advanced32 min studyLesson 321 in Physics

Count physical states in phase space

Normalize classical counting with Planck cells and remove permutation overcounting without confusing dilution with distinguishability.

On this page

What you’ll understand

  • Explain why a classical phase-space integral needs a dimensional cell.
  • Apply the Gibbs factorial separately to each identical species.
  • Use thermal wavelength to diagnose the classical dilute regime.

Counting a continuum of classical positions and momenta is not the same as counting discrete quantum states. A usable statistical measure must specify its units and identify which labeled descriptions represent the same physical configuration. The factors of Planck's constant and the particle factorial solve different problems. We will keep them separate, then derive a gas partition function whose entropy and chemical potential have a consistent normalization. The resulting classical approximation still has a measurable boundary of validity.

Specify a dimensionless measure

A point in the phase space of N particles contains three positions and three momenta per particle. Integrating over that space gives units of action to the power three N, not a pure number. A logarithm used for entropy or free energy must have a dimensionless argument. Dividing by Planck's constant to the same power supplies a semiclassical counting cell. It does not assert that a classical particle follows a grid of discrete positions and momenta.

For identical particles, assigning labels to otherwise identical trajectories counts permutations repeatedly. In the dilute classical limit the symmetric Hamiltonian integral is therefore divided by N factorial. This correction is distinct from the phase-space cell. For several species, divide by a factorial for each identical species rather than the factorial of their combined population. Particles bound to distinct identifiable sites require a different counting problem; the gas factorial must not be inserted into every collection of degrees of freedom.

ZN(T,V)=1N!h3N∫d3Nq d3Np e−βH(q,p),β=1kBT,h=2πℏZ_N(T,V)=\frac{1}{N!h^{3N}}\int d^{3N}q\,d^{3N}p\,e^{-\beta H(q,p)},\qquad \beta=\frac1{k_BT},\qquad h=2\pi\hbar
Zₙ is dimensionless, H is energy in joules, β in J⁻¹, and h in J·s. The formula is the semiclassical Maxwell–Boltzmann approximation for one identical species without resolved internal states.↔ Scroll wide equations horizontally.

Factorize the ideal-gas integral

For a noninteracting monatomic gas, every position integral gives the box volume and every momentum component gives a Gaussian integral. The one-particle partition function is volume divided by thermal wavelength cubed. Raising it to N counts independent labeled contributions; dividing by N factorial then accounts for identical particles. The wavelength packages mass, temperature, and the phase-space normalization. It is a statistical scale, not the wavelength of every individual particle in a thermal sample.

The dimensionless density times thermal wavelength cubed measures whether exchange effects are small. If internal degeneracy supplies several equally accessible states, the relevant occupancy per internal state is reduced by that degeneracy. Dilution makes Maxwell–Boltzmann statistics accurate; it does not turn identical atoms into fundamentally distinguishable quantum particles. When occupancies are appreciable, a single global factorial does not reproduce all Bose enhancement or Pauli exclusion effects, and the quantum occupation counting must replace the dilute approximation.

λT=h2πmkBT,Z1=VλT3,ZN=1N!(VλT3)N,nλT3≪1\lambda_T=\frac{h}{\sqrt{2\pi mk_BT}},\qquad Z_1=\frac{V}{\lambda_T^3},\qquad Z_N=\frac{1}{N!}\left(\frac{V}{\lambda_T^3}\right)^N,\qquad n\lambda_T^3\ll1
λT is in metres, m in kg, n=N/V in m⁻³, and T is positive absolute temperature. The displayed dilution criterion assumes one internal state and negligible interactions.↔ Scroll wide equations horizontally.

Correct counting changes thermodynamic derivatives

Using Stirling's approximation on the factorial makes ideal-gas free energy extensive when volume and particle number grow together at fixed density. Omitting that factorial leaves an artificial entropy of mixing for equal gases and a chemical potential with the wrong size dependence. Pressure at fixed particle number can nevertheless look correct because the missing factorial has no volume derivative. A successful equation of state alone is therefore an incomplete audit of the state-counting convention.

A useful spatial counting check divides a classical uniform ideal gas into two equal observation regions. For a fixed total number, the probability of a given count in the left region is binomial. The combinatorial factor counts choices of positions without requiring experimentally trackable atom labels. This gives a concrete lab connection: the probability instrument computes the exact count distribution for independent positions, while Planck cells, momentum integrals, and quantum exchange corrections remain outside that model.

F≃NkBT[ln⁡(nλT3)−1],μ≃kBTln⁡(nλT3),P=NkBTVF\simeq Nk_BT[\ln(n\lambda_T^3)-1],\qquad \mu\simeq k_BT\ln(n\lambda_T^3),\qquad P=\frac{Nk_BT}{V}
These leading large-N ideal-gas formulas use the Gibbs factor, one internal state, and nλT³≪1. F and μ are in joules and P in pascals; finite-N factorial corrections remain available from the exact Zₙ.↔ Scroll wide equations horizontally.

Separate species and permutation factors

A dilute ideal gas has Z₁=12 at the chosen temperature and volume. Compare the canonical partition function of three identical atoms with that of two atoms of species A and one noninteracting atom of species B, with the same Z₁ for each species.

  1. Three independent labeled one-particle factors give 12³=1728. For three identical atoms divide by 3!=6, obtaining Z=288.

  2. For the mixture divide by 2! for species A and 1! for species B. Its partition function is 1728/2=864, since exchanging different species does not identify the same quantum configuration.

  3. The ratio is three. At this same temperature the mixture free energy is lower by kBT ln3 relative to the identical-particle model with these specified one-particle factors. This comparison concerns different species content, not a claim that relabeling equal atoms changes physical free energy.

The result

The partition functions are 288 for three identical atoms and 864 for the two-species mixture. Species composition determines which factorials apply.

Explore the model

Interactive lab

Interactive lab · probability

Count atoms in half a box

For independent runs Reset and set BOTH success probability p=0.5, trials n=4; then Reset and set p=0.5, n=8. Predict means 2 and 4 and variances 1 and 2 for particles in a half-volume. These are exact binomial probabilities for independent positions, not quantum gas statistics or phase-space integration.

Count atoms in half a boxThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA combinatorial distribution can describe identical dilute atoms without treating arbitrary labels as physical states.

Check your understanding

For NA identical atoms of A and NB identical atoms of a different species B, which factorial removes permutation overcounting in the dilute phase-space integral?

Try it yourself

A one-state gas has nλT³=0.02. Its temperature falls by a factor of four at fixed density. Find the new degeneracy parameter and assess whether exchange effects become more important.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Thermal wavelength varies as T to the power minus one half, so track its cube.

Compare with the solution

The wavelength doubles, and its cube increases eightfold. Thus nλT³ becomes 0.16. Exchange effects become more important, although this is still below unit occupancy. The criterion is an asymptotic small-parameter test, not a universal sharp boundary at one; interactions and internal degeneracy require separate checks.

Take it with you

  • Planck cells remove phase-space units, while Gibbs factorials remove permutation overcounting.
  • Different identical species have separate factorials.
  • The dilute parameter nλT³ controls exchange corrections for a one-state ideal gas.

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