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For a single quantum state, the equilibrium occupations areBoth reduce to $e^{(\mu -ϵ)/(k_{B}T)}$ when $ϵ-\mu \gg k_{B}T$.

For one three-dimensional quadratic branch, the density of states per unit volume isWith ${\lambda }_T=\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$, changing integration variable to $x=ϵ/(k_{B}T)$ givesA degeneracy factor multiplies both densities if additional equivalent branches are present. The first expression is the noncondensed density: a macroscopically occupied ground state must be counted separately.

Bose-Einstein condensation occurs when the excited states can no longer hold all particles and the ground-state occupation becomes macroscopic. At fixed total density,For field-induced triplon condensation, write $\Delta =\nu n-\eta (H-H_g)$, so $ϵ(k)={\hslash }^2{k}^2/(2m)+\Delta$. At fixed reference $n$, the three requested dispersions have gaps $\nu n+\eta H_g$, $\nu n$ and zero respectively. Their curvature is unchanged. Because the actual thermodynamic chemical potential is zero, the Bose distribution requires $\Delta \ge 0$. At $H_c=H_g+\nu n/\eta$, the minimum reaches zero and the noncondensed population saturates, permitting macroscopic occupation.

For the triplon condensation field scaling, the critical density at onset is $n_T=\zeta (3/2)/{\lambda }_T^3$. ThereforeThe low-temperature phase boundary rises from $(T,H)=(0,H_g)$; the condensed phase lies on the higher-field side.

For $H>H_c$, simply continuing the gap below zero would give unphysical Bose occupations. For the condensate density in a shifted-parabola triplon model, added condensate density keeps the minimum at zero: $\nu (n_0+n_T)=\eta (H-H_g)$. At fixed temperature this giveswith $n_0=0$ below onset. At zero temperature it becomes $n_0=\eta (H-H_g)/\nu$ within that model.

This prefactor is conditional on continuing the stated single-density mean-field shift into the condensed phase. A microscopic interacting Bose condensate has collective Bogoliubov excitations $E_k=\sqrt{{ϵ}_k^0({ϵ}_k^0+2g{n}_0)}$, with ${ϵ}_k^0={\hslash }^2{k}^2/(2m)$, which are linear at small $k$. The normal-state shift alone does not fix the condensed interaction closure or identify $g$ with $\nu$. The gap closes and the condensate density grows linearly with excess field in the simple mean-field description; the detailed collective dispersion needs that additional interaction model. A dilute-boson treatment of field-induced magnetic ordering makes this normal/condensed distinction explicitly.^[ref]

At zero applied field, time-reversal symmetry interchanges $M$ and $-M$ without changing the free energy, so the analytic expansion contains only even powers. The term $-{\mu }_{0}HM$ breaks this symmetry at fixed nonzero H. Stationarity and local stability requireFor $T>T_c$ at small field, the cubic term is higher order and $M\simeq {\mu }_{0}H/a(T)$. Thus the susceptibility as defined in the question isThe three requested free-energy sketches and their minima are given separately above.

For the plotted isotherms, divide the equation of state by nonzero M and rearrange:This is an Arrott plot: its slope determines b, and the isotherm through the origin identifies the mean-field Curie temperature.^[ref] Under the paper's stated uniform-field Landau model, use H as supplied on its axis.

The ordinate unit is $(10^4\mathrm{A}{\mathrm{m}}^{-1}{)}^2=10^8{\mathrm{A}}^2{\mathrm{m}}^{-2}$. Reading the original graph gives slopes about $0.076$–$0.080$ ordinate units per unit H/M, so the physical slope is approximately $7.8\times 10^6{\mathrm{A}}^2{\mathrm{m}}^{-2}$. ConsequentlyApproximate horizontal intercepts and the corresponding $a={\mu }_0(H/M{)}_{M^2=0}$ are:

The sign change brackets the transition between 19 and 25 K. A linear fit of $a(T)=\alpha (T-T_c)$ across the four read-offs gives approximately $\alpha =4.8\times 10^{-6}\mathrm{H}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ andInterpolating only the two nearest isotherms also gives about 22 K. These are graph estimates, not exact recovered measurements.

Below $T_c$ at zero field, the stable branches have $M_0^2=-a/b$. Substitution gives the free-energy density relative to the omitted smooth background:The equilibrium excess heat-capacity density is $c_H=-T{\partial }_T^2{F}_{\min }=T{\alpha }^2/(2b)$ below $T_c$ and zero above it. ThereforeThe all-isotherm linear fit gives about $1.5\times 10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. Estimating the slope only between the 19 K and 25 K curves instead gives $\alpha \simeq 6.9\times 10^{-6}\mathrm{H}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and a jump near $3\times 10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. The lines are not exactly parallel and the intercepts are not perfectly linear in T, so a jump of order a few $10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$ is supported; finer precision is not. This is per volume because F is a density; no sample volume is given to convert it to a total heat capacity.

If $b<0$, the quartic truncation is unbounded below. Add a positive sixth-order term $c{M}^6/6$ with $c>0$. At zero field the coexistence conditions giveThus the zero-field transition is generically first order, with a discontinuous spontaneous magnetisation and possible metastability. It occurs before the quadratic coefficient reaches zero. The case $b=0$ is the tricritical boundary of this mean-field model; no value of c is invented.

Write $z=e^{\beta (\mu -ϵ)}$. For one nondegenerate fermionic quantum state, the allowed occupations 0 and 1 give ${\Xi }_F=1+z$. For a bosonic state, summing all nonnegative occupations gives ${\Xi }_B=(1-z{)}^{-1}$, requiring $z<1$ or $\mu <ϵ$. Differentiating the logarithm,The interacting point defect has two distinct spin states, not one fermionic state. Its four microstates have grand-canonical weights $1,z,z,q{z}^2$, with $q=e^{-\beta U}$. The first two single-electron weights are separate up/down spin possibilities; the double state has one electron of each spin. ThusFor $U=0$, $\Xi =(1+z{)}^2$ and $\bar{n}=2z/(1+z)$. The sketch is a single smooth step from 0 to 2, centred on $\mu =ϵ$, where $\bar{n}=1$. As $T\to 0$ its width tends to zero.

For positive U with $\beta U\gg 1$, the lowest grand energies are successively $0$, $ϵ-\mu$ and $2ϵ+U-2\mu$. Away from the two crossings,Finite temperature smooths each step over an energy range of order $k_{B}T$. Near the lower step double occupation is exponentially suppressed, so $\bar{n}\simeq 2z/(1+2z)$. Near the upper step put $y=e^{\beta (\mu -ϵ-U)}$ and neglect the empty state, giving $\bar{n}\simeq 1+y/(2+y)$. Spin degeneracy shifts the half-step positions to $ϵ-k_{B}T\ln 2$ and $ϵ+U+k_{B}T\ln 2$.

For a grand-canonical probability $p_s\propto e^{-\beta (E_s-\mu n_s)}$, differentiation gives ${\partial }_{\mu }{p}_s=\beta (n_s-\bar{n})p_s$. Multiplying by $n_s$ and summing provesFor this defect, direct state counting givesAt $\mu =ϵ$, $z=1$ and thereforeThe same limit gives $\bar{n}=2/3$: empty, up and down states have equal weights, while the double state is suppressed. Replacing this degeneracy by a two-outcome equal-weight model would incorrectly give variance $1/4$ at this chemical potential.

Across the full chemical-potential range the variance has two well-separated narrow peaks and tends to zero on the 0, 1 and 2 plateaux. The limiting shapes are $2z/(1+2z{)}^2$ near the lower step and $2y/(2+y{)}^2$ near the upper one. Each peak has height $1/4$, at the shifted half-step position; the variance at the unshifted crossings is instead $2/9$.