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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.