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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.

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Since posting this question I have come to understand the answer, which I'm posting here for the sake of completeness.


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?

This question has no really satisfactory answer as it needs to be studied on a case by case basis. In reality the reason why we take some steps in a derivation depends of multiple factors. One has to take into account the information that is available or have certain insight in how to tackle the problem. Another possibility is simple trial and error, in which case there is no deeper meaning behind why the steps were made.

In the context of this problem I was given the following information:

$$ \begin{cases} \Delta G(\bar{r}|\bar{r}') = -4\pi \delta(\bar{r} - \bar{r}') \\ \Delta \psi (\bar{r}) = -4\pi \rho(\bar{r})^* \end{cases}, \quad \begin{cases} \psi(\infty) = 0 \\ G(\infty) = 0 \end{cases} $$

*Mind the typo in my original question

Given enough insight, one might try to link the interior (volume integral) with the boundary surface around this interior (surface integral) because of the interesting boundary conditions.

This leads me to the second question.

Again, how? I don't see how Green's theorem and the delta function lead to this equation.

Treating it canonically now means treating it in a way that uses the insight provided in the paragraph above. Green's theorem links a volume integral with a surface integral so we should try to manipulate the equations to satisfy Green's theorem. $$ \int_V \psi \Delta G(\bar{r},\bar{r}') - G\Delta \psi (\bar{r}) d\bar{r} = -4\pi \int_V \psi \delta(\bar{r} - \bar{r}') - G\rho(\bar{r}) d\bar{r} = \int_S (\psi \bar{\nabla} G - G \bar{\nabla}\psi)\cdot \bar{n} dS $$

The surface integral is equal to 0 because of the boundary conditions and the middle integral becomes: $$ -4\pi \psi(\bar{r}') + 4\pi \int_V G(\bar{r}|\bar{r}') \rho(\bar{r}) d\bar{r} $$ This yields the result I was looking for.

After all these years the answer came to me by looking at it for a few minutes, may this question rest in peace for once and for all. Kuddos to Samim Ul Islam whose comment to my question also provides an answer.

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