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Band filling counts occupied one-particle states in an energy band. With spin degeneracy two, a crystal of $N$ primitive unit cells has $2N$ states in each orbital-derived energy band, as in state counting in a Bloch band. Full occupation therefore means two electrons per primitive cell in that band.

In the group velocity approximation, a fully occupied band contributes current densityThe integral vanishes because $E$ is periodic across opposite faces of the Brillouin zone; inversion symmetry is unnecessary.^[ref] A partially occupied band leaves nearby empty states available for ordinary longitudinal conduction. A filled set of bands with a positive band gap instead gives a band insulator at zero temperature.

The vanishing group velocity term does not exclude every electrical response. A filled Chern insulator band can have a transverse quantum Hall effect governed by Berry curvature.^[ref]

Choose orthonormal localized orbitals $|\mathbf{R}⟩$. The on-site energy is $E_0=⟨\mathbf{R}|H|\mathbf{R}⟩$, and the sign convention in the question defines $t(\mathbf{T})=-⟨\mathbf{R}|H|\mathbf{R}+\mathbf{T}⟩$. Bloch state diagonalizes this translationally invariant hopping problem. For real, inversion-symmetric nearest-neighbor amplitudes, $t(\pm a\hat{x})=t_x$ and $t(\pm b\hat{y})=t_y$, so the exponential pairs yieldThe first zone is the rectangle $|k_x|\le \pi /a$, $|k_y|\le \pi /b$, wider in $k_x$ because $a<b$. Use $\Gamma =(0,0)$, $X=(\pi /a,0)$, $Y=(0,\pi /b)$ and $M=(\pi /a,\pi /b)$.

For positive $t_y=t$ and $t_x=2t$, the four corner energies and the requested path areAlong a path parametrized in reduced coordinates $x=k_{x}a$, $y=k_{y}b$, the full curve is simply $E-E_0=-4t\cos x-2t\cos y$. In particular the two appearances of $\Gamma$ and the two diagonal traversals must have consistent endpoint energies.^[ref] A reversed hopping sign changes which point is the band minimum; the usual positive-hopping convention is stated explicitly here.

A magnetic field quantizes closed semiclassical orbits into Landau levels. Their passage through the Fermi energy modulates the density of states and scattering, producing Shubnikov-de Haas oscillations of resistivity. The fundamental frequency for a regular extremal orbit perpendicular to the field isThus two distinct fundamentals ordinarily identify two inequivalent extremal areas, rather than two arbitrary points or two spin directions automatically giving two different geometric areas.

The particle-hole equality of tetragonal extremal orbit areas is a special constraint of the exact half-filled nearest-neighbor band stated here. For $B\parallel z$, its regular closed endpoint sections are the electron contour at $k_{z}c=\pi$ and the hole contour at $k_{z}c=0$ described in b. Translation $(k_{x}a,k_{y}a)\mapsto (k_{x}a+\pi ,k_{y}a+\pi )$ changes the sign of both cosines and maps one enclosed pocket onto the other. Their areas are equal. Direct integration givesConsequently this bare model predictsnot two distinct regular fundamental frequencies. The intermediate $k_{z}c=\pi /2$ contour is a separatrix with a divergent cyclotron period, not another ordinary smooth extremal pocket to insert into the Onsager formula. The larger occupied-area slice at $k_z=0$ is the complement of a hole pocket on the Brillouin torus; counting that complement as a second enclosed planar orbit would give a spurious ratio.

For comparison, an additional small-$k_x,k_y$ parabolic approximation would replace the exact surface by $(k_{x}a{)}^2+(k_{y}a{)}^2\simeq 4+\cos (k_{z}c)$ and suggest a rough belly-to-neck ratio $5/3$. That approximation is uncontrolled at half filling and erases the actual topology; it cannot be asserted as an exact prediction of the supplied cosine band. The observation of genuinely different frequencies in a real material therefore needs further information, such as longer-range hopping, a departure from half filling, extra bands, symmetry breaking or a resolved spin-dependent effect. The universal area-ratio inference is valid, but a nonunit numerical ratio is not determined consistently by the complete stated band assumptions alone.^[ref]

Write the lattice reciprocal vector as $G=2\pi /a$ in one dimension. The nearly-free electron model starts from ${\epsilon }_k={\hslash }^2{k}^2/(2m_e)$ and couples plane waves separated by G. Near a zone boundary, the two-wave Hamiltonian isAt $k=G/2$ the free waves are degenerate and the periodic potential opens a gap $2|V_G|$. A reduced-zone drawing folds k modulo G into $[-G/2,G/2]$, displaying multiple bands at each reduced wavevector. The original plot shows free folded parabolas, avoided crossings and an allowed vertical optical transition. ^[ref]

Including two spin states, filling a free Fermi sphere gives $n=2(4\pi k_F^3/3)/(2\pi {)}^3=k_F^3/(3{\pi }^2)$. A monovalent BCC cell contributes two electrons, soThe sphere therefore lies inside every first-zone face, since the supplied distance is the nearest boundary.

For the shortest BCC reciprocal vector, $G=2\sqrt{2}\pi /a$. In the zero-potential folded-band limit, the energy of a vertical transition from k along G to the neighbouring free branch is ${\hslash }^2(G^2-2Gk)/(2m_e)$. It decreases over the occupied interval and is smallest at $k=k_F$. Thus the monovalent BCC direct optical threshold isThe final state is empty because $G-k_F>k_F$. Finite weak $V_G$ changes the energies and makes optical matrix elements nonzero; an exactly free electron gas by itself has no interband absorption. The expression is a leading weak-potential threshold, not a claim that every peak equals the onset.

Using the supplied lattice constants and bare-electron free bands gives $E_F=(3.24,2.12,1.85)\mathrm{eV}$ and thresholds $(2.06,1.35,1.18)\mathrm{eV}$ for Na,K,Rb. The source plot's broad higher-frequency features shift toward lower energy in the same order as a increases, qualitatively consistent with the predicted $a^{-2}$ scale. Their peak shapes, heights and exact positions need matrix elements, band corrections and scattering; low-frequency intraband conductivity is a separate contribution. No measured trace is reproduced.

Let the atomic spacing be $a$ and define the nearest-neighbour matrix elements $t_{\alpha }=⟨j,p_{\alpha }|H|j+1,p_{\alpha }⟩$, using the same translated orbital convention on every site. ThenFor the opposite sigma and pi hopping signs in a p-orbital chain, consider $p_z$ first: neighbouring lobes above the chain have the same sign, so the in-phase Bloch sum is bonding. Thus $t_{\pi }<0$ andFor $p_x$, the lobes facing one another have opposite signs for equal site coefficients. Bonding therefore needs alternating coefficients, giving $t_{\sigma }>0$ andEnd-on $\sigma$ overlap is normally stronger than side-on $\pi$ overlap, so the $p_x$ band is wider, with bandwidths $4|t_{\sigma }|$ and $4|t_{\pi }|$. The curve orientations require a stated hopping-sign convention; a Hamiltonian written with an explicit minus sign in front of the hopping uses correspondingly opposite parameter signs.^[ref]

For an electron wavepacket in one isolated band under a uniform electric field,Because $ϵ(k+2\pi /a)=ϵ(k)$, acceleration through successive Brillouin zones makes the velocity periodic rather than increasing indefinitely. For $ϵ={ϵ}_0-2J\cos (ka)$,for a positive field along the chosen axis. Integration givesThe excursion amplitude of a Bloch-oscillating wavepacket is $2|J|/(e|E|)$ about its mean and peak-to-peak extent $4|J|/(e|E|)$. The turning of velocity corresponds semiclassically to Bragg reflection near a zone boundary.

Coherent Bloch oscillations require scattering/dephasing times long compared with $T_B$ and a field weak enough to avoid substantial interband tunnelling. Ordinary metals usually scatter too rapidly; semiconductor superlattices and cold atoms in periodic optical potentials provide cleaner realizations. A completely filled band has canceling velocities, so its net current does not oscillate simply because individual wavepackets can.^[ref]

For a spin-degenerate band, each unit cell contributes two available states. An even electron count per cell can fill an integer number of bands; a band insulator additionally needs a global gap between the highest occupied and lowest unoccupied bands. Even filling alone is not sufficient when bands overlap.

For a free two-dimensional gas with spin degeneracy two,The reciprocal vectors of the periodic potential are $Q\hat{\mathbf{x}}$ and $Q\hat{\mathbf{y}}$. Its first Brillouin zone is the square $-Q/2\le k_x,k_y\le Q/2$, of side $Q$ and area $Q^2$; the real-space cell has side $a=2\pi /Q$.

At zero potential, the extended-zone Fermi surface is the circle of radius $k_F$. Since $\sqrt{2}<\alpha <2$, it crosses every square face but misses the corners. For Fermi-surface reconstruction in a weak square potential, weak positive potential opens gaps at the Bragg crossings: the second band has electron pockets at the two inequivalent edge centres $X,Y$, while unfilled regions of the first band form a hole pocket at the zone corner $M$. Opposite edges are identified: four edge half-pockets are two electron pockets, and four corner pieces are one hole pocket.

For the circular-cap carrier count in a square Brillouin zone, put $R=k_F$ and $s=Q/(2R)=\alpha /2$. Each circular cap outside a square face has areaNo two such caps overlap in the specified interval. The total second-band electron area is $A_e=4C$. The first-band hole area is the square area minus the occupied central-circle area, $A_h=Q^2-\pi R^2+4C$. Including spin,Thus at $\alpha =\sqrt{3}$,The carrier densities are the electron- and hole-pocket counts, not the total electron density. The carrier compensation and even band filling in a square lattice follows because their difference obeys $n_e-n_h=n-Q^2/(2{\pi }^2)$, so compensation occurs at $\alpha =\sqrt{\pi }$. These area formulas are the $V_0\to 0^{+}$ limit, before finite-gap reconstruction changes the pocket shapes.^[ref]

The filling per real-space cell is $\nu =n{a}^2=2\pi /{\alpha }^2$. In the given interval, the only even integer filling is $\nu =2$, again requiring $\alpha =\sqrt{\pi }$. At any other fixed filling a full gap cannot yield an ordinary spin-degenerate band insulator in this unchanged periodicity. In particular, the $\alpha =\sqrt{3}$ system retains excess electron carriers even after its holes disappear.

A rough gap threshold comes from comparing different symmetry directions, rather than only a direct zone-face gap. Define $E_Q={\hslash }^2{Q}^2/(2m)$. The zone-face two-plane-wave gap for a cosine potential follows because each cosine couples plane waves by $-V_0/2$, giving the one-dimensional zone-edge levels $E_Q/4\mp V_0/2$. The four-plane-wave splitting at a square-zone corner is the sum of two such pairs. The potential is separable in $x,y$, so to first orderThe indirect gap of a separable square cosine potential is consequentlyAt the compensated filling, this is $V_{0,\min }\sim (\pi /6)E_F^{(0)}$, with $E_F^{(0)}={\hslash }^2{k}_F^2/(2m)$. An insulator needs both the even filling and a positive indirect gap. At the touching threshold the gap is zero; a positive direct gap at every Bragg crossing is not by itself sufficient. The estimate neglects higher-plane-wave corrections and is not an exact critical amplitude.