University Quantum Mechanics II · Identical Particles and Many-Electron Atoms · 7.4

Exchange effects and mean separation in a well

This is the topic where exchange symmetry stops being bookkeeping and starts moving particles about. Two particles that never touch, in the simplest well there is, sit closer together as bosons and further apart as fermions; the Hamiltonian is the same in every case, and only the symmetry of the state has changed.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Put two particles in one infinite well and give them no interaction at all, so that H = h(1) + h(2) and every product ψₙ(x₁)ψₘ(x₂) is an eigenstate with energy K(n² + m²). The only thing left to choose is the exchange symmetry of the state, and it is enough to change what is measured. The ground state costs 2K for distinguishable particles or bosons and 5K for spinless fermions, because antisymmetry empties the doubly occupied level.

Expanding ⟨(x₁ − x₂)²⟩ in a symmetrised state gives the distinguishable answer plus one extra term, ∓2|⟨x⟩_ab|², built from the off-diagonal element ⟨a|x|b⟩; in levels 1 and 2 the rms separation is 0.196a for bosons, 0.321a for distinguishable particles and 0.410a for fermions. Nothing pushed them. The symmetric and antisymmetric states have exactly the same energy, and the density of particles along the well, |ψₐ|² + |ψ_b|², is the same in all three cases; what differs is the joint probability, a correlation of ±0.63 between x₁ and x₂.

The model stops where its two assumptions stop. Add an interaction and the correlation becomes energy, which is how helium's 1s2s triplet comes to lie 0.80 eV below its singlet. And every exchange term is built from the overlap product ψₐ*ψ_b, so when the two orbitals do not overlap the symmetrised state predicts exactly what a product state does, which is why an electron here is never antisymmetrised against one on the Moon.

Simple definition: An exchange effect is a difference between identical and distinguishable particles that comes from the symmetry of the state alone, with no interaction in the Hamiltonian; for positions it appears as a correlation between where the two particles are found.
Example: In levels 1 and 2 of a 1.00 nm well, two spinless fermions have rms separation 0.410 nm, distinguishable particles 0.321 nm and bosons 0.196 nm, yet for electron-mass particles all three pairs have the same energy, 5K = 1.88 eV.

Ground state of two spinless fermionsE₀ = K(1² + 2²) = 5K

Antisymmetry empties the doubly occupied level n = 1, so 3K is paid for symmetry alone: 1.13 eV for electron-mass particles in a 1.00 nm well.

K = π²ħ²/(2ma²) in J, with m the particle mass in kg and a the well width in m; bosons and distinguishable particles have E₀ = 2K

Mean-square separation of a symmetrised pair⟨(x₁ − x₂)²⟩± = ⟨x²⟩ₐ + ⟨x²⟩_b − 2⟨x⟩ₐ⟨x⟩_b ∓ 2|⟨x⟩_ab|²

Derived here. The first three terms are the distinguishable result; symmetry adds only the last, equal in size and opposite in sign for the two kinds of particle.

Upper sign symmetric (bosons), lower antisymmetric (fermions); a ≠ b orthonormal levels; every term in m²

Exchange element in the infinite well⟨x⟩ₙₘ = −8nma/(π²(n² − m²)²) for n + m odd, and 0 for n + m even

Derived by product-to-sum. For even n + m the product ψₙψₘ is even about the centre while x − a/2 is odd, so the element vanishes by parity.

⟨x⟩ₙₘ = ⟨n|x|m⟩ = ∫ψₙ x ψₘ dx, in m; ⟨x⟩₁₂ = −16a/(9π²) = −0.180a

Distinguishable pair in levels n and m⟨(x₁ − x₂)²⟩ = a²[1/6 − (1/n² + 1/m²)/(2π²)]

0.1033a² for levels 1 and 2, an rms of 0.321a. As n and m grow it tends to a²/6, the value for two independent positions spread evenly over the well.

a in m; built from ⟨x⟩ₙ = a/2 and ⟨x²⟩ₙ = a²(1/3 − 1/(2n²π²)), quoted from Quantum Mechanics I

Both particles in one region RP±(R) = Pₐ(R)P_b(R) ± |∫_R ψₐ*ψ_b dx|²

Over the whole well the overlap integral is zero and P = 1. In the left half, for levels 1 and 2: 0.430 for bosons, 0.250 distinguishable, 0.070 for fermions.

Pₐ(R) = ∫_R |ψₐ|² dx; all dimensionless; upper sign for bosons, lower for fermions

Joint density where the particles coincide|Ψ±(x, x)|² = (1 ± 1)|ψₐ(x)ψ_b(x)|²

Bosons are twice as likely to be found together and fermions never are: the exchange hole. A contact force acts only on this line, so it never acts on spinless fermions.

In m⁻²; the factor is 2 for bosons and 0 for fermions, against 1 for distinguishable particles

Symmetry alone moves the ground state from 2K to 5K

Quantum Mechanics I supplies the one-particle problem, quoted here: ψₙ(x) = √(2/a) sin(nπx/a) with Eₙ = Kn² and K = π²ħ²/(2ma²). Leave out any interaction, so H = h(1) + h(2) and ψₙ(x₁)ψₘ(x₂) is an eigenstate with energy K(n² + m²). Distinguishable particles take ψ₁ψ₁ at 2K. Two bosons may too, since ψ₁ψ₁ is already symmetric. Two spinless fermions may not: antisymmetrising ψ₁ψ₁ gives zero (topic 7.3), so the lowest allowed state is (1, 2)− at K + 4K = 5K. At the next level, 5K, distinguishable particles have two states, ψ₁ψ₂ and ψ₂ψ₁; bosons keep only their symmetric combination and fermions only the antisymmetric one, and at 8K the state (2, 2) exists for bosons but not for fermions. For electron-mass particles in a 1.00 nm well K = 0.376 eV, so the ground state is 0.752 eV or 1.880 eV. No charge or coupling constant enters the 3K = 1.13 eV difference, so it is of order α⁰. Two electrons with both spins up behave as spinless fermions: electrons are fermions by the spin-statistics connection (quoted, topic 7.2), and |↑↑⟩ is symmetric, so the spatial factor must be antisymmetric.

The mean-square separation, term by term

Take Ψ± = [ψₐ(x₁)ψ_b(x₂) ± ψ_b(x₁)ψₐ(x₂)]/√2 with a ≠ b orthonormal, and expand ⟨(x₁ − x₂)²⟩ = ⟨x₁²⟩ + ⟨x₂²⟩ − 2⟨x₁x₂⟩. In ⟨x₁²⟩ each cross term carries ⟨b|a⟩ = 0 from the particle the operator does not touch, so ⟨x₁²⟩ = ½(⟨x²⟩ₐ + ⟨x²⟩_b), exactly as for one particle equally likely to be in either level; ⟨x₂²⟩ is the same. In ⟨x₁x₂⟩ the operator touches both particles and nothing is killed: the direct terms give ⟨x⟩ₐ⟨x⟩_b and the cross terms give ±⟨a|x|b⟩⟨b|x|a⟩ = ±|⟨x⟩_ab|². So ⟨(x₁ − x₂)²⟩± = ⟨x²⟩ₐ + ⟨x²⟩_b − 2⟨x⟩ₐ⟨x⟩_b ∓ 2|⟨x⟩_ab|². The first three terms are what the product ψₐ(x₁)ψ_b(x₂) gives for distinguishable particles, so symmetry contributes one term, the same size for bosons and fermions and opposite in sign. This is derived, and it says where to look: one-particle averages never show exchange, and two-particle ones show it only through the off-diagonal element ⟨a|x|b⟩. Every term is a length squared, as it must be.

Levels 1 and 2 in numbers

In the well ⟨x⟩ₙ = a/2, because every |ψₙ|² is symmetric about the centre, and ⟨x²⟩ₙ = a²(1/3 − 1/(2n²π²)), both quoted from Quantum Mechanics I. The distinguishable result is then a²[1/6 − (1/n² + 1/m²)/(2π²)], which for levels 1 and 2 is 0.1033a². The exchange element is derived: product-to-sum turns sin(nπx/a) sin(mπx/a) into two cosines, and ∫₀^a x cos(kπx/a) dx = (a/kπ)²[(−1)^k − 1] then gives ⟨x⟩ₙₘ = −8nma/(π²(n² − m²)²) when n + m is odd and zero when it is even. So ⟨x⟩₁₂ = −16a/(9π²) = −0.1801a and 2|⟨x⟩₁₂|² = 0.0649a². Bosons: 0.0385a², rms 0.196a. Distinguishable: 0.1033a², rms 0.321a. Fermions: 0.1682a², rms 0.410a. In a 1.00 nm well those are 0.196, 0.321 and 0.410 nm. A limiting case checks the distinguishable formula: as n and m grow it tends to a²/6, the value for two independent positions spread evenly across the well, rms 0.408a.

Where the difference lives: the joint density

Write out the joint density: |Ψ±(x₁, x₂)|² = ½[|ψₐ(x₁)ψ_b(x₂)|² + |ψ_b(x₁)ψₐ(x₂)|²] ± Re[ψₐ*(x₁)ψ_b(x₁)ψ_b*(x₂)ψₐ(x₂)]. Integrate over x₂ and the last term dies by orthogonality, so each particle on its own has density ½(|ψₐ|² + |ψ_b|²) whether the pair are bosons, fermions or distinguishable particles put into the two levels at random (in the product ψₐ(x₁)ψ_b(x₂) particle 1 would have |ψₐ|² alone). No measurement on one particle can tell the three apart. The difference sits in where the two are found together. On the line x₁ = x₂ the boson density is 2|ψₐ(x)ψ_b(x)|², twice the distinguishable value, and the fermion density is zero, the exchange hole. Away from the line the same term tilts the whole distribution: in the well both means are a/2, so ±|⟨x⟩₁₂|² = ±0.0324a² is exactly the covariance of x₁ and x₂, and dividing by the variance 0.0517a² of each position gives a correlation coefficient of +0.63 for bosons and −0.63 for fermions. Found on the left, a boson's partner is probably on the left too; a fermion's is probably on the right.

It costs nothing, until something interacts

The states (1, 2)+ and (1, 2)− have exactly the same energy, 5K, although one keeps its particles 0.196a apart and the other 0.410a. A force that pushed the fermions apart would have done work, and the two energies would differ; there is nothing in H to do it. That is why the effect is of order α⁰: no charge enters it. Switch on a repulsion and the correlation becomes energy, since the antisymmetric state keeps its particles out of each other's way. That is the measured pattern of helium's 1s2s configuration: the triplet 2³S₁ lies 19.82 eV above the ground state and the singlet 2¹S₀ at 20.62 eV, 0.80 eV higher (tabulated; topic 7.6 derives its J ± K structure). The correlation itself has been counted directly. In 2007 clouds of metastable ⁴He and ³He atoms, released in one apparatus and timed as they reached a detector, arrived in bunched pairs for the bosonic isotope and in anti-bunched pairs for the fermionic one.

No overlap, no exchange

Every exchange term found so far is built from one product, ψₐ*(x)ψ_b(x), taken at a single point: in ⟨a|x|b⟩, in the overlap over a region ∫_R ψₐ*ψ_b dx, and in the cross term of the joint density. If the two orbitals are nowhere nonzero together, one particle in this well and the other in a well a metre away, that product vanishes at every x, every exchange term vanishes with it, and the symmetrised state predicts exactly what the product state does for anything measured in either well. That is why an electron in the laboratory is never antisymmetrised against one on the Moon: the postulate still holds, and it changes nothing. Real orbitals have exponential tails, so the question becomes how small the overlap is. Two hydrogen 1s orbitals 1.00 nm apart, ρ = R/a₀ = 18.9, overlap by S = e^(−ρ)(1 + ρ + ρ²/3) = 8.6 × 10⁻⁷ (the two-centre integral, quoted), and since both orbitals are positive, the exchange term in any region probability is no larger than about S² ≈ 7 × 10⁻¹³.

Change one variable at a time

Make the relationship visible.

Interactive model

Particle 1 found at x₁0.30 a

Second level m (first is n = 1)2

Drag x₁ across the well: the fermion curve's zero follows it and the boson curve doubles the dashed one there. At x₁ = 0.50 a with m = 2 all three curves merge, since particle 1 sits on the node of ψ₂. Then set m = 3: the rms separations become equal, yet the curves still differ.

RMS x₁ − x₂, BOSONS0.196 a

RMS x₁ − x₂, DISTINGUISHABLE0.321 a

RMS x₁ − x₂, FERMIONS0.410 a

EXCHANGE ELEMENT ⟨x⟩₁ₘ-0.1801 a

Live interpretationRMS x₁ − x₂, BOSONS: 0.196 a. RMS x₁ − x₂, DISTINGUISHABLE: 0.321 a. RMS x₁ − x₂, FERMIONS: 0.410 a. EXCHANGE ELEMENT ⟨x⟩₁ₘ: −0.1801 a

Catch the common trap

Explain before calculating.

Misconception check

The commonest error is to read the fermions' larger separation as the work of a repulsive exchange force and the bosons' smaller one as an attraction. There is no such force here: H = h(1) + h(2), and nothing couples the particles. The test is energy. (1, 2)+ and (1, 2)− have the same energy, 5K, though one holds its particles at 0.196a and the other at 0.410a; a force that moved them apart would have done work, and the work would show up in the energy. Nor is the rise of the fermion ground state from 2K to 5K work done by a force: it is the price of the states the postulate leaves. What exchange changes is the joint probability of where the two are found, the way two cards dealt from one deck match less often than two drawn from separate decks. The exchange energy met in helium and in magnets is the Coulomb repulsion averaged over a correlated state, not the energy of an extra force.

Two identical spin-0 bosons in an infinite well occupy levels 1 and 3, one particle in each, and do not interact. Compared with two distinguishable particles in the same two levels, what does their exchange symmetry change?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo particles, each with the electron's mass, sit in a one-dimensional infinite well of width 0.80 nm and do not interact. Find the ground-state energy if they are distinguishable, if they are spin-0 bosons (hypothetical, chosen for comparison), and if they are electrons with both spins up. Then find the lowest excitation energy in each case.

K = π²ħ²/(2ma²) = h²/(8ma²) = (6.626 × 10⁻³⁴ J s)² ÷ [8 × (9.109 × 10⁻³¹ kg)(0.80 × 10⁻⁹ m)²] = 9.41 × 10⁻²⁰ J = 0.588 eV. Units: J² s² ÷ (kg m²) = J, since J = kg m² s⁻².
With no interaction the pair energy is K(n² + m²). Distinguishable particles and bosons can both use ψ₁(x₁)ψ₁(x₂), which is already symmetric, so E₀ = 2K = 1.18 eV.
Both spins up makes the spin factor symmetric, so the spatial factor must be antisymmetric, and antisymmetrising ψ₁ψ₁ gives zero. The lowest allowed state is (1, 2)−, with E₀ = K(1 + 4) = 5K = 2.94 eV.
Distinguishable particles and bosons next reach 5K, one particle in level 2: two states for distinguishable particles, ψ₁ψ₂ and ψ₂ψ₁, and one symmetric state for bosons. The step costs 3K = 1.76 eV.
For the fermions (2, 2)− does not exist, so the next state is (1, 3)− at 10K, and the step costs 5K = 2.94 eV. No charge enters any of these numbers: the differences are symmetry alone.

AnswerE₀ = 1.18 eV for distinguishable particles and for bosons, and 2.94 eV for the spin-up electrons; the lowest excitation costs 1.76 eV in the first two cases and 2.94 eV for the electrons.

MediumTwo non-interacting particles in an infinite well of width a occupy levels 1 and 2. A detector reports only which half of the well each particle is in. Find the probability that both are in the left half, and that both are on the same side, for distinguishable particles, spin-0 bosons and spinless fermions, and check that each particle alone is still equally likely to be in either half.

Integrate |Ψ±|² with both x₁ and x₂ in a region R. The two direct terms each give Pₐ(R)P_b(R) and the ½ of normalisation halves their sum; the cross terms give ±|I_R|², with I_R = ∫_R ψₐψ_b dx. So P±(R) = Pₐ(R)P_b(R) ± |I_R|², and distinguishable particles have Pₐ(R)P_b(R).
Every |ψₙ|² is symmetric about the centre, so P₁(L) = P₂(L) = ½, and distinguishable particles are both in the left half with probability ¼.
I_L = (2/a)∫₀^(a/2) sin(πx/a) sin(2πx/a) dx. With u = πx/a and sin 2u = 2 sin u cos u this is (4/π)∫₀^(π/2) sin²u cos u du = (4/π)(1/3) = 4/(3π) = 0.4244, so I_L² = 16/(9π²) = 0.1801.
Both in the left half: bosons 0.25 + 0.1801 = 0.430, fermions 0.25 − 0.1801 = 0.070. Reflecting both particles through the centre leaves ψ₁ unchanged and flips ψ₂, so Ψ± only changes sign and both-in-the-right has the same probability. Same side: 0.860 for bosons, 0.500 distinguishable, 0.140 for fermions.
Check for fermions: opposite sides has probability 1 − 0.140 = 0.860, split equally between the two orderings because |Ψ|² is symmetric in the labels, so particle 1 is on the left with probability 0.070 + 0.430 = 0.500. Over the whole well ∫ψ₁ψ₂ dx = 0 and P = 1, as it must be.

AnswerBoth in the left half: 0.430 for bosons, 0.250 for distinguishable particles, 0.070 for fermions; both on the same side: 0.86, 0.50 and 0.14. Each particle alone is in the left half with probability 0.50 in every case.

HardTwo particles of electron mass in a 1.00 nm infinite well feel a weak short-range repulsion V = g δ(x₁ − x₂), with g = 0.050 eV nm. To first order in g, find the ground-state energy of two spin-0 bosons and of two electrons with both spins up, and the energy of the boson state with one particle in each of levels 1 and 2. Explain why the electrons' result is exact.

K = h²/(8mₑa²) = 0.376 eV. Unperturbed: bosons (1, 1) at 2K = 0.752 eV and (1, 2)+ at 5K = 1.880 eV; spin-up electrons (1, 2)− at 5K = 1.880 eV. P commutes with H₀ and V, and each state is the only one of its symmetry at its energy, so non-degenerate first-order theory applies within each symmetry.
The δ function collapses one integral: ⟨V⟩ = g∫|Ψ(x, x)|² dx. Units: g in J m times an integral in m⁻¹ gives J.
Bosons in (1, 1): ∫ψ₁⁴ dx = (4/a²)∫₀^a sin⁴(πx/a) dx = (4/a²)(3a/8) = 3/(2a), so the shift is 1.5g/a = 0.075 eV and E₀ ≈ 0.827 eV.
(1, 2)±: |Ψ±(x, x)|² = (1 ± 1)ψ₁²ψ₂², and ∫ψ₁²ψ₂² dx = (4/a²)(a/4) = 1/a. The direct and exchange terms are each g/a = 0.050 eV; for bosons they add, a shift of 0.100 eV to 1.980 eV, and for the electrons they cancel.
Exactness: any antisymmetric Ψ has Ψ(x, x) = −Ψ(x, x) = 0, so gδ(x₁ − x₂)Ψ = 0 identically and every antisymmetric eigenstate of H₀ keeps its energy to all orders. The boson shifts are under a tenth of the 3K = 1.13 eV level spacing; a numerical second-order sum lowers each by about 0.004 eV.

AnswerBosons: E₀ ≈ 0.827 eV, up from 0.752 eV, and E(1, 2)+ ≈ 1.980 eV; spin-up electrons: E₀ = 1.880 eV exactly, because antisymmetric particles are never found at the same point and a contact force never acts on them.