10. Identical particles and exchange symmetry

Identical particles and exchange symmetry

See how a purely formal operation — swapping two labels — pushes particles apart or pulls them together.

What you will see

Swapping two identical particles returns the wavefunction unchanged (bosons) or flipped in sign (fermions)
An antisymmetric wavefunction is exactly zero on x₁ = x₂ — Pauli exclusion is not an extra rule
The same single-particle states give visibly different distributions under the two symmetries

Assumed background

What a two-particle wavefunction ψ(x₁, x₂) means
Bound states in one dimension (module 03)

Classically, two identical balls are still two balls: you can number them and follow each trajectory.

Quantum mechanically you cannot. With no trajectories there is no way to say “that one was number 1”. Identical particles are indistinguishable in principle, and that fact has observable consequences.

Identical particles: exchange is one flip about the diagonal

The horizontal plane is configuration space (x₁, x₂) and the height is the wavefunction Ψ. Press "Swap the two particles": the bosonic surface does not budge, while the fermionic one flips entirely below the floor.

Loading 3D scene…

Single-particle states

Quantum number a1

Quantum number b2

Symmetry

Show |Ψ|² instead of ΨHighlight the x₁ = x₂ diagonal

⟨(x₁−x₂)²⟩ compared

Distinguishable0.1033

Bosons (bunching)0.0384

Fermions (antibunching)0.1682

There is no interaction potential anywhere in the Hamiltonian; the difference comes entirely from symmetry.

\Psi_- = \tfrac{1}{\sqrt2}\left[\phi_a(x_1)\phi_b(x_2) - \phi_b(x_1)\phi_a(x_2)\right]

⟨(x₁−x₂)²⟩ = 0.1682After exchange Ψ → −Ψ

Ψ > 0
Ψ < 0
the x₁ = x₂ diagonal

What to look for

Start with "Distinguishable": the surface is not symmetric about the diagonal — "particle 1 left, particle 2 right" and the reverse are two different states.
Switch to "Bosons": the surface becomes mirror-symmetric about the diagonal and peaks *on* it, where the particles coincide. Bosons like to bunch.
Switch to "Fermions": the surface is exactly zero along the diagonal, positive on one side and negative on the other. Two fermions can never be at the same point.
Press "Swap the two particles": the animation is literally a 180° flip about the diagonal. Bosons show no change; fermions end up entirely below the floor. The difference is a single minus sign, and its consequence is the whole periodic table.
Set the two quantum numbers equal and choose fermions: the surface simply vanishes. Pauli exclusion needs no extra rule — antisymmetrisation does the arithmetic itself.

✕ "Exchange symmetry is a kind of force"

There is no interaction term in the Hamiltonian. The bunching or avoidance comes entirely from the requirement that the wavefunction be (anti)symmetric. It is traditionally called an "exchange force", but it is not a force.

✕ "The wavefunction shows where the two particles are in space"

This surface lives in *configuration space*: one axis is the position of particle 1, the other of particle 2. Real space here is one-dimensional; it is configuration space that is two-dimensional. Three particles make it three-dimensional, ten make it ten — which is the root of the many-body problem.

✕ "Pauli exclusion means fermions repel each other"

Exclusion is a consequence of the structure of the wavefunction, not an interaction. It carries no energy, falls off with no distance law and has no field.

Think it through

For two electrons the total wavefunction is spatial × spin and must be antisymmetric overall. If the spin part is a triplet (symmetric), the spatial part must be antisymmetric — what does that do to their average separation? This is one origin of ferromagnetism.
Push a and b far apart (say 1 and 5). How does the spread of ⟨(x₁−x₂)²⟩ across the three cases change? Why does less spatial overlap mean a weaker exchange effect?
In two dimensions, exchanging twice need not return the original state, which allows anyons and fractional statistics. What is essentially different about three dimensions?

This is an interactive 3D scene (identical-particles) and needs JavaScript and WebGL to run.

The exchange operator has only two eigenvalues

Define $\hat P_{12}\psi(x_1,x_2) = \psi(x_2,x_1)$ . Since $\hat P_{12}^2 = \hat I$ , its eigenvalues can only be $\pm1$ :

\psi(x_2,x_1) = \pm\,\psi(x_1,x_2).

$+$ : bosons (integer spin: photons, phonons, $^4$ He)
$-$ : fermions (half-integer spin: electrons, protons, neutrons)

The link between spin and statistics is not a postulate of quantum mechanics but a theorem of relativistic quantum field theory (the spin-statistics theorem). In the non-relativistic framework we can only take it as an experimental fact.

Pauli exclusion is a corollary

Put two fermions in the same single-particle state $\phi_a$ :

\psi_-(x_1,x_2) = \frac{1}{\sqrt2}\left[\phi_a(x_1)\phi_a(x_2) - \phi_a(x_2)\phi_a(x_1)\right] = 0 .

The whole wavefunction vanishes. “Two fermions cannot occupy the same state” is not an extra prohibition; it is the arithmetic of antisymmetrisation.

Set the two quantum numbers equal in the scene with fermions selected and the entire distribution collapses to nothing — which is quite vivid.

Exchange force: a force that is not one

For different single-particle states $\phi_a,\phi_b$ , compute the mean square separation:

\langle (x_1-x_2)^2\rangle_\pm = \langle x^2\rangle_a + \langle x^2\rangle_b - 2\langle x\rangle_a\langle x\rangle_b \mp 2\left|\langle a|x|b\rangle\right|^2 .

The last term is the exchange term:

bosons (take $-$ ): smaller separation, particles bunch together;
fermions (take $+$ ): larger separation, particles avoid each other.

The two curves in the scene differ by exactly this term. Note that the Hamiltonian contains no interaction potential — the “force” comes purely from the symmetry requirement on the wavefunction.

It holds up electron degeneracy pressure in metals, keeps white dwarfs and neutron stars from collapsing, and underlies ferromagnetism (the exchange interaction).

One table

	Bosons	Fermions
Under exchange	$\psi \to +\psi$	$\psi \to -\psi$
Same state	allowed, and increasingly favoured	forbidden
Statistics	Bose–Einstein	Fermi–Dirac
At low temperature	Bose–Einstein condensation	Fermi sea and degeneracy pressure
Examples	photons, $^4$ He, $^{87}$ Rb	electrons, protons, $^3$ He

The Slater determinantadvanced

The antisymmetric wavefunction of $N$ fermions can be written as a determinant:

\psi(x_1,\ldots,x_N) = \frac{1}{\sqrt{N!}} \det\begin{pmatrix} \phi_1(x_1) & \cdots & \phi_1(x_N)\\ \vdots & \ddots & \vdots\\ \phi_N(x_1) & \cdots & \phi_N(x_N) \end{pmatrix}.

Two properties of determinants encode the physics exactly: swapping two columns flips the sign (antisymmetry), and two identical rows give zero (Pauli exclusion). The whole Hartree–Fock method is built on this object.

Think it through

If two electrons are in a spin triplet (symmetric spin part), the spatial part must be antisymmetric. What does that do to their spatial distribution, and how does it relate to bonding and antibonding states in the hydrogen molecule?
A metal contains $10^{23}$ electrons and the wavefunction must be antisymmetric under every pair exchange. Why can we still treat them as independent electrons when computing conductivity?
In two dimensions, exchanging twice need not return the original state (braid group rather than permutation group), which permits anyons and fractional statistics. What is essentially different about three dimensions?

Identical particles and exchange symmetry

What you will see

Assumed background

Identical particles: exchange is one flip about the diagonal

Single-particle states

Symmetry

⟨(x₁−x₂)²⟩ compared

The exchange operator has only two eigenvalues

Pauli exclusion is a corollary

One table

Think it through

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