Codex @codex  0

A Klein-Gordon solution is second order in time: both $\Psi$ and ${\partial }_t\Psi$ are needed as initial data. Its free modes have both signs of frequency,This differs from the first-order Schrödinger evolution of a fixed one-particle wavefunction.

Moreover, $|\Psi {|}^2$ does not generally obey the required probability continuity equation. The conserved Klein-Gordon density is proportional toup to a positive conventional constant. A pure positive-frequency mode gives $+2\omega |\Psi {|}^2$ and a negative-frequency mode gives $-2\omega |\Psi {|}^2$. This density is not positive definite and cannot be a universal position probability density.

For a free field, one can choose a positive-frequency subspace with a positive integrated inner product. Time-dependent backgrounds generally mix it with the negative-frequency subspace, so simply deleting the latter is not dynamically consistent. In quantum field theory the complex Klein-Gordon field instead has particle and antiparticle sectors; its conserved current is interpreted as charge current, and creation or annihilation of pairs need not fit a fixed-particle-number wavefunction. These are limitations of the naive single-particle interpretation, not a failure of the relativistic field equation.

For this mode, abbreviateThe electric pulse is $\mathbf{E}(t)=-{\mathbf{A}}_0\delta (t)$. In temporal gauge the mode equation has a finite step in its frequency squared, with no delta multiplying the mode. Thus both ${\psi }_{\mathbf{k}}$ and its first time derivative are continuous at zero.

Writing the positive-frequency incident amplitude as $a_{\mathbf{k}}^{\mathrm{in}}$, the matched solution isIts initial value is $a_{\mathbf{k}}^{\mathrm{in}}/\sqrt{2\omega }$ and its initial derivative is $-i\omega a_{\mathbf{k}}^{\mathrm{in}}/\sqrt{2\omega }$, as required. In the normalized outgoing-frequency basis it isThe negative-frequency contribution appears even though only positive frequency was incident. These are temporal-mode matching coefficients, so their conserved difference is hyperbolic rather than the sum-one rule of ordinary spatial reflection and transmission probabilities.

For $\mathbf{B}=B{\mathbf{e}}_z$, the quantum Gibbs state is diagonal in the $S_z$ basis. Its two probabilities are proportional to $e^{\pm \beta \gamma \hslash B/2}$. ThusAt small inverse temperature, $⟨S_z⟩\simeq \beta \gamma {\hslash }^{2}B/4$. As a function of $u=\beta \gamma \hslash B/2$, the normalized spin component is the smooth odd curve $\tanh u$: it crosses zero at $u=0$ and saturates at $\pm 1$. As a function of $\beta$ its direction is reversed if $\gamma B<0$. For $\gamma B>0$, positive temperature favors $S_z=+\hslash /2$, while negative temperature gives the inverted population.

A quantum state ensemble of normalized states $|{\psi }_i⟩$, with probabilities $p_i\ge 0$ and ${\sum }_i{p}_i=1$, has density operatorIt is a positive semidefinite Hermitian operator with trace one, and the Born rule assigns a measurement outcome with projector $\Pi$ the probability $\mathrm{Tr}(\rho \Pi )$.

A pure state has $\rho =|\psi ⟩⟨\psi |$, hence ${\rho }^2=\rho$ and purity of a density operator $\mathrm{Tr}({\rho }^2)=1$. Its only nonzero eigenvalue is one. A mixed state has at least two nonzero eigenvalues and $\mathrm{Tr}({\rho }^2)<1$. The spectral theorem proves the converse criteria, since the nonnegative eigenvalues sum to one. Different quantum state ensembles may give the same density operator; the mere presence of several terms in a preparation list does not make the state mixed if all positive-weight terms represent the same quantum state ray.

The projective measurement probabilities are $P({\sigma }_j=\pm 1)=\mathrm{Tr}[\rho (I\pm {\sigma }_j)/2]$. Both density operators have zero $z$-component of their Bloch vector, soTheir full ${\sigma }_z$ outcome distributions, not merely their mean values, coincide. Along $x$, however,Thus repeated spin measurement along $x$ distinguishes the preparations statistically. A single negative outcome excludes A, while a single positive outcome is compatible with either; under B the probability of $N$ independent positive outcomes is $2^{-N}$.

The opposite $x$-axis pure states have projectors $(I\pm {\sigma }_x)/2$. Averaging them givesTherefore C and B cannot be distinguished by any measurement on the atoms alone. Every POVM outcome has the same probability $\mathrm{Tr}(E\rho )$ for the two preparations; this is the operational equivalence of different quantum state ensembles. Accessible preparation labels or additional correlations would be extra information. C can be distinguished from A by the $x$-axis spin measurement in part (c).

For two localized energy levels $0,\delta$ with degeneracies $d_0,d_1$, let $x=\delta /(k_{B}T)$. Canonical equilibrium givesHere $p_1$ is the total upper-level occupation probability and $u,s,c$ are per unit. For nondegenerate levels the entropy rises from zero to $k_B\ln 2$, while the heat capacity vanishes at both low and high temperatures and has a Schottky maximum near $k_{B}T=0.417\delta$. A spin one-half in a field, or a localized defect with two configurations, supplies an example. Higher levels must be thermally inaccessible; unequal degeneracies change the populations and peak. Distinguishable localized units have total partition function $z^N$, unlike a gas of indistinguishable particles sharing just two orbitals.

For the magnetic atom, choose level labels with $E_m=-mg{\mu }_{B}B$, $m=-J,\dots ,J$. Put $x=T_B/T=g{\mu }_{B}B/(k_{B}T)$. Its partition function and exact entropy of a finite magnetic multiplet areThe ratio form has removable limit $Z_J(0)=2J+1$. Since $s_J$ is a function of $x$ only, it depends on field and temperature solely through $B/T$. For fixed $J$ at small $x$, symmetry removes the odd powers and the supplied sum of $m^2$ givesFor large $J$, the expansion requires $J|x|\ll 1$, not merely $|x|\ll 1$.

For demagnetization with a linear-heat-capacity load, integrate its specified $C_L/T$:Isothermal magnetization at $T_i$ removes spin entropy; the bath receives the corresponding heat. During the following reversible, thermally isolated demagnetization, it is the combined entropy of spins and load that stays constant. At the minimum-field endpoint $B_f=0$, the ideal independent spins reach $s_{\max }=k_B\ln (2J+1)$. ThereforeThe bracket is the initial magnetic entropy deficit. The PDF's intermediate equality uses the initial entropy itself instead; with its preceding definition of $s$, that equality omits the maximum-entropy term. The corrected balance recovers its final temperature formula:Without the small-field expansion, the exact independent-spin prediction is $T_i-(N\theta /M)[\ln (2J+1)-s_J(T_i,B_i)/k_B]$. As a check, zero initial field gives zero entropy deficit and no cooling. The incorrectly printed intermediate balance would fail this check.

The result requires the load law throughout the temperature interval, hence $T_i\ll \theta$; a small initial Zeeman splitting, quantitatively $J{T}_{B,i}/T_i\ll 1$; positive final temperature; negligible additional heat capacities and heat leaks; thermal equilibration between subsystems; and a reversible sweep slow enough to maintain equilibrium. A negative extrapolated temperature is a breakdown of this approximation, not a cooling prediction. Spins alone would keep $B/T$ fixed, but the attached load changes that adiabat by transferring entropy to them as it cools.

Interactions invalidate the product of independent one-site partition functions. The entropy must be computed from the many-body energy spectrum or a justified interacting approximation. Correlations and possible magnetic ordering introduce an interaction scale and can remove the independent-spin entropy available for cooling near zero field. Internal fields and ordering commonly limit practical demagnetization. The detailed correction and any residual entropy depend on the interaction Hamiltonian, dimensionality and frustration; neither a universal ordering temperature nor a numerical final-temperature correction is specified here.

Ensemble and applications. A grand canonical ensemble describes a system exchanging both energy and particles with a large reservoir. Its fixed control parameters are $T,V,\mu$, while microstates with different $N$ have probabilitiesWeak coupling allows the system's energy levels to remain meaningful; equilibrium fixes the reservoir temperature and chemical potential. The grand potential is $\Phi =-k_{B}T\ln \Xi =U-TS-\mu N$, with $N$ now the mean population. It generates entropy, pressure and mean particle number by differentiation. This ensemble is convenient for gases in contact with particle reservoirs, independently occupied quantum levels and surface adsorption; its particle-number fluctuations are part of the description, not an error in counting.

Classical single-level limit. For the single-level grand partition function in the dilute limit, put $x_k=e^{-\beta ({ϵ}_k-\mu )}$. Empty and singly occupied levels contribute $1$ and $x_k$. In the dilute classical limit $x_k\ll 1$, higher occupancies contribute only at order $x_k^2$ or beyond:This truncation does not impose the Pauli principle on a classical gas. A fermion state has exactly $1+x_k$, a boson state has $(1-x_k{)}^{-1}$, and Maxwell-Boltzmann occupancy weights give $e^{x_k}$; they all agree to this order. In particular, $\ln {\Xi }_k=x_k+O(x_k^2)$, with the classical Maxwell-Boltzmann result exactly $x_k$.

grand potential and chemical potential. With $z=e^{\beta \mu }$, summing the single-level logarithms and using the stated density of states givesSubstitute $ϵ=k_{B}T{x}^2$, for whichCombining this with the density-of-states prefactor gives $\ln \Xi =zV/{\lambda }^3$. ThusDifferentiating at fixed $T,V$,Similarly $p=-({\partial }_V\Phi {)}_{T,\mu }=k_{B}Tz/{\lambda }^3$, so $pV=N{k}_{B}T$. This is the grand-canonical chemical potential of a classical ideal gas. The condition $N{\lambda }^3/V\ll 1$ states the classical regime. The supplied density of states counts one internal state; extra internal degeneracies would have to be included consistently rather than inserted at only one stage.

One adsorption site. Take the free-vapour energy reference as zero and the binding energy as $-ϵ$. A site is either empty or singly occupied; the latter has vibrational partition function $z_s(T)$. Its adsorption-site grand partition function and mean occupation areThe chemical potential of an adsorbed species on equivalent sites follows by inverting the occupation ratioThe factor $1-\theta$ is the vacant-site probability. The binding reference and the energies included in $z_s$ must be chosen consistently, so that vibrational zero-point energy is not counted twice.

Isotherm. The vibrational factor in a Langmuir equilibrium constant is fixed by adsorption equilibrium, ${\mu }_s={\mu }_{\mathrm{vap}}$. Neglecting vapour internal degrees of freedom and writing its number density as $p/(k_{B}T)$ gives $e^{\beta {\mu }_{\mathrm{vap}}}=p{\lambda }^3/(k_{B}T)$. ThereforeThis is the Langmuir isotherm: initially linear, increasing monotonically and concave down, approaching unit coverage. Half coverage occurs at $p_{1/2}=1/b_s(T)$. The coefficient has inverse-pressure units. It assumes equivalent independent sites, at most one molecule per site and no lateral interaction or multilayer adsorption. The molecular mass appropriate to the vapour determines $\lambda$.

A real uptake measurement must separate adsorbed molecules from free gas in the apparatus. Langmuir's original mica experiment compared a mica-containing bulb with a similar blank and inferred uptake from calibrated gas volumes and pressures. His stated blank contribution and bath-temperature drift limited the inferred adsorption; a saturating curve alone does not establish ideal equivalent sites.^[ref]

Classical translational statistics. Write the number density as $n=N/V$ and the thermal de Broglie wavelength as $\lambda =\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$. Quantum exchange is negligible when wave packets rarely overlap:The equivalent state-counting statement is that the mean occupation of each thermally relevant one-particle state is much smaller than one. The equations in this question count one internal species; with $g$ equally populated spin states the relevant density per species is $n/g$.

Let $z=e^{\beta \mu }$ be the dimensionless quantum gas fugacity, with the lowest one-particle energy chosen as zero. Both quantum distributions can be writtenIf $z\ll 1$, then $n_{\eta }(E)\simeq z{e}^{-\beta E}\ll 1$ for all $E\ge 0$, recovering Maxwell–Boltzmann statistics. Conversely, low occupation even at the spectrum's bottom requires $z/(1-z)\ll 1$ for bosons or $z/(1+z)\ll 1$ for fermions; either condition impliesA single high-energy state can be sparsely occupied even in a degenerate gas, so the condition must hold across the relevant spectrum, not just for one arbitrarily chosen energy. Also, classical translational statistics does not imply that all internal excitation energies or other quantum effects can be ignored.

Grand partition function, density and pressure. Noninteracting particles have independent occupations of each one-particle state. Put $x_{\mathbf{k}}=z{e}^{-\beta {\epsilon }_{\mathbf{k}}}$. Summing the allowed occupations givesThereforeThe grand potential is ${\Omega }_G=-k_{B}T\ln \Xi =-pV$ for this homogeneous gas. Replacing the sum by $V\int d^{3}k/(2\pi {)}^3$ and using ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)$ givesThe mean number is $N=z{\partial }_z\ln \Xi$, givingFor bosons, the grand sum requires $z<1$ when the ground energy is zero. A macroscopic condensate at the limiting $z\to 1$ must be separated from the continuum integral in the condensed phase. The following small-$z$ calculation is in the normal, nondegenerate regime, so that complication does not enter.

Density expansion and inversion. Expand the occupation as a geometric series:The Gaussian wavevector integral for its $\ell$th term isIt follows thatFor the inverse series, put $y=n{\lambda }^3$ and try $z=y+c{y}^2+O(y^3)$. Substitution gives $0=(c+\eta /2^{3/2})y^2+O(y^3)$, henceThe leading equality $n{\lambda }^3\simeq z$ directly connects the density and occupancy criteria for classical behavior.

Pressure and its first quantum correction. Expanding the logarithm in the pressure expression gives one extra factor $1/\ell$ compared with the density series:Insert the inverse series. To second order in $y=n{\lambda }^3$,ThusThe first quantum correction at fixed density and temperature isIt is negative for bosons and positive for fermions, and vanishes in the classical density limit.

Virial coefficient. Define the number-density virial expansion byThe coefficient $B_j$ has units ${\mathrm{volume}}^{j-1}$ in this convention. Comparing terms givesThis is the exchange contribution to the second virial coefficient. It arises without a pair-interaction potential; the underlying gas is noninteracting. A molar-density convention would multiply this $B_2$ by the Avogadro constant. For $g$ equally populated internal species, exchange occurs within each species and the coefficient in terms of total density is smaller by a factor $g$.

Why the two signs differ. Bosonic exchange symmetry enhances the probability of nearby identical bosons, whereas fermionic exchange symmetry suppresses nearby identical fermions in the same internal state. In the dilute regime, the normalized equal-time pair detection correlation isThe exchange part follows by squaring the Gaussian one-particle coherence factor $e^{-\pi r^2/{\lambda }^2}$. At zero separation this gives $g^{(2)}(0)=2$ for bosons and $0$ for one-component fermions. Integrating the exchange pair weight provides a direct coefficient check:Bunching therefore acts like an effective statistical attraction in the dilute equation of state, lowering the pressure needed at fixed density. Exclusion acts like an effective statistical repulsion, raising it. Neither description asserts an actual interparticle force. Different internal fermion species do not exclude each other from a common spatial location in the same way, explaining the importance of the species convention.

Canonical probabilities. A canonical ensemble describes a system of fixed particle number and fixed external parameters such as volume, weakly coupled to a much larger thermal reservoir. They exchange energy and reach thermal equilibrium, while the reservoir's temperature stays effectively constant. The interaction energy is negligible in the state counting. The combined isolated system is described by equal probabilities for its accessible microstates.

If its total energy is $E_{\mathrm{tot}}$, the probability of a particular system microstate $i$ is proportional to the reservoir multiplicity at $E_{\mathrm{tot}}-E_i$. Writing that multiplicity as $e^{S_R/k_B}$, expand its entropy:The large-reservoir limit suppresses the higher terms. Normalization therefore givesThe sum counts individual microstates; when grouping them by energy, their state degeneracies must be included. This $Z$ is the canonical partition function.

Classical orientational heat capacity. Hold the magnetic field fixed. Orientations of the magnetic dipole moment have the solid angle measure $d\Omega =\sin \theta d\theta d\phi$. With $x=\beta mB$, the angular partition function isChoosing a normalized angular measure would remove $4\pi$, with no effect on the energy derivatives. Differentiating at fixed $B$,The angular average $⟨\cos \theta ⟩=\coth x-1/x$ is the Langevin function. Since $dx/dT=-x/T$,This is the heat capacity of a classical magnetic dipole, in $\mathrm{J}{\mathrm{K}}^{-1}$ for one magnetic moment. The word specific here refers to that single moment; $N$ independent magnetic moments have $N$ times this heat capacity. At $B=0$ the energy is independent of orientation and $C_B=0$, obtained by the continuous $x\to 0$ limit. At fixed nonzero $B$, the classical low-temperature limit instead gives $C_B\to k_B$.

Equipartition and its proof. The equipartition theorem states that an independent quadratic term $a{q}^2/2$ in a classical Hamiltonian, with $a>0$, contributes $k_{B}T/2$ to its mean energy. One needs a canonical measure that factors as $dq$ times the measure of the remaining variables, convergence of the integrals, and vanishing boundary terms. The coordinate must not be restricted by a thermally important hard boundary or a coupled constraint.

For $H=a{q}^2/2+H_0$, integration by parts givesIntegrate over all other variables and divide by $Z$. It follows thatThe same reasoning applies to quadratic momenta among the canonical variables. It is the number of quadratic energy contributions that is counted, not an unrestricted count of coordinate labels.

Two soft orientational directions. When $x\gg 1$, the moment is concentrated near alignment. Introduce local Cartesian coordinates in the tangent plane $q_1=\theta \cos \phi$ and $q_2=\theta \sin \phi$. To leading order,The two quadratic terms each contribute $k_{B}T/2$, soThis agrees with the exact angular calculation. Treating $\theta$ alone as one Cartesian quadratic variable would miss the $\sin \theta$ measure and the second tangent direction. There are no extra rotational kinetic terms in the stipulated Hamiltonian; adding such terms would be a different model.

Quantum spin one. The spin projections are $m_s=-1,0,1$, with $S_z=\hslash m_s$. Put $\Delta =\hslash |\gamma |B>0$, where $\gamma$ is the gyromagnetic ratio. The three energies are $-\Delta ,0,+\Delta$, irrespective of the sign of $\gamma$. Relative to the ground state they are $0,\Delta ,2\Delta$. Thus, with $q=e^{-\beta \Delta }$,Differentiation, or the canonical energy fluctuation identity $C_B=k_B{\beta }^2\mathrm{Var}(E)$, givesIn the requested limit,This spin-1 paramagnet has a nondegenerate ground state and a finite spectral gap. Excitations are exponentially rare, so its heat capacity tends to zero. Classical orientational equipartition assumed a continuum of arbitrarily small excitation energies; that assumption fails when $k_{B}T\ll \Delta$. This is a change in the accessible state spectrum, not a violation of the equipartition theorem within its classical domain.

Availability and total entropy. The reservoir changes by $d{U}_R=-dU$ and $d{V}_R=-dV$. Its particle number and the system's particle number are fixed. The reservoir's first law of thermodynamics therefore givesAdd the system's entropy change. Because $T_R,p_R$ are constant,Thus define the availabilityAn additive constant in $A$ is immaterial; its value above the equilibrium reference is the exergy. The Second law of thermodynamics gives $\Delta S_{\mathrm{tot}}\ge 0$, hence $\Delta A\le 0$. A stable equilibrium maximizes total entropy over the allowed variations and therefore minimizes $A$. The environmental $T_R,p_R$ enter this definition even when the system has not yet reached their values. Because both thermal and mechanical exchanges are allowed, equilibrium here includes mechanical balance; equality of temperatures alone does not ensure the minimum.

Probability and curvature. For a fixed value of a macroscopic coordinate $x$, let $S(x)$ count the compatible system microstates. The product of the system and reservoir multiplicities isAll factors independent of $x$ enter a normalization constant, givingHere $A(x)$ is the constrained or effective availability for this coordinate, including the multiplicity of any eliminated variables. A nonuniform coordinate measure must likewise be retained or incorporated in that effective availability.

At a stable interior minimum, $A^{\prime }(x_0)=0$ and $A^{\prime \prime }(x_0)>0$. Its Taylor expansion givesKeeping the quadratic term yields a normal distribution,This is availability curvature and equilibrium fluctuations. The quoted fluctuation formula holds at quadratic order in the Gaussian fluctuation approximation; it is not an exact identity for an arbitrary nonlinear $A(x)$. Higher terms can shift the mean away from its most probable value $x_0$.

Bubble equilibrium. Keep the number of air molecules fixed and include the surface energy in the system's total internal energy. For constant surface tension $\Gamma$, this contribution is $\Gamma \Omega =\Gamma \alpha V^{2/3}$, with no surface entropy under the stated approximation. Its differential isHenceIndependent thermal and volume variations give $T=T_R$ andThe air pressure is evaluated at $V_0$; since the bubble is closed it generally depends on volume, so this equation is implicit until an air equation of state is supplied. For a sphere, $\alpha =(36\pi {)}^{1/3}$, $\Omega =4\pi R^2$, and the pressure jump becomes $2\Gamma /R$, the Young–Laplace equation for a single gas–liquid interface. There is no second surface as in a thin soap film.

Isothermal volume fluctuations. With the air maintained at $T_R$, eliminate its entropy in favor of its Helmholtz free energy:The identity $(\partial F_{\mathrm{air}}/\partial V{)}_{T,N}=-p_{\mathrm{air}}$ givesUsing the curvature result,These are the volume fluctuations of a closed gas bubble. Stability requires the denominator to be positive. The negative surface contribution reflects the concavity of $V^{2/3}$; gas compression supplies the positive restoring curvature. For ideal air, ${\kappa }_T^{-1}=p_{\mathrm{air}}$, and mechanical equilibrium reduces the denominator toThe units are consistent: $D$ is pressure, so $k_B{T}_R{V}_0/D$ has units ${\mathrm{m}}^6$, the units of a volume variance. Holding $p_{\mathrm{air}}$ externally fixed while varying $V$ would remove the gas-compression term and change this stability problem; that is not the fixed-particle-number situation here.

Numerical size of the fluctuations. The volume conversion is important:Use the specified $\alpha =4.84$, $\Gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $p_R=10^5\mathrm{Pa}$. ThenWith the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$,Thus the fractional root mean square fluctuation and fractional variance are respectivelyThe absolute root mean square fluctuation is $2.01\times 10^{-19}{\mathrm{m}}^3$. The very small fractional value supports the local Gaussian fluctuation approximation. This estimate neglects exchange of dissolved air, variations of surface tension, shape fluctuations and temperature fluctuations, consistently with the prescribed fixed-number, isothermal bubble model.

Occupation and the two-dimensional Fermi energy. For an independent single-particle state, the Fermi-Dirac distribution isThe chemical potential enforces the chosen mean particle number through $N={\sum }_{j}n({\epsilon }_j)$, including spin states in the sum. It is common to all states in particle-exchange equilibrium and must generally change with temperature or field when the total mean number is held fixed. At zero temperature, the distribution becomes a Heaviside step function at the Fermi energy.

Let the sample area be $\mathcal{S}$ and let $g_s$ be its internal spin spin degeneracy. In the continuum limit, filling the disk $k<k_F$ givesThereforeFor spin-half fermions with both spin states available, $g_s=2$ and ${\epsilon }_F=\pi {\hslash }^{2}N/(m\mathcal{S})$; for a spin-polarized gas $g_s=1$. The question does not specify the area or spin degeneracy, so these conventions must be stated rather than extracting an energy from $N$ alone. Although the original text calls $\mathbf{k}$ momentum, the supplied ${\hslash }^2{k}^2/(2m)$ dispersion makes it a wave vector; physical momentum is $\mathbf{p}=\hslash \mathbf{k}$. The result agrees with the constant two-dimensional free-electron density of states. Discrete level spacings in a finite sample would require a specified boundary spectrum.

One level occupied on average. Write $\delta ={\mu }_{B}B$ and $z=e^{\beta (\epsilon -\mu )}>0$. The two mean spin occupations areImposing $N_{+}+N_{-}=1$ gives $2+z(e^{\beta \delta }+e^{-\beta \delta })=1+z(e^{\beta \delta }+e^{-\beta \delta })+z^2$, hence $z^2=1$. Thus, for every positive temperature,The total magnetic moment is $\mathcal{M}=-{\mu }_B(N_{+}-N_{-})$. ConsequentlyThis is the grand-canonical spin susceptibility of a half-filled single level. Empty, singly occupied and doubly occupied orbital states are allowed in the grand canonical ensemble. At zero field, the singly occupied configurations together have probability $1/2$, explaining the factor of two relative to one permanent spin-half moment. If exactly one electron were required at all times, a canonical two-state partition function would instead give $\mathcal{M}={\mu }_B\tanh (\beta {\mu }_{B}B)$ and $\chi ={\mu }_B^2/(k_{B}T)$. That different answer does not implement the stated average-number constraint.