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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.

Two electrons of opposite spin occupying the same spatial orbital. The mean occupation product measures its probability at a specified site. In the two-electron dimer, the sum over sites is also the probability of any doubly occupied site, since two sites cannot both be doubly occupied.

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”.

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.

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.

The single-band Hubbard model balances electron hopping between localized spatial orbitals against local interaction. A common form isThe first sum is over neighbouring unordered site pairs. The parameter $t$ sets hopping, while $U_H>0$ penalizes double occupancy. At $U_H=0$ electrons occupy independent energy bands; strong repulsion suppresses local charge fluctuations and can favour local-moment behaviour. This competition was the motivation for the model's treatment of narrow-band electron correlation.^[ref]

At one electron per site and $U_H\gg |t|$, virtual double occupancy yields the antiferromagnetic scale $J_{\mathrm{ex}}=4t^2/U_H$. The two-site Hubbard dimer makes this concrete: the singlet state is lowered by $(\sqrt{U_H^2+16t^2}-U_H)/2$, while triplet states have no such shift. Its small-hopping limit gives Hubbard kinetic exchange.

Interaction conventions must match. For dimensionless spin-half operators, $-2U({\mathbf{S}}^2-3n/4)=3U{n}_{\uparrow }{n}_{\downarrow }$, so that spin-square convention has $U_H=3U$, not $U_H=U$.

Spin and spatial exchange symmetry. Write the normalized spin singlet and triplet states asThe singlet is antisymmetric under exchange and has total spin zero; the triplet is symmetric and has total spin one. For orthogonal orbitals defineIdentical spin one-half particles are fermions, so the complete state must be antisymmetric. Thus $C{\chi }_{00}$ is allowed, and $A{\chi }_{1M}$ is allowed for each triplet component. The spatially symmetric same-orbital states $|aa⟩,|bb⟩$ also require the singlet. A same-orbital triplet vanishes by the Pauli exclusion principle.

These products give a spin-adapted basis. More generally the physical state can be writtenA superposition of the singlet and triplet sectors need not factor into one spatial function times one spin function. Factorization holds for a definite-spin sector, as used in the subsequent model. If the original normalized orbitals are not orthogonal, the corresponding $ab\pm ba$ normalization is $\sqrt{2(1\pm |⟨a|b⟩{|}^2)}$; the remainder of this calculation uses the stipulated orthogonal orbitals.

Integral matrix elements and positivity. Let $a(\mathbf{r})$ and $b(\mathbf{r})$ denote the localized spatial orbitals. With fixed proton positions, the one-electron electronic operator isThe proton-proton repulsion is a separate common nuclear energy. The requested one-electron and direct two-electron elements areSince $H_1=h\otimes I$,Likewise,The integrand is nonnegative and is positive on a set of nonzero measure. This is the repulsion cost of double occupancy; the coincidence singularity is integrable for ordinary smooth localized orbitals.

One electron on each center. Suppress the normalized singlet spin factor and use $|\psi ⟩=C$. For each of the spatial products $ab$ and $ba$, the one-body sum has expectation $2E_0$. The cross terms vanish because a one-body hop leaves an orthogonal spectator orbital. The interaction contribution is the direct integral plus the exchange integral, both neglected in the model. ConsequentlyThis state has no same-center double occupancy, but remains an exchange-symmetric spatial superposition rather than assigning a distinguishable electron permanently to each proton.

Two electrons in the bonding orbital. The normalized one-electron bonding orbital is $g=(a+b)/\sqrt{2}$. The trial spatial state is $|\varphi ⟩=|gg⟩$, whose one-body expectation isExpanding the product gives $|\varphi ⟩=(|aa⟩+|ab⟩+|ba⟩+|bb⟩)/2$. The total probability of double occupancy is therefore $1/2$. Keeping only on-site Coulomb repulsion gives the intended interaction truncation,and the interaction expectation is $U/2$. HenceThis truncation must be stated. In particular, mixed Coulomb elements such as $X=⟨aa|H_{1,2}|ab⟩$ do not vanish merely by orbital orthogonality or inversion symmetry. The printed list does not explicitly name them. If they are retained and are equal and real by site symmetry, their contribution to $E_{\varphi }$ is $2X$; the requested expression uses their neglect along with the other off-site interaction terms. No numerical value of $X$ is supplied or invented.

Energy comparison and the bond. Put $x=|t|/U$ with $U>0$ and $t=-|t|$. ThenThe straight line starts at $1/2$ and crosses zero at $x=1/4$. The purely bonding-orbital trial state is better for $x>1/4$, while the singly occupied covalent trial state is better for $x<1/4$.

For localized orbitals of approximately fixed atomic size, the hopping magnitude $|t|$ decreases rapidly, typically exponentially times a polynomial, as the proton separation grows. The on-site integral $U$ remains of atomic Coulomb scale and tends to the isolated-orbital value; it does not decay as the intersite repulsion $1/R$. If the orbitals themselves are allowed to relax, $U$ also changes, so no universal monotonic law for $U(R)$ follows from the data.

At small $|t|/U$, electron correlation suppresses the ionic configurations $aa$ and $bb$, favoring a singlet with roughly one electron on each center. A single doubly occupied bonding orbital has an artificial finite ionic weight even as the centers separate. At large $|t|/U$, delocalization gains enough kinetic energy that the bonding-orbital description becomes useful and charge fluctuations are substantial.

The crossing of two restricted trial energies is not a sharp transition of the actual two-site ground state. To see this without adding empirical data, include both $C$ and the symmetric ionic state $I_{+}=(|aa⟩+|bb⟩)/\sqrt{2}$. This two-site Hubbard dimer reduces toIts ground state is $\cos \theta C+\sin \theta I_{+}$, with $\tan 2\theta =4|t|/U$ and $0\le \theta \le \pi /4$. The ionic probability isIt changes smoothly from zero to $1/2$. At large separation, the electronic lowering is $-4t^2/U$ to leading order and vanishes as hopping vanishes. An absolute molecular binding curve would additionally require $E_0(R)$, the nuclear repulsion and the neglected intersite terms; this simplified comparison alone does not determine an equilibrium bond length.

If two fermions occupied the same complete spin orbital $u$, their antisymmetrized product would beThe zero vector cannot represent a normalized quantum state. This is the Pauli exclusion principle: at most one identical fermion occupies any complete one-particle state. Two electrons may occupy the same spatial orbital with opposite spins, because $a\alpha$ and $a\beta$ are different spin orbitals.

The fixed localized determinant is unsuitable near equilibrium. It does not develop the correct shared bonding orbital and is not a pure singlet: for one alpha and one beta orbital, $⟨S^2⟩/{\hslash }^2=1-|⟨A|B⟩{|}^2$, which is nonzero unless the occupied spatial orbitals coincide. At dissociation it is an equal singlet/triplet mixture, though their energies become degenerate there. Near equilibrium the physical ground state is a bonding singlet. An optimized UHF calculation can recover the restricted, spin-pure solution by making its two occupied spatial orbitals equal; that does not rescue the particular fixed atomic-orbital determinant written here. A spin-adapted multideterminant description is needed to combine good bonding near equilibrium with correct singlet dissociation.

The spin Hilbert space is a tensor product of two two-dimensional spaces and therefore has dimension four. Angular momentum addition decomposes it asThese are three orthogonal triplet states and one singlet state. For identical electrons the complete wavefunction must also satisfy fermionic exchange symmetry: a symmetric triplet spin factor accompanies an antisymmetric spatial factor, whereas the antisymmetric singlet spin factor accompanies a symmetric spatial factor. If both electrons occupy the same spatial orbital, only the singlet spin factor is allowed. The four-dimensional spin space does not assert four allowed spin states for that particular spatial occupation.

Write $\Psi =(|\downarrow \uparrow ⟩+|\uparrow \downarrow ⟩)/\sqrt{2}$. Each term has cancelling $z$ components, so $S_z\Psi =0$. In the formula for $S^2$, the first three terms contribute $(3/2-1/2){\hslash }^2\Psi ={\hslash }^2\Psi$. The two ladder products exchange the two terms and contribute another ${\hslash }^2\Psi$. ThereforeThis spin state is symmetric; the accompanying spatial state must be antisymmetric for two electrons. A labelled product $|{\downarrow }_1{\uparrow }_2⟩$ alone is not an eigenstate of spin exchange: its exchanged vector is orthogonal to it. Thus it cannot supply a definite exchange symmetry when multiplied by a fixed symmetric spatial factor.

This does not prohibit opposite spins in different spatial orbitals. For example, with orthonormal orbitals ${\varphi }_a,{\varphi }_b$, the Slater determinantis a valid antisymmetric two-electron state. What fails is treating persistent particle labels as physically distinguishable identities without the required antisymmetrization.

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.

A single-particle wavefunction used as a mode of a many-particle description. An orbital can include spatial and spin variables; a spatial orbital specifies the spatial part. Orthonormal orbitals define creation and annihilation operators. Occupation of one normalized orbital by every boson gives a Hartree product state, while antisymmetrized occupation by fermions gives a Slater determinant.

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 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.

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.

For real spatial orbitals, $K_{pq}=\iint {\varphi }_p(1){\varphi }_q(1)r_{12}^{-1}{\varphi }_q(2){\varphi }_p(2)d1d2$ in atomic units. It enters the energy of a Slater determinant through antisymmetry and becomes small for well-separated localized orbitals.

A Hartree-Fock ansatz allowing different spatial orbitals for alpha and beta spin. This can lower the energy of a stretched bond, but the determinant need not be an eigenstate of total spin squared.

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.