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A boson is an identical particle whose total multiparticle quantum state is symmetric under exchange of any two particles, whereas a fermion has a total state that is antisymmetric under every exchange. The Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions.

For three distinguishable spin-one particles, the spin Hilbert space has dimensionLetbe the total angular momentum operator. Since each particle haswe haveThe Hamiltonian operator is consequentlyOn the total-spin sector with quantum number $J$, its energy eigenvalue is

The Clebsch-Gordan decomposition may be performed by first coupling particles $1$ and $2$. Their intermediate spin is $j_{12}=0,1,2$, and coupling the third spin givesThus the three spin-one angular-momentum decomposition contains total spin $J=0,1,2,3$ with multiplicities $1,3,2,1$. Multiplying each multiplicity by the multiplet dimension $2J+1$ givesThe degeneracies sum to $27$, as required.

When the particles are indistinguishable, their integer spin makes them bosons. Their common spatial wavefunction is symmetric, so their spin wavefunction must also lie in the symmetric three-spin-one subspace. Its decomposition isHence only the $J=3$ and $J=1$ levels remain, now with one copy of each multiplet:

For distinguishable particles with one-particle Hilbert spaces ${\mathcal{H}}_1,\dots ,{\mathcal{H}}_N$, the multiparticle quantum state belongs toA product of one-particle states gives a product vector, and arbitrary linear combinations give entangled states. A noninteracting Hamiltonian is the sum of one-particle Hamiltonians acting on their respective tensor factors, so product-state energies add.

For two identical particles, the particle exchange operator $P_{12}$ swaps every degree of freedom. Since $P_{12}^2=I$, its eigenvalues are $\pm 1$. Particles whose allowed total states satisfyare bosons, while those satisfyingare fermions. Thus bosonic states are symmetric and fermionic states antisymmetric under interchange. The same condition applies to every transposition in a system of more than two identical particles.

The unit-normalized Fock vacuum is characterized byChoose the relativistic normalization of a one-particle state from the outset:Since any two creation operators commute, any permutation of their order gives exactly the same state vector. Thus these multiparticle quantum states are symmetric under exchange and describe bosons, without a minus sign or any additional exchange phase.

The Hilbert space is the bosonic Fock space, the direct sum of the symmetric $n$-particle Hilbert spaces, including the zero-particle sector. Sharp-momentum kets are distributionally normalized; square-integrable wave packets provide ordinary Hilbert space vectors. For repeated occupation of a discrete mode, a unit-normalized number state uses $(a^{†}{)}^n|0⟩/\sqrt{n!}$. That factorial does not belong in the above labelled continuum-ket convention.

The normal ordering of both linear maps puts every annihilation operator to the right, soIn particular the stated Hamiltonian has already subtracted the free vacuum energy. For a one-particle state, evaluate the commutators using $[AB,C]=A[B,C]+[A,C]B$:Acting on the Fock vacuum therefore givesThe extra factor $\sqrt{2E_{\mathbf{k}}}$ in the chosen relativistic normalization of a one-particle state is a scalar multiplier and does not change either eigenvalue. Applying the same argument to a multiparticle quantum state gives additive total energy and momentum.

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.

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]