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In the spin-degenerate free electron model, a monovalent metal has conduction-electron number density $n$. The density of wave-vector states is $1/(2\pi {)}^3$ per unit real-space volume, with two spin states at each wave vector. Filling a sphere at zero temperature givesFor the Debye model, there are three acoustic polarizations and $3n$ modes per unit volume in total. Replacing their dispersions by $\omega =ck$ with the effective Debye sound speed, mode counting givesThe factor two in the electron count is spin, whereas the factor three in the phonon count is polarization; the two counts are not interchangeable. If branch speeds differ, the Debye effective average obeys $c^{-3}=(c_l^{-3}+2c_t^{-3})/3$ in the isotropic approximation.

Dividing gives $T_F/{\theta }_D=\lambda (3{\pi }^{2}n/2{)}^{1/3}$, where $\lambda =\hslash /(2m_{e}c)$. A monatomic FCC conventional cell contains four atoms, so $n=4/a^3$. ThereforeHere and below $c$ is the speed of sound, not the speed of light. The Debye cutoff counts acoustic modes in a spherical approximation to the Brillouin zone; it is not an electron-state cutoff.

The inverse-susceptibility curve has a Curie–Weiss form above $T_C$ and vanishes on approach to it. The material can have spontaneous magnetization for $T<T_C$, including the zero-temperature limit, whereas for $T>T_C$ its zero-field equilibrium state is paramagnetic. At a continuous transition the spontaneous moment tends to zero at $T_C$; divergence of differential susceptibility there is not a finite saturated permanent moment.

A simple Weiss molecular field model takes the effective internal field intensity to be $H_{\mathrm{eff}}=H+\Lambda M$, with positive coupling $\Lambda$. Let $C$ be the independent-moment Curie constant. Linearizing the single-ion response on the high-temperature side givesThis explains the straight upper branch and its positive temperature intercept. At $T_C$ the zero-field linearized response diverges; finite field, finite size and critical fluctuations limit its use. For a spin-only two-state mean-field model the full equation is $m=\tanh [(T_C/T)m+{\mu }_B{\mu }_{0}H/(k_{B}T)]$, with $m=M/(N{\mu }_B)$; below $T_C$ it has nonzero zero-field solutions.

An ordering scale near $1000\mathrm{K}$ corresponds to $k_B{T}_C\simeq 0.086\mathrm{eV}$, about four orders of magnitude larger than the preceding dipolar pair scale. Exchange interaction can supply such energies: the Pauli antisymmetry of overlapping electron states changes their spatial correlations and hence Coulomb and kinetic energies according to their spin configuration. An effective ferromagnetic coupling can favor parallel moments. It is not simply the classical magnetic field of a neighbouring dipole. For example, with the convention $\mathcal{H}=-J{\sum }_{⟨ij⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$ and $J>0$, the localized-spin mean-field scale is $k_B{T}_C=zJS(S+1)/3$; this illustrates why an atomic electronic interaction, rather than a bare dipolar field, can produce a high transition temperature.

A bulk ferromagnet generally separates into magnetic domains. Within each domain many moments align and the local magnetization can be large, but different domain orientations partially or almost completely cancel the vector sum measured for the entire specimen. This reduces its external stray field and magnetostatic energy. Formation of magnetic domain walls costs exchange and anisotropy energy, so the actual pattern balances these costs with the reduction in stray-field energy.

A strong applied field can move domain walls and rotate moments toward a nearly saturation magnetization. When the field is removed, pinning and the sample's magnetic history determine the remanent moment. Thus a small net moment does not show that local ferromagnetic order has disappeared, or that moments within each domain are all randomly oriented. At finite temperature, thermal spin fluctuations also reduce the local ordered moment below its zero-temperature maximum even in a single-domain specimen. These two reductions—between-domain cancellation and within-domain thermal reduction—should be distinguished.

Take two localized spin orientations with moments $\pm {\mu }_B$ along the applied magnetic field. This assumes an electron-like spin $g$ factor of two. Their energies are $\mp {\mu }_{B}B$, so the one-moment partition function is $Z_1=2\cosh x$, with $x={\mu }_{B}B/(k_{B}T)$. The thermal mean moment is $k_{B}T{\partial }_B\ln Z_1={\mu }_B\tanh x$, and the spin-one-half paramagnetic magnetization isFor $|x|\ll 1$, $M\simeq n{\mu }_B^{2}B/(k_{B}T)$. With the same SI convention $\chi =\partial M/\partial H$ and $B\simeq {\mu }_{0}H$,This Curie law is a zero-field or weak-field susceptibility. At arbitrary field the differential response is $(C/T){\mathrm{sech}}^{2}x$, and $M$ saturates at $n{\mu }_B$; extending the linear law to unlimited field would contradict that finite saturation magnetization.

Within the stipulated localized, one-moment-per-atom spin model, the number density follows directly from mass density and molar mass:The face-centered cubic lattice is consistent with four atoms per conventional unit cell; no additional factor four is needed once $n$ has been obtained from the density. The Curie constant is $C=0.71549\mathrm{K}$. Identifying the model's $\theta =\lambda C$ with the specified Curie temperature givesAt very low temperature, a fully ordered single domain has $M_s=n{\mu }_B=8.4764\times 10^5\mathrm{A}{\mathrm{m}}^{-1}$. Its Weiss molecular field isThe cancellation of $n$ in the middle expression provides a useful check. The ordinary magnetostatic scale is only ${\mu }_0{M}_s\simeq 1.065\mathrm{T}$, smaller by the factor $\lambda$. Thus $B_{\mathrm{eff}}$ is not a plausible ordinary self-generated dipolar magnetic field of the nickel moments. It represents the exchange interaction energy in field units, not an additional Maxwell field of that magnitude. This does not assert that such large externally generated fields are forbidden. Nickel is an itinerant ferromagnet; the imposed one-${\mu }_B$-per-atom model is an estimate rather than its measured microscopic moment.

Expand a single-electron wavefunction as $|\psi ⟩={\sum }_n{\psi }_n|n⟩$. Project the Schrödinger equation onto $⟨n|$ to obtainTranslation symmetry suggests the Bloch state ${\psi }_n=N^{-1/2}{e}^{ikna}$ on a periodic chain of $N$ sites. Substitution givesTake the conventional positive hopping $t>0$. The first Brillouin zone is $-\pi /a\le k<\pi /a$; the band has minimum ${ϵ}_0-2t$ at $k=0$ and maximum ${ϵ}_0+2t$ at the zone edges. Its width is $4t$. A negative hopping interchanges the extrema but does not change the density of states after replacing $t$ by $|t|$.

There are $N$ allowed wavevectors and two spin states per wavevector. For the total spin-summed density of states $D(E)$, each interior band energy has two roots, $\pm k$. Using the spacing $2\pi /(Na)$ givesThus the density of states per atom isIt is symmetric about ${ϵ}_0$ and has integrable inverse-square-root Van Hove singularities at both band edges. Its integral over the band is two states per atom. The plot labels the dispersion extrema and both singular edges.

The first Brillouin zone is the squareAt half filling, translation of the wavevector by $\mathbf{Q}=(\pi /a,\pi /a)$ exchanges the two band energies:Thus the same energy-pair argument as in part c givesAt zero temperature, the Fermi surface satisfies $\cos (k_{x}a)+\cos (k_{y}a)=0$. Within the square Brillouin zone this is the diamondwhose vertices are $(\pm \pi /a,0)$ and $(0,\pm \pi /a)$. The occupied region is the diamond's interior, containing the band minimum at the zone centre. It has half the Brillouin zone area, consistent with one electron per atom and two spin states per wavevector.

An $f$ shell has orbital quantum number $l=3$ and seven orbital states with $m_l=-3,-2,\dots ,3$. The first of Hund's rules maximizes total spin subject to the Pauli exclusion principle. Seven electrons occupy the seven orbitals singly with parallel spins, givingTheir orbital projections sum to zero, so the half-filled shell also has total orbital angular momentum $L=0$, total $J=7/2$, and Landé g-factor $g_J=2$. These further values identify the magnetic moment used below.

Let the sample area be $A$, set the band touching energy to zero, and write $m^{\ast }=m_e^{\ast }=m_h^{\ast }$. Assume one electronic valley in each energy band with two spin states. The near-edge parabolic dispersions areUnder periodic boundary conditions, each spin occupies reciprocal-space area $(2\pi {)}^2/A$. Counting the quantum states inside a circle givesIts derivative is the constant two-dimensional free-electron density of states. Defining $D=A{m}^{\ast }/(\pi {\hslash }^2)$, the signed-energy density of states isThe total electronic density of states is therefore $D$ on either side of zero, not $2D$ at every signed energy. If instead one counts positive-energy excitations, the electron and hole species each have density of states $D$, giving $2D$ when summed. A common additional valley degeneracy multiplies all these quantities by that factor. The effective mass approximation assumes thermally sampled energies remain within the near-edge regions of both energy bands.

Equal effective masses and equal spin and valley degeneracies give electron-hole symmetry. In the intrinsic semiconductor, charge neutrality requires equal numbers of conduction band electrons and valence band holes. It therefore pins the chemical potential to the band touching point, $\mu =0$. The Fermi-Dirac distribution givesMeasured relative to the filled valence band at zero temperature, the thermal internal energy includes both species. Removing a valence band electron at $-E$ costs the positive hole energy $E$, just as adding a conduction band electron at $E$ does. HenceDifferentiate at fixed area and fixed total electronic particle number, allowing creation of electron-hole pairs:There is no conflict with the equipartition theorem. The normalization $N_c$ counts only the excited conduction band electrons, while the heat capacity includes both electrons and holes; moreover, $N_c\propto T$ is not fixed. In fact $\Delta U/N_c={\pi }^2{k}_{B}T/(6\ln 2)$, and differentiating $N_c(T)$ supplies another equal contribution to the heat capacity. Finally, $\mu =0$ gives occupation $1/2$ at the touching point, so the excitation gas is not in the classical dilute limit even though $k_{B}T$ exceeds the vanishing band gap.

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.

In the hydrogenic donor state approximation, the screened Coulomb potential energy is $-e^2/(4\pi {ϵ}_0{ϵ}_{r}r)$ and the kinetic term uses $m^{\ast }=0.067m_e$. Rescaling the hydrogen atom gives the effective Bohr radius of a hydrogenic donor and the positive binding energyThereforeThe radius here is the exponential length of the ground-state wavefunction, not its root-mean-square radius. With $\psi (r)=(\pi a_b^3{)}^{-1/2}{e}^{-r/a_b}$, radial integration gives $⟨r^2⟩=3a_b^2$ and $⟨x^2+y^2⟩=2a_b^2$.

For dilute independent donors of number density $n$, a bound spin one-half electron has magnetic moment magnitude ${\mu }_s=|g|{\mu }_B/2$, with the Bohr magneton defined using the bare electron mass. Its weak-field Curie law response isFor the orbital response, expand the effective mass Hamiltonian $(\mathbf{p}+e\mathbf{A}{)}^2/(2m^{\ast })$ in a uniform magnetic field, using $\mathbf{A}=\mathbf{B}\times \mathbf{r}/2$. The spherical ground state has zero orbital angular momentum; the quadratic term raises its energy byDifferentiating this energy twice and using $B\simeq {\mu }_{0}H$ givesThe donor-only magnetic susceptibility cancels whenThe negative sign of $g=-0.45$ drops out of the linear magnetic susceptibility. This crossover is far below $E_b/k_B\simeq 63.6\mathrm{K}$, consistent with donor binding. It assumes sufficiently weak magnetic field, negligible exchange between donors and a thermal scale above any inter-donor splitting; sufficiently dilute donors can satisfy those assumptions. The background diamagnetism from part (b) is excluded from this cancellation.

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.

For a long chain of length $L$ with periodic boundary conditions, each spin has $Ldk/(2\pi )$ quantum states in a wavenumber interval $dk$. Including two spin states, a full energy band contains $2L/a$ electrons. The state counting in a Bloch band therefore gives the energy densityThe integral of $1+y^2$ is $8/3$. For $ϵ>0$, the even square-root integral isFor example, this follows by setting $u=2y$, applying the supplied antiderivative, and evaluating both endpoints; the lower-end logarithm contributes $\ln ϵ$. HenceThe continuous zero-coupling limit is $\mathcal{U}(0)=2E_{\mathrm{ZB}}/(3a)$. This agrees with directly integrating the one-dimensional free electron model up to $k_F=\pi /a$ at two electrons per unit cell.

Without the Peierls distortion, adjacent chemical atoms are separated by $a/2$. The primitive unit cell contains one chemical atom and contributes one electron. A energy band with two spin states accommodates two electrons per primitive unit cell, so the first energy band is half filled. Its true first Brillouin zone extends to $\pm 2\pi /a$, while the Fermi points are at $\pm \pi /a$.

Describing the same chain with a doubled unit cell of length $a$ merely folds the half-filled energy band into two branches that touch at those Fermi points. The unbroken lattice translation by $a/2$ prevents an artificial band gap there. A Peierls distortion breaks that smaller lattice translation and supplies the missing Fourier coefficient at $2k_F=G$; the lower folded energy band can then be completely filled and separated from the upper energy band. Thus the zero-band gap result at $\delta =0$ agrees with the original one-electron band filling.

Let $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant, $T$ the absolute temperature, and $\mu$ the common equilibrium chemical potential. The conduction band minimum and valence band maximum are ${ϵ}_c$ and ${ϵ}_v$; their separation is the band gap $E_g={ϵ}_c-{ϵ}_v$. The positive parameters $m_e,m_h$ in the supplied density of states functions are the band effective masses of electrons and electron holes, not necessarily the bare electron mass.

The electron occupation is the Fermi-Dirac distribution $f(ϵ)=[e^{\beta (ϵ-\mu )}+1{]}^{-1}$. The appropriate statistical factors are $f_c=f$ for electrons and $f_v=1-f$ for electron holes. In a nondegenerate semiconductor, ${ϵ}_c-\mu \gg k_{B}T$ and $\mu -{ϵ}_v\gg k_{B}T$, so the Boltzmann approximation givesFor the conduction band, set $u=ϵ-{ϵ}_c$ and then $\xi =\beta u$. The supplied integral yieldsFor the valence band, set $u={ϵ}_v-ϵ$ to obtainThe effective density of states factors $N_c,N_v$ include the two spin states already present in the supplied density of states functions. Multiplication cancels the chemical potential and gives the semiconductor mass-action lawHere $n_i=\sqrt{N_c{N}_v}{e}^{-E_g/(2k_{B}T)}$ is the intrinsic carrier concentration. Equality $n=p=n_i$ additionally requires an electrically neutral intrinsic semiconductor. The product law also applies to an equilibrium doped semiconductor while both populations remain nondegenerate. The derivation assumes parabolic energy bands, the stated spin degeneracy, and a single common chemical potential.

For a single impurity level in a crystal with allowed occupations zero and one and equal occupied/empty statistical weights, the Fermi-Dirac distribution givesSince $n_{\mathrm{full}}=N_D{f}_D$ and $n_{\mathrm{empty}}=N_D(1-f_D)$,The same result follows from the two grand canonical ensemble weights $1$ and $e^{-\beta ({ϵ}_D-\mu )}$.

The equal-statistical-weight convention is necessary for the printed formula when $n_{\mathrm{full}}$ and $n_{\mathrm{empty}}$ count whole donor impurities. More generally, donor charge-state degeneracy gives $n_{\mathrm{full}}/n_{\mathrm{empty}}=(g_{\mathrm{full}}/g_{\mathrm{empty}})e^{\beta (\mu -{ϵ}_D)}$. An ordinary singly occupied donor with two spin orientations and a nondegenerate empty state has ratio $2e^{\beta (\mu -{ϵ}_D)}$; at $\mu ={ϵ}_D$ its occupied-to-empty ratio is two, a concrete counterexample to treating the printed expression as universal. The following part uses the paper's ratio-one convention.

For a fixed open-shell electron configuration in the LS coupling regime, Hund's rules select the free-ion ground term in the following order:

The first two rules come from the Coulomb interaction between electrons within the partially filled subshell, including its exchange interaction. Parallel spins require an antisymmetric spatial wavefunction, reducing close approach and often lowering the repulsive Coulomb interaction energy. Among the allowed high-spin terms, the angular dependence of the Coulomb interaction generally favours the largest $L$; a simple magnetic alignment argument alone does not derive this second ordering rule.

The weaker spin-orbit coupling then splits a selected $L,S$ term. In LS coupling, its term-level energy is proportional toThe effective sign appropriate to less-than-half versus more-than-half filling gives the third rule. These rules assume the electrostatic term separation dominates spin-orbit coupling; strong crystal field splitting or other coupling regimes require further analysis.

The given electron configuration has an inert closed $[\mathrm{Xe}]$ core and two $4f$ electrons. The Aufbau principle identifies the occupied subshells; the free-ion term within the open $4f^2$ configuration is then selected by Hund's rules. A closed core contributes $L=S=0$.

Each $f$ electron has orbital angular momentum quantum number $l=3$. Place the two spins parallel, so $S=1$. The Pauli exclusion principle then requires distinct orbital projections. The largest allowed total projection is $M_L=3+2=5$, obtained from $m_l=3$ and $m_l=2$, giving the maximum allowed $L=5$. A larger $M_L=6$ would place two parallel-spin electrons in the same atomic orbital and is forbidden.

The $f$ subshell holds fourteen electrons; $f^2$ is less than half filled. The third of Hund's rules therefore selects $J=L-S=4$. HenceThe term symbol uses $2S+1=3$ and the spectroscopic letter $H$ for $L=5$. The praseodymium ion retains an orbital contribution, so a spin-only estimate would be inappropriate.

Hund's rules maximizes total spin within the partially filled shell: parallel spins require an antisymmetric spatial wavefunction, reducing Coulomb overlap through exchange. Among terms with that spin, the second rule selects the largest total orbital angular momentum. The third rule, from spin-orbit coupling within the selected term, gives $J=|L-S|$ below half filling and $J=L+S$ above half filling, with the half-filled case treated directly.

For seven $4f$ electrons, put one parallel-spin electron in each $m_l=-3,-2,-1,0,1,2,3$ orbital. Their orbital projections sum to zero and form the half-filled $L=0$ shell; all seven spins add. ThusThere is no ambiguity from applying the below- or above-half-filled formula because $L=0$. The absence of a first-order orbital moment is important for the weak magnetic anisotropy discussed in c. These are isolated-ion term rules; a crystal field or strong covalency can alter other ions' orbital behavior, but the half-filled Eu${}^{2+}$ shell is unusually robust.