University Physics V · Multi-Electron Atoms · 12.1
Identical Particles & the Exchange Operator
The labels you write on two electrons are yours, not nature's. Here you meet the operator that audits them, learn what its ±1 eigenvalues do and do not prove, and see exactly which line of the standard argument is a theorem and which is an axiom smuggled in beside it.
Build the model
Connect the measurement to the mechanism.
A two-particle wavefunction carries a label nature never issued: it calls one particle 1 and the other 2, and no measurement can ever return that name. The operator that audits the label is the transposition P̂₁₂, which swaps the two slots of the tensor product. It is Hermitian and satisfies P̂₁₂² = Î, so it is unitary too and its spectrum is exactly +1 and −1; on the span of |αβ⟩ and |βα⟩ it is σₓ, with eigenvectors (|αβ⟩ ± |βα⟩)/√2.
Because a Hamiltonian for identical particles is symmetric under the swap — any pair interaction written in |r₁ − r₂| included — [Ĥ, P̂₁₂] = 0, and exchange character is a constant of the motion. That is the whole yield of the dynamics, and it is less than it looks: a conserved quantity is preserved, not chosen. Nothing in the Schrödinger equation forbids a state that is 20% symmetric and 80% antisymmetric; it merely keeps it that way.
Confining nature to the totally symmetric sector or the totally antisymmetric sector, under every transposition at once, is a separate postulate — and for N ≥ 3 a strong one, because the transpositions do not commute with each other and only two irreducible representations of SN are one-dimensional. That axiom is the price non-relativistic quantum mechanics pays; relativistic field theory repays it through spin-statistics, and two-dimensional systems, where exchange is a braid rather than a permutation, decline to pay at all.
- Simple definition
- The exchange operator P̂ᵢⱼ swaps the labels of two identical particles in a state, and because it is Hermitian and squares to the identity its only eigenvalues are +1 for a symmetric state and −1 for an antisymmetric one.
- Example
- In the ordered basis (|αβ⟩, |βα⟩) with ⟨α|β⟩ = 0, P̂₁₂ is the matrix with rows (0, 1) and (1, 0) — literally σₓ — so its characteristic polynomial is λ² − 1, its eigenvalues are +1 and −1, and no third value is available.
Turns a worry about names into an operator you can diagonalise alongside Ĥ.
Acts on the tensor product H ⊗ H; dimensionless. It permutes the slots, never the coordinates inside one slot.
Exchange parity is a two-valued label, not an adjustable dial — but this fixes the spectrum, not the state.
The two together also make P̂₁₂ unitary, since P̂₁₂†P̂₁₂ = P̂₁₂² = Î.
Splits any two-particle state into a symmetric part and an antisymmetric part, and weighs each one.
Each is a projector, since Ŝ₊² = Ŝ₊ and Ŝ₋² = Ŝ₋; ⟨P̂₁₂⟩ = ‖Ŝ₊|Ψ⟩‖² − ‖Ŝ₋|Ψ⟩‖².
Whatever exchange character a state starts with it keeps; no dynamics can push it into a sector.
Holds whenever Ĥ is unchanged by 1 ↔ 2, including any pair interaction written in |r₁ − r₂|.
The extra axiom the dynamics cannot supply, and the root from which exclusion later follows.
Integer spin takes +1, half-integer spin −1. The demand is on all N(N−1)/2 transpositions at once.
Exclusion appears here first — antisymmetry kills a repeated orbital before any statement about occupancy.
For α = β the minus combination vanishes identically; the plus one is |αα⟩ and needs no √2.
A name no apparatus returns
Write ψ(r₁, r₂) and you have already done something nature did not: you named one electron 1 and the other 2. No apparatus returns that name. An electron carries no serial number, and a detector reports a click at a place, not an identity. The formal version of that complaint is an operator: P̂₁₂|α⟩₁|β⟩₂ = |β⟩₁|α⟩₂, or in the position representation (P̂₁₂ψ)(r₁, r₂) = ψ(r₂, r₁). Notice what it swaps. It exchanges the two slots of the tensor product H ⊗ H, not the coordinates inside one slot, and it is defined by its action on product states and then extended linearly to every superposition. Notice also what it does not require. The swap map can be written down for any two particles whose one-particle spaces are the same space — an electron and a proton included, both living in L²(R³) ⊗ C². What makes a pair identical is the extra fact that Ĥ is unchanged by the swap, and only then does the swap commute with the dynamics. So the opening move of this topic is to convert a philosophical worry about names into a linear operator you can diagonalise beside the Hamiltonian.
P̂₁₂ is σₓ in disguise
Two properties do all the work. Hermiticity follows from relabelling the dummy integration variables in ⟨Φ|P̂₁₂Ψ⟩ = ∫∫ Φ*(r₁, r₂) ψ(r₂, r₁) d³r₁ d³r₂, so P̂₁₂† = P̂₁₂ and its eigenvalues are real. Involution is even cheaper: swapping the slots twice restores them, so P̂₁₂² = Î exactly, which forces λ² = 1 and hence λ = ±1. The two facts together make P̂₁₂ unitary as well, since P̂₁₂†P̂₁₂ = Î. Now take orthonormal orbitals α ≠ β and restrict to the two-dimensional span of |αβ⟩ and |βα⟩. In that ordered basis P̂₁₂ carries zeros on the diagonal and ones off it — it is σₓ, the same matrix you diagonalised for spin along the x axis. Its eigenvectors are therefore the familiar pair (|αβ⟩ ± |βα⟩)/√2, and the projectors onto the two sectors are Ŝ₊ = ½(Î + P̂₁₂) and Ŝ₋ = ½(Î − P̂₁₂), built exactly like ½(Î ± σₓ), with Ŝ₊ + Ŝ₋ = Î and Ŝ₊Ŝ₋ = 0.
A commutator conserves; it does not select
For identical particles Ĥ is symmetric under the swap: the kinetic term is a sum over particles, an external potential is the same function for each, and a pair interaction written in |r₁ − r₂| is untouched by relabelling. So [Ĥ, P̂₁₂] = 0, and Ehrenfest's theorem gives d⟨P̂₁₂⟩/dt = 0. That is genuinely useful — exchange parity is a good quantum number, and a state prepared antisymmetric stays antisymmetric under any symmetric interaction, however strong. But read carefully what it does not say. Take |Ψ⟩ = (3|αβ⟩ − |βα⟩)/√10, a perfectly legitimate vector in the two-particle space. Its symmetric weight is ‖Ŝ₊|Ψ⟩‖² = 0.20, its antisymmetric weight 0.80, and ⟨P̂₁₂⟩ = 0.20 − 0.80 = −0.60. Every one of those numbers is frozen for all time by the very commutator that was supposed to help. Conservation preserves what it is handed; it never manufactures an eigenstate. If nature contains no such state, something outside the Schrödinger equation has to say so.
For three particles the transpositions stop commuting
With N = 2 there is one transposition and the story is a two-level system. With N ≥ 3 the permutation group SN is non-abelian, and the postulate changes shape. Write |αβγ⟩ for particle 1 in α, 2 in β, 3 in γ. Then P̂₁₂P̂₂₃|αβγ⟩ = P̂₁₂|αγβ⟩ = |γαβ⟩, while P̂₂₃P̂₁₂|αβγ⟩ = P̂₂₃|βαγ⟩ = |βγα⟩. Those are different states, so [P̂₁₂, P̂₂₃] ≠ 0 and no basis diagonalises every transposition at once. Demanding one simultaneous ±1 eigenvalue for all of them is therefore a real restriction, not bookkeeping: it can be met only inside a one-dimensional irreducible representation of SN, and SN has exactly two, the trivial one and the sign one. For three distinct orbitals the six orderings span six dimensions, decomposing as 6 = 1 + 1 + 2 × 2 — one totally symmetric state, one totally antisymmetric state, and a two-dimensional mixed-symmetry representation appearing twice. The postulate keeps two of those six dimensions and throws away four.
Where the postulate comes from, and where it breaks
Non-relativistic quantum mechanics cannot derive the choice; relativistic quantum field theory can. Pauli's spin-statistics theorem shows that Lorentz invariance, a Hamiltonian bounded below, and microcausality — local observables commuting at spacelike separation — force integer-spin fields to be quantised with commutators and half-integer-spin fields with anticommutators, which is precisely the symmetric and the antisymmetric sector. A composite counts by its constituents, as long as nothing probes its internal structure: helium-4 carries 2 protons, 2 neutrons and 2 electrons, six fermions in all, so it is a boson and goes superfluid at 2.17 K, while helium-3 carries 2 protons, 1 neutron and 2 electrons, five fermions, so it is a fermion and must pair before it can flow — which it manages only in the millikelvin range, at 0.93 mK on the saturated vapour curve and 2.5 mK at the melting pressure. The postulate also has a boundary. In two dimensions the configuration space of two identical particles is not simply connected, exchange is a braid carrying a winding number rather than a permutation, and the exchange phase may be any e(iθ). Those are anyons — not a loophole in the algebra, a different topology.
When indistinguishability actually changes an answer
Symmetrising costs nothing when the one-particle states do not overlap, because every cross term carries an overlap integral that has already vanished — two electrons in two different laboratories may be labelled without error. The scale that decides is the thermal de Broglie wavelength Λ = h/√(2πmkBT) set against the mean spacing n(−1/3). Argon at 300 K and one atmosphere has Λ = 0.016 nm against a spacing of 3.4 nm, so nΛ³ ≈ 1 × 10⁻⁷ and Maxwell–Boltzmann counting is safe. Liquid helium-4 has a spacing of 0.36 nm and Λ = 0.59 nm at 2.17 K, giving nΛ³ = 4.5; an ideal Bose gas at that density would condense at 3.13 K, against the measured λ-point of 2.17 K. Electrons at 300 K have Λ = 4.3 nm, some seventeen times the 0.25 nm spacing of atoms in a metal — which is why conduction electrons are degenerate at room temperature, and why two electrons on one atom can never be treated as distinguishable.
Change one variable at a time
Make the relationship visible.
Start at mixing zero, the labelled product state, and watch the filled dot sitting on the dashed coincidence line. Push the mixing to minus one and that dot drops to exactly zero — for every x₂ and every second orbital, the Fermi hole forced by antisymmetry alone. Push it to plus one and the dot lifts off zero again as ⟨P₁₂⟩ runs from −1 to +1.
EXCHANGE ⟨P₁₂⟩0.00
SYMMETRIC WEIGHT0.50
ANTISYMMETRIC WEIGHT0.50
Ψ AT x₁ = x₂ (SCALED)0.66
Live interpretationEXCHANGE ⟨P₁₂⟩: 0.00. SYMMETRIC WEIGHT: 0.50. ANTISYMMETRIC WEIGHT: 0.50. Ψ AT x₁ = x₂ (SCALED): 0.66
Catch the common trap
Explain before calculating.
Two identical particles have a Hamiltonian symmetric under exchange, so [Ĥ, P̂₁₂] = 0, and P̂₁₂² = Î. Using only those two facts, what can you say about the state |Ψ⟩ = (3|αβ⟩ − |βα⟩)/√10 prepared at t = 0, with α and β orthonormal?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTwo identical particles occupy orthonormal orbitals α and β. Normalise |Ψ⟩ = N(|αβ⟩ + 2|βα⟩) and |Φ⟩ = N(|αβ⟩ + 2i|βα⟩), then find ⟨P̂₁₂⟩ and the symmetric and antisymmetric weights for each. The two differ only in the relative phase of the second coefficient.
- Work in the ordered basis (|αβ⟩, |βα⟩), which is orthonormal, so ⟨Ψ|Ψ⟩ = |c₁|² + |c₂|². For (1, 2) that is 1 + 4 = 5, and for (1, 2i) it is 1 + |2i|² = 5 as well, so both states take N = 1/√5.
- P̂₁₂ swaps the components: (c₁, c₂) ↦ (c₂, c₁). Hence ⟨P̂₁₂⟩ = (c₁*c₂ + c₂*c₁)/(|c₁|² + |c₂|²) = 2 Re(c₁*c₂)/5.
- Real case: 2 Re[(1)(2)]/5 = 4/5 = 0.80. Complex case: 2 Re[(1)(2i)]/5 = 0, because 2i is purely imaginary. Same magnitudes, different relative phase, different answer.
- Weights for |Ψ⟩: Ŝ₊|Ψ⟩ = ½N(3, 3) with squared norm (9 + 9)/(4 × 5) = 0.90, and Ŝ₋|Ψ⟩ = ½N(−1, 1) with squared norm (1 + 1)/(4 × 5) = 0.10. Check: 0.90 + 0.10 = 1 and 0.90 − 0.10 = 0.80, matching step 3.
- Weights for |Φ⟩: Ŝ₊|Φ⟩ = ½N(1 + 2i)(1, 1), squared norm 2|1 + 2i|²/(4 × 5) = 2 × 5/20 = 0.50; Ŝ₋|Φ⟩ = ½N(1 − 2i)(1, −1), squared norm 2 × 5/20 = 0.50. Check: sum 1, difference 0, again matching step 3.
- Neither vector is an eigenvector of P̂₁₂, so a symmetric Ĥ freezes each split exactly where it is. Note what actually moved the answer: not the sizes of the coefficients, which were identical, but the relative phase between the two labellings.
AnswerN = 1/√5 for both. The real combination has ⟨P̂₁₂⟩ = 0.80 and splits 0.90 symmetric / 0.10 antisymmetric; the purely imaginary one has ⟨P̂₁₂⟩ = 0 and splits 0.50 / 0.50. The relative phase of the two labellings, not the coefficient magnitudes, sets the exchange character — and the postulate, not the dynamics, forbids both states.
MediumTwo identical non-interacting particles occupy the n = 1 and n = 2 orbitals of a one-dimensional infinite well of width L. Using ⟨x²⟩₁ = L²(1/3 − 1/2π²), ⟨x²⟩₂ = L²(1/3 − 1/8π²), ⟨x⟩₁ = ⟨x⟩₂ = L/2 and the cross element ⟨x⟩₁₂ = ⟨1|x|2⟩ = −16L/(9π²), compare the rms separation for a symmetric spatial state, an antisymmetric spatial state, and a labelled pair.
- For |Ψ±⟩ = (|1,2⟩ ± |2,1⟩)/√2 with orthonormal orbitals, ⟨(x₁ − x₂)²⟩ = ⟨x²⟩₁ + ⟨x²⟩₂ − 2⟨x⟩₁⟨x⟩₂ ∓ 2|⟨x⟩₁₂|². Only that last term knows about the symmetry.
- Numbers: ⟨x²⟩₁ = (0.33333 − 0.05066)L² = 0.28267L², ⟨x²⟩₂ = (0.33333 − 0.01267)L² = 0.32067L², and 2⟨x⟩₁⟨x⟩₂ = 2(L/2)(L/2) = 0.50000L².
- Labelled pair, with no exchange term: 0.28267 + 0.32067 − 0.50000 = 0.10334L², so the rms separation is √0.10334 L = 0.3215L.
- Exchange term: ⟨x⟩₁₂ = −16L/(9π²) = −0.18013L, so 2|⟨x⟩₁₂|² = 2(0.032446)L² = 0.06489L².
- Symmetric: 0.10334 − 0.06489 = 0.03845L², rms 0.1961L. Antisymmetric: 0.10334 + 0.06489 = 0.16823L², rms 0.4102L.
- Ratio 0.4102/0.1961 = 2.09. The Hamiltonian is identical in all three calculations — the entire effect is the ± sign in the spatial state, so which number applies to two electrons is decided by their spin state, not by the word 'fermion'.
Answerrms separation 0.196L for the symmetric spatial state, 0.321L labelled, 0.410L antisymmetric: a factor of 2.09 between the two sectors, with no interaction term anywhere in Ĥ.
HardThree identical particles occupy three distinct orthonormal orbitals α, β and γ. Show that the transpositions fail to commute, and count how much of the six-dimensional space the symmetrisation postulate keeps.
- Write |αβγ⟩ for particle 1 in α, 2 in β, 3 in γ. The six orderings of α, β and γ span a six-dimensional subspace, one basis vector per element of S₃.
- Apply the transpositions in both orders: P̂₂₃|αβγ⟩ = |αγβ⟩ and then P̂₁₂|αγβ⟩ = |γαβ⟩; the other way round, P̂₁₂|αβγ⟩ = |βαγ⟩ and then P̂₂₃|βαγ⟩ = |βγα⟩.
- |γαβ⟩ ≠ |βγα⟩, so [P̂₁₂, P̂₂₃] ≠ 0. No basis diagonalises every transposition at once, so for N ≥ 3 the eigenvalue of exchange is not one well-defined number.
- A state can still be a simultaneous eigenvector of all transpositions if the representation it spans is one-dimensional. S₃ has exactly two of those: the trivial one, +1 for every transposition, and the sign one, −1 for every transposition.
- Dimension count: 6 = 1 (totally symmetric) + 1 (totally antisymmetric) + 2 × 2, the last term being a two-dimensional mixed-symmetry representation that appears twice.
- So the postulate keeps 2 of the 6 dimensions and discards 4 — the parastatistics no experiment has produced. That discard is the content of the postulate, and no commutator implies it.
Answer[P̂₁₂, P̂₂₃] ≠ 0, since P̂₁₂P̂₂₃|αβγ⟩ = |γαβ⟩ while P̂₂₃P̂₁₂|αβγ⟩ = |βγα⟩. Of the six dimensions the postulate keeps two — one boson state, one fermion state — and throws four away.