Not for all but only $\vec{r}$ on the line defect (see the original question part).

I also understand the meaning of $\Gamma G \Gamma G^{\dagger}$. But why should be acted on the states of the electrodes ?

I understand $|-\rangle$ is odd under reflection. But I could not thence deduce $\langle\vec{r}|-\rangle = 0$ explicitly using the expressions given above.

Thanks for your post. I understand your point. But how do you show that from the expression (given in the original post) of $|-\rangle$?

There is a very important element which is missing in that picture. That is the Pauli exclusion principle: it is crucial to see that this principle produces an effective two-dimensional surface for the motion of a pair of two added electrons. It is because of this two-dimensionality that they could bind in a bound state given any weak attraction. This is not true for higher-dimension.

@Trimok: Thanks a lot. I'll read the paper. I also think there is something wrong in that equation. The factor $1/\omega$ seems lost.

@Trimok: Thanks for your response. $<X>_{\omega}$ is the Fourier transform of the average of $X$. $V$ is volume. You can click the link to have a look at the whole lecture.

There are various ways to show that explicitly. You may take the Fourier transform of the Poisson equation. Then you see $\tilde{G}(\vec{k}) = \varepsilon^{-1}_0k^{-2}$, where $\tilde{G}$ is the Fourier transform of $G$ and $k=|\vec{k}|$. Transforming back , you obtain $G(\vec{r}) = \varepsilon^{-1} \int d^3\vec{k} \frac{e^{i\vec{k}\vec{r}}}{k^2}$. To perform this integral, you can write $d^3\vec{k} = k^2\sin(\theta)dk d\theta d\varphi$ and $\vec{k}\vec{r} = kr\cos(\theta)$, where we have chosen the $\vec{r}$ along z-axis. Now you can use standard formula to get the final result.

(1) By 'obviously satisfying ...', I mean it on the physics not mathematics: Coulomb's law is a solution to Possion's equation (thanks for correcting me for using Laplace equation) in the presence of a point charge. (2) I would have used the symbol by the questioner, but I do not know how to make it in Latex. (3) Thanks for pointing out that the missing $d^3\vec{r}'$.