# How to obtain the quasiparticle equation from Dyson equation?

The problem is formulated as follows:

1. Dyson equation for zero temperature Green's function: $$$$\left[ i\dfrac{\partial}{\partial t_1} - h(\vec{r}_1) \right] G(1,2)-\int d3 \Sigma(1,3)G(3,2)=\delta(1,2) \tag{1}$$$$ in which $$\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: $$$$G(\vec{r}_1,\vec{r}_2;\omega) = \sum_i \dfrac{\psi_i^{QP}(\vec{r}_1)\psi_i^{QP*}(\vec{r}_2)}{\omega-E_i^{QP}} \tag{3}$$$$ Insert $$(3)$$ into (2) one can obtain the following quasiparticle equation: $$$$\left[ -\dfrac{1}{2}\nabla^2+V_H(\vec{r})+V_{ext}(\vec{r}) \right] \psi_i^{QP}(\vec{r}) + \int \Sigma(\vec{r},\vec{r}';E_i^{QP})\psi_i^{QP}(\vec{r}')d\vec{r}' =E_i^{QP} \psi_i^{QP}(\vec{r}) \tag{4}$$$$

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:

• This paper by Sham and Kohn and this paper by Layzer citing Morse and Fesbach argue that $G(\vec{r},\vec{r'};E)$ admits diagonal like expansion in terms of bi-orthogonal basis of integro-differential operator appearing in Eq.(4) above. – Sunyam May 29 '19 at 12:51
• @Sunyam Have you go through the details? (By the way, thanks for your nice help so many times) – Jack May 29 '19 at 13:15
• The papers cited above are bit sketchy and hand-way, have to check more details like when does the integro-differential operator $\int d\vec{r}_3\left[-\omega - h(\vec{r}_1)\right]\delta(\vec{r}_1,\vec{r}_3)-\Sigma(\vec{r}_1,\vec{r}_3;\omega)$ admit bi-orthonormal eigenbasis. – Sunyam May 29 '19 at 13:36
• You might find Sec. 3.6 (pg. no. 153) of Datta's book and this paper by W van Haeringen, B Farid and D Lenstra a useful read. Also see my comment on hft's answer. – Sunyam Jun 8 '19 at 23:08
• @Sunyam: Could you provide an answer based on what you have read? – Jack Jun 9 '19 at 13:51

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

• Seems there is a problem related to the definition of Fourier transform to cancel $\omega \psi_i(\vec{r}_1)$, I will fix it. Many thanks. – Jack Jun 7 '19 at 13:49
• Yes, that it right. Your LHS fourier transform should have (w - h(r)) not (-w - h(r)). I have updated my answer to reflect this. – hft Jun 7 '19 at 16:49
• @hft You seem to have assumed that the integral operator (which parametricaly depends on $\omega$) $\int d\vec{r}'\left[\left(\omega - h(\vec{r})\right)\delta(\vec{r},\vec{r}')-\Sigma(\vec{r},\vec{r}')\right]\boldsymbol{\cdot}$ has an orthonormal basis which is not always guaranteed (as this operator is not Hermitian or Normal in general), however it might admit bi-orthonormal basis (not sure if this is guaranteed always). So the above arguments need to be generalized taking account of this. – Sunyam Jun 8 '19 at 22:59