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The Gibbs factor for systems that exchange particles, the Fermi-Dirac and Bose-Einstein distributions, their classical limit, and the quantum concentration that decides when they matter.
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By the end of this lesson you will be able to derive the Fermi-Dirac and Bose-Einstein distributions from the Gibbs factor, find the occupancy of a state for fermions, bosons and classical particles, and decide from the quantum concentration whether a gas is degenerate.
You know the Boltzmann factor, the chemical potential as the energy cost of adding a particle, and that identical particles cannot be told apart. From modern physics you know that electrons obey the Pauli exclusion principle and that photons do not. This lesson turns those facts into the two distributions that replace the Boltzmann distribution when particles crowd together.
| Term | What it means |
|---|---|
| Gibbs factor | $e^{-(E - \mu N)/kT}$, the relative probability of a state with energy $E$ and $N$ particles, for a system exchanging both with a reservoir. |
| Grand partition function | $\mathcal{Z} = \sum e^{-(E - \mu N)/kT}$, the sum of Gibbs factors. |
| Fermion | A particle with half-integer spin, such as an electron, proton or neutron; at most one per quantum state. |
| Boson | A particle with integer spin, such as a photon or a helium-4 atom; any number per state. |
| Fermi-Dirac distribution | $\bar{n}_{\text{FD}} = 1/(e^{(\epsilon - \mu)/kT} + 1)$. |
| Bose-Einstein distribution | $\bar{n}_{\text{BE}} = 1/(e^{(\epsilon - \mu)/kT} - 1)$. |
| Quantum concentration | $n_Q = (2\pi mkT/h^2)^{3/2}$; quantum statistics matter when $n \gtrsim n_Q$. |
| Degenerate | Describes a gas so dense or cold that $n \gg n_Q$ and quantum statistics dominate. |
Take a single quantum state of energy $\epsilon$ as the system, and let it exchange both energy and particles with a large reservoir at temperature $T$ and chemical potential $\mu$. Repeating the Boltzmann-factor argument with particles exchanged as well, the probability of each occupancy is proportional to the Gibbs factor $e^{-(E - \mu N)/kT}$, and with $N$ particles in the state $E = N\epsilon$.
Fermions — electrons, protons, neutrons — obey the exclusion principle, so the state holds $0$ or $1$. The Gibbs sum has two terms, $\mathcal{Z} = 1 + e^{-(\epsilon - \mu)/kT}$, and the average occupancy is
$$\bar{n}_{\text{FD}} = \frac{1}{e^{(\epsilon - \mu)/kT} + 1}.$$
Bosons — photons, helium-4 atoms — can pile into the state in any number, so the sum is a geometric series, $\mathcal{Z} = 1/(1 - e^{-(\epsilon - \mu)/kT})$, and
$$\bar{n}_{\text{BE}} = \frac{1}{e^{(\epsilon - \mu)/kT} - 1}.$$
The two differ from each other, and from the classical Boltzmann occupancy $e^{-(\epsilon - \mu)/kT}$, only by $\pm 1$ in the denominator. For states far above $\mu$ the exponential is huge, the $\pm 1$ is negligible, and both reduce to Boltzmann. For states near or below $\mu$ they part company: fermion occupancies never exceed one, and boson occupancies can grow without limit.
When do the quantum forms matter? Roughly, when the particles are crowded enough that there are about as many of them as available low-energy states — when the concentration $n$ approaches the quantum concentration $n_Q = (2\pi mkT/h^2)^{3/2}$.
Another way: picture
Picture single-particle states as seats in a theater. Classical particles are so few that nobody ever competes for a seat, and it does not matter whether two could share. Fermions are ticket holders at a sold-out show: one per seat, the cheap seats full to a sharp edge, the expensive ones empty. Bosons are friends who would all rather squeeze into the same favorite seat, and when cold enough, a huge crowd of them does.
Another way: steps
Checks. The fermion occupancy is symmetric about $\mu$: a state $\Delta$ above $\mu$ is filled with probability $p$, one $\Delta$ below with probability $1 - p$. The boson occupancy is always larger than the classical one, and the fermion occupancy smaller. For photons the boson formula with $\mu = 0$ is the Planck distribution of the next lesson. And any occupancy above one for fermions, or a boson $\mu$ above the lowest state, signals an error.
The derivation mirrors the Boltzmann factor's. The system and reservoir share a fixed total energy $U_0$ and particle number $N_0$. The probability of a system state with energy $E$ and $N$ particles is proportional to the reservoir's multiplicity, $e^{S_R/k}$, evaluated at $U_0 - E$ and $N_0 - N$. Expanding the reservoir's entropy to first order in both,
$$S_R(U_0 - E, N_0 - N) \approx S_R(U_0, N_0) - \frac{E}{T} + \frac{\mu N}{T},$$
using $\partial S/\partial U = 1/T$ and $\partial S/\partial N = -\mu/T$. The first term is common to every state, so the probability is proportional to $e^{-(E - \mu N)/kT}$. The chemical potential appears as the reservoir's price for each particle, exactly as temperature sets its price for energy. Normalizing by the grand partition function $\mathcal{Z} = \sum e^{-(E - \mu N)/kT}$ turns the factors into probabilities, and the fermion and boson distributions follow by summing over the allowed occupancies of one state.
The quantum concentration $n_Q = (2\pi mkT/h^2)^{3/2}$ is roughly one particle per cube of its thermal de Broglie wavelength. Air at room temperature has $n \approx 2.5 \times 10^{25}$ m⁻³ and, for nitrogen, $n_Q \approx 1.4 \times 10^{32}$ m⁻³: the ratio is $10^{-7}$, so every molecule has millions of states to itself and quantum statistics are irrelevant.
Electrons are about fifty thousand times lighter than nitrogen molecules, so their $n_Q$ at $300$ K is only $1.25 \times 10^{25}$ m⁻³, while the conduction electrons in copper number $8.5 \times 10^{28}$ m⁻³. The ratio is nearly $7000$: the electron gas in every metal is deeply degenerate at room temperature, and would stay so up to tens of thousands of kelvin. That single fact explains why metals conduct, why their electrons contribute almost nothing to their heat capacity, and why their properties barely change with temperature.
Liquid helium-4 near $2$ K has a ratio of about $5$, and it is a boson: that is where superfluidity appears.
The boson distribution has a consequence worth seeing directly. When photons are already present in a state, the probability that another photon enters it is enhanced by a factor $1 + \bar{n}$ — the bunching behind the $-1$ in the denominator. For atoms emitting light, this is stimulated emission: an atom is more likely to emit into a mode already full of photons, and the new photon is an exact copy of the ones there. A laser exploits it by filling one mode of a cavity with enormous numbers of photons, so that almost every emission goes into that mode. Einstein predicted stimulated emission in 1917 from exactly this statistical argument, four decades before the first laser was built.
Copper has one free electron per atom, $8.5 \times 10^{28}$ per cubic meter, filling states up to a Fermi energy of $7.0$ eV. At room temperature, $kT = 0.026$ eV. A state $0.1$ eV below the Fermi energy is filled with probability $1/(e^{-3.87} + 1) = 0.98$; one $0.1$ eV above with probability $0.02$. Only within a few $kT$ of the Fermi energy is the occupancy far from zero or one.
That has a striking consequence. To absorb heat or respond to an electric field, an electron must move into an empty state of slightly higher energy, and deep below the Fermi energy every nearby state is already full. So only the electrons within about $kT$ of the Fermi energy — about $kT/E_F \approx 0.4$ percent of them — take part. Equipartition would give the electron gas a heat capacity of $\tfrac{3}{2}R$ per mole; the measured electronic heat capacity of copper at room temperature is about a hundred times smaller, and it was exactly this puzzle that Sommerfeld solved in 1927 by applying Fermi-Dirac statistics.
The same statistics govern every transistor. In silicon, the probability that a conduction-band state $0.56$ eV above the chemical potential is occupied is about $e^{-21.7} \approx 4 \times 10^{-10}$ at room temperature, and doping adjusts the chemical potential to change that by factors of millions.
Helium-4 atoms are bosons, and liquid helium at atmospheric pressure has a density of $2.18 \times 10^{28}$ atoms per cubic meter. At $2$ K, its quantum concentration is about $4.3 \times 10^{27}$ m⁻³, so $n/n_Q \approx 5$: the atoms' de Broglie waves overlap, and Bose-Einstein statistics take over.
Below $2.17$ K, the lambda point, a macroscopic fraction of the atoms falls into the lowest quantum state together, and the liquid becomes a superfluid: it flows through channels a few atoms wide with no viscosity, creeps up the walls of its container in a film, and carries heat so efficiently that it stops boiling abruptly. Helium-3, with the same chemistry but half-integer spin, is a fermion, and does nothing of the kind until below $0.0025$ K, where its atoms pair up into bosons.
Superfluid helium is a working engineering material: the superconducting magnets of the Large Hadron Collider are cooled by about $100$ tonnes of helium, most of it superfluid at $1.9$ K, chosen precisely because its superfluid heat transport keeps the magnets uniformly cold.
It is tempting to think of Fermi-Dirac and Bose-Einstein statistics as corrections that matter only near absolute zero, in exotic laboratory systems. What decides it is not the temperature alone but the ratio $n/n_Q$, which depends on how crowded and how light the particles are. For air it is $10^{-7}$ and quantum statistics never matter; for the electrons in any metal it is thousands at room temperature, and Fermi-Dirac statistics govern everything, including why a copper wire conducts and why its heat capacity is almost entirely due to the lattice.
The other trap is to apply the Boltzmann distribution to states below the chemical potential. There $e^{-(\epsilon - \mu)/kT}$ exceeds one, which for fermions is impossible. The classical form is only the limit for states well above $\mu$, where occupancies are small; wherever occupancies are near one, the $\pm 1$ in the denominator decides the physics.
A state lies $kT$ above the chemical potential. Find the exponential.
$x = \dfrac{\epsilon - \mu}{kT} = 1, \qquad e^{x} = 2.718$
The distance from $\mu$ in units of $kT$.
Find the fermion occupancy.
$\bar{n}_{\text{FD}} = \dfrac{1}{2.718 + 1} = 0.269$
Plus one: exclusion.
Find the boson occupancy.
$\bar{n}_{\text{BE}} = \dfrac{1}{2.718 - 1} = 0.582$
Minus one: bunching.
Find the classical occupancy.
$\bar{n}_{\text{MB}} = e^{-1} = 0.368$
Neither correction.
Compare the three results.
$0.269 < 0.368 < 0.582$
At only $kT$ above $\mu$ the three differ substantially; far above $\mu$ they converge.
A single state can hold $0$ or $1$ fermion. Write the Gibbs factor of each occupancy.
$N = 0: \ 1; \qquad N = 1: \ e^{-(\epsilon - \mu)/kT}$
Energy $0$ and $\epsilon$, with $0$ and $1$ particles.
Add the terms together.
$\mathcal{Z} = 1 + e^{-(\epsilon - \mu)/kT}$
The grand partition function of the state.
Write the average occupancy.
$\bar{n} = \dfrac{0 \times 1 + 1 \times e^{-(\epsilon - \mu)/kT}}{\mathcal{Z}}$
Each occupancy weighted by its probability.
Multiply top and bottom by $e^{(\epsilon - \mu)/kT}$.
$\bar{n} = \dfrac{1}{e^{(\epsilon - \mu)/kT} + 1}$
The Fermi-Dirac distribution.
Check the state at the chemical potential.
$\epsilon = \mu: \quad \bar{n} = \dfrac{1}{1 + 1} = \dfrac{1}{2}$
Half filled.
Check the symmetry about $\mu$.
$\bar{n}(\mu - \Delta) = \dfrac{1}{e^{-\Delta/kT} + 1} = 1 - \dfrac{1}{e^{\Delta/kT} + 1} = 1 - \bar{n}(\mu + \Delta)$
The chance of an empty state below $\mu$ equals the chance of a filled one the same distance above.
Write the quantum concentration for electrons.
$n_Q = \left(\dfrac{2\pi m_ekT}{h^2}\right)^{3/2}$
One electron per thermal de Broglie cube.
Evaluate the bracket at $300$ K.
$\dfrac{2\pi \times 9.11 \times 10^{-31} \times 1.381 \times 10^{-23} \times 300}{(6.626 \times 10^{-34})^2} = 5.40 \times 10^{16}\ \text{m}^{-2}$
SI units throughout.
Raise it to the power three halves.
$n_Q = (5.40 \times 10^{16})^{3/2} = 1.25 \times 10^{25}\ \text{m}^{-3}$
Small, because electrons are light.
Find copper's conduction electron concentration.
$n = \dfrac{8.96 \times 10^{3}\ \text{kg/m}^3}{0.06355\ \text{kg/mol}} \times 6.02 \times 10^{23} = 8.49 \times 10^{28}\ \text{m}^{-3}$
One free electron per atom.
Divide the two quantities.
$\dfrac{n}{n_Q} = \dfrac{8.49 \times 10^{28}}{1.25 \times 10^{25}} \approx 6800$
Thousands of electrons compete for every thermally accessible state.
Draw the conclusion.
$n \gg n_Q \quad\Rightarrow\quad \text{Fermi-Dirac statistics}$
The electrons fill states up to a high Fermi energy regardless of temperature.
Compare with helium gas at room conditions.
$\dfrac{n}{n_Q} = \dfrac{2.45 \times 10^{25}}{7.8 \times 10^{30}} = 3 \times 10^{-6}$
Helium atoms are far heavier and far fewer: a perfectly classical gas.
Say what temperature would make copper's electrons classical.
$n_Q \propto T^{3/2}: \quad 6800^{2/3} \times 300 \approx 1.1 \times 10^{5}\ \text{K}$
Far above copper's melting point, so metals' electrons are always quantum.
Write the Bose-Einstein occupancy.
$\bar{n} = \dfrac{1}{e^x - 1}$
Minus one for bosons.
Evaluate the exponential.
$e^{\ln 2} = 2$
The exponential undoes the logarithm.
Divide the two quantities.
A single quantum state lies well below the chemical potential of a gas at $600$ K. What is its average occupancy if the particles are fermions, and if they are bosons?
Complete the worked solution: a single quantum state can hold zero or one fermion. When filled, its Gibbs factor is $e^{-(\epsilon - \mu)/kT} = 0.6$. Find the Gibbs sum and the probabilities that it is filled and empty.
Add the Gibbs factors of the two occupancies.
$\mathcal{Z} = 1 + 0.6 =$ z
Empty contributes $1$; filled contributes its Gibbs factor.
Divide the filled term by the sum.
$P_{\text{filled}} = \dfrac{0.6}{\mathcal{Z}} =$ p
This is the average occupancy, the Fermi-Dirac distribution.
Subtract from one for the empty state.
$P_{\text{empty}} = 1 - P_{\text{filled}} =$ q
The state is either filled or empty.
Particles at $136$ K occupy single-particle states of energy $\epsilon$ with chemical potential $\mu$. Match each to its average occupancy.
| $1/(e^{(\epsilon - \mu)/kT} + 1)$ | $1/(e^{(\epsilon - \mu)/kT} - 1)$ | $e^{-(\epsilon - \mu)/kT}$ | $\tfrac{1}{2}$ | |
|---|---|---|---|---|
| fermions | ||||
| bosons | ||||
| either kind, far above $\mu$ | ||||
| a fermion state at $\epsilon = \mu$ |
A state lies at $x = (\epsilon - \mu)/kT = 0.5$, where $e^x = 1.6487$. Fill in its average occupancy for fermions, bosons and the classical limit, to four decimal places.
| fermions | bosons | classical | |
|---|---|---|---|
| this state |
In a gas of fermions at $150$ K, a state lies $20$ meV above the chemical potential (a negative value means below). What is its average occupancy?
Answer:
For conduction electrons in silver at 300 K, the particle concentration is $5.86 \times 10^{28}$ m⁻³ and the quantum concentration is $n_Q = 1.255 \times 10^{25}$ m⁻³. What is $n/n_Q$, and so are quantum statistics needed? Give the ratio.
Answer:
In a copper wire at $300$ K, what is the probability that an electron state $0.1$ eV from the Fermi energy (positive above, negative below) is occupied? Use $kT = 0.02585$ eV.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A state lies at $x = (\epsilon - \mu)/kT = 1$, where $e^x = 2.7183$. Fill in its average occupancy for fermions, bosons and the classical limit, to four decimal places.
| fermions | bosons | classical | |
|---|---|---|---|
| this state |
You can find how many fermions or bosons occupy a state and decide when quantum statistics matter. Explain to someone why the electrons in a copper wire are a quantum gas at room temperature while the air around it is not.
15. Your turn: a boson state lies at $x = (\epsilon - \mu)/kT = \ln 2$. What is its average occupancy?, step 3
$\bar{n} = \dfrac{1}{2 - 1} = 1$
One boson on average; a fermion state here would hold $1/3$.