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

  • $\begingroup$ Could you clarify the notation/problem? As written, your Hamiltonian is 4-by-4 and already diagonal. But perhaps there's summation over $k$ implied OR it could be a Hamiltonian of a site in a Hubbard model or a QD that is coupled to leads - to have the idea why you need to transform it. $\endgroup$
    – Roger V.
    Oct 19, 2021 at 6:26
  • $\begingroup$ I want a canonical transformation that maps $H$ to $H'$. Allowed operations are (i) adding terms $\propto b^\dagger b$ and (ii) unitary rotations. I am interested in this problem because it is a minimal model of a transformation relevant to superconductivity (in the more general problem one must also contend with mode/band labels, but their inclusion here only adds unnecessary complications). $\endgroup$ Oct 19, 2021 at 14:23

2 Answers 2


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.

  • $\begingroup$ Yeah this is what I want to do. Unfortunately your result relies on the additional (unstated) assumption that $[A,A^\dagger] = 0$, and thus does not apply to my case. As mentioned in the question, this condition does hold when decoupling in the density channel (so that $A = A^\dagger$), and the transformation given equates to the Lang-Firsov transformation. $\endgroup$ Oct 21, 2021 at 18:28

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.

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    $\begingroup$ Maybe I have misunderstood your point, but mean field theory makes the additional assumption that $b^\dagger \approx \mathrm{cons.}$, which I do not want to make in this case. Hubbard-Stratonovich is an exact transformation, whereas mean field theory is an approximation. $\endgroup$ Oct 18, 2021 at 21:03
  • $\begingroup$ You literally want to transform the Hamiltonian in the OP, treating it as 4-by-4 Hamiltonian (I thought it was just a schematic representation for a more complex problem)? $\endgroup$
    – Roger V.
    Oct 19, 2021 at 4:45
  • $\begingroup$ You understood the question correctly the first time, it is a schematic (in reality there are mode/band labels etc to contend with too). However a correct answer should work in this simple case nonetheless. $\endgroup$ Oct 19, 2021 at 14:13
  • $\begingroup$ @ComptonScattering With a mode/band one could use some kind of bosonization-like approach (canonical/operator, not as a path integral) - exact or not depends on the shape of the band and the number of dimensions. I am thinking along the lines of the derivation in the paper by Schotte&Schotte. $\endgroup$
    – Roger V.
    Oct 19, 2021 at 14:32

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