Canonical ensemble

At fixed temperature, the canonical ensemble assigns a state of energy $E_s$ probability proportional to $e^{-\beta E_s}$.

The canonical partition functionnormalizes Boltzmann probabilities and generates equilibrium thermodynamic quantities.

The trace of imaginary-time evolution is the canonical partition function of a quantum system. In a coordinate basis it is the integral of the diagonal Euclidean path integral kernel. Inserting a symmetry operator instead gives a weighted thermal trace.

For two equal-frequency quantum harmonic oscillators, the ordinary thermal trace is $r/(1-r{)}^2$ and the trace with the parity operator is $r/(1+r{)}^2$, where $r=e^{-\beta \omega }$. Total excitation $N$ has degeneracy $N+1$ and parity $(-1{)}^N$. Therefore the traces expand as ${\sum }_{n\ge 1}(\pm 1{)}^{n-1}n{r}^n$. The alternating weights come from parity insertion in a bosonic state space, not fermionic statistics.

Canonical differentiation gives

Canonical energy variance equals ${\partial }_{\beta }^2\log Z=k_B{T}^2{C}_V$.

Forput $a=\kappa R^2/2$. The normalized classical one-particle partition function isIts mean total and potential energies are

If $a\gg k_{B}T$, the wall is thermally inaccessible and $⟨E⟩\sim 2k_{B}T$, as four quadratic terms predict. If $a\ll k_{B}T$, the spatial distribution is nearly uniform in the disk andso only the two kinetic quadratic terms contribute at leading order.

For $N$ indistinguishable noninteracting classical particles, $Z_N=z_1^N/N!$, where $z_1$ is the normalized one-particle phase-space integral.

The three-dimensional translational momentum integral is $1/{\lambda }^3$ per unit volume, where

For energy $pc+U(x)$,If $U(r)=r^{2n}/V^{2n/3}$, thenConsequently $pV=N{k}_{B}T$, while the mean energy and variance are $NA{k}_{B}T$ and $NA(k_{B}T{)}^2$ with $A=3+3/(2n)$.

The radial density is proportional to $r^2{e}^{-r^{2n}/(V^{2n/3}{k}_{B}T)}$, whose mode is

In the classical dilute regime, a monatomic ideal gas has entropy

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.

For $N$ independent spins with $s_z\in \{-1,0,1\}$ and one-spin Hamiltonian $-\mu B{s}_z$,Writing $x=\beta \mu B$, its heat capacity is

For $N$ noninteracting units with energies $-ϵ,0,ϵ$, put $x=\beta ϵ$. Thenand

Negative absolute temperature can occur only for a system with an energy spectrum bounded above. Since $\beta =\partial S/\partial E<0$, higher-energy states are more populated than lower-energy states.

Instantaneously reversing a Hamiltonian $H(B)$ satisfying $H(-B)=-H(B)$ leaves populations unchanged but rewrites them as a canonical distribution with ${\beta }^{\prime }=-\beta$. When the partition function is unchanged, the free energy changes sign and the heat capacity is unchanged.

A negative-temperature system is hotter than every positive-temperature system. If heat $\delta Q$ passes from it to a positive-temperature body, the total entropy change is

Each independent quadratic term in a classical Hamiltonian contributes $k_{B}T/2$ to the mean energy.

In the dilute Maxwell--Boltzmann regime, a species has number density proportional to $g(m{k}_{B}T{)}^{3/2}{e}^{(\mu -m{c}^2)/(k_{B}T)}$.

At inverse temperature $\beta =1/(k_{B}T)$, the Boltzmann factor $e^{-\beta E}$ is the relative canonical ensemble weight of a microstate with energy $E$.