  We consider a system of two species of bosons of equal mass, with
interactions $U^{a}(|x|)$ and $U^{x}(|x|)$ for bosons of the same and different
species respectively. We present a rigorous proof -- valid when the Hamiltonian
does not include a species switching term -- showing that, when
$U^{x}(|x|)>U^{a}(|x|)$, the ground state is fully "polarized" (consists of
atoms of one kind only). In the unpolarized phase the low energy excitation
spectrum corresponds to two linearly dispersing modes that are even a nd odd
under species exchange. The polarization instability is signaled by the vani
shing of the velocity of the odd modes.
