Symmetric and Antisymmetric States

The exchange geometry of this article is now photographed routinely. Few-fermion experiments in Heidelberg (Holten et al., 2021) imaged Pauli crystals: correlation patterns of three-to-six trapped ⁶Li atoms whose only structuring agent is the antisymmetric sign — the exchange hole of the hero figure, in laboratory pixels. Hanbury Brown–Twiss experiments with ultracold atoms measure the boson huddling as a bunching peak in pair correlations and the fermionic mirror as an antibunching dip. And the same matrix element $X_{ab}$ runs quantum hardware: exchange-based two-qubit gates in silicon double quantum dots switch on the overlap between neighboring electrons deliberately, letting the $J\pm K$ energetics (owned by Exchange Forces and Exchange Energy) execute logic. The sign this article built is, quite literally, a clock rate.

Plot the joint probability density $|\psi(x_1,x_2)|^2$ over the square of configurations of two particles in a one-dimensional box, one in the ground state and one in the first excited state. For the symmetric combination the density piles up on the diagonal $x_1=x_2$: the two particles are most likely found in the same region, huddled. For the antisymmetric combination the density is exactly zero along the entire diagonal — a trench called the exchange hole — and the probability mass is exiled to the off-diagonal corners: one particle left, one particle right.

The trench is not carved by repulsion and the ridge is not built by attraction: the two-body potential here is identically zero. Both features are interference between the two ways of assigning particles to states — the "direct" $\psi_a(1)\psi_b(2)$ and the "exchanged" $\psi_b(1)\psi_a(2)$. A symmetric state adds the amplitudes and interferes constructively where the assignments agree; an antisymmetric state subtracts them and cancels exactly where a swap changes nothing — the diagonal $x_1=x_2$. Permutation Symmetry and the Exchange Operator proved there is no third option for two particles, and which sector a species occupies is dictated by its spin (Bosons and Fermions, with the deep reason in Spin-Statistics Theorem). What remains — and what this article owns — is to build the states explicitly, get their normalization exactly right, and turn the sign into a number a detector can read.

Start from two orthonormal single-particle states $\psi_a$, $\psi_b$ and two identical particles. The raw product $\psi_a(x_1)\psi_b(x_2)$ answers the illegal question "which particle is in $a$?", so it cannot be physical. Projecting it into the two legal sectors — the symmetrizer and antisymmetrizer of Permutation Symmetry and the Exchange Operator — and normalizing gives the only two states nature accepts:

where: $\psi_\pm$ — the symmetric ($+$, bosonic) and antisymmetric ($-$, fermionic) two-particle states; $x_i$ — all coordinates of particle $i$; the prefactor $1/\sqrt2$ assumes $\langle a|b\rangle=0$. Swapping $x_1\leftrightarrow x_2$ returns $+\psi_+$ and $-\psi_-$: these are exchange eigenstates, and the eigenvalue is the physics. When the two single-particle states are not orthogonal — overlapping wavepackets, non-orthogonal orbitals in a molecule — the innocent-looking $1/\sqrt2$ is wrong, and the correct prefactor carries the overlap:

where: $\langle a|b\rangle=\int\psi_a^*\psi_b\,\mathrm{d}x$ — the single-particle overlap. Two limits read the formula for you: orthogonal states restore $1/\sqrt2$; and as $\psi_b\to\psi_a$ the symmetric denominator goes to $\sqrt4=2$, so the state becomes $\psi_a(x_1)\psi_a(x_2)$ — already normalized, with no residual $\sqrt2$ — while the antisymmetric denominator goes to zero precisely because the numerator dies faster: $\psi_-$ ceases to exist.

For two particles these two states are the whole story: Permutation Symmetry and the Exchange Operator shows the two-particle space splits cleanly as symmetric ⊕ antisymmetric, so every physical state is a superposition within one sector. The systematic constructors for $N$ particles — determinants for fermions, permanents for bosons — are owned by Slater Determinants and Permanents and Symmetrized Boson States.

Here is the theorem that makes the sign measurable — the mean-square separation $\langle(x_1-x_2)^2\rangle$ in three preparations: distinguishable particles in a product state, identical bosons in $\psi_+$, identical fermions in $\psi_-$ (orthonormal $a$, $b$):

where: $\langle x\rangle_s,\langle x^2\rangle_s$ — moments in the single-particle state $s$; $X_{ab}=\langle\psi_a|\,x\,|\psi_b\rangle=\int\psi_a^*\,x\,\psi_b\,\mathrm{d}x$ — the cross matrix element between the two occupied states. The entire statistical effect is one term: symmetric states subtract $2|X_{ab}|^2$ (huddling), antisymmetric states add it (repelling). No potential appears anywhere in the derivation — this is interference, not force.

For the box pair of the hero figure ($\psi_1,\psi_2$ of a well of width $L$ — the states of Infinite Square Well, which writes the width as $a$), every ingredient is exact: $\langle x\rangle_1=\langle x\rangle_2=L/2$, $\langle x^2\rangle_n=L^2\!\left(\tfrac13-\tfrac{1}{2n^2\pi^2}\right)$, and the cross element — the same $X_{12}$ that drives the swinging centroid of Expectation Values — is

where: the sign is convention (it enters equation (3) squared), and the magnitude $16/9\pi^2\approx0.18$ says the two clouds overlap substantially. The three separations come out $L^2(\tfrac16-\tfrac{5}{8\pi^2})\mp\tfrac{512}{81\pi^4}L^2$: numerically $0.038\,L^2$ (bosons), $0.103\,L^2$ (distinguishable), $0.168\,L^2$ (fermions). One sign, a factor of 4.4 in mean-square separation.

One subtlety the ladder hides: the exchange term can vanish by symmetry even when the clouds overlap completely. For the box pair $(1,3)$ both states are even about the center, so $X_{13}=0$ exactly — and the two mean-square separations are equal. The statistics has not disappeared (the antisymmetric state keeps its exchange hole); it has merely gone invisible to this observable. $X_{ab}$ is a transition matrix element, and it obeys selection rules.

For real electrons the coordinates in equation (1) include spin, and the exchange sign applies to the total state. If the state factorizes into a spatial part and a spin part, the fermionic minus sign can be paid by either factor:

where: $\Phi$ — spatial wavefunction, built exactly as in equation (1) from orbitals; $\chi$ — two-spin state. Which spin states are symmetric and which antisymmetric is settled canon: Singlet and Triplet States constructs the four two-spin states and proves the space-spin correlation theorem — electron pairs in a symmetric spatial state must be spin singlets, pairs in an antisymmetric spatial state must be triplets — taking the symmetrization requirement itself as input. This article is that input: equation (1) is the postulated structure that article consumed. Crossed with the Coulomb repulsion, the correlation splits helium into its ortho and para ladders — the story told there and quantified in The Helium Atom. The energy bookkeeping deserves its own honesty box. Because $\psi_\pm$ shifts where the particles are, any interaction $V(x_1,x_2)$ acquires sign-dependent expectation values — the direct-plus-exchange structure $J\pm K$. That machinery, the exchange integral $K$, and the phrase "exchange force" belong to Exchange Forces and Exchange Energy; this article establishes the prior fact that the correlation exists with $V=0$. There is no force of exchange — there is geometry of exchange.

Example 1 — the box pair, end to end. Two identical particles occupy the $n=1$ and $n=2$ states of a box of width $L=1\,\mathrm{nm}$. Find the rms separation for bosons, fermions, and distinguishable particles.

Step 1. Moments (exact, from the standard box integrals): $\langle x\rangle_1=\langle x\rangle_2=L/2$; $\langle x^2\rangle_1=L^2(\tfrac13-\tfrac{1}{2\pi^2})=0.28267\,L^2$; $\langle x^2\rangle_2=L^2(\tfrac13-\tfrac{1}{8\pi^2})=0.32067\,L^2$.

Step 2. Distinguishable baseline: $\langle(x_1-x_2)^2\rangle_d=\langle x^2\rangle_1+\langle x^2\rangle_2-2(L/2)^2=L^2\big(\tfrac16-\tfrac{5}{8\pi^2}\big)=0.10334\,L^2$.

Step 3. Exchange term: $X_{12}=-16L/9\pi^2=-0.18011\,L$, so $2X_{12}^2=512L^2/81\pi^4=0.06489\,L^2$ — 63% of the baseline. This is not a small correction.

Step 4. Apply the theorem: bosons $0.10334-0.06489=0.03845\,L^2$; fermions $0.10334+0.06489=0.16823\,L^2$.

Step 5. At $L=1$ nm the rms separations are $\sqrt{0.03845}=\mathbf{0.196\ nm}$ (bosons), $\mathbf{0.321\ nm}$ (distinguishable), $\mathbf{0.410\ nm}$ (fermions). A factor 2.1 in rms distance between the two identical-particle answers, from one sign.

Example 2 — two ions in a trap and the 1 : 2 : 3 law. Two ⁴⁰Ca⁺ ions sit in the $n=0$ and $n=1$ states of a harmonic trap with $\omega=2\pi\times1\,\mathrm{MHz}$ (states of Quantum Harmonic Oscillator: Analytic Solution). Compare the three mean-square separations.

Step 1. Both states are centered: $\langle x\rangle_0=\langle x\rangle_1=0$. The second moments are $\langle x^2\rangle_0=\hbar/2m\omega$ and $\langle x^2\rangle_1=3\hbar/2m\omega$, so the distinguishable answer is $\langle(x_1-x_2)^2\rangle_d=2\hbar/m\omega$.

Step 2. The cross element is the oscillator's fundamental dipole element: $X_{01}=\sqrt{\hbar/2m\omega}$, so $2X_{01}^2=\hbar/m\omega$ — exactly half the distinguishable answer.

Step 3. The theorem then delivers integers: bosons $2\hbar/m\omega-\hbar/m\omega=\hbar/m\omega$; fermions $2\hbar/m\omega+\hbar/m\omega=3\hbar/m\omega$. The three answers stand in the ratio $\mathbf{1:2:3}$ — exact, for any mass and any frequency.

Step 4. Numbers: $\hbar/m\omega=2.53\times10^{-16}\,\mathrm{m}^2$ gives rms separations $\mathbf{15.9\ nm}$ (bosons), $\mathbf{22.5\ nm}$ (distinguishable), $\mathbf{27.5\ nm}$ (fermions); the cross element itself is $X_{01}=11.2$ nm. Trapped-ion and quantum-gas-microscope experiments resolve length differences far smaller than these.

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §5.1 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §7.1–7.4 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 10 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. II, Ch. XIV · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 12.

Provenance. All algebra is standard. A verifier should reconfirm: $P_{12}^2=\mathbb{1}\Rightarrow\lambda=\pm1$ (Sakurai §7.1, Griffiths §5.1); the normalisation $\langle\psi_\pm|\psi_\pm\rangle=2(1\pm|\langle a|b\rangle|^2)$; the projector identities $P_S+P_A=\mathbb{1}$, $P_S^2=P_S$; the direct/exchange decomposition $\langle V\rangle_\pm=J\pm K$ (Griffiths §5.1.2, §7.2); and the determinant/permanent generalisations. No numerical constants are quoted; all results are structural and textbook-verifiable in Griffiths Ch. 5, Sakurai Ch. 7, Shankar Ch. 10, Cohen-Tannoudji Vol. II Ch. XIV. — Editorial note for this rebuild: the base's exchange-energy derivation ($\langle V\rangle_\pm=J\pm K$) is honored here as a pointer only — that machinery is owned by Exchange Forces and Exchange Energy; the projector algebra it invokes is owned by Permutation Symmetry and the Exchange Operator; the determinant generalisation by Slater Determinants. Every number added by this rebuild (box and oscillator moments, $X_{12}=-16L/9\pi^2$, the $1:2:3$ law, all separations) is recomputed in the committed verification script.

Builds on: Indistinguishability of Identical Particles · Permutation Symmetry and the Exchange Operator · Tensor product of Hilbert spaces. Leads to: Bosons and Fermions · Spin-Statistics Theorem · Pauli Exclusion Principle · Slater Determinants · Singlet and Triplet States in Two-Electron Systems · Exchange Forces and Exchange Energy.