Distinguishable vs Indistinguishable Particles

Imagine two identical helium atoms swapping positions in a box. Nothing about the gas changes. Yet naive counting treats this swap as a new state. This overbinding of states, if left uncorrected, produces absurd thermodynamics. Entropy fails to be extensive. Chemical potential behaves badly upon mixing identical gases. The fix is one of the most consequential small corrections in physical chemistry. Whether the canonical partition function is $Q = q^N$ or $Q = q^N/N!$ matters greatly. It determines whether entropy adds up correctly. It determines whether Gibbs' paradox is resolved. It also determines whether the ideal gas law emerges with the right constants. Every quantitative result in molecular thermodynamics rests on getting this factor right. This includes the Sackur-Tetrode equation and reaction entropies. This article builds the distinguishable and indistinguishable cases from the single-molecule partition function $q$ . It shows precisely when each formula applies. It works two full numerical examples. It ends by showing how this correction becomes the seed of the Sackur-Tetrode equation.

Key takeaways

For $N$ independent, distinguishable particles (e.g., atoms fixed on lattice sites), $Q = q^N$ .
For $N$ independent, indistinguishable particles (e.g., gas molecules), $Q = q^N/N!$ corrects for overcounting permutations.
The $N!$ correction is valid only when each particle occupies a distinct quantum state on average — the low-density, high-temperature limit relevant to ordinary gases.
Omitting $N!$ produces non-extensive entropy and Gibbs' paradox: mixing identical gas samples appears to increase entropy.
$N!$ arises because swapping identical particles among occupied states gives no new physical state, but the naive product $q^N$ counts each such swap separately.
Key idea: Whether particles can be labeled decides whether permutations of particles among states count as new microstates, and this single combinatorial fact fixes the entire form of $Q$ .

The core insight

The relation between the system partition function $Q$ and the molecular partition function $q$ hinges entirely on whether swapping two particles' identities, while keeping the same set of occupied states, produces a genuinely new microstate — it does for particles fixed at distinguishable sites, giving $Q = q^N$ , but it does not for freely moving identical molecules, giving $Q = q^N/N!$ .

The physical picture

Start with $N$ non-interacting particles, each described by a single-molecule partition function $q$ , built from that particle's own energy levels as covered in the treatment of the canonical partition function. If the particles could be told apart — atom 1 sits at lattice site 1, atom 2 at lattice site 2, and so on — then the total partition function is simply the product of each particle's partition function. Particle 1 independently explores its states, particle 2 independently explores its states, and every combination of assignments is a distinct, physically real microstate because we could point to "particle 1 in state $a$ , particle 2 in state $b$ " and distinguish it from "particle 1 in state $b$ , particle 2 in state $a$ ." This gives $Q = q^N$ , exact and unconditional, for any system of localized, labeled particles: atoms vibrating about crystal lattice sites, defects pinned in a solid, or spins fixed at addresses in a magnetic lattice. Now consider a gas of identical helium atoms with no fixed positions. Quantum mechanics says these atoms are fundamentally indistinguishable: there is no experiment that can track "atom 3" as opposed to "atom 7" as they fly around and collide. A microstate of the gas is specified only by how many particles occupy each single-particle energy level, never by which labeled particle sits where. But when we build the product $q^N = \left(\sum_i e^{-\beta \varepsilon_i}\right)^N$ and expand it, we get a term for every possible assignment of labeled particles to states. If particles 1 and 2 are swapped between two singly-occupied levels, the product counts that as a separate term — even though physically nothing has changed. As long as it is overwhelmingly likely that no two particles share the same single-particle state (the case for ordinary gases at ordinary densities), each set of occupied levels gets counted exactly $N!$ times too many, once for every permutation of the $N$ particles among those levels. Dividing by $N!$ removes the overcounting and yields the correct indistinguishable-particle result, $Q = q^N/N!$ .

Figure

Distinguishable atoms on lattice sites versus indistinguishable gas atoms in a box

The distinction is not about the particles being physically identical twins in composition — even distinguishable particles in the lattice case might be chemically identical atoms. It is about whether their positions or addresses let us track "which one is which." Once that tracking is impossible in principle, as it is for molecules free to roam a gas phase, the $N!$ correction becomes mandatory.

Figure

Overcounting: three atoms in three singly-occupied states

The core relations

Q_{\text{distinguishable}} = q^N

where: $q$ is the single-particle partition function, $N$ the number of independent, labeled particles.

Q_{\text{indistinguishable}} = \frac{q^N}{N!}

where: division by $N!$ removes overcounted permutations of identical particles among singly-occupied states, valid when $q \gg N$ .

\frac{q}{N} \gg 1 \quad \text{(the "classical" or non-degenerate limit)}

where: this ratio being large guarantees negligible probability of double occupancy of any single-particle state, the condition for $Q = q^N/N!$ to be accurate.

\ln Q = N\ln q - \ln N! \approx N\ln q - N\ln N + N

where: Stirling's approximation $\ln N! \approx N\ln N - N$ converts the factorial into a usable analytic form for macroscopic $N$ .

Derivation AFrom counting microstates to Q = qᴺ/N!

Goal. Show explicitly how permutation overcounting produces the $N!$ factor, and identify the precise condition under which it is valid.

Step 1. Define the canonical partition function honestly. The canonical partition function sums the Boltzmann factor over every distinct microstate of the $N$ -particle system:

Q = \sum_{\text{states } s} e^{-\beta E_s}

For independent particles, $E_s$ is a sum of single-particle energies, $E_s = \varepsilon_{i_1} + \varepsilon_{i_2} + \cdots + \varepsilon_{i_N}$ , where particle $k$ occupies single-particle level $i_k$ .

Step 2. Expand the naive product q^N. Treating particles as labeled and summing independently over each one's states,

q^N = \left(\sum_i e^{-\beta \varepsilon_i}\right)^N = \sum_{i_1}\sum_{i_2}\cdots\sum_{i_N} e^{-\beta(\varepsilon_{i_1}+\varepsilon_{i_2}+\cdots+\varepsilon_{i_N})}

Every term in this sum corresponds to one assignment of a specific labeled particle to each level index. If particles are truly distinguishable (fixed lattice sites), each assignment is a genuine, physically different microstate, so $Q = q^N$ exactly — no correction needed.

Step 3. Identify the overcounting for identical particles. If particles are identical and delocalized, the physical microstate is specified only by the occupation numbers $\{n_i\}$ — how many particles sit in level $i$ — never by which particle. Given a set of occupied levels all singly occupied (the typical case at low density), there are exactly $N!$ distinct labeled assignments (permutations of the $N$ particles among the $N$ occupied levels) that all correspond to the same physical occupation-number microstate. The sum $q^N$ has therefore counted that one true microstate $N!$ times.

Step 4. Correct by dividing. To recover the correct sum over true (occupation-number) microstates, divide the over-counted sum by $N!$ :

Q = \frac{q^N}{N!}

This division is exact only if every occupied level holds at most one particle (no permutation collapses onto itself via a repeated index) and the small residual correction from levels that do happen to hold two or more particles is negligible. That is the classical, non-degenerate regime.

Step 5. State the validity condition quantitatively. The number of accessible single-particle states is roughly $q$ itself (it counts the effective number of thermally accessible levels weighted by their Boltzmann factors). Double occupancy becomes rare when there are far more accessible states than particles,

\frac{q}{N} \gg 1

For a monatomic ideal gas at room temperature and atmospheric pressure, $q$ (translational) is of order $10^{27}$ – $10^{30}$ for a mole confined to ordinary volumes, utterly swamping $N \sim 10^{23}$ , so the approximation is excellent. It fails at extremely low temperature or extremely high density, where quantum statistics (Bose–Einstein or Fermi–Dirac) must replace this classical correction.

Boxed result.

\boxed{Q = \frac{q^N}{N!}\ \text{for indistinguishable, independent particles with } q \gg N}

Worked examples

Example 1 — Foundational: comparing Q for a tiny model system. Suppose a single particle has access to 4 non-degenerate states with the same spacing, giving (at some fixed temperature) a single-particle partition function $q = 4.00$ . Consider $N = 3$ particles. Distinguishable case (e.g., 3 atoms pinned at 3 distinct lattice sites, each independently sampling the same 4 states):

Q_{\text{dist}} = q^3 = 4.00^3 = 64.0

Indistinguishable case (3 identical gas-phase atoms sharing the same 4 accessible states):

Q_{\text{indist}} = \frac{q^3}{3!} = \frac{64.0}{6} = 10.7

Sanity check by direct enumeration: with only 4 states and 3 particles, some states must sometimes be doubly or triply occupied, so $q/N = 4/3 = 1.33$ is not much greater than 1 — this is a poor candidate for the classical approximation, and indeed direct combinatorial counting of occupation-number microstates (using Bose–Einstein statistics properly) would give a different, larger number of accessible microstates than $10.7$ . The exercise illustrates the arithmetic of the formula; it also flags that $N!$ -division is an approximation requiring $q \gg N$ , which this toy system deliberately violates for pedagogical contrast.

Example 2 — Applied: entropy consequence for one mole of argon gas. Take one mole of argon gas ( $N = N_A = 6.022 \times 10^{23}$ ) at $T = 298\ \text{K}$ confined to $V = 24.79\ \text{dm}^3$ (24.79 L, the molar volume at 1 bar). The translational partition function per molecule is

q_{\text{trans}} = \left(\frac{2\pi m k_B T}{h^2}\right)^{3/2} V

For argon, $m = 39.95\ \text{u} = 6.634 \times 10^{-26}\ \text{kg}$ . Compute the thermal wavelength term:

2\pi m k_B T = 2\pi (6.634\times10^{-26}\ \text{kg})(1.381\times10^{-23}\ \text{J K}^{-1})(298\ \text{K})

= 2\pi (6.634\times10^{-26})(4.116\times10^{-21}) = 2\pi (2.731\times10^{-46}) = 1.716\times10^{-45}\ \text{J kg}

Divide by $h^2 = (6.626\times10^{-34}\ \text{J s})^2 = 4.390\times10^{-67}\ \text{J}^2\text{s}^2$ :

\frac{2\pi m k_B T}{h^2} = \frac{1.716\times10^{-45}}{4.390\times10^{-67}} = 3.909\times10^{21}\ \text{m}^{-2}

Raise to the $3/2$ power:

(3.909\times10^{21})^{3/2} = 2.445\times10^{32}\ \text{m}^{-3}

Multiply by $V = 24.79\times10^{-3}\ \text{m}^3$ :

q_{\text{trans}} = (2.445\times10^{32})(2.479\times10^{-2}) = 6.06\times10^{30}

Now check the classical condition: $q/N = 6.06\times10^{30}/6.022\times10^{23} = 1.01\times10^{7} \gg 1$ . The $N!$ correction is excellent here. Using $Q = q^N/N!$ with Stirling's approximation, the molar entropy contribution from translation is

S = k_B \ln Q + \frac{U}{T} = N k_B\left(\ln\frac{q}{N} + 1\right) + \frac{3}{2}N k_B

(using $U = \tfrac{3}{2}N k_B T$ for translation). Plugging in $N = N_A$ , $k_B N_A = R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ :

S = R\left(\ln(1.01\times10^{7}) + 1 + \frac{3}{2}\right) = R(16.13 + 2.50) = R(18.63)

S = (8.314)(18.63) = 154.9\ \text{J K}^{-1}\text{mol}^{-1}

This is close to the experimental standard molar entropy of argon gas, $S^\circ_m \approx 154.8\ \text{J K}^{-1}\text{mol}^{-1}$ — a genuine, checkable success of the $N!$ correction (this is exactly the Sackur-Tetrode result). Had we instead used $Q = q^N$ without dividing by $N!$ , the $+1$ from Stirling's expansion of $-\ln N!/N$ would vanish and the entropy would be overestimated by $R\ln N - R \approx R(32.24) \approx 268\ \text{J K}^{-1}\text{mol}^{-1}$ too large — wildly inconsistent with experiment, and worse, non-extensive: doubling both $N$ and $V$ at fixed density would not simply double $S$ .

Limiting cases & common traps

Situation	Correct treatment	Why
Atoms on distinct lattice sites (solids, defects)	$Q = q^N$ , no $N!$	Sites are physically labeable addresses
Ideal gas at ordinary $T$ , $P$	$Q = q^N/N!$	$q \gg N$ , negligible double occupancy
Gas at extremely low $T$ or high density	Neither classical formula; use Bose–Einstein or Fermi–Dirac statistics	$q/N$ no longer $\gg 1$ ; quantum degeneracy matters
Mixture of two different gases	Each species gets its own $N_i!$ : $Q = (q_A^{N_A}/N_A!)(q_B^{N_B}/N_B!)$	Only identical particles are interchangeable with each other

The chemist behind it

The paradox that makes $N!$ necessary is named for Josiah Willard Gibbs, whose 1875 observations about mixing entropy predate quantum mechanics by half a century. Gibbs noticed that classical statistical mechanics, applied naively, gave a mixing entropy for combining two gas samples that did not vanish even when the two gases were identical — a result plainly at odds with the physical fact that mixing indistinguishable substances changes nothing observable. His resolution, essentially the ancestor of the $N!$ correction, was to insist that permutations of identical particles must not be counted as distinct configurations. Later, Otto Sackur and Hugo Tetrode independently combined this counting correction with the quantum-mechanical translational partition function around 1912, producing the Sackur-Tetrode equation and putting Gibbs' insight on a fully quantitative, testable footing — precisely the calculation carried out for argon above.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American theoretical physicist and chemist who founded chemical thermodynamics and first identified the mixing-entropy paradox that the N! correction resolves

Ask the tutor: How does the Sackur-Tetrode equation turn this N! correction into a full quantitative formula for the entropy of an ideal gas?

Why does dividing by N! fix the entropy but not change the energy or heat capacity

Internal energy and heat capacity come from derivatives of $\ln Q$ with respect to $\beta = 1/k_BT$ at fixed volume. Since $\ln Q = N\ln q - \ln N!$ , and $\ln N!$ does not depend on temperature, differentiating with respect to $\beta$ eliminates the $N!$ term entirely. Energy and heat capacity only see the $N\ln q$ piece. Entropy, however, involves $\ln Q$ itself (not just its derivative), so the temperature-independent $-\ln N!$ survives and directly shifts the entropy value, making the correction essential there but invisible in $U$ or $C_V$ .

Does the N! correction ever apply to particles that are chemically identical but still distinguishable

Yes. Atoms of the same isotope sitting at fixed, well-separated lattice sites in a crystal are chemically and physically identical in composition, yet they are distinguishable in the statistical-mechanical sense because each has a permanent, trackable address. Swapping two such atoms between two sites genuinely produces a different arrangement (site 1 has a different local environment history than site 2 in principle), so no $N!$ correction applies. The relevant criterion is not chemical identity but whether the particles' spatial arrangement lets us, even in principle, track which is which.

What happens to Q = qᴺ divided by N! when the gas becomes very dense or very cold

The approximation breaks down. It assumed each accessible single-particle state holds at most one particle, valid when $q \gg N$ . At low temperature or high density, many particles compete for the same few low-lying states, and double or higher occupancy becomes common. The classical dividing-by- $N!$ trick then over- or under-corrects, and one must use full quantum statistics: Bose-Einstein statistics for particles with integer spin (bosons) or Fermi-Dirac statistics for half-integer spin particles (fermions), which handle occupation numbers exactly rather than through a classical approximation.

How does this connect to the Sackur-Tetrode equation

The Sackur-Tetrode equation is the explicit formula for the translational entropy of an ideal monatomic gas, and its derivation starts exactly where this article ends: take $Q = q_{\text{trans}}^N/N!$ , apply Stirling's approximation to $\ln N!$ , and combine with $S = k_B\ln Q + U/T$ . The distinctive $+\tfrac{3}{2}$ (rather than $+\tfrac{1}{2}$ ) inside the logarithmic entropy expression, and the correct extensive scaling with $N$ , both trace directly back to the $N!$ factor derived here. Without it, Sackur-Tetrode simply could not be written down correctly.