= Solution
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, so
$$
\boxed{p_n=1/\Omega\text{ on the shell},\qquad S=k_B\log\Omega.}
$$
For the canonical case impose both normalization and $\sum p_nE_n=U$. Stationarity of $-k_B\sum p_n\log p_n-\lambda(\sum p_n-1)-k_B\beta(\sum p_nE_n-U)$ gives $p_n\propto e^{-\beta E_n}$. Thus
$$
\boxed{p_n=e^{-\beta E_n}/Z,\qquad Z=\sum_ne^{-\beta E_n},\qquad\beta=(k_BT)^{-1}.}
$$
The 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 $z=\epsilon/(k_BT)$ and $Z_1=1+2\cosh z$. The full <partition function> is $Z_1^N$ because the particles occupy labeled lattice sites. Differentiation gives
$$
\boxed{\overline E=-\frac{2N\epsilon\sinh z}{1+2\cosh z},\qquad C=Nk_Bz^2\frac{2\cosh z+4}{(1+2\cosh z)^2}.}
$$
The heat capacity also equals $N\operatorname{Var}(E_1)/(k_BT^2)$, confirming positivity. As $T\to+\infty$, the three occupations become equal:
$$
\boxed{\overline E\sim-\frac{2N\epsilon^2}{3k_BT}\to0,\qquad C\sim\frac{2N\epsilon^2}{3k_BT^2}\to0.}
$$
As $T\downarrow0$, all particles occupy their unique lowest state:
$$
\boxed{\overline E=-N\epsilon+N\epsilon e^{-z}+O(e^{-2z}),\qquad C\sim Nk_Bz^2e^{-z}\to0.}
$$
A <negative temperature> is possible because the total energy is bounded above. An equilibrated population inversion has more particles at $+\epsilon$ than at $-\epsilon$. For example probabilities $(p_-,p_0,p_+)=(1,2,4)/7$ correspond to $T=-\epsilon/(k_B\log2)$ and mean energy $3N\epsilon/7$. 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: \b[the system obeys the third law of thermodynamics]. The negative-temperature branch likewise approaches the unique highest-energy state as $T\to0^-$; it does not introduce residual ground-state entropy.
Back to article page
