Ensembles

In the previous lesson, we saw that entropy counts microstates. But how do we actually compute thermodynamic quantities from microscopic physics? The answer is the ensemble: a collection of mental copies of a system, one for each possible microstate, weighted by probability. By averaging over the ensemble, we extract macroscopic physics from microscopic laws.

Different physical situations call for different ensembles. The choice depends on what the system can exchange with its environment: nothing, energy, or both energy and particles.

The Microcanonical Ensemble

The simplest case: an isolated system with fixed energy $E$, volume $V$, and particle number $N$. No exchange of anything with the outside world.

The fundamental postulate of statistical mechanics states that all microstates with the correct energy are equally probable:

$$P_i = \begin{cases} 1/\Omega(E) & \text{if } E_i = E \\ 0 & \text{otherwise} \end{cases}$$

where $\Omega(E)$ is the number of microstates with energy $E$. The entropy is $S = k_B \ln \Omega(E)$, and all thermodynamics follows.

While conceptually simple, the microcanonical ensemble is often difficult to work with in practice because fixing the energy exactly is mathematically awkward. A much more powerful approach is to let the energy fluctuate.

The Canonical Ensemble

Consider a system in thermal contact with a large heat bath at temperature $T$. The system can exchange energy with the bath, but particle number and volume remain fixed. What is the probability of finding the system in microstate $i$ with energy $E_i$?

The derivation is elegant. If the system has energy $E_i$, the bath has energy $E_{\text{total}} - E_i$. The number of bath microstates is $\Omega_{\text{bath}}(E_{\text{total}} - E_i)$. Since all microstates of the combined system are equally likely:

$$P_i \propto \Omega_{\text{bath}}(E_{\text{total}} - E_i) = e^{S_{\text{bath}}(E_{\text{total}} - E_i)/k_B}$$

Expanding the entropy for small $E_i$ compared to the bath energy:

$$S_{\text{bath}}(E_{\text{total}} - E_i) \approx S_{\text{bath}}(E_{\text{total}}) - \frac{\partial S_{\text{bath}}}{\partial E} E_i = \text{const} - \frac{E_i}{T}$$

giving $P_i \propto e^{-E_i / k_B T} = e^{-\beta E_i}$.

The Partition Function

The partition function $Z$ is the most powerful object in statistical mechanics. It is simply the sum of Boltzmann factors over all microstates:

Why is $Z$ so powerful? Because derivatives of $\ln Z$ give thermodynamic quantities directly:

Average energy:

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

Energy fluctuations:

$$\langle (\Delta E)^2 \rangle = \frac{\partial^2 \ln Z}{\partial \beta^2} = k_B T^2 C_V$$

Entropy:

$$S = k_B \left(\ln Z + \beta \langle E \rangle\right)$$

Pressure:

$$P = k_B T \frac{\partial \ln Z}{\partial V}$$

Knowing $Z$ means knowing everything about the equilibrium thermodynamics of the system.

Example: The Two-Level System

Consider a single particle with two states: ground state (energy $0$) and excited state (energy $\epsilon$). The partition function is:

$$Z = 1 + e^{-\beta \epsilon}$$

The average energy is:

$$\langle E \rangle = \frac{\epsilon \, e^{-\beta\epsilon}}{1 + e^{-\beta\epsilon}} = \frac{\epsilon}{e^{\beta\epsilon} + 1}$$

At low temperature ($\beta\epsilon \gg 1$), $\langle E \rangle \to 0$: the particle is frozen in the ground state. At high temperature ($\beta\epsilon \ll 1$), $\langle E \rangle \to \epsilon/2$: both states are equally occupied.

Helmholtz Free Energy

The partition function is directly related to the Helmholtz free energy $F$:

The relationship $F = -k_B T \ln Z$ is one of the most important equations in physics. It connects the microscopic (the partition function, summing over all microstates) to the macroscopic (free energy, a thermodynamic potential).

From $F$, all thermodynamic quantities follow:

$$S = -\frac{\partial F}{\partial T}\bigg|_V, \quad P = -\frac{\partial F}{\partial V}\bigg|_T, \quad E = F + TS$$

The Grand Canonical Ensemble

When the system can exchange both energy and particles with a reservoir, we need the grand canonical ensemble. This is characterized by temperature $T$ and chemical potential $\mu$, which controls the average particle number.

The grand partition function sums over all particle numbers and all microstates for each:

The chemical potential $\mu$ controls the average number of particles, just as temperature controls the average energy. The associated thermodynamic potential is the grand potential:

$$\Phi = -k_B T \ln \Xi = F - \mu N = E - TS - \mu N$$

For an ideal gas, $\Phi = -PV$, giving the equation of state directly.

Thermodynamic Potentials

Each ensemble naturally produces a different thermodynamic potential. These are related by Legendre transforms:

Potential	Definition	Natural Variables	Ensemble
Energy $E$	$E$	$S, V, N$	Microcanonical
Helmholtz $F$	$E - TS$	$T, V, N$	Canonical
Gibbs $G$	$E - TS + PV$	$T, P, N$	Isothermal-isobaric
Grand $\Phi$	$E - TS - \mu N$	$T, V, \mu$	Grand canonical

The Gibbs free energy $G$ is particularly important for chemistry and phase transitions, since laboratory experiments typically fix temperature and pressure. At equilibrium, $G$ is minimized. Chemical reactions proceed in the direction that lowers $G$.

The Equipartition Theorem

A beautiful consequence of the canonical ensemble: each quadratic degree of freedom in the energy contributes $\frac{1}{2}k_B T$ to the average energy.

For a monatomic ideal gas, each molecule has 3 translational degrees of freedom ($\frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}mv_z^2$), giving:

$$\langle E \rangle = \frac{3}{2}Nk_B T, \quad C_V = \frac{3}{2}Nk_B$$

For a diatomic gas, adding 2 rotational degrees of freedom gives $C_V = \frac{5}{2}Nk_B$. But experiments show that at very low temperatures, rotational modes "freeze out", a quantum effect that equipartition cannot explain. This was one of the early hints that classical statistical mechanics was incomplete.

Equivalence of Ensembles

A remarkable result: in the thermodynamic limit ($N \to \infty$), all ensembles give identical results for thermodynamic quantities. Energy fluctuations in the canonical ensemble are tiny:

$$\frac{\sqrt{\langle (\Delta E)^2 \rangle}}{\langle E \rangle} \sim \frac{1}{\sqrt{N}}$$

For $N \sim 10^{23}$, relative fluctuations are $\sim 10^{-12}$: utterly negligible. The system effectively has a well-defined energy even though it can exchange energy with a bath. This is why choosing the most convenient ensemble is always safe for macroscopic systems.

Key Insights

• The microcanonical ensemble describes isolated systems where all microstates of given energy are equally probable
• The canonical ensemble describes systems at fixed temperature, with the Boltzmann distribution $P_i \propto e^{-\beta E_i}$
• The partition function $Z$ encodes all equilibrium thermodynamics via $F = -k_B T \ln Z$
• The grand canonical ensemble handles open systems with variable particle number
• Each ensemble has a natural thermodynamic potential: $E$, $F$, $G$, or $\Phi$
• Equipartition gives $\frac{1}{2}k_B T$ per quadratic degree of freedom (classically)
• All ensembles agree in the thermodynamic limit, fluctuations vanish as $1/\sqrt{N}$

Looking Ahead

The ensembles we've developed are built on classical physics. But at low temperatures or high densities, quantum effects become essential. In the next lesson we'll see how identical quantum particles, bosons and fermions, lead to fundamentally different statistics:

• Bose-Einstein statistics: particles that love to share quantum states
• Fermi-Dirac statistics: particles that obey the exclusion principle
• Dramatic consequences: superfluidity, metals, white dwarfs, and the blackbody spectrum

The partition function framework remains; only the counting of states changes.