Representing Green function as a coherent state path integral

Question

I am working through the problem "self-consistent T-matrix approximation" in Altland and Simons (second edition) pg 234. One of the steps involves representing the Green function as a coherent state path integral and performing the Gaussian integral over the Grassman field, and the solution given in the book is quite brief. I would like more details so I can follow the derivation.

I've taken the below from this physics overflow link, which summarizes well the statement of the problem (but is ultimately asking about different aspects of the problem).

The Hamiltonian of the problem is $\hat H = \hat H_0 + \hat H_{\text{imp}} \; $, where $$\hat H_0 = \sum_{\mathbf{k}} \epsilon_{\mathbf{k}} c_{\mathbf{k}}^\dagger c_{\mathbf{k}},$$ $$\hat H_{imp} = v_0 a^d \displaystyle\sum_{i=1}^{N_{imp}} c^{\dagger}(\mathbf{R_i})c(\mathbf{R_i}),$$ and $N_{\text{imp}} \; $ is the number of impurities. We want to compute the single-particle Green function $G_n = \langle \langle c^{\dagger}_n (\mathbf{r}) c_n(\mathbf{r'}) \rangle \rangle_{imp} \; $, where $n$ is the Matsubara frequency index and $\langle \cdots \rangle_{imp} \equiv \frac{1}{L^d} \int \prod_i d^d\mathbf{R_i}$ is the configurational average over all impurity coordinates.

In the first line of the solution given on page 236:

Representing the Green function as a coherent state path integral and performing the Gaussian integral over the Grassman field, we obtain the formal result $\hat{G}_n = (i \omega_n - \hat{H}_0 - \hat{H}_{\text{imp}} \; )^{-1}$.

I would like some details about how to do this. For example, are $c(r), c_k$ the same operators except one is in real space and the other in momentum space, or do they represent different operators entirely?

Answer 1

I am not sure is this is all you need, but the answer to your question is yes, the operators are the same, just expressed in real vs k-space: $$ c({\bf{r}})=\frac{1}{\sqrt{V}}\sum_k c_k e^{-i \bf{k r}},\quad c({\bf k})=\frac{1}{\sqrt{2\pi}}\int c(r) e^{i \bf{k r}}d{\mathbf r} $$ (the sign convention was picked up from (2.24))

As of the "formal correct result", you can demonstrate in general case, that if the Green's function is defined as $$ G_{n,\alpha\beta}=\langle \bar{c}_{n\alpha} c_{n\beta} \rangle, $$ (here the averaging is defined through the path integral and Grassman variables), and your Hamiltonian is $$ {\bf{H}}=\sum_{\alpha\beta}H_{\alpha\beta}c_\alpha^{\dagger} c_\beta, $$ then it is possible to show (see the equation right after Eq. 4.22) that the Green function $$ G_{n,\alpha\beta} \, \equiv \frac{1}{Z} \int Dc D\bar c\, \bar{c}_{n\alpha} \; c_{n\beta} \; e^{\sum_{\, n\alpha\beta} \; \; {\bar{c}_{\, n\alpha} \; (i\omega_n \; \delta_{\alpha\beta} } \; \; -H_{\alpha\beta} \; ) \, c_{\beta n}}, $$ if treated as a matrix with indices $\alpha$ and $\beta$, can be expressed as
$$ G_{n,\alpha\beta}=G_n=(i\omega_n-H)^{-1}, $$ which is essentially how the "technical solution" is obtained.