Identical Particles and Exchange Symmetry

For a system of $N$ identical particles, the wavefunction must satisfy:

$\psi(\ldots, \mathbf{r}_i, \ldots, \mathbf{r}_j, \ldots) = \pm\psi(\ldots, \mathbf{r}_j, \ldots, \mathbf{r}_i, \ldots)$

• Bosons (integer spin): symmetric ($+$ sign). Any number can occupy the same state.
• Fermions (half-integer spin): antisymmetric ($-$ sign). Pauli exclusion: no two fermions can occupy the same state.

For two particles, the properly symmetrised states are:

$\psi_S = \frac{1}{\sqrt{2}}\left[\psi_a(1)\psi_b(2) + \psi_b(1)\psi_a(2)\right] \quad \text{(bosons)}$

$\psi_A = \frac{1}{\sqrt{2}}\left[\psi_a(1)\psi_b(2) - \psi_b(1)\psi_a(2)\right] \quad \text{(fermions)}$

Even without an explicit interaction potential, the requirement of (anti)symmetry leads to an effective exchange interaction. For two electrons in a box, the probability of finding them close together differs between the triplet (spatially antisymmetric, spin symmetric) and singlet (spatially symmetric, spin antisymmetric) states:

$|\psi_{\text{triplet}|^2 = 0 \quad \text{when} \mathbf{r}_1 = \mathbf{r}_2}$

$|\psi_{\text{singlet}|^2 > 0 \quad \text{when} \mathbf{r}_1 = \mathbf{r}_2}$

The triplet state keeps electrons apart (effective repulsion), while the singlet allows them to be close. This is the origin of the Hund”s first rule: parallel spins are energetically favourable for atoms because the exchange interaction lowers the Coulomb repulsion.

The helium Hamiltonian (ignoring nuclear motion):

$\hat{H} = -\frac{\hbar^2}{2m_e}\left(\nabla_1^2 + \nabla_2^2\right) - \frac{2e^2}{4\pi\varepsilon_0 r_1} - \frac{2e^2}{4\pi\varepsilon_0 r_2} + \frac{e^2}{4\pi\varepsilon_0|\mathbf{r}_1 - \mathbf{r}_2|}$

Ground state (parahelium): Both electrons in the $1s$ orbital with opposite spins (singlet). The spatial part is symmetric: $\psi_{100}(\mathbf{r}_1)\psi_{100}(\mathbf{r}_2)$.

First-order perturbation theory for the electron-electron repulsion:

$E^{(1)} = \frac{5}{4}\frac{e^2}{4\pi\varepsilon_0 a_0} = \frac{5}{2}\times 13.6\ \text{eV} = 34.0\ \text{eV}$

The unperturbed ground state energy is $E^{(0)} = 2 \times (-54.4\ \text{eV}) = -108.8$ eV (two electrons in $Z = 2$ Coulomb potential). Including perturbation: $E \approx -108.8 + 34.0 = -74.8$ eV. The experimental value is $-79.0$ eV.

Excited states: When one electron is excited to $1s\,nl$The spin configuration matters:

• Parahelium (singlet, $S = 0$): symmetric spatial, antisymmetric spin. Lower energy for given configuration.
• Orthohelium (triplet, $S = 1$): antisymmetric spatial, symmetric spin. Higher energy.

The exchange integral $K$ and direct integral $J$:

$J = \iint |\psi_a(1)|^2\frac{e^2}{4\pi\varepsilon_0 r_{12}}|\psi_b(2)|^2\, d^3r_1 d^3r_2$

$K = \iint \psi_a^*(1)\psi_b^*(2)\frac{e^2}{4\pi\varepsilon_0 r_{12}}\psi_b(1)\psi_a(2)\, d^3r_1 d^3r_2$

The energy splitting between singlet and triplet is $2K$With the triplet lower by $2K$.

For $N$ fermions, the antisymmetric wavefunction is efficiently written as a Slater determinant:

$\Psi(1, 2, \ldots, N) = \frac{1}{\sqrt{N!}}\begin{vmatrix} \phi_1(1) & \phi_2(1) & \cdots & \phi_N(1) \\ \phi_1(2) & \phi_2(2) & \cdots & \phi_N(2) \\ \vdots & \vdots & \ddots & \vdots \\ \phi_1(N) & \phi_2(N) & \cdots & \phi_N(N) \end{vmatrix}$

Properties:

• Swapping any two rows (particles) changes the sign
• If any two columns (orbitals) are identical, the determinant vanishes (Pauli exclusion)
• The normalisation is correct if the spin-orbitals $\phi_i$ are orthonormal

• Spin-statistics connection: Particles with integer spin are bosons; half-integer spin are fermions. No exceptions in 3+1 dimensions.
• Exchange energy: $E_{\text{exchange}} = J \mp K$ where the sign depends on the spin configuration. The triplet (parallel spins) has energy $J - K$ and the singlet (antiparallel) has $J + K$.
• Slater determinant size: For $N$ particles each with $M$ available states, the Hilbert space dimension is $\binom{M}{N}$ for fermions versus $M^N$ for bosons.
• Pfaffian for pairs: For an even number of fermions, the antisymmetric state can also be written as a Pfaffian, which is computationally efficient for specific pairing structures.

• Forgetting normalisation: The symmetrisation prefactor $1/\sqrt{2}$ in two-particle states is essential. Without it, the states are not normalised and probability conservation fails.
• Confusing exchange with interaction: The exchange splitting $2K$ arises from symmetry requirements, not from an explicit interaction potential between particles.
• Assuming all particles are fermions or bosons: Composite particles can be either. For example, $^4$He atoms (2 protons, 2 neutrons, 2 electrons) are bosons, while $^3$He atoms are fermions.
• Neglecting spin in antisymmetrisation: The full two-particle wavefunction (spatial $\times$ spin) must be antisymmetric for fermions. Using only the spatial part leads to incorrect results.

• Electron gas in metals: The Pauli exclusion principle forces electrons into progressively higher energy states, creating the Fermi sea. This accounts for the electronic specific heat and the stability of matter.
• White dwarf and neutron star stability: Electron degeneracy pressure (from the Pauli principle) supports white dwarfs against gravitational collapse. Neutron degeneracy pressure supports neutron stars.
• Bose-Einstein condensation: Below a critical temperature, a macroscopic fraction of bosons occupies the lowest energy state, producing superfluidity and coherent emission (atom lasers).
• Magnetic ordering: Hund’s rules and exchange interactions determine whether a material is ferromagnetic or antiferromagnetic. The exchange integral $K > 0$ favours parallel alignment (ferromagnetism).

Consider three electrons confined to a one-dimensional box of length $L$. The single-particle energies are $E_n = n^2\pi^2\hbar^2/(2mL^2)$. The lowest configuration has two electrons in $n = 1$ (opposite spins) and one in $n = 2$.

The spatial part of the wavefunction must be antisymmetric under exchange of any two electrons. Using the Slater determinant with orbitals $\phi_1$, $\phi_1$, $\phi_2$ (where $\phi_1$ and $\phi_2$ are the spatial wavefunctions of the box), the antisymmetric state is:

$\Psi = \frac{1}{\sqrt{3!}}\begin{vmatrix} \phi_1(r_1)\alpha(1) & \phi_1(r_1)\beta(1) & \phi_2(r_1)\alpha(1) \\ \phi_1(r_2)\alpha(2) & \phi_1(r_2)\beta(2) & \phi_2(r_2)\alpha(2) \\ \phi_1(r_3)\alpha(3) & \phi_1(r_3)\beta(3) & \phi_2(r_3)\alpha(3) \end{vmatrix}$

The total energy is $E_1 + E_1 + E_2 = 6\pi^2\hbar^2/(2mL^2)$. The exchange splitting between the two possible spin configurations (total $S = 1/2$) depends on the exchange integral $K_{12}$ between the $n = 1$ and $n = 2$ states.