Zero-temperature filling of finite fermion levels ($E_0(N)={\sum }_{\alpha =1}^N{ϵ}_{\alpha }$)

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”.