Delta Function

The Dirac delta is a distribution concentrated at one point. Its defining property is not a pointwise formula but an action on a test function:

$$\int_{-\infty}^{\infty} f(x)\delta(x-a)\,dx = f(a).$$

This single rule underlies continuous-basis normalization, Fourier inversion, Green-function sources, density-of-states calculations, and jump conditions at singular interactions.

Definition as a distribution

Let $f$ be a smooth test function, for example a compactly supported smooth function. The delta distribution at $a$ is the linear functional

$$\langle\delta_a,f\rangle=f(a).$$

The integral notation

$$\langle\delta_a,f\rangle = \int f(x)\delta(x-a)\,dx$$

is useful shorthand for this action. It does not imply that $\delta(x-a)$ is an ordinary function with an assignable value at each point.

The support of $\delta_a$ is the single point $\{a\}$: it annihilates every test function that vanishes near $a$. No locally integrable function has this property together with the sifting rule. The familiar picture of a spike with unit area is an intuition supplied by regularizing sequences, not the definition.

For the general language of test functions and distributional equality, see Distributions.

Sifting and multiplication

If the integration region contains $a$ and $f$ is continuous there, then

$$\int f(x)\delta(x-a)\,dx=f(a).$$

More generally,

$$f(x)\delta(x-a) = f(a)\delta(x-a)$$

as distributions. To verify the identity, apply both sides to a test function $g$:

$$\begin{aligned} \langle f\delta_a,g\rangle &= \langle\delta_a,fg\rangle\\ &= f(a)g(a)\\ &= \langle f(a)\delta_a,g\rangle. \end{aligned}$$

Useful special cases include

$$(x-a)\delta(x-a)=0$$

and

$$x\delta(x)=0.$$

Products of a distribution with a smooth function are defined this way. Products such as $\delta(x)^2$ are not defined in classical distribution theory without extra regularization or a larger algebra.

If the support point lies exactly at the boundary of an integration interval, the result depends on how the restricted distribution or regularization is defined. A factor of one half is common in symmetric limiting prescriptions, but it is not a universal extra rule attached to the delta distribution.

Translation, scaling, and units

Translation moves the support:

$$\delta_a(x)=\delta(x-a).$$

For nonzero real $c$,

$$\delta(cx) = \frac{1}{\lvert c\rvert}\delta(x).$$

More generally,

$$\delta(cx-b) = \frac{1}{\lvert c\rvert} \delta\left(x-\frac{b}{c}\right).$$

The absolute value is required because reversing orientation cannot make a positive sifting measure negative.

Dimensions follow from the same rule. If $x$ has dimensions $[x]$, then

$$[\delta(x-a)]=[x]^{-1},$$

so that $\delta(x-a)\,dx$ is dimensionless. Thus $\delta(x)$, $\delta(p)$, and $\delta(E)$ have different physical units.

The delta distribution is even:

$$\delta(-x)=\delta(x).$$

This is the scaling rule with $c=-1$.

Nonlinear change of variables

Let $g$ be smooth and suppose its zeros $x_i$ are isolated and simple:

$$g(x_i)=0, \qquad g'(x_i)\ne0.$$

Then

$$\delta(g(x)) = \sum_i \frac{\delta(x-x_i)} {\lvert g'(x_i)\rvert}.$$

To understand the Jacobian, restrict to a neighborhood where $g$ is one-to-one and set $u=g(x)$. The sifting rule in $u$ contributes the inverse absolute derivative at the root. Summing the separate root neighborhoods gives the formula.

For example, if $a>0$,

$$\delta(x^2-a^2) = \frac{1}{2a} \left[ \delta(x-a)+\delta(x+a) \right].$$

The simple-root hypothesis is essential. If $g'(x_i)=0$, the displayed formula diverges and cannot be used. An expression such as $\delta(x^2)$ does not define a distribution by this rule.

Multidimensional delta

In $d$ dimensions,

$$\int_{\mathbb R^d} f(\mathbf x) \delta^{(d)}(\mathbf x-\mathbf a) \,d^d x = f(\mathbf a).$$

For an invertible linear map $A$,

$$\delta^{(d)}(A\mathbf x) = \frac{1}{\lvert\det A\rvert} \delta^{(d)}(\mathbf x).$$

For a smooth coordinate change $\mathbf x=\mathbf x(\mathbf u)$ with Jacobian $J=\partial\mathbf x/\partial\mathbf u$, the delta acquires the reciprocal volume Jacobian. In spherical coordinates, away from coordinate singularities,

$$\begin{aligned} \delta^{(3)} \left( \mathbf r-\mathbf r_0 \right) &= \frac{\delta(r-r_0)}{r^2}\\ &\quad\times \frac{ \delta(\theta-\theta_0) \delta(\varphi-\varphi_0) }{\sin\theta}. \end{aligned}$$

The denominator cancels the spherical volume element $r^2\sin\theta\,dr\,d\theta\,d\varphi$. Coordinate periodicity and points on the polar axis require charts or an invariant formulation rather than blind use of this expression.

Distributional derivatives

The derivative is defined by moving differentiation onto the test function:

$$\langle\delta_a',f\rangle = -f'(a).$$

Equivalently,

$$\int f(x)\delta'(x-a)\,dx = -f'(a).$$

Higher derivatives satisfy

$$\int f(x)\delta^{(n)}(x-a)\,dx = (-1)^n f^{(n)}(a).$$

The minus sign is the distributional version of integration by parts. It assumes the test-function framework has already removed endpoint terms.

A useful identity is

$$(x-a)\delta'(x-a) = -\delta(x-a).$$

Indeed, acting on $f$ gives

$$\begin{aligned} \left\langle (x-a)\delta_a',f \right\rangle &= \left\langle \delta_a',(x-a)f \right\rangle\\ &= -\left. \frac{d}{dx} \left[ (x-a)f(x) \right] \right|_{x=a}\\ &= -f(a). \end{aligned}$$

Jumps and cusp conditions are developed in Distributional Derivatives.

Regularizing sequences

A family of ordinary functions can converge to $\delta$ in the distributional sense. Two standard examples are

$$\delta_\epsilon^{\mathrm G}(x) = \frac{1}{\sqrt{\pi}\epsilon} e^{-x^2/\epsilon^2}$$

and

$$\delta_\epsilon^{\mathrm L}(x) = \frac{1}{\pi} \frac{\epsilon}{x^2+\epsilon^2},$$

with $\epsilon>0$. Each has unit integral. For a test function $f$,

$$\lim_{\epsilon\to0^+} \int f(x)\delta_\epsilon(x)\,dx = f(0).$$

The two families have different shapes and tails but the same distributional limit. This nonuniqueness is healthy: a distribution is characterized by its action, not by a preferred microscopic spike.

There is no ordinary pointwise limit equal to a function. Away from zero the regularizers tend to zero, while their values near zero grow without bound. Area is redistributed into a narrowing region.

Fourier representations

With the momentum convention,

$$\delta(x-x') = \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} e^{ip(x-x')/\hbar}\,dp.$$

The dual identity is

$$\delta(p-p') = \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} e^{ix(p-p')/\hbar}\,dx.$$

Both are distributional statements. They mean that inserting the kernel under an integral reproduces the test function.

The Fourier transform of a translated delta is a phase:

$$\mathcal F_\hbar \left[ \delta(x-a) \right](p) = \frac{1}{\sqrt{2\pi\hbar}} e^{-ipa/\hbar}.$$

Conversely,

$$\mathcal F_\hbar[1](p) = \sqrt{2\pi\hbar}\,\delta(p)$$

as a tempered distribution. This pair expresses reciprocal localization: a point source contains all Fourier modes, while a constant contains only zero momentum.

Convolution identity

For a sufficiently regular function $f$,

$$\begin{aligned} (\delta_a*f)(x) &= \int_{-\infty}^{\infty} \delta(x-y-a)f(y)\,dy\\ &= f(x-a). \end{aligned}$$

Thus $\delta_0$ is the identity element for convolution:

$$\delta*f=f.$$

Differentiation commutes with convolution in the distributional sense:

$$\delta' * f=f'.$$

The complete transform and convolution rules are collected in Convolution.

Continuous quantum bases

Position eigenkets obey

$$\langle x\vert x'\rangle = \delta(x-x')$$

and the formal completeness relation

$$\int_{-\infty}^{\infty} \lvert x\rangle\langle x\rvert\,dx = I.$$

Momentum eigenkets similarly satisfy

$$\langle p\vert p'\rangle = \delta(p-p'), \qquad \int \lvert p\rangle\langle p\rvert\,dp = I.$$

Applying position completeness to a state gives

$$\begin{aligned} \psi(x) &= \langle x\vert\psi\rangle\\ &= \int \langle x\vert x'\rangle \langle x'\vert\psi\rangle\,dx'\\ &= \int \delta(x-x')\psi(x')\,dx'. \end{aligned}$$

These kets are generalized eigenvectors rather than elements of the ordinary Hilbert space. See Generalized Eigenvectors and Plane Waves and Delta Normalization.

The probability density $\lvert\psi(x)\rvert^2$ is ordinary when $\psi$ is a normalizable state. Delta normalization belongs to the basis, not to the probability density of a localized physical state.

Density-of-states Jacobians

Delta functions often convert a conservation law into an integral over allowed states. For a one-dimensional free particle,

$$E(p)=\frac{p^2}{2m}.$$

At fixed $E>0$, define

$$p_E=\sqrt{2mE}.$$

The two simple roots are $p=\pm p_E$, so

$$\begin{aligned} \delta\left( E-\frac{p^2}{2m} \right) &= \frac{m}{p_E}\delta(p-p_E)\\ &\quad+ \frac{m}{p_E}\delta(p+p_E). \end{aligned}$$

Therefore,

$$\begin{aligned} I_E[F] &= \int_{-\infty}^{\infty} F(p) \delta\left( E-\frac{p^2}{2m} \right)dp\\ &= \frac{m}{p_E} \left[ F(p_E)+F(-p_E) \right]. \end{aligned}$$

The factor $m/p_E$ is the inverse slope $\lvert dE/dp\rvert^{-1}$ and is the local density-of-states Jacobian. At $E=0$ the roots merge and are not simple, so the formula must not be used by direct substitution.

Point sources and jump conditions

A delta source in a differential equation creates a finite jump in a lower derivative. For example, integrating

$$-a\,y''(x) +g\delta(x-x_0)y(x) = F(x)$$

through $x_0$, with $F$ locally integrable, gives

$$y'(x_0^+)-y'(x_0^-) = \frac{g}{a}y(x_0).$$

The sign follows from the displayed differential equation. Continuity of $y$ is a separate domain assumption and is appropriate for the standard one-dimensional delta potential. The full quantum model belongs in Delta-Function Potential.

Dirac comb

A periodic delta array is

$$\operatorname{III}_L(x) = \sum_{n\in\mathbb Z} \delta(x-nL).$$

It samples a test function on a lattice:

$$\left\langle \operatorname{III}_L,f \right\rangle = \sum_{n\in\mathbb Z}f(nL),$$

when the sum is meaningful. Its Fourier transform is another comb on the reciprocal lattice. This is the distributional core of the Poisson Summation Formula.

A reliable delta workflow

1. Identify the integration variable and the units of its delta.
2. Locate every root of the delta argument inside the integration region.
3. Check that each root is simple before using the Jacobian formula.
4. Include the absolute derivative at every root.
5. If derivatives of delta occur, move them onto the test function with the correct sign.
6. Treat Fourier representations and basis normalizations distributionally.
7. Keep regularization choices explicit when products or boundary values are involved.

Common mistakes

• Assigning a finite or infinite numerical value to $\delta(0)$.
• Treating the delta as an ordinary function outside an integral or pairing.
• Forgetting the absolute value in the scaling Jacobian.
• Keeping only one root of $g(x)=0$ in $\delta(g(x))$.
• Applying the simple-root formula when $g'(x_i)=0$.
• Giving $\delta(x)$ no units.
• Confusing a Dirac delta with a Kronecker delta.
• Assuming a delta at an integration endpoint always contributes one half.
• Squaring a delta distribution as though products of distributions were automatic.
• Losing the minus sign in the action of $\delta'$.
• Calling a delta-normalized basis vector a normalized physical state.
• Using a Fourier integral representation as an ordinary convergent integral.

Exercises

1. For $a>0$, evaluate

   $$\int_{-\infty}^{\infty} f(x)\delta(x^2-a^2)\,dx.$$

Solution

The roots of $g(x)=x^2-a^2$ are $x_\pm=\pm a$, and

$$\lvert g'(x_\pm)\rvert=2a.$$

Therefore,

$$\delta(x^2-a^2) = \frac{1}{2a} \left[ \delta(x-a)+\delta(x+a) \right],$$

and

$$\begin{aligned} I_a &= \int_{-\infty}^{\infty} f(x)\delta(x^2-a^2)\,dx\\ &= \frac{f(a)+f(-a)}{2a}. \end{aligned}$$

2. Prove the distributional identity

   $$x\delta'(x)=-\delta(x).$$

Solution

Apply the left side to a test function $f$:

$$\begin{aligned} \langle x\delta',f\rangle &= \langle\delta',xf\rangle\\ &= -\left. \frac{d}{dx} \left[ xf(x) \right] \right|_{x=0}\\ &= -f(0). \end{aligned}$$

The right side acts as

$$\langle-\delta,f\rangle=-f(0).$$

Since the actions agree on every test function, the distributions are equal.

3. Show that the Gaussian family

   $$\delta_\epsilon(x) = \frac{1}{\sqrt{\pi}\epsilon} e^{-x^2/\epsilon^2}$$

   converges distributionally to $\delta(x)$.

Solution

For a smooth compactly supported test function $f$, set $x=\epsilon u$ and write the pairing as $I_\epsilon[f]$. Then

$$\begin{aligned} I_\epsilon[f] &= \int_{-\infty}^{\infty} f(x)\delta_\epsilon(x)\,dx\\ &= \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} f(\epsilon u)e^{-u^2}\,du. \end{aligned}$$

For each fixed $u$, $f(\epsilon u)\to f(0)$. The integrand is dominated by $\lVert f\rVert_\infty e^{-u^2}$, which is integrable. Dominated convergence therefore gives

$$\begin{aligned} \lim_{\epsilon\to0^+} I_\epsilon[f] &= \frac{f(0)}{\sqrt{\pi}} \int_{-\infty}^{\infty} e^{-u^2}\,du\\ &= f(0). \end{aligned}$$

This is exactly convergence to $\delta$ in the distributional sense.

4. Let $E(p)=p^2/(2m)$ and $E>0$. Evaluate

   $$\int_{-\infty}^{\infty} p^2\delta\left( E-\frac{p^2}{2m} \right)dp.$$

Solution

The roots are $p=\pm p_E$, where $p_E=\sqrt{2mE}$. At either root,

$$\left\lvert \frac{d}{dp} \left( E-\frac{p^2}{2m} \right) \right\rvert = \frac{p_E}{m}.$$

Hence

$$\begin{aligned} \int p^2\delta\left( E-\frac{p^2}{2m} \right)dp &= \frac{m}{p_E} \left( p_E^2+p_E^2 \right)\\ &= 2mp_E\\ &= 2m\sqrt{2mE}. \end{aligned}$$

The answer has units of momentum cubed divided by energy, as required by $p^2\,dp\,\delta(E)$.

References

• I. M. Gel’fand and G. E. Shilov, Generalized Functions, Vol. 1: Properties and Operations, Academic Press, 1964.
• F. G. Friedlander and M. Joshi, Introduction to the Theory of Distributions, 2nd ed., Cambridge University Press, 1998.
• M. J. Lighthill, Introduction to Fourier Analysis and Generalised Functions, Cambridge University Press, 1958.
• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• A. Bohm, Quantum Mechanics: Foundations and Applications, 3rd ed., Springer, 1993.