Skip to content

The Dirac Delta

Source lecture(s): PHY621 Ch. 3

Intuition

The delta "function" is the continuous analogue of the Kronecker delta: the thing that picks out one value from a continuum, exactly as \(\delta_{ij}\) picks one term from a sum.

\[\sum_j \delta_{ij}a_j = a_i \qquad\longleftrightarrow\qquad \int \delta(x - x')f(x')\,dx' = f(x)\]

It is not a function. There is no function that is zero everywhere except one point and still integrates to 1. It is a distribution — an object defined entirely by what it does inside an integral, and any statement about \(\delta(x)\) that is not eventually integrated against a test function is meaningless.

Where it comes from in this course

Push Fourier series to an infinite interval and the sum becomes an integral; the completeness relation of the discrete basis becomes

\[\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ik(x-x')}\,dk = \delta(x - x')\]

That identity is Fourier inversion. Substituting it into \(f(x) = \int\delta(x-x')f(x')dx'\) reproduces the transform pair, and it is the continuous version of \(\sum_n|n\rangle\langle n| = \hat 1\) — see Dirac notation.

Properties worth knowing

\[\delta(ax) = \frac{1}{|a|}\delta(x), \qquad \delta(g(x)) = \sum_i\frac{\delta(x - x_i)}{|g'(x_i)|}\ \text{(simple roots }x_i)\]
\[\int\delta'(x)f(x)\,dx = -f'(0), \qquad x\,\delta(x) = 0, \qquad \frac{d}{dx}\Theta(x) = \delta(x)\]

The derivative rule comes from integrating by parts and throwing the derivative onto the test function — which is how all derivatives of distributions are defined. The last identity (the step function's derivative) is the one you use constantly in Green's functions.

As a limit

\(\delta\) is the limit of any normalised sequence that narrows:

\[\delta(x) = \lim_{\epsilon\to0}\frac{1}{\epsilon\sqrt{\pi}}e^{-x^2/\epsilon^2} = \lim_{\epsilon\to0}\frac{1}{\pi}\frac{\epsilon}{x^2+\epsilon^2} = \lim_{L\to\infty}\frac{\sin Lx}{\pi x}\]

The limit must be taken after integrating against the test function, never before. The Lorentzian form is the one that appears in scattering theory and in the Landau prescription; the \(\sin Lx/\pi x\) form is what makes Fourier inversion work.

Why physicists reach for it

  • Point sources. A point charge is \(\rho = q\delta^3(\mathbf{r})\); a point mass, an impulse, a delta-function potential.
  • Green's functions. The whole method is: solve \(\mathcal{L}G = \delta\), then build the response to any source by superposition, \(u = \int G f\). The delta is the "unit kick" whose response you record once and reuse forever.
  • Convolution. \(\delta\) is the identity: \(f * \delta = f\).
  • Orthonormality of continuous bases. \(\langle x|x'\rangle = \delta(x-x')\).

Common mistakes

  • Evaluating \(\delta(0)\). It is not a number. If your algebra produces \(\delta(0)\) you have usually squared a delta, which is undefined.
  • Forgetting the Jacobian. \(\delta(g(x))\) picks up \(1/|g'|\) at each root; missing it is a common factor-of-something error in scattering and in changes of variable.
  • Taking the limit too early. The sequence definitions only make sense under the integral.
  • Squaring it. \(\delta(x)^2\) is meaningless. If you need \(|\langle x|\psi\rangle|^2\) that is fine; \(\delta^2\) is not.

Knowledge graph position

Prerequisites: Fourier series, inner product spaces. Leads to: Fourier transform, Green's functions, distribution theory, quantum mechanics.

Quiz

Q1 (computational). Evaluate \(\int_{-\infty}^{\infty}\delta(2x - 6)\,x^2\,dx\).

Answer

Write \(g(x) = 2x-6\), root at \(x = 3\), \(|g'| = 2\). So \(\delta(2x-6) = \tfrac12\delta(x-3)\) and the integral is \(\tfrac12(3)^2 = 4.5\). Forgetting the \(1/|g'|\) would give 9.

Q2 (conceptual). In what sense is \(\frac{1}{2\pi}\int e^{ik(x-x')}dk = \delta(x-x')\) the same statement as Fourier inversion?

Answer

Substituting it into \(f(x) = \int\delta(x-x')f(x')dx'\) and exchanging the order of integration gives exactly the forward-then-inverse transform. The completeness of the plane waves and the invertibility of the transform are one fact stated two ways.

Q3 (MCQ). The Dirac delta is best understood as:

  • (a) a function that is infinite at zero
  • (b) a distribution, defined by its action on test functions inside an integral
  • (c) a limit of Gaussians, evaluated pointwise
  • (d) a discrete Kronecker delta
Answer

(b). (a) is not well-defined; (c) fails because the limit must be taken after integration; (d) is the discrete analogue, not the same object.