The Three-Dimensional Dirac Delta Function

The punchline identity is quietly present inside atomic spectra. The hyperfine interaction between an electron and the proton's magnetic moment contains a contact term proportional to $\delta^3(\mathbf{r})$ — born of exactly the concentrated-divergence structure of dipole fields — so the splitting depends on $|\psi(0)|^2$, the probability of the electron sitting on the proton. For hydrogen's ground state that contact term produces the 21-centimetre line, the radio-astronomy workhorse that mapped the Milky Way's spiral arms and now anchors searches for the cosmic dark ages. A delta function, measured across the galaxy. On the computational frontier, point sources are a daily engineering problem: numerical relativity treats black holes and neutron stars in the small-body limit as delta sources whose fields must be regularized on a finite grid — a discretized version of this article's smearing derivation — and molecular-dynamics codes handle millions of point charges whose $\delta^3$ densities are deliberately smeared (Ewald and particle-mesh methods) to make the long-range Coulomb sums converge. In both, the physics of the point and the art of the regulator are exactly the two derivations above, industrialized.

Watch the paradox assemble itself. Draw the field $\hat{\mathbf{r}}/r^2$: arrows everywhere pointing away from the origin, thinning with distance exactly as fast as spheres grow. Compute its divergence at any point you like — zero. Divergence measures whether a point manufactures flux, and out here no point does: whatever enters a small cell leaves it. Yet the field is unmistakably flowing outward from somewhere. Put a sphere of any radius around the origin and count: flux $=\frac{1}{r^2}\times4\pi r^2=4\pi$, always. Sources cannot hide in the bulk — we checked every point. There is exactly one point we could not check.

The one-dimensional delta was built for exactly this shape of fact: zero everywhere, finite total. Its three-dimensional sibling is the product of one spike per Cartesian direction, $\delta^3(\mathbf{r})=\delta(x)\delta(y)\delta(z)$ — zero unless all three coordinates vanish simultaneously, unit integral over any volume containing the origin, units of $1/\mathrm{m}^3$ (a volume density of pure concentration). Everything you drilled in one dimension carries over verbatim: sampling, ranges, Jacobians. What is new is the geometry — and the payoff, which is nothing less than making Maxwell's first equation true at the location of every charge in the universe. There is a second way to see the same concentration, and it is the honest analyst's way: regularize. Smooth the singularity by replacing $1/r$ with $1/\sqrt{r^2+a^2}$ — physically, smear the point source into a ball of radius $\sim a$. The divergence is then an ordinary function: a peak of height $\sim1/a^3$ and width $\sim a$, with total integral exactly $-4\pi$ independent of $a$. Shrink $a$: the peak narrows, grows, and preserves its integral — the defining choreography of a delta family, now in three dimensions.

The definition assembles three one-dimensional spikes, one per coordinate:

where: $\mathrm{d}\tau=\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$; the triple integral factorizes into three unit integrals. All one-dimensional machinery lifts immediately: the sifting property becomes $\int f(\mathbf{r})\,\delta^3(\mathbf{r}-\mathbf{a})\,\mathrm{d}\tau=f(\mathbf{a})$ when the volume contains $\mathbf{a}$ (zero otherwise); scaling picks up one Jacobian factor per dimension, $\delta^3(k\mathbf{r})=\delta^3(\mathbf{r})/|k|^3$; and the units are $1/\mathrm{m}^3$ — exactly what a volume charge density $\rho=q\,\delta^3(\mathbf{r}-\mathbf{r}')$ needs to carry $\mathrm{C/m^3}$. One three-dimensional novelty deserves its own display, because it is a standard trap: in curvilinear coordinates the delta must divide by the Jacobian, so that its volume integral — taken with the honest volume element — still gives 1:

where: the prefactor cancels the $r^2\sin\theta$ of $\mathrm{d}\tau$, leaving three clean one-dimensional sift integrals. (For a spike at the origin the angular coordinates are undefined and the useful form is $\delta^3(\mathbf{r})=\frac{\delta(r)}{4\pi r^2}$ — the $4\pi r^2$ spreads the unit source over the vanishing sphere.) Forgetting the Jacobian here is the three-dimensional cousin of forgetting the $1/|k|$ in one dimension, and it silently miscounts every source it touches. Now the main event.

Because $\hat{\mathbf{r}}/r^2=-\nabla(1/r)$ (the gradient computed in the spherical-coordinates article), the punchline has an immediate second costume, this time starring the Laplacian:

where: $\mathbf{r}'$ — the source location. Equation (3) is the seed of everything called a Green function in this track: it says $-1/4\pi|\mathbf{r}-\mathbf{r}'|$ is the potential response to a unit delta source, so the response to any source $\rho$ is the superposition $V=\frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\mathrm{d}\tau'$ — Coulomb's law, re-derived as one delta identity plus linearity. E2 will industrialize this.

Which identity answers which question

1 · Sifting drill in three dimensions (confidence builder)

Evaluate: (a) $\int(r^2+3)\,\delta^3(\mathbf{r}-\mathbf{a})\,\mathrm{d}\tau$ over all space, where $\mathbf{a}=(1,2,2)$ m; (b) the same integral over a sphere of radius 2 m centred at the origin; (c) $\int_{\text{all space}}e^{-r}\,\nabla\cdot(\hat{\mathbf{r}}/r^2)\,\mathrm{d}\tau$.

Step 1. (a) Sample at the spike: $r^2\to|\mathbf{a}|^2=1+4+4=9$, so the answer is $9+3=\boxed{12}$.

Step 2. (b) Locate the spike first: $|\mathbf{a}|=3$ m — outside the sphere of radius 2 m. The integrand vanishes throughout the region: $\boxed{0}$.

Step 3. (c) Use the punchline identity: $\nabla\cdot(\hat{\mathbf{r}}/r^2)=4\pi\delta^3(\mathbf{r})$, so the integral samples $e^{-r}$ at the origin: $4\pi e^{0}=\boxed{4\pi\approx12.6}$. One line — this is the identity working as a tool rather than a theorem.

2 · Double audit, and a point charge in an external potential (exam level)

(a) Recompute example 1(c) without the delta identity, by integration by parts, and confirm the answers agree. (b) A point charge $q=2.0\ \mathrm{nC}$ sits at $\mathbf{r}_0$ with $|\mathbf{r}_0|=0.10$ m in an external potential $V_{\mathrm{ext}}(\mathbf{r})=\kappa r^2$ with $\kappa=500\ \mathrm{V/m^2}$. Write its charge density and compute its potential energy $W=\int\rho\,V_{\mathrm{ext}}\,\mathrm{d}\tau$.

Step 1. (a) Move the divergence off the singular field with the product rule $\nabla\cdot(f\mathbf{v})=f\,\nabla\cdot\mathbf{v}+\mathbf{v}\cdot\nabla f$, integrated over a large ball of radius $R$: $\int e^{-r}\,\nabla\cdot\mathbf{v}\,\mathrm{d}\tau=\oint e^{-r}\,\mathbf{v}\cdot\mathrm{d}\mathbf{a}-\int\mathbf{v}\cdot\nabla(e^{-r})\,\mathrm{d}\tau$.

Step 2. The surface term: on the big sphere, $e^{-R}\times\frac{1}{R^2}\times4\pi R^2=4\pi e^{-R}\to0$ as $R\to\infty$. The gradient: $\nabla e^{-r}=-e^{-r}\hat{\mathbf{r}}$, so the remaining integral is $+\int e^{-r}\,\frac{1}{r^2}\,\mathrm{d}\tau=\int_0^\infty e^{-r}\frac{1}{r^2}\,4\pi r^2\,\mathrm{d}r=4\pi\int_0^\infty e^{-r}\mathrm{d}r=\boxed{4\pi}$.

Step 3. The two routes agree exactly: the delta identity in one line, honest calculus in four — and the by-parts route never touched the singularity, because the $r^2$ of the volume element tamed the $1/r^2$ of the field. Passing this kind of double audit is what "the identity is correct" means operationally.

Step 4. (b) The density is $\rho(\mathbf{r})=q\,\delta^3(\mathbf{r}-\mathbf{r}_0)$ — units check: C × m⁻³. The energy integral sifts:

The delta converted "energy of a point charge" — conceptually awkward, since the charge occupies no volume — into a perfectly ordinary volume integral that evaluates itself. That conversion, repeated with fields instead of external potentials, is how all of electrostatic energy bookkeeping will run in E1.

References. Griffiths, Introduction to Electrodynamics (4th ed.), §1.5 · Jackson, Classical Electrodynamics (3rd ed.), §1.7 and Eq. (1.31) · Purcell & Morin, Electricity and Magnetism (3rd ed.), Appendix on delta functions · Zangwill, Modern Electrodynamics, §1.5

Provenance. Equations cross-checked against Griffiths Introduction to Electrodynamics 4th ed. §1.5 (Eqs. 1.98-1.102), Jackson Classical Electrodynamics 3rd ed. §1.7 Eq. (1.31), and Zangwill Modern Electrodynamics §1.5. A verifier should: (1) recompute $\nabla\!\cdot(\hat{\mathbf{r}}/r^2)=0$ for $r\neq0$ via the spherical divergence formula; (2) confirm the sphere flux $=4\pi$ independent of radius; (3) verify the regularized limit $\nabla^2(r^2+a^2)^{-1/2}=-3a^2(r^2+a^2)^{-5/2}\to-4\pi\delta^3$ including the substitution integral $\int_0^\infty r^2(r^2+a^2)^{-5/2}\mathrm{d}r=1/3a^2$; (4) check the scaling property $\delta^3(a\mathbf{r})=|a|^{-3}\delta^3(\mathbf{r})$ by change of variables. The 21 cm hyperfine contact-term statement is standard (e.g. Griffiths Quantum Mechanics, hydrogen hyperfine structure). Worked-example numbers on this page (12; 0; $4\pi$ by two independent routes; $W=10$ nJ for $q=2$ nC in $V=\kappa r^2$) are illustrative arithmetic recomputed from the boxed identities.

Builds on: The Divergence of a Vector Field · The Divergence (Gauss) Theorem · Volume Integrals · Second Derivatives and the Laplacian Leads to: Poisson's and Laplace's Equations · Gauss's Law · The Electric Potential