Chemistry · Lesson

Microstates, ensembles and partition functions

State what is fixed before assigning microscopic probabilities. Count accessible microstates, include energy-level degeneracy once and distinguish a single-unit partition function from the whole-system sum.

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By the end, you can

Count states under a specified fixed-energy constraint.
Distinguish microcanonical, canonical and grand-canonical ensembles.
Normalize Boltzmann weights and use the correct particle-counting model.

A macrostate restricts the accessible microstates

A microstate specifies the microscopic configuration represented by the model. An ensemble is a probability description of possible states under stated constraints, rather than many physical copies simultaneously occupying the same system. The macrostate may leave several microstates accessible.

For four distinguishable fixed sites, each in state 0 or 1, there are 2⁴ = 16 unconstrained configurations. If each excitation costs ε and the total energy is exactly 2ε, only the six configurations with two excited sites are accessible. Under an explicitly equiprobable microcanonical model each has probability 1/6. The entropy is k_B ln 6, not k_B ln 16, where k_B is the Boltzmann constant.

A fixed-energy constraint selects six of sixteen configurations

Microstate	Excited sites	Energy	Probability
0011	2	2ε	1/6
0101	2	2ε	1/6
0110	2	2ε	1/6
1001	2	2ε	1/6
1010	2	2ε	1/6
1100	2	2ε	1/6

Four distinguishable fixed two-level sites with total energy exactly 2ε. Each row is one permitted microstate, not one energy level. Equal probabilities are an explicit microcanonical assumption. Other configurations belong to different energies and are excluded by this model constraint.

The reservoir changes what can fluctuate

A microcanonical model fixes energy, volume and particle number. A canonical model fixes T,V,N while allowing energy exchange with a temperature reservoir. A grand-canonical model fixes T,V and the reservoir’s particle chemical potential while allowing both energy and particle number to fluctuate.

These descriptions can agree for many ordinary bulk systems in a suitable thermodynamic limit, but finite, critical or nonadditive systems can differ. A stationary distribution alone is not proof of thermodynamic equilibrium; a driven system can sustain currents while its average observables remain constant.

Normalize the appropriate statistical weights

For a canonical whole-system model, Q = Σⱼ exp(−βEⱼ), β = 1/(k_B T), and pⱼ = exp(−βEⱼ)/Q. Here k_B is the Boltzmann constant and the sum is over microstates. If instead summing grouped energy levels, multiply each level’s weight by its degeneracy, the number of microstates at that energy. Do not count the degeneracy twice.

A single localized two-level unit with energies 0 and ε has q = 1 + exp(−βε) when both levels are nondegenerate. For N independent distinguishable sites, Q = qᴺ. The whole-system Q is not the same object as the single-site q, even though factorization makes their relation simple.

Mobile identical particles need different counting

In the dilute classical Maxwell–Boltzmann approximation for independent identical mobile particles, Q ≈ qᴺ/N!, rather than the distinguishable-site expression. The factorial corrects overcounted particle labels. It is not the exact general Bose or Fermi partition function. Classical dilute occupancy and weak interactions are separate assumptions.

Grand-canonical weights are exp[−β(Eⱼ − μ_particle Nⱼ)]. The chemical potential in this expression is energy per particle. If using a molar μ, the term is μ_molar Nⱼ/N_A, where N_A is the Avogadro constant. Mixing a molar potential with a bare particle count would give an inconsistent energy basis.

Worked examples

Count before applying an entropy formula

Four fixed distinguishable sites each have two states, but the energy constraint permits exactly two excitations. Find Ω and the model entropy.

Choose which two of the four sites are excited.
The six choices are 0011, 0101, 0110, 1001, 1010 and 1100.
With equal permitted-state probabilities, Ω = 6 and S = k_B ln 6.
Using k_B = 1.380649 × 10⁻²³ J K⁻¹ gives about 2.47 × 10⁻²³ J K⁻¹ for this four-site system.

Ω = 6 and S ≈ 2.47 × 10⁻²³ J K⁻¹. Counting all 16 states would ignore the supplied fixed-energy constraint.

Normalize grouped levels without counting twice

At one stipulated temperature, take three energy levels 0, ε, 2ε with degeneracies 1, 2, 1 and ε/(k_B T) = ln 2. Find q and each energy-level probability.

The Boltzmann factors are 1,1/2 and 1/4.
Include each level’s degeneracy once: q = 1 +2(1/2)+1(1/4) = 9/4.
The level probabilities are 1/q,1/q and(1/4)/q.
Each of the two individual microstates at energy ε has probability(1/2)/q = 2/9.

q = 9/4. The grouped level probabilities are 4/9,4/9 and 1/9; the two separate states at ε each have probability 2/9.

Try it yourself

Which quantity can fluctuate in a canonical T,V,N model: energy or particle number?
Show the answer and reasoning
Energy.
The temperature reservoir exchanges energy while N is fixed.
A level has degeneracy 3 and Boltzmann factor 1/4. What weight does it contribute to a level-grouped partition sum?
Show the answer and reasoning
3/4.
Multiply the factor by degeneracy once.
For two independent distinguishable fixed sites each with q = 2, what is Q?
Show the answer and reasoning
4.
Use q²; the mobile identical-particle factorial belongs to a different counting model.
What energy basis is required for μ in exp[−β(E − μN)] with N a particle count?
Show the answer and reasoning
Energy per particle.
A molar chemical potential must be converted using N_A.

What to carry forward

Count states under the stated constraints.
Name the fixed quantities and allowed fluctuations.
Normalize state or degeneracy-weighted level sums consistently.
Keep localized-site, identical-particle and molar bases distinct.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.