Indistinguishable Particles and N!

Introduction

The canonical partition function Q counts states of a whole system, while q counts states of one molecule. For independent molecules the two must be connected, and the connection is almost — but not quite — the obvious one. Getting it right requires facing a genuinely quantum idea: identical particles cannot be labelled. The resulting factor of 1/N! may look like a small correction, but ln N! for a mole is about 3 × 10²⁵, and leaving it out gives entropies that are not even extensive.

Core explanation

Distinguishable, independent particles. Suppose N particles do not interact and each can be identified, for instance because it sits at a particular site in a crystal. The system energy is E = ε(1) + ε(2) + … + ε(N), where each term is the energy of one labelled particle. Summing over all independent choices of each particle's state gives, by the same factorisation argument used for molecular modes,

Q = q × q × … × q = q^N

This applies to localised systems: the Einstein model of a crystal, spins fixed on lattice sites, or molecules adsorbed at distinct surface sites.

Indistinguishable particles. In a gas, identical molecules wander freely and swap places. Quantum mechanically, a state in which molecule A is in state a and molecule B is in state b is the same state as the reverse assignment. The product q^N counts each such arrangement N! times — once for every permutation of the labels. Dividing out the overcounting gives

Q = q^N/N!

When is this exact enough? The N! correction is correct only if almost every molecule occupies a different one-particle state. If two molecules share a state, swapping them does not create a new term in q^N, and dividing by N! overcorrects. The condition is that the number of thermally accessible translational states greatly exceeds N:

q T/N = V/(NΛ³) ≫ 1

Equivalently, the average distance between molecules must be much larger than the thermal wavelength Λ = h/(2πmkT)^½. For ordinary gases near room temperature this ratio is around 10⁵–10⁷, so the approximation is excellent. It fails for very light particles at very low temperature and high density — liquid helium or electrons in metals — where Bose–Einstein or Fermi–Dirac statistics are needed.

Consequences. Taking logarithms with Stirling's approximation:

ln Q = N ln q − ln N! ≈ N ln q − N ln N + N = N ln(q/N) + N

The N! term does not depend on temperature, so it does not affect U or C V. It does affect entropy, Helmholtz energy and chemical potential, and it is what makes those properties extensive.

Gibbs paradox. Without the N!, removing a partition between two samples of the same gas at equal T and p would appear to increase the entropy, which is absurd. With the N!, the entropy of mixing is zero for identical gases and R ln 2 per mole of mixture (for equal amounts) only when the two gases are different.

Formulae

Localised, distinguishable: Q = q^N. Gas, indistinguishable: Q = q^N/N!. Stirling: ln N! ≈ N ln N − N. Validity: V/(NΛ³) ≫ 1, with V/N = kT/p for a perfect gas.

Step-by-step reasoning

To decide which relation to use:

1. Ask whether the particles are identical. 2. If identical, ask whether they are localised (on sites) or free (gas). 3. Localised identical particles: Q = q^N, because sites label them. 4. Free identical particles: Q = q^N/N!, provided V/(NΛ³) ≫ 1. 5. If V/(NΛ³) is near or below 1, use quantum statistics instead.

Visual explanation

Draw two boxes labelled "state a" and "state b", and two molecules. With labelled molecules there are two drawings: A in a with B in b, and B in a with A in b. Now erase the labels: the two drawings become identical. For N molecules in N different states, N! labelled drawings collapse into one.

Real-world analogy

Counting the ways to seat three named guests on three chairs gives 3! = 6 arrangements. If the guests are three identical mannequins, every arrangement looks the same: there is only one. Dividing 6 by 3! corrects the count, as long as no two mannequins try to share a chair.

Real-world example

Electrons in a metal violate the condition V/(NΛ³) ≫ 1: they are light and densely packed. Classical statistics would predict a large electronic heat capacity of (3/2)R per mole of electrons, yet measured metals show only a tiny electronic contribution at room temperature. Fermi–Dirac statistics, which properly handle multiple occupancy restrictions, explain this.

Why?

Why does a crystal use q^N when its atoms are also identical? Each atom is tied to a particular lattice site, and the sites themselves are distinguishable by position. Exchanging two atoms would not correspond to a distinct thermal state that the system samples, so the site label does the distinguishing.

Common misconception

"The N! is a classical correction with no deep meaning." In classical mechanics particles can always be tracked, so there is no reason to divide. The N! is a quantum-mechanical requirement: identical particles are fundamentally indistinguishable. Gibbs introduced it before quantum theory only because the entropy otherwise made no sense.

Worked example

Question: Test the validity of Q = q^N/N! for helium at 298 K and 1 bar. (m = 6.65 × 10⁻²⁷ kg.)

Reasoning: Λ = h/(2πmkT)^½ = 6.626 × 10⁻³⁴/(2π × 6.65 × 10⁻²⁷ × 4.115 × 10⁻²¹)^½ = 5.05 × 10⁻¹¹ m, so Λ³ = 1.29 × 10⁻³¹ m³. V/N = kT/p = 4.115 × 10⁻²¹/1.00 × 10⁵ = 4.12 × 10⁻²⁶ m³. Ratio = 4.12 × 10⁻²⁶/1.29 × 10⁻³¹ ≈ 3.2 × 10⁵.

Answer: V/(NΛ³) ≈ 3 × 10⁵ ≫ 1, so the approximation holds well, even for the lightest noble gas.

Quick check

1. Should Q = q^N or Q = q^N/N! be used for CO molecules adsorbed at fixed surface sites? Explain briefly. Answer: Q = q^N, because each molecule is localised at a distinguishable site, so no permutation overcounting occurs.

Exam focus

Be able to explain the origin of 1/N!, state the condition V/(NΛ³) ≫ 1, and use Stirling's approximation to give ln Q = N ln(q/N) + N. A standard question asks which properties are affected by the N! (S, A, G, μ) and which are not (U, C V, p).

Advanced insight

As V/(NΛ³) approaches 1, bosons and fermions diverge from classical behaviour in opposite ways. Bosons pile into low states, leading to Bose–Einstein condensation, first achieved with ultracold rubidium atoms in 1995. Fermions resist sharing states and exert a degeneracy pressure that supports white dwarf stars against gravity.

Summary

For independent distinguishable (localised) particles Q = q^N; for independent indistinguishable particles in a gas Q = q^N/N!. The N! removes the overcounting of relabelled states and is valid when V/(NΛ³) ≫ 1, true for ordinary gases. Using Stirling's approximation, ln Q = N ln(q/N) + N. The factor alters S, A and μ but not U or C V, and it resolves the Gibbs paradox.

Practice questions

1. Use Stirling's approximation to write ln Q for a perfect gas in terms of q and N. Answer: ln Q = N ln q − (N ln N − N) = N ln(q/N) + N. 2. Why does the N! not affect the internal energy? Answer: U depends on the temperature derivative of ln Q, and ln N! is independent of temperature. 3. How does V/(NΛ³) change if the temperature of a gas is lowered at constant pressure? Answer: It decreases, because V/N = kT/p falls in proportion to T and Λ³ grows as T^(−3/2), so the ratio scales as T^(5/2). 4. What goes wrong with the entropy if the N! is omitted for a gas? Answer: The entropy is no longer extensive, and mixing two samples of the same gas would wrongly appear to increase entropy (the Gibbs paradox).