Canonical Ensemble

The canonical ensemble describes a system that can exchange energy with a thermal environment while its particle number and external control parameters are fixed. Its standard macroscopic controls are

$$T,\qquad V,\qquad N,$$

with Hamiltonian $H_N(V,\lambda,\ldots)$ acting on the fixed-$N$ Hilbert space.

The equilibrium state is

$$\rho_{N,\beta} = \frac{ e^{-\beta H_N} }{ Z_N },$$

where

$$Z_N(\beta,V,\lambda,\ldots) = \operatorname{Tr}_{\mathcal H_N} e^{-\beta H_N}.$$

The subscript $N$ is often suppressed, but the trace domain is part of the ensemble definition. The general operator properties of this state live in Thermal Density Operators. Here the focus is fixed-$T,N,V$ thermodynamics.

Helpful background. Use thermal density operators for the spectral Gibbs state and ensemble comparison for the trace-space choice.

Fixed temperature and fluctuating energy

Quantity	Canonical status
temperature $T$	externally fixed
volume $V$ and other controls	externally fixed
particle number $N$	fixed exactly
energy $E$	fluctuates
density operator	$\rho_{N,\beta}$
thermodynamic potential	Helmholtz free energy $F$

The ensemble describes a probability distribution over energy eigenstates in one fixed-particle-number sector. It does not imply that particles are exchanged with the environment. Particle exchange belongs to the grand-canonical ensemble.

Boltzmann weights from a large reservoir

Let a system exchange energy with a much larger bath. Neglect their interaction energy in the equilibrium state count and assign equal probabilities in the composite energy shell. If the system occupies one eigenstate of energy $E_n$, its weight is proportional to the bath’s compatible state count $W_B(E_{\mathrm{tot}}-E_n)$.

Choose a reference system energy $U_*$ and write $\delta E_n=E_n-U_*$ and $E_B^*=E_{\mathrm{tot}}-U_*$. Expanding the smooth bath entropy about $E_B^*$ gives

$$S_B(E_B^*-\delta E_n) =S_B(E_B^*)-\frac{\delta E_n}{T} -\frac{(\delta E_n)^2}{2T^2 C_B}+\cdots.$$

Here $T$ and $C_B=(\partial E_B/\partial T)$ are evaluated at the reference bath state. Dividing by $k_{\mathrm B}$ and exponentiating yields

$$p_n\propto e^{-\beta E_n} \exp\!\left[-\frac{(\delta E_n)^2}{2k_{\mathrm B}T^2 C_B}+\cdots\right].$$

The reference-dependent constant cancels in normalization. An effectively infinite bath makes the correction negligible throughout the energies the system appreciably samples. For a stable bath with $C_B>0$, the quadratic correction suppresses excursions relative to the ideal Boltzmann weights. Expanding about a reference energy makes this criterion independent of the energy zero.

This derivation assumes an equilibrium shell assignment and meaningful subsystem energies. Weak contact can permit equilibration while contributing negligible equilibrium interaction energy. When that approximation fails, the reduced state involves a Hamiltonian of mean force.

The Partition Function as a Generating Object

The logarithm of the fixed-sector trace determines equilibrium energy, entropy, and free energy. The following derivatives hold with the other external parameters fixed; see partition functions for spectral evaluation and cumulants.

Define

$$F(T,V,N,\lambda,\ldots) = -k_{\mathrm B}T\ln Z_N.$$

Equivalently,

$$\beta F = -\ln Z_N.$$

The internal energy is

$$U = \langle H_N\rangle = -\frac{\partial}{\partial\beta} \ln Z_N,$$

for a Hamiltonian with no explicit temperature or $\beta$ dependence.

Another useful identity is

$$U = \frac{\partial}{\partial\beta} \left( \beta F \right).$$

The entropy follows from

$$S = -\left( \frac{\partial F}{\partial T} \right)_{V,N,\lambda}$$

and equals

$$S = k_{\mathrm B} \left( \ln Z_N+\beta U \right).$$

These relations use the natural logarithm.

Helmholtz Free Energy

The Helmholtz free energy is the thermodynamic potential appropriate to fixed $T$, $V$, and $N$:

$$F = U-TS.$$

At positive temperature, the Gibbs state minimizes the state functional

$$\mathcal F_T(\rho) = \operatorname{Tr}(\rho H_N) - TS(\rho)$$

over normalized density operators on $\mathcal H_N$.

The equilibrium value is

$$\mathcal F_T(\rho_{N,\beta}) = F = -k_{\mathrm B}T\ln Z_N.$$

Free energy balances energy against entropy. At positive temperature, low energy and large entropy both lower this functional. For a normalizable negative-temperature Gibbs state the variational inequality reverses; see the relative-entropy identity.

Generalized Forces

Let the Hamiltonian depend on an external parameter $\lambda$. Even when $H$ and $\partial_\lambda H$ do not commute, cyclicity of the trace gives

$$\frac{\partial F}{\partial\lambda} = \left\langle \frac{\partial H}{\partial\lambda} \right\rangle_\beta.$$

If the physical generalized force is defined by

$$\mathcal X_\lambda = -\frac{\partial H}{\partial\lambda},$$

then

$$\langle\mathcal X_\lambda\rangle = -\frac{\partial F}{\partial\lambda}.$$

Examples include:

$$P = -\left( \frac{\partial F}{\partial V} \right)_{T,N}$$

for pressure and

$$M = -\left( \frac{\partial F}{\partial B} \right)_{T,N}$$

for magnetization when the Hamiltonian contains the appropriate magnetic-field coupling.

Sign conventions follow the definition of the Hamiltonian term. They should be checked rather than memorized independently of $H$.

Differential Form

For a simple system,

$$dF = -S\,dT - P\,dV + \mu\,dN.$$

The canonical ensemble holds $N$ fixed during fluctuations, but $\mu$ can still be defined by comparing neighboring fixed-$N$ systems:

$$\mu = \left( \frac{\partial F}{\partial N} \right)_{T,V}$$

in a thermodynamic or smoothly interpolated description.

At finite integer $N$, addition and removal chemical potentials are naturally finite differences:

$$\begin{aligned} \mu_+ &= F_{N+1}-F_N, \\ \mu_- &= F_N-F_{N-1}. \end{aligned}$$

They need not coincide in a finite system or across a gap.

Energy Fluctuations

The second $\beta$ derivative of $\ln Z_N$ is

$$\frac{\partial^2}{\partial\beta^2} \ln Z_N = \langle H_N^2\rangle - \langle H_N\rangle^2.$$

Therefore

$$\operatorname{Var}(E) = -\frac{\partial U}{\partial\beta}.$$

For a temperature-independent Hamiltonian,

$$C_V \equiv \left( \frac{\partial U}{\partial T} \right)_{V,N} = k_{\mathrm B}\beta^2 \operatorname{Var}(E).$$

Thus

$$C_V\geq0$$

in the canonical ensemble under these assumptions.

This does not prove that every definition of heat capacity in every ensemble is nonnegative. Microcanonical and nonadditive long-range systems can require separate analysis.

Energy distributions and concentration

For distinct energy values $E_a$ with degeneracy $g_a$,

$$P_\beta(E_a)=\frac{g_a e^{-\beta E_a}}{Z_N}.$$

The degeneracy multiplies the probability of the energy value, not the probability of each individual eigenstate. The competition between increasing state count and decreasing Boltzmann weight determines the typical energy.

For regular additive systems away from criticality, $C_V\propto N$, so the root-mean-square fluctuation of energy per particle is

$$\frac{\sqrt{\operatorname{Var}(E)}}{N}\propto N^{-1/2}.$$

The ratio to the mean energy has this scaling only if the chosen energy density is nonzero; it depends on the arbitrary energy zero. Concentration of energy density is the physically useful statement. It helps establish ensemble equivalence under additional locality and regularity assumptions, but global fluctuations still distinguish the ensembles.

Near a critical point the scaling can change. At first-order coexistence the energy distribution can be bimodal, so a Gaussian approximation need not describe it.

Density of states and the dominant energy

The exact density of states is $\omega_N(E)=\sum_a g_a\delta(E-E_a)$, with units of inverse energy. Then

$$Z_N(\beta)=\int dE\,\omega_N(E)e^{-\beta E}.$$

For a smooth approximation, define a dimensionless entropy using a fixed reference energy $\varepsilon_0$:

$$S_\omega(E)=k_{\mathrm B}\ln[\varepsilon_0\omega_N(E)],\qquad Z_N\simeq\int\frac{dE}{\varepsilon_0} e^{S_\omega(E)/k_{\mathrm B}-\beta E}.$$

An interior dominant maximum satisfies $S_\omega'(E_*)=1/T$. If it is sufficiently sharp and unique, $F\simeq E_*-T S_\omega(E_*)$ at leading extensive order. At coexistence several contributions may compete; a stationary point is not enough to identify the dominant one. The microcanonical entropy conventions explain the relation to finite-window counts.

Factorization for Independent Subsystems

Suppose

$$H = H_A\otimes\mathbb I_B + \mathbb I_A\otimes H_B$$

on

$$\mathcal H = \mathcal H_A\otimes\mathcal H_B.$$

The terms commute, so

$$e^{-\beta H} = e^{-\beta H_A} \otimes e^{-\beta H_B}.$$

Therefore

$$Z_{AB} = Z_AZ_B,$$

and

$$F_{AB} = F_A+F_B.$$

This factorization assumes a tensor-product decomposition with independent Hamiltonians. Identical-particle symmetrization, global particle-number constraints, or interactions can prevent a naive product of one-particle partition functions.

Canonical Ensemble and Identical Particles

For $N$ identical particles, the trace must be taken over the appropriate symmetric or antisymmetric $N$-particle subspace:

$$Z_N^{(\pm)} = \operatorname{Tr}_{\mathcal H_N^{(\pm)}} e^{-\beta H_N}.$$

One should not begin with an $N$-fold labeled-particle trace and forget to impose exchange symmetry.

For a dilute classical gas, a factor of $1/N!$ emerges in the appropriate limit. For a quantum-degenerate gas, Bose–Einstein or Fermi–Dirac statistics must be treated directly.

The canonical ensemble fixes $N$ exactly. It can still describe identical particles, occupation numbers, and exchange correlations.

Finite-Size Analyticity

For a finite-dimensional Hamiltonian $H(g)$ analytic in a real parameter $g$,

$$Z_N(\beta,g) = \operatorname{Tr} e^{-\beta H(g)} > 0$$

at finite positive temperature. Therefore

$$F_N(\beta,g) = -\frac{1}{\beta} \ln Z_N(\beta,g)$$

is analytic under the stated assumptions.

Finite systems can have sharp peaks, rapid crossovers, and bimodal distributions. A genuine thermal nonanalyticity requires an infinite-system or another singular limit.

Zero temperature is different: taking $\beta\to\infty$ can expose a finite-system level crossing in the ground-state energy.

Temperature-Dependent Effective Hamiltonians

The familiar identity

$$U = -\partial_\beta\ln Z$$

assumes $H$ has no explicit $\beta$ dependence. If an effective Hamiltonian is written as $H(\beta)$, then

$$-\frac{\partial}{\partial\beta} \ln Z = \left\langle H + \beta \frac{\partial H}{\partial\beta} \right\rangle.$$

The extra term cannot be dropped.

Temperature-dependent effective Hamiltonians can arise after integrating out degrees of freedom or fitting phenomenological parameters. Their thermodynamic consistency must be derived from the underlying model rather than assumed from formal resemblance.

Example: One Two-Level System

Let

$$H = \Delta|1\rangle\langle1|, \qquad \Delta>0.$$

With

$$x = \beta\Delta,$$

the partition function is

$$Z_1 = 1+e^{-x}.$$

The free energy and internal energy are

$$\begin{aligned} F_1 &= -k_{\mathrm B}T \ln(1+e^{-x}), \\ U_1 &= \frac{\Delta}{ e^x+1 }. \end{aligned}$$

The energy variance is

$$\operatorname{Var}(E) = \Delta^2 \frac{e^x}{ (1+e^x)^2 }.$$

Therefore

$$\frac{C_1}{k_{\mathrm B}} = x^2 \frac{e^x}{ (1+e^x)^2 }.$$

The heat capacity vanishes as $T\to0$ and $T\to\infty$ and has a finite-temperature Schottky peak.

Example: Independent Two-Level Sites

For $L$ distinguishable noninteracting copies,

$$H_L = \sum_{j=1}^{L} \Delta |1\rangle_j\langle1|.$$

Factorization gives

$$Z_L = (1+e^{-x})^L.$$

Consequently,

$$\begin{gathered} F_L = L F_1, \\ U_L = L U_1, \\ C_L = L C_1. \end{gathered}$$

The energy standard deviation grows as $\sqrt L$, while the mean energy grows as $L$ at fixed finite temperature. Relative energy fluctuations therefore scale as $L^{-1/2}$.

Example: Harmonic Oscillator

For

$$H = \hbar\omega \left( a^\dagger a+\frac12 \right),$$

define

$$x = \beta\hbar\omega.$$

The partition function is

$$Z = \frac{ e^{-x/2} }{ 1-e^{-x} }.$$

The internal energy is

$$U = \hbar\omega \left( \frac12 + \frac{1}{e^x-1} \right).$$

The heat capacity is

$$\frac{C}{k_{\mathrm B}} = x^2 \frac{e^x}{ (e^x-1)^2 }.$$

As $T\to0$, $U\to\hbar\omega/2$ and $C\to0$. At high temperature, $U$ approaches $k_{\mathrm B}T$ and $C$ approaches $k_{\mathrm B}$.

Exercises

Shift the energy zero

If $H'=H+C\mathbb I$, determine how $\rho_\beta$, $Z$, $F$, $U$, $S$, and $C_V$ change.

Solution

The partition function changes by

$$Z' = e^{-\beta C}Z.$$

The normalized state is unchanged:

$$\rho_\beta' = \rho_\beta.$$

The free energy and internal energy shift:

$$F'=F+C, \qquad U'=U+C.$$

The entropy is

$$S' = \frac{U'-F'}{T} = S,$$

and $C_V=\partial U/\partial T$ is unchanged when $C$ is temperature independent.

Independent-site factorization

For $L$ independent two-level sites with energies $0$ and $\Delta$, derive $Z_L$, $F_L$, and the probability that exactly $m$ sites are excited.

Solution

Each site has

$$Z_1 = 1+e^{-\beta\Delta}.$$

Independence gives

$$Z_L = Z_1^L, \qquad F_L = L F_1.$$

There are $\binom{L}{m}$ configurations with $m$ excitations, each of energy $m\Delta$. Hence

$$\Pr(m) = \binom{L}{m} \frac{ e^{-\beta m\Delta} }{ (1+e^{-\beta\Delta})^L }.$$

This is a binomial distribution with single-site excitation probability $1/(e^{\beta\Delta}+1)$.

Fluctuation identity

Show that

$$\operatorname{Var}(E) = \frac{\partial^2}{\partial\beta^2} \ln Z.$$

Solution

First,

$$\frac{\partial}{\partial\beta} \ln Z = -\langle H\rangle.$$

Differentiating again,

$$\begin{aligned} \frac{\partial^2}{\partial\beta^2} \ln Z &= \frac{1}{Z} \operatorname{Tr} \left( H^2e^{-\beta H} \right) - \left[ \frac{1}{Z} \operatorname{Tr} \left( He^{-\beta H} \right) \right]^2 \\ &= \langle H^2\rangle - \langle H\rangle^2. \end{aligned}$$

Noncommuting generalized force

For $H(\lambda)=\Delta\sigma_z+\lambda\sigma_x$ with fixed $\Delta>0$, calculate $F$ and verify $\partial_\lambda F=\langle\sigma_x\rangle$ even though $[H,\sigma_x]\ne0$.

Solution

Since $H^2=r^2 I$, where $r=\sqrt{\Delta^2+\lambda^2}$, its eigenvalues are $\pm r$. Thus

$$F=-\frac1\beta\ln[2\cosh(\beta r)],\qquad \partial_\lambda F=-\frac{\lambda}{r}\tanh(\beta r).$$

The density operator is $\rho=\tfrac12[I-\tanh(\beta r)H/r]$. Using $\operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij}$ gives the same result for $\langle\sigma_x\rangle$. Cyclicity of the trace justifies the first derivative without replacing a noncommuting operator exponential by scalar differentiation.

References

• H. B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed., Wiley (1985).
• K. Huang, Statistical Mechanics, 2nd ed., Wiley (1987).
• M. Kardar, Statistical Physics of Particles, Cambridge University Press (2007).
• L. D. Landau and E. M. Lifshitz, Statistical Physics, Part 1, 3rd ed., Butterworth–Heinemann (1980).
• R. K. Pathria and P. D. Beale, Statistical Mechanics, 3rd ed., Elsevier (2011).
• L. E. Reichl, A Modern Course in Statistical Physics, 4th ed., Wiley–VCH (2016).