A question regarding Dyson equation and Green's function When knowing the self-energy, I can derive the many-body Green's function as
$$ G(\mathbf{r}_i,t_i ; \mathbf{r}_f,t_f) = G_0(\mathbf{r}_i,t_i ; \mathbf{r}_f,t_f) + \int \mathrm{d} \mathbf{r}_1 \int \mathrm{d} t_1 \int \mathrm{d} \mathbf{r}_2 \int \mathrm{d} t_2 G_0(\mathbf{r}_i,t_i ; \mathbf{r}_1,t_1) \Sigma(\mathbf{r}_1,t_1 ; \mathbf{r}_2,t_2) G(\mathbf{r}_2,t_2 ; \mathbf{r}_f,t_f)$$
I would like to demonstrate that the very simple property
$$ G(\mathbf{r}_i,t_i ; \mathbf{r}_f,t_f) = \int \mathrm{d} \mathbf{r}_1 G(\mathbf{r}_i,t_i ; \mathbf{r}_1,t^*) G(\mathbf{r}_1,t^* ; \mathbf{r}_f,t_f) $$
holds for the $G$ as defined from the Dyson equation, for any $t^*$ such that $t_i < t^* < t_f$.
This is elementary to show in terms of diagrams (two connected bold lines make a single, longer bold line) but I struggle to demonstrate it analytically...
 A: From the definition of Green function
$$
\psi({\bf r},t)=\int d^3x'dt'G({\bf r},t;{\bf r}',t')\psi({\bf r}',t').
$$
Add an intermediate step as
$$
\psi({\bf r},t)=\int d^3x'dt'G({\bf r},t;{\bf r}',t')
\int d^3x''dt''G({\bf r}',t';{\bf r}'',t'')\psi({\bf r}'',t'').   
$$
Just note that the integration variable is a dummy index and exchange $t'\rightarrow t''$. This gives
$$
\psi({\bf r},t)=\int d^3x''dt''G({\bf r},t;{\bf r}'',t'')
\int d^3x'dt'G({\bf r}'',t'';{\bf r}',t')\psi({\bf r}',t').   
$$
One has, after exchanging the integration order,
$$
\psi({\bf r},t)=\int  d^3x'dt'd^3x''dt''G({\bf r},t;{\bf r}'',t'')
G({\bf r}'',t'';{\bf r}',t')\psi({\bf r}',t').   
$$
But this must be equal to the integral we started from and so
$$
   G({\bf r},t;{\bf r}',t')=\int d^3x''dt''G({\bf r},t;{\bf r}'',t'')
G({\bf r}'',t'';{\bf r}',t').
$$
This can be easily obtained by path integrals (see Feynman and Hibbs, pag.37).
ADDED AFTER OP's COMMENT: I will introduce the Fourier transform as
$$
   G({\bf r},t;{\bf r}',t')=\int d^3x''dt''\int\frac{d^4p}{(2\pi)^4}e^{ip\cdot x''}G({\bf r},t;p)\int\frac{d^4p'}{(2\pi)^4}e^{ip'\cdot x''}G(p';{\bf r}',t').
$$
This yields
$$
   G({\bf r},t;{\bf r}',t')=\int\frac{d^4p}{(2\pi)^4}\int\frac{d^4p'}{(2\pi)^4}
   G({\bf r},t;p)G(p';{\bf r}',t')\int d^3x''dt''e^{i(p+p')\cdot x''}
$$
or
$$
   G({\bf r},t;{\bf r}',t')=\int\frac{d^4p}{(2\pi)^4}
   G({\bf r},t;p)G(-p;{\bf r}',t').
$$
Taking the Fourier transform with respect to $({\bf r},t)$, one has
$$
   G(k;{\bf r}',t')=\int\frac{d^4p}{(2\pi)^4}G(k;p)G(-p;{\bf r}',t').
$$
