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With downward position as the vertical coordinate, the full potential energy isThe equilibrium tensions are $2k(a-\ell )=6mg$, $k(b-\ell )=3mg$ and $k(c-\ell )=mg$. On expanding, these relations cancel every term linear in the displacements, leaving $V=k{x}^2+k(x-y{)}^2/2+k(y-z{)}^2/2+V_0$. Newton's second law therefore givesFor a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, set $s=m{\omega }^2/k$. The generalized eigenvalue problem isIts determinant is $2(1-s)[3(1-s{)}^2-2]$, so the angular frequencies and mode directions areThe square roots of these positive values give the requested frequencies. Only the lowest mode has all components of the same sign, so all three particles oscillate in phase at the lowest frequency. The mode shapes are shown below.

For release from rest with $Y$ remaining at equilibrium, chooseThis excites only the middle-frequency mode, so $y(t)=0$ for all time. It is a stationary middle mass in a three-mass spring chain: the equal and opposite adjacent spring forces cancel on $Y$.

Introduce a Lagrange multiplier for the constraint $G=1$ and vary $F-\lambda G$ with $y\mapsto y+ϵ\eta$, where $\eta (a)=\eta (b)=0$. The first variation isThe fundamental lemma of the calculus of variations gives the Sturm-Liouville equation $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$. Multiplying by $y$, integrating by parts and using the endpoint conditions gives $F=\lambda G$. Therefore the normalization implies $\lambda =F$. Assume the coefficient and function regularity needed for these variations and integrations; the condition $w>0$ makes the constraint derivative nonzero at a normalized function.

For the radial problem, set $\sigma =r\psi$. A direct differentiation gives ${\sigma }^{\prime \prime }=r{\psi }^{\prime \prime }+2{\psi }^{\prime }$, soFinite $\psi (0)$ supplies the first boundary value. The Rayleigh quotient is $R[\sigma ]={\int }_0^1{\sigma }^{\prime 2}dr/{\int }_0^1{\sigma }^{2}dr$. For the two trials, ${\sigma }_1=r(1-r)$ and ${\sigma }_2=r(1-r^2)$, the numerator-denominator pairs are $(1/3,1/30)$ and $(4/5,8/105)$, respectively. HenceThe exact eigenvalues are $n^2{\pi }^2$, so $E_{\min }={\pi }^2\simeq 9.86960$. Both trials give upper bounds by the Rayleigh-Ritz variational principle, and the first estimate is more accurate, since it is smaller. Trial admissibility uses the Dirichlet energy form for $\sigma$; the first trial's nonzero radial derivative at the origin does not invalidate this variational calculation.

To improve, add independent trial functions and minimize the quotient over their span. Here merely combining the two supplied trials gives no improvement: their two-dimensional stiffness and mass matrices areThe lowest Ritz value is still ten. Instead add the symmetric quartic $\tau =r^2(1-r{)}^2$. On the span of ${\sigma }_1,\tau$, the generalized eigenvalue problem hasIts determinant is $(E^2-112E+1008)/529200$, so the least root givesThis is strictly better than both original estimates. The symmetry obstruction in a polynomial Rayleigh-Ritz trial space explains why the cubic addition failed while the quartic helps. It is also the shifted-interval version of the polynomial Rayleigh-Ritz trial space for a one-dimensional box.

Let $R[y]=F[y]/G[y]$, where $G[y]>0$ for a nonzero admissible real function. Writing $\lambda =R[y]$, its first variation isChoose boundary conditions making the boundary term vanish, for example homogeneous Dirichlet boundary conditions, homogeneous Neumann boundary conditions, or a corresponding mixed pair. Arbitrary interior variations then give the Sturm-Liouville problemMultiplying this equation by $y$ and integrating by parts, with $[py{y}^{\prime }{]}_{\alpha }^{\beta }=0$, yieldsThus eigenfunctions are stationary directions of the quotient. The lowest eigenvalue is its global minimum; higher eigenfunctions need not be unrestricted maxima or minima. General Robin boundary conditions require including their boundary energy in the quotient unless the displayed endpoint products also vanish.

The Rayleigh-Ritz method chooses admissible trial functions depending on finitely many coefficients and minimizes this quotient over them. Expansion in a complete orthonormal basis with weight $w$ shows $R[y]\ge {\lambda }_0$, so the trial minimum is an upper bound on the true lowest eigenvalue. In a finite basis, the minimization becomes a generalized eigenvalue problem for the stiffness and weighted mass matrices. Enlarging the trial space can only lower or preserve this upper bound.

Choose generalized coordinates $q_i$ measured from a static equilibrium point, and assume a conservative system with no velocity-dependent forces. Keeping terms of degree two in displacements and velocities gives the small oscillation LagrangianHere $A$ is the positive-definite symmetric mass matrix evaluated at equilibrium, and $B$ is the symmetric Hessian matrix of the potential energy there. The linear potential term is zero by equilibrium, and an additive constant has been dropped. Coordinate-dependent corrections to the kinetic matrix contribute only at higher order.

The Euler-Lagrange equations are $A\ddot{\mathbf{q}}+B\mathbf{q}=0$. For a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, they give the generalized eigenvalue problem $(B-{\omega }^{2}A)\mathbf{v}=0$. A nonzero mode shape exists exactly whenThis is a degree-$n$ polynomial in ${\omega }^2$. At a stable equilibrium its roots are nonnegative; zero roots correspond to zero modes.

With the stated parameters, the mass and restoring matrices areThe generalized eigenvalue problem givesTherefore the two normal angular frequencies areThe second row of the modal equation is $-{\omega }^{2}x+(1-{\omega }^2)\varphi =0$. It gives the mode shapesEach normal mode has arbitrary small amplitude $a$ and phase $\delta$. The bob's horizontal displacement is $X=x+\varphi$ to first order. Thus in the lower mode, bob and cart move in phase, with bob amplitude $1+\sqrt{2}$ times the cart amplitude. In the higher mode they move oppositely, because $X=(1-\sqrt{2})x$; the bob amplitude is $\sqrt{2}-1$ times the cart amplitude. Its vertical displacement is quadratic in the angle and is absent at this order.

For an ordinary conservative mechanical system, choose generalized coordinates $q_i$ measured from a static equilibrium point. Assume the kinetic energy is quadratic in velocities and there are no velocity-dependent forces. Expanding about equilibrium, the linear term in the potential energy vanishes. Retaining only terms of total degree two gives the small oscillation LagrangianAn irrelevant constant has been dropped. Coordinate dependence of $M$ contributes only terms of third or higher total degree. The matrices $M,K$ are symmetric, $M$ is positive definite for independent coordinates, and $K$ is positive semidefinite at a stable equilibrium. The Euler-Lagrange equations areA normal mode has a fixed mode shape $\mathbf{v}$ and one time factor: $\mathbf{q}=\mathbf{v}Q(t)$. Its angular frequency $\omega$ solves the generalized eigenvalue problem $K\mathbf{v}={\omega }^{2}M\mathbf{v}$, with $\ddot{Q}+{\omega }^{2}Q=0$. Thus $\omega$ is the normal angular frequency, and the frequency in cycles per unit time is $\omega /(2\pi )$. A zero mode satisfies $K\mathbf{v}=0$; its time dependence is $Q=A+Bt$, describing an unopposed displacement or uniform drift rather than a restoring oscillation.

Label the equal masses $1,2$ and the mass $2m$ by $3$, in cyclic order. Let their equilibrium angles be $0,2\pi /3,4\pi /3$, with angular displacements $q_1,q_2,q_3$. Because the hoop has unit radius, its three arc lengths areThe springs themselves are constrained to the hoop, so these are arc lengths. If their common natural length is ${\ell }_0$, Hooke's law gives $V=\frac{k}{2}{\sum }_i({\ell }_i-{\ell }_0{)}^2$. Since the arc lengths have fixed sum $2\pi$, all terms linear in the $q_i$ cancel. Dropping the constant,Thus the unspecified common natural length does not affect the motion. The mass and restoring matrices in the generalized eigenvalue problem areExchange of the two equal masses separates symmetric and antisymmetric normal modes. Direct multiplication gives the complete setThe first is rigid rotation of all three beads: every arc length stays fixed, giving a zero mode. In the second, the equal masses move oppositely while the heavy bead stays fixed; the net force on that bead cancels. In the third, the equal masses move together and the heavy bead moves oppositely with equal angular amplitude. Both nonzero normal modes have zero total angular momentum. These are the normal modes of three beads connected by springs on a circular hoop with $M=2m$.

For the added trial term $q_n=x^n{e}^{-Ax}$, the left Dirichlet boundary condition is automatic for a positive integer $n$. At the right endpoint,To preserve $y^{\prime }(1)=0$ for an arbitrary variational coefficient $C$, choose $\boxed{A=n}$. Then minimize the Rayleigh quotient of $x{e}^{-x}+C{x}^n{e}^{-nx}$ over $C$.

Since $C=0$ remains available, this optimization cannot worsen the original upper bound. For $n=1$, the added function is a multiple of the first trial, so the quotient does not change: an independent choice $n\ge 2$ is needed for extra freedom. More generally, several such functions form the exponential-polynomial trial space for a mixed Sturm-Liouville problem, and the Rayleigh-Ritz method reduces minimization to a small generalized eigenvalue problem.

For two bars, after subtracting the gravitational equilibrium,A normal mode $\mathbf{\eta }=\mathbf{v}{e}^{i\omega t}$ requires the generalized eigenvalue problem $(K-{\omega }^{2}M)\mathbf{v}=0$. With $z={\omega }^2$,Thus, for $m_2>0$,Both squared angular frequencies are positive. A corresponding eigenvector is $(1,2-{\omega }_j^2{)}^T$, from the first row of the matrix equation.

For sufficiently smooth functions satisfying the same boundary conditions, integration by parts gives the boundary formThe stated endpoint relation annihilates this expression. This proves the required self-adjoint differential operator identity; for the usual regular Sturm-Liouville problem, separated real boundary conditions give a self-adjoint realization on its domain.

In this problem the transformed diffusive flux is $J=-e^{-\lambda t}p{y}^{\prime }$. Thus the physical boundary conditions areequivalently $y^{\prime }(\pm L)=0$ when the endpoint values of $p$ are positive. In particular $p{f}^{\prime }=p{g}^{\prime }=0$ at the endpoints, which makes the boundary form vanish directly. The inner product used here is the ordinary integral; the weight $w$ enters the generalized eigenvalue problem, not this boundary-form calculation.

Order points along each $j$-line before moving to the next $k$-line. The splitting hasand $N$ couples the two neighboring $k$-lines. Factor $T_n$ once with the Thomas algorithm; each iteration then needs $n$ tridiagonal solves and the neighboring-line right sides, costing $\boxed{O(n^2)}$ for $n^2$ unknowns. Known boundary values are moved to the right side; the error eigenmodes have homogeneous boundary values.

The generalized eigenvalue problem iswith $U=0$ at the boundary indices. Put ${\theta }_r=r\pi /(n+1)$, ${\theta }_s=s\pi /(n+1)$. The sine addition formula gives $U_{j+1,k}+U_{j-1,k}=2\cos {\theta }_r{U}_{j,k}$ and the corresponding identity in $k$. ThusThese tensor-product sine vectors form a basis, so they give the whole spectrum. With $c=\cos (\pi /(n+1))\in [0,1)$,The iteration therefore converges.

It is not a uniformly effective multigrid smoother. Take $r=1$ and $s=n$: the error is slowly varying along a line but highly oscillatory between lines, and its multiplier is $-c/(2-c)\to -1$ as the grid is refined. It changes sign at each step while retaining almost its full amplitude. This line-Jacobi Fourier spectrum distinguishes convergence on each fixed grid from the high-frequency damping needed by the multigrid method.

A normal mode has all displacement coordinates proportional to a common time dependence, with fixed relative amplitudes. For a nonzero frequency the time dependence is sinusoidal. A zero-frequency mode is a free displacement or uniform drift rather than an oscillation.

Use the printed mass ordering $M,m,M$. Displacements are measured from the unstretched equilibrium positions, givingEquivalently,This is the generalized eigenvalue problem for normal modes of three masses connected by two equal springs. Substitute $\mathbf{x}=\mathbf{u}{e}^{i\omega t}$ to obtain $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. Reflection symmetry gives an antisymmetric vector $(1,0,-1)$ and symmetric vectors $(1,b,1)$. The former has ${\omega }^2=k/M$. For a symmetric vector, the end and centre equations give $M{\omega }^2=k(1-b)$ and $m{\omega }^{2}b=2k(b-1)$. Their solutions are $b=1$, $\omega =0$, or $b=-2M/m$, ${\omega }^2=k/M+2k/m$. HenceThese are all three eigenmodes. The last two keep the centre of mass fixed and are orthogonal in the mass-weighted inner product. The general zero-mode contribution is $(A+Bt)(1,1,1)$, accounting for both a centre-of-mass shift and a conserved total momentum. There are no fixed end springs in this system.

The original amplitude profiles illustrate $M=2m$; the symmetric internal shape is rescaled for display. The formula above gives every positive mass ratio.

Two further physical examples are standing-wave modes of a string fixed at both ends, with spatial shapes $\sin (n\pi x/L)$ and frequencies $n\pi c/L$, and electrical normal modes of two coupled LC circuits, in which the capacitor charges oscillate at fixed relative amplitudes. Independent normal mode coordinates simplify both systems in the same way as the spring-chain eigenvectors.

Why degenerate perturbation theory is needed. If several unperturbed states have equal or nearly equal energy, a perturbation can mix them at leading order. In that subspace, diagonalize the perturbation matrix rather than using nondegenerate formulas with small energy denominators. The resulting eigenvectors identify the combinations selected by the coupling, and the eigenvalues give the leading splitting. Other remote states can then be included perturbatively.

For the linear combination of atomic orbitals description of molecular hydrogen, equivalent atomic $1s$ states on the two nuclei supply the nearly degenerate starting functions. In an orthonormal two-state model, equal diagonal energy $E_a$ and real hopping $t$ give $E_a\pm t$ and symmetric/antisymmetric combinations. For actual overlapping atomic functions with overlap $S$ and Hamiltonian coupling $h$, one instead solves a generalized eigenvalue problem:The bonding state has enhanced density between the nuclei and the antibonding state a separating node. Two opposite-spin electrons occupying the bonding molecular orbital describe the simplest closed-shell hydrogen-molecule bond. Electron-electron interaction and nuclear repulsion are needed to calculate a realistic total binding energy; the two-state splitting alone is not the full molecular energy.

In the nearly-free electron model, the analogous degenerate states are plane waves $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$ at a Bragg scattering. A Fourier coefficient $U_{\mathbf{G}}$ of the periodic potential couples them. At exact degeneracy $E_B$, diagonalizing their two-by-two matrix givesThe associated standing waves place electronic density differently relative to the ionic potential. These are the same gap-opening combinations illustrated in Q1(b). The labels of the two starting basis states survive as mixed orbital or plane-wave character, rather than independent crossing eigenstates.

Origin of broad and narrow bands. Spatially extended s/p-derived orbitals have substantial overlap and therefore large hopping amplitudes and broad bands. More localized d orbitals overlap less strongly and produce narrower bands. A multiorbital tight-binding calculation assigns the on-site orbital energies and hopping matrices, forms Bloch sums, and diagonalizes the resulting matrix at each wavevector. Different hopping ranges and orbital symmetry determine the bandwidths and allowed interband coupling. The two-band description retains one broad combination and one nearly flat d-derived combination; a real transition metal contains several d orbitals and more complicated dispersions.

Hybridized eigenvalues and eigenstates. Let $|{\psi }_{\mathbf{k}}⟩=a|{\varphi }_{\mathbf{k}}⟩+b|{\chi }_{\mathbf{k}}⟩$, in an orthonormal basis. Hermitian operator fixes the reverse coupling to be $V^{\ast }$. Projection onto the two components givesSolving this quadratic yields the band hybridization dispersionFor nonzero $V$, one eigenvector representation is $(a,b)\propto (V,E-E_1)$, followed by normalization. Thus the wavefunctions exchange broad-band and d-band character through the avoided crossing; the levels are equal mixtures at the uncoupled crossing.

At $V=0$, the parabola crosses the horizontal level at $k=\pm k_0$, with $k_0=\sqrt{2m{E}_d}/\hslash$. For small nonzero $|V|$, the curves avoid crossing with minimum direct separation $2|V|$. The lower branch follows the parabola below the crossing and flattens toward $E_d$ above it; the upper does the opposite. The lower branch satisfies $E_{-}<E_d$ everywhere and tends to $E_d$ from below as $k\to \infty$, while the upper branch always lies above $E_d$. This is the perfectly flat-band continuum approximation, with zone-boundary intersections excluded as requested.

Fixed particle number and the larger Fermi sphere. Denote the uncoupled broad-band Fermi radius by $k_0$ to distinguish it from the new one $K$. With two spin states, the broad-band number density at $V=0$ is $k_0^3/(3{\pi }^2)$. The stipulated equal population in the flat sector makes the total density twice that value. At zero temperature and any nonzero hybridization, all occupied states fit into the monotonic lower branch below $E_d$, while the upper branch is empty. ThereforeThis is the stated factor increase. It uses fixed total particle number, equal spin counting, an exactly flat unhybridized band and enough available states below any Brillouin-zone restriction. In generic pairs of dispersive bands both hybridized pockets can remain occupied; then Fermi-volume conservation in two hybridized electron bands fixes a sum of volumes rather than this one radius. The exactly flat band is degenerate at $V=0$, so its initial occupation must be specified, as it is here.

Fermi-velocity suppression. Set $\Delta (k)=E_1(k)-E_d$. The exact derivative of the lower branch isAt $k=K$, $E_1(K)=2^{2/3}{E}_d$ and $\Delta (K)=(2^{2/3}-1)E_d>0$. For $|V|\ll \Delta (K)$, the square bracket divided by two is $|V{|}^2/\Delta (K{)}^2+O(|V{|}^4/\Delta (K{)}^4)$. Since group velocity is ${\hslash }^{-1}dE/dk$ and the broad-band reference velocity is $v_F^{(1)}=\hslash k_0/m$,so the flat-band hybridization Fermi-velocity suppression isThe reference velocity is that of the original broad-band Fermi sphere, not the broad-band slope at the larger $K$; omitting the radius change loses the factor $2^{1/3}$.

The low-temperature qualification becomes more stringent as $V\to 0$: the new chemical potential approaches $E_d$ with $E_d-\mu \simeq |V{|}^2/\Delta (K)$, and the relevant thermal energy must be small compared with this hybridization scale. At fixed nonzero temperature the singular flat-band redistribution is not automatically resolved for arbitrarily small coupling. A small but finite narrow-band dispersion can also change the occupation topology.

Density of states and magnetism. For an isotropic three-dimensional branch with spin degeneracy two,A small Fermi velocity therefore produces a large density of states. The localized, partially filled d-derived bands in iron, cobalt and nickel place many states close to the Fermi level; broad s/p-derived conduction bands in copper, silver and gold coexist with filled d bands lying below it. This explains the substantially larger Fermi-level density of states of the former group within the band picture. An exact factor of ten requires the actual band parameters and is not fixed by the two-band toy model.

A high density of states lowers the kinetic-energy cost of spin polarization. A repulsive exchange interaction can then satisfy the Stoner criterion and stabilize ferromagnetism. With a single-spin density of states $g_s$ and an on-site interaction normalization, the criterion is $U{g}_s(E_F)>1$; with spin-summed $g=2g_s$, it is $Ug(E_F)>2$. The large d-state contribution favors this instability in Fe, Co and Ni. The noble metals do not satisfy the corresponding instability condition in the same ordinary state; this does not imply that they have no weak magnetic response whatsoever.

Adopt the usual orthogonal tight-binding approximation, so the three localized spatial states are orthonormal. Each site has two neighbors, with the same real hopping matrix element. For coefficients $c_n$, the Schrödinger equation at site $n$ isThe constant-coefficient state has $c_1=c_2=c_3$, giving energy $E_0+2t$. Its normalization therefore yieldsTo verify that it is the ground state, the three discrete ring momenta are $\theta =0,\pm 2\pi /3$, with $c_n\propto e^{i\theta (n-1)}$ and energies $E_0+2t\cos \theta$. The other two states both have energy $E_0-t$. Since $t<0$, $E_0+2t<E_0-t$ and the gap is $3|t|$. This is the three-site tight-binding ring spectrum.

The spatial ground state is unique and has equal probability $1/3$ on every site; its common overall phase is arbitrary. Spin would supply a separate twofold degeneracy when no spin-dependent terms are included, but does not change the requested spatial state. If the localized functions were instead nonorthogonal raw atomic orbitals, their overlap matrix would be additional data needed for normalization and a generalized eigenvalue problem.

Two atomic orbitals. In the linear combination of atomic orbitals approximation, use $|\psi ⟩=c_A|A⟩+c_B|B⟩$, where the two basis functions are localized atomic orbitals. Projecting $\hat{H}|\psi ⟩=E|\psi ⟩$ onto each orbital gives the generalized eigenvalue problemHere the scalar overlap in the matrix is $S=⟨A|B⟩$, while $H_{ab}=⟨a|\hat{H}|b⟩$. Normalization requires $c^{†}Sc=1$, with $S$ now denoting the full overlap matrix. For equivalent real orbitals with diagonal energy $E_a$, real hopping $t$, and real overlap $s$, the eigenstates and energies areFor the usual negative hopping convention, the symmetric combination is the lower bonding orbital and the antisymmetric combination is the higher antibonding orbital. In an orthonormal approximation, $s=0$ and the result reduces to a two-by-two Hermitian matrix eigenproblem.

Hybridized bands. At each $\mathbf{k}$, take the two Bloch basis states to be orthonormal, as implicit in the stated band model, and write ${ϵ}_k={\hslash }^2{k}^2/(2m)$. The Hamiltonian isFor $|{\psi }_{\mathbf{k}}⟩=a_{\mathbf{k}}|{\varphi }_{\mathbf{k}}⟩+b_{\mathbf{k}}|{\chi }_{\mathbf{k}}⟩$, the eigenvalue condition isso $(E-{ϵ}_k)(E-E_d)-|V{|}^2=0$. Thus band hybridization givesFor nonzero $V$, a convenient explicit normalized pair of eigenstates isA phase convention can make $V$ real, but the eigenvalues depend only on its modulus. These are exact within the truncated two-band model, and approximate as a description of a material.

Dispersions and limits. For $V=0$, the parabola crosses the flat band atFor $V\ne 0$ the crossing is avoided, with a direct hybridization gap $2|V|$ at $k_0$. The lower and upper branches exchange orbital character through this avoided crossing. At the origin their energies areso for $|V|\ll E_d$ the small-$k$ expansions areAt large $k$, where ${ϵ}_k-E_d\gg |V|$,The lower branch approaches $E_d$ from below and becomes flat; the upper branch approaches the free-electron parabola from above. At zero coupling the unsorted orbital bands remain exactly flat and parabolic. Sorting them by energy instead gives $E_{-}=\min ({ϵ}_k,E_d)$ and $E_{+}=\max ({ϵ}_k,E_d)$, which cross at $k_0$ without mixing.

Fermi wavevector and velocity. A partially occupied perfectly flat band has chemical potential $E_d$. Thus before mixing, the occupied broad band has radius $k_0$ defined above. For spin degeneracy two, its particle density is $n_1=k_0^3/(3{\pi }^2)$; equal population of the flat sector gives total density $n=2n_1$.

For any nonzero mixing, $E_{-}(k)<E_d$ at finite $k$, while $E_{+}(k)>E_d$. Moreover $E_{-}(k)$ increases monotonically. At zero temperature and this fixed density, a single occupied sphere in the lower branch therefore accommodates all particles, and the upper branch is empty. Fermi-volume conservation in two hybridized electron bands givesThe new radius is about $26\%$ larger than the old broad-band radius. The strictly flat uncoupled band has no uniquely determined Fermi radius of its own; its prescribed population is what enters the count.

The radial group velocity is $v_{-}(k)={\hslash }^{-1}d{E}_{-}/dk$. Differentiating the exact dispersion givesAt $k_F$, put $\delta =2^{2/3}-1$, so ${ϵ}_{k_F}-E_d=\delta E_d$. Relative to the original broad-band Fermi velocity $v_F^{(1)}=\hslash k_0/m$,Expanding for $|V|\ll E_d$ gives the flat-band hybridization Fermi-velocity suppressionThe coefficient is about $3.65$. The enlarged occupied volume coexists with a small velocity because the new Fermi surface lies on the almost-flat part of the lower branch. This result assumes fixed particle density, sufficient available states before any Brillouin zone boundary, and temperature low enough to resolve energy differences of order $|V{|}^2/E_d$. The limit $V\to 0$ at a fixed nonzero temperature need not give this sharp redistribution. Giving the narrow band a finite dispersion can also leave both pockets occupied, so the result is specific to the ideal model.

Use one orthonormalized localized atomic orbital $|j⟩$ at each position $ja$. In the nearest-neighbour tight-binding model, translation invariance gives the on-site energy and real hopping integralThe Bloch state isSince $H|j⟩=E_0|j⟩+t|j-1⟩+t|j+1⟩$, shifting the two summation indices givesThe hopping integral is an energy, and its sign sets whether $k=0$ is a minimum or maximum. $E_0$ is the diagonal matrix element in the crystal Hamiltonian, which can differ from an isolated-atom energy.

The orthogonal tight-binding approximation matters. If raw neighbouring orbitals retain a nonzero orbital overlap $s$ in their normalization, projection instead gives a generalized eigenvalue problem and the nonorthogonal nearest-neighbour tight-binding band $E(k)=[E_0+2h\cos (ka)]/[1+2s\cos (ka)]$. The stated cosine band follows after orthonormalization and nearest-neighbour truncation, or after neglecting that overlap metric while retaining the hopping it motivates.

Choose one atomic orbital ${\varphi }_A$ and ${\varphi }_B$ on the two atoms and approximate a molecular orbital by the linear combination of atomic orbitalsProjecting the Schrödinger equation onto the two orbitals gives a generalized eigenvalue problem $H\mathbf{c}=ES\mathbf{c}$, where $H_{ij}=⟨{\varphi }_i|\hat{H}|{\varphi }_j⟩$ and $S_{ij}=⟨{\varphi }_i|{\varphi }_j⟩$. In an orthonormal basis the matrix isThe coupling mixes the atomic states and produces a lower bonding molecular orbital and upper antibonding molecular orbital. For equivalent real orbitals with overlap $s$ the normalized combinations are $({\varphi }_A\pm {\varphi }_B)/\sqrt{2(1\pm s)}$; the precise energy ordering follows from the Hamiltonian matrix elements. This restricted orbital basis is an approximation to the molecule's wavefunction, rather than a claim that exact isolated-atom orbitals remain unchanged in the molecule.

The equation requiring a homogeneous matrix system for an eigenvector to have a nonzero solution. In an orthonormal basis, $(H-EI)c=0$ requires $\det (H-EI)=0$. For a nonorthonormal basis with overlap matrix $S$, it becomes $\det (H-ES)=0$, a generalized eigenvalue problem. The nearly-free electron model uses this condition on a resonant plane wave subspace.