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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 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 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.

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.