Can one do something like a Hubbard-Stratonovich transformation that decouples the Cooper channel without Field theory?
In other words, is there a sense (which can be made precise without appealing to field theory) in which a Hamiltonian $$ H = \sum_\sigma \omega c_\sigma^\dagger c_\sigma + g c_\uparrow^\dagger c_\downarrow^\dagger c_\downarrow c_\uparrow $$ is equivalent to a second Hamiltonian $$ H' = \sum_\sigma \omega c_\sigma^\dagger c_\sigma + \omega' b^\dagger b + g' c_\uparrow^\dagger c_\downarrow^\dagger b + g' c_\downarrow c_\uparrow b^\dagger. $$
I can see that the answer is yes in the density-density channel, where the Hubbard-Stratonovich transformation appears to have the same effect as a Lang-Firsov transformation. Unfortunately applying the same approach in the Cooper channel seems to be spoiled by the fact that $[c_\uparrow^\dagger c_\downarrow^\dagger, c_\downarrow c_\uparrow] \neq 0$.