University Quantum Mechanics II · Identical Particles and Many-Electron Atoms · 7.1

The exchange operator: why [P, H] = 0 forces no symmetry

Swapping the labels of two identical particles is an operator with eigenvalues ±1 that commutes with every Hamiltonian they can have, and it is tempting to conclude that nature must therefore pick a sign. It need not. Seeing exactly what the commutator guarantees, and what it leaves open, is what makes the next postulate necessary.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Two particles live on Unit 4's tensor-product space, described by one wavefunction Ψ(r₁, r₂) whose square gives the joint probability |Ψ|² d³r₁ d³r₂. The labels 1 and 2 name its arguments, and the exchange operator P swaps them. Swapping twice changes nothing, so P² = 1: P is Hermitian and unitary with eigenvalues +1 and −1, and every state splits into a symmetric and an antisymmetric part with weights (1 ± ⟨P⟩)/2.

For identical particles no observable may depend on the labels, so every observable commutes with P, the Hamiltonian included, and two consequences must be kept apart. The dynamics conserve exchange symmetry exactly: a state of definite symmetry keeps it, both weights stay fixed, and no symmetric operator, however strong, connects the two sectors. But the dynamics do not require a symmetry.

With the particles in different orbitals, ψₘ(r₁)ψₙ(r₂) and ψₙ(r₁)ψₘ(r₂) have the same energy, every combination of them solves the Schrödinger equation, and an interaction that splits the pair keeps both its symmetric and its antisymmetric member. Which sector nature uses is a further postulate, the next topic's. The argument needs particles identical in every property, since unequal masses give PHP ≠ H, and the swap of positions alone commutes with H only while spin is ignored, an approximation good to relative order α².

Simple definition: The exchange operator P swaps the labels of two particles, PΨ(r₁, r₂) = Ψ(r₂, r₁); because P² = 1 its only eigenvalues are +1, for symmetric states, and −1, for antisymmetric ones.
Example: Two electron-mass particles in a 1.00 nm box in ψ₁(x₁)ψ₂(x₂): P turns this into ψ₂(x₁)ψ₁(x₂), an orthogonal state of the same energy 5E₁ = 1.88 eV, so ⟨P⟩ = 0 and the product is exactly half symmetric, half antisymmetric.

The exchange operatorPΨ(r₁, r₂) = Ψ(r₂, r₁), so P² = 1 and the eigenvalues of P are ±1

Derived in one line: PΨ = λΨ gives Ψ = P²Ψ = λ²Ψ. With spin, P swaps every label a particle carries, the spin projection included.

Ψ in m⁻³ for two particles in three dimensions, since |Ψ|² d³r₁ d³r₂ is a probability; r₁, r₂ in m; P is dimensionless.

Symmetric and antisymmetric weightsw_S = (1 + ⟨P⟩)/2 and w_A = (1 − ⟨P⟩)/2, with ⟨P⟩ = ⟨Ψ|P|Ψ⟩

⟨P⟩ is the overlap of a state with its own swap. A product of orthogonal orbitals has ⟨P⟩ = 0, so it is exactly half of each.

Dimensionless: the squared norms of (1 ± P)Ψ/2 for a normalised Ψ, with w_S + w_A = 1.

What identical means for the HamiltonianPHP = H, that is [P, H] = 0, for H = h(1) + h(2) + W(|r₁ − r₂|)

It fails for unequal masses: P turns p̂₁²/2m₁ + p̂₂²/2m₂ into p̂₂²/2m₁ + p̂₁²/2m₂. Only for identical particles is exchange a symmetry of every Hamiltonian they can have.

h(i) = p̂ᵢ²/2m + V(rᵢ), with one mass m and one potential V for both particles; H, h and W in J.

Exchange character is conservedd⟨P⟩/dt = (i/ħ)⟨[H, P]⟩ = 0

Ehrenfest's theorem from Quantum Mechanics I, applied to P: whatever mix of symmetric and antisymmetric the preparation gave is kept for all time.

ħ = 1.055 × 10⁻³⁴ J s; ⟨P⟩ is dimensionless, so the rate is in s⁻¹. It holds for a time-dependent H(t) symmetric at every t.

Symmetric operators never connect the sectors⟨Ψ_S|O|Ψ_A⟩ = 0 whenever [P, O] = 0

Insert POP = O and the element equals (+1)(−1) times itself. No field, collision or interaction that treats the particles alike turns one sector into the other, at any order.

PΨ_S = +Ψ_S and PΨ_A = −Ψ_A; O is any symmetric operator, in its own SI unit.

Exchange degeneracycos θ ψₘ(r₁)ψₙ(r₂) + sin θ ψₙ(r₁)ψₘ(r₂) has energy εₘ + εₙ for every θ

The Schrödinger equation accepts every θ, so it cannot pick the symmetry. Only when m = n is the level single, and then its state is symmetric by force.

εₘ, εₙ one-particle energies in J; each ψ in m⁻³⁄²; θ a dimensionless mixing angle. For m ≠ n the state has ⟨P⟩ = sin 2θ.

Labels are arguments of Ψ, not readings of a detector

Unit 4 put two particles on the tensor-product space: one wavefunction Ψ(r₁, r₂), with |Ψ(r₁, r₂)|² d³r₁ d³r₂ the probability of finding particle 1 in d³r₁ and particle 2 in d³r₂. That is a pure number, so Ψ carries m⁻³ in three dimensions and m⁻¹ for two particles on a line. The labels 1 and 2 name the two arguments. For an electron and a proton they name something physical, because a detector can sort the particles by mass or charge. For two electrons nothing can: a counter that fires at r reports a particle at r, never which one. The formalism writes the labels anyway, and this unit is about what that unobservable bookkeeping may and may not do.

Swapping twice does nothing, so the eigenvalues are ±1

Define P by PΨ(r₁, r₂) = Ψ(r₂, r₁), or on product kets P|a⟩|b⟩ = |b⟩|a⟩. Two swaps restore every state, so P² = 1, and if PΨ = λΨ then Ψ = P²Ψ = λ²Ψ, so λ = +1 or −1. Renaming the integration variables in ∫∫ Ψ′*(r₁, r₂)Ψ(r₂, r₁) d³r₁ d³r₂ shows ⟨Ψ′|PΨ⟩ = ⟨PΨ′|Ψ⟩, so P is Hermitian, and with P² = 1 it is unitary too. The projectors (1 + P)/2 and (1 − P)/2 add to 1 and multiply to zero, so every state splits uniquely into a symmetric and an antisymmetric part, with weights (1 ± ⟨P⟩)/2. For ψₘ(r₁)ψₙ(r₂) with orthonormal orbitals, ⟨P⟩ = |⟨ψₘ|ψₙ⟩|² = 0: an ordinary product is exactly half of each. For particles with spin the spin projection is one more label, and P swaps it along with the position.

Identical particles make every observable symmetric

Conjugating by P moves an operator from one particle to the other: Px̂₁P = x̂₂ and Pp̂₁P = p̂₂. If no measurement can tell the particles apart, no observable may depend on the labels, so every observable obeys POP = O, which is [P, O] = 0. The position of particle 1 alone fails that test and is not an observable of two electrons; x̂₁ + x̂₂, (x̂₁ − x̂₂)² and the number of particles in a region pass it. The Hamiltonian passes automatically: h(1) + h(2) + W(|r₁ − r₂|), with one mass and one potential in both h, is unchanged by the swap. Give the particles different masses and it fails, since P turns p̂₁²/2m₁ + p̂₂²/2m₂ into p̂₂²/2m₁ + p̂₁²/2m₂. One consequence follows at once: ⟨PΨ|O|PΨ⟩ = ⟨Ψ|POP|Ψ⟩ = ⟨Ψ|O|Ψ⟩, so a state and its swap make identical predictions for everything measurable.

A symmetry the dynamics keep exactly

Because [P, H] = 0, P commutes with every power of H and so with U(t) = e^(−iHt/ħ). A state with PΨ(0) = ±Ψ(0) keeps PΨ(t) = UPΨ(0) = ±Ψ(t), and a general state keeps both weights, since (1 ± P)Ψ(t)/2 = U(1 ± P)Ψ(0)/2 and U preserves norms. Stronger, inserting POP = O gives ⟨Ψ_S|O|Ψ_A⟩ = (+1)(−1)⟨Ψ_S|O|Ψ_A⟩, so a symmetric operator never connects the sectors, at any order. Helium is the measured check. Its Coulomb Hamiltonian ignores spin, so the swap of positions alone commutes with it, as does the dipole operator −e(r₁ + r₂). The spatially antisymmetric 1s2p level at 20.96 eV should not radiate to the symmetric 1s² ground state, and the tabulated rate of its 59.1 nm line is 1.8 × 10² s⁻¹, against 1.8 × 10⁹ s⁻¹ for the 58.4 nm line of the symmetric 1s2p level at 21.22 eV. The leak comes from spin-dependent terms of relative order α², quoted from relativistic quantum mechanics, which mix the two levels by roughly the measured fine-structure spread over their 0.25 eV gap: (1.3 × 10⁻⁴ eV / 0.25 eV)² ≈ 3 × 10⁻⁷, the right order.

Exchange degeneracy: the equation cannot choose

Without interaction H₀ = h(1) + h(2), and if hψₙ = εₙψₙ then Φ = ψₘ(r₁)ψₙ(r₂) and its swap PΦ share the energy εₘ + εₙ. For m ≠ n they are orthogonal, and in the basis Φ, PΦ the exchange operator is Quantum Mechanics I's σₓ while H₀ is εₘ + εₙ times the identity. [P, H₀] = 0 promises that common eigenstates exist, (Φ ± PΦ)/√2, but the level also holds cos θ Φ + sin θ PΦ for every θ, with ⟨P⟩ = sin 2θ anywhere from −1 to 1. For two electron-mass particles in a 1.00 nm box, ψ₁ψ₂ and ψ₂ψ₁ share 5E₁ = 1.88 eV. A symmetric interaction W does act: [P, W] = 0 makes W commute with σₓ inside the pair, so W = J I + K σₓ, with I the 2 × 2 identity, J = ⟨Φ|W|Φ⟩ and K = ⟨Φ|W|PΦ⟩, and the good states are (Φ ± PΦ)/√2 at εₘ + εₙ + J ± K to first order. It chooses the basis, not the sector: both survive. Only for m = n is the level single, and Φ = PΦ is symmetric by force.

Same energy, different physics, and no law yet to choose

The members of that level are not one state written differently. For two identical particles sharing 5E₁ in a box, a two-line integral gives the probability that both are found in the left half as ¼ + 0.180 sin 2θ, where 0.180 = 16/(9π²): 0.25 for the plain product, 0.43 for the symmetric combination and 0.07 for the antisymmetric one. It depends on θ only through ⟨P⟩, as every symmetric observable must, because cross terms between the sectors vanish: what can be measured is the two weights and the two normalised parts. Those are exactly what the dynamics conserve. So the Schrödinger equation carries forward whatever exchange character a state starts with and has no mechanism to set it. Were every weight allowed, identical particles could show any both-left probability from 0.07 to 0.43. For real identical particles that are spinless, or share one spin state, experiment finds only the ends, never a mixture, a fact the next topic turns into the symmetrisation postulate, with each species' sign fixed by the spin-statistics connection, quoted from relativistic quantum field theory and not derived here.

Change one variable at a time

Make the relationship visible.

Interactive model

Mixing angle θ30 °

Particle detected at c0.25 L

Upper level n2

Compare θ = 0° with 90°: the labels swap and the picture does not, because only ⟨P⟩ = sin 2θ is visible. At 45° the partner density at the detected particle doubles, at 135° it vanishes, and the energy never moves. At c = 0.50 with n = 2 nothing changes: ψ₂ has a node there.

ENERGY AT EVERY MIXING ANGLE5 E₁

EXCHANGE MEAN ⟨P⟩0.87

SYMMETRIC WEIGHT0.93

PARTNER DENSITY AT DETECTOR7.46 L⁻²

Live interpretationENERGY AT EVERY MIXING ANGLE: 5 E₁. EXCHANGE MEAN ⟨P⟩: 0.87. SYMMETRIC WEIGHT: 0.93. PARTNER DENSITY AT DETECTOR: 7.46 L⁻²

Catch the common trap

Explain before calculating.

Misconception check

[P, H] = 0 is often read as a proof that identical particles must be symmetric or antisymmetric: commuting operators share eigenstates, so the energy eigenstates must be eigenstates of P. That confuses ‘can be chosen’ with ‘must be’. Commuting guarantees a basis of common eigenstates; it does not make every energy eigenstate one of them. When two particles occupy different orbitals the level is exchange degenerate, and cos θ ψₘ(r₁)ψₙ(r₂) + sin θ ψₙ(r₁)ψₘ(r₂) solves the Schrödinger equation for every θ. Where a level is single, as for two particles sharing the lowest orbital or after an interaction splits the pair, its state is forced to be symmetric or antisymmetric, but the spectrum then contains both kinds, and a superposition of a symmetric and an antisymmetric eigenstate is a legal solution whose weights never change. Conservation constrains how a state evolves, never which states may start.

Two identical spinless particles are prepared in Ψ(0) = (√3/2)ψ₁(x₁)ψ₂(x₂) + ½ψ₂(x₁)ψ₁(x₂), with ψ₁ and ψ₂ orthonormal. A weak interaction W(|x₁ − x₂|) is then switched on and left on. What is the probability of finding the pair in the antisymmetric sector at a later time t?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo identical spinless particles share a one-dimensional harmonic trap with ħω = 3.0 meV and do not interact. For the level with n₁ + n₂ = 2, find the energy, list the product states, count the symmetric and antisymmetric states, and give a normalised state of that energy that is neither, with its ⟨P⟩.

Each particle has (nᵢ + ½)ħω, so the pair has (n₁ + n₂ + 1)ħω = 3ħω = 9.0 meV. The product states, first ket for particle 1, are |0⟩|2⟩, |2⟩|0⟩ and |1⟩|1⟩: the level is threefold degenerate.
P exchanges |0⟩|2⟩ and |2⟩|0⟩ and leaves |1⟩|1⟩ alone. The symmetric states are |1⟩|1⟩ and (|0⟩|2⟩ + |2⟩|0⟩)/√2; the only antisymmetric one is (|0⟩|2⟩ − |2⟩|0⟩)/√2. Two symmetric, one antisymmetric.
Check: the trace of P over the level counts the product states it leaves unchanged, here 1, and must equal (+1)(2) + (−1)(1) = 1. It does.
Take Ψ = (|0⟩|2⟩ + |1⟩|1⟩)/√2. Every term has n₁ + n₂ = 2, so H₀Ψ = 3ħωΨ: an energy eigenstate at 9.0 meV. But PΨ = (|2⟩|0⟩ + |1⟩|1⟩)/√2 is neither Ψ nor −Ψ, and ⟨P⟩ = ⟨Ψ|PΨ⟩ = ½(0 + 0 + 0 + 1) = 0.50, only the |1⟩|1⟩ terms overlapping. The weights are 0.75 symmetric and 0.25 antisymmetric.
Limiting case: the lowest level, n₁ + n₂ = 0, holds only |0⟩|0⟩, so there [P, H₀] = 0 does force a symmetric state. Degeneracy is what leaves room for anything else.

AnswerE = 9.0 meV, threefold degenerate: two symmetric states and one antisymmetric. (|0⟩|2⟩ + |1⟩|1⟩)/√2 has the same energy, ⟨P⟩ = 0.50, and weights 0.75 and 0.25.

MediumA double-well trap holds two identical spinless particles. A labelled calculation puts particle 1 in φ_L, a normalised Gaussian centred at x = −d/2, and particle 2 in φ_R, centred at +d/2, each of position spread σ = 0.20 nm, with d = 0.50 nm; their overlap is s = e^{(−d²/8σ²)}. Find ⟨P⟩, the two weights and the normalised symmetric and antisymmetric states, and check both limits of d.

PΨ = φ_L(x₂)φ_R(x₁), so ⟨P⟩ = ∫∫ φ_L(x₁)φ_R(x₂)φ_R(x₁)φ_L(x₂) dx₁ dx₂ = (∫ φ_Lφ_R dx)² = s² for real orbitals. Each φ carries m⁻¹⁄², so s is dimensionless, as a weight must be.
d²/(8σ²) = (0.50 nm)² ÷ (8 × (0.20 nm)²) = 0.78125, so s = e^(−0.78125) = 0.458 and ⟨P⟩ = s² = 0.210.
Weights: w_S = (1 + 0.210)/2 = 0.605 and w_A = (1 − 0.210)/2 = 0.395, which sum to 1.
Normalised parts: (φ_Lφ_R + φ_Rφ_L)/√(2(1 + s²)) = (φ_Lφ_R + φ_Rφ_L)/1.555 and (φ_Lφ_R − φ_Rφ_L)/√(2(1 − s²)) = (φ_Lφ_R − φ_Rφ_L)/1.257. The familiar 1/√2 holds only for orthogonal orbitals.
Limits: for d ≫ σ, s → 0, the weights go to ½ each and 1/√2 returns. For d → 0, s → 1: both particles share one orbital, the product is already symmetric, w_S → 1 and the antisymmetric part vanishes.
Any Hamiltonian that treats the particles alike, whatever the barrier or the interaction, keeps 0.605 and 0.395 fixed for all time.

Answer⟨P⟩ = s² = 0.210, so w_S = 0.605 and w_A = 0.395. The normalised states are (φ_Lφ_R ± φ_Rφ_L) divided by 1.555 and by 1.257, not by √2.

HardTwo identical spinless particles of electron mass in a box of width L = 1.00 nm interact weakly through W = g δ(x₁ − x₂), with g = 8.0 × 10⁻³⁰ J m. Use [P, W] = 0 to find the first-order energies and good states of the level at 5E₁. Show that the antisymmetric energy is exact, and follow a solution that starts as ψ₁(x₁)ψ₂(x₂).

E₁ = h²/(8mL²) = 0.376 eV, so the unperturbed level is 5E₁ = 1.880 eV, holding Φ = ψ₁(x₁)ψ₂(x₂) and PΦ. W is symmetric, so inside the pair it commutes with P = σₓ and must be J I + K σₓ, with J = ⟨Φ|W|Φ⟩ and K = ⟨Φ|W|PΦ⟩. The good states are (Φ ± PΦ)/√2, shifted by J ± K.
The delta function sets x₁ = x₂ in both integrals, so J = K = g ∫ ψ₁²ψ₂² dx = g(4/L²)(L/π)(π/4) = g/L = 8.0 × 10⁻²¹ J = 0.050 eV, an energy as J m ÷ m must be.
E_S = 1.880 + 2 × 0.050 = 1.98 eV and E_A = 1.880 + 0 = 1.88 eV. Even the larger shift, 0.10 eV, is small beside the 1.13 eV gap to the nearest other levels, so first order is reliable.
The antisymmetric result is exact: any antisymmetric Ψ has Ψ(x, x) = −Ψ(x, x) = 0, so WΨ_A = 0 identically, and the antisymmetric eigenstates of H₀ stay eigenstates of H₀ + W with unchanged energy, at every order.
Write Φ = (Ψ_S + Ψ_A)/√2. Each part only gains its own phase, so the probability of PΦ at time t is sin²(ΔE t/2ħ) with ΔE = 2g/L = 0.0999 eV, reaching 1 at t = h/(2ΔE) = 20.7 fs. The symmetric weight stays ½ throughout: the labels trade places, the sectors do not.
Check: as g → 0 the splitting vanishes, the transfer time diverges, and the exchange degeneracy returns.

AnswerE_S = 1.98 eV and E_A = 1.88 eV (exact), with good states (Φ ± PΦ)/√2. The solution that starts as ψ₁(x₁)ψ₂(x₂) becomes ψ₂(x₁)ψ₁(x₂) after 21 fs, its symmetric weight fixed at 0.50.