quantum field theory - Inverse Green's Function identity in derivation of Hedin's equations

I'm trying to work through a derivation of Hedin's Equations in Effect of Interaction on One-Electron States by Hedin and Lundqvist (1969) and I've come across an identity that is given without much explanation that I'd like to understand.

We start with an equation of motion with single particle Green's functions G:

$\left[ i\frac{\partial}{\partial t_1} - h(1) - \phi(1) - V_H(1)\right] G(1,2) - i\int v(1^{+},3) \frac{\delta G(1,2)}{\delta \phi(3)} d3 = \delta(1,2)$

where $1=(x_1,t_1)$, $h$ is the one-electron kinetic energy and ion interaction, $\phi$ is a small perturbing field that is set to zero at the end, $V_H$ is the Hartree potential, and $v$ is the Coulomb interaction.

They then insert the identity

$\frac{\delta G(1,2)}{\delta\phi(3)} = -\int G(1,4) \frac{\delta G^{-1}(4,5)}{\delta\phi(3)} G(5,2) d(4,5)$

which is then used to define the self energy

$\Sigma(1,2) = -i \int v(1^+,3) G(1,4)\frac{\delta G^{-1}(4,2)}{\delta\phi(3)} d(3,4)$.

I know this identity is related to the definition of the inverse Green's function

$\int G(1,3) G^{-1}(3,2) d(3) = \delta(1,2)$

but I can't figure out how to get from this definition to the identity that is inserted into the equation of motion. Any ideas?

Thank you!

Answer

Take the functional derivative of the functional identity

$$\mathbf{\int d(3)G_{}^{}(1,3)G_{}^{-1}(3,2)=\delta_{}^{}(1,2)} \tag{1}$$ with respect to $\phi(4)$ to get $$\int d(3)\frac{\delta G_{}^{}(1,3)}{\delta\phi_{}^{}(4)}G_{}^{-1}(3,2)+\int d(3)G_{}^{}(1,3)\frac{\delta G_{}^{-1}(3,2)}{\delta\phi_{}^{}(4)}=0. \tag{2}$$ In going from $(1) \rightarrow (2)$, chain rule for functional differentiation and independence of $\delta_{}^{}(1,2)$ on $\phi$ are used.

Right multiply Eq.$(2)$ on both sides with $G_{}^{}(2,5)$ and integrating over variables $(2)$ to get

$$\int d(2)\int d(3)\frac{\delta G_{}^{}(1,3)}{\delta\phi_{}^{}(4)}G_{}^{-1}(3,2)G_{}^{}(2,5)+\int d(2)\int d(3)G_{}^{}(1,3)\frac{\delta G_{}^{-1}(3,2)}{\delta\phi_{}^{}(4)}G_{}^{}(2,5)=0. \tag{3}$$ Finally use Eq.$(1)$ to get the desired identity $$\mathbf{\frac{\delta G_{}^{}(1,5)}{\delta\phi_{}^{}(4)}=-\int d(2)\int d(3)G_{}^{}(1,3)\frac{\delta G_{}^{-1}(3,2)}{\delta\phi_{}^{}(4)}G_{}^{}(2,5)}. \tag{4}$$