Degeneracy and the Partition Function

Introduction

Many molecular energy levels are made up of several distinct states that happen to have exactly the same energy. A rotating diatomic molecule in level J has 2J + 1 states, differing in the orientation of its angular momentum. Counting such states correctly is essential: miss a degeneracy and a calculated entropy or population can be wrong by a large factor. This page shows how degeneracy enters the partition function.

Core explanation

States and levels. The partition function is fundamentally a sum over states. When several states share an energy, their Boltzmann factors are identical, so we can group them. If level j has energy εⱼ and contains gⱼ states,

q = Σ levels gⱼ exp(−βεⱼ)

The degeneracy gⱼ simply counts how many times the same Boltzmann factor appears.

Level populations. The fraction of molecules in the whole level j is

nⱼ/N = gⱼ exp(−βεⱼ)/q

and the ratio of populations of two levels is

nⱼ/nᵢ = (gⱼ/gᵢ) exp[−(εⱼ − εᵢ)/kT]

Competition between degeneracy and energy. The Boltzmann factor always favours the lower level, but a larger degeneracy favours the upper one. When Δε is small compared with kT, the degeneracy wins and the upper level can hold more molecules than the lower. Each individual state of the upper level is still less populated than each state of the lower level.

High- and low-temperature limits. As T → 0 only the ground level survives, so q → g₀. As T → ∞ every Boltzmann factor tends to 1 and the level populations become proportional to their degeneracies.

Sources of degeneracy.

- Rotation: a linear rotor level J has gⱼ = 2J + 1. - Atomic electronic levels: a level with total angular momentum quantum number J has g = 2J + 1. Spin–orbit coupling splits a term into levels with different J. - Spin: an unpaired electron has two spin states; in the absence of a magnetic field they are degenerate. - Three-dimensional motion: in a cubic box, different combinations of quantum numbers (nx, ny, nz) can give the same energy.

Lifting degeneracy. An external field can split degenerate states. In a magnetic field, the two spin states of an electron separate in energy (the Zeeman effect); the partition function then becomes a sum of two different Boltzmann factors. This is the basis of magnetic resonance spectroscopy.

Formulae

q = Σ gⱼ exp(−εⱼ/kT); nⱼ/N = gⱼ exp(−εⱼ/kT)/q; nⱼ/nᵢ = (gⱼ/gᵢ) exp(−Δε/kT); atomic level g = 2J + 1; rotor level g = 2J + 1.

Step-by-step reasoning

To evaluate q for a set of degenerate levels:

1. List each level with its energy above the ground level and its degeneracy. 2. Calculate the Boltzmann factor exp(−εⱼ/kT) for each level. 3. Multiply each factor by its degeneracy. 4. Add the terms to obtain q. 5. Divide each term by q to find the fraction of molecules in that level.

Visual explanation

Draw each level as a stack of short horizontal lines, one line per state: one line for a non-degenerate level, three for a triply degenerate one. Colour each line by its Boltzmann factor. The population of a level is the total colour in its stack, so a tall stack of pale lines can outweigh a single dark line below it.

Real-world analogy

Imagine a theatre where the front row has four seats but the balcony has forty. Every individual balcony seat is less popular than a front-row seat, yet the balcony may still hold more people overall because it has so many more seats. Degeneracy is the number of seats in each row.

Real-world example

The fluorine atom's ground term is split by spin–orbit coupling into a ²P₃/₂ level (g = 4) and a ²P₁/₂ level (g = 2) about 404 cm⁻¹ higher. At high temperatures, such as in flames or plasmas, the upper level holds a substantial fraction of the atoms, which changes the thermodynamic functions and the spectroscopic intensities used to measure temperature.

Why?

Why must degeneracy be included in the entropy? Entropy counts the accessible microstates. A molecule in a level of degeneracy g has g distinct states available, so even at T = 0 a degenerate ground level contributes k ln g per molecule. Ignoring degeneracy would undercount the microstates and hence the entropy.

Common misconception

"A higher level always has a smaller population." This is true for individual states, but not necessarily for levels. With a sufficiently large degeneracy and a spacing that is small relative to kT, an upper level can be more heavily populated, as happens for rotational levels of molecules at room temperature.

Worked example

Question: Calculate the electronic partition function of fluorine atoms at 298 K and the fraction in the upper ²P₁/₂ level, using g₀ = 4, g₁ = 2 and a splitting of 404 cm⁻¹.

Reasoning: ε/kT = 404/207 = 1.95, so exp(−1.95) = 0.142. Then q = 4 + 2 × 0.142 = 4.28. The fraction in the upper level is 0.284/4.28 = 0.066.

Answer: q el ≈ 4.28, and about 6.6% of the atoms are in the ²P₁/₂ level at 298 K.

Quick check

1. Two levels have degeneracies 1 and 3 and are separated by 0.50 kT. Which level holds more molecules? Answer: The upper level, since the ratio is 3 × exp(−0.50) = 1.8, greater than 1.

Exam focus

Always check whether a question refers to states or to levels, and include degeneracy factors when comparing levels. Know the standard degeneracies: 2J + 1 for rotor levels and atomic levels, 2 for an unpaired electron spin, and ground-level degeneracy as the low-temperature limit of q.

Advanced insight

Degeneracy is intimately linked to symmetry. States are degenerate when a symmetry operation transforms one into another without changing the energy; the degeneracies of molecular states match the dimensions of the irreducible representations of the molecule's point group. Breaking the symmetry, for instance by a field or a distortion, splits the level and changes the partition function.

Summary

Grouping states of equal energy gives q = Σ gⱼ exp(−εⱼ/kT). Level populations carry the degeneracy factor, so nⱼ/nᵢ = (gⱼ/gᵢ) exp(−Δε/kT). Degeneracy can make an upper level more populated than a lower level, although each individual state is still less populated. As T → 0, q → g₀; as T → ∞, level populations become proportional to degeneracy.

Practice questions

1. A system has a non-degenerate ground level and a triply degenerate level 300 cm⁻¹ higher. Calculate q at 298 K. Answer: q = 1 + 3 exp(−300/207) = 1 + 3 × 0.235 = 1.70. 2. For the system in question 1, what fraction of molecules is in the upper level? Answer: 0.705/1.70 ≈ 0.41, so about 41% of the molecules are in the upper level. 3. The NO molecule has two electronic levels, each doubly degenerate, separated by 121 cm⁻¹. Calculate q el at 298 K. Answer: q = 2 + 2 exp(−121/207) = 2 + 2 × 0.557 = 3.11. 4. What is the ratio of populations of levels J = 1 and J = 0 of a rotor in the high-temperature limit, and why? Answer: It tends to 3, because the Boltzmann factors approach 1 and the ratio becomes g₁/g₀ = 3/1.