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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?

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    $\begingroup$ As long as you're asking for details about the $\delta$-function, I feel obliged to point out that there are all sorts of caveats with saying $\delta(0) = \infty$. While this may help physical intuition, mathematically the most natural interpretation of that equation would still leave the integral as zero, since (Lebesgue) integrals never depend on a single point's value. Probably best just to think of it as an object with the appropriate integration properties. $\endgroup$ – user10851 Aug 9 '12 at 0:22
  • $\begingroup$ Following this discussion - what are the dimension of the Heaviside "step" function? $\endgroup$ – E Be Aug 10 '16 at 17:46
  • $\begingroup$ @Udi Behar for Heaviside step see physics.stackexchange.com/q/274380/45664 $\endgroup$ – user45664 Oct 11 '16 at 21:40
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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 $n$.

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Let $x$ be dimensionless and Using the property $\delta (ax)=\frac{1}{|a|}\delta (x)$ we see that indeed the dimension of a Dirac delta is the dimension of the inverse of its argument.

One reoccurring example is eg $\delta(p'-p)$ where $p$ denotes momentum, this delta has dimension of inverse mass in natural units.

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