How to obtain the quasiparticle equation from Dyson equation? The problem is formulated as follows:


*

*Dyson equation for zero temperature Green's function:
\begin{equation}
\left[
i\dfrac{\partial}{\partial t_1} - h(\vec{r}_1)
\right]
G(1,2)-\int d3 \Sigma(1,3)G(3,2)=\delta(1,2)
\tag{1}
\end{equation}
in which
\begin{equation*}
h(\vec{r}_1)\equiv-\dfrac{\nabla^2}{2}+V_H(\vec{r}_1)+V_{ext}(\vec{r}_1)
\qquad
1\equiv(\vec{r}_1,t_1)
\end{equation*}

*Take Fourier transformation to the energy domain:
\begin{align}
\left[
-\omega - h(\vec{r}_1)
\right]
G(\vec{r}_1,\vec{r}_2;\omega)-\int d\vec{r}_3
\Sigma(\vec{r}_1,\vec{r}_3;\omega)
G(\vec{r}_3,\vec{r}_2;\omega)
=
\delta (\vec{r}_1-\vec{r}_2) \tag{2}
\end{align}

*The zero temperature Green's function under quasiparticle approximation can be represented as:
\begin{equation}
G(\vec{r}_1,\vec{r}_2;\omega)
=
\sum_i
\dfrac{\psi_i^{QP}(\vec{r}_1)\psi_i^{QP*}(\vec{r}_2)}{\omega-E_i^{QP}}
\tag{3}
\end{equation}
Insert $(3)$ into (2) one can obtain the following quasiparticle equation:
\begin{equation}
\left[
-\dfrac{1}{2}\nabla^2+V_H(\vec{r})+V_{ext}(\vec{r})
\right]
\psi_i^{QP}(\vec{r})
+
\int \Sigma(\vec{r},\vec{r}';E_i^{QP})\psi_i^{QP}(\vec{r}')d\vec{r}'
=E_i^{QP}
\psi_i^{QP}(\vec{r})
\tag{4}
\end{equation}
How can I complete the final step? I cannot build any connection between $(2)$ and $(3)$.
This problem is related to $(6)$, $(7)$ and $(8)$.
For completeness, the Fourier transform of $(1)$ is presented:


 A: 
How can I complete the final step? I cannot build any connection between (2) and (3).

You need to use the completeness of the eigenfunctions
$$
\sum_i \psi_i(\vec r_1)\psi_i^*(\vec r_2) = \delta(\vec r_1 - \vec r_2)
$$
After plugging in the above completeness relationship on the RHS and your definition of the Green's function on the LHS, the resulting equation is:
$$
\sum_i \psi_i^*(\vec r_2)\frac{1}{(\omega - E_i)}\left \{(\omega - h(\vec r_1))\psi_i(\vec r_1) - \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3)\right\}
=\sum_i \psi_i(\vec r_1)\psi_i^*(\vec r_2)
$$
Note, In the above equation I have corrected an error in the Fourier transform of the LHS in your original statement of the problem. Your $(-\omega -h(\vec r))$ has been changed to $(\omega - h(\vec r))$. This is because the $i\partial/\partial t$ transforms to $\omega$ not $-\omega$.
The coefficients of $\psi_i^*(\vec r_2)$ have to be equal by completeness so:
$$
\frac{1}{(\omega - E_i)}\left \{(\omega - h(\vec r_1))\psi_i(\vec r_1) - \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3)\right\}
=\psi_i(\vec r_1)
$$
Or
$$
(\omega - h(\vec r_1))\psi_i(\vec r_1) - \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3)
=\psi_i(\vec r_1)(\omega - E_i)
$$
Cancel the $\omega \psi_i(\vec r_1)$ from both sides and then multiply both sides by $-1$ to get:
$$
h(\vec r_1)\psi_i(\vec r_1) + \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3)
=\psi_i(\vec r_1)E_i
$$
