Dirac delta function

The Dirac delta function is a function introduced in 1930 by P. A. M. Dirac in his seminal book on quantum mechanics. A physical model that visualizes a delta function is a mass distribution of finite total mass M&mdash;the integral over the mass distribution. When the distribution becomes smaller and smaller, while M is constant, the mass distribution shrinks to a point mass, which by definition has zero extent and yet has a finite-valued integral equal to total mass M. In the limit of a point mass the distribution becomes a Dirac delta function.

Heuristically, the Dirac delta function can be seen as an extension of the Kronecker delta from integral indices (elements of  $$\mathbb{Z}$$ ) to real indices (elements of $$\mathbb{R}$$ ). Note that the Kronecker delta acts as a "filter" in a summation:
 * $$ \sum_{i=m}^n \; f_i\; \delta_{ia} = \begin{cases} f_a & \quad\hbox{if}\quad a\in[m,n] \sub\mathbb{Z}  \\ 0   & \quad \hbox{if}\quad a \notin [m,n]. \end{cases} $$

In analogy, the Dirac delta function &delta;(x&minus;a) is defined by (replace i by x and the summation over i by an integration over x),
 * $$ \int_{a_0}^{a_1} f(x) \delta(x-a) \mathrm{d}x = \begin{cases} f(a) & \quad\hbox{if}\quad  a\in[a_0,a_1] \sub\mathbb{R},   \\ 0   & \quad \hbox{if}\quad a \notin [a_0,a_1]. \end{cases} $$

The Dirac delta function is not an ordinary well-behaved map $$\mathbb{R} \rightarrow \mathbb{R}$$, but a distribution, also known as an improper or generalized function. Physicists express its special character by stating that the Dirac delta function makes only sense as a factor in an integrand ("under the integral"). Mathematicians say that the delta function is a linear functional on a space of test functions.

Properties
Most commonly one takes the lower and the upper bound in the definition of the delta function equal to $$-\infty$$ and $$ \infty$$, respectively. From here on this will be done.
 * $$ \begin{align} \int_{-\infty}^{\infty} \delta(x)\mathrm{d}x &= 1, \\ \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ikx} \mathrm{d}k &= \delta(x) \\ \delta(x-a) &= \delta(a-x), \\ (x-a)\delta(x-a) &= 0, \\ \delta(ax) &= |a|^{-1} \delta(x) \quad (a \ne 0), \\ f(x) \delta(x-a) &= f(a) \delta(x-a), \\ \int_{-\infty}^{\infty} \delta(x-y)\delta(y-a)\mathrm{d}y &= \delta(x-a) \end{align} $$

The physicist's proof of these properties proceeds by making proper substitutions into the integral and using the ordinary rules of integral calculus. The delta function as a Fourier transform of the unit function f(x) = 1 (the second property) will be proved below. The last property is the analogy of the multiplication of two identity matrices,
 * $$ \sum_{j=1}^n \;\delta_{ij}\;\delta_{jk} = \delta_{ik}, \quad i,k=1,\ldots, n. $$



Delta-convergent sequences
There exist families of regular functions F&alpha;(x) of which the family members differ by the value of a single parameter &alpha;. An example of such a family is formed by the family of Gaussian functions F&alpha;(x) = exp(&minus;&alpha;x&sup2;), where the different values of the single parameter &alpha; distinguish the different members. When all members are linearly normalizable, i.e., the following integral is finite irrespective of &alpha;,
 * $$ -\infty < \; \int_{-\infty}^{\infty} F_\alpha(x) \mathrm{d}x \; < \infty $$

and all members peak around x = 0, then the family may form a delta-convergent sequence.

Block functions
The simplest example of a delta-convergent sequence is formed by the family of block functions, characterized by positive &Delta;,
 * $$ B_\Delta(x)\equiv \begin{cases} 0 &\quad\hbox{if}\quad x< -\Delta/2 \\ \frac{1}{\Delta} &\quad\hbox{if}\quad -\Delta/2 \le x\le \Delta/2 \\ 0 &\quad\hbox{if}\quad x> \Delta/2 \\ \end{cases} $$

In Fig. 1 the block function B&Delta; is shown in red. Evidently, the area (width times height) under the red curve is equal to unity, irrespective of the value of &Delta;,
 * $$ \int_{-\infty}^{\infty} B_\Delta(x) \mathrm{d}x  = 1 < \infty. $$

Let the arbitrary function f(x) (blue in Fig. 1) be continuous (no jumps) and finite in the neighborhood of x=0. When &Delta; becomes very small, and the block function very narrow (and necessarily very high because width times height is constant) the product f(x) B&Delta;(x) becomes in good approximation equal to f(0) B&Delta;(x). The narrower the block the better the approximation. Hence for &Delta; going to zero,
 * $$ \lim_{\Delta\rightarrow0} \int_{-\infty}^{\infty} f(x)\;B_\Delta(x) \mathrm{d}x = f(0)\lim_{\Delta\rightarrow0} \int_{-\infty}^{\infty} B_\Delta(x) \mathrm{d}x = f(0), $$

which may be compared with the definition of the delta function,
 * $$ \int_{-\infty}^{\infty} f(x)\;\delta(x) \mathrm{d}x = f(0) . $$

This shows that the family of block functions converges to the Dirac delta function for decreasing parameter &Delta;; the family forms a delta-convergent sequence:
 * $$ \lim_{\Delta \rightarrow 0}B_\Delta(x) \rightarrow \delta(x). $$

Note: We integrated over the whole real axis. Obviously this is not necessary, we could have excluded the zero-valued wings of the block function and integrated only over the hump in the middle, from &minus;&Delta;/2 to +&Delta;/2. In mathematical texts, as e.g. Ref. , this refinement in the integration limits is included in the definition of the delta-convergent sequence. That is, it is required that the integrals over the two wings vanish in the limit. Because the delta-convergent sequences encountered in physical applications usually satisfy this condition, we omit the more exact mathematical definition.

Gaussian functions
Consider the family,
 * $$ F_\alpha(x) = \frac{1}{2\sqrt{\pi\alpha}} e^{-x^2/(4\alpha)}\quad\hbox{with}\quad \int_{-\infty}^\infty F_\alpha(x)\mathrm{d} x = 1. $$

As is shown in Fig. 2 the functions peak around x = 0 and become narrower for decreasing &alpha;. Hence the family of Gaussian functions forms a delta-convergent sequence,
 * $$ \lim_{\alpha \rightarrow 0} F_\alpha (x) \rightarrow \delta(x). $$



Lorentz-Cauchy functions
The family of functions shown in Fig. 3
 * $$ F_\epsilon(x) = \frac{1}{\pi}\frac{\epsilon}{x^2+\epsilon^2}\quad\hbox{with}\quad \int_{-\infty}^{\infty} F_\epsilon(x)\mathrm{d}x = 1 $$

forms a delta-convergent sequence,
 * $$ \lim_{\epsilon \rightarrow 0} F_\epsilon(x) \rightarrow \delta(x). $$



Sinc functions
The family of functions (often called sinc functions) shown in Fig. 4 is
 * $$ F_\nu(x) = \frac{1}{\pi} \frac{\sin\nu x}{x}\quad\hbox{with}\quad \int_{-\infty}^{\infty} F_\nu(x)\mathrm{d}x = 1. $$

This family converges to the delta function for increasing &nu;
 * $$ \lim_{\nu \rightarrow \infty} F_\nu(x) \rightarrow \delta(x). $$

This limit leads readily to the Fourier integral representation of the delta function:
 * $$ \int^\nu_{-\nu} e^{ikx} \mathrm{d}k = \frac{1}{ix} \left[ e^{ikx} \right]_{-\nu}^{\nu} = \frac{2\sin \nu x}{x}, $$

so that
 * $$ \int^\infty_{-\infty} e^{ikx} \mathrm{d}k = \lim_{\nu \rightarrow \infty } \frac{2\sin \nu x}{x} = 2\pi\delta(x). $$

The Dirac delta function is the Fourier transform of the unit function f(x) = 1.

Derivatives of the delta function
Consider a differentiable function f(x) that vanishes at plus and minus infinity. Integrate by parts
 * $$ \begin{align} \int_{-\infty}^{\infty} \frac{d f(x)}{dx} \delta(x) \mathrm{d}x &= f(x)\delta(x)\Big|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} f(x) \frac{d \delta(x)}{dx} \mathrm{d}x \\  &= - \int_{-\infty}^{\infty} f(x) \frac{d \delta(x)}{dx} \mathrm{d}x. \end{align} $$

In the same way as one proves the turnover rule  and Hermiticity of the quantum mechanical momentum operator $$p_x = -i\hbar d/dx$$, we showed here that d/dx is anti-Hermitian,
 * $$ \left(\frac{d}{dx}\right)^\dagger = - \frac{d}{dx}. $$

Indeed, when we write the integral as an inner product, it follows from partial integration and the vanishing of f(x) on the integration limits that
 * $$ \langle\; \frac{df}{dx}\; |\; \delta \;\rangle \equiv \langle\; f\;|\; \left(\frac{d}{dx}\right)^\dagger \delta \; \rangle = - \langle\; f\;|\; \frac{d\delta}{dx} \; \rangle . $$

This turnover rule is used as the definition of the derivative of the delta function,
 * $$ \langle\; f\;|\; \frac{d\delta}{dx} \; \rangle \equiv - \langle \frac{df}{dx}\; |\; \delta \;\rangle \quad\Longleftrightarrow\quad \int_{-\infty}^{\infty} f(x) \frac{d \delta(x)}{dx} \mathrm{d}x \equiv - \int_{-\infty}^{\infty} \frac{d f(x)}{dx} \delta(x) \mathrm{d}x = - f'(0), $$

where the prime indicates the first derivative of f(x). According to the definition of the delta function the first derivative is evaluated in x = 0. Using m times the turnover rule, it follows that the mth derivative of the delta function is defined by
 * $$ \int_{-\infty}^{\infty} f(x)\; \delta^{(m)}(x) \mathrm{d}x \equiv \int_{-\infty}^{\infty} f(x) \frac{d^m \delta(x)}{dx^m} \mathrm{d}x = (-1)^m \frac{d^m f(0)}{dx^m} \equiv (-1)^m f^{(m)}(0). $$

Properties of the derivative

 * $$ \begin{align} \delta^{(m)}(x) &= (-1)^m \delta^{(m)}(-x)\\ \int_{-\infty}^\infty \delta^{(m)}(x-y)\delta^{(n)}(y-a)\mathrm{d}y &= \delta^{(m+n)}(x-a) \\ x^{m+1} \delta^{(m)}(x)&= 0\\ \end{align} $$

These results can be proved by making the substitution x &rarr; &minus;x and use of the turnover rule for d/dx (see above).

Primitive
As shown here the primitive (also known as antiderivative, or undetermined integral) of the Dirac delta is the Heaviside step function H(x),
 * $$ H'(x) \equiv \frac{d H(x)}{dx} = \delta(x). $$

Clearly,
 * $$ \int_{-\infty}^{\infty} H(x) f'(x) \mathrm{d}x = -f(0). $$

The Dirac delta function in three dimensions
Write the delta function in three-dimensional space as &delta;(r), with r = (x,y, z), and define,
 * $$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\mathbf{r})\delta(\mathbf{r})\, \mathrm{d}x \mathrm{d}y\mathrm{d}z = f(\mathbf{0}). $$

From which follows more generally,
 * $$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\mathbf{r} )\delta(\mathbf{r}-\mathbf{r}_0)\, \mathrm{d}x \mathrm{d}y\mathrm{d}z = f(\mathbf{r}_0). $$

The three-dimensional delta function can be factorized
 * $$ \delta(\mathbf{r}-\mathbf{r}_0) = \delta(x-x_0)\; \delta(y-y_0)\;\delta(z-z_0). $$

In spherical polar coordinates
 * $$ \begin{align} \delta(\mathbf{r}-\mathbf{r}_0) &= \frac{1}{r_0^2\sin\theta_0} \delta(r-r_0)\; \delta(\theta-\theta_0)\;\delta(\phi-\phi_0) \\ &= \frac{1}{r_0^2} \delta(r-r_0)\; \delta(\cos\theta-\cos\theta_0)\;\delta(\phi-\phi_0) \\ \end{align} \qquad\qquad\qquad\qquad (1) $$

Proof of equation (1)

Write
 * $$ \mathbf{r} = \mathbf{r}(r,\theta, \phi) = \begin{pmatrix} r\sin\theta\cos\phi \\ r\sin\theta\sin\phi \\ r\cos\theta \\ \end{pmatrix} $$

The Jacobian (Jacobi determinant) of this transformation from Cartesian coordinates to spherical polar coordinates is
 * $$ J(r,\theta,\phi) = r^2\sin\theta \, $$

Consider
 * $$ \begin{alignat}{2} \iiint\; \delta(r-r_0)&\delta(\theta-\theta_0)\delta(\phi-\phi_0)\;f(\mathbf{r})\; \mathrm{d}x \mathrm{d}y\mathrm{d}z \\ &= \iiint\; \delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi-\phi_0)\;f\big(\mathbf{r}(r,\theta,\phi)\big)\; J(r,\theta,\phi)\;\mathrm{d}r \mathrm{d}\theta\mathrm{d}\phi \\ &= f\big(\mathbf{r}(r_0,\theta_0,\phi_0)\big)\; J(r_0,\theta_0,\phi_0) = J(r_0,\theta_0,\phi_0) f(\mathbf{r}_0)\\ &= J(r_0,\theta_0,\phi_0) \iiint\; \;\delta(\mathbf{r}-\mathbf{r}_0)\;f(\mathbf{r})\; \mathrm{d}x \mathrm{d}y\mathrm{d}z \end{alignat} $$

so that
 * $$ \delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi-\phi_0) = J(r_0,\theta_0,\phi_0)\; \delta(\mathbf{r}-\mathbf{r}_0)\, $$

and
 * $$ \delta(\mathbf{r}-\mathbf{r}_0) = J(r_0,\theta_0,\phi_0)^{-1} \delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi-\phi_0)    \, $$

The last line in equation (1) follows from the chain rule.

The following useful and frequently applied property is proved here,
 * $$ \nabla^2\frac{1}{r} = - 4\pi \delta(\mathbf{r}), $$

where &nabla;2 is the Laplace operator in three-dimensional Cartesian coordinates and r is the length of r.