Particle exchange operator ($P_{12}$)

For two particles, the exchange operator sends $|\alpha ⟩\otimes |\beta ⟩$ to $|\beta ⟩\otimes |\alpha ⟩$. It has eigenvalues $+1$ on symmetric states and $-1$ on antisymmetric states.

Physical states of identical particles have definite symmetry under every particle exchange operator: symmetric for bosons and antisymmetric for fermions.

For two equal-mass particles with orthonormal coordinates $y_{\pm }=(x_1\pm x_2)/\sqrt{2}$ confined to $|y_{\pm }|<a$, the Hamiltonian is the sum of two width-$2a$ infinite square well Hamiltonians. Exchange preserves $y_{+}$ and reverses $y_{-}$. The relative-coordinate box mode is even for odd index $s$ and odd for even index $s$. Pure spatial bosonic exchange symmetry therefore allows odd $s$, while pure spatial fermionic exchange symmetry allows even $s$.

The bosonic ground mode is $(r,s)=(1,1)$, while its first excited mode $(2,1)$ is degenerate in energy with the fermionic ground mode $(1,2)$. Angular spin degrees of freedom, if present, must be included in the full exchange symmetry rather than assigned the spatial rule automatically. The mean relative distance is $⟨|y_{-}|⟩=a(1/2-2/{\pi }^2)$ in the lowest even mode and $a/2$ in the lowest odd mode, obtained by integrating the normalized squared cosine and sine. Multiplication by $\sqrt{2}$ gives the particle separation.

A measurement addressing a spatial mode can prepare an opposite-spin product in the mode description without violating fermionic antisymmetry. For orthogonal orbitals ${\varphi }_a,{\varphi }_b$, the conditional mode state with spin up in $a$ and spin down in $b$ corresponds to the antisymmetric particle-coordinate state $[{\varphi }_a(1){\varphi }_b(2)|\uparrow \downarrow ⟩-{\varphi }_b(1){\varphi }_a(2)|\downarrow \uparrow ⟩]/\sqrt{2}$. Exchange of mode spin factors is different from full exchange of the identical particles. Only the latter must remain antisymmetric.

For free fermions on a ring, each one-particle orbital has integer angular momentum label $l$ and two spin states. Two particles occupy the $l=0$ orbital in a unique singlet state at zero kinetic energy. At the next distinct energy ${\hslash }^2/(2m{R}^2)$, orbitals $0,+1$ or $0,-1$ can be combined with one singlet state or three triplet state spin states, giving total quantum degeneracy eight. The lowest triplet state has six states and its antisymmetric spatial factor vanishes when the particles coincide.

The operator ${\hat{P}}_{12}$ interchanges all one-particle variables for particles 1 and 2. Since ${\hat{P}}_{12}^2=\hat{I}$, a state of definite exchange symmetry has eigenvalue $+1$ or $-1$.

The complete state of identical bosons is symmetric under exchange, while that of identical fermions is antisymmetric. The condition includes spin and other internal variables as well as spatial coordinates.

Bosons are identical particles whose total multiparticle state is symmetric under exchange. In three-dimensional relativistic quantum theory they have integer spin.

A boson with spin one. A massive vector boson has three physical particle polarizations. A complex Proca field describes a charged particle and its antiparticle, each with three particle polarizations; the free classical field has three complex dynamical components after its constraints are enforced.

Bosonic exchange symmetry is the symmetric exchange statistics of identical bosons. Commuting creation operators build the symmetric bosonic Fock space; a normalized one-particle mode has occupation states $(a^{†}{)}^n|0⟩/\sqrt{n!}$ for every nonnegative integer $n$. At thermal equilibrium this symmetry leads to Bose-Einstein statistics and the Bose-Einstein distribution. Exchange symmetry itself does not require thermal equilibrium.

Place one noninteracting spin-zero boson in each of two orthonormal orbitals $a,b$. Its normalized symmetric spatial wavefunction is $[a({\mathbf{r}}_1)b({\mathbf{r}}_2)+b({\mathbf{r}}_1)a({\mathbf{r}}_2)]/\sqrt{2}$. At coincident positions the probability density is $2|a(\mathbf{r}){|}^2|b(\mathbf{r}){|}^2$, twice the labelled distinguishable-product value. Coincidence at an exact point is understood as a density or the limit of small detector volumes. If both particles occupy the same orbital, the normalized product is already symmetric and this comparison gives factor one instead; the factor two is not a universal property of every two-boson state.

Commuting creation operators make multiparticle states invariant under exchange of their labels. With $[a,a^{†}]=1$, the normalized number states $(a^{†}{)}^r|0⟩/\sqrt{r!}$ exist for all $r\ge 0$, giving unrestricted bosonic occupation numbers. The single-mode thermal sum is $Z={\sum }_{r\ge 0}{e}^{-\beta Er}=(1-e^{-\beta E}{)}^{-1}$ and the mean occupation is $(e^{\beta E}-1{)}^{-1}$. Thus the exchange symmetry and the Bose-Einstein distribution are two expressions of the same bosonic algebra.

Fermions are identical particles whose total multiparticle state is antisymmetric under exchange. In three-dimensional relativistic quantum theory they have half-integer spin.

The complete wavefunction of identical fermions changes sign on exchange of two particles, including their spatial and internal coordinates. The Pauli exclusion principle follows because an antisymmetrized product with two identical complete spin orbitals is zero. Antisymmetry concerns the total state; a spin-antisymmetric singlet state can have a symmetric spatial factor.

The antisymmetric exchange statistics of identical fermions. Interchanging two one-particle labels reverses the total state, and fermionic creation operators anticommute. Occupation of a one-particle mode is therefore zero or one, giving the Pauli exclusion principle; thermal equilibrium leads to the Fermi-Dirac distribution. Color, flavor, spin and orbital factors must together have the required antisymmetry, even if one factor is symmetric.

The lowest even quantum harmonic oscillator quantum state has density $\rho (y)=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$, and the lowest odd quantum state has density $2y^2\rho (y)/{\tau }^2$. Since ${\rho }^{\prime }=-2y\rho /{\tau }^2$, integration by parts gives ${\mathbb{E}}_{\mathrm{F}}[f]={\mathbb{E}}_{\mathrm{B}}[f+y{f}^{\prime }]$ whenever the boundary terms vanish. For $f(y)=y^n$, this proves the displayed identity. Odd moments vanish in both quantum states by parity; even moments quantify the larger spatial spread in the odd quantum state. In particular the relative-coordinate variance is three times as large, and the odd density vanishes at particle coincidence.

The convention $F_j(\eta )=\Gamma (j+1{)}^{-1}{\int }_0^{\infty }{x}^j/[1+e^{x-\eta }]dx$ is useful for parabolic-band carrier densities. In three dimensions, a spin-degenerate single valley has $n=2[m{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}{F}_{1/2}[(\mu -E_c)/(k_{B}T)]$. Other conventions omit the gamma normalization, so prefactors must be stated.

For equal-mass conduction and valence edges at $0,-E_g$ and fixed $\mu =0$, $n=N(T)F_{1/2}(0)$ and $p=N(T)F_{1/2}[-E_g/(k_{B}T)]$, where $N(T)=2[m{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. Other states pinning the chemical potential can compensate the unequal carrier charges; imposing intrinsic charge neutrality in addition would change the problem.

Expanding the Fermi denominator as an alternating exponential series gives $F_{1/2}(0)=(1-2^{-1/2})\zeta (3/2)\simeq 0.765147$. Thus a band with chemical potential pinned exactly at its lower edge has $n=F_{1/2}(0)N_c$, rather than the classical result $N_c$.

For two identical spin-one-half fermions with no additional antisymmetric internal factor, the antisymmetric spin-one-half singlet state combines with an even-$\ell$ symmetric orbital state, while the symmetric spin-one-half triplet state combines with an odd-$\ell$ antisymmetric orbital state.

The spatial ground orbital of a three-dimensional quantum harmonic oscillator has $l=0$ and energy $3\hslash {\omega }_0/2$; its first excited shell has $l=1$ and energy $5\hslash {\omega }_0/2$. Two identical electrons both in the ground orbital must form the singlet state, giving one state at total energy $3\hslash {\omega }_0$.

At energy $4\hslash {\omega }_0$, one electron occupies each shell. There are two ground-shell spin orbitals and six first-excited-shell spin orbitals, hence twelve Slater determinants. Equivalently, $0\otimes 1$ gives $L=1$; symmetric spatial combinations with the singlet state give three states, while antisymmetric spatial combinations with the triplet states give nine. Distinct occupied orbitals allow both spatial exchange symmetries, so both spin sectors must be kept.

For the interaction $\lambda \mathbf{S}\cdot \mathbf{L}$ built from the total two-electron spin and orbital angular momentum, set $\mathbf{J}=\mathbf{L}+\mathbf{S}$. The spin-dot-product eigenvalue identity then gives the shift in the title. In the first excited two-electron oscillator manifold, the singlet has $L=1,S=0,J=1$ and zero shift with multiplicity three. The triplet has $L=S=1$, with $J=0,1,2$ shifts $-2\lambda {\hslash }^2,-\lambda {\hslash }^2,+\lambda {\hslash }^2$ and multiplicities $1,3,5$.

The multiplicities sum to twelve, and their weighted shift sums to zero. The specified interaction preserves both $L^2$ and $S^2$; it is distinct from summing separate single-electron spin-orbit interactions. Its shifts are exact within this model's invariant oscillator manifold, with weak coupling ensuring the original level ordering remains useful.

For well width $a$, put $\epsilon ={\pi }^2{\hslash }^2/(2m{a}^2)$ and $J=\lambda {\hslash }^2$ when the interaction is $-\lambda {\mathbf{S}}_1\cdot {\mathbf{S}}_2$. The spin-dot-product eigenvalue gives exchange energies $3J/4$ for the singlet state and $-J/4$ for the triplet states. Identical fermions require symmetric spatial orbitals with the singlet and antisymmetric orbitals with the triplet. Two particles in the same orbital can therefore only be singlets.

For $0<J\ll \epsilon$, the three lowest distinct energy levels are the orbital-$11$ singlet at $2\epsilon +3J/4$, orbital-$12$ triplet at $5\epsilon -J/4$, and orbital-$12$ singlet at $5\epsilon +3J/4$, with multiplicities $1,3,1$. Distinguishable particles instead allow an orbital-$11$ triplet, making the lowest three levels $2\epsilon -J/4$, $2\epsilon +3J/4$, and $5\epsilon -J/4$, with multiplicities $3,1,6$. The sixfold level includes both orbital orderings. For negative $J$, the spin-sector orderings reverse within each fixed orbital sum.

For two identical fermions of mass $m$ in the same one-dimensional quantum harmonic oscillator, use center of mass $X=(x_1+x_2)/2$ and separation $r=x_1-x_2$. The corresponding oscillator masses are $2m$ and reduced mass $m/2$, with the same frequency $\omega$. If $N,n$ are their occupation numbers, the unperturbed energy is $\hslash \omega (N+n+1)$. The fermionic exchange symmetry requires even $n$ with a spin-one-half singlet state and odd $n$ with a spin-one-half triplet state.

For a delta potential $\lambda \delta (r)$ the first-order shift is $\lambda |{\phi }_n^{\mathrm{rel}}(0){|}^2$. Every odd relative wavefunction vanishes at contact, so these states remain exact unaffected eigenstates of the spatial problem. For the even relative ground state, the oscillator length is $\sqrt{2}\ell$, where $\ell =\sqrt{\hslash /(m\omega )}$, and the shift is $\lambda /(\sqrt{2\pi }\ell )$, independent of $N$. This separates interaction-sensitive singlets from contact-insensitive triplets.

The Pauli exclusion principle says that each complete one-particle state can have fermionic occupation number zero or one. “Complete” includes spin and any other internal labels, as well as the spatial wavefunction. Thus two electrons can share a spatial orbital with opposite spin projections: these are two different spin orbitals. Equal energy alone does not mean equal quantum state, because degenerate energy levels contain several independent one-particle states.

This distinction explains how an electron shell can accommodate several electrons while exclusion still allows only one per complete one-particle state. It also makes a finite model of fermionic state counting surprisingly simple: two spatial orbitals with two spin choices each give four complete one-particle states, and exactly six two-electron basis states. The following diagram enumerates them; its boxes represent occupation numbers, not locations in space.

Pauli's Exclusion principle and quantum mechanics by Wolfgang Pauli is his retrospective account, delivered on 13 December 1946 for the 1945 Nobel Prize. On printed pp. 29–30, he describes proposing an extra two-valued electron property in autumn 1924 and formulating exclusion in 1925. The crucial clue linked counts of resolved energy eigenstates in alkali spectra under a magnetic field with the populations of closed electron shells. Resolving every label turned a complicated shell capacity into one electron per complete one-particle state. This historical route preceded the modern spin interpretation and the antisymmetric wavefunction formulation; it was not originally a deduction from a Slater determinant.

When spin does not split the orbital energies, a fermionic orbital can contain $2s+1$ particles with distinct spin projections. Fill these complete states in order of increasing energy to find a noninteracting many-particle ground state. Pauli exclusion does not require different spatial orbitals for particles with different available spin states. Imposed spin polarization or interactions can change the filling.

One identical fermion in each of two input modes remains one in each output after a lossless two-mode unitary transformation. The fully occupied two-mode state transforms by the determinant phase. Pauli exclusion prevents double occupation of either specified mode.

Suppression of a transition when its required final fermion state is already occupied. In a low-temperature Fermi gas, an otherwise allowed collision must place outgoing particles in empty states; occupation restricts its available phase space. This does not mean that fermions cease interacting.

In 1925, Wolfgang Pauli connected two puzzles: alkali metal doublets and their anomalous Zeeman effect, and the capacities of closed electron shells. His paper assigned the extra two-valued property to the optically active electron, rather than an angular momentum of the closed core. Its Paschen-Back effect classification used four quantum numbers; these historical labels should not be identified literally with modern $n,\ell ,m_{\ell },m_s$. The exclusion rule forbade matching all four. This connected resolved one-electron states with closed-group populations. An additional test was the missing triplet counterpart of the equivalent two-$s$-electron ground state in alkaline earth metals. Pauli explicitly claimed counts of allowed states, not their energies or ordering.^[ref]

The modern question is especially concrete: if a Quantum Hamiltonian has a lowest one-particle state, why cannot every electron occupy it? Minimizing energy answers which of the allowed quantum states is preferred; the Pauli exclusion principle determines which quantum states are allowed in the first place. Spectral lines then test energy differences between allowed multiparticle quantum states. A line count is not itself a direct count of occupied spin orbitals: transitions also depend on selection rules and the prepared quantum state populations. The observed anomalous Zeeman effect supplied constraints for the historical classification; the two-particle antisymmetrization below supplies the modern mathematical explanation of exclusion.

Choose the one-particle states of a hydrogen atom, or a central-field description of a chemical atom, with labelsThe principal quantum number $n$ identifies the electron shell. The orbital angular momentum label $\ell$ identifies the electron subshell; the magnetic quantum number $m_{\ell }$ labels $2\ell +1$ independent spatial orbitals. Each has two spin one-half along an axis choices, hence $2(2\ell +1)$ spin orbitals. Applying the Pauli exclusion principle once per spin orbital givesThese quantum numbers label wavefunctions and spin projections, not electron trajectories. They need not resolve exact energy eigenstates in the presence of spin-orbit interaction; changing to coupled angular momentum eigenstates preserves the count of independent one-particle states within this shell model.

For example, $n=3$ allows $\ell =0,1,2$, conventionally $3s,3p,3d$. Their spatial orbital counts are $1,3,5$; their spin orbital counts are $2,6,10$. The sum is eighteen. The algebra behind “one per complete state” is short: two-particle antisymmetrization gives $u\wedge u=(u\otimes u-u\otimes u)/\sqrt{2}=0$, and the zero vector cannot be a normalized quantum state. Opposite-spin electrons in the same spatial orbital have distinct complete states, so their antisymmetric product survives.

This supplies a structural link to the periodic table: finite electron subshell capacities make repeated patterns of occupied electron configurations possible. It does not predict the exact ground state electron configuration or energy of a many-electron chemical atom. Different electron subshells overlap in energy, and electron interactions change their ordering. In particular, the periodic table period lengths are not simply successive $2n^2$ values: the third periodic table period has eight chemical elements, although the full $n=3$ electron shell has eighteen available states. A capacity counts available spin orbitals; an actual filling order requires a Quantum Hamiltonian.

A spatial orbital is a normalized spatial wavefunction used in a one-particle description. An atomic orbital is the corresponding notion for a chemical atom; a molecular orbital belongs to a molecule. A spatial orbital is not a trajectory and does not by itself specify spin.

When spatial and spin factors separate, a spin orbital combines a spatial orbital ${\varphi }_a$ with a normalized spin state ${\chi }_{\sigma }$. For an electron, choose spin one-half along an axis with $\sigma =\uparrow ,\downarrow$. The two spin orbitals $u_{a\uparrow }$ and $u_{a\downarrow }$ are orthogonal even though their spatial orbital is identical, since their inner product contains $⟨{\chi }_{\uparrow }|{\chi }_{\downarrow }⟩=0$. When spatial and spin variables do not separate, the complete one-particle state is instead a multicomponent wavefunction $u(\mathbf{r},s)$; exclusion still applies to that complete quantum state.

The particle exchange operator $P_{12}$ swaps the complete particle slots, including spin. Its antisymmetric projector is $A_{12}=(I-P_{12})/2$. For orthonormal one-particle states $u,v$, the normalized multiparticle quantum state isBut attempting the same construction with $v=u$ gives the zero vector:The zero vector cannot be normalized to a quantum state. This is exclusion as cancellation of probability amplitudes, rather than a repulsive force. More generally, for normalized $u,v$ the squared norm of the numerator divided by $\sqrt{2}$ is $1-|⟨u|v⟩{|}^2$; the expression has unit norm only when $u,v$ are orthogonal. An overall quantum phase difference between $u,v$ describes the same one-particle state and also gives zero. Pauli explicitly connects exclusion with a vanishing antisymmetric wavefunction.^[ref]

A Slater determinant extends two-particle antisymmetrization to $N$ fermions. Here $x_i$ contains position and spin or other internal variables. Distinct orthonormal one-particle states give the normalized wavefunction displayed in the title. Repeating a one-particle state repeats a column of the determinant, so the wavefunction vanishes. Exchanging particle slots exchanges rows and reverses its sign. A general fermionic quantum state can be a quantum superposition of Slater determinants; it need not be one Slater determinant.

Rules expressing Hamiltonian matrix elements between Slater determinants through one- and two-electron orbital integrals. A two-body Hamiltonian cannot connect determinants differing in more than two spin-orbitals.

Exclusion constrains complete one-particle states, not a collection of little spatial boxes. In particular, two opposite-spin electrons in the same spatial orbital haveThe spatial factor is symmetric; the spin-one-half singlet state supplies the required antisymmetry. At ${\mathbf{r}}_1={\mathbf{r}}_2$, this wavefunction need not vanish for opposite spin labels. For equal spin labels it does vanish. In a continuous position description, an exact coincident point has zero measure anyway; the useful comparison concerns joint probability densities or detection in finite regions. Equal positions do not imply equal complete quantum states.

The fermionic occupation numbers $n_{\alpha }=0,1$ refer to a chosen orthonormal basis of one-particle states. A quantum superposition need not have definite occupation numbers in that basis, and an expectation value $⟨n_{\alpha }⟩$ can be fractional. For example, one electron in $(u+v)/\sqrt{2}$ has mean occupation number $1/2$ in each of two orthogonal spin orbitals, without splitting into two particles. The canonical anticommutation relations give $(a_{\alpha }^{†}{)}^2=0$ and make $a_{\alpha }^{†}{a}_{\alpha }$ a projector, so its expectation value stays between zero and one in every normalized quantum state.

For $M$ complete energy eigenstates of a one-particle Quantum Hamiltonian for identical noninteracting fermions, order their energies ${ϵ}_1\le \cdots \le {ϵ}_M$. At fixed particle number $0\le N\le M$, the Quantum Hamiltonian and the particle-number constraint areThe ground state occupies the $N$ lowest complete one-particle states. A short minimization argument suffices: moving an occupant from a higher energy to an empty lower energy decreases the total energy. This filling describes the ground state at temperature $T=0$; exclusion by itself also applies to excited quantum states and at positive temperature. The underlying thermal description is the Fermi-Dirac distribution.^[ref]

If the last occupied energy belongs to a group of $g$ one-particle states at the same energy and $q$ of them must be occupied, with all lower states filled, the ground state subspace has dimension $\left(\genfrac{}{}{0ex}{}{g}{q}\right)$. Exclusion does not select a particular basis or a particular quantum superposition in that subspace. The factor of two for an electron spatial orbital assumes both spin projections are included; a model restricted to one spin projection has only one state per spatial orbital. A magnetic field can split the degenerate energy levels, and interactions can change the lowest-energy electron configuration. Thus this finite noninteracting filling rule is not a complete prescription for interacting chemical atoms.

Choosing $N$ different complete one-particle states from an orthonormal basis of size $M$ gives $\left(\genfrac{}{}{0ex}{}{M}{N}\right)$ independent Slater determinants. This is the binomial coefficient because only the occupied subset matters: permutations change the sign of the Slater determinant, not the physical quantum state. Summing over particle numbers gives $2^M$ states in the finite fermionic Fock space.

For the diagram's two spatial orbitals $A,B$, take energies $0,\Delta$ with $\Delta >0$, each independent of spin. In the complete-state order $(A\uparrow ,A\downarrow ,B\uparrow ,B\downarrow )$, the six $N=2$ occupation number strings are $1100,1010,1001,0110,0101,0011$. Their total energies are $0,\Delta ,\Delta ,\Delta ,\Delta ,2\Delta$, giving energy multiplicities $1,4,1$. For $N=3$, the ground state energy is $\Delta$ with two independent ground states, according to which spin orbital in $B$ is occupied. Across $N=0,1,2,3,4$, the counts are $1,4,6,4,1$, summing to $16$. These are exact counts for this model, not measured atomic energies.

Consider an original noninteracting model with three orthonormal spatial orbitals $A,B,C$ at energies $0,\Delta ,3\Delta$, where $\Delta >0$. Both spin projections are available at each energy, so there are six spin orbitals. The Quantum Hamiltonian is $H=h\otimes I+I\otimes h$, with no interaction between particles and no dependence on spin. Comparing its allowed ground states isolates the effect of fermionic exchange symmetry.

For two distinguishable spin-one-half particles, the spatial ground state is $A\otimes A$ and any of four independent two-spin states is possible: $E_0=0$ and ground-state degeneracy four. For two identical electrons, both can still occupy $A$, using $A\uparrow$ and $A\downarrow$, but there is just one allowed ground state: the symmetric spatial factor times the spin-one-half singlet state,Exclusion has reduced the ground-state degeneracy without raising the ground state energy.

Now restrict both particles to the same spin projection $\uparrow$. Distinguishable particles can still both occupy $A\uparrow$, so $E_0=0$. Identical electrons must occupy different spatial orbitals; the lowest choice is $A,B$, withThe spin factor is a spin-one-half triplet state, and the spatial factor is antisymmetric. These are ground states within the restricted parallel-spin sector, not the unrestricted two-electron ground state. Both restricted models have ground-state degeneracy one.

The first excited two-electron energy in the unrestricted model is $\Delta$, with four independent quantum states: one occupant in $A$, one in $B$, and two spin choices for each. A basis adapted to particle exchange operators separates these into one spin-one-half singlet state with symmetric spatial factor and three spin-one-half triplet states with antisymmetric spatial factor. All four have equal energy in this model. Singlet versus triplet does not by itself determine an energy gap; an interaction or a spin-dependent term is needed to split them.

As a further count, the full two-electron space has dimension $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$. Its energies in units of $\Delta$ and corresponding multiplicities are $(0,1,2,3,4,6)$ and $(1,4,1,4,4,1)$. For distinguishable spin-one-half particles the ordered tensor product instead has dimension $6^2=36$, with multiplicities $(4,8,4,8,8,4)$ at those same energies. These are exact model counts, not atomic measurements. The higher restricted ground state energy follows from removing forbidden quantum states; the Quantum Hamiltonian contains no additional “exclusion force”.

The 2007 paper Comparison of the Hanbury Brown-Twiss effect for bosons and fermions by T. Jeltes and others compares polarized helium-3 fermions with helium-4 bosons. The chemical atoms occupy a metastable atomic level. Switching off magnetic trapping of neutral atoms releases them onto a microchannel plate detector with a delay-line anode, 63 cm below. Position and arrival-time readouts give three-dimensional pair separations by time-of-flight atom detection. About 1000 clouds per isotope supply histograms; normalization removes the broad cloud shape.

The pair detection correlation dips below one for helium-3 and rises above one for helium-4. The same apparatus, matched temperatures near $0.5\mu \mathrm{K}$, nearly identical trapping potentials, negligible interactions, and magnetic field gradient checks of spin polarization support an exchange-statistics interpretation. Finite detector resolution suppresses contrast; trap switch-off and resolution uncertainties limit quantitative agreement. A diverging laser lens deepens the dip, but its aberrations preclude a precise lens-model test. This detects exchange correlations, rather than measuring occupation of one selected spin orbital.^[ref]

The pair detection correlation divides joint detection probability by the product of single detection probabilities. The baseline $g^{(2)}=1$ means independent detection; $g^{(2)}<1$ means suppressed pairs. Its value depends on the prepared quantum state and detector resolution, so exclusion alone does not require every experimental curve to reach zero.

The experiment above combines a microchannel plate detector with a delay-line anode to record individual atom arrivals. The apparatus is described in the paper's Fig. 1 and Methods.

The spin-statistics theorem associates integer-spin particles with bosonic exchange symmetry and half-integer-spin particles with fermionic exchange symmetry in relativistic quantum field theory under the standard locality and positivity assumptions.

A spin-one parent cannot decay into two identical spin-zero particles in the same internal state while preserving rotational symmetry and Bose exchange symmetry. Their final total angular momentum is orbital $L=1$, which is antisymmetric under interchange. Bose symmetry requires even $L$ instead. An energetically allowed decay with matching parity can therefore remain forbidden independently of the interaction's coupling strength.

Restrictions linking the exchange symmetry of nuclear spin and spatial wavefunctions. For identical nuclei, the total wavefunction must be antisymmetric for fermions and symmetric for bosons. Electronic and vibrational exchange parity must be included before assigning rotational parity.

Exchanging two identical bound atoms exchanges their constituent fermions. An atom with mass number $A$ and atomic number $Z$ contains $A$ nucleons and $Z$ electrons, so its exchange sign is $(-1{)}^{A+Z}$ when its internal state is preserved. An odd count gives a composite fermion; an even count gives a composite boson. This is equivalently the integer-versus-half-integer total-spin distinction in the Spin-statistics theorem. For example, neutral lithium-6 has nine fermionic constituents and is a fermion, while neutral lithium-7 has ten and is a boson. Counting nucleons alone instead classifies the nucleus, not the neutral atom. This effective composite-particle description requires energies and separations where the bound atom remains intact.

For two spin-one particles, the spin tensor product has a six-dimensional symmetric subspace and a three-dimensional antisymmetric subspace. A symmetric spatial state combines with symmetric spin, while an antisymmetric spatial state combines with antisymmetric spin, so that the total bosonic state remains symmetric.

If $E_n=E_0+n\Delta$, the two-particle level $2E_0+N\Delta$ has degeneracyEach unequal orbital pair contributes six symmetric-spin states and three antisymmetric-spin states; an equal orbital pair exists only for even $N$ and contributes six states.