= Two-pair component of polarization-entangled down-conversion
{title2=$|\Psi_2\rangle=(K^\dagger)^2|0\rangle/(2\sqrt3)$}
For two collected spatial modes $a,b$, each with polarizations $H,V$, put $K^\dagger=a_H^\dagger b_V^\dagger+a_V^\dagger b_H^\dagger$. The one-pair state is a <Bell state>. The normalized two-pair component is
$$
|\Psi_2\rangle=\frac{|2H_a,2V_b\rangle+|H_aV_a,H_bV_b\rangle+|2V_a,2H_b\rangle}{\sqrt3}.
$$
This follows by expanding $(K^\dagger)^2$ and using the <bosonic creation operator> identity $(a^\dagger)^n|0\rangle=\sqrt{n!}|n\rangle$. The three orthogonal normalized <Fock states> have equal probability $1/3$. A tensor product of independently labelled pairs is a different multimode description; identifying its labels without renormalization changes its norm.
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