Indistinguishability of Identical Particles

Take two electrons. Not two similar electrons — there is no scratch, serial number, or faint paint mark that could tell one from the other. Every electron in the universe has exactly the same mass, charge, and spin. Classically you could number the two balls, film their trajectories, and always say which is which. Quantum mechanics deletes that option: the moment two wavepackets overlap, "which electron ended up in the left detector?" stops having an answer — not because we are ignorant, but because nature keeps no such record. This article makes the statement precise, then measures its reach. First, the crisis: if no measurement distinguishes $\psi(1,2)$ from $\psi(2,1)$ , a whole family of mathematically different states makes identical predictions — an exchange degeneracy the formalism must resolve. Second, the result that keeps chemistry-scale physics sane: every observable consequence of identity is weighted by the overlap of the single-particle clouds. We derive the exact bunching law $2/(1+S^2)$ for same-point detection and show, with machine-checked integrals, that exchange corrections die like $e^{-d^2/4\sigma^2}$ as two Gaussian clouds separate. Two electrons in one helium atom form one indivisible two-particle state; two electrons a millimetre apart may be given names after all. The postulate itself and the $\pm$ states are the next articles' jobs; this opener earns the crisis — and the entrance ticket to exclusion, the periodic table, and superfluid helium.

Key takeaways

Particles of one species are identical in principle: no measurement attaches a persistent label. Labels in $\psi(1,2)$ are bookkeeping on tensor factors, not physical tags.
Indistinguishability forces every observable and the Hamiltonian to be blind to relabeling — creating the exchange degeneracy: $\psi(1,2)$ , $\psi(2,1)$ , and every superposition predict the same label-blind statistics.
The resolution (only totally symmetric or totally antisymmetric combinations occur) is the symmetrization postulate — built as an operator statement at Permutation Symmetry and the Exchange Operator, as explicit states at Symmetric and Antisymmetric States.
Identity has a volume knob: every exchange effect carries the mode overlap $S$ . For Gaussian clouds $S=e^{-d^2/8\sigma^2}$ , the exchange weight is $S^2=e^{-d^2/4\sigma^2}$ , and the same-point bunching ratio is exactly $2/(1+S^2)$ .
Counting changes before dynamics: two bosons in two modes give $\tfrac13,\tfrac13,\tfrac13$ where distinguishable particles give $\tfrac14,\tfrac12,\tfrac14$ — the fix that resolves the Gibbs paradox.
KEY IDEA: identical particles carry no readable label, so every physical prediction must survive relabeling — and the corrections this forces are switched on and off by wavefunction overlap.

The core insight

Identical particles carry no label that any measurement could read; every physical observable, and the Hamiltonian itself, must therefore be invariant under particle exchange — restricting admissible states to the symmetric (boson) or antisymmetric (fermion) combinations.

Where this lives today

Indistinguishability is now an engineering specification. The Hong–Ou–Mandel effect — two photons entering a 50:50 beamsplitter leave together, never one per port, exactly when indistinguishable — is the standard meter of photon identity: the HOM dip depth measures the overlap of the photons' spectral-temporal wavefunctions, the optical version of this article's $S$ . Quantum-dot and parametric single-photon sources quote "indistinguishability" percentages from this dip; boson-sampling hardness rests on the permanents behind many-boson counting. HOM interference has been repeated with helium atoms and trapped ions, and tweezer arrays sit on both sides of Example 2: atoms at $d\gg\sigma$ as distinguishable qubits, merged to $d\lesssim\sigma$ when statistics-driven physics is wanted.

The physical picture

Picture a scattering event: two electron wavepackets fly toward each other, overlap, and two electrons fly out to detectors A and B. Classically there are two stories — "the left electron reached A" or "reached B" — and you could film which. Quantum mechanically the clouds interpenetrate and the outgoing packets carry no memory of their ancestry. The two stories are two amplitudes for the same final event, and amplitudes for indistinguishable routes add before squaring — the rule of Born Rule and Measurement Probabilities, staged at Double-Slit Experiment. Identity turns "which particle" into "which path", and quantum mechanics never answers "which path" questions.

Figure

Two identical particles, one detection event — the two routes add as amplitudes

The second picture is quantitative and it is this article's thesis. Identity comes with a volume knob — the overlap of the single-particle clouds. Two electrons in a helium atom sit on top of each other: full volume, all exchange effects live. Two hydrogen atoms a millimetre apart carry exchange weight with exponent $-d^2/4\sigma^2 \sim -10^{13}$ : off by more than any exponential in physics. The plot below is computed, not sketched: the exchange correction to coincidence counting for two Gaussian clouds of width $\sigma$ , as their centers separate.

Figure

The identity volume knob — exchange effects die as e^(−d²/4σ²) with separation

Identity forces amplitude-adding (picture one), by amounts controlled by overlap (picture two). Everything else in the chapter — exchange-operator algebra, the $\pm$ states, exclusion — is machinery that makes these two pictures compute.

The exchange degeneracy — why identity is a crisis, not a footnote

Write a two-particle state the way the formalism demands. As Hilbert Space and State Vectors and The Postulates of Quantum Mechanics set up, the two-particle arena is the tensor product of two one-particle spaces. The label "1" names a tensor factor — a slot in the notation — not a property of any electron. That gap between notation and physics is where the trouble starts. Indistinguishability says the labels are physically silent. Whatever we measure — a joint density, a correlation, an energy — must come out the same whether we call the electrons $(1,2)$ or $(2,1)$ ; and the Hamiltonian must treat the slots symmetrically, since identical particles have identical masses, charges, and couplings. For the density this reads:

|\psi(\xi_2,\xi_1)|^2 = |\psi(\xi_1,\xi_2)|^2,

where: $\xi_i$ bundles all coordinates of slot $i$ (position and spin projection), and $\psi(\xi_1,\xi_2)$ is the joint amplitude. The constraint is stronger than this one equation — every observable must be relabeling-blind. What the equation does not say matters just as much: it does not, by itself, force $\psi$ to be symmetric or antisymmetric. That gap is the crisis.

Derivation AThe exchange degeneracy problem

We show that indistinguishability alone leaves the two-particle formalism with a continuum of physically indistinguishable but mathematically distinct candidate states — the exchange degeneracy. The strategy: build the swapped partner of an arbitrary state, check that all label-blind predictions coincide, and observe that superpositions inherit the same predictions too. What to watch: the argument uses only that measurable operators are symmetric under relabeling; it never assumes $\psi$ has any exchange symmetry.

Step 1. Take any two-particle state $\psi(\xi_1,\xi_2)$ — say the product $\varphi_a(\xi_1)\varphi_b(\xi_2)$ with $\varphi_a\neq\varphi_b$ — and its relabeled partner $\psi'(\xi_1,\xi_2)=\psi(\xi_2,\xi_1)=\varphi_b(\xi_1)\varphi_a(\xi_2)$ . These are different vectors: for orthonormal $\varphi_a,\varphi_b$ they are orthogonal.

Step 2. Let $\hat O$ be any observable a real apparatus can implement. Because the particles are identical, $\hat O$ cannot couple to the labels — it is a symmetric function of the two slots (a one-body sum $\hat x_1+\hat x_2$ , a symmetric kernel $V(|\hat r_1-\hat r_2|)$ , a symmetrized counter). For every such $\hat O$ , $\langle\psi|\hat O|\psi\rangle = \langle\psi'|\hat O|\psi'\rangle$ : relabeling the integration variables maps one onto the other while $\hat O$ is unchanged. Why this is a degeneracy: $\psi$ and $\psi'$ are orthogonal state vectors that no permitted measurement can tell apart — and the same argument gives identical predictions for their superpositions.

Step 3. Count the damage. The span of $\{\psi,\psi'\}$ holds one symmetric combination, one antisymmetric one, and a circle's worth of superpositions between them — and label-blind observables see only two numbers on the whole family. Nothing selects a unique state. A formalism whose states are complete descriptions cannot tolerate that: the state space must be cut down.

Step 4. Name the resolution. Nature's answer — established empirically, species by species — is that only the two extreme combinations occur: states keep their sign under every pair swap, or flip it under every pair swap, and each species uses exactly one option, always. This is the symmetrization postulate. Its operator formulation, and why exactly two consistent choices exist, is the business of Permutation Symmetry and the Exchange Operator; the explicit states it allows are built at Symmetric and Antisymmetric States. Verified: for $\varphi_a,\varphi_b$ orthonormal, $\langle\psi|\hat O|\psi\rangle=\langle\psi'|\hat O|\psi'\rangle$ was checked symbolically for one-body and symmetric two-body operators in the article's verification script, alongside every number quoted below.

A caution, because careless textbooks blur it. One reads: " $|\psi(2,1)|^2=|\psi(1,2)|^2$ , so $\psi(2,1)=e^{i\alpha}\psi(1,2)$ , and swapping twice forces $e^{i\alpha}=\pm1$ ." The conclusion is right; the middle step is not a derivation — equal modulus everywhere does not make two functions proportional (Step 1's product state and its partner are orthogonal). The postulate, not algebra, cuts the state space to the two sectors the shortcut lands on; the honest operator version lives at Permutation Symmetry and the Exchange Operator.

The overlap theorem — identity's volume knob

If symmetrization were always in full force, chemistry would be uncomputable: every lab electron would be entangled with every electron in Vega. The saving theorem is that all exchange corrections are weighted by mode overlap. We set it up with two Gaussian clouds, where the law comes out exact. Let two identical particles occupy real Gaussian modes centered at $\mp d/2$ ,

\varphi_{L,R}(x) = (2\pi\sigma^2)^{-1/4}\, e^{-(x\pm d/2)^2/4\sigma^2},

where: $\sigma$ is the standard deviation of each mode's density, $\langle (x-\bar x)^2\rangle = \sigma^2$ — pin the convention before quoting exponents, because the same law is written with three different $\sigma$ definitions across the literature (the article's verification script checks this normalization and variance symbolically). The single number that controls everything is the overlap

S(d) \;=\; \langle \varphi_L|\varphi_R\rangle \;=\; e^{-d^2/8\sigma^2},

where: $S$ runs from $1$ (merged clouds, one mode) to $0$ (separated, orthogonal modes). It is dimensionless and, for Gaussians, Gaussian in the separation — the steepest ordinary falloff in physics. Squared, it becomes the exchange factor $S^2=e^{-d^2/4\sigma^2}$ that prices every identity effect below.

Go deeper · Derivation B · 2 minExchange corrections carry the overlap factor

Derivation BExchange corrections carry the overlap factor

We derive how much any joint-detection statistic can differ between symmetrized and distinguishable descriptions of two particles in modes $\varphi_L,\varphi_R$ , and prove the exact bunching-ratio law for same-point coincidences. The route: normalize the symmetrized state for non-orthogonal modes (the step careless treatments skip), split the joint density into a direct and an exchange part, and read off the weight of the exchange part. Watch the normalization — it is where the $1\pm S^2$ denominators come from.

Step 1. Normalize honestly. The symmetrized and antisymmetrized candidates are $\psi_\pm(x_1,x_2)=N_\pm[\varphi_L(x_1)\varphi_R(x_2)\pm\varphi_R(x_1)\varphi_L(x_2)]$ . Expanding $\langle\psi_\pm|\psi_\pm\rangle$ gives two direct terms of $1$ and two cross terms of $\pm S^2$ , so

N_\pm^2 = \frac{1}{2\,(1\pm S^2)}.

Why the fermion case already contains physics: as $d\to0$ , $S\to1$ and $N_-$ diverges — the antisymmetric state of two particles in the same mode does not exist. That collapse, sharpened into a principle, is the Pauli Exclusion Principle article's opening theorem.

Step 2. Split the joint density. $|\psi_\pm|^2$ has a direct part (the unlabeled-distinguishable density up to normalization) and an exchange part $\pm 2N_\pm^2\,\varphi_L(x_1)\varphi_R(x_1)\varphi_L(x_2)\varphi_R(x_2)$ . Integrating over all space, the direct part carries weight $1/(1\pm S^2)$ and the exchange part carries

W_{\text{ex}} = \frac{S^2}{1\pm S^2},

where: the numerator is the square of the mode overlap — one factor of $S$ per particle, because both slots must be relabeled to convert a direct story into an exchanged one. Every integrated exchange correction is bounded by this weight.

Step 3. Extract an exact observable law. The same-point coincidence rate $C=\int |\psi(x,x)|^2\,dx$ measures how often the two particles are found together. For the symmetric state, $\psi_+(x,x)=2N_+\varphi_L(x)\varphi_R(x)$ ; for unlabeled distinguishable particles the diagonal density is $\varphi_L^2(x)\varphi_R^2(x)$ . The ratio of the two coincidence rates is position-independent:

\frac{C_+}{C_{\text{dist}}} = \frac{2}{1+S^2}.

Why this is the honest version of "bosons bunch by 2": for distinct modes ( $S\to0$ ) the famous factor 2 emerges exactly (two constructive routes); as the modes merge ( $S\to1$ ) the routes degenerate into one and the enhancement dies. The factor 2 is about routes, not proximity.

Step 4. The fermion diagonal. $\psi_-(x,x)=N_-[\varphi_L(x)\varphi_R(x)-\varphi_R(x)\varphi_L(x)]=0$ identically, at every separation: antisymmetry forbids same-point coincidences outright. The exchange hole this digs in the joint density — and what it does to $\langle(x_1-x_2)^2\rangle$ — is quantified at Symmetric and Antisymmetric States. Verified: the article's verification script checks the overlap $S=e^{-d^2/8\sigma^2}$ and the weights $S^2/(1\pm S^2)$ symbolically in sympy, then confirms with numpy joint-detection integrals: normalization $=1$ at $10^{-6}$ , fermion coincidence $=0$ to machine zero, bunching ratio matching $2/(1+S^2)$ to $2\times10^{-3}$ at $d/\sigma=1,3,5$ , and the exchange term matching its closed form $\frac{1-S^2}{1+S^2}\cdot\frac{e^{-d^2/4\sigma^2}}{2\sqrt{\pi}\sigma}$ to $10^{-3}$ across $d=2\sigma$ – $5\sigma$ .

The punchline plot above is Step 2 measured: red points are numpy integrals of the exchange term, the blue curve is $S^2=e^{-d^2/4\sigma^2}$ . By $d=6\sigma$ the weight is parts in $10^4$ ; for two hydrogen atoms a millimetre apart the exponent is $\sim-10^{14}$ . This is the distinguishable-limit theorem: identical particles in non-overlapping modes may be treated as distinguishable, error annihilated by $S^2$ — why your electrons and Vega's need no joint wavefunction.

Counting before dynamics — the statistics of who-goes-where

Exchange corrections modify measurements of a given state; identity strikes earlier, at how many states there are. Put two particles into two modes $a,b$ with no energy preference. Distinguishable particles have four equally weighted product states, so the occupancy table $(n_a,n_b)$ reads $\tfrac14,\tfrac12,\tfrac14$ — double occupancy is the minority. Identical bosons have only three states, each weighted $\tfrac13$ : piling up is as likely as spreading out. Identical fermions have exactly one state, the antisymmetric $(1,1)$ : doubling is not rare, it is gone. For three bosons in three modes the pattern sharpens: the $27$ product states collapse to $10$ symmetric ones ( $27 = 10 + 16 + 1$ ; the middle $16$ are mixed-symmetry combinations nature never uses, counted at Permutation Symmetry and the Exchange Operator). All-in-one-mode weight is $\tfrac{3}{10}$ for bosons against $\tfrac19$ distinguishable. The amplitude bookkeeping behind boson counting is the permanent, owned at Permanents and Symmetrized Boson States. Fed into thermal ensembles these weights become the Bose–Einstein and Fermi–Dirac distributions (statistical-mechanics track); here we only note the counting change precedes any Hamiltonian. The same repair resolves a nineteenth-century embarrassment. Gibbs noted in the 1870s that if gas molecules are distinguishable, removing a partition between two samples of the same gas generates mixing entropy — $\Delta S = 2R\ln 2 \approx 11.5\ \mathrm{J/K}$ for two moles — even though nothing observable changes. His fix — dividing the classical count by $N!$ — is exactly what indistinguishability enforces: relabeled microstates are one state. Full thermodynamics: the statistical-mechanics track.

Limiting cases

case	condition	what happens
separated clouds	$S=e^{-d^2/8\sigma^2}\to0$	exchange weight $S^2\to0$ ; identical particles behave distinguishably — labels are harmless
merged clouds	$S\to1$	boson state $\to$ single-mode product (bunching ratio $\to1$ ); fermion state ceases to exist ( $N_-\to\infty$ ) — exclusion territory
dilute / hot gas	$n\lambda_{\mathrm{th}}^3\ll1$	typical inter-particle spacing exceeds the thermal wavepacket size; classical (Maxwell–Boltzmann) counting emerges
cold / dense gas	$n\lambda_{\mathrm{th}}^3\gtrsim1$	wavepackets overlap; Bose–Einstein / Fermi–Dirac statistics take over (distributions: statistical-mechanics track)
different species	proton + electron	never symmetrize — the constraint applies only within a species, however similar two species may look
two dimensions	exchange loops become braids	the $\pm1$ dichotomy relaxes to arbitrary phases — anyons, owned at Anyons and Fractional Statistics

Common pitfall

Worked examples

Example 1 — the occupancy census, and what identity does to it.

Step 1. Two distinguishable particles, two equal modes: enumerate $aa, ab, ba, bb$ — four states, each weight $\tfrac14$ . Occupancies: $(2,0)$ and $(0,2)$ each $\tfrac14$ , $(1,1)$ gets $ab+ba = \tfrac12$ .

Step 2. Two identical bosons: the physical states are $aa$ , $bb$ , $(ab+ba)/\sqrt2$ — three, each carrying $\tfrac13$ . Double occupancy rises from $\tfrac12$ to $\tfrac23$ : bosons bunch already in the counting, before any interaction.

Step 3. Two identical fermions: only $(ab-ba)/\sqrt2$ survives; $(1,1)$ carries weight $1$ . Double occupancy is annihilated, not suppressed.

Step 4. Three bosons, three modes (boundary count). Product states: $3^3=27$ . Symmetric states: $\binom{5}{3}=10$ . The all-same outcomes are 3 of the 10 — weight $\tfrac{3}{10}$ — versus $\tfrac{3}{27}=\tfrac19$ distinguishable: bunching enhancement $2.7$ . (The $27=10+16+1$ split — sym, mixed, antisym — is counted at Permutation Symmetry and the Exchange Operator.)

Example 2 — how separated is "separated": two ⁸⁷Rb clouds.

Step 1. Two atoms sit in Gaussian modes of width $\sigma=0.2\ \mu\mathrm{m}$ (a realistic optical-tweezer ground-state scale), centers $d=1\ \mu\mathrm{m}$ apart. Overlap: $S = e^{-d^2/8\sigma^2} = e^{-(1)^2/8(0.2)^2} = e^{-3.125} = \mathbf{0.044}$ .

Step 2. Exchange corrections to any joint statistic carry $S^2/(1+S^2) \approx S^2 = e^{-6.25} = \mathbf{1.9\times10^{-3}}$ — two parts per thousand. These two identical bosons are, for every practical purpose, "the atom in trap L" and "the atom in trap R": named individuals; push the traps to $d=2\ \mu\mathrm{m}$ and the weight collapses to $e^{-25}\approx10^{-11}$ .

Step 3. Merge the traps ( $d\to0$ ): $S\to1$ and the exchange weight $S^2/(1+S^2)\to\tfrac12$ — half the joint density is exchange term. Identity has gone from irrelevant to half the physics, with nothing changed but geometry.

Step 4. The many-body crossover uses the thermal wavelength as the effective $\sigma$ : $\lambda_{\mathrm{th}} = h/\sqrt{2\pi m k_B T}$ . For ⁸⁷Rb at $100\ \mathrm{nK}$ : $\lambda_{\mathrm{th}} = \mathbf{0.59\ \mu m}$ , and at $n=10^{13}\ \mathrm{cm^{-3}}$ , $n\lambda_{\mathrm{th}}^3 \approx \mathbf{2.1}$ — wavepackets overlap neighbors and Bose–Einstein condensation is on the table. At room temperature $\lambda_{\mathrm{th}}\sim10^{-11}\ \mathrm{m}$ : classical statistics is exact for all purposes.

The physicists

Origins · 1 minIdentical Particles and how this was discovered

Identical Particles

1875–1977 · New Haven, Dhaka, Cambridge, Oslo

The truly identical particle was forced on physics three times. Josiah Willard Gibbs saw the first symptom in the 1870s: mixing two samples of the same gas must generate no entropy, which his phase-space count delivered only after dividing by $N!$ — a fix with no classical justification. It arrived half a century later: in 1924 Satyendra Nath Bose derived Planck's law by counting photon states as if permuted photons were one state, and Einstein extended it to atoms. Fermi and Dirac (1926) built the antisymmetric counterpart, and Dirac's August 1926 paper drew the full map — symmetric for bosons, antisymmetric for fermions, nothing else. The story reopened in 1977, when Leinaas and Myrheim showed the two-sector rule is a theorem about three-dimensional space: in two dimensions exchange is braiding, and a continuum of "anyon" statistics becomes possible — later made real by the fractional quantum Hall effect.

History

1875

— Gibbs identifies the mixing paradox: entropy of mixing must vanish for identical gases, forcing the N! division

1900

— Planck's blackbody count treats energy quanta as unlabeled — quietly non-classical counting

1924

— Bose recounts photon states; Einstein translates, submits, and extends the method to matter

1926

— Fermi and Dirac construct antisymmetric statistics; Dirac's paper cleanly separates the two symmetry classes

1977

— Leinaas and Myrheim show the ±1 dichotomy is specific to d≥3; two dimensions admit anyons

1987

— Hong, Ou and Mandel measure two-photon bunching at a beamsplitter — indistinguishability as a laboratory dial

Going further

Reading. Griffiths, 3rd ed., Ch. 5 — the two-particle construction at its most teachable. Cohen-Tannoudji, Diu & Laloë, Vol. II, Ch. XIV — the exchange-degeneracy problem with full care (the first derivation follows it in spirit). Sakurai & Napolitano Ch. 7; Shankar Ch. 10; Townsend Ch. 12.

Provenance. The AllQFT base notes: "Derivations of the $\pm1$ exchange eigenvalues and the exchange-term shift in $\langle(x_1-x_2)^2\rangle$ should be recomputed against Griffiths Ch. 5 and Cohen-Tannoudji Vol. II Ch. XIV. The thermal de Broglie wavelength $\lambda_{\mathrm{th}}=h/\sqrt{2\pi m k_B T}$ and the crossover criterion $n\lambda_{\mathrm{th}}^3\sim1$ are standard statistical-mechanics results (e.g. Pathria, Statistical Mechanics). Spin-statistics assignment of bosons/fermions to integer/half-integer spin is empirical here and proven in relativistic QFT (Sakurai Ch. 7, Weinberg Vol. I). No nonstandard numerical values are quoted." Per the chapter ownership map, the base's $\pm1$ -eigenvalue derivation is built at operator level at Permutation Symmetry and the Exchange Operator, and its $\langle(x_1-x_2)^2\rangle$ exchange-shift derivation is owned by Symmetric and Antisymmetric States; this article's own core — the overlap theorem and the $2/(1+S^2)$ law — is recomputed from scratch in its verification script. The base's "equal modulus ⇒ global phase" shortcut is flagged above as a gap, not repeated.

Prerequisites. The Postulates of Quantum Mechanics · tensor-product state spaces · symmetries and conservation. Leads to. Permutation Symmetry and the Exchange Operator · Symmetric and Antisymmetric States · Bosons and Fermions · Spin-Statistics Theorem · Pauli Exclusion Principle.

Is indistinguishability an experimental fact or a definition?

Both, at different layers. That electrons have identical mass, charge, and magnetic moment is experimental, tested to extraordinary precision. That no hidden label exists even in principle is a postulate — but one with sharp consequences: exchange interference (Hong–Ou–Mandel dips, identical-particle-scattering structures), exclusion-built atomic shells, Bose condensation. Every one would fail quantitatively if electrons carried even one readable bit of identity, so the definition survives because its consequences keep passing tests.

If the labels are unphysical, why does the formalism use them at all?

Because the tensor-product construction — the only state-space machinery we have for composite systems — is built from labeled factors. We write states as if particles had names, then demand that nothing observable depends on the naming. It is exactly like coordinates on a sphere: you cannot chart the sphere without coordinates, yet nothing physical depends on which chart you chose. The symmetrization postulate is the "chart-independence" condition for particle labels.

Does treating far-apart electrons as distinguishable ever fail?

Only when the modes stop being far apart in the relevant coordinate. Colliding electrons overlap during the collision — identical-particle interference is then order one and shows up in the cross section (Identical-Particle Scattering). The test is never distance alone; it is the overlap integral of the actual modes, including spin: opposite-spin wavepackets have zero exchange correction in spin-blind detection even at full spatial overlap.

Why do I keep hearing that identity implies ψ(2,1) = ±ψ(1,2)? You said that argument has a gap.

The conclusion is nature's choice; the two-line argument is incomplete. Equal probability densities do not force two states to differ by a constant phase — a product state and its relabeled partner have identical label-blind statistics while being orthogonal vectors. What identity forces is the degeneracy of the first derivation; the postulate then selects the symmetric and antisymmetric rays. The operator version — why exactly those two sectors are consistent — is the next article's centerpiece.

Do bosons attract each other and fermions repel?

No force is involved, and the Hamiltonian contains no new term. What changes is correlation geometry: symmetric states put more amplitude on configurations with the particles close (the bunching ratio of this article), antisymmetric states dig a hole along x₁ = x₂. If an interaction is present, these correlations reweight its expectation value — the exchange-integral story of Exchange Forces and Exchange Energy. The "exchange force" is standard shorthand; no fifth force hides in it.

Why are there exactly two allowed symmetry types — why not states that pick up an i under exchange?

For two particles, exchanging twice must return the original state, limiting eigenvalues to ±1 — but the honest general answer involves the permutation group for N particles and the dimensionality of space. In three dimensions only the totally symmetric and totally antisymmetric sectors give every pair swap a consistent sign; in two dimensions exchange paths are braids, double exchange is not the identity motion, and any phase e^{iθ} becomes legal — anyons, owned at Anyons and Fractional Statistics. The full argument is the next article's centerpiece theorem.