An interactive demo created by Claude Opus 5.5, based on the lecture notes by Sang Hoon Lee (Chapter 7, Section 7.2, Bosons and Fermions)

Bosons and Fermions

Quantum statistics is the study of dense systems in which two or more identical particles have a reasonable chance of wanting to occupy the same single-particle state. Whether they are allowed to share it changes everything.

Counting states

Take \(N\) noninteracting particles and a handful of single-particle states, all with energy zero. Then every Boltzmann factor is \(e^{-\beta E} = 1\) and the partition function simply counts system states, \(Z = \Omega\). With 2 particles and 5 states, distinguishable particles give \(Z = 5 \times 5 = 25\), but the shortcut for indistinguishable particles from Section 6.6 gives \(Z_1^N/N! = 5^2/2! = 12.5\). Half a state?

The shortcut assumes the particles are always in different states. Choose who you are counting and see what really happens.

Distinguishable
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\(Z_1^N\)
The \(1/N!\) shortcut
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\(Z_1^N/N!\), not always a whole number
Bosons
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\(\dbinom{Z_1+N-1}{N}\), any occupancy allowed
Fermions
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\(\dbinom{Z_1}{N}\), at most one per state

When the particles have plenty of room

When the number of available single-particle states is much greater than the number of particles, \(Z_1 \gg N\), the chance of any two particles wanting the same state is negligible. Then all three quantum-correct counts approach the \(1/N!\) shortcut, and it just doesn’t matter whether the particles are bosons or fermions. The chart shows each count divided by \(Z_1^N/N!\) for the current \(N\).

Bosons ÷ shortcut Fermions ÷ shortcut Dot marks the current \(Z_1\)

Who may share a state

Some types of particles can occupy the same state, while others can’t. Bosons have integer spin (\(0, 1, 2, \ldots\) in units of \(\hbar\)): photons, pions, helium-4 atoms. Fermions have half-integer spin (\(\tfrac12, \tfrac32, \ldots\)): electrons, protons, neutrons, neutrinos, helium-3 atoms.

Fermions obey the Pauli exclusion principle. They avoid sharing a state not because they physically repel each other, but because of quantum mechanics. That is why choosing Fermions above crosses out every doubly occupied state, leaving 10 of the 15 boson states in the lecture example.

When does it matter?

For a nonrelativistic ideal gas, \(Z_1 = V Z_{\text{int}}/v_Q\), where \(Z_{\text{int}}\) is some reasonably small number and the quantum volume is roughly the cube of the average de Broglie wavelength:

\[ v_Q = \ell_Q^3 = \left(\frac{h}{\sqrt{2\pi m k_{\mathrm{B}} T}}\right)^3 \]

So \(Z_1 \gg N\) means

\( \displaystyle \frac{V}{N} \gg v_Q \)

The average distance between particles must be much greater than their average de Broglie wavelength. Gases become quantum when they are very dense, very cold, or made of very light particles. Each gray halo below has the diameter \(\ell_Q\).

Spacing \((V/N)^{1/3}\)
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\(\ell_Q\)
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\(V/(N v_Q)\)
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Neutron-star values are rough (interior temperature about \(10^8\ \mathrm{K}\), nuclear density). Photons in a hot oven are quantum too, but they are massless, so they need the separate treatment that comes later in the chapter.

The distribution functions

Use the Gibbs factor, with a twist: let the system be one single-particle state of energy \(\varepsilon\), and the reservoir be all the other single-particle states. If the state holds \(n\) particles, its energy is \(n\varepsilon\), so

\( \displaystyle P(n) = \frac{1}{\mathcal{Z}}\, e^{-(n\varepsilon - \mu n)/k_{\mathrm{B}}T} \)\( \displaystyle = \frac{1}{\mathcal{Z}}\, e^{-n(\varepsilon - \mu)/k_{\mathrm{B}}T} \)

where \(\mathcal{Z}\) sums the Gibbs factors over all possible \(n\). For fermions, \(n = 0\) or \(1\):

\( \displaystyle \mathcal{Z} = 1 + e^{-(\varepsilon-\mu)/k_{\mathrm{B}}T}, \)\( \displaystyle \bar{n} = 0\cdot P(0) + 1\cdot P(1) = \frac{e^{-(\varepsilon-\mu)/k_{\mathrm{B}}T}}{1 + e^{-(\varepsilon-\mu)/k_{\mathrm{B}}T}} \)
\( \displaystyle \bar{n}_{\mathrm{FD}} = \frac{1}{e^{(\varepsilon-\mu)/k_{\mathrm{B}}T} + 1} \) Fermi–Dirac

For bosons, \(n = 0, 1, 2, \ldots\), and the Gibbs sum is a geometric series in \(x = e^{-(\varepsilon-\mu)/k_{\mathrm{B}}T}\). It converges only if \(x < 1\), so only states with \(\varepsilon > \mu\) are allowed:

\[ \mathcal{Z} = \sum_{n=0}^{\infty} x^n = \frac{1}{1 - e^{-(\varepsilon-\mu)/k_{\mathrm{B}}T}} \]

With \(u \equiv (\varepsilon-\mu)/k_{\mathrm{B}}T\) and the result of Problem 7.6,

\[ \bar{n} = \frac{k_{\mathrm{B}}T}{\mathcal{Z}}\frac{\partial \mathcal{Z}}{\partial \mu} = -\frac{1}{\mathcal{Z}}\frac{\partial \mathcal{Z}}{\partial u} = \frac{e^{-u}}{1 - e^{-u}} \]
\( \displaystyle \bar{n}_{\mathrm{BE}} = \frac{1}{e^{(\varepsilon-\mu)/k_{\mathrm{B}}T} - 1} \) Bose–Einstein

For comparison, Boltzmann statistics gives \(\bar{n} = N P(s) = (N/Z_1)\, e^{-\varepsilon/k_{\mathrm{B}}T}\), and with \(\mu = -k_{\mathrm{B}}T\ln(Z_1/N)\) from Problem 6.44:

\( \displaystyle \bar{n}_{\mathrm{Boltzmann}} = e^{-(\varepsilon-\mu)/k_{\mathrm{B}}T} \)
Fermi–Dirac Bose–Einstein Boltzmann Shaded: \(\varepsilon < \mu\), forbidden for bosons
Temperature controls the sharpness of the step.
\(u = (\varepsilon-\mu)/k_{\mathrm{B}}T\)
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\(\bar{n}_{\mathrm{FD}}\)
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\(\bar{n}_{\mathrm{BE}}\)
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\(\bar{n}_{\mathrm{Boltzmann}}\)
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Inside one state

The probe state’s occupation probabilities \(P(n)\). A fermion state is either empty or full. A boson state follows a geometric distribution, \(P(n) = (1-x)\,x^n\), whose tail grows longer as \(\varepsilon\) approaches \(\mu\) from above.

The classical limit

\(\bar{n}_{\mathrm{FD}} \to 1\) when \(\varepsilon \ll \mu\) and \(\bar{n}_{\mathrm{FD}} = \tfrac12\) exactly at \(\varepsilon = \mu\). \(\bar{n}_{\mathrm{BE}} \to \infty\) as \(\varepsilon \to \mu\) from above. But when \(\varepsilon - \mu \gg k_{\mathrm{B}}T\), the \(\pm 1\) in the denominator is negligible and all three distributions converge to \(e^{-(\varepsilon-\mu)/k_{\mathrm{B}}T}\): the classical limit. Switch on the logarithmic axis and move the probe to the right to watch them merge.