= Normal-ordered boson pair count
{title2=$(a^\dagger)^2a^2=n(n-1)$}
For a <bosonic> mode satisfying the <canonical commutation relation> $[a,a^\dagger]=1$, the <number operator> $n=a^\dagger a$ obeys $n^2=a^\dagger aa^\dagger a=(a^\dagger)^2a^2+n$. Thus a two-body on-site interaction is proportional to $n(n-1)$, with no self-interaction. In a <fixed-number two-mode condensate> with probability $q$ of occupying this mode, $\langle n(n-1)\rangle=N(N-1)q^2$ by the <binomial distribution>.
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