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Identical Particles and Exchange Symmetry

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

ψ(,ri,,rj,)=±ψ(,rj,,ri,)\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:

ψS=12[ψa(1)ψb(2)+ψb(1)ψa(2)](bosons)\psi_S = \frac{1}{\sqrt{2}}\left[\psi_a(1)\psi_b(2) + \psi_b(1)\psi_a(2)\right] \quad \text{(bosons)}

ψA=12[ψa(1)ψb(2)ψb(1)ψa(2)](fermions)\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:

ψtriplet2=0whenr1=r2|\psi_{\text{triplet}|^2 = 0 \quad \text{when} \mathbf{r}_1 = \mathbf{r}_2}

ψsinglet2>0whenr1=r2|\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):

H^=22me(12+22)2e24πε0r12e24πε0r2+e24πε0r1r2\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 1s1s orbital with opposite spins (singlet). The spatial part is symmetric: ψ100(r1)ψ100(r2)\psi_{100}(\mathbf{r}_1)\psi_{100}(\mathbf{r}_2).

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

E(1)=54e24πε0a0=52×13.6 eV=34.0 eVE^{(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×(54.4 eV)=108.8E^{(0)} = 2 \times (-54.4\ \text{eV}) = -108.8 eV (two electrons in Z=2Z = 2 Coulomb potential). Including perturbation: E108.8+34.0=74.8E \approx -108.8 + 34.0 = -74.8 eV. The experimental value is 79.0-79.0 eV.

Excited states: When one electron is excited to 1snl1s\,nlThe spin configuration matters:

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

The exchange integral KK and direct integral JJ:

J=ψa(1)2e24πε0r12ψb(2)2d3r1d3r2J = \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=ψa(1)ψb(2)e24πε0r12ψb(1)ψa(2)d3r1d3r2K = \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 2K2KWith the triplet lower by 2K2K.

Worked Example 9.1: Helium $1s2s$ States

For the 1s2s1s\,2s configuration of helium:

J1s,2s=e24πε0ψ1s(1)21r12ψ2s(2)2d3r1d3r2J_{1s,2s} = \frac{e^2}{4\pi\varepsilon_0}\int |\psi_{1s}(1)|^2\frac{1}{r_{12}}|\psi_{2s}(2)|^2\, d^3r_1 d^3r_2

K1s,2s=e24πε0ψ1s(1)ψ2s(2)1r12ψ2s(1)ψ1s(2)d3r1d3r2K_{1s,2s} = \frac{e^2}{4\pi\varepsilon_0}\int \psi_{1s}^*(1)\psi_{2s}^*(2)\frac{1}{r_{12}}\psi_{2s}(1)\psi_{1s}(2)\, d^3r_1 d^3r_2

Evaluating these (using the multipole expansion 1/r12=lr<l/r>l+1Pl(cosθ)1/r_{12} = \sum_l r_<^l/r_>^{l+1}\,P_l(\cos\theta)):

J1s,2s0.42 Ry=5.7 eVJ_{1s,2s} \approx 0.42\ \text{Ry} = 5.7\ \text{eV}

K1s,2s0.032 Ry=0.43 eVK_{1s,2s} \approx 0.032\ \text{Ry} = 0.43\ \text{eV}

The singlet (parahelium) has energy E=E0+J+KE = E_0 + J + K And the triplet (orthohelium) has E=E0+JKE = E_0 + J - K.

The splitting: EsingletEtriplet=2K0.86E_{\text{singlet} - E_{\text{triplet} = 2K \approx 0.86}} eV. This is the exchange splitting.

The orthohelium 23S2^3S state is metastable: it cannot decay to the ground state by electric dipole transition (because ΔS=0\Delta S = 0 for E1 transitions, and the ground state is a singlet). Its lifetime is 104\sim 10^4 s.

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

Ψ(1,2,,N)=1N!ϕ1(1)ϕ2(1)ϕN(1)ϕ1(2)ϕ2(2)ϕN(2)ϕ1(N)ϕ2(N)ϕN(N)\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 ϕi\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: Eexchange=JKE_{\text{exchange}} = J \mp K where the sign depends on the spin configuration. The triplet (parallel spins) has energy JKJ - K and the singlet (antiparallel) has J+KJ + K.
  • Slater determinant size: For NN particles each with MM available states, the Hilbert space dimension is (MN)\binom{M}{N} for fermions versus MNM^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 1/21/\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 2K2K 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, 4^4He atoms (2 protons, 2 neutrons, 2 electrons) are bosons, while 3^3He 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>0K > 0 favours parallel alignment (ferromagnetism).

Consider three electrons confined to a one-dimensional box of length LL. The single-particle energies are En=n2π22/(2mL2)E_n = n^2\pi^2\hbar^2/(2mL^2). The lowest configuration has two electrons in n=1n = 1 (opposite spins) and one in n=2n = 2.

The spatial part of the wavefunction must be antisymmetric under exchange of any two electrons. Using the Slater determinant with orbitals ϕ1\phi_1, ϕ1\phi_1, ϕ2\phi_2 (where ϕ1\phi_1 and ϕ2\phi_2 are the spatial wavefunctions of the box), the antisymmetric state is:

Ψ=13!ϕ1(r1)α(1)ϕ1(r1)β(1)ϕ2(r1)α(1)ϕ1(r2)α(2)ϕ1(r2)β(2)ϕ2(r2)α(2)ϕ1(r3)α(3)ϕ1(r3)β(3)ϕ2(r3)α(3)\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 E1+E1+E2=6π22/(2mL2)E_1 + E_1 + E_2 = 6\pi^2\hbar^2/(2mL^2). The exchange splitting between the two possible spin configurations (total S=1/2S = 1/2) depends on the exchange integral K12K_{12} between the n=1n = 1 and n=2n = 2 states.

flowchart TD
A[10_Identical Particles And Exchange Symmetry] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

Identical particles are truly indistinguishable, not just similar. Swapping two electrons must give the same physics, which forces their wavefunction to be antisymmetric. This creates the exchange interaction, an effective force arising purely from symmetry, not from any physical interaction. The Pauli exclusion principle is a consequence: two fermions cannot occupy the same state because the antisymmetric wavefunction would vanish. Bosons, with symmetric wavefunctions, crowd together instead. This explains why electrons fill up energy levels in atoms, why white dwarfs resist collapse, and why laser light is coherent.

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Ensure you have mastered the prerequisite material before attempting this advanced content.

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Ensure you have mastered the prerequisite material before attempting this advanced content.