopic systems make sense. The critical temperature, critical field, Bravais lattice spacing, individual correlation-length and magnetic susceptibility amplitudes, and analytic free-energy backgrounds are not generally universal. Thus two models can share critical exponents without sharing their transition temperature or the absolute size of their response.

Use ${\beta }_T=1/(k_{B}T)$ for inverse temperature, to distinguish it from the order-parameter critical exponent $\beta$. In a translation-invariant pure thermodynamic phase, the connected two-point correlation function isFor finite correlation length, at $|r|\gg \xi$ it decays exponentially, possibly with an algebraic prefactor; ${\xi }^{-1}$ is the asymptotic exponential decay rate. Below an ordered transition the pure phase is selected by taking the thermodynamic limit before $h\to 0^{+}$ or $h\to 0^{-}$. A zero-field symmetric mixture of the two ordered phases has a nondecaying connected term and does not describe the usual pure-phase magnetic susceptibility.

Let $\mathcal{F}=-k_{B}T\log Z$ denote extensive physical free energy and $S={\sum }_r{\sigma }_r$. Since the field contributes $-hS$ to the statistical Hamiltonian, differentiation givesExpanding the variance and using translation invariance proves the correlation-function susceptibility sum rule:The sum includes $r=0$. If the source is instead defined as the dimensionless quantity ${\beta }_{T}h$, the corresponding derivative omits the displayed inverse-temperature factor.

A real-space renormalization group step of length factor $b>1$ uses a local blocking kernel $B_b({\sigma }^{\prime },\sigma )\ge 0$ satisfying ${\sum }_{{\sigma }^{\prime }}{B}_b({\sigma }^{\prime },\sigma )=1$. For example, a deterministic majority-spin rule assigns one coarse Ising spin to each block; tied blocks can be resolved with equal probabilities. Define the blocked statistical Hamiltonian byThe spacing becomes $a^{\prime }=ba$. All generated interactions, including multispin terms and the field-independent constant, must be retained for an exact transformation. Summing over ${\sigma }^{\prime }$ uses normalization to prove $Z(u^{\prime },C^{\prime },N^{\prime })=Z(u,C,N)$. Iteration gives $N_p=b^{-pD}N$ and $Z(u_p,C_p,N_p)=Z(u_0,C_0,N)$.

The coarse probability law is the exact marginal law of these block variables. Indeed, for any coarse observable $A$, its expected value is the microscopic expected value of ${\sum }_{{\sigma }^{\prime }}{B}_b({\sigma }^{\prime },\sigma )A({\sigma }^{\prime })$. A blocking rule that retains the long-wavelength order-parameter mode, with source and observable transformations tracked, therefore retains long-distance response and correlation functions. Their field normalization may change, while their physical length scales and critical singularities do not. Partition-function equality alone would not justify this assertion for an arbitrary kernel that discards all slow variables. Truncating the generated operator family is likewise an approximation.

For the free-energy recursion use dimensionless free energy per site,in the thermodynamic limit. Exact partition-function invariance gives $F(u_0,C_0)=b^{-pD}F(u_p,C_p)$. If one step sends $C$ to $C^{\prime }=b^{D}C+c(u)$, the coefficient $c(u)$ records the identity-operator contribution from eliminated fluctuations and normalization. ConsequentlyRepeated substitution derives the additive free-energy recursion under blocking:In physical energy units, divide this convention by ${\beta }_T$. The additive term is essential for absolute free energies even when it does not affect the leading critical powers.

A renormalization-group fixed point satisfies $R_b(u_{\ast })=u_{\ast }$. Linearize in scaling coordinates $v_i$ so that $v_i^{\prime }=b^{y_i}{v}_i+O(v^2)$. A relevant operator has $y_i>0$, an irrelevant operator has $y_i<0$, and a marginal operator has $y_i=0$ with behaviour decided by nonlinear terms. Thus the eigenvalues of the discrete linear map are $b^{y_i}$; the sign criterion applies to the generator exponents $y_i$, not to the discrete multipliers. The critical surface is the stable manifold on which every relevant scaling field has been tuned away. Irrelevant coordinates flow to the renormalization-group fixed point along it. A repulsive renormalization-group trajectory leaves the renormalization-group fixed point along an unstable relevant direction, ultimately entering a noncritical phase or another scaling regime.

To separate the singular free-energy density from its regular background, suppose the eliminated-mode contribution $g$ is analytic near the renormalization-group fixed point. Look for an analytic background $a(u)$ satisfying $a(u)=b^{-D}a(R_{b}u)+g(u)$. Then $F_s=f-a$ satisfies $F_s(u)=b^{-D}{F}_s(R_{b}u)$. In linear thermal and field coordinates, an analytic monomial $g_{mn}{t}^m{h}^n$ has background coefficientprovided the denominator is nonzero. This explicitly explains when analytic subtraction of an inhomogeneous renormalization recursion works. The printed inhomogeneous equation may describe the free-energy function before this subtraction, or a singular part defined only up to analytic terms. The inhomogeneous sum need not be small compared with the full free energy: an analytic constant can dominate a vanishing singular contribution. It can be omitted in deriving leading singular powers after regular subtraction, provided no renormalization-group free-energy resonance, marginal logarithm or dangerously irrelevant coupling obstructs the homogeneous critical asymptotics. At a renormalization-group free-energy resonance $m{\lambda }_t+n{\lambda }_h=D$, additive logarithms can remain, so simply dropping the term would miss them.

Suppose there are exactly two relevant scaling fields, thermal $t$ and magnetic $h$, with positive exponents ${\lambda }_t,{\lambda }_h$, and the remaining fields approach a regular limit. Physical reduced temperature and magnetic field agree with these coordinates to leading order, up to nonuniversal metric factors. The homogeneous scaling hypothesis for critical phenomena isas an asymptotic scaling equation. Choose $\ell =|t{|}^{-1/{\lambda }_t}$ to make the thermal argument $+1$ or $-1$. Defining $f_{\pm }(z)=F_s(\pm 1,z)$ givesThe two functions describe respectively $t>0$ and $t<0$: approaching the transition from disordered and ordered sides need not give the same function. In particular the ordered zero-field function has one-sided derivatives associated with spontaneous magnetization. Metric factors multiply this formula in different microscopic normalizations.

The correlation length in Bravais lattice units obeys $\xi (t,h)=\ell \xi ({\ell }^{{\lambda }_t}t,{\ell }^{{\lambda }_h}h)$. At $h=0$, the same choice of $\ell$ gives $\nu =1/{\lambda }_t$. Two thermal derivatives of $F_s(t,0)$ give $2-\alpha =D/{\lambda }_t$. One and two field derivatives give, respectively, $\beta =(D-{\lambda }_h)/{\lambda }_t$ and $\gamma =(2{\lambda }_h-D)/{\lambda }_t$, assuming the relevant branch amplitudes are nonzero. At $t=0$, choose $\ell =|h{|}^{-1/{\lambda }_h}$; then $M\propto \mathrm{sgn}(h)|h{|}^{(D-{\lambda }_h)/{\lambda }_h}$, so $\delta ={\lambda }_h/(D-{\lambda }_h)$. HenceSimple algebra now gives the Rushbrooke scaling relation $\alpha +2\beta +\gamma =2$ and the hyperscaling relation $\alpha =2-D\nu$. Also $\beta \delta ={\lambda }_h/{\lambda }_t=\beta +\gamma$, giving the Widom scaling relation. These statements refer to leading singular contributions with the homogeneous two-field scaling assumptions. Above the upper critical dimension, the quartic coupling is dangerously irrelevant because it is still needed to stabilize the ordered phase; ordinary mean-field critical exponents then satisfy Rushbrooke and Widom but generally not hyperscaling. Marginal logarithms also require qualifications at the upper critical dimension.

For the connected-correlation scaling form $G(r)=|r{|}^{-(D-2+\eta )}{f}_G(|r|/\xi )$, the function $f_G(x)$ tends to a finite nonzero value as $x\to 0$ and decays exponentially at large $x$, up to powers of $x$, in a massive short-range phase. The scaling ansatz alone does not fix that prefactor. If the usual massive Gaussian field correlation tail $G(r)\propto r^{-(D-1)/2}{e}^{-r/\xi }$ applies, matching the powers gives $f_G(x)\sim B{x}^{(D-3)/2+\eta }{e}^{-x}$, with a correlation-length-dependent amplitude in $G$ fixed by matching the critical scaling form.

In physical coordinates the lattice sum has site-density factor $a^{-D}$. Its long-distance singular part therefore giveswhere ${\Omega }_{D-1}$ is the area of the unit sphere in $D$ dimensions. For $\eta <2$, finite $f_G(0)$, an integrable exponential tail and a nonzero integral, the last factor tends to a finite nonzero constant as $\xi \to \infty$. Microscopic distances contribute a regular background. Thus ${\chi }_s\propto {\xi }^{2-\eta }\propto |t{|}^{-(2-\eta )\nu }$, proving the Fisher scaling relationEquivalently the field has scaling dimension $(D-2+\eta )/2$, so its conjugate field has ${\lambda }_h=(D+2-\eta )/2$, which gives the same identity from the exponent formulas above.

A partition function sums the Boltzmann weights of all states: $Z={\sum }_s{e}^{-{\beta }_{T}H(s)}$ for discrete states, or the corresponding integral for continuous fields. It normalizes equilibrium probabilities. Its physical Helmholtz free energy is $\mathcal{F}=-k_{B}T\log Z$, and its free-energy density is $\mathcal{F}/V$. With a momentum ultraviolet cutoff ${\Lambda }_0$, a scalar-field example isThe normalization of the functional measure and field-independent energy terms must be tracked when discussing absolute free energy.

Choose a smaller cutoff $\Lambda$ and split the field into retained modes ${\varphi }_{<}$ with $|q|<\Lambda$ and eliminated modes ${\varphi }_{>}$ with $\Lambda <|q|<{\Lambda }_0$. Define the Wilsonian coarse-grained statistical Hamiltonian bySubstitution into the partition function proves $Z=\int \mathcal{D}{\varphi }_{<}{e}^{-{\beta }_T{H}_{\Lambda }[{\varphi }_{<}]}$ for every intermediate cutoff. A source retained in this definition similarly preserves long-wavelength connected correlation functions and response. Thus the coefficients in a Hamiltonian density must run with $\Lambda$: the dependence induced by eliminated modes compensates the dependence of the remaining field integral. Holding the original bare coefficients fixed while changing the cutoff would generally change the system. Exact integration generates all interactions allowed by symmetry, including higher powers, derivative terms and an additive constant; a local derivative expansion or a truncation to a few couplings is an extra approximation.

Cutoff independence makes arbitrary intermediate coarse-graining scales unphysical. It allows calculations at the scale of the observable, avoids repeatedly resolving irrelevant short-distance details and exposes the renormalization-group fixed points responsible for universality. A microscopic cutoff, such as a Bravais lattice spacing, can still label a physical microscopic model; taking it to infinity as a continuum limit requires tuning appropriate bare couplings. This is distinct from lowering an intermediate Wilsonian cutoff in a fixed model.

To interpret the zero-cutoff limit, define $M=V^{-1}\int \varphi (x)d^{D}x$, the spatially uniform mode of the order parameter. Keep this mode fixed and integrate the others, rather than integrating over $M$ too. Write $\varphi =M+\eta$, with $\int \eta =0$, and define the constrained order-parameter free energy byIn a finite periodic box, once $\Lambda$ is below the smallest nonzero momentum, all nonconstant modes have been eliminated. The remaining effective energy is $H_{\Lambda }[M]=V{U}_{\Lambda }(M)$, so its defining integral is exactly $Z_M$. Therefore, with consistent normalization,When $\mathcal{H}(\Lambda ,M)$ denotes this uniform effective Hamiltonian density, this is the stated identification of $F(M)$ with its zero-cutoff limit. It requires a free-energy density constrained at fixed $M$, not the unconstrained extensive free energy; all integrated fluctuations and additive constants are included, the zero mode in field theory is retained, and a density for the uniform configuration is well defined. The finite-volume argument avoids an interchange of the thermodynamic and cutoff limits without justification. On a homogeneous pure-phase branch one may use a local potential description, while a thermodynamic limit allowing phase separation can convexify the constrained potential. A bare nonconvex Landau polynomial is accordingly a homogeneous-branch approximation, not a claim that the exact equilibrium Legendre potential stays nonconvex.

In a field $h$, the remaining integral is $Z(h)=\int dM{e}^{-{\beta }_{T}V[F_c(M)-hM]}$, up to the consistently chosen zero-mode measure. The thermodynamic saddle then gives $f(h)={\inf }_M\{F_c(M)-hM\}$ and selects the equilibrium magnetization. Landau theory assumes that the relevant uniform potential can be approximated by a regular analytic expansion in $M$, with symmetry-constrained coefficients, and minimizes it. At tree level the integrated fluctuations are neglected, recovering the familiar bare polynomial. With fluctuations retained, the coefficients run and can become singular near criticality; the zero-cutoff identification of constrained free energy alone does not justify mean-field critical exponents.

For an ordinary short-range scalar thermodynamic critical point, the leading interaction is quartic. Absorb inverse temperature and the gradient coefficient into field normalization and writeScale $x=b{x}^{\prime }$ and $\varphi (x)=b^{-(D-2)/2}{\varphi }^{\prime }(x^{\prime })$ to leave the gradient term invariant. The quartic term changes to $g{b}^{4-D}\int {\varphi }^{\prime 4}/4!$, so its engineering scaling exponent is $4-D$. It is irrelevant at the Gaussian fixed point for $D>4$, relevant for $D<4$, and marginal for $D=4$. HenceThe loop expansion makes the breakdown concrete. A quartic-vertex correction is proportional to $g^2{I}_2(r)$, whereFor $0<D<4$, rescaling $q=\sqrt{r}k$ gives $I_2(r)\sim C_D{r}^{(D-4)/2}$, diverging as $r\to 0$. At $D=4$, radial integration giveswhich is logarithmically divergent. Above four dimensions its infrared limit is finite; the leading cutoff-dependent constant can be absorbed into a coupling renormalization. The effective expansion parameter is thus of order $g{\xi }^{4-D}$ below four, or $g\log (\Lambda \xi )$ at four, since $\xi \sim r^{-1/2}$. It eventually becomes large for any fixed bare interaction in the lower-dimensional critical limit, defeating the uncorrected mean-field expansion. The Ginzburg criterion gives the same threshold by comparing correlation-volume fluctuations with the squared mean order parameter.

Below four dimensions an interacting renormalization-group fixed point can replace the Gaussian fixed point and change critical exponents, provided a continuous transition exists. At four dimensions the interaction is marginal and produces logarithmic corrections: the leading power exponents can remain their mean-field values, so “breakdown” there means that pure mean-field power laws omit those corrections, rather than necessarily that every power exponent changes. Above four the stabilizing quartic interaction is dangerously irrelevant for the ordered phase, explaining why mean-field thermodynamic exponents need not obey naive hyperscaling with the spatial dimension.

Translate the scattering solution for the atom at the origin by $\mathbf{a}$. Its incident wave becomes $e^{i\mathbf{k}\cdot (\mathbf{r}-\mathbf{a})}$; multiplying the whole solution by $e^{i\mathbf{k}\cdot \mathbf{a}}$ restores the prescribed incident wave. At large $r$,Thus the outgoing amplitude acquires the phase $e^{i(\mathbf{k}-{\mathbf{k}}^{\prime })\cdot \mathbf{a}}$, with ${\mathbf{k}}^{\prime }=k\hat{\mathbf{r}}$.

In the single-scattering approximation, amplitudes from lattice sites add coherently. If $\mathbf{Q}=\mathbf{k}-{\mathbf{k}}^{\prime }$,For a rectangular block of a Bravais lattice with $N_j$ sites along primitive vectors ${\mathbf{a}}_j$ and $N=N_1{N}_2{N}_3$, the geometric series givesEach ratio has its limiting value $N_j$ when its argument is a multiple of $2\pi$. Thus the peaks have height $N$ and widths of order $1/N_j$ in those phase variables, at $\mathbf{Q}\cdot {\mathbf{a}}_j\in 2\pi \mathbb{Z}$, exactly the reciprocal lattice vectors.

With the incident-wave normalization used here, the Born approximation isFor the stated attractive three-dimensional delta potential,independent of scattering angle. This is the formal first Born amplitude, not a claim that an unregularized three-dimensional delta interaction defines an exact scattering Hamiltonian.

The 60-degree scattering angle is twice the Bragg angle, so $\theta =30^{\circ }$. First-order elastic Bragg scattering condition gives $2d\sin \theta =\lambda$, henceHigher orders would give $d=n\lambda$; the smallest, usual first-order spacing is the likely answer.

Bloch theorem states that energy eigenfunctions in a Bravais lattice potential may be chosen asThe wavevector is defined modulo a reciprocal lattice vector, so one can restrict it to the first Brillouin zone. This follows because lattice translations commute with the Hamiltonian and have simultaneous eigenvalues $e^{i\mathbf{k}\cdot \mathbf{l}}$.

In one dimension write $V(x)={\sum }_G{V}_G{e}^{iGx}$ with $G=2\pi j/a$ and expand a Bloch state as ${\psi }_k={\sum }_G{c}_G{e}^{i(k+G)x}$. The Schrödinger equation becomesFor zero potential, the parabolas $E_G^0(k)={\hslash }^2(k+G{)}^2/(2m)$ fold into the Brillouin zone. A weak potential shifts a nondegenerate level by $V_0$ at first order, with second-order correction ${\sum }_{G\ne 0}|V_G{|}^2/(E_0^0-E_G^0)$ where the denominators stay away from zero. This expansion fails near crossings, where degenerate perturbation theory is required.

At the first zone boundary, put $G=2\pi /a$ and $k=G/2+q$. The nearly degenerate waves with momenta $\hslash k$ and $\hslash (k-G)$ are coupled by $V_G$. Retaining their two-dimensional subspace givesDiagonalizing producesAt $q=0$ there is a gap $\boxed{\Delta E=2|V_G|}$: the two standing-wave combinations sample the periodic potential differently. Similar Bragg degeneracies occur at $k=j\pi /a$, with gaps controlled to first order by the corresponding Fourier components. If that component vanishes, a gap need not open at first order. Between the gaps the continuous energy eigenvalues form energy bands, periodically represented in the reduced-zone scheme. For the first zone edge, the curvature of the upper and lower branches gives a minimum and a maximum respectively for a sufficiently weak nonzero potential, so the avoided crossing is a genuine local forbidden energy interval.

At zero temperature electrons occupy states up to the Fermi energy. The Fermi surface is the set $E_n(\mathbf{k})=E_F$ separating occupied from unoccupied states within partially filled bands (Fermi points in one dimension). A partially filled band supplies arbitrarily nearby empty states, allowing occupations to respond to a weak electric field and conduct. Completely filled bands give no net current from such a redistribution; if the next empty band is separated by a positive band gap, the system is a band insulator. Band overlap can instead yield a metal even at an integer band filling. At finite temperature an insulator can conduct through thermally excited electrons and holes.

A three-dimensional Bravais lattice is the set of all integer combinations of three linearly independent primitive vectors. Its reciprocal lattice isand the first Brillouin zone is the Wigner-Seitz cell of the origin in reciprocal space.

For the stated FCC lattice, the primitive-cell volume isThe reciprocal-basis formula givesA reciprocal primitive cell, and hence the Brillouin zone, has volume

In the first Born approximation, with momentum transfer $q=k-k^{\prime }$, the scattering amplitude isFor the spherical shell, align the polar axis with $q$ and use its radial Dirac delta function:Therefore

Translation of each atomic potential multiplies its transform by a phase, so the crystal lattice structure factor givesFor an infinite lattice, the lattice Poisson summation formula turns the sum intoThus the amplitude vanishes away from reciprocal-lattice momentum transfers.

In elastic scattering, $|k|=|k^{\prime }|$ and henceThe two shortest relevant reciprocal-vector lengths of this FCC lattice areBoth are kinematically available when $k>2\pi /a$. Assigning their scattering angles ${\theta }_2$ and ${\theta }_1$, respectively, gives

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.

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.

TakeTheir integer span consists exactly of the cubic points together with the body-centre points, so the BCC set is a Bravais lattice. Using the inverse matrix supplied in the question, reciprocal primitive vectors areThese generate a face-centred cubic lattice; thus the reciprocal lattice of BCC is FCC, with the displayed scale.

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.

Let the Bravais lattice have primitive vectors $a_1,a_2,a_3$, and putThe reciprocal lattice has the basisThe scalar triple product and cross product identities giveConsequently every $q={\sum }_j{n}_j{b}_j$ satisfies $q\cdot l\in 2\pi \mathbb{Z}$ for every lattice point $l={\sum }_i{m}_i{a}_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be integers. This proves that the displayed vectors generate ${\Lambda }^{\ast }$.

In the Born approximation, suppose that the crystal potential is the sum of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its Fourier transform at the momentum transfer $Q=k-k^{\prime }$ factors asThus the single-atom scattering amplitude is multiplied by the crystal lattice structure factor $\Delta (Q)$. Writingseparates the sum into three finite geometric series. For $l_i=-L_i/2,\dots ,L_i/2$ this givesWhen the $L_i$ are large, a factor has a sharp maximum when ${\alpha }_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in {\Lambda }^{\ast }$. These are the reciprocal-lattice peaks of a finite crystal.

For the stated body-centered cubic lattice basis,the reciprocal-basis formula givesEquivalently,The shortest nonzero reciprocal vectors have squared Euclidean norm $2(2\pi /a{)}^2$, so