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The potential has period $2\pi$, so the first Brillouin zone is $-1/2\le k<1/2$. At its boundary the free-particle statesare degenerate, with energyThe relevant Fourier coefficient is supplied by $V_0\cos x$:The term $-V_0\cos 2x$ changes wavevector by two and has no matrix element within this degenerate pair. Thus degenerate perturbation theory gives the matrixand the two band-edge energiesThe nearly-free electron model therefore predicts the lowest band gap

Let $G_n=2\pi n/a$. The free states $|k⟩$ and $|k-G_n⟩$ are degenerate at the Bragg point $k=G_n/2=n\pi /a$. Sincedegenerate perturbation theory restricts the Hamiltonian near this crossing toWriting $k=n\pi /a+\kappa$ and diagonalizing gives the nearly-free electron dispersion near a one-dimensional band gapThus the two continuous bands avoid crossing, and at $\kappa =0$ their separation isThis is the nearly-free electron model. The dispersion relation is the relation $E=E(k)$ between energy and Bloch wavevector.

For the specified potential,Henceand all other nonconstant Fourier coefficients vanish. The gaps are thereforeTheir centre energies are respectivelyThe extended-zone sketch consists of the free-particle parabolas shifted upward by $V_0$, with avoided crossings of these two widths at the listed Bragg points; elsewhere they meet to this order because the corresponding Fourier coefficients vanish.

The degenerate $E_0=9/2$ subspace has ordered basisUsing the ladder actions,Thus degenerate perturbation theory asks us to diagonalizeIts normalized eigenvectors and first-order energies areThere is no order-$\lambda$ admixture from other unperturbed levels.

In fact, $H^{\prime }$ preserves $n+2m$, so this two-dimensional eigenspace of $H_0$ is invariant. Directly,Thereforefor every real $\lambda$, with no omitted higher-order terms. This is exact diagonalization within a degenerate subspace, specifically the resonant two-to-one oscillator coupling.

For $H_0=E_2{B}_2^{†}{B}_2$, the eigenspaces of energies $0$ and $E_2$ are each two-dimensional because $n_1$ is free. Degenerate perturbation theory requires diagonalizing $E_1{B}_1^{†}{B}_1$ within each space. It is already diagonal in the product basis, with eigenvalues $0,E_1$. Therefore the first-order eigenvectors remainand their energies are $0,E_1,E_2,E_2+E_1$, respectively. These are the exact answers, so all higher corrections vanish.

Within each degenerate unperturbed eigenspace of fixed $n_2$, the perturbation has matrixIts eigenvectors are $|+{⟩}_1$ and $|-{⟩}_1$, with first-order shifts $+g$ and $-g$. Thus degenerate perturbation theory gives exactly the states and energies found in part (i); again, all higher corrections vanish because the perturbation commutes with the second-oscillator Hamiltonian and is exactly diagonalized within each degenerate block.

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.

mass as band curvature. In a periodic potential, a weakly forced electron is usefully described by a wave packet confined to one energy band. Its group velocity isDifferentiating the velocity givesThis effective-mass tensor describes the band response, not a change of the electron's bare rest mass. Along a principal direction, a steep upward curvature gives a small positive mass and a flatter curvature a larger one. Near a band maximum the electron curvature and effective mass are negative; the corresponding missing-electron description uses positive electron hole masses. Anisotropy produces distinct principal masses, and strong nonparabolicity makes them momentum dependent. The single-band semiclassical description assumes fields weak and slowly varying enough to avoid substantial interband transitions.

Lattice symmetry and Bloch's theorem. Use the active spatial translation operator $T_{\mathbf{a}}\psi (\mathbf{r})=\psi (\mathbf{r}-\mathbf{a})$. A lattice-periodic scalar potential satisfies $V(\mathbf{r}+\mathbf{a})=V(\mathbf{r})$, while the kinetic term is translation invariant. Therefore $[H,T_{\mathbf{a}}]=0$. The lattice translations also commute with one another, so they can be diagonalized simultaneously with $H$, including within a degenerate energy subspace.

For a common eigenstate, write $T_{\mathbf{a}}\psi =t(\mathbf{a})\psi$. Unitarity gives $|t(\mathbf{a})|=1$, and the group law gives $t(\mathbf{a}+\mathbf{b})=t(\mathbf{a})t(\mathbf{b})$. On primitive Bravais vectors ${\mathbf{a}}_j$, choose $t({\mathbf{a}}_j)=e^{-i\mathbf{k}\cdot {\mathbf{a}}_j}$. Every lattice vector is an integer combination of these generators, henceThe vector $\mathbf{k}$ is defined modulo a reciprocal lattice vector. The band index labels the different energy eigenstates with the same translation character. Define $u_{n\mathbf{k}}(\mathbf{r})=e^{-i\mathbf{k}\cdot \mathbf{r}}{\psi }_{n\mathbf{k}}(\mathbf{r})$. The phase factors cancel under a lattice translation, proving Bloch's theorem in its usual form:This is the Translation-character proof of Bloch theorem; the commutator guarantees a basis of such states, even when an arbitrary superposition of different translation characters is not itself a translation eigenstate.

Periodic Hamiltonian. Acting once with momentum gives $\mathbf{p}(e^{i\mathbf{k}\cdot \mathbf{r}}u)=e^{i\mathbf{k}\cdot \mathbf{r}}(\mathbf{p}+\hslash \mathbf{k})u$. Applying it again and cancelling the common plane-wave factor yieldsThe Bloch Hamiltonian acts on lattice-periodic functions in a unit cell. Expanding its square separates a zero-momentum Hamiltonian, a term linear in $\mathbf{k}$ and an identity term quadratic in $\mathbf{k}$.

Expansion about the zone centre. Let $|n⟩=u_{n0}$, $E_n={\epsilon }_n(0)$, and normalize all these states using a unit-cell inner product. For the real scalar Hamiltonian in the question, a nondegenerate zone-centre eigenfunction can be chosen real. Thenbecause opposite unit-cell faces cancel. Equivalently, time-reversal symmetry forbids a nonzero electric current in this nondegenerate spinless state. This explains why no linear energy term appears; it should not simply be omitted without a symmetry or extremum argument.

The first-order energy shift is therefore ${\hslash }^2{k}^2/(2m)$. In the second-order sum the quadratic term is proportional to the identity and has no off-diagonal matrix elements. Only the linear momentum vertex contributes. Second-order nondegenerate perturbation theory gives the momentum matrix-element expansion of band energyHermitian operator makes the two reciprocal matrix elements complex conjugates. All other periodic eigenstates at $\mathbf{k}=0$ enter the sum. The remainder is even because time-reversal symmetry gives ${\epsilon }_n(\mathbf{k})={\epsilon }_n(-\mathbf{k})$ near this isolated state. At a degeneracy the denominators signal the need for degenerate perturbation theory instead.

mass ratio and the printed formula. Write $p_{nl}^i=⟨n|p_i|l⟩$. Differentiating the quadratic energy yieldsFor a specified unit direction $\mathbf{e}$, the resulting scalar mass ratio isIn an isotropic three-dimensional band, averaging equivalent directions givesCoupling to lower bands contributes positively to inverse mass; coupling to higher bands contributes negatively. Their signs and matrix elements determine the actual curvature.

The original PDF prints $m^{\ast }={\hslash }^2[{\nabla }_{\mathbf{k}}^2{\epsilon }_n{]}^{-1}$. Taken literally as a three-dimensional Laplacian, this misses a factor of three for an isotropic scalar mass. The correct identities areThe printed expression would return $m/3$ even for the free dispersion ${\hslash }^2{k}^2/(2m)$. It is correct in one dimension, or if its second derivative is interpreted directionally. For completeness, its literal three-dimensional trace ratio would instead be $3+(2/m){\sum }_{l\ne n}|⟨n|\mathbf{p}|l⟩{|}^2/(E_n-E_l)$; that is the trace of the dimensionless inverse-mass tensor, not the usual scalar $m/m^{\ast }$.

In the nearly-free electron model, a weak periodic potential has Fourier components $V_{\mathbf{G}}$ that couple plane waves at $\mathbf{k}$ and $\mathbf{k}-\mathbf{G}$. Far from a degeneracy their admixture gives small perturbative energy shifts. At a Bragg plane, however,so degenerate perturbation theory is required. In one dimension the first such crossing is at $k=\pm G/2$, the boundary of the first Brillouin zone.

Within the degenerate pair the eigenvalues are ${ϵ}_{G/2}\pm |V_G|$. ThusThe eigenstates are standing-wave combinations of counterpropagating waves. Their charge densities sit differently relative to the ionic potential, giving different potential energies and removing the free-electron degeneracy. The resulting avoided crossing produces continuous allowed energy bands separated by a forbidden interval. A Fourier component or symmetry-allowed coupling is needed: geometry alone does not require every possible gap to be nonzero.

The unperturbed $n=2$ level is degenerate. The useful point is that a central perturbation is already diagonal within its usual orbital eigenbasis:Angular orthogonality removes the possible mixing of $2s$ and $2p$, and also the mixing of different magnetic sublevels. Therefore the matrix required by degenerate perturbation theory needs no further diagonalization: its diagonal entries are precisely the expectation values above. It would be incorrect to justify the calculation by claiming that the unperturbed spectrum was nondegenerate.

For $b=0$ the energy needs only $n$. For a nonzero central confinement term the energy generally needs $n,l$, but remains independent of $m$. The magnetic quantum number is still needed to specify an individual orbital eigenstate, rather than just its energy.

Put $X={\sum }_s{a}_s^{†}{b}_s$ and $Y=X^{†}$, and take real hopping amplitude $t$ and real $\mathcal{E}$. Write the anti-Hermitian ansatz with its ordering visible:Using $Yf(\Delta )=f(\Delta +2)Y$, its commutator with $\mathcal{E}$ isBoth terms equal their required hopping coefficients $-t$ ifThe shifted expression is then $f(\Delta +2)=t/[\mathcal{E}(\Delta +2)-\mathcal{E}(\Delta )]$. For a transition from difference $d$ to $d+2$, the left-ordered factor evaluates as $f(d+2)=t/[\mathcal{E}(d+2)-\mathcal{E}(d)]$, the final-minus-initial energy denominator.

The formula describes occupation-dependent virtual hopping on actual transitions with nonzero energy gaps. Perturbation theory also requires the hopping matrix element, including its bosonic occupation factors, to be small compared with that gap. The resonance obstruction to perturbative elimination at a zero denominator on a nonzero hop requires degenerate perturbation theory. Values on unreachable states whose hopping amplitude vanishes need not determine the generator.

For a divalent square Bravais lattice, let $E_X={\hslash }^2{\pi }^2/(2m_e{a}^2)$ and real periodic potential coefficients $V_{10}<V_{11}<0$. The leading two-wave edge levels are $E_X\pm V_{10}$. The four-wave corner levels are $2E_X+V_{11}\pm 2V_{10}$ and a doublet at $2E_X-V_{11}$. Comparing the lowest corner level with the upper edge level gives the displayed indirect band gap. The first spin-degenerate energy band can hold the two electrons per primitive unit cell; it becomes isolated when $-E_X-V_{11}-3V_{10}>0$ in this truncation. This follows from resonant degenerate perturbation theory, and higher plane waves can shift the threshold.

A hopping process through an intermediate configuration of different unperturbed energy, contributing to a low-energy effective description through perturbation theory. If an intermediate state costs $\Delta >0$, second-order nondegenerate perturbation theory gives a contribution $-|⟨a|T|0⟩{|}^2/\Delta$. Its amplitude is of order $t/\Delta$ and its energy correction of order $-t^2/\Delta$; a degenerate low-energy sector requires the corresponding degenerate perturbation theory before identifying the ground state.