The Molecular Partition Function

Introduction

The Boltzmann distribution tells us that each state carries a weight exp(−εᵢ/kT). To turn these weights into actual fractions, we must divide by their sum. That sum turns out to be far more than a bookkeeping device. Called the molecular partition function , q, it encodes how molecules are spread over their states, and every thermodynamic property of an ideal gas can be derived from it.

Core explanation

Definition. The molecular partition function is

q = Σᵢ exp(−βεᵢ), with β = 1/kT

where the sum runs over every individual state i of one molecule, and εᵢ is the energy of that state measured from the ground state. The name comes from the German Zustandssumme , "sum over states", which describes it exactly.

Normaliser. The fraction of molecules in state i is

pᵢ = exp(−βεᵢ)/q

Summing over all states gives Σ pᵢ = q/q = 1, so q is precisely the factor that converts Boltzmann factors into probabilities.

Properties. q is dimensionless and always at least 1 when energies are measured from the ground state, because the ground state contributes exp(0) = 1 and every other state contributes a positive amount. It depends on temperature and on the spacing of the energy levels (and, for translation, on volume).

A uniform ladder. Consider non-degenerate levels at 0, ε, 2ε, 3ε, … — a good model for a harmonic vibration. Then

q = 1 + e^(−βε) + e^(−2βε) + … = 1/(1 − e^(−βε))

because the terms form a geometric series with ratio e^(−βε) < 1. At low temperature (βε large), q → 1. At high temperature (βε small), e^(−βε) ≈ 1 − βε, so q ≈ 1/(βε) = kT/ε.

The zero of energy. If all energies are shifted by a constant ε₀, every term is multiplied by e^(−βε₀), so q is multiplied by the same factor. Populations are unchanged, because the factor cancels in pᵢ. Thermodynamic functions such as U and G are shifted by the corresponding constant, which is why tabulated values always specify a reference. By convention the ground state is set at zero.

Mean energy. Because the Boltzmann factors depend on β, differentiating q generates energy-weighted sums. The mean energy of a molecule is

⟨ε⟩ = Σ εᵢpᵢ = −(1/q)(∂q/∂β) = −∂ ln q/∂β

This is the first glimpse of how thermodynamic properties are obtained by differentiating the logarithm of a partition function.

Formulae

q = Σᵢ exp(−εᵢ/kT); pᵢ = exp(−εᵢ/kT)/q; uniform ladder q = 1/(1 − e^(−ε/kT)); ⟨ε⟩ = −∂ ln q/∂β.

Step-by-step reasoning

To evaluate a partition function numerically:

1. List the energies of the states relative to the ground state. 2. Calculate βεᵢ = εᵢ/kT for each state in consistent units. 3. Evaluate each Boltzmann factor and add them, stopping when the terms become negligible. 4. Divide individual Boltzmann factors by q to obtain the populations.

Visual explanation

Plot q against T for a uniform ladder. The curve starts flat at q = 1 near absolute zero, bends upward once kT becomes comparable to ε, and then rises almost linearly along the line kT/ε. The bend marks the temperature at which the excited states "switch on".

Real-world analogy

Think of q as the total of the ballot papers cast in an election. Each candidate's share of the vote is their tally divided by the total. The Boltzmann factors are the tallies, q is the total, and the population of each state is its share.

Real-world example

Spectroscopists use partition functions to convert measured line intensities into temperatures. In a flame or a stellar atmosphere, the intensity of a transition depends on the population of the lower state, which equals its Boltzmann factor divided by q. Measuring several lines lets the temperature of a star's surface be determined remotely.

Why?

Why is q so powerful when it is merely a sum? Because it depends on T through every term, its derivatives with respect to temperature automatically produce population-weighted averages. A single function therefore contains the internal energy, heat capacity and entropy, and each is extracted by a standard mathematical operation rather than a separate calculation.

Common misconception

"q is the number of molecules in the sample." q refers to a single molecule and counts weighted states, not particles. It is dimensionless and independent of how many molecules are present (though translational q depends on the volume of the container).

Worked example

Question: A uniform ladder has spacing ε with ε/kT = 1.00. Calculate q and the fractions in the lowest three states.

Reasoning: q = 1/(1 − e^(−1)) = 1/(1 − 0.368) = 1/0.632 = 1.58. Then p₀ = 1/1.58 = 0.632, p₁ = e^(−1)/1.58 = 0.233 and p₂ = e^(−2)/1.58 = 0.086.

Answer: q = 1.58; the populations are 0.632, 0.233 and 0.086, with the remaining 0.049 spread over higher states.

Quick check

1. What value does q approach as T approaches absolute zero for a molecule with a non-degenerate ground state? Answer: q approaches 1, because only the ground state contributes a non-negligible Boltzmann factor.

Exam focus

You must be able to write q as a sum over states, sum a geometric series for the uniform ladder, and use pᵢ = exp(−βεᵢ)/q. Note clearly that the sum is over states; when levels are degenerate, each must be counted as many times as its degeneracy.

Advanced insight

The same mathematical object appears in probability theory as a generating function. Just as −∂ ln q/∂β gives the mean energy, the second derivative ∂² ln q/∂β² gives the variance of the energy, which is linked to the heat capacity. Thus the partition function contains not only average properties but also the size of energy fluctuations.

Summary

The molecular partition function q = Σ exp(−εᵢ/kT) is the sum of Boltzmann factors over all states of one molecule. Dividing by q normalises the Boltzmann distribution. q equals 1 at T = 0 (non-degenerate ground state) and grows with temperature. For a uniform ladder, q = 1/(1 − e^(−ε/kT)). Derivatives of ln q yield average properties such as the mean energy.

Practice questions

1. Calculate q at 298 K for a uniform ladder with spacing 500 cm⁻¹. Answer: ε/kT = 500/207 = 2.42; q = 1/(1 − e^(−2.42)) = 1/(1 − 0.089) ≈ 1.10. 2. Show that the uniform-ladder partition function approaches kT/ε at high temperature. Answer: For small βε, e^(−βε) ≈ 1 − βε, so q = 1/(1 − 1 + βε) = 1/(βε) = kT/ε. 3. A molecule has non-degenerate states at 0, 100 and 300 cm⁻¹ only. Calculate q at 298 K. Answer: q = 1 + e^(−100/207) + e^(−300/207) = 1 + 0.617 + 0.235 = 1.85. 4. How does adding a constant ε₀ to the energy of every state change q and the populations? Answer: Every Boltzmann factor, and therefore q, is multiplied by exp(−ε₀/kT), but the populations are unchanged because this common factor cancels in exp(−βεᵢ)/q.