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—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:

$\sum _{i=m}^{n}\;f_{i}\;\delta _{ia}={\begin{cases}f_{a}&\quad {\hbox{if}}\quad a\in [m,n]\subset \mathbb {Z} \\0&\quad {\hbox{if}}\quad a\notin [m,n].\end{cases}}$

In analogy, the Dirac delta function δ(x−a) is defined by (replace i by x and the summation over i by an integration over x),

$\int _{a_{0}}^{a_{1}}f(x)\delta (x-a)\mathrm {d} x={\begin{cases}f(a)&\quad {\hbox{if}}\quad a\in [a_{0},a_{1}]\subset \mathbb {R} ,\\0&\quad {\hbox{if}}\quad a\notin [a_{0},a_{1}].\end{cases}}$

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.

${\begin{aligned}\int _{-\infty }^{\infty }\delta (x)\mathrm {d} x&=1,\\{\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{ikx}\mathrm {d} k&=\delta (x)\\\delta (x-a)&=\delta (a-x),\\(x-a)\delta (x-a)&=0,\\\delta (ax)&=|a|^{-1}\delta (x)\quad (a\neq 0),\\f(x)\delta (x-a)&=f(a)\delta (x-a),\\\int _{-\infty }^{\infty }\delta (x-y)\delta (y-a)\mathrm {d} y&=\delta (x-a)\end{aligned}}$

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,

$\sum _{j=1}^{n}\;\delta _{ij}\;\delta _{jk}=\delta _{ik},\quad i,k=1,\ldots ,n.$

Fig. 1. Block ("boxcar") function (red) times regular function f(x) (blue).

Delta-convergent sequences

There exist families of regular functions F_α(x) of which the family members differ by the value of a single parameter α. An example of such a family is formed by the family of Gaussian functions F_α(x) = exp(−αx²), where the different values of the single parameter α distinguish the different members. When all members are linearly normalizable, i.e., the following integral is finite irrespective of α,

$-\infty <\;\int _{-\infty }^{\infty }F_{\alpha }(x)\mathrm {d} x\;<\infty$

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 Δ,

$B_{\Delta }(x)\equiv {\begin{cases}0&\quad {\hbox{if}}\quad x<-\Delta /2\\{\frac {1}{\Delta }}&\quad {\hbox{if}}\quad -\Delta /2\leq x\leq \Delta /2\\0&\quad {\hbox{if}}\quad x>\Delta /2\\\end{cases}}$

In Fig. 1 the block function B_Δ is shown in red. Evidently, the area (width times height) under the red curve is equal to unity, irrespective of the value of Δ,

$\int _{-\infty }^{\infty }B_{\Delta }(x)\mathrm {d} x=1<\infty .$

Let the arbitrary function f(x) (blue in Fig. 1) be continuous (no jumps) and finite in the neighborhood of x=0. When Δ becomes very small, and the block function very narrow (and necessarily very high because width times height are constant) the product f(x) B_Δ(x) becomes in good approximation equal to f(0) B_Δ(x). The narrower the block the better the approximation. Hence for Δ going to zero,

$\lim _{\Delta \rightarrow 0}\int _{-\infty }^{\infty }f(x)\;B_{\Delta }(x)\mathrm {d} x=f(0)\lim _{\Delta \rightarrow 0}\int _{-\infty }^{\infty }B_{\Delta }(x)\mathrm {d} x=f(0),$

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 Δ; the family forms a delta-convergent sequence:

$\lim _{\Delta \rightarrow 0}B_{\Delta }(x)\rightarrow \delta (x).$

Fig. 2. Gaussian functions.

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 −Δ/2 to +Δ/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 α. Hence the family of Gaussian functions forms a delta-convergent sequence,

$\lim _{\alpha \rightarrow 0}F_{\alpha }(x)\rightarrow \delta (x).$

Fig. 3. Lorentz-Cauchy functions

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,

$\lim _{\epsilon \rightarrow 0}F_{\epsilon }(x)\rightarrow \delta (x).$

Fig. 4. Sinc functions.

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 ν

$\lim _{\nu \rightarrow \infty }F_{\nu }(x)\rightarrow \delta (x).$

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

${\begin{aligned}\int _{-\infty }^{\infty }{\frac {df(x)}{dx}}\delta (x)\mathrm {d} x&=f(x)\delta (x){\Big |}_{-\infty }^{\infty }-\int _{-\infty }^{\infty }f(x){\frac {d\delta (x)}{dx}}\mathrm {d} x\\&=-\int _{-\infty }^{\infty }f(x){\frac {d\delta (x)}{dx}}\mathrm {d} x.\end{aligned}}$

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,

$\langle \;f\;|\;{\frac {d\delta }{dx}}\;\rangle \equiv -\langle {\frac {df}{dx}}\;|\;\delta \;\rangle \quad \Longleftrightarrow \quad \int _{-\infty }^{\infty }f(x){\frac {d\delta (x)}{dx}}\mathrm {d} x\equiv -\int _{-\infty }^{\infty }{\frac {df(x)}{dx}}\delta (x)\mathrm {d} x=-f'(0),$

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

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

These results can be proved by making the substitution x → −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),

$H'(x)\equiv {\frac {dH(x)}{dx}}=\delta (x).$

Clearly,

$\int _{-\infty }^{\infty }H(x)f'(x)\mathrm {d} x=-f(0).$

References

• A. Messiah, Quantum Mechanics, vol. I. North Holland, Amsterdam (1967), Appendix A