The Dirac Delta Function

⚡ Key Takeaways

The Dirac delta function $\delta(x)$ is a generalized function (distribution) that is zero everywhere except at $x=0$ , where it is "infinite" in such a way that its total integral equals one. Its defining property is the sifting property: $\int_{-\infty}^{\infty} f(x)\,\delta(x-a)\,dx = f(a)$ . It appears throughout physics and engineering wherever a quantity is concentrated at a single point — point masses, point charges, instantaneous impulses, and as the identity element of convolution. Introduced by Paul Dirac (1927) in quantum mechanics and made rigorous by Laurent Schwartz (1945–1950) via the theory of distributions.

💡 The Core Insight

The delta function is defined entirely by what it does inside an integral, not by its pointwise values. To know $\delta$ is to know that it "samples" or "sifts out" the value of any well-behaved function at a single point. Everything else — its scaling rules, its derivatives, its Fourier transform — follows from this one sifting property. Stop thinking of $\delta(x)$ as a graph and start thinking of it as an operation.

DefinitionThe Dirac Delta as a Distribution

The Dirac delta $\delta$ is the continuous linear functional on the space $\mathcal{D}$ of smooth, compactly supported test functions $\varphi$ defined by

1

\quad \langle \delta,\varphi\rangle \;=\; \varphi(0).

Derivation hint: This is the definition — $\delta$ assigns to each test function $\varphi$ its value at the origin. Linearity and continuity (in the topology of $\mathcal{D}$ ) are immediate. Symbols: $\varphi$ is an infinitely differentiable function vanishing outside a bounded set; $\langle \cdot,\cdot\rangle$ denotes the action of a distribution on a test function. The shifted version $\delta(x-a)$ satisfies $\langle \delta(\cdot - a),\varphi\rangle = \varphi(a)$ .

TheoremDefining Properties of the Delta Function

For suitable functions $f$ and constants $a$ , the delta satisfies:

4

\quad \delta(-x) = \delta(x) \quad(\text{even}),\qquad \delta(ax) = \frac{1}{|a|}\,\delta(x).

Derivation hint: Substitute $u=ax$ in the sifting integral; the Jacobian $1/|a|$ and the modulus from reversing limits when $a<0$ produce the factor $1/|a|$ .

5

\quad x\,\delta(x) = 0, \qquad f(x)\,\delta(x-a) = f(a)\,\delta(x-a).

Derivation hint: Test both sides against an arbitrary $\varphi$ : $\langle x\delta,\varphi\rangle = \langle \delta, x\varphi\rangle = (x\varphi)|_{x=0}=0$ .

6

\quad \delta\big(g(x)\big) = \sum_{i}\frac{\delta(x-x_i)}{|g'(x_i)|},\quad \text{where } g(x_i)=0,\ g'(x_i)\neq 0.

Derivation hint: Near each simple root $x_i$ , linearize $g(x)\approx g'(x_i)(x-x_i)$ and apply the scaling rule [4] to each piece.

Figure

The delta function as a limit of unit-area spikes (Gaussian nascent deltas)

ProofThe Delta as a Limit of Nascent Delta Functions

We construct $\delta$ as a limit of ordinary functions and verify the sifting property.

Step 1 — Choose a nascent delta. Let $\delta_\varepsilon(x) = \dfrac{1}{\varepsilon\sqrt{\pi}}\,e^{-x^2/\varepsilon^2}$ , a Gaussian of width $\sim\varepsilon$ .

7

\quad \int_{-\infty}^{\infty}\delta_\varepsilon(x)\,dx = \frac{1}{\varepsilon\sqrt{\pi}}\int_{-\infty}^{\infty} e^{-x^2/\varepsilon^2}\,dx = 1 \quad \text{for all } \varepsilon>0.

Derivation hint: Substitute $u=x/\varepsilon$ and use the Gaussian integral $\int e^{-u^2}du = \sqrt{\pi}$ .

Step 2 — Apply to a continuous test function. For any $f$ continuous at $0$ ,

8

\quad \int_{-\infty}^{\infty} f(x)\,\delta_\varepsilon(x)\,dx = \frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty} f(\varepsilon u)\,e^{-u^2}\,du.

Derivation hint: Same substitution $u=x/\varepsilon$ , $dx=\varepsilon\,du$ , which cancels the prefactor's $\varepsilon$ .

Step 3 — Take the limit. As $\varepsilon\to 0$ , $f(\varepsilon u)\to f(0)$ pointwise, and dominated convergence gives

9

\quad \lim_{\varepsilon\to0}\int_{-\infty}^{\infty} f(x)\,\delta_\varepsilon(x)\,dx = f(0)\cdot\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty} e^{-u^2}\,du = f(0).

Derivation hint: Pull out $f(0)$ and recognize the remaining normalized Gaussian integral as $1$ .

Step 4 — Identify the limit. The functional $f\mapsto \lim_{\varepsilon\to0}\int f\,\delta_\varepsilon$ equals $f\mapsto f(0)$ , which is exactly $\delta$ in [1]. The same limit results for rectangular, Lorentzian, or $\mathrm{sinc}$ nascent deltas — so $\delta$ is independent of the approximating family. $\blacksquare$

∎

PA

Paul Adrien Maurice Dirac

1902–1984

Dirac introduced the delta function in his 1927 paper and his 1930 monograph The Principles of Quantum Mechanics as a tool to handle continuous spectra and to express orthonormality of position eigenstates, $\langle x | x'\rangle = \delta(x-x')$ . He called it an "improper function" and was fully aware it was not a function in the classical sense, justifying it by the consistency of the results it produced. Trained as an electrical engineer before turning to physics, Dirac prized formal manipulation that "worked," and his confident, almost cavalier use of $\delta$ challenged mathematicians for two decades — culminating in Laurent Schwartz's rigorous theory of distributions, which vindicated Dirac entirely.

🧠 Intuition

Picture pouring exactly one liter of water onto a flat table. Spread thin, it covers a wide area at low height — that is a broad nascent delta. Now confine it to a smaller and smaller puddle: the height shoots up while the volume stays fixed at one liter. The delta is the idealized limit — infinitely tall, infinitely thin, total "amount" exactly one. The crucial mental shift is that you never care about the height; you only care about how much "stuff" gets deposited when you integrate. When you multiply $\delta(x-a)$ by a function $f$ and integrate, the spike acts like a probe that asks, "What is $f$ doing right here at $a$ ?" and reports back $f(a)$ , ignoring $f$ everywhere else. This is why $\delta$ is the perfect language for anything point-like: a point charge has all its charge at one location, an idealized hammer-strike delivers all its impulse in one instant.

ExampleSifting and the Scaling Rule

Given: Evaluate $\displaystyle I = \int_{-\infty}^{\infty} (3x^2 + 2)\,\delta(2x - 4)\,dx$ .

Step 1 — Locate the root. The argument $2x-4 = 0$ at $x = 2$ .

Step 2 — Apply the scaling rule [6] with $g(x)=2x-4$ , $g'(x)=2$ , so $\delta(2x-4) = \frac{1}{|2|}\delta(x-2)$ .

Step 3 — Sift.

10

\quad I = \frac{1}{2}\int_{-\infty}^{\infty}(3x^2+2)\,\delta(x-2)\,dx = \frac{1}{2}\,(3\cdot 4 + 2).

Derivation hint: Factor out $1/2$ , then the sifting property replaces $x$ by $2$ in the polynomial.

Step 4 — Compute. $\frac{1}{2}(12 + 2) = \frac{14}{2}$ .

\boxed{I = 7}\quad(\text{dimensionless})

ExampleImpulse on a Mass — Physical Units

Given: A mass $m = 0.50\ \text{kg}$ at rest receives a force $F(t) = J\,\delta(t - t_0)$ with impulse magnitude $J = 3.0\ \text{N}\,\cdot\,\text{s}$ . Find the velocity just after the impulse.

Step 1 — Newton's law. $m\,\dfrac{dv}{dt} = F(t) = J\,\delta(t-t_0)$ .

Step 2 — Integrate across the impulse over $[t_0^-, t_0^+]$ :

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\quad m\big(v(t_0^+) - v(t_0^-)\big) = \int_{t_0^-}^{t_0^+} J\,\delta(t-t_0)\,dt = J.

Derivation hint: The integral of $\delta$ over any interval containing its spike is $1$ ; the smooth velocity term contributes the jump on the left.

Step 3 — Solve. With $v(t_0^-)=0$ : $v(t_0^+) = J/m = (3.0\ \text{N}\,\cdot\,\text{s})/(0.50\ \text{kg})$ .

\boxed{v(t_0^+) = 6.0\ \text{m/s}}

The delta encodes the instantaneous nature of an idealized collision: finite velocity change in zero time.

⚠️ Common mistake

$\delta(x)$ is not a function and $\delta(0)$ is not a number. Writing " $\delta(0) = \infty$ " is a heuristic, not a value you may use in arithmetic. Likewise, products of deltas at the same point are undefined: $\delta(x)^2$ has no meaning in distribution theory, because distributions cannot generally be multiplied. Statements like $\int \delta(x)^2\,dx$ are ill-defined (and the heuristic "answer" $\delta(0)$ is precisely why). Always interpret $\delta$ through its action on a test function, never by evaluating it pointwise.

CounterexampleWhy No Ordinary Function Can Be the Delta

Suppose $\delta$ were an ordinary (Lebesgue-integrable) function equal to $0$ for all $x\neq 0$ . Then it differs from the zero function only at the single point $x=0$ , a set of measure zero. By the standard theory of Lebesgue integration, any function that is zero almost everywhere has integral zero:

12

\quad \int_{-\infty}^{\infty}\delta(x)\,dx = 0 \neq 1.

Derivation hint: Modifying a function on a measure-zero set never changes its Lebesgue integral; hence the unit-area requirement is impossible for a genuine function. This contradiction with [3] proves $\delta$ cannot live among ordinary functions — it requires the larger framework of distributions.

connection

The delta function ties together many areas. It is the derivative of the <span class="wiki-link-missing">Heaviside Step Function</span>: $H'(x) = \delta(x)$ in the distributional sense, since integrating $\delta$ produces the unit jump. It is the identity element of <span class="wiki-link-missing">Convolution</span>: $(f * \delta)(x) = f(x)$ , making it the natural starting point for Green's Functions, which are responses to a delta-source. Its Fourier Transform is the constant $1$ (and conversely the transform of $1$ is $2\pi\,\delta$ ), revealing that an infinitely localized signal contains all frequencies equally. The rigorous home of $\delta$ is the Theory of Distributions, and in Measure Theory it corresponds exactly to the Dirac measure $\delta_a(A) = 1$ if $a\in A$ , else $0$ .

🔬 Where this appears

⏳ History

- 1822 — Joseph Fourier writes the Fourier integral theorem in a form implicitly containing the integral representation of the delta.

- 1828 — George Green introduces "Green's functions," responses to point sources, foreshadowing delta methods in potential theory.

- 1880s — Gustav Kirchhoff and Oliver Heaviside use impulse-like quantities informally in physics and electrical engineering.

- 1894 — Heaviside's operational calculus manipulates step and impulse functions, drawing sharp criticism from rigor-minded mathematicians.

- 1927 — Paul Dirac introduces

\delta(x)

explicitly in his quantum mechanics papers, listing its key properties.

- 1930 — Dirac's The Principles of Quantum Mechanics popularizes the "improper function"

\delta

worldwide.

- 1936 — Sergei Sobolev develops generalized functions while studying hyperbolic PDEs, a key precursor to distributions.

- 1945–1950 — Laurent Schwartz formalizes the theory of distributions, rigorously defining

\delta

as a continuous linear functional; he earns the Fields Medal (1950) partly for this work.

- 1955 — Israel Gelfand and Georgi Shilov publish Generalized Functions, systematizing the subject for physics and applied mathematics.

open-problem

While the linear theory of distributions is complete, multiplying distributions remains fundamentally problematic: products like $\delta\cdot\delta$ or $\delta\cdot H$ are not defined within Schwartz's framework, a serious obstacle in nonlinear PDEs, quantum field theory (where products of field operators produce divergences requiring renormalization), and general relativity (defining distributional curvature for singular spacetimes). Colombeau algebras and related theories of "generalized functions" attempt to allow consistent multiplication, but they sacrifice some natural properties and have not achieved universal adoption. Deciding the "right" framework for nonlinear operations on singular objects — and giving rigorous meaning to formal manipulations physicists perform daily — remains an active area at the intersection of analysis and mathematical physics.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also