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For $N$ distinguishable noninteracting spins, let a fraction $\alpha$ occupy the level $+ϵ$ and the remainder occupy $-ϵ$. The Stirling formula givesThe relation $1/T=\partial S/\partial E$ yieldsPositive temperature corresponds to $0\le \alpha <1/2$; a population inversion $1/2<\alpha \le 1$ has negative temperature.

Let two systems exchange energy inside an otherwise isolated composite system, so $E_1+E_2=E$ is fixed. By the Second law of thermodynamics, equilibrium maximizes the total entropyAt an interior maximum,Thus $T_1=T_2$, which is entropy maximization under thermal contact and the condition for thermal equilibrium.

Stability requires this entropy maximum to be locally strict, so each ordinary subsystem has a concave entropy-energy relation. Sinceone has $\partial T/\partial E>0$, or equivalentlyThis is the entropy concavity and positive heat capacity criterion.

Now let $N_{\uparrow }=\alpha N$ and $N_{\downarrow }=(1-\alpha )N$. Because the spins are distinguishable, the number of microstates is the binomial coefficientThe Boltzmann entropy $S=k_B\log \Omega$ and the Stirling formula give, to leading order for large $N$,The total energy isThereforeSolving for the up-spin fraction gives the independent spin-one-half two-level system resultFor ordinary positive absolute temperature,with $\alpha \to 0$ as $T\to 0^{+}$ and $\alpha \to 1/2$ as $T\to +\infty$. Since the spectrum is bounded above, the population-inverted range $1/2<\alpha \le 1$ is also mathematically possible and corresponds to negative temperature; $\alpha =1/2$ is the infinite-temperature state.

Substitution givesand direct differentiation yieldsThus the energy increases with temperature throughout either finite-temperature branch, in particular throughout the required positive-temperature range.

Because the one-particle spectrum is bounded above, a negative temperature is possible. A population-inverted configuration with more particles in the $+ϵ$ level than in the $-ϵ$ level haswhich requires $\beta <0$. Equivalently, above the maximum-entropy energy $E=0$, adding energy reduces the number of compatible microstates, so $\partial S/\partial E=1/T<0$. The limiting configuration with every particle at $+ϵ$ has $E=Nϵ$, zero entropy, and $T\to 0^{-}$.

Immediately after the reversal the state populations have not changed, but the new Hamiltonian satisfiesThusso the same state is canonical for the reversed field withThis is the magnetic-field reversal and negative temperature construction. Sincethe partition function is unchanged. Therefore

A gas has kinetic energy unbounded above. Its partition function diverges for $\beta <0$, so it cannot have a normalizable negative temperature canonical state and cannot satisfy the zeroth-law equilibrium condition ${\beta }_{\mathrm{gas}}={\beta }^{\prime }<0$.

If an amount $\delta Q>0$ flows from the spins into a positive-temperature gas, the total entropy change isThe second law therefore selects heat flowThis is the general ordering expressed by heat flow from negative to positive temperature.