# Why can I not use the Dirac delta function in the classical potential expansion?

I would like to verify my understanding on why it is that we can represent a charge density $$\rho(\mathbf{r})$$ as delta functions.

If we start from $$\nabla^2V=-4\pi\rho(\mathbf{r})\tag{1}$$ then we conclude that $$V=\frac{1}{|\mathbf{r}-\mathbf{r}'|}+\Gamma(\mathbf{r}, \mathbf{r}')\tag{2}$$ where $$\nabla^2\Gamma=0$$. We then find that [plugging (2) in for (1)] $$-4\pi\delta(\mathbf{r}-\mathbf{r}')=-4\pi\rho(\mathbf{r})$$ or, equivalently, $$\rho(\mathbf{r})=\delta(\mathbf{r}-\mathbf{r}').$$ This result is intuitive to me. The charge density $$\rho$$ is the charge per unit volume, zero everywhere, but has some value at $$\mathbf{r}$$. The specific value can be obtained by recognizing that, if $$\mathbf{r}$$ is the only non-vanishing point, it must have a small volume (as to not interfere with other points). Hence, $$\rho(\mathbf{r})=\lim_{\mathrm{volume}\to0}\frac{\mathrm{charge}}{\mathrm{volume}}\to\pm\infty$$

But how could this be so if $$V=\int d^3r'\,\frac{\rho(\mathbf{r})}{|\mathbf{r}-\mathbf{r}'|}=\int d^3r'\,\frac{\delta(\mathbf{r}-\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\to\pm\infty$$

What assumptions have I made that are incorrect? (Is it that $$\rho(\mathbf{r})$$ is only a function of $$\mathbf{r}$$ and not $$\mathbf{r}'$$ so in fact $$\rho(\mathbf{r})=\delta(\mathbf{r})\neq \delta(\mathbf{r}-\mathbf{r}')$$?)

• Not sure how you conclude (2) from (1)? Commented Feb 27, 2021 at 15:33
• @NDewolf the Laplacian of $1/|\mathbf{r}-\mathbf{r}'|$ is $-4\pi\delta(\mathbf{r}-\mathbf{r}')$. Commented Feb 27, 2021 at 15:34
• So you're deriving (equation 4) that $\rho = \delta$ by already assuming it? Commented Feb 27, 2021 at 15:39
• @NDewolf From (2) and (1) we obtain the third equation and hence the fourth one. Commented Feb 27, 2021 at 15:40
• You can assume (2) and use it with (1) to obtain $\rho(r) = \delta(r-r')$, but you cannot just conclude (2) from (1). Commented Feb 27, 2021 at 15:44

The potential $$V_0(\textbf{r},\textbf{r}') = \frac{1}{|\textbf{r}-\textbf{r}'|}$$ is a solution to (1) only if $$\rho_0(\textbf{r}.\textbf{r}') = \delta(\textbf{r} - \textbf{r}')$$ for some fixed $$\mathbf{r}'$$. Also note that $$\nabla^2$$ acts with respect to the unprimed coordinates only.
Because the equation is linear, we can use this to construct a general solution to an arbitrary charge distribution $$\rho(\textbf{r})$$ (which now is a given function, and not a Dirac delta). First, we multiply the equation by $$\rho(\textbf{r}')$$. Here $$\textbf{r}'$$ is fixed, so $$\rho(\textbf{r}')$$ is just a number. This yields $$$$\rho(\textbf{r}') \nabla^2 V_0(\textbf{r},\textbf{r}') = - \rho(\textbf{r}') 4\pi \delta(\textbf{r} - \textbf{r}').$$$$ Now integrate both sides of the equation with respect to the primed coordinates over all of space. This gives $$$$\int d\textbf{r}' \rho(\textbf{r}') \nabla^2 V_0(\textbf{r},\textbf{r}') = -4\pi \int d\textbf{r}' \rho(\textbf{r}') \delta(\textbf{r}-\textbf{r}') = - 4\pi \rho(\textbf{r}).$$$$ Now on the left-hand side of the equation, we note that the integral is with respect to the primed coordinates while the differential operator $$\nabla^2$$ acts with respect to the unprimed coordinates. These are independent, so we can move the Laplacian outside the integral. Then we obtain $$$$\nabla^2 \int d\textbf{r}' \rho(\textbf{r}') V_0(\textbf{r},\textbf{r}') = -4\pi \rho (\textbf{r})$$$$ But this means that the function $$$$V(\textbf{r}) = \int d\textbf{r}' \rho(\textbf{r}') V_0(\textbf{r},\textbf{r}')$$$$ solves the equation $$\nabla^2 V(\textbf{r}) = - 4\pi \rho(\textbf{r})$$. This is the origin of the integral formula that you have at the end.
• Please don't use MathJax for formatting of text (such as \textit) -- use the Markdown constructs (like *italics* or _italics_) instead. Commented Feb 27, 2021 at 16:07