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For a one-dimensional free-electron gas of length $L$, allowed wave vectors have spacing $2\pi /L$. Including two spin states and occupied wave vectors from $-k_F$ to $+k_F$,The ground-state kinetic energy is found by integrating rather than assigning every electron the Fermi energy:Charge neutrality gives $N_e=ZN$ because each positive ion has charge $+Ze$. Thus $K_e=ZN{E}_F/3$. Since the ion separation is $2R$, $L\simeq 2RN$ and $n_e=Z/(2R)$. These are one-dimensional number densities and state counts; using a three-dimensional $k_F\propto n^{1/3}$ relation here would change the model. The results concern the ground-state Fermi-gas contribution, neglecting electron interactions beyond the supplied electrostatic energy.

Using $c=2700\mathrm{m}{\mathrm{s}}^{-1}$ gives the electron-acoustic recoil lengthThe temperature ratio therefore fixesIt follows that $n=4/a^3=9.48\times 10^{28}{\mathrm{m}}^{-3}$, $k_F=1.4105\times 10^{10}{\mathrm{m}}^{-1}$ and $k_D=1.7771\times 10^{10}{\mathrm{m}}^{-1}$. Substitution givesThe resulting Fermi energy is about $7.58\mathrm{eV}$ and the temperature ratio is indeed $240$. These values follow from the supplied effective sound speed and free-electron/Debye approximations; they are not independent precision measurements of copper's parameters.

Pauli exclusion principle prevents most filled electronic states from changing occupation at $T\ll T_F$. Only a shell of width of order $k_{B}T$ around the Fermi energy participates. Since the density of states near that energy is of order $n/E_F$ per volume and energy, the fraction of active electrons is of order $k_{B}T/E_F=T/T_F$. Their typical change of energy is of order $k_{B}T$, so the thermal energy increment per volume scales asDifferentiating with respect to temperature gives the electronic heat capacityThe simple shell argument fixes the linear law, not its coefficient. A Sommerfeld expansion using the spin-summed free-electron density of states $g(E_F)=3n/(2E_F)$ gives $C_e=({\pi }^2/3)g(E_F)k_B^{2}T$, hence $A={\pi }^2/2$, consistent with the supplied value. The result excludes strong band-structure or interaction corrections and requires a degenerate electron population.

In the free electron model, conduction electrons occupy independent plane-wave states with energy ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)$ in a uniform mean background. Crystal-periodic band structure and explicit electron–electron correlations are omitted. Electrons still obey the Pauli exclusion principle and Fermi–Dirac statistics; free does not mean distinguishable classical particles. A scattering time can be added phenomenologically to describe transport.

The Fermi energy is the highest occupied single-particle energy at zero temperature, measured here from the bottom of the parabolic band. The Fermi temperature is the corresponding scale $T_F=E_F/k_B$, not the actual thermal temperature of the sample. Metallic electron densities are large, so filling all distinct states up to the required momentum gives Fermi energies of several electronvolts, equivalent to tens of thousands of kelvin or more. At ordinary temperatures, only a narrow region around the Fermi surface has appreciably changing occupations.

A room-temperature metal and its lattice are nevertheless at room temperature, not at $T_F$. High occupied kinetic energies are present even in the zero-temperature ground state and cannot all be liberated as heat without violating the allowed occupancy structure. A metal below skin temperature often feels cold because it transfers heat rapidly out of the skin; its thermal conductivity and thermal effusivity matter, not the value of its Fermi temperature. At the same actual temperature, a poorly conducting material can therefore feel warmer while neither material is thermodynamically hotter.

The density of states $g(E)$ counts one-electron states per energy interval, including both electron spin orientations. It may be a total or a per-volume quantity; $U$ must have the matching normalization. The Fermi-Dirac distributionis the occupation probability of such a state. In the fixed-volume free electron model, the integration variable $E$ and the density of states do not depend on $T$. The occupation depends explicitly on $T$ and implicitly through the chemical potential $\mu (T)$ needed to preserve electron number. Thermal expansion or a changing electronic band structure would require additional effects outside this model.

At $T\ll T_F$, only states within order $k_{B}T$ of the Fermi energy change their occupation. There are order $g(E_F)k_{B}T$ such states, each carrying an excitation energy of order $k_{B}T$. Thus $U(T)-U(0)\propto g(E_F)(k_{B}T{)}^2$ and the electronic heat capacity is proportional to $g(E_F)k_B^{2}T$. Using the supplied coefficient,The shell argument fixes the scaling, not the numerical coefficient; the controlled Sommerfeld expansion fixes that coefficient at fixed electron number. Filled states far below $E_F$ cannot accept arbitrary thermal excitations because of the Pauli exclusion principle.

The chemical potential is the change in Helmholtz free energy when particles are added at fixed temperature and volume:Equivalently, $dU=TdS-PdV+\mu dN$ identifies it as the derivative of internal energy at fixed entropy and volume. In equilibrium, systems that exchange particles have equal chemical potentials.

The Fermi energy is the zero-temperature chemical potential, or, for independent electrons, the highest occupied single-particle energy in the zero-temperature limit. Its numerical value depends on the chosen energy reference; the Fermi energy need not be measured from the bottom of a band unless that convention is specified.

Monovalency gives one electron per atom, so the two-spin-state band is half filled. At zero temperature, occupying $-k_F<k<k_F$ gives $2(Na/2\pi )(2k_F)=N$. HenceRelative to the band minimum, the Fermi energy is $2t$.

The density of states obeys $g({ϵ}_0+x)=g({ϵ}_0-x)$. At $\mu ={ϵ}_0$, the Fermi-Dirac distribution satisfiesPairing the energies above and below the band centre therefore leaves exactly half of the two available states per atom occupied, at every positive temperature. Increasing $\mu$ strictly increases the mean particle number, so this solution at fixed half filling is unique. ConsequentlyThis particle-hole symmetry of a half-filled bipartite band argument uses the symmetric tight binding band and fixed half filling; it is not a general temperature-independence property of chemical potential in metals with additional bands or an asymmetric density of states.

For a nondegenerate two-dimensional electron gas, the two quadratic momentum terms supply $k_{B}T/2$ each by the equipartition theorem. There is no lattice restoring potential energy for a free electron. ThusOn cooling below the Fermi temperature, the Pauli exclusion principle restricts the thermally active electrons to within approximately $k_{B}T$ of the Fermi energy. The linear low-temperature heat capacity of a two-dimensional Fermi gas isThe heat capacity therefore decreases from its classical value as quantum degeneracy becomes important. The particle number of electrons and sample area are held fixed when differentiating the internal energy.

For the degenerate electron gas, let $D_g=Am/(\pi {\hslash }^2)$ include both spin states. The Sommerfeld expansion givesFor the gapless symmetric intrinsic semiconductor, part (d) gave the same coefficient with $D_s=A{m}^{\ast }/(\pi {\hslash }^2)$. Thus for equal sample areas and equal spin and valley degeneracies,The equality often intended in this comparison requires the same density of states. The source sets $m_e^{\ast }=m_h^{\ast }$ but does not explicitly equate them to the electron gas mass $m$. Without that additional assumption the ratio is $m^{\ast }/m$, or more generally $D_s/D_g$ if sample areas or degeneracy factors differ. For example, $m^{\ast }=2m$ with equal areas satisfies the stated equal-effective mass condition in the semiconductor and gives ratio two. The filled Fermi sea does not add a factor $N_e/N_c$: only the thermal shell near its Fermi energy contributes to its heat capacity.

Use the stipulated noninteracting impurity band approximation, with one donated electron per donor and two available spin states per donor orbital. The energy band is half filled. Approximating its spin-summed density of states per volume near the Fermi energy by $D_v(E_F)\simeq 2n/W$ gives the zero-temperature Pauli paramagnetismWith $\alpha =1$,A factor of order one depends on the density of states shape; using the conduction band $g$-factor rather than silently setting $g=2$ matters here.

To locate the comparison with the magnitude of the undoped Larmor-Langevin susceptibility, put $x=n^{-1/3}/a_b>1$. Numerically,For the stated atomic-scale estimate $R=a_0$ in part (b), equality with $|{\chi }_0|=2.599\times 10^{-5}$ requiresThus the dilute-branch crossover iswell below $a_b^{-3}=9.45\times 10^{23}{\mathrm{m}}^{-3}$. Within this model the positive spin response exceeds the negative background magnitude for lower dopant densities on this dilute branch: the exponential narrowing of $W$ wins over the loss of donors. Indeed $d\ln (e^x/x^3)/dx=1-3/x>0$ for $x>3$. This is not a claim of ferromagnetism.

The degenerate Pauli paramagnetism calculation requires $k_{B}T\ll W$, with an unimportant order-one filling factor. At the estimated crossing,The weak-field condition also requires $|g|{\mu }_{B}B\ll W$. For a different assumed atomic $R$, use $|{\chi }_0|=2.599\times 10^{-5}(R/a_0{)}^2$ in the displayed equation for $x$. An order-of-magnitude background $|{\chi }_0|\sim 10^{-5}$ gives the same density order and a crossover scale of order $10^{-5}\mathrm{K}$. At higher temperature than this extremely narrow bandwidth permits, the $n/W$ enhancement cannot be used; the independent-donor Curie law is the appropriate thermal scale if interactions remain negligible. At very small dopant density, disorder, localization and Coulomb interactions can also invalidate the assumed noninteracting impurity band. The formal divergence as $n\to 0$ is therefore a limit of the stated zero-temperature band model, not a prediction for isolated donors at a fixed nonzero temperature.

One orbital per primitive cell supplies two states per wave vector when spin is included. A monovalent metal therefore half-fills this band. Nearest-neighbor hopping on the bipartite rectangular lattice has $E(\mathbf{k}+\mathbf{Q})=2E_0-E(\mathbf{k})$ for $\mathbf{Q}=(\pi /a,\pi /b)$, fixing the half-filled Fermi energy atFor $t_x=2t_y>0$, the two Fermi surface obeyThey are corrugated open sheets in an extended-zone description, with $|k_{x}a|$ ranging from $\pi /3$ to $2\pi /3$. They are not a small circular pocket centered at the band bottom. The same $\mathbf{Q}$ maps one Fermi line onto its nested partner. A density-wave or Peierls-like modulation at that wave vector can open additional gaps at the Fermi energy. The unperturbed single-orbital band does not acquire those new gaps without a symmetry-breaking modulation or another coupling; ordinary lattice Bragg effects are already built into its dispersion.

For the three-dimensional tetragonal version set $t_x=t_y=t$, $t_z=t/2$, $a=b<c$. The corresponding half-filled surface isIt is a corrugated periodic surface, not a sphere. At $k_z=\pi /c$ its regular closed contour is electron-like around $\Gamma$, with $\cos (k_{x}a)+\cos (k_{y}a)=1/2$; at $k_z=0$ the closed contour is hole-like around $M$, with the sum $-1/2$. The topology switches through the square-lattice separatrix at $k_{z}c=\pm \pi /2$. Drawing one simple weakly warped cylindrical pocket across every section misses that feature of the half-filled cosine band.

Angle-resolved photoemission can directly map the dispersion and occupied Fermi surface. Angular quantum-oscillation measurements test its closed extremal cross sections and cyclotron masses; momentum-density probes such as positron annihilation or Compton scattering give a complementary bulk test. The different techniques constrain different features rather than merely providing interchangeable pictures.

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]

Periodic boundary conditions give wavevectors $(2\pi /L)(n_x,n_y)$. In the continuum, large-box limit each state occupies area $(2\pi /L{)}^2$ in wavevector space. A circular annulus therefore contains $[L^2/(2\pi {)}^2]2\pi kdk$ spinless states. Since $ϵ={\hslash }^2{k}^2/(2m)$, the density of states isAt finite $L$ the exact spectrum is discrete; this formula is its continuum density.

With electron spin the two Zeeman branches are ${ϵ}_{\mathrm{kin}}\mp \mu H$. The requested calculation includes this spin coupling and neglects orbital coupling to the field; for a perpendicular field, Landau levels would change the result. It is appropriate for a field parallel to the ideal two-dimensional plane, or for the stated spin-only model. Taking $\mu H\ge 0$, the lower branch starts at $-\mu H$, and the upper one at $\mu H$, soConstruct the ground state by filling the lowest one-electron levels subject to the Pauli exclusion principle. If the Fermi energy ${ϵ}_F$ exceeds $\mu H$, the two populations are $N_{+}=g_0({ϵ}_F+\mu H)$ and $N_{-}=g_0({ϵ}_F-\mu H)$. ConsequentlyThe ground-state energy, including the Zeeman term, follows by integrating the energy over each occupied branch:

After its giant phases, the Sun ejects much of its envelope and leaves a carbon–oxygen white dwarf. The remnant mass lies below the Chandrasekhar limit. Its support is chiefly electron degeneracy pressure: the Pauli exclusion principle forces electrons into distinct momentum states even at low temperature, giving $P\propto {\rho }^{5/3}$ when they are nonrelativistic. Consequently pressure support need not cease when nuclear burning ends. The white dwarf subsequently cools and fades.

The process called inverse beta decay here is electron capture, $e^{-}+p\to n+{\nu }_e$, becomes possible for free particles when the electron's total energy exceeds approximately $(m_n-m_p)c^2\simeq 2.6m_e{c}^2$, corresponding to a kinetic energy about $1.6m_e{c}^2$. Electron capture on nuclei has thresholds modified by nuclear binding energies. In a solar-mass progenitor's ordinary white dwarf remnant, the Fermi energy is below the relevant capture thresholds, so the electron population providing support is stable. At much higher density, electron capture can reduce the electron density and pressure and promote collapse; this is why stability against electron capture matters in assessing degeneracy support. The Sun ends as a cooling white dwarf, rather than a neutron star.

The chemical potential is $\mu =(\partial E/\partial N{)}_{S,V}$, equivalently $(\partial F/\partial N{)}_{T,V}$: it is the free-energy cost of adding a particle at fixed temperature and volume. With units $k_B=1$, a large reservoir has entropy $S_R(E_{\mathrm{tot}}-E,N_{\mathrm{tot}}-N)=S_R^0-E/T+\mu N/T$ to first order. Since the number of compatible reservoir states is proportional to its exponential, the grand canonical ensemble assignsFor independent fermions, a one-particle state of energy $\epsilon$ has occupation zero or one. Its partition factor is $1+e^{-(\epsilon -\mu )/T}$, so the mean occupation is the Fermi-Dirac distribution $[e^{(\epsilon -\mu )/T}+1{]}^{-1}$. Multiplication by the given density of states yields $dN=C{\epsilon }^{1/2}d\epsilon /[e^{(\epsilon -\mu )/T}+1]$.

At zero temperature the occupation is a step function, filled below $\mu$ and empty above it, so $\mu ={\epsilon }_F$ and $N=(2C/3){\epsilon }_F^{3/2}$. At fixed particle number, the low-temperature shift of $\mu$ is $O(T^2/{\epsilon }_F)$ and does not affect the leading estimate. In the narrow interval of width $T$ above ${\epsilon }_F$, replace the density of states by $C\sqrt{{\epsilon }_F}$ and put $u=(\epsilon -{\epsilon }_F)/T$. ThenEquivalently, the fraction above the Fermi energy is asymptotic to $(3\log 2/2)T/{\epsilon }_F$.

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.

Apply the preceding temperature profile at the envelope base. Introduce the dimensionless parameterThen $R/r_b-1=q$, so the geometrical thickness of a white-dwarf radiative envelope isThe smallness to be established is therefore $\mathcal{R}{T}_c\ll GM/R$, up to the numerical factor $17/4$.

That ordering follows from the cold, strongly electron-degenerate white dwarf model. Throughout the mass-bearing core, $k{T}_c\ll E_F$, where $E_F$ is the Fermi energy. For nonrelativistic electron degeneracy pressure,The stellar hydrostatic equation gives a characteristic core $P_{\mathrm{deg}}/\rho$ of order $GM/R$. Consequentlyfor fixed composition with order-one molecular-weight ratios, and $s\ll R$. Equivalently, the matched base mass density is much lower than a characteristic core mass density, since ${\rho }_b/{\rho }_{\mathrm{core}}\sim [\mathcal{R}{T}_c/(GM/R){]}^{3/2}$. The degeneracy ordering applies in the bulk core; at the matching point degeneracy is becoming marginal.

The inequalities $T_c\gg T_{\mathrm{eff}}$ and $k{T}_c\ll E_F$ express different facts. The first justifies ignoring surface-temperature terms, while the second makes the envelope geometrically thin. If the core were hot enough that $q$ were not small, the exact expression $s/R=q/(1+q)$ would not justify a thin-layer approximation.

Bosons have symmetric many-particle states, so any number may occupy the same one-particle state. At low temperature, an ideal Bose gas therefore places a macroscopic population in its lowest one-particle state; below the critical temperature this is Bose-Einstein condensation.

Fermions have antisymmetric many-particle states and obey the Pauli exclusion principle, so each one-particle state can hold at most one fermion of each available internal label. At zero temperature, a noninteracting Fermi gas fills distinct states in increasing order of energy up to the Fermi energy. Its many-particle ground state is thus the filled Fermi sea.

At zero temperature the Fermi-Dirac distribution is the indicator of $0\le \epsilon <E_F$. HenceSolving for the Fermi energy gives

For a parabolic two-dimensional electron gas with one valley and two spin states, the density of states per area is $g_{2D}=m^{\ast }/(\pi {\hslash }^2)$. Integrating from the subband edge to the zero-temperature Fermi energy gives $n_s=g_{2D}{E}_F$, henceHere $m^{\ast }=0.067(9.109\times 10^{-31})=6.103\times 10^{-32}\mathrm{kg}$, and the energy is measured from the occupied subband bottom. Equivalently $k_F=\sqrt{2\pi n_s}=1.3722\times 10^8{\mathrm{m}}^{-1}$. ThereforeThe spin factor is already included in $g_{2D}$; counting it again would halve the energy incorrectly.