$\rm Tr[log( )]$ calculation to go from BCS to Ginzburg-Landau It seems like calculating the effective action $|\Delta|^2 + Tr[ln(G^{-1})]$ give the Ginzburg Landau action.
\begin{equation}
G^{-1} =\begin{pmatrix} i\partial_t - H & \Delta \\
\Delta^* & i\partial_t + H
\end{pmatrix}
\end{equation}
Does any body know how to take the approximation of $Tr[ln(G^{-1})]$?
I am not sure how to get $|\Delta|^4$ term and $|\nabla \Delta|^2$ term.
(which are $|\psi|^4$ and $|\nabla \psi|^2$ in typical textbooks.)
 A: What I am writing below is taken from Altland and Simons' book "Condensed Matter field theory".
Let us first define the operators
$$
\hat{G}_0^{-1} = \left(
\begin{array}{cc}
i\partial_t - H && 0 \\
0 && i\partial_t + H
\end{array}
\right);
\;\;\;\;\;\;\;\;
\hat{\Delta} = \left(
\begin{array}{cc}
0 && \Delta \\
\Delta^* && 0
\end{array}
\right).
$$
You can write the inverse Green function in terms of these operators as
$$
\hat{G}^{-1} = \hat{G}^{-1}_0 + \hat{\Delta}
$$
$$
\hat{G}^{-1} = \hat{G}_0^{-1} \left(1 + \hat{G}_0 \hat{\Delta} \right).
$$
$$
\text{Tr}[\log{(\hat{G}^{-1})}] = \text{Tr}[\log{(\hat{G}_0^{-1})} ] + \text{Tr}[\log{(1 + \hat{G}_0 \hat{\Delta})} ]
$$
The first term represents the contribution to the action due to the non-interacting electron gas; while the second term contains the dependence on the complex field $\Delta$ which, at the lowest orders in $\Delta$, is the Ginzburg-Landau action.
With this simple trick you can expand the logarithm in powers of $\hat{\Delta}$ and obtain the Ginzburg-Landau action by truncating at the desired order:
$$
\text{Tr}[\log{(1 + \hat{G}_0\hat{\Delta})}] = \sum_{n=1}^{\infty} {(-1)^{n+1} \over n} \text{Tr}[ (\hat{G}_0\hat{\Delta})^{n} ]
$$
One can prove that only terms with an even value of $n$ survive after taking the trace, so the action is reduced to
$$
- \sum_{k=1}^{\infty} {1 \over 2k} \text{Tr}[ (\hat{G}_0\hat{\Delta})^{2k} ]
$$
Now we can build the Ginzburg-Landau theory by keeping only terms of order $k=1,2$. For instance, let's focus on $k=1$: $- {1 \over 2} \text{Tr}[\hat{G}_0 \hat{\Delta} \hat{G}_0 \hat{\Delta}]$
and let's switch to Matsubara-frequency and momentum space $k = (i\omega_n , \vec{k})$, where the Green function and the $\Delta$ field can be written as
$$
\hat{G}_0(k, p) = \left(
\begin{array}{cc}
G_0(k) & 0 \\
0 & -G_0(-k)
\end{array}
\right) \delta_{k, p}
\;\;\;\;\;
\hat{\Delta}(k, p) = \left(
\begin{array}{cc}
0 & \Delta(k-p) \\
\Delta^*(k-p) & 0
\end{array}
\right).
$$
Expanding the matrix product and taking the trace over momenta we get
$$
- {1 \over 2} \sum_{k_1\, k_2\, k_3\, k_4} 2 G_{0,11}(k_1, k_2) \Delta(k_2, k_3) G_{0,22}(k_3, k_4) \Delta^*(k_4, k_1)
$$
$$
\sum_{k_1, k_3} G_0(k_1) \Delta(k_1 - k_3) G_0(-k_3) \Delta^*(k_3 - k_1)
$$
$$
k_1 = k, \;\;\; k_1 - k_3 = q, \;\;\;\;\;\;
\sum_{k,q} G_0(k) \Delta(q) G_0(-k+q) \Delta^*(-q)
= \sum_{q} \chi(q) \Delta^*(-q) \Delta(q)
$$
where $\chi(q) = \sum_k G_0(k) G_0(-k+q)$ and we have used the shortcut $\sum_k = {T \over V} \sum_{i\omega_n} \sum_{\vec{k}}$ ($T$ is the temperature and $V$ the volume). Now the evaluation of $\chi(q)$ is a difficult task, however one can prove that at lowest order in $q$ we can expand this function as $\chi(q) \sim a + b q^2$ and the term $\propto \sum_q q^2 \Delta^*(-q) \Delta(q)$ rewritten in real space is the gradient term $\sim \int d^3x |\nabla \Delta|^2$
