I am interested in mean-field theories in the path integral formalism. However, I have a technical problem by evaluating the stationary phase approximation (mean-field approximation).
After the Hubbard-Stratonovich transformation and integrating out the fermionic degree of freedom we have a action functional of the form
$$ S = c\sum_{q}\phi_{q}V^{-1}\left(q\right)\phi_{-q} - \text{tr}\ln\left(G^{-1}\right) $$
where c is a constant, $V^{-1}\left(q\right)$ is the inverse interaction potential and $G$ is the Greens operator for the free fermions interacting with the bosonic field $\phi$.
Two examples can be found in the book by Altland & Simons. In the case of a interacting electron gas. Here $G$ has the form
$$ \left(G^{-1}\right)_{kq} = \left(-i\omega_{n} + \frac{k^{2}}{2m} - \mu\right)\delta_{kq} + \frac{i}{\beta V}\phi_{q-k} $$
Then applying the stationary phase approximation we will obtain
$$ \frac{\delta S}{\delta \phi_{q}} = cV^{-1}\left(q\right)\phi_{-q} + \frac{2i}{\beta V}\sum_{q_{1}}G_{q_{1},q_{1}-q} = 0 $$
A general solution of these equation is not known. But I found that in the case of a homogeneous mean-field $\phi_{q-k} = \bar{\phi}$, that the Green's function can be written as
$$ G_{q_{1},q_{1}-q}\frac{1}{-i\omega_{n} + \frac{k^{2}}{2m} - \mu + \frac{i}{\beta V}\bar{\phi}} $$
My question is why it is not
$$ G_{q_{1},q_{1}-q}\frac{1}{\left(-i\omega_{n} + \frac{k^{2}}{2m} - \mu\right)\delta_{q_{1},q_{1}-q} + \frac{i}{\beta V}\bar{\phi}} $$
which would cancel the sum $\sum_{q_{1}}$?
The second example from the book by Altland & Simons is a superconductor. Here the action function reads
$$ S_{\text{BCS}} = \sum_{Q}\phi_{Q}^{\dagger}\left(\frac{g}{\beta V}\right)^{-1}\phi_{Q} - \text{tr}\ln\left(\left(G_{\text{BCS}}^{-1}\right)\right) $$
with $\left(G_{\text{BCS}}^{-1}\right)_{k,q} = \begin{pmatrix} \left(-i\omega + \epsilon_{k}\right)\delta_{k,q} & \phi_{k-q} \\ \phi_{q-k}^{\dagger} & \left(-i\omega - \epsilon_{k}\right)\delta_{k,q} \end{pmatrix}$. The mean-field equation is then given by
$$ \frac{\delta S_{\text{BCS}}}{\delta \phi_{Q}^{\dagger}} = \left(\frac{g}{\beta V}\right)^{-1}\phi_{Q} - \sum_{kq}\text{tr}_{2\times 2}\left(\left(G_{\text{BCS}}\right)_{kq} \frac{\delta}{\delta \phi_{Q}^{\dagger}}\left(G_{\text{BCS}}^{-1}\right)_{qk}\right) $$
Because of
$$ \frac{\delta}{\delta \phi_{Q}^{\dagger}}\left(G_{\text{BCS}}^{-1}\right)_{qk} = \begin{pmatrix} 0 & 0 \\ \delta_{q-k,Q} & 0 \end{pmatrix} $$
and
$$ \left(G_{\text{BCS}}\right)_{k,q} = \frac{-1}{\left(\omega_{n}^{2} + \epsilon_{k}\right)\Delta_{kq} + \phi_{k-q}\phi_{q-k}^{\dagger}}\begin{pmatrix} \left(-i\omega + \epsilon_{k}\right)\delta_{k,q} & \phi_{k-q} \\ \phi_{q-k}^{\dagger} & \left(-i\omega - \epsilon_{k}\right)\delta_{k,q} \end{pmatrix} $$
This leads to
$$ \left(\frac{g}{\beta V}\right)^{-1}\phi_{Q} - \sum_{kq}\frac{\phi_{k-q}\delta_{q-k,Q}}{\left(\omega_{n}^{2} + \epsilon_{k}^{2}\right)\delta_{kq} + \phi_{k-q}\phi_{q-k}^{\dagger}} $$
Then, again under the assumption of a homogeneous mean-field $\phi_{k-q} = \Delta$ the gap equation must follows
$$ \left(\frac{g}{\beta V}\right)^{-1}\Delta - \sum_{k}\frac{\Delta}{\omega_{n}^{2} + \epsilon_{k}^{2} + \left|\Delta\right|^{2}} = 0 $$
But for me is the step to the gap equation not clear. Maybe somebody can explain to me why due to the assumption of a homogeneous mean-field the two mean-field equations are valid?
