Fermi Systems

The **Fermi-Dirac distribution function** governs the statistical occupation of single-particle quantum states for any system of identical fermions (particles with half-integer spin, obeying the Pauli exclusion principle).

1. Derivation via the Grand Canonical Ensemble

Consider a single-particle state $i$ with energy $\epsilon_i$. Due to the Pauli exclusion principle, the state can be occupied by either $n_i = 0$ or $n_i = 1$ fermion. The grand canonical partition function for this single state is: $$\Xi_i = \sum_{n_i \in \{0, 1\}} e^{-\beta(\epsilon_i - \mu)n_i} = 1 + e^{-\beta(\epsilon_i - \mu)}$$ The mean occupation number (probability of the state being occupied) is: $$f(\epsilon_i) = \langle n_i \rangle = -\frac{1}{\beta}\frac{\partial \ln \Xi_i}{\partial \epsilon_i} = \frac{e^{-\beta(\epsilon_i - \mu)}}{1 + e^{-\beta(\epsilon_i - \mu)}} = \frac{1}{e^{(\epsilon_i - \mu)/(k_B T)} + 1}$$ This is the celebrated **Fermi-Dirac Distribution Function**.

2. Behavior at Absolute Zero ($T = 0\text{ K}$)

At $T = 0\text{ K}$, the chemical potential defines the **Fermi Energy**: $\mu(0) \equiv \epsilon_F$. $$\lim_{T \to 0} \frac{\epsilon - \epsilon_F}{k_B T} = \begin{cases} -\infty & \text{if } \epsilon < \epsilon_F \\ +\infty & \text{if } \epsilon > \epsilon_F \end{cases}$$ Therefore: $$f(\epsilon) = \begin{cases} 1 & \text{if } \epsilon < \epsilon_F \\ 0 & \text{if } \epsilon > \epsilon_F \end{cases}$$ At absolute zero, the distribution is an exact Heaviside step function: all quantum states below $\epsilon_F$ are $100\%$ completely occupied, while all states above $\epsilon_F$ are strictly empty.

3. Thermal Broadening at Finite Temperature ($T > 0$)

At any finite temperature $T > 0$:

• At $\epsilon = \mu$, $f(\mu) = \frac{1}{e^0 + 1} = \frac{1}{2}$ regardless of temperature! The chemical potential is always the exact energy level where the occupation probability is $50\%$.
• Thermal excitation only affects states within a narrow energy window of width $\sim 2 k_B T$ to $4 k_B T$ around $\mu$.
• For $\epsilon - \mu \gg k_B T$, $f(\epsilon) \approx e^{-(\epsilon - \mu)/k_B T}$, decaying into the classical Maxwell-Boltzmann tail.

We now consider an ideal gas of $N$ non-interacting spin-1/2 fermions (electrons, neutrons) confined in a container of volume $V$.

1. Quantum Density of States for Spin-1/2 Particles

In 3D reciprocal wavevector space ($k$-space), the volume occupied by one spatial orbital with periodic boundary conditions is $(2\pi/L)^3 = 8\pi^3 / V$. Taking into account electron spin degeneracy $g_s = 2s + 1 = 2$ (spin-up and spin-down): $$g(k) \, dk = 2 \times \frac{V}{(2\pi)^3} \, 4\pi k^2 dk = \frac{V}{\pi^2} k^2 dk$$ For non-relativistic fermions, energy is $\epsilon = \frac{\hbar^2 k^2}{2m} \implies k = \frac{\sqrt{2m\epsilon}}{\hbar}$, and $dk = \frac{1}{2\hbar}\sqrt{\frac{2m}{\epsilon}} d\epsilon$: $$g(\epsilon) \, d\epsilon = \frac{V}{\pi^2} \left(\frac{2m\epsilon}{\hbar^2}\right) \frac{1}{2\hbar}\sqrt{\frac{2m}{\epsilon}} d\epsilon = \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \epsilon^{1/2} d\epsilon$$ The density of states grows as the square root of energy: $g(\epsilon) \propto \epsilon^{1/2}$.

2. Normalization Conditions

The total number of particles $N$ and total internal energy $U$ are given by: $$N = \int_0^\infty g(\epsilon) f(\epsilon) \, d\epsilon = \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \int_0^\infty \frac{\epsilon^{1/2} d\epsilon}{e^{(\epsilon - \mu)/k_B T} + 1}$$ $$U = \int_0^\infty \epsilon \, g(\epsilon) f(\epsilon) \, d\epsilon = \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \int_0^\infty \frac{\epsilon^{3/2} d\epsilon}{e^{(\epsilon - \mu)/k_B T} + 1}$$

At absolute zero, fermions pack into the lowest available quantum states up to a sharp energy cutoff termed the **Fermi Energy** $\epsilon_F$.

1. Derivation of the Fermi Energy

Setting $T = 0\text{ K}$, where $f(\epsilon) = 1$ for $\epsilon \le \epsilon_F$ and $0$ for $\epsilon > \epsilon_F$: $$N = \int_0^{\epsilon_F} g(\epsilon) \, d\epsilon = \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \int_0^{\epsilon_F} \epsilon^{1/2} d\epsilon = \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \frac{2}{3} \epsilon_F^{3/2}$$ $$N = \frac{V}{3\pi^2} \left( \frac{2m \epsilon_F}{\hbar^2} \right)^{3/2}$$ Solving explicitly for $\epsilon_F$ in terms of particle number density $n = N/V$: $$\epsilon_F = \frac{\hbar^2}{2m} (3\pi^2 n)^{2/3}$$

2. Fermi Wavevector and Fermi Momentum

In $k$-space, occupied states fill a sphere of radius $k_F$ called the **Fermi Sphere**: $$k_F = (3\pi^2 n)^{1/3}$$ The corresponding **Fermi Momentum** is: $$p_F = \hbar k_F = \hbar (3\pi^2 n)^{1/3}$$ The boundary in momentum space separating occupied from unoccupied states at $T = 0\text{ K}$ is the **Fermi Surface**.

3. Ground-State Total Energy and Zero-Point Pressure

The total kinetic energy of the Fermi gas at $T = 0\text{ K}$ is: $$U_0 = \int_0^{\epsilon_F} \epsilon \, g(\epsilon) \, d\epsilon = \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \frac{2}{5} \epsilon_F^{5/2} = \frac{3}{5} N \epsilon_F$$ The average energy per particle at absolute zero is $\langle \epsilon \rangle = \frac{3}{5}\epsilon_F > 0$. Even at absolute zero, fermions possess substantial kinetic energy due to quantum confinement and Pauli exclusion, exerting an enormous **Zero-Point Degeneracy Pressure**: $$P_0 = -\frac{\partial U_0}{\partial V} = -\frac{3}{5} N \frac{\partial \epsilon_F}{\partial V} = \frac{2}{5} \frac{N}{V} \epsilon_F = \frac{2}{3} \frac{U_0}{V} = \frac{2}{5} n \epsilon_F$$

To understand why quantum effects dominate electron behavior at room temperature, we introduce the **Fermi Temperature**.

1. Definition of Fermi Temperature ($T_F$)

The Fermi temperature is defined as: $$T_F \equiv \frac{\epsilon_F}{k_B} = \frac{\hbar^2}{2m k_B} (3\pi^2 n)^{2/3}$$

2. Characteristic Numerical Values for Real Metals

Consider metallic Copper (Cu):

• Electron density: $n \approx 8.49 \times 10^{28}\text{ electrons/m}^3$
• Fermi energy: $\epsilon_F = \frac{(1.055 \times 10^{-34})^2}{2(9.109 \times 10^{-31})} [3\pi^2 (8.49 \times 10^{28})]^{2/3} = 1.125 \times 10^{-18}\text{ J} \approx 7.03\text{ eV}$
• Fermi temperature: $T_F = \frac{1.125 \times 10^{-18}\text{ J}}{1.381 \times 10^{-23}\text{ J/K}} \approx 81,600\text{ K}$

3. The Extreme Quantum Degeneracy of Metals

Because $T_F \approx 80,000\text{ K}$ is vastly higher than room temperature ($T = 300\text{ K}$), the ratio is: $$\frac{T}{T_F} \approx \frac{300}{80,000} \approx 0.0037 \ll 1$$ Even glowing white-hot molten steel ($T \sim 1800\text{ K}$) has $T/T_F \sim 0.02 \ll 1$. Conduction electrons in metals are **permanently and deeply in their quantum degenerate ground state** under all terrestrial conditions!

Because electrons are packed into states up to $\epsilon_F$, electrons at the Fermi surface move with tremendous speeds.

1. The Fermi Velocity ($v_F$)

The speed of an electron residing on the Fermi surface is: $$v_F = \frac{p_F}{m} = \frac{\hbar k_F}{m} = \sqrt{\frac{2\epsilon_F}{m}}$$ For Copper ($\epsilon_F = 7.03\text{ eV}$): $$v_F = \sqrt{\frac{2 \times (7.03 \times 1.602 \times 10^{-19}\text{ J})}{9.109 \times 10^{-31}\text{ kg}}} = 1.57 \times 10^6\text{ m/s}$$ This is approximately $0.5\%$ of the speed of light ($c$)! Even at absolute zero, electrons zip through the atomic crystal lattice at over $1,500\text{ km/second}$.

2. Mean Velocity of Free Electrons

The mean speed $\langle v \rangle$ of electrons throughout the Fermi sphere at $T = 0\text{ K}$ is: $$\langle v \rangle = \frac{1}{N} \int_0^{\epsilon_F} \sqrt{\frac{2\epsilon}{m}} \, g(\epsilon) \, d\epsilon = \frac{1}{N} \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt{\frac{2}{m}} \int_0^{\epsilon_F} \epsilon \, d\epsilon = \frac{3}{4} v_F$$ For Copper, $\langle v \rangle = 0.75 \times 1.57 \times 10^6\text{ m/s} = 1.18 \times 10^6\text{ m/s}$.

To evaluate thermodynamic quantities at temperatures $T \ll T_F$, we use the **Sommerfeld Expansion**.

1. The Sommerfeld Lemma

For any smooth function $H(\epsilon)$ that vanishes at $\epsilon = 0$: $$\int_0^\infty H(\epsilon) f(\epsilon) \, d\epsilon = \int_0^\mu H(\epsilon) \, d\epsilon + \frac{\pi^2}{6}(k_B T)^2 H'(\mu) + \frac{7\pi^4}{360}(k_B T)^4 H'''(\mu) + \dots$$

2. Temperature Dependence of Chemical Potential $\mu(T)$

Applying the Sommerfeld expansion to the particle number $N = \int_0^\infty g(\epsilon) f(\epsilon) d\epsilon$: $$N = \int_0^\mu g(\epsilon) d\epsilon + \frac{\pi^2}{6}(k_B T)^2 g'(\mu)$$ Since $N = \int_0^{\epsilon_F} g(\epsilon) d\epsilon$, and $g(\epsilon) \propto \epsilon^{1/2} \implies g'(\mu) = \frac{1}{2\mu} g(\mu)$: $$\mu(T) \approx \epsilon_F \left[ 1 - \frac{\pi^2}{12} \left( \frac{T}{T_F} \right)^2 \right]$$ The chemical potential shifts downward slightly with increasing temperature.

3. Electronic Heat Capacity of Metals ($C_V^{\text{el}}$)

Expanding total energy $U(T)$: $$U(T) \approx U_0 + \frac{\pi^2}{6} g(\epsilon_F) (k_B T)^2 = \frac{3}{5}N\epsilon_F + \frac{\pi^2}{4} N k_B \frac{T^2}{T_F}$$ Differentiating with respect to temperature gives the celebrated **Sommerfeld Linear Heat Capacity**: $$C_V^{\text{el}} = \frac{\partial U}{\partial T} = \frac{\pi^2}{2} N k_B \left( \frac{T}{T_F} \right) = \gamma T$$ where $\gamma = \frac{\pi^2}{2} \frac{N k_B}{T_F}$ is the **Sommerfeld constant**.

4. Resolution of the Classical Heat Capacity Catastrophe

Classical physics predicted that conduction electrons should contribute $C_V = \frac{3}{2} N k_B$, which was contradicted by experiments showing total heat capacity in metals at room temperature was dominated by phonons ($3R$). Quantum statistics explains this: only a tiny fraction of electrons—those within $k_B T$ of the Fermi surface (a fraction $\sim T / T_F \approx 0.004$)—can be thermally excited. The remaining $99.6\%$ of electrons are locked in lower states by the Pauli exclusion principle and cannot absorb thermal energy!

When an external magnetic field $\mathbf{B} = B\hat{\mathbf{z}}$ is applied to a free electron gas, classical mechanics (the Bohr-van Leeuwen theorem) asserts that thermal equilibrium orbital magnetism is identically zero. Lev Landau (1930) showed that quantum mechanics produces a purely orbital diamagnetic response.

1. Quantized Landau Energy Levels

In a uniform magnetic field $B$, classical cyclotron orbits are quantized into discrete **Landau levels**: $$\epsilon(n, p_z) = \left( n + \frac{1}{2} \right) \hbar \omega_c + \frac{p_z^2}{2m}$$ where $\omega_c = \frac{e B}{m}$ is the **cyclotron frequency**, and $n = 0, 1, 2, \dots$. Each Landau level has an enormous macroscopic degeneracy per unit area: $$g_L = \frac{e B}{h} = \frac{1}{2\pi \ell_B^2}$$ where $\ell_B = \sqrt{\hbar / eB}$ is the magnetic length.

2. Landau Diamagnetic Susceptibility

Summing over the discrete Landau levels in the grand potential and expanding for weak fields ($k_B T \gg \hbar \omega_c$ or $\epsilon_F \gg \hbar \omega_c$): $$\Phi_G(B) \approx \Phi_G(0) + \frac{V}{6} \mu_B^2 \left( \frac{\partial n}{\partial \mu} \right) B^2$$ The resulting magnetization $M = -\frac{1}{V}\frac{\partial \Phi_G}{\partial B}$ yields the **Landau Diamagnetic Susceptibility**: $$\chi_{\text{Landau}} = -\frac{1}{3} \mu_B^2 g(\epsilon_F) = -\frac{1}{3} \chi_{\text{Pauli}}$$ Quantized orbital motion creates an opposing diamagnetic moment exactly one-third the magnitude of spin paramagnetism.

Wolfgang Pauli (1927) explained why the conduction electrons in metals exhibit a small, temperature-independent paramagnetic susceptibility.

1. Zeeman Splitting in the Conduction Band

Each electron possesses an intrinsic magnetic dipole moment $\boldsymbol{\mu} = -g \mu_B \mathbf{s} \approx -2 \mu_B \mathbf{s}$, where $\mu_B = \frac{e\hbar}{2m} = 9.274 \times 10^{-24}\text{ J/T}$ is the Bohr magneton. In an external field $B$, Zeeman interaction splits the energy levels: $$\epsilon_\uparrow = \epsilon - \mu_B B \quad (\text{spin parallel to } B), \quad \epsilon_\downarrow = \epsilon + \mu_B B \quad (\text{spin antiparallel})$$

2. Fermi Surface Asymmetry

In equilibrium, both spin sub-bands must fill to the identical chemical potential $\epsilon_F$. Consequently, spin-up states expand while spin-down states contract: $$N_\uparrow = \frac{1}{2} \int_0^{\epsilon_F + \mu_B B} g(\epsilon) d\epsilon \approx \frac{N}{2} + \frac{1}{2} g(\epsilon_F) \mu_B B$$ $$N_\downarrow = \frac{1}{2} \int_0^{\epsilon_F - \mu_B B} g(\epsilon) d\epsilon \approx \frac{N}{2} - \frac{1}{2} g(\epsilon_F) \mu_B B$$ The net excess of parallel spins is: $$\Delta N = N_\uparrow - N_\downarrow = g(\epsilon_F) \mu_B B$$

3. The Pauli Susceptibility Formula

The macroscopic magnetic moment is $M = \mu_B \Delta N = \mu_B^2 g(\epsilon_F) B$. The magnetic susceptibility per unit volume is: $$\chi_{\text{Pauli}} = \frac{\mu_0 M}{B} = \mu_0 \mu_B^2 g(\epsilon_F) = \frac{3 n \mu_0 \mu_B^2}{2 \epsilon_F}$$ Notice that:

1. $\chi_{\text{Pauli}}$ is **strictly independent of temperature** to leading order (unlike classical Curie paramagnetism $\chi \propto 1/T$).
2. Combining Pauli paramagnetism with Landau diamagnetism gives the net response: $$\chi_{\text{total}} = \chi_{\text{Pauli}} + \chi_{\text{Landau}} = \left(1 - \frac{1}{3}\right) \chi_{\text{Pauli}} = \frac{2}{3} \chi_{\text{Pauli}} > 0$$ The electron gas in normal metals remains weakly paramagnetic overall.

Thermionic emission is the thermally induced flow of charge carriers over a surface potential barrier (used in vacuum tubes, electron microscopes, and cathode ray tubes).

1. The Escape Condition and Work Function

Inside a metal, electrons occupy states up to the Fermi energy $\epsilon_F$. The minimum energy required to liberate an electron from the Fermi level into vacuum at rest is the **Work Function** $\Phi$ (typically $4.5\text{ eV}$ for tungsten). An electron at surface $z = 0$ can escape into vacuum only if its kinetic energy perpendicular to the surface exceeds the barrier: $$\frac{p_z^2}{2m} \ge \epsilon_F + \Phi$$

2. Derivation of the Emission Current Density

The emission current density $J$ is obtained by integrating the flux $v_z = p_z / m$ over all escape states: $$J = 2 e \left(\frac{1}{h^3}\right) \int_{-\infty}^\infty dp_x \int_{-\infty}^\infty dp_y \int_{p_{z,\min}}^\infty dp_z \, \frac{p_z}{m} \frac{1}{e^{(\epsilon - \mu)/k_B T} + 1}$$ Because $\epsilon - \mu \ge \Phi \gg k_B T$, the Fermi-Dirac distribution reduces to the Maxwell-Boltzmann approximation $e^{-(\epsilon - \epsilon_F)/k_B T}$: $$J = \frac{2 e}{m h^3} e^{\epsilon_F / k_B T} \left( \int_{-\infty}^\infty e^{-p_x^2 / 2m k_B T} dp_x \right)^2 \int_{p_{z,\min}}^\infty p_z e^{-p_z^2 / 2m k_B T} dp_z$$ The transverse integrals yield $(2\pi m k_B T)$. Evaluating the $p_z$ integral: $$\int_{p_{z,\min}}^\infty p_z e^{-p_z^2 / 2m k_B T} dp_z = m k_B T \exp\left( -\frac{\epsilon_F + \Phi}{k_B T} \right)$$

3. The Richardson-Dushman Equation

Combining terms gives the famous **Richardson-Dushman Law**: $$J = A T^2 \exp\left( -\frac{\Phi}{k_B T} \right)$$ where $A$ is the universal **Richardson constant**: $$A = \frac{4\pi m e k_B^2}{h^3} = 1.20173 \times 10^6\text{ A/(m}^2\cdot\text{K}^2)$$

When an intermediate-mass star (such as the Sun) exhausts its nuclear fuel, it collapses under self-gravity until halted by the **electron degeneracy pressure** of its completely degenerate electron gas, forming a **White Dwarf star**.

1. Non-Relativistic Equilibrium

In a star of mass $M$ and radius $R$, the gravitational potential energy is: $$U_{\text{grav}} = -\frac{3}{5} \frac{G M^2}{R}$$ For non-relativistic degenerate electrons ($p_F \ll m_e c$), the kinetic energy scales as: $$U_{\text{kin}} = \frac{3}{5} N \epsilon_F \propto N \frac{\hbar^2}{m_e} \left(\frac{N}{R^3}\right)^{2/3} \propto \frac{\hbar^2 N^{5/3}}{m_e R^2}$$ Minimizing total energy $E = U_{\text{kin}} + U_{\text{grav}}$ with respect to $R$: $$\frac{dE}{dR} = -\frac{2 C_1}{R^3} + \frac{C_2 G M^2}{R^2} = 0 \implies R \propto M^{-1/3}$$ Remarkably, a heavier white dwarf is physically smaller!

2. Relativistic Core Collapse and the Chandrasekhar Limit

As stellar mass increases, the star contracts and core density soars, driving the Fermi momentum into the ultra-relativistic regime: $$p_F = \hbar (3\pi^2 n)^{1/3} \gg m_e c$$ For ultra-relativistic electrons, energy is linear in momentum: $\epsilon = p c$. The kinetic degeneracy energy becomes: $$U_{\text{kin}}^{\text{rel}} \propto N p_F c \propto \hbar c \frac{N^{4/3}}{R}$$ Notice that both gravitational and relativistic kinetic energy scale as $1/R$: $$E_{\text{total}} = \left( A \hbar c N^{4/3} - B G M^2 \right) \frac{1}{R}$$ If gravity exceeds degeneracy pressure ($B G M^2 > A \hbar c N^{4/3}$), no stable equilibrium radius exists: the star collapses indefinitely! Equating the two terms yields the **Chandrasekhar Mass Limit** (Subrahmanyan Chandrasekhar, 1930): $$M_{\text{Ch}} = \frac{\omega_3^0}{4\pi} \left( \frac{h c}{G} \right)^{3/2} \left( \frac{1}{\mu_e m_p} \right)^2 \approx 1.44 M_\odot$$ Any stellar core exceeding $1.44$ solar masses cannot be supported by electron degeneracy and must collapse into a neutron star or black hole.