I am stuck from the mass operator to vertex function in the derivation of Hedin's equations. The problem could be organized as follows:

  1. Mass operator: $$M(1,2)=i\hbar\int d(34)v(1^+,3)\dfrac{G_1(1,4)}{\delta U(3)}G_1^{-1}(4,2)$$ with $1=(x,t)$, $U$ is the external perturbation and $G_1$ is the single particle Green's function.

  2. The total classical potential $V$: $$V(1) \equiv U(1)-i\hbar\int d3v(1,3)G_1(3,3^+)$$

Regarding alternatively $G_1$ as the functional of $V$ instead of $U$ and using chain rule and the following relation $$\dfrac{\delta F[h(\zeta);x,y]}{\delta h(z)}=-\int d\xi d\eta F[h(\zeta);x,\xi]\dfrac{\delta F^{-1}[h(\zeta);\xi,\eta]}{\delta h(z)}F[h(\zeta);\eta,y]$$ one can derive $$M(1,2)=-i\hbar\int d(345)v(1^+,3)G_1(1,4)\dfrac{\delta G_1^{-1}(4,2)}{\delta V(5)} \dfrac{\delta V(5)}{\delta U(3)} \tag{1}$$ from which one can introduce the vertex function $$\Gamma(1,2,3)\equiv \dfrac{\delta G_1^{-1}(1,2)}{\delta V(3)}=\delta(1,2)\delta(1,3)+\dfrac{\delta M(1,2)}{\delta V(3)} \tag{2}$$

How can I prove the equation $(1)$ and derive the equation $(2)$?

This problem is related to this paper. [(3.16)--(3.17)]

Thanks in advance.

  • 1
    $\begingroup$ Quick hint: Use this followed by functional chain rule to get Eq. (1). Now observe $G_{}^{-1}(1,2)=G_{0}^{-1}(1,2)-V(1)\delta(1,2)-\Sigma(1,2)$ to get Eq. (2). $\endgroup$ – Sunyam May 25 at 18:59

Finally, I have gone through all the details.

The first part:

Chain rule: \begin{equation} \dfrac{\delta G_1(1,4)}{\delta U(3)} = \int d5 \dfrac{\delta G_1(1,4)}{\delta V(5)} \dfrac{\delta V(5)}{\delta U(3)} \end{equation}

Functional identity: \begin{equation} \dfrac{\delta G_1(1,4)}{\delta V(5)} = -\int d67 G_1(1,7) \dfrac{\delta G_1^{-1}(7,6)}{\delta V(5)} G_1(6,4) \end{equation}

Insert them into the mass operator: \begin{align*} M(1,2) & = - i\hbar \int d34 v(1^+,3) \int d567 G_1(1,7) \dfrac{\delta G_1^{-1}(7,6)}{\delta V(5)} G_1(6,4) \dfrac{\delta V(5)}{\delta U(3)} G_1^{-1}(4,2) \\ & = - i\hbar \int d3567 v(1^+,3) G_1(1,7) \dfrac{\delta G_1^{-1}(7,6)}{\delta V(5)} \dfrac{\delta V(5)}{\delta U(3)} \delta(6,2) \\ & = - i\hbar \int d357 v(1^+,3) G_1(1,7) \dfrac{\delta G_1^{-1}(7,2)}{\delta V(5)} \dfrac{\delta V(5)}{\delta U(3)}\\ & = - i\hbar \int d345 v(1^+,3) G_1(1,4) \dfrac{\delta G_1^{-1}(4,2)}{\delta V(5)} \dfrac{\delta V(5)}{\delta U(3)} \end{align*}

The second part:

The inverse single-particle Green's fucntion: \begin{equation*} G_1^{-1}(1,2) = G_1^{(0)-1}(1,2) - U(1) \delta(1,2) - \Sigma(1,2) \end{equation*}

The external potential $U$: \begin{equation*} U(1) = V(1)+i\hbar \int d3 v(1,3)G_1(3,3^+) \end{equation*}

The self energy: \begin{align*} \Sigma(1,2) = \Sigma_H(1,2) + M(1,2) \end{align*}

Hartree self-energy: \begin{align*} \Sigma_{H}(1,2) = - \delta(1,2) i \hbar \int d3 v(1,3) G_1(3,3^+) \end{align*}

\begin{align*} & \Gamma(1,2;3) \equiv - \dfrac{\delta G_1^{-1}(1,2)}{\delta V(3)} = - \dfrac{ \delta \left[ G_1^{(0)-1}(1,2) - U(1)\delta(1,2) - \Sigma(1,2) \right] }{\delta V(3)}\\ & = 0 + \delta(1,2) \dfrac{\delta U(1)}{\delta V(3)} + \dfrac{\delta \Sigma_H(1,2)}{\delta V(3)} + \dfrac{\delta M(1,2)}{\delta V(3)}\\ & = \delta(1,2)\delta(1,3) + \delta(1,2) i \hbar \int d3 v(1,3) \dfrac{\delta G_1(3,3^+)}{\delta V(3)} - \delta(1,2) i \hbar \int d3 v(1,3) \dfrac{\delta G_1(3,3^+)}{\delta V(3)} + \dfrac{\delta M(1,2)}{\delta V(3)}\\ & = \delta(1,2)\delta(1,3) + \dfrac{\delta M(1,2)}{\delta V(3)} \end{align*}

  • $\begingroup$ Looks like you are double counting Hartree self energy by including it both in $U$ and $\Sigma$? $\endgroup$ – Sunyam Jun 6 at 17:58
  • $\begingroup$ @Sunyam Hartree self-energy appeared only in $\Sigma$, $U$ is viewed as the external perturbation term to using the functional derivative technique and can be diminished in the last stage by taking $U \rightarrow 0$. $\endgroup$ – Jack Jun 7 at 13:14

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