Two-pair component of polarization-entangled down-conversion ($|{\Psi }_2⟩=(K^{†}{)}^2|0⟩/(2\sqrt{3})$)

For two collected spatial modes $a,b$, each with polarizations $H,V$, put $K^{†}=a_H^{†}{b}_V^{†}+a_V^{†}{b}_H^{†}$. The one-pair state is a Bell state. The normalized two-pair component isThis follows by expanding $(K^{†}{)}^2$ and using the bosonic creation operator identity $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$. 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.

For an ideal undepleted pump and one collected pair of spatial modes, the two orthogonal polarization-pair creation terms generate two commuting two-mode squeezed vacua. If $\lambda ={\tanh }^{2}r$, either squeezed pair has geometric occupation probability $(1-\lambda ){\lambda }^k$. Adding their pair counts gives $n+1$ equal-weight possibilities, hence $P_n=(n+1)(1-\lambda {)}^2{\lambda }^n$. Thus $P_1=2(1-\lambda {)}^2\lambda$ and $P_2=3(1-\lambda {)}^2{\lambda }^2\simeq 3P_1^2/4$ at weak gain. Different unresolved temporal or spectral modes need not obey this single-mode formula.