Identical Particles and Exchange Symmetry

Derivation

Identical Particles and Exchange Symmetry

D-229 Home PU-301 Threads matter · symmetry · chance Depends on Spectral Theorem for Hermitian Observables, Addition of Angular Momenta

Statement

For a system of \(N\) indistinguishable particles whose Hamiltonian is invariant under every particle interchange, the physical state must be a simultaneous eigenstate of every pair-exchange operator \(\hat{P}_{ij}\) with a single common eigenvalue: \(+1\) (totally symmetric, bosons) or \(-1\) (totally antisymmetric, fermions). The antisymmetric \(N\)-fermion state built from orthonormal orbitals \(\{\varphi_k\}\) is the Slater determinant \(\Psi=\frac{1}{\sqrt{N!}}\det[\varphi_i(x_j)]\), and the electron–electron interaction splits a two-orbital configuration into \(E_\pm=\varepsilon_a+\varepsilon_b+J\pm K\), where the sign of the exchange integral \(K\) is tied by antisymmetry to the total spin.

Why it matters

Exchange symmetry is not an approximation but a superselection rule: it removes the vast majority of the naive tensor-product Hilbert space and leaves only the symmetric or antisymmetric sector. This single restriction produces the Pauli exclusion principle, the shell structure of atoms, the rigidity of matter, the ferromagnetic alignment of spins, and the entire distinction between Fermi and Bose statistics.

The exchange integral \(K\) is a purely quantum, spin-independent electrostatic energy that nonetheless behaves as if it were a spin–spin coupling, because antisymmetry correlates the spatial and spin parts of the wavefunction. It is the microscopic origin of Hund's first rule and of the Heisenberg exchange constant in magnetism.

Assumptions

The particles are strictly identical.If any intrinsic label (mass, charge, flavour) distinguishes them, \(\hat P_{ij}\) is not a symmetry, the state need not be an eigenstate of exchange, and exclusion and exchange energy vanish. The Hamiltonian commutes with every \(\hat P_{ij}\).If \(H\) contained a term acting asymmetrically on the labels, exchange eigenvalue would not be conserved and symmetric/antisymmetric sectors would mix under time evolution. Spin–statistics connection holds: half-integer spin \(\Rightarrow\) antisymmetric, integer spin \(\Rightarrow\) symmetric.Dropping this (as in non-relativistic QM, where it is an extra postulate) leaves both sectors formally allowed; the empirical periodic table and blackbody spectrum would be unexplained. It is a theorem only in relativistic QFT with locality and positive energy. The full Hilbert space is exhausted by one-dimensional representations of the permutation group in 3D.In two dimensions this fails: braid-group anyons with fractional statistics appear, and the symmetric/antisymmetric dichotomy is no longer complete. Orbitals used to build the determinant are orthonormal single-particle states.If non-orthogonal, the normalisation \(1/\sqrt{N!}\) and the clean \(J\pm K\) split acquire overlap corrections (Löwdin terms).

Derivation

\[ \hat P_{12}\,\Psi(x_1,x_2)=\Psi(x_2,x_1) \]

Define the exchange operator on the two-particle wavefunction (\(x\) collects space and spin). Identity of the particles means this relabelling is a symmetry operation. A

\[ \hat P_{12}^{\,2}\,\Psi(x_1,x_2)=\hat P_{12}\,\Psi(x_2,x_1)=\Psi(x_1,x_2)\quad\Rightarrow\quad \hat P_{12}^{\,2}=\hat I \]

Swapping twice returns the original arrangement. The operator squares to the identity. A

\[ \hat P_{12}\Psi=\lambda\Psi\;\Rightarrow\;\lambda^2=1\;\Rightarrow\;\lambda=\pm 1 \]

Eigenvalues of an operator whose square is the identity can only be \(\pm1\). \(\hat P_{12}\) is also Hermitian and unitary, so its eigenvalues are real and lie on the unit circle. A

\[ [\,H,\hat P_{12}\,]=0 \]

Because \(H(1,2)=H(2,1)\) for identical particles, \(H\) is unchanged by the relabelling, so it commutes with \(\hat P_{12}\). By the spectral theorem for commuting Hermitian observables, a common eigenbasis exists and the exchange parity is a conserved quantum number. B

\[ \{\hat P_{ij}\}\ \text{generate}\ S_N,\quad \text{physical states span a 1D irrep} \]

All transpositions must share one eigenvalue on a physical state (adjacent transpositions are conjugate in \(S_N\), forcing equal eigenvalues). The only one-dimensional representations of \(S_N\) are the trivial (\(+1\), symmetric) and the sign (\(-1\), antisymmetric). This is the symmetrization postulate. C

\[ \Psi_{\text{F}}(x_1,x_2)=\tfrac{1}{\sqrt2}\big[\varphi_a(x_1)\varphi_b(x_2)-\varphi_a(x_2)\varphi_b(x_1)\big] \]

Impose \(\lambda=-1\) for fermions on a product of two orbitals; the antisymmetric combination is the unique (up to phase) fermionic two-body state. A

\[ \Psi_{\text{F}}(x_1,\dots,x_N)=\frac{1}{\sqrt{N!}}\begin{vmatrix}\varphi_1(x_1)&\cdots&\varphi_1(x_N)\\ \vdots&\ddots&\vdots\\ \varphi_N(x_1)&\cdots&\varphi_N(x_N)\end{vmatrix} \]

Generalise: a determinant changes sign under exchange of any two columns (particles), so it is automatically totally antisymmetric. This is the Slater determinant; \(1/\sqrt{N!}\) normalises it for orthonormal orbitals. B

\[ \varphi_a=\varphi_b\;\Rightarrow\;\text{two equal rows}\;\Rightarrow\;\Psi_{\text{F}}=0 \]

A determinant with two identical rows vanishes: two fermions cannot occupy the same single-particle state. This is the Pauli exclusion principle, obtained without any extra postulate. A

\[ \Psi=\psi_{\text{space}}(\mathbf r_1,\mathbf r_2)\,\chi_{\text{spin}}(1,2),\qquad \psi_\pm=\tfrac{1}{\sqrt2}\big[\varphi_a(\mathbf r_1)\varphi_b(\mathbf r_2)\pm\varphi_a(\mathbf r_2)\varphi_b(\mathbf r_1)\big] \]

For spin-independent \(H\), factor the state. Total antisymmetry pairs a symmetric space part \(\psi_+\) with the antisymmetric singlet spin state, and antisymmetric \(\psi_-\) with the symmetric triplet (this pairing follows from Clebsch–Gordan addition of two spin-\(\tfrac12\): \(\mathbf 2\otimes\mathbf 2=\mathbf 3_S\oplus\mathbf 1_A\)). B

\[ \langle\psi_\pm|\,V(|\mathbf r_1-\mathbf r_2|)\,|\psi_\pm\rangle=J\pm K \]

Insert \(\psi_\pm\) into the expectation of the pair interaction. The cross terms of the antisymmetrised product generate the exchange contribution; the sign follows the spatial symmetry. B

\[ J=\!\iint\! |\varphi_a(\mathbf r_1)|^2\,V\,|\varphi_b(\mathbf r_2)|^2\,d^3r_1\,d^3r_2 \]

\[ K=\!\iint\! \varphi_a^{*}(\mathbf r_1)\varphi_b^{*}(\mathbf r_2)\,V\,\varphi_b(\mathbf r_1)\varphi_a(\mathbf r_2)\,d^3r_1\,d^3r_2 \]

Read off the two integrals. \(J\) (direct/Coulomb) is the electrostatic energy of the two charge clouds; \(K\) (exchange) has the orbitals swapped between the coordinates and has no classical analogue. For real orbitals and \(V>0\), \(K\ge 0\). C

\[ E_\pm=\varepsilon_a+\varepsilon_b+J\pm K,\qquad \Delta E_{\text{S–T}}=E_+-E_-=2K \]

Add the one-body energies \(\varepsilon_a+\varepsilon_b\). The singlet (\(\psi_+\)) sits at \(+K\), the triplet (\(\psi_-\)) at \(-K\); their separation is \(2K\). For \(K>0\) the triplet lies lower — Hund's first rule. A

Result

\[ \boxed{\;\Psi_{\text{F}}=\frac{1}{\sqrt{N!}}\det[\varphi_i(x_j)],\qquad E_\pm=\varepsilon_a+\varepsilon_b+J\pm K,\qquad \Delta E_{\text{S–T}}=2K\;} \]

Reading. Nature admits only the fully symmetric (boson) or fully antisymmetric (fermion) sector; fermions are described by Slater determinants, which vanish when two orbitals coincide (Pauli). Even though the Coulomb interaction knows nothing about spin, antisymmetry forces the spatial part to depend on the spin state, so the two-electron energy carries a spin-dependent piece \(\pm K\). The exchange integral \(K\) — a positive electrostatic quantity — pushes the parallel-spin (triplet) configuration below the antiparallel (singlet) one by \(2K\).

Units check. \(V=e^2/4\pi\varepsilon_0 r\) has units of energy (J); \(|\varphi|^2\,d^3r\) is dimensionless (normalised probability), so \(J\) and \(K\) integrate energy \(\times\) dimensionless \(=\) energy. \(\varepsilon_a,\varepsilon_b\) are single-particle energies (J), so \(E_\pm\) and \(2K\) are energies. Consistent.

Limiting cases

Non-overlapping orbitals (\(\varphi_a,\varphi_b\) spatially disjoint): \(K\to 0\), singlet and triplet degenerate, exchange effects disappear — distant electrons behave classically.
Contact interaction \(V=g\,\delta(\mathbf r_1-\mathbf r_2)\): \(J=K\) exactly, so the triplet interaction energy \(J-K=0\) (the Pauli hole excludes coincidence), while the singlet feels \(2J\).
Same orbital, \(\varphi_a=\varphi_b\): only the singlet spin state survives; the triplet spatial part vanishes identically (exclusion), so no exchange splitting exists for a doubly-occupied orbital.
Bosonic limit: the sign flips to \(+1\), determinants become permanents, and equal-orbital occupation is enhanced rather than forbidden (Bose bunching).
\(N\to\infty\) free fermions: the antisymmetry produces the Fermi sea; the exchange term becomes the Fock exchange energy \(\propto n^{4/3}\) of the electron gas.

Breaks when

Two spatial dimensions. The permutation group is replaced by the braid group; anyons with fractional exchange phase \(e^{i\theta}\) (\(0<\theta<\pi\)) appear, and the strict \(\pm1\) dichotomy no longer holds (fractional quantum Hall regime).
Spin–orbit coupling / relativistic terms. When \(H\) does not commute with total spin, the clean space–spin factorisation of step 9 fails; states are labelled by \(j\), and singlet/triplet are no longer good quantum numbers, mixing the \(\pm K\) structure.
Strong correlation. A single Slater determinant is a poor state when interactions dominate the level spacing (near-degenerate configurations, Mott insulators); multi-determinant (configuration-interaction) expansions are required and \(E_\pm=\varepsilon_a+\varepsilon_b+J\pm K\) is only a leading estimate.
Non-orthogonal orbitals. The \(1/\sqrt{N!}\) normalisation and the tidy \(J\pm K\) split acquire overlap-matrix corrections; naive use gives wrong energies.

Failure modes

"Exchange force is a real force." \(K\) is electrostatic energy re-bookkept by antisymmetry, not a new fundamental interaction. There is no exchange field in the Hamiltonian.
Pairing the wrong spin with the wrong space. Students attach the symmetric spatial part to the triplet. It is the singlet (antisymmetric spin) that goes with symmetric space, so total antisymmetry holds.
Forgetting spin in the exclusion count. "Two electrons cannot be in the \(1s\) orbital" — they can, with opposite spins, because the full state (space\(\times\)spin) is what must be antisymmetric.
Dropping the \(1/\sqrt{N!}\). Omitting the normalisation gives energies off by factors of \(N!\) in expectation values.
Sign error in \(K\). Writing \(E=\ldots-K\) for the singlet; the singlet takes \(+K\).
Applying the symmetrization postulate in 2D. Assuming only bosons/fermions exist when anyons are possible.
Treating \(J\) and \(K\) as independent free parameters rather than as fixed integrals over the chosen orbitals.

Discussion

The deepest point is that indistinguishability is enforced at the level of states, not dynamics. Once \(\hat P_{ij}\) commutes with \(H\) and with all observables (all physical operators are symmetric in the labels), the exchange parity becomes a superselection charge: no measurement and no time evolution can move a state between the symmetric and antisymmetric sectors. The choice of sector is therefore a property of the particle species, fixed once and for all, and this is exactly what the spin–statistics theorem supplies in the relativistic theory.

Exchange energy is the quiet engine behind chemistry and magnetism. Hund's first rule — maximise total spin — is just the statement \(K>0\), so aligning spins forces an antisymmetric spatial wavefunction that keeps electrons apart and lowers the Coulomb energy. Coarse-graining the same integral over neighbouring atoms yields the Heisenberg exchange constant \(J_{\text{ex}}\) and hence the sign of ferromagnetic versus antiferromagnetic ordering. The rigidity of solids and the electron degeneracy pressure supporting white dwarfs and neutron stars are the same antisymmetry seen at bulk scale.

At the formal level, the Slater determinant is the ground state of the antisymmetric Fock space, and its structure is what second quantisation makes automatic: fermionic creation operators obey \(\{\hat a_i^\dagger,\hat a_j^\dagger\}=0\), so \((\hat a_i^\dagger)^2=0\) reproduces Pauli exclusion algebraically, and \(\langle\text{vac}|\hat a_N\cdots\hat a_1\,\hat a_1^\dagger\cdots\hat a_N^\dagger|\text{vac}\rangle\) reconstructs the determinant. The exchange integral \(K\) reappears as the Fock term in Hartree–Fock theory, the leading correction that Hartree's mean-field product misses precisely because it ignores antisymmetry.

Common misconceptions. Exchange symmetry does not make particles interact through a mysterious "statistics force"; it constrains which states exist. Two electrons in a triplet are not held apart by a force but by the geometry of an antisymmetric wavefunction (the Fermi/Pauli hole). And identical does not mean "very similar" — it means physically no operator can tell them apart, a strict condition that fails the moment any distinguishing label is introduced.

Worked examples

Example 1 — Singlet–triplet splitting of helium \(1s2s\).

\[ E_{\text{singlet}}=E_0+J+K,\qquad E_{\text{triplet}}=E_0+J-K \]

The \(1s2s\) configuration of He gives a symmetric (para, \(2\,^1\!S\)) and antisymmetric (ortho, \(2\,^3\!S\)) spatial state. A

\[ \Delta E=E_{\text{singlet}}-E_{\text{triplet}}=2K \]

Subtract; the common part \(E_0+J\) cancels, leaving twice the exchange integral. A

\[ K=\frac{\Delta E}{2}=\frac{0.80\ \text{eV}}{2}=0.40\ \text{eV} \]

Insert the measured separation of the \(2\,^1\!S\) (\(20.62\) eV) and \(2\,^3\!S\) (\(19.82\) eV) levels above the ground state, \(\Delta E=0.80\) eV. A

\[ K_{1s2s}\approx 0.40\ \text{eV},\qquad \text{triplet lies }0.80\ \text{eV below singlet} \]

Reading. The purely electrostatic exchange integral is \(0.40\) eV; because \(K>0\) the ortho (triplet) state is the lower — orthohelium is more stable than parahelium in this configuration, and no photon connects them at first order (the \(2\,^3\!S\) metastable state). Units check. \(\Delta E\) in eV, halved, gives \(K\) in eV. Consistent.

Example 2 — Two electrons in a 1D box with a contact interaction.

\[ \varphi_n(x)=\sqrt{\tfrac{2}{L}}\sin\!\frac{n\pi x}{L},\qquad V=g\,\delta(x_1-x_2),\qquad \varepsilon_n=\frac{n^2\pi^2\hbar^2}{2mL^2} \]

Take the lowest two orbitals \(n=1,2\) and a short-range contact interaction of strength \(g\) (units eV·nm). A

\[ J=g\!\int_0^L\!|\varphi_1|^2|\varphi_2|^2\,dx=K=g\!\int_0^L\!\varphi_1\varphi_2\varphi_2\varphi_1\,dx \]

For a delta interaction the direct and exchange integrands are identical (both equal \(g\,\varphi_1^2\varphi_2^2\)), so \(J=K\). B

\[ J=g\,\frac{4}{L^2}\!\int_0^L\!\sin^2\!\frac{\pi x}{L}\,\sin^2\!\frac{2\pi x}{L}\,dx=g\,\frac{4}{L^2}\cdot\frac{L}{4}=\frac{g}{L} \]

Use \(\sin^2\theta=\tfrac12(1-\cos2\theta)\); all oscillating terms integrate to zero over \([0,L]\), leaving \(L/4\). Hence \(J=K=g/L\). B

\[ E_{\text{triplet}}=\varepsilon_1+\varepsilon_2+J-K=\varepsilon_1+\varepsilon_2,\qquad E_{\text{singlet}}=\varepsilon_1+\varepsilon_2+2\frac{g}{L} \]

Insert \(J=K\): the triplet's interaction energy cancels (Pauli hole keeps the electrons from coinciding), the singlet feels the full \(2g/L\). A

\[ \frac{\hbar^2}{2mL^2}=\frac{(1.055\times10^{-34})^2}{2(9.11\times10^{-31})(10^{-9})^2}=6.11\times10^{-21}\ \text{J}=0.0381\ \text{eV} \]

\[ \varepsilon_1=\pi^2(0.0381)=0.376\ \text{eV},\quad \varepsilon_2=4\varepsilon_1=1.505\ \text{eV},\quad \varepsilon_1+\varepsilon_2=1.88\ \text{eV} \]

Numbers for \(L=1\) nm and an electron. Take \(g=1\) eV·nm so \(g/L=1\) eV. A

\[ E_{\text{triplet}}=1.88\ \text{eV},\qquad E_{\text{singlet}}=1.88+2.0=3.88\ \text{eV},\qquad \Delta E_{\text{S–T}}=2\ \text{eV} \]

Reading. The parallel-spin (triplet) pair pays no contact-interaction penalty because antisymmetry forbids the electrons from sitting on top of each other, while the singlet pays \(2g/L\). The exchange effect is here maximal: the whole interaction energy is spin-selective. Units check. \(\varepsilon_n\propto\hbar^2/mL^2\) is energy; \(g/L=\) (eV·nm)/nm \(=\) eV. Both terms in eV. Consistent.

Problems

Show that the exchange operator \(\hat P_{12}\) is Hermitian and unitary, and hence that its eigenvalues are \(\pm1\).
Solution
Hermiticity: \(\langle\Phi|\hat P_{12}\Psi\rangle=\int\Phi^*(x_1,x_2)\Psi(x_2,x_1)\,dx_1dx_2\). Relabel dummy variables \(x_1\leftrightarrow x_2\): \(=\int\Phi^*(x_2,x_1)\Psi(x_1,x_2)\,dx_1dx_2=\langle\hat P_{12}\Phi|\Psi\rangle\), so \(\hat P_{12}^\dagger=\hat P_{12}\). From \(\hat P_{12}^2=\hat I\) we have \(\hat P_{12}^{-1}=\hat P_{12}=\hat P_{12}^\dagger\), so it is unitary. A Hermitian operator with \(\hat P^2=\hat I\) satisfies \(\lambda^2=1\) on eigenstates, giving real eigenvalues \(\lambda=\pm1\).
Write the two-electron Slater determinant for orbitals \(\varphi_a,\varphi_b\), expand it, and show it vanishes when \(\varphi_a=\varphi_b\).
Solution
\(\Psi=\frac{1}{\sqrt2}\begin{vmatrix}\varphi_a(x_1)&\varphi_a(x_2)\\ \varphi_b(x_1)&\varphi_b(x_2)\end{vmatrix}=\frac{1}{\sqrt2}[\varphi_a(x_1)\varphi_b(x_2)-\varphi_a(x_2)\varphi_b(x_1)]\). Setting \(\varphi_a=\varphi_b=\varphi\): \(\Psi=\frac{1}{\sqrt2}[\varphi(x_1)\varphi(x_2)-\varphi(x_2)\varphi(x_1)]=0\). Two electrons cannot occupy the same single-particle state — Pauli exclusion.
A two-orbital configuration has direct integral \(J=11.4\) eV and exchange integral \(K=1.2\) eV, with \(\varepsilon_a+\varepsilon_b=-45.0\) eV. Find the singlet and triplet energies and the splitting, and state which is the ground state.
Solution
\(E_{\text{singlet}}=\varepsilon_a+\varepsilon_b+J+K=-45.0+11.4+1.2=-32.4\) eV. \(E_{\text{triplet}}=\varepsilon_a+\varepsilon_b+J-K=-45.0+11.4-1.2=-34.8\) eV. Splitting \(\Delta E=2K=2.4\) eV. Since \(K>0\), the triplet at \(-34.8\) eV is the ground state (Hund's rule).
Repeat Worked Example 2 for orbitals \(n=1,3\) in the 1D box with contact interaction \(V=g\delta(x_1-x_2)\). Compute \(J=K\) and the singlet/triplet interaction energies.
Solution
\(J=K=g\frac{4}{L^2}\int_0^L\sin^2\frac{\pi x}{L}\sin^2\frac{3\pi x}{L}\,dx\). Using \(\sin^2\theta=\tfrac12(1-\cos2\theta)\), the product is \(\tfrac14[1-\cos\frac{2\pi x}{L}-\cos\frac{6\pi x}{L}+\cos\frac{2\pi x}{L}\cos\frac{6\pi x}{L}]\); the last term \(=\tfrac12[\cos\frac{4\pi x}{L}+\cos\frac{8\pi x}{L}]\). All cosines integrate to zero over \([0,L]\), leaving \(\int=L/4\). Thus \(J=K=g/L\) again. Triplet interaction \(=J-K=0\); singlet interaction \(=2g/L\). (For any \(n\neq m\) the contact integral gives \(g/L\).)
Using Clebsch–Gordan addition of two spin-\(\tfrac12\) particles, write the singlet and triplet spin states and pair each with the correct spatial symmetry so the total two-electron state is antisymmetric. For \(K>0\), which total spin has the lower energy?
Solution
\(\mathbf 2\otimes\mathbf 2=\mathbf 3\oplus\mathbf 1\). Triplet (\(S=1\), symmetric): \(|\!\uparrow\uparrow\rangle,\ \tfrac{1}{\sqrt2}(|\!\uparrow\downarrow\rangle+|\!\downarrow\uparrow\rangle),\ |\!\downarrow\downarrow\rangle\). Singlet (\(S=0\), antisymmetric): \(\tfrac{1}{\sqrt2}(|\!\uparrow\downarrow\rangle-|\!\downarrow\uparrow\rangle)\). Total antisymmetry requires: symmetric spin (triplet) \(\times\) antisymmetric space \(\psi_-\); antisymmetric spin (singlet) \(\times\) symmetric space \(\psi_+\). Energies \(E(\psi_\pm)=\varepsilon_a+\varepsilon_b+J\pm K\), so the triplet (\(\psi_-\)) has \(-K\) and, for \(K>0\), the \(S=1\) state lies lower.