Analysing decoupling channels in Hubbard-Stratonovich transformation I have an action defined in terms of fermionic fields $c$ and $d$ that looks like
$$S = - \bar{d}(t)\bar{c}(t') V(t,t') d(t')c(t)$$
where $V$ is an interaction matrix.
Then performing Hubbard-Stratonovich transformation via the exchange channel, say, (see Altland and Simons' Condensed Matter Field Theory (2nd ed.), sec. 6.2.) would introduce an auxiliary (exchange) field defined as 
$$\phi_1=\langle \bar{d}(t) d(t') \rangle.$$
Alternately we may choose to let the auxiliary field be
$$\phi_2=\langle \bar{c}(t') c(t) \rangle.$$
My question is, how do we choose which $\phi$ to use? Are there any existing physical systems that would choose one over the other?
If we choose to use both $\phi_1$ and $\phi_2$, this would be like decoupling via the same channel twice. Hence would we need a multiplicative factor of 1/2 somewhere? (Again, are there any physical systems that require this procedure?)
(This is a follow-up question on my other post: Hubbard-Stratonovich transformation and decoupling channels)
 A: It seems that there is no reason to discreminate the $c$ and $d$ fields when just the interaction term is given.
I think a more good notation would be helpful to you.
Instead of treating $c$ and $d$ fields separately, let us introduce a two-component spinor $\psi=(c,d)^T$. Then, the interaction you give as an example is written as
$$
S=-V \psi_1^\dagger \psi_2^\dagger \psi_2 \psi_1 = -\frac{V}{2}\sum_{i,j} \psi_i^\dagger \psi_j^\dagger \psi_j \psi_i \approx -\frac{V}{2}(\psi^\dagger\psi)(\psi^\dagger\psi)
$$
where in the last eqaulity, I ignored some unimportant terms involving just two fermionic operators.
Introducing the Hubbard-Stratonovich bosonic fields $\phi$ which will be identified with $\langle\psi^\dagger\psi\rangle $ at the level of the saddle point approximation, we obtain
$$
S=\frac{1}{2V}\phi^2+\phi\psi^\dagger\psi
$$
Note that no distinction has been made between the $c$ and $d$ field up to this point.
Now, if we add the single-particle Hamiltonian of $c$ and $d$, for example, $$H_0=
\begin{pmatrix}
\varepsilon_c & 0 \\
0 & \varepsilon_d
\end{pmatrix},
$$
the distinction between $c$ and $d$ fields appear:
$$
\langle c^\dagger c \rangle = \frac{1}{\varepsilon_c+\phi} \\
\langle d^\dagger d \rangle = \frac{1}{\varepsilon_d+\phi}
$$
with $\phi=-V(\langle c^\dagger c \rangle+\langle d^\dagger d \rangle)$.
When $\varepsilon_c \neq \varepsilon_d$, $\langle c^\dagger c \rangle \neq \langle d^\dagger d \rangle$ is expected.
