Slater determinant

{title2=$A_{12}=(I-P_{12})/2$}

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> is
$$
|u\wedge v\rangle=\frac{|u\rangle\otimes|v\rangle-|v\rangle\otimes|u\rangle}{\sqrt2}.
$$
But attempting the same construction with $v=u$ gives the zero vector:
$$
A_{12}(|u\rangle\otimes|u\rangle)=\frac{|u\rangle\otimes|u\rangle-|u\rangle\otimes|u\rangle}{2}=0.
$$
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 $\sqrt2$ is $1-|\langle u|v\rangle|^2$; 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>.<Exclusion principle and quantum mechanics by Wolfgang Pauli>[p. 31, antisymmetric functions with repeated states]{ref}

\Image[/pauli-antisymmetric-state-pairs.png]
{title=Antisymmetrization removes equal-state pairs and pairs the twelve remaining ordered products into six fermionic basis states}


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 Incoming links: Two-particle antisymmetrization