Physics · Molecular physics and thermodynamics · Year 4

Quantum statistics

Why do the electrons in a metal contribute almost nothing to its heat capacity?

The situation

Each copper atom contributes one free electron. According to classical physics these electrons should add another $\frac{3}{2} R$ to the heat capacity — but experiment shows no such thing. In 1995 physicists cooled rubidium atoms below 200 nK and “gathered” them into a single quantum state. And in the sky, stars the size of the Earth — white dwarfs — are not crushed under their own weight. The secret of all three phenomena is the same: whether the particles are fermions or bosons. How does it work?

Topic

Quantum statistics: fermions and bosons. Electron gas in a metal and a condensate of laser-cooled atoms

Key ideas

Fermi–Dirac and Bose–Einstein
Fermi energy
Debye T³ law

Formulas

\overline{n} (ε) = \frac{1}{e^{(ε - μ) / k T} \pm 1}, {\overline{n}}_{M B} = e^{- (ε - μ) / k T}

The Fermi–Dirac (+) and Bose–Einstein (−) distributions; for $e^{(ε - μ) / k T} ≫ 1$ both reduce to Maxwell–Boltzmann.

E_{F} = \frac{ℏ^{2}}{2 m} {(3 π^{2} n)}^{2 / 3}, C_{e} = \frac{π^{2}}{2} N k \frac{T}{T_{F}}

A degenerate electron gas: the Fermi energy and the electronic heat capacity ( $T ≪ T_{F}$ ).

C_{V} = 9 N k {(\frac{T}{θ_{D}})}^{3} \int_{0}^{θ_{D} / T} \frac{x^{4} e^{x} d x}{{(e^{x} - 1)}^{2}} \overset{\overset{}{T ≪ θ_{D}}}{\to} \frac{12 π^{4}}{5} N k {(\frac{T}{θ_{D}})}^{3}

Debye theory; at high temperature the Dulong–Petit law $C = 3 R$ .

k T_{C} = ℏ \overline{ω} {(\frac{N}{ζ (3)})}^{1 / 3} \approx 0.94 ℏ \overline{ω} N^{1 / 3}, \frac{N_{0}}{N} = 1 - {(\frac{T}{T_{C}})}^{3}

An ideal Bose gas in a harmonic trap: the condensation temperature and the condensate fraction. For a uniform gas $T_{C} = \frac{2 π ℏ^{2}}{m k} {(\frac{n}{2.612})}^{2 / 3}$ .

M_{C h} \approx \frac{5.83}{μ_{e}^{2}} M_{⊙} \approx 1.46 M_{⊙} (μ_{e} = 2), R \propto M^{- 1 / 3}

A white dwarf: the greatest mass that degeneracy pressure can support (the Chandrasekhar limit); in the non-relativistic regime a heavier dwarf is smaller.

Theory

Fermions and bosons

Particles with half-integer spin are fermions (electron, proton): by the Pauli exclusion principle no more than one can occupy a state, $\overline{n} \leq 1$ . Particles with integer spin are bosons (photon, phonon, ⁸⁷Rb atom): they “like” to gather in one state. If the particles are sparse and hot ( $n λ^{3} ≪ 1$ , where $λ$ is the de Broglie wavelength) the difference disappears — Maxwell–Boltzmann statistics. The fluctuations differ too: for bosons $⟨ Δ n^{2} ⟩ = \overline{n} (1 + \overline{n})$ , for fermions $\overline{n} (1 - \overline{n})$ — in a photon gas these fluctuations set the limit of sensitive instruments. In non-equilibrium gases — as in a semiconductor under illumination or in a laser — the distribution departs from these formulas.

A degenerate electron gas, the Fermi sphere

At $T = 0$ the electrons fill the states from the lowest up to $E_{F}$ : in $k$ -space this is the Fermi sphere (the dots on the screen are the allowed states). For copper $E_{F}$ = 7.0 eV and $T_{F} \approx 81 600$ K: at room temperature $k T = 0.026$ eV is very small, and the gas is strongly degenerate. Only a layer of thickness $\sim k T$ near the Fermi surface can be excited. The density of states is $g (ε) \propto \sqrt{ε}$ . In a crystal the energy is divided into bands: if a band is partly filled — a metal; a full band and a wide forbidden gap — an insulator; a narrow forbidden gap — a semiconductor (in detail in the experiment “Bands in solids”).

Electronic heat capacity and magnetic properties

Classical theory gives $\frac{3}{2} R$ per mole of electrons; in fact $C_{e} = γ T$ , $γ = π^{2} R / (2 T_{F})$ — for copper 0.50 mJ/(mol·K²) in the free-electron model and 0.695 mJ/(mol·K²) in experiment (the difference is the effective mass). Because the lattice contribution is $T^{3}$ , $γ T$ is noticeable only at a few kelvin: if we plot $C / T$ against $T^{2}$ we get a straight line, and its intercept on the $C / T$ axis is $γ$ . For the same reason only the electrons near $E_{F}$ can turn their spins: Pauli paramagnetism is small and independent of temperature; orbital motion gives Landau diamagnetism (a fraction $- \frac{1}{3}$ of the Pauli paramagnetism).

Heat capacity of solids: Dulong–Petit, Einstein, Debye

The classical equipartition theorem: $\frac{1}{2} k T$ for every quadratic degree of freedom (by the virial theorem, the kinetic and potential energies in an oscillation are equal) — for $3 N$ oscillators $C = 3 R \approx 24.9$ J/(mol·K) (Dulong–Petit). At low temperature $C$ decreases — the energy is quantised. Einstein (1907): all atoms have the same frequency — $C$ falls exponentially (very fast). Debye (1912): the vibrations are phonons, the density of states is $\propto ω^{2}$ up to $ω_{D}$ ; the result is $C \propto T^{3}$ , where $θ_{D}$ is the Debye temperature. The photon gas is similar: the energy is $\propto T^{4}$ (Planck's formula, the experiment “Thermal radiation and Planck”). In a diatomic gas the rotational and vibrational degrees of freedom likewise “freeze out”. A harmonic lattice would not expand on heating and would conduct heat infinitely well — it is anharmonicity that makes phonons scatter and the solid expand.

Bose–Einstein condensation

For $T < T_{C}$ a macroscopic fraction of the bosons falls into the lowest state. In a harmonic trap $N_{0} / N = 1 - {(T / T_{C})}^{3}$ . In the experiment the trap is switched off, the cloud expands in free flight for a few milliseconds and its shadow is photographed: a sharp peak on top of a broad thermal cloud — the condensate. In the tight direction the condensate expands faster (the uncertainty principle), so its shape is “inverted”. The model is an ideal gas without interactions, with a Gaussian thermal cloud; a real condensate is wider because of interactions.

White dwarfs and the Chandrasekhar limit

What holds a white dwarf up against gravity is not thermal pressure but the degeneracy pressure of the electrons (the Pauli principle). In a non-relativistic gas $P \propto ρ^{5 / 3}$ , giving $R \propto M^{- 1 / 3}$ — a heavier dwarf is smaller! As the mass increases the electrons become relativistic ( $x = p_{F} / m_{e} c > 1$ ), $P \propto ρ^{4 / 3}$ , and a maximum mass appears: $M_{C h} \approx 1.46 M_{⊙}$ ( $μ_{e} = 2$ , often quoted as “1.4”). A remnant heavier than this becomes a neutron star or a black hole. In the experiment the equation of hydrostatic equilibrium is solved numerically with the fully relativistic equation of state.

Did you know?

In 1995 a JILA (USA) group cooled a vapour of rubidium-87 atoms to about 170 nK and made the first atomic Bose–Einstein condensate. In 2001 E. Cornell and C. Wieman (JILA), together with W. Ketterle (MIT), who made a condensate from sodium atoms, were awarded the Nobel Prize for this discovery.
Sirius B, the companion of the star Sirius, has a mass about equal to that of the Sun and a size slightly smaller than the Earth. A teaspoonful of its matter would weigh about 10 tonnes on Earth.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

More experiments in this section

Thermal expansion — Grade 10
Refrigerator and heat pump — Grade 10
Real gases — Grade 10
Maxwell–Boltzmann distribution — Grade 10
Transport phenomena — Year 1
Phase transition and fluctuations — Year 4