University Quantum Mechanics II · Identical Particles and Many-Electron Atoms · 7.2

The symmetrisation postulate and spin-statistics

The previous lesson proved that exchange symmetry, once present, survives; it never said which symmetry, or that there must be one. This lesson supplies the postulate that decides, marks the one link to spin that has to be imported from field theory, and derives why an atom built entirely of fermions can be a boson.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

On its own, quantum mechanics cannot say what state two identical particles in orbitals a and b are in. The kets |ab⟩, |ba⟩ and every superposition α|ab⟩ + β|ba⟩ have the same energy, no symmetric observable tells |ab⟩ from |ba⟩, and yet the superpositions predict different things: the chance of finding both at one point is twice the distinguishable value for α = β and zero for α = −β. The symmetrisation postulate ends this exchange degeneracy by decree: the state of N identical particles must be an eigenvector of every exchange Pᵢⱼ, with eigenvalue +1 for bosons and −1 for fermions.

Two parts of that are theorems. Pᵢⱼ² = 1 limits the eigenvalue to ±1, and P₁₃ = P₁₂P₂₃P₁₂ forces one sign on every pair. The assumption is that a common eigenvector is required at all, and from three particles on it bites: of the 6 states that three particles in three different orbitals can form, 4 have mixed symmetry and are forbidden to every species.

Which species takes which sign is the spin-statistics connection, ε = (−1)^(2s), proved only in relativistic quantum field theory and used here as an empirical rule without known exception. Derived here: swapping two bound composites swaps their constituents pair by pair, so a composite of n fermions carries (−1)ⁿ. So helium-4 is a boson and helium-3 a fermion, while the atoms stay too cold and too far apart for their insides to matter.

Simple definition: The symmetrisation postulate requires a state of identical particles to be unchanged by every pairwise exchange (bosons) or to change sign under every one (fermions); the spin-statistics connection, quoted from field theory, makes integer-spin particles bosons and half-integer-spin ones fermions.
Example: Three identical particles in three different orbitals have 3! = 6 independent product states. The postulate keeps one combination, the symmetric one for bosons or the antisymmetric one for fermions, and forbids the 4 mixed-symmetry states to both.

The symmetrisation postulatePᵢⱼ|ψ⟩ = ε|ψ⟩ for every pair i ≠ j, with ε = +1 or ε = −1

Postulated: [Pᵢⱼ, H] = 0 only preserves a symmetry the state already has. ε = +1 defines bosons and ε = −1 fermions, one value per species.

Pᵢⱼ swaps all coordinates of particles i and j, position and spin together; Pᵢⱼ and ε are dimensionless

One sign for every pairP₁₃ = P₁₂P₂₃P₁₂, so ε₁₃ = ε₁₂ε₂₃ε₁₂ = ε₂₃

Derived: any swap is another swap relabelled, so all share one sign, and a permutation built from k swaps multiplies the state by εᵏ.

εᵢⱼ is the eigenvalue of Pᵢⱼ on a state that is an eigenvector of every exchange; Pᵢⱼ² = 1 makes each εᵢⱼ = ±1

What three particles losed³ = (d + 2)(d + 1)d/6 + d(d − 1)(d − 2)/6 + 2d(d² − 1)/3

Derived by counting. For two particles the symmetric and antisymmetric counts already sum to d², so mixed symmetry first appears at N = 3; at d = 3 it holds 16 of 27 states.

Three particles sharing d orthonormal one-particle states; the terms count symmetric, antisymmetric and mixed states

The spin-statistics connection, quotedε = (−1)^(2s)

Quoted from relativistic quantum field theory, where Lorentz invariance, a lowest energy and causality force it. Non-relativistic quantum mechanics can state it but not prove it.

s is the spin quantum number: s = 0, 1, 2, … gives ε = +1 (bosons) and s = ½, 3/2, … gives ε = −1 (fermions)

Exchange sign of a bound compositeε_AB = (−1)^(n_F)

Derived: swapping composites A and B is n_F fermion swaps, each worth −1. For a neutral atom the neutron number N decides: ⁴He (N = 2) is a boson, ³He (N = 1) a fermion.

n_F counts the fermions in each composite; bosons inside contribute +1. A neutral atom has n_F = 2Z + N

Spin states of two identical nucleiN_S = (2I + 1)(I + 1) symmetric and N_A = (2I + 1)I antisymmetric

Derived: the 2I + 1 states with m₁ = m₂ are symmetric and the rest pair into ± combinations. Rotational levels inherit the ratio (I + 1)/I, and which J gets it reveals ε.

I is the nuclear spin quantum number; N_S + N_A = (2I + 1)², every spin state of the pair

A smaller state space, which the dynamics never leave

From the previous topic, in one line: every observable of identical particles commutes with each exchange Pᵢⱼ, so the dynamics conserve exchange symmetry but never choose it, and exchange-degenerate states such as α|ab⟩ + β|ba⟩ make different predictions. The symmetrisation postulate chooses by shrinking the space. The states of N identical particles are not the whole N-fold tensor product but only its totally symmetric subspace, for bosons, or its totally antisymmetric one, for fermions. The conservation law makes such a rule safe to impose: no Hamiltonian and no measurement of identical particles leads out of either subspace, so a restriction imposed once holds for ever. It is still a bold claim: whole families of states must be missing. An ¹⁶O nucleus has spin 0 and is a boson, and O₂'s ground electronic state changes sign when its nuclei are swapped (quoted from molecular spectroscopy). A rotational level N multiplies that by (−1)^N, two bosons need +1 overall, so only odd N may exist. Every even-N level is indeed absent from O₂'s spectrum, and searches for them have found none.

The postulate, and the two parts of it that are theorems

Written as an equation, the postulate demands that the state of N identical particles be an eigenvector of every exchange, Pᵢⱼ|ψ⟩ = ε|ψ⟩, and it calls ε = +1 bosons and ε = −1 fermions. Two parts follow by algebra. Swapping twice restores every coordinate, so Pᵢⱼ² = 1 and ε = ±1, never a phase such as i. And every pair shares the sign. Follow a product state: P₁₂ turns |xyz⟩ into |yxz⟩, P₂₃ then gives |yzx⟩, and P₁₂ again gives |zyx⟩, which is P₁₃|xyz⟩. So P₁₃ = P₁₂P₂₃P₁₂, and on a common eigenvector ε₁₃ = ε₁₂ε₂₃ε₁₂ = ε₂₃. Any permutation is a string of swaps, so it multiplies the state by ε to the number of swaps: the cyclic relabelling 1 → 2 → 3 → 1 is two swaps and leaves even a fermion state unchanged. Treating exchange as relabelling is safe in three dimensions; in two, particles carried around one another can pick up other phases, and the anyons of fractional quantum Hall systems do. What no algebra supplies is the demand itself.

Three particles: the states the postulate forbids

For two particles the postulate only picks a side: (1 + P₁₂)/2 and (1 − P₁₂)/2 split every state into a symmetric and an antisymmetric part, and nothing is left over. From three particles on, something is. Place three particles in different orbitals a, b and c. Their 3! = 6 product states hold one symmetric combination, one antisymmetric, and 4 dimensions of mixed symmetry. A sample is |M⟩ = (2|abc⟩ + 2|bac⟩ − |acb⟩ − |bca⟩ − |cab⟩ − |cba⟩)/√12, which P₁₂ leaves alone, while ⟨M|P₂₃|M⟩ = −½: P₂₃ carries it into a different state, 120° away. Mixed states come in planes on which P₁₂, P₂₃ and P₁₃ act as three mirrors at 60° to one another, and no direction in such a plane is kept or reversed by all three. Particles allowed into them would obey parastatistics, a consistent alternative in which a whole plane of vectors describes one physical state. The postulate excludes them, and no known particle needs them. With d states per particle, 2d(d² − 1)/3 of the d³ three-particle states are mixed: 16 of 27 when d = 3.

Which sign: the spin-statistics connection is imported

The postulate says each species has one sign; it does not say which. Measurement does: electrons, protons, neutrons, neutrinos and quarks, all spin ½, are fermions, while photons and the W and Z (spin 1), and pions and the Higgs boson (spin 0), are bosons. The rule ε = (−1)^(2s) is the spin-statistics connection, and this course quotes it. Fierz in 1939 and Pauli in 1940 proved it in relativistic quantum field theory, from Lorentz invariance, an energy bounded below, and the requirement that measurements at spacelike separation cannot disturb one another. Non-relativistic quantum mechanics has an energy bounded below but neither Lorentz invariance nor a light cone to define spacelike separation, and in it a Hamiltonian for spinless fermions is perfectly consistent, so no derivation inside this course can be complete. The familiar heuristic, that a 2π rotation gives e^{(−2πiJ_z/ħ)} = (−1)^(2j) (Unit 5) and an exchange resembles half a turn of the pair about its midpoint, reproduces the pattern without proving it. The rule has no known exception: searches for Pauli-forbidden X-ray lines from electrons in copper find none.

Composites: count the fermions

A helium-4 atom holds two protons, two neutrons and two electrons. Exchange two such atoms, A and B, and every constituent of A trades places, spin and all, with its counterpart in B: the exchange is a product of six commuting swaps, two of protons, two of neutrons and two of electrons. The postulate applied to each species makes every swap worth −1, so the atoms exchange with (−1)⁶ = +1 and are bosons; helium-3, one neutron short, gives (−1)⁵ = −1. In general a composite of n_F fermions has ε = (−1)^(n_F), however many bosons it holds, and for a neutral atom n_F = 2Z + N, so the neutron number decides. The rule agrees with spin-statistics, as it must: spin-½ constituents and integer orbital angular momenta add to a half-integer total exactly when n_F is odd, so ⁴He has J = 0 and ³He has J = ½. The rule once overturned a model of the nucleus. In 1929 the rotational Raman lines of N₂ showed nitrogen-14 nuclei to be bosons of spin 1, impossible for 14 protons and 7 electrons, 21 fermions, and natural for 7 protons and 7 neutrons.

Where treating a composite as one particle stops

The sign (−1)^(n_F) for a whole-composite swap is exact. What is approximate is treating the composite as a single particle with no insides, because the full state must also change sign when one electron of atom A is swapped with one of atom B while everything else stays put. Those partial swaps give terms weighted by the overlap of the two atoms' wavefunctions, and they vanish when the atoms are far apart compared with their size and too slow to excite one another. Helium-4 meets both conditions by a wide margin. Its first excited level lies 19.8 eV up, against k_B T = 1.87 × 10⁻⁴ eV at the 2.17 K superfluid transition, a ratio of 1.1 × 10⁵; and in the liquid the atoms sit 0.36 nm apart, nine times the ⟨r⟩ = 0.040 nm of a hydrogenic 1s orbital with Z = 2. Push two atoms together and the picture fails: their steep short-range repulsion arises from the electrons' antisymmetry across both atoms, which no point-boson model contains. Helium-3 shows no transition near 2 K and turns superfluid only in the millikelvin range, once its atoms pair up into two-fermion units.

Change one variable at a time

Make the relationship visible.

Interactive model

Direction θ in the mixed plane0 °

Weight m in the mixed plane1.00

Allowed sector s: +1 sym, −1 antisym1

At m = 1 and θ = 0° the state is symmetric under P₁₂, yet the other two bars sit at −0.50. Sweep θ: no angle brings all three bars to ±1, because their sum stays 0. Then slide m to 0, and only there do all three reach s.

⟨P₁₂⟩1.00

⟨P₂₃⟩-0.50

⟨P₁₃⟩-0.50

SUM, ±3 WHEN ALLOWED0.00

Live interpretation⟨P₁₂⟩: 1.00. ⟨P₂₃⟩: −0.50. ⟨P₁₃⟩: −0.50. SUM, ±3 WHEN ALLOWED: 0.00

Catch the common trap

Explain before calculating.

Misconception check

The tempting shortcut runs: swapping twice restores the state, so P² = 1, so every state of identical particles is symmetric or antisymmetric, and the postulate is a theorem. The algebra proves much less. P² = 1 says that an eigenvalue of P, if the state has one, is +1 or −1; it does not make the state an eigenvector. For two particles the gap is easy to miss, because every state is a sum of a symmetric and an antisymmetric part and the postulate merely keeps one. For three it cannot be missed: (2|abc⟩ + 2|bac⟩ − |acb⟩ − |bca⟩ − |cab⟩ − |cba⟩)/√12 is unchanged by P₁₂, yet ⟨P₂₃⟩ = −½, so it is not an eigenvector of P₂₃ at all, although P₂₃² = 1 holds as always. Demanding an eigenvector of every exchange is a genuine postulate, and tying the sign to spin is a second input, borrowed from field theory.

Potassium-40 has Z = 19, 21 neutrons and nuclear spin I = 4. Gases of neutral ⁴⁰K atoms have been cooled below 1 μK. Which statement gives both the statistics the atoms obey under exchange and a reason that holds for every composite?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyUse the composite rule to decide whether each of these obeys Bose or Fermi statistics, then test each verdict against the measured angular momentum of its ground state: the helium-3 atom, the ³He⁺ ion, the deuterium atom, and the alpha particle ⁴He²⁺.

The rule derived in this lesson is ε = (−1)^(n_F): count every proton, neutron and electron, each a spin-½ fermion.
Helium-3 atom: 2 + 1 + 2 = 5 fermions, so ε = −1, a fermion. Check: its two 1s electrons pair to J = 0, so the atom's angular momentum is the nuclear spin ½, a half-integer, as spin-statistics requires.
³He⁺ ion: removing one electron leaves 2 + 1 + 1 = 4 fermions, so ε = +1, a boson. Check: nuclear spin ½ and electron J = ½ combine to F = 0 or 1, both integers; the measured 8.67 GHz ground-state hyperfine line joins those two levels.
Deuterium atom: 1 + 1 + 1 = 3 fermions, so ε = −1, a fermion. Check: deuteron spin 1 and electron J = ½ give F = ½ or 3/2, and the measured 327 MHz hyperfine line joins them.
Alpha particle: 2 + 2 + 0 = 4 fermions, so ε = +1, a boson, and its measured nuclear spin is 0.

Answer³He atom: fermion (5 fermions, J = ½). ³He⁺: boson (4, F = 0 or 1). Deuterium atom: fermion (3, F = ½ or 3/2). Alpha particle: boson (4, I = 0). Removing one electron turns helium-3 from a fermion into a boson.

MediumThree identical spinless particles occupy three orthonormal orbitals a, b and c, one each. Find the fractions of the product state |abc⟩ lying in the symmetric, antisymmetric and mixed-symmetry subspaces. Repeat for |χ⟩ = (|abc⟩ + |bac⟩)/√2, find ⟨χ|P₂₃|χ⟩, and say what a boson or a fermion state could keep of |χ⟩.

The symmetriser S = (1/6)Σ P and the antisymmetriser A = (1/6)Σ (−1)ᵏP, summed over the six permutations P with k swaps each, are projectors: S² = S and A² = A. So the fraction of |φ⟩ in each subspace is ⟨φ|S|φ⟩ or ⟨φ|A|φ⟩.
For |abc⟩ only the identity returns |abc⟩; the other five permutations give orthogonal product states. So ⟨abc|S|abc⟩ = ⟨abc|A|abc⟩ = 1/6, and the mixed fraction is 1 − 2/6 = 2/3.
For |χ⟩, S absorbs any permutation, so S|bac⟩ = S|abc⟩ and S|χ⟩ = √2 S|abc⟩, giving ⟨χ|S|χ⟩ = 2 × 1/6 = 1/3. A|bac⟩ = −A|abc⟩, so A|χ⟩ = 0, and the mixed fraction is again 2/3.
P₂₃ sends |abc⟩ to |acb⟩ and |bac⟩ to |bca⟩, both orthogonal to |χ⟩, so ⟨χ|P₂₃|χ⟩ = 0. P₂₃ keeps each subspace to itself, so no cross terms appear: 0 = (1/3)(+1) + (2/3)⟨P₂₃⟩ₘ, and the mixed part has ⟨P₂₃⟩ₘ = −½, its overlap with its own swapped copy.
The postulate lets a boson state keep only the symmetric third, renormalised to (|abc⟩ + |bac⟩ + |acb⟩ + |bca⟩ + |cab⟩ + |cba⟩)/√6. No fermion state can be made from |χ⟩ at all, because it is already symmetric in particles 1 and 2.

Answer|abc⟩: 1/6 symmetric, 1/6 antisymmetric, 2/3 mixed. |χ⟩: 1/3, 0 and 2/3, with ⟨χ|P₂₃|χ⟩ = 0 and ⟨P₂₃⟩ = −½ in its mixed part. A boson keeps the symmetric third; a fermion keeps nothing.

HardIn their ground electronic state (¹Σg⁺, unchanged by swapping the nuclei), H₂ and D₂ have rotational levels E_J = hcBJ(J + 1), with B = 59.3 cm⁻¹ for H₂ and 29.9 cm⁻¹ for D₂. Protons have spin ½ and deuterons spin 1. Find which nuclear-spin states each J pairs with, the high-temperature ratio of the two kinds, and the equilibrium fraction of each gas in even-J levels at 77.4 K.

Statistics: protons are spin-½ fermions, a quoted spin-statistics result. A deuteron is one proton and one neutron, two fermions, so the composite rule makes it a boson, consistent with its integer spin 1.
Swapping the nuclei reverses the internuclear vector, which multiplies every spherical harmonic of degree J by (−1)^J; the ¹Σg⁺ electronic state contributes +1 (quoted). Nuclear-spin states: (2I + 1)(I + 1) symmetric and (2I + 1)I antisymmetric, so 3 and 1 for H₂, 6 and 3 for D₂.
H₂ must change sign: even J pairs with the 1 antisymmetric spin state (para), odd J with the 3 symmetric ones (ortho). D₂ must not: even J pairs with the 6 symmetric states (ortho), odd J with the 3 antisymmetric (para). At high temperature: 3 : 1 odd to even for H₂ and 2 : 1 even to odd for D₂, the ratios measured in normal hydrogen and deuterium.
Rotational temperatures θ = hcB/k_B = (1.4388 cm K)B, a temperature since cm K × cm⁻¹ = K, are 85.3 K for H₂ and 43.0 K for D₂. With Zₑ = Σ(2J + 1)e^{(−J(J+1)θ/T)} over even J and Zₒ the same sum over odd J, H₂ at 77.4 K (θ/T = 1.1023) has Zₑ = 1 + 5e^(−6.614) = 1.0067 and Zₒ = 3e^(−2.2046) = 0.3309.
H₂ even-J (para) fraction: 1.0067/(1.0067 + 3 × 0.3309) = 0.504. D₂ (θ/T = 0.5558): Zₑ = 1 + 5e^(−3.335) + 9e^(−11.12) = 1.1782 and Zₒ = 3e^(−1.1116) + 7e^(−6.670) = 0.9960, so the even-J (ortho) fraction is 6 × 1.1782/(6 × 1.1782 + 3 × 0.9960) = 0.703.
Limits: as T grows the two sums become equal and the fractions tend to 1/4 and 2/3; as T → 0 both gases settle into J = 0, pure para-H₂ and pure ortho-D₂. At 20.4 K the same sums give 0.998 and 0.978, the compositions the liquids relax to. Conversion needs a nuclear spin flip, so without a catalyst it is slow and quickly cooled gas keeps its high-temperature ratio.

AnswerH₂: even J with the nuclear singlet (para), odd J with the triplet (ortho), 3 : 1 at high T. D₂: even J with 6 symmetric states, odd J with 3, 2 : 1. At 77.4 K the even-J fractions are 0.504 (H₂) and 0.703 (D₂).