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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.

The microcanonical ensemble describes an isolated system at fixed energy, volume and particle number. The canonical ensemble describes fixed particle number and volume with heat exchange at a prescribed temperature. The grand canonical ensemble additionally allows particles to exchange with a reservoir, prescribing temperature, volume and chemical potential. Their respective equilibrium probabilities are uniform on the accessible energy shell, proportional to $e^{-\beta E_n}$, and proportional to $e^{-\beta (E_n-\mu N_n)}$. For stable short-range systems in the thermodynamic limit, concentration of relative energy and particle-number fluctuations and concavity of entropy make their bulk predictions equivalent; phase coexistence or nonadditive interactions require care.

For the microcanonical case restrict the allowed states to the fixed-energy shell of size $\Omega$, imposing $p_n=0$ outside it. Maximize the Gibbs entropy subject to $\sum p_n=1$. A Lagrange multiplier gives $-k_B(\log p_n+1)+\lambda =0$ on the shell, soFor the canonical case impose both normalization and $\sum p_n{E}_n=U$. Stationarity of $-k_B\sum p_n\log p_n-\lambda (\sum p_n-1)-k_B\beta (\sum p_n{E}_n-U)$ gives $p_n\propto e^{-\beta E_n}$. ThusThe last identification follows from $\partial S/\partial U=1/T$. Entropy is strictly concave in the probabilities, so these stationary distributions are maxima.

For the independent three-level particles, put $z=ϵ/(k_{B}T)$ and $Z_1=1+2\cosh z$. The full partition function is $Z_1^N$ because the particles occupy labeled lattice sites. Differentiation givesThe heat capacity also equals $N\mathrm{Var}(E_1)/(k_B{T}^2)$, confirming positivity. As $T\to +\infty$, the three occupations become equal:As $T\downarrow 0$, all particles occupy their unique lowest state:

A negative temperature is possible because the total energy is bounded above. An equilibrated population inversion has more particles at $+ϵ$ than at $-ϵ$. For example probabilities $(p_{-},p_0,p_{+})=(1,2,4)/7$ correspond to $T=-ϵ/(k_B\log 2)$ and mean energy $3Nϵ/7$. This is a macroscopic occupation pattern; an isolated individual microstate is not itself assigned a temperature. At the positive zero-temperature limit the ground state is unique, so the entropy tends to zero: the system obeys the third law of thermodynamics. The negative-temperature branch likewise approaches the unique highest-energy state as $T\to 0^{-}$; it does not introduce residual ground-state entropy.

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.

Specify the magnetic coupling before fixing its sign: let the magnetic moment be $\mathbf{\mu }=\gamma \mathbf{S}$, where $\gamma$ is a signed gyromagnetic ratio. The Zeeman effect gives $H_Z=-\gamma \mathbf{B}\cdot \mathbf{S}$. For spin one-half, $\mathbf{S}=\hslash \mathbf{\sigma }/2$, so with $B=|\mathbf{B}|$ and $u=\beta \gamma \hslash B/2$,This is the quantum Gibbs state. Its eigenvalues are $(1\pm \tanh u)/2$, so its purity is $(1+{\tanh }^{2}u)/2$. For nonzero level splitting it approaches a pure state as $|u|\to \infty$: positive inverse temperature selects the lower energy, while negative inverse temperature selects the upper one. It is a maximally mixed state at $u=0$, including infinite temperature or zero magnetic splitting. At zero field the Gibbs state remains $I/2$; lowering temperature alone does not select a unique pure state from that degeneracy.

Negative temperature is meaningful for an internally equilibrated spin sector with a bounded energy spectrum and population inversion. It must be sufficiently isolated from unbounded motional degrees of freedom: the oscillator's unbounded spectrum, for instance, makes its partition sum diverge at negative inverse temperature. The two-level spin partition sum itself stays finite for either sign.

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.

Temperature labels thermal equilibrium: two internally equilibrated bodies have the same temperature when contact permitting only heat exchange produces no net heat transfer. The zeroth law of thermodynamics makes the comparison transitive. If bodies A and B separately reach thermal equilibrium with the same thermometer without appreciably disturbing either body, they agree with each other. This is what lets a gas thermometer compare a metal block with water, despite their different internal energies and heat capacities. Insulation alone makes no such comparison: it can prevent heat transfer between bodies at different temperatures. ^[ref]

The zeroth law of thermodynamics establishes equality comparisons, but does not choose a numerical scale. Thermodynamic temperature supplies an absolute scale consistent with Carnot's theorem and entropy. For a simple compressible system at thermodynamic equilibrium with fixed particle number $N$ and composition,Here $U$ is internal energy, $S$ is entropy, and $V$ is volume; the second equality assumes the derivative can be inverted. Temperature is a thermodynamic conjugate variable to entropy and belongs to the intensive and extensive thermodynamic quantities as an intensive quantity. Its partial derivative tells us how much internal energy must change per unit change of entropy under the stated constraints. More general systems require their other mechanical and material variables to be held fixed too. ^[ref]

For an infinitesimal quantity $\delta Q>0$ of heat transferred from A to B, with no work or matter exchange, the equilibrium entropy changes giveFor ordinary positive temperatures, spontaneous heat transfer from $T_A>T_B$ increases total entropy; thermal equilibrium has $T_A=T_B$. This comparison uses changes in entropy, rather than the total internal energy contained in either body. A small hot object can have less internal energy than a large cool one. See entropy maximization under thermal contact for the equilibrium criterion and negative temperature for the special case of bounded energy spectra.

Express absolute thermodynamic temperature $T$ in kelvin, and a Celsius reading $t_C$ in degree Celsius:The offset is exact: $300\mathrm{K}$ corresponds to $26.85{}^{\circ }\mathrm{C}$, and $400\mathrm{K}$ to $126.85{}^{\circ }\mathrm{C}$. Their absolute temperature ratio is $4/3$; dividing their Celsius readings gives a different number. Thus gas thermometer pressure ratios and Carnot's theorem require thermodynamic temperature. The modern SI definition of the kelvin fixes the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$; a reference bath realizes or transfers the scale rather than defining that constant.^[ref]

Temperature is not universally the mean kinetic energy of a particle. In a classical, nonrelativistic ideal gas at thermal equilibrium, the equipartition theorem gives the mean translational kinetic energy relative to the gas's bulk motion:Here $m$ is particle mass, $\mathbf{v}$ its velocity, and $\mathbf{u}$ the mean flow velocity. In this regime, different particle masses at the same temperature have equal mean translational kinetic energies but different characteristic speeds. A moving gas can also carry bulk kinetic energy without being hotter in its rest frame. Internal molecular modes and quantum statistics require further care; the equipartition theorem has classical assumptions, whereas the equilibrium entropy definition applies more widely. ^[ref]