Mathematical description of Hubbard-Stratonovich transformation For an interacting quantum system, Hubbard-Stratonovich transformation is often used to decouple interaction terms in the action, by introducing a complex (bosonic) scalar field. This scalar field couples, like a potential, to the fermionic fields, such that the action is quadratic in the fermionic fields. 
For example in the case of a superconducting system the Fock space $\mathcal{F}$ reduced to $\mathcal{W} = \mathcal{H}\oplus\mathcal{H}^{\star}$, where $\mathcal{H}$ (the so-called Nambu space) is the single-particle Hilbert space and $\mathcal{H}^{\star}$ the dual space (see Ref Symmetry classes of disordered fermions by P. Heinzner, A. Huckleberry, M.R. Zirnbauer).
However, this is just the fermionic Hilbert space. In the total Hilbert space there must be also a bosonic part for the scalar field. 
From my point of view the full Hilbert space must be equal to:
$$
\mathcal{H}_{\text{Fermion}}\oplus\mathcal{H}_{\text{Boson}}
$$
where $\mathcal{H}_{\text{Fermion}}$ is the single-particle Hilbert space for the fermionic field and $\mathcal{H}_{\text{Boson}}$ the Hilbert spacefor the bosonic field.
My question is: is this correct so far? I.e. that the Hubbard-Stratonovich transformation can be written as
$$
\mathcal{F}_{\text{Fermion}}\to\mathcal{H}_{\text{Fermion}}\oplus\mathcal{H}_{\text{Boson}}
$$
Then, maybe a very stupid question, but if this is correct is there a relation to supersymmetric systems? Because the Hilbert space in the supersymmetric system is defined as above or not?
 A: The Hilbert space based on the perturbation theory is in general ill-defined in systems with interactions. Precisely, it doesn't see bound states appearing when we take into account the non-perturbative properties of the theory. An example is the Hydrogen atom, which doesn't exist in perturbation theory of the QED. In two words, the perturbation theory based Hilbert space is schematically defined as
$$
H_{\text{naive}} \simeq H_{\text{free}},
$$ 
where $H_{\text{free}}$ includes only non-interacting free states, while the correct Hilbert space is
$$
\tag 1 H_{\text{correct}} \simeq H_{\text{free}}\oplus H_{\text{bounded states}}
$$
The Cooper pair, which is Your scalar in the superconductor, is already included in $(1)$. This particularly answers on Your question.
Now we have to understand the meaning of the H-S transformation. To do this note that we typically reflect the structure of the Hilbert space in the construction of the Lagrangian operator. In the perturbation theory it is constructed from the creation-destruction fields $\hat{\Psi}(x)$ with operators $\hat{a}^{\dagger}(\mathbf p)$. They are defined in a way such that without interactions the Fock state
$$
\langle 0|\hat{\Psi}(x)
$$ 
doesn't have non-zero matrix elements with many particle states. But since the Hilbert space is defined to be the direct sum of all possible Fock states, then the bounded states are of course invisible. 
There is the simple way to modify the theory for including of such bounded states (i.e., to make them visible). Suppose that our bounded state is annihilated by the operator $\hat{\Phi}(\hat{\Psi})$. In order to include the bound state $\Phi$ in the theory, one can to include the lagrangian term
$$
\tag 2 L' = (\hat{\Psi} - \hat{\Phi}(x))^{2},
$$ 
which in fact is very similar to the Hubbard-Stratonovich transformation.
Another situation appears when there is non-zero vacuum average value $\langle \text{vac} |\hat{\Phi}(x) |\text{vac}\rangle$. Then one may treat $\Phi(x)$ as the classical external field. The transformation generated by $(2)$ is thus the mapping from the interacting theory to the free theory with motion in external field.
