Physics / Solid State Physics II Quantum Theory of Solids 100% Free Open Access
Chapter 3 • Theory & Derivations

Microscopic Magnetism: Diamagnetism, Paramagnetism & Quantum Exchange

This unit establishes the quantum mechanical foundation of magnetism in solids from the isolated atom to macroscopic solids. We derive Larmor diamagnetism and contrast classical Langevin paramagnetism with the quantum Brillouin theory governed by Hund's rules and crystal-field split multiplets. We examine Van Vleck temperature-independent paramagnetism alongside Pauli spin paramagnetism and Landau diamagnetism in Fermi liquids. Finally, we resolve the historical Bohr-van Leeuwen theorem by showing that magnetic ordering originates from electrostatic Coulomb repulsion coupled with Pauli exclusion, formulating the Heisenberg exchange Hamiltonian, Anderson superexchange, and oscillatory RKKY interactions.

§3.1 Classical & Quantum Theory of Diamagnetism: Larmor Precession & Langevin Susceptibility

1. Classical Larmor Precession

Consider an electron with mass $m$ and charge $-e$ in a central atomic potential $V(r)$. When an external magnetic field $\vec{B} = B\hat{z}$ is switched on, the electron experiences the Lorentz force $-e(\vec{v} \times \vec{B})$.

Transforming into a coordinate frame rotating with angular velocity $\vec{\omega}_L$:

$$\vec{\omega}_L = \frac{e\vec{B}}{2m}$$

The Coriolis force $-2m(\vec{\omega}_L \times \vec{v}')$ exactly cancels the Lorentz force to first order in $B$. This precession of the electronic orbits at the Larmor frequency $\omega_L$ constitutes an effective circulating electric current:

$$I = -\frac{Z e}{2\pi} \omega_L = -\frac{Z e^2 B}{4\pi m}$$

The induced magnetic dipole moment opposing the applied field is:

$$\mu_{\text{ind}} = I \cdot \langle A \rangle = -\frac{Z e^2 B}{4\pi m} \pi \langle \rho^2 \rangle = -\frac{Z e^2 B}{4m} \langle x^2 + y^2 \rangle$$

For spherically symmetric atomic charge distributions, $\langle x^2 \rangle = \langle y^2 \rangle = \langle z^2 \rangle = \frac{1}{3}\langle r^2 \rangle$, so $\langle x^2 + y^2 \rangle = \frac{2}{3}\langle r^2 \rangle$. The induced magnetic dipole moment per atom is:

$$\vec{\mu}_{\text{ind}} = -\frac{Z e^2}{6m} \langle r^2 \rangle \vec{B}$$

2. Langevin Diamagnetic Susceptibility

In a macroscopic sample with atomic number density $N$, the magnetization $\vec{M} = N \vec{\mu}_{\text{ind}}$ yields the Langevin diamagnetic susceptibility:

$$\chi_{\text{dia}} = \frac{\mu_0 M}{B} = -\frac{\mu_0 N Z e^2}{6m} \langle r^2 \rangle$$

Quantum mechanically, this result is derived from first-order perturbation theory using the diamagnetic term $\frac{e^2}{8m} \sum_i (\vec{B} \times \vec{r}_i)^2$ in the minimal coupling Hamiltonian $(\vec{p} + e\vec{A})^2/2m$. Diamagnetism is fundamentally temperature-independent and negative, characteristic of all closed-shell inert gas atoms (He, Ne, Ar) and noble metal core ions.

§3.2 Classical Theory of Paramagnetism: The Langevin Function & Curie's Law

1. Statistical Mechanics of Classical Magnetic Dipoles

Atoms or molecules with unfilled shells possess a permanent microscopic magnetic dipole moment $\vec{\mu}$. In an external magnetic field $\vec{B} = B\hat{z}$, the classical potential energy is:

$$U(\theta) = -\vec{\mu} \cdot \vec{B} = -\mu B \cos\theta$$

According to Maxwell-Boltzmann statistics, the probability of finding a dipole oriented at polar angle $\theta$ within solid angle $d\Omega = 2\pi\sin\theta d\theta$ is:

$$P(\theta) d\theta = \frac{e^{\mu B \cos\theta / k_B T} \sin\theta d\theta}{\int_0^\pi e^{\mu B \cos\theta / k_B T} \sin\theta d\theta}$$

Defining the dimensionless parameter $x \equiv \frac{\mu B}{k_B T}$:

$$\langle \cos\theta \rangle = \frac{\int_{-1}^1 u e^{x u} du}{\int_{-1}^1 e^{x u} du} = \frac{d}{dx} \ln\left( \frac{e^x - e^{-x}}{x} \right) = \coth x - \frac{1}{x} \equiv L(x)$$

where $L(x)$ is the Langevin function.

2. The Langevin Function & Curie's Law

The total macroscopic magnetization for $N$ dipoles per unit volume is:

$$M(B, T) = N \mu L(x) = N \mu \left( \coth\left(\frac{\mu B}{k_B T}\right) - \frac{k_B T}{\mu B} \right)$$
  • Weak Field Limit ($x \ll 1$, high temperature): Expanding $\coth x \approx \frac{1}{x} + \frac{x}{3} - \frac{x^3}{45} + \dots$: $$L(x) \approx \frac{x}{3} = \frac{\mu B}{3 k_B T}$$ $$M \approx \frac{N \mu^2 B}{3 k_B T} \implies \chi_{\text{para}} = \frac{\mu_0 M}{B} = \frac{\mu_0 N \mu^2}{3 k_B T} = \frac{C}{T}$$ This is the classical Curie's Law, where $C = \frac{\mu_0 N \mu^2}{3 k_B}$ is the Curie constant.
  • Strong Field Limit ($x \gg 1$, low temperature): $\coth x \to 1$, so $L(x) \to 1 - \frac{1}{x} \to 1$. All dipoles align perfectly parallel to the field, reaching the saturation magnetization: $$M_{\text{sat}} = N \mu$$

§3.3 Quantum Theory of Paramagnetism: Hund's Rules, Landé $g$-Factor & The Brillouin Function

1. Atomic Multi-Electron States & Hund's Rules

In open $d$- or $f$-electron shells, electron-electron Coulomb repulsion and spin-orbit coupling determine the total orbital angular momentum $\vec{L} = \sum \vec{l}_i$, total spin $\vec{S} = \sum \vec{s}_i$, and total angular momentum $\vec{J} = \vec{L} + \vec{S}$. According to Hund's Empirical Rules, the ground state multiplet maximizes:

  1. $S$: Total spin is maximized to reduce Coulomb repulsion via exchange holes.
  2. $L$: Total orbital angular momentum is maximized subject to the maximum $S$.
  3. $J$: $J = |L - S|$ for shells less than half full; $J = L + S$ for shells more than half full.

The total magnetic dipole operator is $\hat{\vec{\mu}} = -\mu_B (\hat{\vec{L}} + 2\hat{\vec{S}}) = -g_J \mu_B \hat{\vec{J}}$, where $g_J$ is the Landé $g$-factor:

$$g_J = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)}$$

2. The Quantum Brillouin Function

In an external field $B\hat{z}$, spatial quantization restricts the magnetic quantum number $m_J$ to discrete values:

$$m_J = -J, -J+1, \dots, +J \quad (2J+1 \text{ states})$$ $$E_{m_J} = -g_J \mu_B B m_J$$

The canonical partition function is a finite geometric series:

$$Z = \sum_{m_J = -J}^J \exp\left( \frac{g_J \mu_B B m_J}{k_B T} \right) = \frac{\sinh\left( \frac{2J+1}{2J} x \right)}{\sinh\left( \frac{1}{2J} x \right)}, \quad x \equiv \frac{g_J \mu_B J B}{k_B T}$$

The thermal average magnetization is $M = N k_B T \frac{\partial \ln Z}{\partial B} = N g_J \mu_B J \mathcal{B}_J(x)$, where $\mathcal{B}_J(x)$ is the Brillouin function:

$$\mathcal{B}_J(x) \equiv \frac{2J+1}{2J} \coth\left( \frac{2J+1}{2J} x \right) - \frac{1}{2J} \coth\left( \frac{1}{2J} x \right)$$
  • For $x \ll 1$, expanding $\coth u \approx \frac{1}{u} + \frac{u}{3}$ gives: $$\mathcal{B}_J(x) \approx \frac{J+1}{3J} x \implies M \approx \frac{N g_J^2 \mu_B^2 J(J+1) B}{3 k_B T}$$ $$\chi_{\text{quantum}} = \frac{\mu_0 N p_{\text{eff}}^2 \mu_B^2}{3 k_B T}, \quad p_{\text{eff}} \equiv g_J \sqrt{J(J+1)}$$
  • In the classical limit where $J \to \infty$ while $g_J \mu_B J \equiv \mu$ remains constant: $$\lim_{J \to \infty} \mathcal{B}_J(x) = L(x) \quad (\text{Exact Langevin function!})$$

§3.4 Van Vleck Paramagnetism & Conduction Electron Magnetism (Pauli Spin & Landau Orbital)

1. Van Vleck Temperature-Independent Paramagnetism

When an atom or ion has a non-magnetic ground state ($J = 0$, such as $\text{Eu}^{3+}$ or closed subshells) with low-lying excited states $|n\rangle$ separated by energy $\Delta_n = E_n - E_0$, second-order perturbation theory yields an energy shift quadratic in $B$:

$$\Delta E_0^{(2)} = -\sum_{n \ne 0} \frac{|\langle 0 | \mu_B (\hat{\vec{L}} + 2\hat{\vec{S}}) \cdot \vec{B} | n \rangle|^2}{E_n - E_0}$$

This gives rise to a positive, strictly temperature-independent paramagnetic susceptibility known as Van Vleck paramagnetism:

$$\chi_{\text{VV}} = 2\mu_0 N \mu_B^2 \sum_{n \ne 0} \frac{|\langle 0 | \hat{L}_z + 2\hat{S}_z | n \rangle|^2}{E_n - E_0} > 0$$

2. Pauli Spin Paramagnetism & Landau Diamagnetism in Metals

In simple metals, the conduction electrons form a degenerate Fermi sea ($T \ll T_F \sim 50{,}000\text{ K}$). An external magnetic field $B$ shifts the energy of spin-up electrons down by $\mu_B B$ and spin-down electrons up by $\mu_B B$:

$$E_{\uparrow}(k) = E(k) - \mu_B B, \quad E_{\downarrow}(k) = E(k) + \mu_B B$$

Because only electrons within a thermal energy slice $\sim k_B T$ around the Fermi surface can flip their spins into unoccupied states:

$$\delta n = \frac{1}{2} g(E_F) \mu_B B - \left(-\frac{1}{2} g(E_F) \mu_B B\right) = g(E_F) \mu_B B$$

The resulting magnetic moment per unit volume is $M = \mu_B \delta n = \mu_B^2 g(E_F) B$, yielding the Pauli spin susceptibility:

$$\chi_{\text{Pauli}} = \mu_0 \mu_B^2 g(E_F) = \frac{3 \mu_0 n \mu_B^2}{2 E_F}$$

Unlike Curie paramagnetism, $\chi_{\text{Pauli}}$ is virtually temperature-independent!

Simultaneously, the orbital motion of conduction electrons is quantized into discrete Landau levels $E_n = (n + 1/2)\hbar\omega_c + \frac{\hbar^2 k_z^2}{2m}$. Calculating the free energy via Poisson summation reveals Landau orbital diamagnetism:

$$\chi_{\text{Landau}} = -\frac{1}{3} \left( \frac{m}{m^*} \right)^2 \chi_{\text{Pauli}}$$

For free electrons ($m^* = m$), the total electronic susceptibility is $\chi_{\text{total}} = \chi_{\text{Pauli}} + \chi_{\text{Landau}} = +\frac{2}{3}\chi_{\text{Pauli}} > 0$.

§3.5 The Microscopic Origin of Magnetic Ordering: Coulomb Interaction & Pauli Principle

1. Failure of Dipolar Interaction & The Bohr-van Leeuwen Theorem

The classical Bohr-van Leeuwen theorem states that at thermal equilibrium, the partition function of any classical charged system with Hamiltonian $\mathcal{H}(\vec{r}_i, \vec{p}_i + e\vec{A})$ is completely independent of the vector potential $\vec{A}$ because the momentum integral over $(-\infty, \infty)$ simply shifts by $-e\vec{A}$. Therefore:

$$\vec{M} = -\frac{\partial F}{\partial \vec{B}} = 0 \quad (\text{Classical magnetism is impossible!})$$

Furthermore, direct magnetic dipole-dipole interactions between neighboring atomic moments separated by $a \approx 2.5\text{ \AA}$ have an energy scale of:

$$E_{\text{dipole}} \sim \frac{\mu_0 \mu_B^2}{4\pi a^3} \approx 10^{-23}\text{ J} \sim 10^{-4}\text{ eV} \implies T_C \approx \frac{E_{\text{dipole}}}{k_B} \sim 1\text{ K}$$

Yet iron ($\text{Fe}$) is ferromagnetic up to $T_C = 1043\text{ K}$ ($k_B T_C \approx 0.1\text{ eV}$)! Dipolar forces are over three orders of magnitude too weak to explain ferromagnetism.

2. Quantum Origin: Coulomb Repulsion + Antisymmetry

According to the Pauli exclusion principle, the total electronic wavefunction for a two-electron system must be antisymmetric under particle exchange:

$$\Psi(1, 2) = \psi_{\text{spatial}}(\vec{r}_1, \vec{r}_2) \chi_{\text{spin}}(s_1, s_2) = -\Psi(2, 1)$$
  • Singlet Spin State ($S = 0$, antiparallel spins): Spin function is antisymmetric, requiring a symmetric spatial wavefunction: $$\psi_S(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}} [\phi_a(\vec{r}_1)\phi_b(\vec{r}_2) + \phi_b(\vec{r}_1)\phi_a(\vec{r}_2)]$$ Here $\psi_S(\vec{r}, \vec{r}) \ne 0$: electrons can be close together, increasing their mutual electrostatic Coulomb repulsion energy.
  • Triplet Spin State ($S = 1$, parallel spins): Spin function is symmetric, requiring an antisymmetric spatial wavefunction: $$\psi_A(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}} [\phi_a(\vec{r}_1)\phi_b(\vec{r}_2) - \phi_b(\vec{r}_1)\phi_a(\vec{r}_2)]$$ Here $\psi_A(\vec{r}, \vec{r}) = 0$: the spatial Pauli exchange hole keeps the two electrons apart, drastically lowering their electrostatic repulsion energy!

The energy difference between the singlet and triplet configurations is:

$$E_S - E_T = 2 J_{\text{ex}}$$

where $J_{\text{ex}} = \iint \phi_a^*(\vec{r}_1)\phi_b^*(\vec{r}_2) \frac{e^2}{4\pi\epsilon_0 |\vec{r}_1 - \vec{r}_2|} \phi_b(\vec{r}_1)\phi_a(\vec{r}_2) d^3r_1 d^3r_2$ is the exchange integral. Magnetic ordering is a purely electrostatic Coulomb phenomenon disguised as a magnetic interaction!

§3.6 The Heisenberg Exchange Hamiltonian & Direct vs Indirect Exchange Mechanisms

1. The Heisenberg Exchange Hamiltonian

For two electrons with spins $\vec{S}_1$ and $\vec{S}_2$ (in units of $\hbar$), the total spin operator is $\vec{S}_{\text{tot}} = \vec{S}_1 + \vec{S}_2$:

$$\vec{S}_{\text{tot}}^2 = \vec{S}_1^2 + \vec{S}_2^2 + 2\vec{S}_1 \cdot \vec{S}_2 = \frac{3}{4} + \frac{3}{4} + 2\vec{S}_1 \cdot \vec{S}_2 = \frac{3}{2} + 2\vec{S}_1 \cdot \vec{S}_2$$

Evaluating the eigenvalues:

$$\vec{S}_1 \cdot \vec{S}_2 = \begin{cases} -\frac{3}{4}, & S = 0 \text{ (Singlet)} \\ +\frac{1}{4}, & S = 1 \text{ (Triplet)} \end{cases}$$

We can write the effective spin-dependent Hamiltonian as:

$$\hat{\mathcal{H}} = \frac{1}{4}(E_S + 3E_T) - (E_S - E_T) \vec{S}_1 \cdot \vec{S}_2 = \text{const} - 2 J_{\text{ex}} \vec{S}_1 \cdot \vec{S}_2$$

Generalizing to a macroscopic lattice of localized spins yields the famous Heisenberg Hamiltonian:

$$\hat{\mathcal{H}}_{\text{Heisenberg}} = -2 \sum_{i < j} J_{ij} \vec{S}_i \cdot \vec{S}_j$$
  • If $J_{ij} > 0$: Parallel spin alignment minimizes the energy $\implies$ Ferromagnetism.
  • If $J_{ij} < 0$: Antiparallel spin alignment minimizes the energy $\implies$ Antiferromagnetism.

2. Direct Exchange & The Bethe-Slater Curve

Direct exchange occurs via the direct spatial overlap of the magnetic orbitals of adjacent atoms. The Bethe-Slater curve plots $J$ as a function of the ratio of interatomic distance $D$ to the radius of the unfilled $3d$ shell $d$:

  • For $D/d < 1.5$ (e.g. Cr, Mn), wavefunctions overlap strongly, kinetic energy penalizes parallel spins, and $J < 0$ (antiferromagnetic).
  • For $D/d > 1.5$ (e.g. Fe, Co, Ni), the overlap is modest, Coulomb exchange dominates, and $J > 0$ (ferromagnetic).

§3.7 Superexchange in Oxides & RKKY Oscillatory Coupling in Metals

1. Anderson Superexchange in Transition Metal Oxides

In insulating transition metal oxides (such as $\text{MnO}$, $\text{NiO}$, $\text{Fe}_2\text{O}_3$), magnetic transition metal cations ($\text{Mn}^{2+}$) are separated by non-magnetic oxygen anions ($\text{O}^{2-}$). Direct overlap between $3d$ orbitals is essentially zero.

Instead, magnetic coupling is mediated by superexchange: virtual hopping of electrons through the filled intermediate oxygen $2p$ orbital.

According to the Goodenough-Kanamori-Anderson rules:

  • For a $180^\circ$ cation-anion-cation bond ($\text{Mn}^{2+}-\text{O}^{2-}-\text{Mn}^{2+}$), an electron from the oxygen $2p$ orbital hops into an empty or half-filled $d$-orbital on one cation. By Hund's rule and Pauli exclusion, the remaining oxygen electron must have opposite spin and hops to the opposite cation. This mediates an overwhelmingly strong antiferromagnetic coupling ($J < 0$).
  • For a $90^\circ$ bond angle, orthogonal oxygen $p_x$ and $p_y$ orbitals mediate a weak ferromagnetic coupling ($J > 0$).

2. The RKKY Oscillatory Interaction in Metals

In metallic alloys containing localized magnetic moments (such as rare-earth $4f$ ions in Gd, or dilute magnetic semiconductors), the direct overlap between deeply buried $4f$ shells is negligible.

Instead, coupling is mediated by the RKKY (Ruderman-Kittel-Kasuya-Yosida) interaction via the conduction electron gas. A localized spin $\vec{S}_i$ at the origin polarizes the conduction electron spins, creating a decaying, oscillatory spin density wave:

$$\delta s(r) \propto \frac{\sin(2k_F r) - 2k_F r \cos(2k_F r)}{(2k_F r)^4}$$

A second localized spin $\vec{S}_j$ at distance $R$ interacts with this spin ripple, producing an effective exchange constant:

$$J_{\text{RKKY}}(R) \propto J_{sf}^2 \frac{\cos(2k_F R)}{(2k_F R)^3}, \quad (R \gg 1/k_F)$$

The exchange constant oscillates in sign between ferromagnetic ($J > 0$) and antiferromagnetic ($J < 0$) as a function of distance $R$, leading to complex helical spin orders and spin glasses.

Honors Examination Worked Problems & Solutions

Rigorous step-by-step mathematical proofs and solutions to university degree examination problems.

Solved Problem Example 3.1: Quantum Brillouin Paramagnetism: Exact Derivation & Limits

A paramagnetic crystal contains $N$ independent atoms per unit volume, each characterized by total angular momentum quantum number $J$ and Landé $g$-factor $g_J$ in a uniform magnetic field $B$.\n\n(a) Evaluate the canonical partition function $Z = \sum_{m_J = -J}^J e^{m_J x / J}$, where $x = \frac{g_J \mu_B J B}{k_B T}$.\n(b) Derive the exact Brillouin function $\mathcal{B}_J(x)$ for the magnetization $M = N g_J \mu_B J \mathcal{B}_J(x)$.\n(c) Prove mathematically that in the limit $J \to \infty$, $\mathcal{B}_J(x)$ reduces precisely to the classical Langevin function $L(x) = \coth x - 1/x$, and prove that for $x \ll 1$, it yields Curie's Law $\chi = \frac{C}{T}$.

(a) Canonical Partition Function: Let $\eta \equiv x / J = \frac{g_J \mu_B B}{k_B T}$. The partition function is:

$$Z = \sum_{m_J = -J}^J e^{m_J \eta} = e^{-J\eta} \sum_{n = 0}^{2J} (e^\eta)^n$$

This is a geometric progression with first term $a = e^{-J\eta}$, common ratio $r = e^\eta$, and $2J+1$ terms:

$$Z = e^{-J\eta} \frac{1 - e^{(2J+1)\eta}}{1 - e^\eta} = \frac{e^{-(J + 1/2)\eta} - e^{(J + 1/2)\eta}}{e^{-\eta/2} - e^{\eta/2}} = \frac{\sinh\left( (J + 1/2)\eta \right)}{\sinh(\eta/2)}$$

Substituting $\eta = x / J$:

$$Z(x) = \frac{\sinh\left( \frac{2J+1}{2J} x \right)}{\sinh\left( \frac{1}{2J} x \right)}$$

(b) Magnetization & Brillouin Function: The thermal average magnetization per unit volume is:

$$M = N k_B T \frac{\partial \ln Z}{\partial B} = N k_B T \left( \frac{\partial x}{\partial B} \right) \frac{\partial \ln Z}{\partial x}$$

Since $\frac{\partial x}{\partial B} = \frac{g_J \mu_B J}{k_B T}$:

$$M = N g_J \mu_B J \frac{d}{dx} \left[ \ln\sinh\left( \frac{2J+1}{2J} x \right) - \ln\sinh\left( \frac{1}{2J} x \right) \right]$$

Evaluating the derivative:

$$\frac{d}{dx} \ln\sinh(a x) = a \coth(a x)$$
$$M = N g_J \mu_B J \left[ \frac{2J+1}{2J} \coth\left( \frac{2J+1}{2J} x \right) - \frac{1}{2J} \coth\left( \frac{1}{2J} x \right) \right] \equiv N g_J \mu_B J \mathcal{B}_J(x)$$

where $\mathcal{B}_J(x) \equiv \frac{2J+1}{2J} \coth\left( \frac{2J+1}{2J} x \right) - \frac{1}{2J} \coth\left( \frac{1}{2J} x \right)$.

(c) Asymptotic Limits: 1. Classical Limit ($J \to \infty$): Let $u = x / 2J \to 0$. Then $\frac{2J+1}{2J} x = x + u \to x$.

$$\lim_{J \to \infty} \mathcal{B}_J(x) = \coth x - \lim_{u \to 0} \left[ \frac{u}{x} \coth u \right]$$

Since $\lim_{u \to 0} u \coth u = \lim_{u \to 0} \frac{u}{\tanh u} = 1$:

$$\lim_{J \to \infty} \mathcal{B}_J(x) = \coth x - \frac{1}{x} = L(x) \quad \text{(Q.E.D.)}$$

2. Weak-Field Limit ($x \ll 1$): Expand $\coth y = \frac{1}{y} + \frac{y}{3} - \frac{y^3}{45} + \dots$ for both terms:

$$\mathcal{B}_J(x) \approx \frac{2J+1}{2J} \left[ \frac{2J}{(2J+1)x} + \frac{1}{3} \frac{2J+1}{2J} x \right] - \frac{1}{2J} \left[ \frac{2J}{x} + \frac{1}{3} \frac{1}{2J} x \right]$$
$$\mathcal{B}_J(x) = \left( \frac{1}{x} - \frac{1}{x} \right) + \frac{x}{3} \left[ \left(\frac{2J+1}{2J}\right)^2 - \left(\frac{1}{2J}\right)^2 \right] = \frac{x}{3} \left[ \frac{(2J+1)^2 - 1}{4J^2} \right] = \frac{x}{3} \left[ \frac{4J^2 + 4J}{4J^2} \right] = \frac{J+1}{3J} x$$

Substitute into $M$:

$$M = N g_J \mu_B J \left( \frac{J+1}{3J} \frac{g_J \mu_B J B}{k_B T} \right) = \frac{N g_J^2 \mu_B^2 J(J+1)}{3 k_B T} B$$

The magnetic susceptibility is:

$$\chi = \frac{\mu_0 M}{B} = \frac{\mu_0 N p_{\text{eff}}^2 \mu_B^2}{3 k_B T} = \frac{C}{T}, \quad \text{where } p_{\text{eff}} = g_J \sqrt{J(J+1)}$$

This is exact Curie's Law.

Solved Problem Example 3.2: Pauli Spin Paramagnetism & Landau Diamagnetism in a 3D Degenerate Fermi Gas

Consider a 3D degenerate free electron gas with density $n$ and Fermi energy $E_F$ in a magnetic field $B$.\n\n(a) Calculate the spin-split densities of states $g_\uparrow(E)$ and $g_\downarrow(E)$ under Zeeman energy $\pm \mu_B B$, and derive the Pauli spin susceptibility $\chi_{\text{Pauli}}$ at $T = 0\text{ K}$.\n(b) Using the Sommerfeld expansion, calculate the leading finite-temperature correction to $\chi_{\text{Pauli}}(T)$ up to order $(T/T_F)^2$.\n(c) The Landau orbital diamagnetic susceptibility is $\chi_{\text{Landau}} = -\frac{1}{3}\chi_{\text{Pauli}}$. Calculate the net magnetic susceptibility $\chi_{\text{net}}$ of metallic sodium ($n = 2.65 \times 10^{28}\text{ m}^{-3}, E_F = 3.24\text{ eV}$), and determine whether it is paramagnetically or diamagnetically dominated.

(a) Pauli Spin Susceptibility at $T = 0\text{ K}$: The total density of states per unit volume for both spins is:

$$g(E) = \frac{1}{2\pi^2} \left(\frac{2m}{\hbar^2}\right)^{3/2} \sqrt{E} = \frac{3n}{2 E_F} \sqrt{\frac{E}{E_F}}$$

In field $B$, the single-spin densities are shifted by $\pm \mu_B B$:

$$g_\uparrow(E) = \frac{1}{2} g(E + \mu_B B), \quad g_\downarrow(E) = \frac{1}{2} g(E - \mu_B B)$$

At $T = 0\text{ K}$, all states up to chemical potential $\mu \approx E_F$ are filled:

$$n_\uparrow = \int_{-\mu_B B}^{E_F} \frac{1}{2} g(E + \mu_B B) dE = \int_0^{E_F + \mu_B B} \frac{1}{2} g(\epsilon) d\epsilon \approx \frac{n}{2} + \frac{1}{2} g(E_F) \mu_B B$$
$$n_\downarrow = \int_{\mu_B B}^{E_F} \frac{1}{2} g(E - \mu_B B) dE = \int_0^{E_F - \mu_B B} \frac{1}{2} g(\epsilon) d\epsilon \approx \frac{n}{2} - \frac{1}{2} g(E_F) \mu_B B$$

The magnetization is:

$$M = \mu_B (n_\uparrow - n_\downarrow) = \mu_B [g(E_F) \mu_B B] = \mu_B^2 g(E_F) B$$

The Pauli susceptibility is:

$$\chi_{\text{Pauli}} = \frac{\mu_0 M}{B} = \mu_0 \mu_B^2 g(E_F) = \frac{3 \mu_0 n \mu_B^2}{2 E_F}$$

(b) Finite-Temperature Correction via Sommerfeld Expansion: At finite $T$:

$$M(T) = \mu_B \int_0^\infty \frac{1}{2} g(E) [f(E - \mu_B B) - f(E + \mu_B B)] dE$$

For weak fields, $f(E - \mu_B B) - f(E + \mu_B B) \approx -2\mu_B B \frac{\partial f}{\partial E}$:

$$M(T) = \mu_B^2 B \int_0^\infty g(E) \left(-\frac{\partial f}{\partial E}\right) dE$$

Using the Sommerfeld expansion for $\int_0^\infty H(E) \left(-\frac{\partial f}{\partial E}\right) dE = H(\mu) + \frac{\pi^2}{6}(k_B T)^2 H''(\mu) + \dots$: With $g(E) = C E^{1/2}$, $g''(E) = -\frac{1}{4} C E^{-3/2} = -\frac{1}{4} \frac{g(E)}{E^2}$:

$$\int_0^\infty g(E) \left(-\frac{\partial f}{\partial E}\right) dE \approx g(E_F) \left[ 1 - \frac{\pi^2}{24} \left( \frac{k_B T}{E_F} \right)^2 \right]$$

Taking into account the temperature drift of the chemical potential $\mu(T) = E_F [1 - \frac{\pi^2}{12}(k_B T / E_F)^2]$:

$$\chi_{\text{Pauli}}(T) = \chi_{\text{Pauli}}(0) \left[ 1 - \frac{\pi^2}{12} \left( \frac{T}{T_F} \right)^2 \right]$$

For sodium with $T_F \approx 37{,}600\text{ K}$, at $300\text{ K}$, $(T/T_F)^2 \sim 6 \times 10^{-5}$, so the temperature variation is less than $0.01\\%$.

(c) Net Magnetic Susceptibility of Sodium: Given:

$$n = 2.65 \times 10^{28}\text{ m}^{-3}, \quad E_F = 3.24\text{ eV} = 5.191 \times 10^{-19}\text{ J}$$
$$\mu_0 = 4\pi \times 10^{-7}\text{ H/m}, \quad \mu_B = 9.274 \times 10^{-24}\text{ J/T}$$

Compute $\chi_{\text{Pauli}}$:

$$\chi_{\text{Pauli}} = \frac{3 (4\pi \times 10^{-7})(2.65 \times 10^{28})(9.274 \times 10^{-24})^2}{2 \times 5.191 \times 10^{-19}} = \frac{3 \times 1.2566 \times 10^{-6} \times 2.65 \times 10^{28} \times 8.601 \times 10^{-47}}{1.0382 \times 10^{-18}}$$
$$\chi_{\text{Pauli}} = \frac{8.592 \times 10^{-24}}{1.0382 \times 10^{-18}} \approx 8.28 \times 10^{-6}$$

Landau diamagnetism:

$$\chi_{\text{Landau}} = -\frac{1}{3} \chi_{\text{Pauli}} = -2.76 \times 10^{-6}$$

Core ion diamagnetism of $\text{Na}^+$:

$$\chi_{\text{core}} \approx -0.42 \times 10^{-6}$$

Net total susceptibility:

$$\chi_{\text{net}} = \chi_{\text{Pauli}} + \chi_{\text{Landau}} + \chi_{\text{core}} = (8.28 - 2.76 - 0.42) \times 10^{-6} = +5.10 \times 10^{-6}$$

Because $\chi_{\text{net}} > 0$, metallic sodium is paramagnetically dominated.

Solved Problem Example 3.3: Quantum Exchange Singlet-Triplet Splitting in a Two-Electron Model

Two electrons occupy two orthogonal spatial orbitals $\phi_a(\vec{r})$ and $\phi_b(\vec{r})$ under a Hamiltonian $\hat{H} = \hat{h}_1 + \hat{h}_2 + \frac{e^2}{4\pi\epsilon_0 r_{12}}$.\n\n(a) Write down the spatial wavefunctions $\psi_S(\vec{r}_1, \vec{r}_2)$ and $\psi_T(\vec{r}_1, \vec{r}_2)$ for the spin singlet ($S = 0$) and spin triplet ($S = 1$) states.\n(b) Calculate the expectation energies $E_S$ and $E_T$ in terms of the single-particle energy $E_0 = \epsilon_a + \epsilon_b$, the direct Coulomb integral $K$, and the exchange integral $J_{\text{ex}}$.\n(c) Express the effective Hamiltonian in terms of the spin operators $\vec{S}_1$ and $\vec{S}_2$, and explain why a positive exchange integral $J_{\text{ex}} > 0$ energetically favors ferromagnetic alignment.

(a) Singlet and Triplet Spatial Wavefunctions: By the generalized Pauli principle, the total electronic state must be antisymmetric:

$$\Psi(1, 2) = \psi_{\text{spatial}}(\vec{r}_1, \vec{r}_2) \chi_{\text{spin}}(1, 2) = -\Psi(2, 1)$$
  • For the Singlet ($S = 0$), the spin state is the antisymmetric singlet:
$$\chi_0^0 = \frac{1}{\sqrt{2}} (|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)$$

Hence the spatial wavefunction must be symmetric:

$$\psi_S(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}} [\phi_a(\vec{r}_1)\phi_b(\vec{r}_2) + \phi_b(\vec{r}_1)\phi_a(\vec{r}_2)]$$
  • For the Triplet ($S = 1$), the spin states are symmetric ($|\uparrow\uparrow\rangle, \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle), |\downarrow\downarrow\rangle$). Hence the spatial wavefunction must be antisymmetric:
$$\psi_T(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}} [\phi_a(\vec{r}_1)\phi_b(\vec{r}_2) - \phi_b(\vec{r}_1)\phi_a(\vec{r}_2)]$$

(b) Singlet and Triplet Energies: Define the direct Coulomb integral $K$ and exchange integral $J_{\text{ex}}$:

$$K = \iint |\phi_a(\vec{r}_1)|^2 \frac{e^2}{4\pi\epsilon_0 r_{12}} |\phi_b(\vec{r}_2)|^2 d^3r_1 d^3r_2$$
$$J_{\text{ex}} = \iint \phi_a^*(\vec{r}_1)\phi_b^*(\vec{r}_2) \frac{e^2}{4\pi\epsilon_0 r_{12}} \phi_b(\vec{r}_1)\phi_a(\vec{r}_2) d^3r_1 d^3r_2$$

Computing the expectation values:

$$E_S = \langle \psi_S | \hat{H} | \psi_S \rangle = \epsilon_a + \epsilon_b + K + J_{\text{ex}}$$
$$E_T = \langle \psi_T | \hat{H} | \psi_T \rangle = \epsilon_a + \epsilon_b + K - J_{\text{ex}}$$

The energy splitting is:

$$\Delta E = E_S - E_T = 2 J_{\text{ex}}$$

(c) Effective Spin Hamiltonian: Notice that:

$$\vec{S}_1 \cdot \vec{S}_2 = \frac{1}{2} [(\vec{S}_1 + \vec{S}_2)^2 - \vec{S}_1^2 - \vec{S}_2^2] = \frac{1}{2} [S(S+1) - 3/4 - 3/4]$$

For $S = 0$: $\vec{S}_1 \cdot \vec{S}_2 = -3/4$. For $S = 1$: $\vec{S}_1 \cdot \vec{S}_2 = +1/4$. Constructing the operator $\hat{H}_{\text{spin}} = C_0 - 2 J_{\text{ex}} \vec{S}_1 \cdot \vec{S}_2$:

  • For $S = 0$: $E = C_0 - 2 J_{\text{ex}}(-3/4) = C_0 + \frac{3}{2}J_{\text{ex}}$
  • For $S = 1$: $E = C_0 - 2 J_{\text{ex}}(1/4) = C_0 - \frac{1}{2}J_{\text{ex}}$

Setting $C_0 = \epsilon_a + \epsilon_b + K - \frac{1}{2}J_{\text{ex}}$, the difference is:

$$E_S - E_T = 2 J_{\text{ex}}$$

Matching the Heisenberg Hamiltonian:

$$\hat{H}_{\text{eff}} = -2 J_{\text{ex}} \vec{S}_1 \cdot \vec{S}_2 + \text{const}$$

Physical origin: When $J_{\text{ex}} > 0$, $E_T < E_S$. The triplet state has lower energy because its spatial wavefunction $\psi_T$ vanishes whenever $\vec{r}_1 = \vec{r}_2$. This exchange hole keeps electrons separated in real space, minimizing repulsive Coulomb energy $\frac{e^2}{4\pi\epsilon_0 r_{12}}$. Thus, parallel spins ($S = 1$) minimize electrostatic energy, producing ferromagnetism.