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

$$ \frac{\delta}{\delta \phi_{Q}^{\dagger}}\left(G_{\text{BCS}}^{-1}\right)_{qk} = \begin{pmatrix} 0 & 0 \\ \delta_{q-k,Q} & 0 \end{pmatrix} $$


$$ \left(G_{\text{BCS}}\right)_{k,q} = \frac{-1}{\left(\omega_{n}^{2} + \epsilon_{k}\right)\Delta_{kq} + \phi_{k-q}\phi_{q-k}^{\dagger}}\begin{pmatrix} \left(-i\omega + \epsilon_{k}\right)\delta_{k,q} & \phi_{k-q} \\ \phi_{q-k}^{\dagger} & \left(-i\omega - \epsilon_{k}\right)\delta_{k,q} \end{pmatrix} $$

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?


I think your general intuition is correct here. The assumption of a homogenous mean-field solution corresponds to setting $$\phi_q = V\beta \Delta \delta_{q,0}.$$ In both cases then the Green's function becomes diagonal in momentum (and frequency) space $$G_{k,k'} = G_k \delta_{k, k'},$$ which as you've noted cancels the sum over $q$. I hope that helps somewhat.

Just as a note, Altland and Simons at times defines the Green's function with two different sign conventions $$ G^{-1} \sim -i \omega_n + \xi_k$$ and $$G^{-1} \sim i \omega_n - \xi_k.$$ However, I have only ever seen that latter used anywhere, so it's probably best to stick with that one.

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