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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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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. –  Chris White Aug 9 '12 at 0:22

1 Answer 1

up vote 9 down vote accepted

Yes. The delta function always has the same dimensions as the inverse of its argument. You can read this from its definition, your first equation. So in one dimension $\delta (x)$ has dimensions of inverse of length, in three spatial dimensions $\delta^{3}(\vec x)$ or simply $\delta(\vec x)$ has dimension of inverse of volume, and in n spatial dimension $\delta^{(n)}(\vec p)$ has dimensions of inverse of momentum to the n.

Let me remark that $$\delta^{3}(\vec x)=\delta(x)\delta(y)\delta(z)$$

where the cartesian components of $\vec x$ are $(x,y,z)$.

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