Multi-Electron Atoms & Chemical Bonding in Solids
Quantum mechanics of many-electron systems: identical particles, exchange symmetry, Pauli exclusion principle, para- and ortho-helium, Hartree and Hartree-Fock self-consistent field methods, atomic multiplets, Hund's rules, characteristic X-ray spectra, and LCAO molecular orbital theory of covalent bonds.
§4.1 Identical Particles, Permutation Symmetry & Pauli Principle
1. Indistinguishability & The Permutation Operator
In quantum mechanics, identical particles (such as electrons) are fundamentally indistinguishable. Consider a system of two identical particles described by coordinate vectors $\xi_1 = (\vec{r}_1, \sigma_1)$ and $\xi_2 = (\vec{r}_2, \sigma_2)$ combining spatial and spin coordinates. The permutation operator $\hat{P}_{12}$ exchanges the particles:
Because $\hat{P}_{12}^2 = \hat{I}$, its eigenvalues are $\lambda = \pm 1$. The Symmetrization Postulate partitions all physical particles in nature into two disjoint classes:
- Bosons (Integer Spin $S = 0, 1, 2, \dots$): Symmetric wavefunctions under exchange:
$$\Psi(\xi_2, \xi_1) = + \Psi(\xi_1, \xi_2)$$
- Fermions (Half-Integer Spin $S = 1/2, 3/2, \dots$): Antisymmetric wavefunctions under exchange:
$$\Psi(\xi_2, \xi_1) = - \Psi(\xi_1, \xi_2)$$
2. The Pauli Exclusion Principle & Slater Determinants
Electrons are spin-$1/2$ fermions, so the total wavefunction of an $N$-electron system must be completely antisymmetric under the exchange of any pair of electrons $i$ and $j$. If the electrons occupy single-particle spin-orbitals $\chi_{\alpha}(\xi) = \phi_a(\vec{r}) \chi_s(\sigma)$, the total antisymmetric wavefunction is represented by a Slater Determinant:
If two electrons occupy the identical quantum state ($\alpha_1 = \alpha_2$), two columns of the determinant are identical, causing $\Psi \equiv 0$. This establishes the Pauli Exclusion Principle: no two electrons can occupy the same quantum state simultaneously.
§4.2 The Helium Atom: Exchange Symmetry, Para- & Ortho-Helium
1. Helium Hamiltonian & Electron-Electron Repulsion
The non-relativistic Hamiltonian of the neutral Helium atom ($Z = 2$) with nucleus at the origin is:
The total two-electron wavefunction factors into spatial and spin parts: $\Psi(\xi_1, \xi_2) = \psi(\vec{r}_1, \vec{r}_2) \chi_{\text{spin}}(1, 2)$. Total antisymmetry requires:
- Singlet State ($S = 0$, Para-Helium): Antisymmetric spin $\chi_{0,0} = \frac{1}{\sqrt{2}}(\alpha_1 \beta_2 - \beta_1 \alpha_2)$ coupled to a symmetric spatial wavefunction:
$$\psi_+(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}} [\phi_a(\vec{r}_1)\phi_b(\vec{r}_2) + \phi_b(\vec{r}_1)\phi_a(\vec{r}_2)]$$
- Triplet State ($S = 1$, Ortho-Helium): Symmetric spin ($M_S = +1, 0, -1$) coupled to an antisymmetric spatial wavefunction:
$$\psi_-(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}} [\phi_a(\vec{r}_1)\phi_b(\vec{r}_2) - \phi_b(\vec{r}_1)\phi_a(\vec{r}_2)]$$
2. Direct and Exchange Integrals
Evaluating the expectation value of the electron-electron Coulomb repulsion $\hat{V}_{ee}$ in first-order perturbation theory:
where:
- Direct Coulomb Integral $J$: Classical electrostatic repulsion between the two electron charge clouds:
$$J = \iint |\phi_a(\vec{r}_1)|^2 \frac{e^2}{4\pi\varepsilon_0 r_{12}} |\phi_b(\vec{r}_2)|^2 d^3r_1 d^3r_2 > 0$$
- Exchange Integral $K$: Purely quantum mechanical energy shift arising from the overlap of the two single-particle orbitals:
$$K = \iint \phi_a^*(\vec{r}_1) \phi_b^*(\vec{r}_2) \frac{e^2}{4\pi\varepsilon_0 r_{12}} \phi_b(\vec{r}_1) \phi_a(\vec{r}_2) d^3r_1 d^3r_2 > 0$$
The total energy eigenvalues are: $E_{\text{singlet}} = E_0 + J + K$ and $E_{\text{triplet}} = E_0 + J - K$. The triplet state (ortho-helium) lies lower in energy than the singlet state (para-helium) by the exchange splitting $\Delta E = 2K$. In the triplet state, $\psi_-(\vec{r}, \vec{r}) = 0$, meaning electrons avoid each other in space, reducing Coulomb repulsion.
§4.3 Hartree & Hartree-Fock Self-Consistent Field (SCF) Methods
1. The Central Field Approximation
In multi-electron atoms and solids with $N$ electrons, the exact many-body Schrödinger equation cannot be solved analytically. In the central field approximation, each electron moves independently in an effective spherically symmetric potential $V_{\text{eff}}(r)$ created by the nucleus plus the spherically averaged charge distribution of all other $N-1$ electrons:
2. The Hartree Self-Consistent Field (SCF) Method
Douglas Hartree (1928) formulated an iterative variational procedure. Assuming a trial product wavefunction $\Psi = \phi_1(\vec{r}_1)\phi_2(\vec{r}_2)\dots\phi_N(\vec{r}_N)$, electron $i$ experiences an electrostatic potential produced by the charge density $\rho_j(\vec{r}') = -e |\phi_j(\vec{r}')|^2$ of all other electrons:
The self-consistent iteration cycle proceeds as follows:
- Guess an initial set of radial wavefunctions $\{\phi_j^{(0)}\}$.
- Compute the electronic charge density $\rho^{(0)}(\vec{r})$ and the Hartree potential $V_i^{(0)}(\vec{r})$.
- Solve the single-particle Schrödinger equations to obtain a new set of wavefunctions $\{\phi_j^{(1)}\}$.
- Repeat steps 2–3 iteratively until the input and output potentials converge within a numerical tolerance: $|V^{(k+1)} - V^{(k)}| < \delta$.
3. The Hartree-Fock Method & Non-Local Exchange Potential
The simple Hartree product fails to satisfy the Pauli principle. Fock (1930) replaced the product with a fully antisymmetric Slater determinant, introducing an additional non-local exchange potential $V_{\text{ex}}$:
The exchange operator acts only between electrons with parallel spins, creating a surrounding depletion zone known as the Fermi hole (or exchange hole).
§4.4 Hund's Rules, Characteristic X-Rays & Molecular LCAO Bonding
1. Hund's Rules for Atomic Ground States
For multi-electron open subshells ($p^n, d^n, f^n$), electrostatic repulsion and spin-orbit coupling determine the energetic ordering of spectroscopic term symbols $^{2S+1}L_J$ via Hund's three empirical rules:
- Rule 1 (Maximize Spin Multiplicity $S$): The ground state term has the maximum total spin $S$ permitted by the Pauli exclusion principle (minimizes Coulomb repulsion by maximizing exchange stabilization).
- Rule 2 (Maximize Total Orbital Angular Momentum $L$): For a given $S$, the ground state has the maximum total orbital angular momentum $L$ (electrons orbit in the same sense, minimizing spatial close encounters).
- Rule 3 (Spin-Orbit Coupling $J$):
- If the subshell is less than half full, the lowest energy level has minimum total angular momentum: $J = |L - S|$.
- If the subshell is more than half full, the lowest energy level has maximum total angular momentum: $J = L + S$.
2. Characteristic X-Ray Spectra & Moseley's Law
When high-energy electrons eject an inner-core atomic electron, vacancies are filled by radiative transitions from outer shells, producing sharp characteristic X-ray lines:
- $K_\alpha$ line: transition from $n = 2$ ($L$-shell) to $n = 1$ ($K$-shell).
- $K_\beta$ line: transition from $n = 3$ ($M$-shell) to $n = 1$ ($K$-shell).
- $L_\alpha$ line: transition from $n = 3$ ($M$-shell) to $n = 2$ ($L$-shell).
Henry Moseley (1913) discovered that the frequency $\nu$ of characteristic X-rays is related linearly to the atomic number $Z$:
For the $K_\alpha$ line, $b = 1$ (the remaining $1s$ electron screens one unit of nuclear charge), yielding:
3. Molecular Orbital Theory & LCAO Covalent Bonding
In solids, atomic orbitals overlap to form extended molecular orbitals. In the simplest diatomic system ($\text{H}_2^+$ and $\text{H}_2$), Linear Combination of Atomic Orbitals (LCAO) yields two molecular orbitals from atomic hydrogen $1s$ states $\phi_A$ and $\phi_B$:
- Bonding Orbital ($\sigma_g$): $\psi_+ = \frac{1}{\sqrt{2(1+S)}} (\phi_A + \phi_B)$. Large electron probability density $|\psi_+|^2$ builds up between the two positively charged nuclei, screening their Coulomb repulsion and lowering the total electronic energy.
- Antibonding Orbital ($\sigma_u^*$): $\psi_- = \frac{1}{\sqrt{2(1-S)}} (\phi_A - \phi_B)$. Nodal plane ($\psi = 0$) midway between the nuclei pushes electron density away, increasing nuclear repulsion and raising the energy.
In the excited $(1s)(2s)$ configuration of the Helium atom, the direct Coulomb integral is $J = 8.78 \text{ eV}$ and the exchange integral is $K = 1.19 \text{ eV}$. The unperturbed two-electron energy level is $E_0 = -59.38 \text{ eV}$. (a) Calculate the total energies of the singlet state ($1^1S_0$, para-helium) and the triplet state ($2^3S_1$, ortho-helium). (b) Determine the exchange splitting energy $\Delta E$ and explain physically why the triplet state lies lower in energy.
In the singlet state, the spatial wavefunction is symmetric (J + K), while in the triplet state it is antisymmetric (J - K).
Evaluate E_singlet.
Evaluate E_triplet.
The exchange splitting equals twice the exchange integral.
E_{\text{singlet}} = -49.41 \text{ eV}, \quad E_{\text{triplet}} = -51.79 \text{ eV}, \quad \Delta E = 2.38 \text{ eV}
Use Hund's rules to determine the ground state spectroscopic term symbol $^{2S+1}L_J$ for: (a) Carbon (neutral atom, valence subshell $2p^2$). (b) Iron ($\text{Fe}^{2+}$ ion, valence subshell $3d^6$).
Rule 1: Maximize S. Place both electrons with parallel spins in different m_l orbitals: m_s = +1/2, +1/2.
Rule 2: Maximize L using available m_l in {+1, 0, -1}. Placing electrons in m_l = +1 and m_l = 0 gives L = 1.
Rule 3: The 2p subshell is less than half full (2 of 6 electrons), so J = |L - S| = 0. The ground state is ^3P_0.
Rule 1: Place 5 electrons spin-up in m_l = +2, +1, 0, -1, -2 and the 6th electron spin-down in m_l = +2. S = 2.
Rule 2: L = 2 (D state). Rule 3: The 3d subshell is more than half full (6 of 10 electrons), so J = L + S = 4. The ground state is ^5D_4.
\text{Carbon: } {^3P_0}, \quad \text{Iron (Fe}^{2+}\text{): } {^5D_4}
Copper has atomic number $Z = 29$. The Rydberg constant is $R_\infty = 1.09737 \times 10^7 \text{ m}^{-1}$. Using Moseley's law with screening constant $b = 1.0$ for the $K_\alpha$ transition: (a) Calculate the cyclical frequency $\nu_{K_\alpha}$ of the emitted X-ray photon. (b) Determine the wavelength $\lambda_{K_\alpha}$ in Angstroms and the photon energy in $\text{keV}$.
The K_alpha transition originates from n=2 to n=1 with screening factor b=1.
Substitute c, R_infinity, and Z=29 into the formula.
Compute lambda = c / nu. This closely matches the experimental Cu K_alpha value of 1.541 Angstroms.
Convert wavelength to energy in keV.
\nu_{K_\alpha} = 1.934 \times 10^{18} \text{ Hz}, \quad \lambda_{K_\alpha} = 1.550 \text{ Å}, \quad E = 8.00 \text{ keV}
Solved University Examination Problems
Step-by-step mathematical solutions to classic university honors examination questions.