Hubbard-Stratonovich transformation without field theory

Question

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 $[c_\uparrow^\dagger c_\downarrow^\dagger, c_\downarrow c_\uparrow] \neq 0$.

Answer 1

Consider a Hamiltonian of the form $$ H = H_0 - A^{\dagger}A, $$ where $H_0$ is anything you want, but you would like to transform your problem into something linear in $A$ and $A^{\dagger}$. Let's first add an independent oscillator degree of freedom, writing instead $$ H = H_0 + \omega b^{\dagger}b - A^{\dagger} A, $$ where $[b,b^{\dagger}] = 1$, and $[b,A] = [b,H_0] = 0$. Adding this independent "spectating" oscillator will not change any correlation functions involving the operators contained in the original $H$ (this is the sense in which the theories are related at the level of the partition function in the field theoretic Hubbard-Stratonovich transformation).

Now take the unitary transformation $$ H' = U^{\dagger} H U, $$ with $$ U = \exp\left[ \frac{1}{\sqrt{\omega}} \left( b A^{\dagger} - A b^{\dagger} \right) \right]. $$ This takes $b \rightarrow b - \frac{1}{\sqrt{\omega}} A$ and $b^{\dagger} \rightarrow b^{\dagger} - \frac{1}{\sqrt{\omega}} A^{\dagger}$. So the transformed Hamiltonian is $$ H' = H_0 + \omega b^{\dagger} b - \sqrt{\omega} \left(b A^{\dagger} + b^{\dagger} A \right), $$ as required.

Answer 2

Weiss mean field theory. Habbard-Stratonivich is a fancy (but also systematic) way to do mean field theory.

If you necessarily want it in operator form, one could probably design some approach along the lines of bosonization or Schrieffer-Wolff procedure or even slave particles (slave-boson, drone-fermion, Schwinger bison) - all are doable without path Integral.