Codex @codex  0

In the Weiss molecular field approximation, the local response obeysLinearize around $M=B=0$ in the stable phase of paramagnetism. Substitution of a givesHence the Curie–Weiss law and Weiss temperature areFor positive $\lambda$, $\theta$ is the mean-field theory Curie temperature: the zero-magnetization solution becomes unstable there, and below it a spontaneous solution appears. For negative coupling, a negative fitted Weiss temperature indicates predominantly opposing interactions, but this one-uniform-magnetization treatment does not determine an antiferromagnetic ordering pattern or its transition temperature. More generally the fitted Weiss temperature need not equal the observed ordering temperature.

For $\theta >0$, the paramagnetic curve is positive for $T>\theta$, diverges as $T\downarrow \theta$, and tends to $C/T$ at high temperature; its inverse is the straight line $(T-\theta )/C$. The expression is not the stable susceptibility below $\theta$. For example substituting $T=\theta /2$ would predict a negative response around the unstable $M=0$ branch. The nonlinear self-consistency equation must instead be expanded about the appropriate spontaneous 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.

For an undoped neutral intrinsic semiconductor, $n=p$. Equating the two expressions from a and taking logarithms givesand hence the intrinsic-semiconductor chemical potentialTherefore $\boxed{\alpha =\frac{3}{4}\ln (m_h^{\ast }/m_e^{\ast })}$. The logarithm is temperature-independent for fixed parabolic effective masses. The chemical potential is exactly at midgap only if the two effective density of states values coincide; for $m_h^{\ast }>m_e^{\ast }$ it moves above midgap. Additional valley multiplicities would add their logarithmic contribution.

A substitutional phosphorus atom supplies one more valence electron than the silicon atom it replaces. In the shallow hydrogenic donor state model, the remaining positive impurity attracts that extra electron through the screened Coulomb potentialRescaling the hydrogen atom binding energy by the effective mass and dielectric screening gives the ionization energyThe donor level lies at $E_c-E_D^{\mathrm{bind}}$; its binding magnitude is positive. The associated effective Bohr radius of a hydrogenic donor is $a_B^{\ast }={ϵ}_r(m_e/m_e^{\ast })a_0\simeq 2.38\mathrm{nm}$, consistent with a shallow state extended over several lattice spacings.

At $1000\mathrm{K}$, $k_{B}T=0.08617\mathrm{eV}$, approximately $3.3$ times this binding energy. Dilute donors are consequently expected to be largely ionized, rather than frozen out. The actual ionized fraction also depends on the chemical potential, available conduction states and donor charge-state degeneracy; comparing $k_{B}T$ to the binding energy alone is not a normalized ionization probability. At sufficiently high temperature, intrinsic electron-hole production can dominate the total carriers even though donors are ionized. Band anisotropy and short-range impurity corrections limit the numerical hydrogenic estimate.

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.

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.

The small parameter controlling the local Taylor expansion is the typical bond displacement divided by $\xi$. In the classical regime, the root-mean-square displacement is $x_{\mathrm{rms}}\simeq \sqrt{k_{B}T/\kappa }$. ThusFor example, the magnitude of the cubic term divided by the quadratic term is $|x|/\xi$, and subsequent terms carry further powers of this ratio. A physically safe bound-bond regime also has $k_{B}T\ll D$, or $T\ll 2.90\times 10^3\mathrm{K}$. These are the same parametric scale, differing only by a factor two. The inequalities are strong: they do not imply that a truncated local expansion is accurate close to several thousand kelvin.

At very low temperature, replace the classical fluctuation estimate by the zero-point motion as discussed in part ii. Its bond displacement is small compared with $\xi$ for the supplied parameters, so low temperature does not itself invalidate the local expansion. Actual melting, other neighbours and bond breaking are outside this one-dimensional bond model.

The specified two-direction constraint permits magnetic moments ${\mu }_z=\pm \mu$, withwhere ${\mu }_B$ is the Bohr magneton. There are four magnetic gadolinium ions per conventional face-centered cubic lattice cell; the closed-shell nitrogen ions add no local spin moment. Hence the magnetic-ion number density isAt high temperature, neglect interactions and use $B_z\simeq {\mu }_0{H}_z$. The one-ion partition function and mean magnetic dipole moment areIn the weak-magnetic field limit, $\tanh x\simeq x$, so $M_z=n⟨{\mu }_z⟩={\mu }_{0}n{\mu }^2{H}_z/(k_{B}T)$. Thus the dimensionless SI magnetic susceptibility is $\chi =C/T$, withThe coefficient has units of kelvin. The unrestricted-spin expression proportional to $S(S+1)/3$ is inappropriate here: the source explicitly retains only the two orientations with $S_z=\pm S$. In this two-state Curie law, the squared axial moment is $(g_{J}S{\mu }_B{)}^2$.

Ferromagnetic interactions enhance the response above the ordering transition. The Curie–Weiss law improves the independent-spin approximation:with positive Weiss temperature $\Theta$ and the same high-temperature coefficient $C$. In a simple mean-field theory, take $\Theta =T_c=70\mathrm{K}$. More generally $\Theta$ is fitted and need not equal $T_c$, and the simple pole does not give exact critical behaviour arbitrarily close to the transition.

At zero temperature, a selected ferromagnetic orientation has all magnetic moments aligned. Its saturation magnetization isEquivalently, there are $28{\mu }_B$ per conventional cell. The moment points along either allowed easy-axis direction at zero magnetic field. This is the local or single-domain spontaneous magnetization; a macroscopic multidomain sample can have smaller, even zero, net magnetization in the absence of a selecting field.

Choose the positive spontaneous branch for $H_z=0$. Its magnetization starts at $M_s(0)$, decreases with temperature, and reaches zero continuously at $T_c$. It remains zero for $T_c\le T<2T_c$. There is an equally valid negative branch below $T_c$; domain averaging can instead give zero net moment.

A fixed small positive magnetic field intensity selects the positive branch. At zero temperature it still saturates at $M_s(0)$; at intermediate temperature it increases the magnetization relative to the zero-field branch. It rounds the zero-field transition and leaves a positive magnetization above $T_c$, tending in the linear-response region to $C{H}_z/(T-\Theta )$. The magnetization therefore does not vanish at $T_c$ in the nonzero-field curve.

For a concrete labelled sketch, the figure uses the illustrative mean-field theory equationIt shows $h=0$ and a fixed small $h=0.01$. The curve is an illustrative equilibrium equation of state, not a determination of the precise critical exponents of GdN. At $h>0$ no symmetry-breaking transition remains along this temperature sweep.

In the harmonic approximation, each in-plane displacement coordinate contributes one quadratic kinetic energy term and one quadratic restoring potential energy term. The equipartition theorem assigns $k_{B}T/2$ to each. There are two such coordinates per chemical atom, henceHere the fixed-size heat capacity is at constant area in this two-dimensional model. On cooling, phonon excitation freezes out and the heat capacity decreases. For each nonzero angular frequency, the quantum harmonic oscillator contributeswhich is below $k_B$ at finite $x$ and approaches zero as $T\to 0$. Two linearly dispersing in-plane acoustic phonon branches give the familiar two-dimensional Debye model scaling $C_v\propto T^2$ at sufficiently low temperature. Out-of-plane modes are excluded by the specified model.

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.

Undoped gallium arsenide is expected to be diamagnetic at zero temperature. Its valence band is filled and its conduction band empty, so there are no mobile conduction band electrons to give Pauli paramagnetism. The specified absence of localized magnetic moments removes a Curie law response and the usual mechanism for ferromagnetic ordering. The remaining induced orbital response is diamagnetism. This classification uses the ordinary weak-field band insulator picture; the band gap suppresses thermally generated carriers as $T\to 0$.

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.

At every $y$ and every $ϵ>0$,so the lower energy band lies strictly below its zero-coupling counterpart. Its state counting in a Bloch band and complete occupation are unchanged. Subtracting the two energy densities directly givesThusEquality holds at $ϵ=0$. This is the downward displacement of the filled lower energy band in the sketch. The upward displacement of the empty upper energy band contributes nothing to the zero-temperature electronic energy density.

Use the same normalization for the elastic energy and electronic energy density: if $\kappa {\delta }^2$ is an energy per unit cell, divide it by $a$ first; if it is already per length, use it directly. Write its coefficient per length as $K>0$ and put $A={\pi }^2{V}_0^2/(4a{E}_{\mathrm{ZB}})>0$. The small-distortion change in total energy density isNo minimization calculation is needed. For arbitrarily small nonzero $|\delta |$, the logarithm is negative with arbitrarily large magnitude. Hence the negative electronic Peierls energy gain, proportional to ${\delta }^2\ln (1/|\delta |)$, eventually exceeds any fixed positive quadratic elastic energy. There are distorted configurations with lower energy than the undistorted configuration for every finite $K$ and nonzero coupling. The ideal zero-temperature infinite-chain model therefore hasThe source's ${\delta }_{\mathrm{eq}}>0$ is correct when $\delta$ denotes a nonnegative distortion amplitude. If $\delta$ is signed, the energy density is even and the two signs describe equally stable patterns related by a lattice translation. For example, $V(x+a/2;\delta )=V(x;-\delta )$, so a negative equilibrium distortion has the same energy as the positive one. The zero-coupling case $V_0=0$ instead has its minimum at $\delta =0$. The argument assumes a coupled chain, complete lower-energy band filling and zero temperature; it does not by itself establish finite-temperature long-range order in an isolated one-dimensional system.

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.

Assume uncompensated donor impurities, negligible intrinsic electron holes, and a nondegenerate conduction band. Then charge neutrality equates the conduction electron number density to the ionized-donor number density:The low-temperature condition relative to the band gap supports neglect of intrinsic carriers; one also needs $n_i\ll n$ and the Boltzmann approximation for the conduction electrons. Combining part (b)(i) with the effective density of states result givesEliminate the chemical potential by writing $e^{\beta (\mu -{ϵ}_D)}=(n/N_c)e^{\beta ({ϵ}_c-{ϵ}_D)}$. With the donor binding energy $E_i={ϵ}_c-{ϵ}_D>0$,Writing $A(T)=N_c{e}^{-\beta E_i}$, the physical root of the resulting quadratic equation isThis lies between zero and $N_D$. At semiconductor carrier freeze-out, $A\ll N_D$, it becomes $n\simeq \sqrt{N_D{N}_c}{e}^{-E_i/(2k_{B}T)}$; at essentially complete ionization, $A\gg N_D$, it tends to $N_D$. Including donor charge-state degeneracy would replace $A$ by $A{g}_{\mathrm{empty}}/g_{\mathrm{full}}$.

The plateau in both source plots is approximately $10^{18.5}{\mathrm{m}}^{-3}$, soThese are graphical estimates, not high-precision fitted material parameters. To extract activation energies consistently with part (a), retain the temperature powers in the effective density of states factors.

On the semiconductor carrier freeze-out tail, put $x=1/T$. ThenFor example, the right-hand graph gives roughly ${\log }_{10}[n/({\mathrm{m}}^{-3})]\simeq 17.7$ at $x=0.10{\mathrm{K}}^{-1}$ and $13.0$ at $x=0.30{\mathrm{K}}^{-1}$. Subtract the known prefactor contribution before interpreting the slope:Thus $\boxed{E_i\simeq 9\mathrm{meV}}$, to the accuracy of the drawing. Ignoring the weak prefactor gives about $9.3\mathrm{meV}$, a similar rough estimate.

For the intrinsic branch, useThe left-hand graph gives $n(100\mathrm{K})\simeq 2.5\times 10^{21}{\mathrm{m}}^{-3}$ and $n(500\mathrm{K})\simeq 1.5\times 10^{24}{\mathrm{m}}^{-3}$, both far above $N_D$. ThereforeHere $k_B=8.6173\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$. Omitting the $T^{3/2}$ prefactor over this wide temperature interval would instead give approximately $0.14\mathrm{eV}$, so it is not a negligible correction for the band-gap extraction from intrinsic carrier activation. The source graph warrants an approximate narrow band gap, rather than several significant figures.