Three-Dimensional Dirac Delta Function

The three-dimensional Dirac delta function, $\delta({\bf r}-{\bf r}')$ , has the property

$$\delta({\bf r}-{\bf r}') =0 \mbox{\hspace{0.5cm}for\hspace{0.5cm}${\bf r}\neq {\bf r}'$} .$$ (21)

In addition, however, the function is singular at ${\bf r}={\bf r}'$ in such a manner that

$$\int_V \delta({\bf r}-{\bf r}') \,dV = 1.$$ (22)

Here, $V$ is any volume that contains the point ${\bf r}={\bf r}'$ . (Also, $dV$ is an element of $V$ expressed in terms of the components of ${\bf r}$ , but independent of the components of ${\bf r}'$ .) It follows that

$$\int_V f({\bf r})\,\delta({\bf r}-{\bf r}')\,dV = f({\bf r}'),$$ (23)

where $f{\bf (r})$ is an arbitrary function that is well behaved at ${\bf r}={\bf r}'$ . It is also easy to see that

$$\delta({\bf r}'-{\bf r}) = \delta({\bf r}-{\bf r}').$$ (24)

We can show that

$$\nabla^{\,2}\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)=-4\pi\,\delta({\bf r}-{\bf r}').$$ (25)

(Here, $\nabla^{\,2}$ is a Laplacian operator expressed in terms of the components of ${\bf r}$ , but independent of the components of ${\bf r}'$ .) We must first prove that

$$\nabla^{\,2}\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)=0 \mbox{\hspace{0.5cm}for\hspace{0.5cm}${\bf r}\neq {\bf r}'$} ,$$ (26)

in accordance with Equation (21). If $R=\vert{\bf r}-{\bf r}'\vert$ then this is equivalent to showing that

$$\frac{1}{R^{\,2}}\,\frac{d}{dR}\!\left[R^{\,2}\,\frac{d}{dR}\!\left(\frac{1}{R}\right)\right]=0$$ (27)

for $R>0$ , which is indeed the case. (Here, $R$ is treated as a radial spherical coordinate.) Next, we must show that

$$\int_V \nabla^{\,2}\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)dV =- 4\pi,$$ (28)

in accordance with Equations (22) and (25). Suppose that $S$ is a spherical surface, of radius $R$ , centered on ${\bf r}={\bf r}'$ . Making use of the definition $\nabla^{\,2}\phi \equiv \nabla\cdot\nabla\phi$ , as well as the divergence theorem, we can write

$$\int_V \nabla^{\,2}\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)dV = \int_V \nabla\cdot\nabla\left(\frac{1}{\vert{\bf r}-{\bf r}'\ve... ... = \int_S \nabla\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)\cdot d{\bf S}$$

$$= 4\pi\,R^{\,2}\,\frac{d}{dR}\!\left(\frac{1}{R}\right) = -4\pi.$$ (29)

(Here, $\nabla$ is a gradient operator expressed in terms of the components of ${\bf r}$ , but independent of the components of ${\bf r}'$ . Likewise, $d{\bf S}$ is a surface element involving the components of ${\bf r}$ , but independent of the components of ${\bf r}'$ .) Finally, if $S$ is deformed into a general surface (without crossing the point ${\bf r}={\bf r}'$ ) then the value of the volume integral is unchanged, as a consequence of Equation (26). Hence, we have demonstrated the validity of Equation (25).