How to obtain the quasiparticle equation from Dyson equation?

Question

The problem is formulated as follows:

1. 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*}

2. 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}

3. 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:

Answer 1

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: $$ h(\vec r_1)\psi_i(\vec r_1) + \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3) =\psi_i(\vec r_1)E_i $$