= Boson-pair realization of SU(1,1)
{title2=$K_+=a^\dagger b^\dagger,\ K_-=ab,\ K_3=(N_a+N_b+1)/2$}
Pair creation and annihilation in two independent oscillator modes obey $[K_-,K_+]=2K_3$ and $[K_3,K_\pm]=\pm K_\pm$. The <Hermitian> components $K_1=(K_++K_-)/2$ and $K_2=(K_+-K_-)/(2i)$ satisfy $[K_1,K_2]=-iK_3$. Multiplying all three <Hermitian> components by $i$ gives the real Lie bracket signs of <SU(1,1)>. Both occupations change equally, so their difference is conserved. This realizes the algebra used in two-mode squeezed states.
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