Motivation
How does indistinguishability reshape a joint position measurement?
For identical particles, the two labelings of a pair of occupied orbitals describe the same physical arrangement. The total wavefunction is constrained to be symmetric for bosons or antisymmetric for fermions under exchange of both coordinates. Spatially overlapping orbitals generate interference between the two assignments, so the joint probability density differs from a simple product. The antisymmetric wavefunction must vanish when both particles are measured at the same position for a spinless two-fermion state. The effect can be explored quantitatively by changing the separation and width of two normalized orbitals.
Prerequisites: Identical particles, product states, exchange operators, Gaussian integrals and squared wavefunction amplitudes.
Computed physics
Live observable readouts
Two-body spatial correlations
Compare symmetric and antisymmetric joint densities
The two labels refer only to coordinates in the display; the physical probabilities respect exchange of x₁ and x₂.
Left: normalized symmetric two-particle wavefunction; right: normalized antisymmetric wavefunction. Density is plotted over coordinates x₁,x₂ in oscillator-width units and is symmetric about the diagonal in both cases.
Pauli exclusion and coincidence
Probability density of coincident positions
Along x₁=x₂ the antisymmetric amplitude cancels exactly, while the symmetric state can have enhanced spatial coincidence.
The curves show P(x,x) as a density per squared length, not a finite detector-bin probability. Ideal identical spinless fermions have a node for every x on the diagonal.
Derive the result
Normalize exchanged orbital assignments
Let φa(x),φb(x) be individually normalized real Gaussian orbitals with centers at −d/2 and +d/2, both having probability-density variance σ². Their scalar overlap is S=〈φa|φb〉=exp[−d²/(8σ²)]. The two distinguishable labelings have amplitudes φa(x₁)φb(x₂) and φb(x₁)φa(x₂).
The normalization depends on S because these assignment amplitudes are not orthogonal when the orbitals overlap. Under coordinate exchange x₁↔x₂, the bosonic amplitude Ψ+ is unchanged and the fermionic Ψ− acquires a minus sign. Their squared densities both respect coordinate exchange, but have different interference terms.
For x₁=x₂=x the antisymmetric numerator cancels exactly, so Ψ−(x,x)=0. If the orbitals become identical, S→1 and the antisymmetric numerator vanishes everywhere, expressing the impossibility of placing two identical spinless fermions into the same spatial orbital. Symmetric orbital occupations remain allowed and can be enhanced by overlap.
Exactly soluble limits and sanity checks
At large separation d≫σ the overlap S approaches zero and normalization tends to 1/√2 for either exchange symmetry. At d→0, the symmetric two-particle state becomes φ(x₁)φ(x₂), whereas the normalized antisymmetric construction is undefined at exact orbital coincidence. The slider stops at positive d to avoid assigning a spurious finite value to a Pauli-forbidden state.
What assumptions define this model?
Two identical spinless noninteracting particles described by real normalized Gaussian orbital wavefunctions in one spatial dimension. The model concerns a prepared two-orbital state rather than a thermal many-body ensemble; spin, interactions, scattering phases and wave-packet time evolution are outside the scope.
Self-directed physics exercises
Three independent investigations
- Identify symmetry without particle labels. Choose moderate separation and compare densities at (x₁,x₂) and (x₂,x₁). Verify equal probabilities for either statistics even though the fermionic amplitude changes sign.
- Investigate the diagonal. Move the coincidence position x and compare bosonic and fermionic diagonal densities. Explain the node of the antisymmetric wavefunction and the role of orbital overlap.
- Vary orbital overlap. Reduce d/σ. Track S and the normalization denominators. Explain why the fermionic expression becomes singular as the two one-particle orbitals become identical.
Self-check: For identical spinless fermions, what is Ψ−(x,x) when the two orbital states are linearly independent?
Extension: combine a spin singlet with an antisymmetric spin triplet and examine how the spatial symmetry requirements change for electrons.
Scientific meaning
Conclusions, caveats and further reading
Exchange statistics constrain the wavefunction's symmetry, not simply classical particle-label counting. The interference of two orbital assignments produces distinct bosonic and fermionic position correlations. This is a spatial counterpart of bosonic enhancement and fermionic Fock-space signs.
Corresponding lecture-note sources: Identical-particle permutation symmetry · Symmetric and antisymmetric states.