# What are the units or dimensions of the Dirac delta function?

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?

• 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.
– user10851
Commented Aug 9, 2012 at 0:22
• Following this discussion - what are the dimension of the Heaviside "step" function?
– E Be
Commented Aug 10, 2016 at 17:46
• @Udi Behar for Heaviside step see physics.stackexchange.com/q/274380/45664 Commented Oct 11, 2016 at 21:40

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 $n$.
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.