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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.

The eigenvalue of a mode number operator counts excitations in that mode. Bosonic occupation numbers range over all nonnegative integers; fermionic occupation numbers are zero or one.

With no pairing, the Hamiltonian is diagonal in the fermionic occupation numbers. The Pauli exclusion principle allows each momentum mode to be either empty or occupied once. Without a fixed fermion number constraint, occupying a mode lowers the energy precisely when its single-particle energy is negative. Therefore, with $|0⟩$ the Fock vacuum, a ground state isChoose a fixed ordering of the fermionic creation operators; changing it alters only the overall phase. Any zero-energy modes may be occupied or empty without changing the ground-state energy.

To identify the occupied modes, complete the square:Here $\beta$ is the dispersion parameter, not inverse temperature. If ${\beta }^2<4\mu$, every mode has positive energy and the unique ground state is the Fock vacuum, with $E_{\mathrm{GS}}=0$. If ${\beta }^2=4\mu$, there are still no negative-energy modes, but $p=\pm \sqrt{\beta /2}$ have zero energy if they lie on the allowed momentum grid. Each such mode doubles the ground-state degeneracy.

For ${\beta }^2>4\mu$, defineBoth are positive, and the occupied Fermi sea consists of the two intervals $p_{-}<|p|<p_{+}$. With $p=2\pi n/L$, the requested finite-size ground-state energy is explicitlyThe factor two counts opposite momenta. Endpoints are excluded from this sum because they have zero energy; when an endpoint is allowed, its optional occupation adds ground-state degeneracy. A narrow negative-energy interval can contain no allowed finite-size momenta, in which case the sum is empty. The zero-momentum mode is always empty since $ϵ(0)=\alpha \mu >0$.

Write $Q=c^{†}c+d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Expanding the number operators and restoring the displayed order with the canonical anticommutation relations givesThe minus sign in the first line follows from $c^{†}{d}^{†}+dc=-(cd+d^{†}{c}^{†})$. Substitution into the Hamiltonian operator yieldsLet $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$. A convenient real Bogoliubov transformation hasThese obey $u^2+v^2=1$, $u^2-v^2=\epsilon /E$ and $2uv=\Delta /E$. Equivalently $u=\cos \theta$, $v=\sin \theta$ with $0<\theta <\pi /4$ and $\tan (2\theta )=\Delta /\epsilon$. The coefficient of $P$ vanishes, while the coefficient of $Q$ is $E$. The constant isThus the diagonalized two-mode fermion pairing Hamiltonian isThe two transformed fermionic occupation numbers independently take values zero or one. The energy eigenvalues and their eigenspace dimensions are thereforeThe positive quasiparticle energy $E$ makes the transformed vacuum state the unique ground state. The constant $\mu$ is essential for the correct ground-state energy.

As a separate sign and spectrum check, use the original fermionic Fock space basis. Both singly occupied states are energy eigenstates with energy $\epsilon$, since the pairing terms annihilate them. In the ordered even-fermion parity basis $(|0⟩,a^{†}{b}^{†}|0⟩)$,The off-diagonal signs follow from $b^{†}{a}^{†}=-a^{†}{b}^{†}$ and $ab{a}^{†}{b}^{†}|0⟩=-|0⟩$. Its characteristic polynomial is $z^2-2\epsilon z-{\Delta }^2$, giving the remaining two eigenvalues $\epsilon \pm E$, in agreement with the Bogoliubov transformation.

The PDF's displayed zero-mode term has no derivative. Read literally, $d_0\cdot d_0$ vanishes for classical Grassmann variables and supplies no zero-mode symplectic structure. The subsequent canonical-algebra requests therefore require the standard kinetic term $\frac{i}{2}{d}_0\cdot {\dot{d}}_0$. We use that intended correction explicitly; the rest of the displayed action fixes the nonzero-mode normalization.

The Ramond level operator, with vacuum-annihilating normal ordering, isThe canonical oscillator relations, for transverse indices $I,J$, areThe hermiticity convention is ${\alpha }_k^{†}={\alpha }_{-k}$ and $d_k^{†}=d_{-k}$. The commuting bosonic zero mode is supplied by the center-of-mass momentum. For $k>0$, define $a_k^I={\alpha }_k^I/\sqrt{k}$ and $f_k^I=d_k^I$. Their bosonic occupation numbers are $0,1,2,\dots$ and their fermionic occupation numbers are $0,1$. Therefore, on the Fock space generated from an oscillator vacuum,Equivalently, creation operators raise the level by $k$, since $[N_R,{\alpha }_{-k}^I]=k{\alpha }_{-k}^I$ and $[N_R,d_{-k}^I]=k{d}_{-k}^I$. The zero modes commute with $N_R$ and do not change the level. The multiplier imposesso the $N=0$ states are massless. In the Ramond sector the bosonic and fermionic oscillator zero-point contributions cancel, consistently with the stated zero intercept. The massless ground states are spacetime spinors, as the Ramond zero-mode Clifford algebra now shows.

Normalize ${\gamma }^I=\sqrt{2}{\hat{d}}_0^I$. ThenLet $v=|0⟩\ne 0$. Each positive-frequency bosonic annihilator commutes with ${\gamma }^I$, while each fermionic annihilator anticommutes with it. Applying either annihilator to ${\gamma }^{I}v$ therefore gives zero. Thus all eight ${\gamma }^{I}v$ are oscillator vacua. For real $a_I$, the hermitian operator $A={\sum }_I{a}_I{\gamma }^I$ satisfies $A^2=({\sum }_I{a}_I^2)1$, soThis proves the real independence of Clifford-generated vectors, and hence their linear independence over $\mathbb{R}$.

The same argument applies to the nonzero vacuum ${\gamma }^{1}v$, because $({\gamma }^1{)}^2=1$. It gives eight real-linearly independent oscillator vacua ${\gamma }^I{\gamma }^{1}v$. They include $v$ itself, at $I=1$. Products of two zero modes preserve vacuum annihilation just as products of one do.

For the chirality matrix ${\gamma }_9={\gamma }^1{\gamma }^2\cdots {\gamma }^8$, reversing eight anticommuting factors introduces $(-1{)}^{8\cdot 7/2}=1$. HenceMoving any ${\gamma }^I$ through the other seven factors also gives ${\gamma }_9{\gamma }^I=-{\gamma }^I{\gamma }_9$. If ${\gamma }_{9}v=v$, thenThe first collection has negative chirality; the second has positive chirality. If $u_{+}$ and $u_{-}$ have these respective chiralities, hermiticity gives $⟨u_{+},u_{-}⟩=⟨{\gamma }_9{u}_{+},u_{-}⟩=⟨u_{+},{\gamma }_9{u}_{-}⟩=-⟨u_{+},u_{-}⟩$, so they are orthogonal. Combining the two real-independent collections therefore gives at least sixteen real-linearly independent oscillator vacua, eight in each chirality.

The real qualification in the question matters: the particular eight vectors generated from an arbitrary complex $v$ need not be independent over $\mathbb{C}$. Nevertheless the dimension bound from paired Clifford involutions also follows from the full Clifford algebra. Define four commuting hermitian involutions $J_j=i{\gamma }^{2j-1}{\gamma }^{2j}$, $j=1,\dots ,4$. Their joint spectral projections preserve the vacuum space, so it contains a nonzero common eigenvector $w$. Multiplication by ${\gamma }^{2j-1}$ flips the eigenvalue of $J_j$ and leaves the other three eigenvalues unchanged. The sixteen products obtained by independently choosing whether to apply these four odd-indexed Gamma matrices to $w$ consequently have distinct joint eigenvalue quadruples. They are nonzero, mutually orthogonal oscillator vacua. Thus the unprojected vacuum space also has complex dimension at least sixteen, with eight states of each chirality in the minimal representation. A further chiral projection is an additional physical restriction, not part of the oscillator-vacuum conditions here.

In second quantization, expand the annihilating quantized field in a complete orthonormal basis of one-particle states, $\hat{\psi }(x)={\sum }_{\alpha }{\phi }_{\alpha }(x)a_{\alpha }$. At equal times, the canonical commutation relation for bosons and the canonical anticommutation relations for fermions are, respectively,andHere $\delta$ is a Dirac delta distribution. Multiplying by the basis wave functions and integrating gives $[a_{\alpha },a_{\beta }^{†}]={\delta }_{\alpha \beta }$ for bosons, or $\{a_{\alpha },a_{\beta }^{†}\}={\delta }_{\alpha \beta }$ for fermions; the two-annihilator and two-creator brackets vanish in the corresponding convention.

The mode number operator and total particle number operator areThe bosonic occupation numbers are $0,1,2,\dots$. For fermions, $a_{\alpha }^2=0$ and the canonical anticommutation relations imply $n_{\alpha }^2=n_{\alpha }$, so the fermionic occupation numbers are only $0$ and $1$, as required by the Pauli exclusion principle.

Keeping only two modes later is a projection: the projected field bracket is ${\sum }_{j=1}^2{\phi }_j(x){\phi }_j^{\ast }(x^{\prime })$, rather than the full Dirac delta distribution.

With no pairing, the Hamiltonian is diagonal in the fermionic occupation numbers. The Pauli exclusion principle allows each momentum mode to be either empty or occupied once. Without a fixed fermion number constraint, occupying a mode lowers the energy precisely when its single-particle energy is negative. Therefore, with $|0⟩$ the Fock vacuum, a ground state isChoose a fixed ordering of the fermionic creation operators; changing it alters only the overall phase. Any zero-energy modes may be occupied or empty without changing the ground-state energy.

To identify the occupied modes, complete the square:Here $\beta$ is the dispersion parameter, not inverse temperature. If ${\beta }^2<4\mu$, every mode has positive energy and the unique ground state is the Fock vacuum, with $E_{\mathrm{GS}}=0$. If ${\beta }^2=4\mu$, there are still no negative-energy modes, but $p=\pm \sqrt{\beta /2}$ have zero energy if they lie on the allowed momentum grid. Each such mode doubles the ground-state degeneracy.

For ${\beta }^2>4\mu$, defineBoth are positive, and the occupied Fermi sea consists of the two intervals $p_{-}<|p|<p_{+}$. With $p=2\pi n/L$, the requested finite-size ground-state energy is explicitlyThe factor two counts opposite momenta. Endpoints are excluded from this sum because they have zero energy; when an endpoint is allowed, its optional occupation adds ground-state degeneracy. A narrow negative-energy interval can contain no allowed finite-size momenta, in which case the sum is empty. The zero-momentum mode is always empty since $ϵ(0)=\alpha \mu >0$.

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.