In three dimensions, the Dirac delta function $\delta^3 (\textbf{r}) = \delta(x) \delta(y) \delta(z)$ is defined by the volume integral:
$$\int_{\text{all space}} \delta^3 (\textbf{r}) \, dV = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \delta(x) \delta(y) \delta(z) \, dx \, dy \, dz = 1$$
where
$$\delta(x) = 0 \text{ if } x \neq 0$$
and
$$\delta(x) = \infty \text{ if } x = 0$$
and similarly for $\delta(y)$ and $\delta(z)$.
Does this mean that $\delta^3 (\textbf{r})$ has dimensions of reciprocal volume?
As an example, a textbook that I am reading states:
For a collection of $N$ point charges we can define a charge density
$$\rho(\textbf{r}) = \sum_{i=1}^N q_i \delta(\textbf{r} – \textbf{r}_i)$$
where $\textbf{r}_i$ and $q_i$ are the position and charge of particle $i$, respectively.
Typically, I would think of charge density as having units of charge per volume in three dimensions: $(\text{volume})^{-1}$. For example, I would think that units of $\frac{\text{C}}{\text{m}^3}$ might be possible SI units of charge density. If my assumption is true, then $\delta^3 (\textbf{r})$ must have units of $(\text{volume})^{-1}$, like $\text{m}^{-3}$ for example. Is this correct?
Best Answer
Yes. The Dirac delta always has the inverse dimension of its argument. You can read this from its definition, your first equation. So in one dimension $\delta(x)$ has dimensions of inverse length, in three spatial dimensions $\delta^{(3)}(\vec x)$ (sometimes simply written $\delta(\vec x)$) has dimension of inverse volume, and in $n$ dimensions of momentum $\delta^{(n)}(\vec p)$ has dimensions of inverse momentum to the power of $n$.