I have the following equations:
$$ \quad\Delta G(\bar{r}|\bar{r}') = -4\pi\delta(\bar{r}-\bar{r}') \\ \tag1$$$$ \quad\Delta\Psi(\bar{r}) = -4\pi\delta(\bar{r}-\bar{r}')\tag 2$$$$ \quad \\ \bar{r} \in V \\ \Psi(\infty) = 0 \text{ (Free space)} $$
Now in my notes it says that we treat 1 and 2 canonically, i.e. we do:
$$\int_V [\Psi.\Delta G - G.\Delta\Psi]d\bar{r} $$
So my first question: what happened here? What does "canonically" mean in this context? I know the word means something like standard, or not random, but I don't see why someone would do this step. I see that it allows us to use Green's 2nd integral theorem, but how was this known a priori?
Then it says that after applying Green's theorem and the definition of the (dirac) delta function, we find:
$$\Psi(\bar{r}') = \int_V G(\bar{r}|\bar{r}')\rho(\bar{r}) d\bar{r} \\ \Rightarrow \Psi(\bar{r}) = \int_V G(\bar{r}|\bar{r}')\rho(\bar{r}') d\bar{r'} \text{ (symmetry)} $$
Again, how? I don't see how Green's theorem and the delta function lead to this equation.
I did however find this:
$$ \Delta\Psi(\bar{r}) = -4\pi\rho(\bar{r}) \\ \Rightarrow \Delta\Psi(\bar{r}) = \int -4\pi\delta(\bar{r}-\bar{r}')\rho(\bar{r}')d\bar{r}'\quad^*\\ \Rightarrow \Delta\Psi(\bar{r}) = \int\Delta G(\bar{r}|\bar{r}')\rho(\bar{r}')d\bar{r}'\\ \Rightarrow \Psi(\bar{r}) = \int_V G(\bar{r}|\bar{r}')\rho(\bar{r}') d\bar{r'} $$
I don't know if my derivation is correct, but it does seem to give me the result I'm looking for. Is there any fundamental difference between the 2 methods?
$$^* \rho(\bar{r}) = \int \delta(\bar{r}-\bar{r}')\rho(\bar{r}')d\bar{r}' $$ Writing a potential density as a superposition of point densities.