The Boltzmann Distribution

Heat a gas of iodine molecules and more of them start vibrating vigorously. Cool it down and almost all of them settle into the ground vibrational state. This shift in populations is not random. It follows one of the most useful formulas in physical chemistry: the Boltzmann distribution. The Boltzmann distribution tells us how molecules share out a fixed amount of total energy among their available quantum states at thermal equilibrium. It underlies reaction rates, spectroscopic intensities, heat capacities, and equilibrium constants. Almost every quantitative prediction in statistical thermodynamics traces back to this single exponential law. This article states the distribution, explains each symbol carefully, and shows how it converts an energy-level diagram into real numbers. You will see how populations depend on temperature. You will see how they depend on degeneracy. Two worked examples show the arithmetic in full, from a two-level spin system to a set of rotational levels.

Key takeaways

The Boltzmann distribution gives the relative population of an energy level in a system at thermal equilibrium.
Population ratio between two levels depends only on their energy gap and the temperature: $N_i/N_j = e^{-(\varepsilon_i-\varepsilon_j)/kT}$ .
Degenerate levels get an extra factor equal to their degeneracy $g_i$ , since each state within a level is equally likely.
Low temperature crowds molecules into the lowest levels; high temperature spreads population across many levels.
The partition function $q$ is simply the normalizing sum that turns these ratios into absolute fractional populations.
Key idea: at thermal equilibrium, the fraction of molecules in a state falls off exponentially with that state's energy, scaled by $kT$ .

The core insight

At thermal equilibrium the population of a state depends only on its energy relative to $kT$ , so that states within about $kT$ of the ground state are well populated while states far above it are exponentially suppressed.

The physical picture

Imagine a huge number of identical molecules — a mole of them, say — sharing a fixed total energy at a fixed temperature. Each molecule can only occupy one of a discrete ladder of energy levels: translational, rotational, vibrational, or electronic. At any instant, collisions are constantly redistributing energy among molecules, kicking some up to higher levels and letting others relax down. Statistical thermodynamics assumes this exchange is fast enough, and the system large enough, that the population of each level settles into a stable time-averaged value: the equilibrium distribution. You already know from counting microstates and configurations that the overwhelmingly most probable way to distribute energy among distinguishable particles is not the most "even" one, nor the most "extreme" one, but a specific exponential profile. The Boltzmann distribution is the statement of exactly what that profile looks like. It says the population $N_i$ of a nondegenerate level of energy $\varepsilon_i$ , relative to the population $N_0$ of the ground level ( $\varepsilon_0 = 0$ ), is

\frac{N_i}{N_0} = e^{-\varepsilon_i/kT}

Two features deserve attention immediately. First, the ratio depends only on the energy above the ground state, not on the absolute zero of energy you happen to choose — shifting all energies by a constant leaves every ratio unchanged. Second, temperature sets the "currency exchange rate" between energy and probability: $kT$ is the natural energy unit of thermal populations. At room temperature (298 K), $kT \approx 4.11 \times 10^{-21}\text{ J}$ , equivalent to about $2.57$ kJ mol⁻¹ or $208$ cm⁻¹ in spectroscopic units — a useful benchmark for judging whether a given energy gap will be "thermally accessible." If a level is degenerate — several distinct quantum states sharing the same energy — each of those states is equally probable, so the whole level is more heavily populated in proportion to its degeneracy $g_i$ . This gives the general working form of the distribution used throughout thermodynamics and spectroscopy.

Figure

Populations of an energy ladder collapsing as temperature falls

The core relations

\frac{N_i}{N_0} = \frac{g_i}{g_0}\, e^{-(\varepsilon_i-\varepsilon_0)/kT}

where: $N_i$ , $N_j$ are level populations, $g_i$ is the degeneracy of level $i$ , $\varepsilon_i$ are level energies measured from a common zero, $k$ is Boltzmann's constant, $T$ is absolute temperature.

\frac{N_i}{N_j} = \frac{g_i}{g_j}\, e^{-(\varepsilon_i-\varepsilon_j)/kT}

where: this is the same law written between any two levels $i$ and $j$ , useful when neither is the ground state.

\frac{N_i}{N} = \frac{g_i e^{-\varepsilon_i/kT}}{q}, \qquad q = \sum_i g_i\, e^{-\varepsilon_i/kT}

where: $N$ is the total number of molecules, $q$ is the molecular partition function (the normalizing sum over all levels), previewed here and developed fully once you meet the partition function itself.

Worked examples

Example 1 — a two-level spin system (foundational). Consider a collection of non-interacting spin-½ nuclei in a magnetic field of 11.7 T (typical of a 500 MHz NMR spectrometer). The two nondegenerate levels are separated by

\Delta\varepsilon = h\nu = (6.626\times10^{-34}\,\text{J s})(500\times10^6\,\text{s}^{-1}) = 3.313\times10^{-25}\,\text{J}

At $T = 298$ K, $kT = (1.381\times10^{-23}\,\text{J K}^{-1})(298\,\text{K}) = 4.116\times10^{-21}\,\text{J}$ . The exponent is

\frac{\Delta\varepsilon}{kT} = \frac{3.313\times10^{-25}\,\text{J}}{4.116\times10^{-21}\,\text{J}} = 8.048\times10^{-5}

So the population ratio of the upper (β) to lower (α) spin state is

\frac{N_\beta}{N_\alpha} = e^{-8.048\times10^{-5}} = 0.9999195

This tiny departure from unity — about 80 parts per million — is exactly why NMR signals are so weak compared with, say, electronic or vibrational transitions: the two levels are almost equally populated because $\Delta\varepsilon \ll kT$ . Sanity check: since the gap is minuscule compared with $kT$ , we expect the ratio very close to 1, which matches. Increasing the field or lowering $T$ would push this ratio further from unity and boost NMR sensitivity, exactly as spectroscopists exploit in cryoprobes.

Example 2 — rotational level populations of HCl (applied). For a diatomic rotor, energy levels are $\varepsilon_J = BJ(J+1)$ with degeneracy $g_J = 2J+1$ , where $B$ is the rotational constant. For H³⁵Cl, $B = 10.59\ \text{cm}^{-1}$ . Convert to joules using $hc$ :

B = (10.59\,\text{cm}^{-1})(6.626\times10^{-34}\,\text{J s})(2.998\times10^{10}\,\text{cm s}^{-1}) = 2.103\times10^{-22}\,\text{J}

At $T = 298$ K, $kT = 4.116\times10^{-21}$ J, so $B/kT = 0.05110$ . We want the ratio of the population in $J=3$ to that in $J=0$ (the ground rotational state). The energies are $\varepsilon_0 = 0$ and $\varepsilon_3 = B(3)(4) = 12B = 2.524\times10^{-21}$ J. Degeneracies are $g_0 = 1$ and $g_3 = 7$ .

\frac{N_3}{N_0} = \frac{g_3}{g_0}\,e^{-\varepsilon_3/kT} = 7 \times e^{-2.524\times10^{-21}/4.116\times10^{-21}}

= 7 \times e^{-0.6132} = 7 \times 0.5417 = 3.79

So at room temperature nearly four times as many HCl molecules sit in $J=3$ as in $J=0$ — the degeneracy factor $2J+1$ dominates at low $J$ even though the exponential is working against population. Sanity check: this behavior — population rising then falling with $J$ — is exactly the textbook shape of a rotational population distribution, with the maximum near

J_{\max} \approx \sqrt{\frac{kT}{2B}} - \frac{1}{2} \approx \sqrt{\frac{1}{2(0.05110)}} - 0.5 \approx 3.13 - 0.5 \approx 2.6

consistent with the population still climbing between $J=0$ and $J=3$ .

Limiting cases & common traps

Situation	What happens	Why
$\varepsilon_i \ll kT$	$N_i/N_0 \to g_i/g_0$ , nearly equal populations	exponential term is close to 1
$\varepsilon_i \gg kT$	$N_i/N_0 \to 0$	level essentially unpopulated
$T \to 0$	all population collapses into the ground level(s)	exponential penalty becomes infinite for any $\varepsilon_i>0$
$T \to \infty$	populations tend toward ratios of degeneracies only	thermal energy swamps the energy gaps
Degenerate levels ignored	populations underestimated	forgetting the $g_i$ factor is the single most common error

The chemist behind it

Ludwig Boltzmann built the statistical foundations connecting microscopic mechanics to macroscopic thermodynamics in the 1870s, arguing that entropy itself measures the number of microscopic arrangements consistent with a given macroscopic state. His distribution law emerged from asking which arrangement of energy among particles is overwhelmingly the most probable — the same maximization argument you meet when deriving the distribution from statistical weights. His ideas were fiercely contested in his lifetime by scientists skeptical of atoms as real objects, yet they now underpin virtually every quantitative link between molecular structure and bulk thermodynamic behavior.

Origins · 1 minLudwig Boltzmann and how this was discovered

Ludwig Boltzmann

1844–1906 · Austrian physicist who founded statistical mechanics and gave entropy its microscopic, probabilistic meaning

Ask the tutor: Want to see how maximizing the statistical weight actually derives this exponential law from first principles?

Does the Boltzmann distribution apply to just one molecule or to a whole sample?

It describes the equilibrium population averaged over an enormous number of molecules (or over long times for one molecule undergoing many collisions). A single molecule constantly hops between energy levels as it exchanges energy in collisions. The Boltzmann distribution tells you the fraction of time it spends in each level, which equals the fraction of a large ensemble of identical molecules found in that level at any instant. Both viewpoints — ensemble average and time average — give the same numbers at equilibrium.

Why does kT appear rather than just T in the exponential?

Energy and temperature have different units, so a genuine comparison needs an energy scale built from $T$ . Multiplying by Boltzmann's constant $k$ converts temperature into an energy, $kT$ , which can be directly compared with the level spacing $\varepsilon_i$ . The ratio $\varepsilon_i/kT$ is dimensionless, exactly what an exponent requires. This is why $kT$ shows up throughout thermodynamics as the "natural" thermal energy unit at temperature $T$ .

How is the Boltzmann distribution related to the partition function?

The partition function $q$ is simply the sum over all levels of their Boltzmann factors, $q = \sum_i g_i e^{-\varepsilon_i/kT}$ . It acts as a normalizing constant so that fractional populations $N_i/N$ sum to exactly 1. Once you know $q$ , the Boltzmann distribution converts directly into absolute populations rather than just ratios, and $q$ itself becomes the gateway to computing energy, entropy, and other thermodynamic functions from molecular data.

Can the Boltzmann distribution predict negative populations or populations above 100 percent?

No — every Boltzmann factor $e^{-\varepsilon_i/kT}$ is strictly between 0 and 1 for $\varepsilon_i > 0$ , and degeneracies are positive integers, so $N_i$ always comes out as a nonnegative number less than or equal to $N$ . If a calculation yields something outside this range, check for an arithmetic sign error in the exponent or a units mismatch between $\varepsilon_i$ and $kT$ , which is by far the most common source of such nonsensical results.

Is the Boltzmann distribution only valid at thermal equilibrium?

Yes, strictly speaking. It describes populations once a system has settled into thermal equilibrium at a well-defined temperature $T$ , with energy freely exchanging among levels. Systems driven far from equilibrium — a laser-pumped gas, a rapidly quenched glass, a population inverted by a pulse of radiation — do not follow this distribution until they relax back toward equilibrium, though the same mathematical machinery often still describes each subsystem locally.