The Dirac Delta Function

Having already used the concept of a Dirac's ``delta function'' several times, we shall now introduce it in a way which lets us see how this concept fits into the familiar realm of functions and integrals.

Definition: Let $\delta_\alpha (x)$ be a set of $\alpha$ -parameterized functions with the following properties: for any ``well-behaved'' (in a sense which depends upon the context) function

1. $\lim\limits_{\alpha\to 0}\int^\infty_{-\infty} f(x)\delta_\alpha (x)\,dx = f(0)$
2. 1. $\lim\limits_{\alpha\to 0}\int^{-\varepsilon}_{-\infty} f(x) \delta_\alpha (x)\,dx = 0$
   2. $\lim\limits_{\alpha\to 0}\int^\infty_\varepsilon f(x) \delta_\alpha (x)\,dx = 0$
   for all $\varepsilon >0$ .

Then Dirac's ``delta function'' is defined by

$$\delta (x)\colon\int^\infty_{-\infty}f(x)\delta (x)\,dx\equiv \lim_{\alpha \to 0}\int^\infty_{-\infty} f(x)\delta_\alpha (x)\,dx~~(=f(0))$$ (222)

or

$$\textrm{\lq\lq }\delta (x)=\lim_{\alpha\to 0}\delta_\alpha (x)\textrm{''}~~.$$

Remark 1. We have put the last equation in quotes because, strictly speaking, $\delta (x)$ is only defined when integrated against a ``well-behaved'' function. In other words, the frequently quoted definition ``this function equals zero everywhere, except at $x=0$ where it is infinite so that its integral is one'' is in conflict with the classical definition of a function and integral. Indeed $\delta (x) =0$ $\forall\,x\not= 0$ and $\lim\limits_{\varepsilon\to 0} \int^\varepsilon_{-\varepsilon} \delta (x)\,dx\not= 0$ violates the classical definiton of a function.

Remark 2. Equations (2) in the definition do not imply that

$$\lim_{\alpha\to 0}\delta_\alpha (x'-x)=0~~\qquad~~\textrm{for}~~x' \not= x\,.$$

Example 4 (below) is an instance.

It is easy to come up with examples of parametrized functions that give rise to $\delta (x)$ .

1. The Impulse Function
   $$\delta_\alpha (x) \equiv\left\{ \begin{array}{lcl} \display... ...ert x\vert > \displaystyle \frac{\alpha}{2} \end{array}\right.$$

2. The Gaussian function
   $$\delta_\alpha (x) \equiv\frac{1}{\alpha\sqrt{\pi}} ~ e^{-\displaystyle {x^2}/{\alpha^2}}$$

3. The ``Lorentz line'' function
   $$\delta_\alpha (x)= \frac{1}{\pi}~{\frac{\alpha}{x^2+\alpha^2}}$$

4. The $\hbox{sinc}$ function

   $$\delta_\alpha (x) = \frac{\sin\frac{\pi}{\alpha} x}{\pi x} \equiv \frac{1}{\alpha} \hbox{sinc} \frac{\pi}{\alpha} x$$ (223)

   
   
   All of them satisfy
   1. $\int^\infty_{-\infty}\delta_\alpha (x)\,dx = 1$ $\forall\,\alpha >0$ .
   2. $\lim\limits_{\alpha\to 0}\int^\infty_{-\infty} f(x)\delta_\alpha (x)\,dx = f(0)$ .
   3. Examples 1-3 are characterized by $\lim\limits_{\alpha\to 0}\delta_\alpha (x)=0$ whenever $x\not= 0.$
      However, for example 4 one has $\lim\limits_{\alpha\to 0}\delta_\alpha (x)\not= 0$ whenever $x\not= 0$ .

Exercise 22.1 (DIRAC DELTA AS A LIMITING WAVE PACKET)
Show that

$$\lim\limits_{w\to\infty} \frac{\sin 2\pi wx}{\pi x}\,,\quad w > 0$$

is a representation of the Dirac $\delta\!$ -function.
Discussion:

Let

$$\delta_w(x)={\sin 2\pi wx\over {\pi x}}$$

and let $f(x)$ be a function which is piecewise continuous on $[-a,a]$ , in particular,

$$\lim\limits_{x\to 0^+}f(x)=f(0^+)$$

$$\lim\limits_{x\to 0^-}f(x)=f(0^-)$$

To show that

$$\lim\limits_{w\to\infty}\delta_w(x)=\delta(x)\quad (\hbox{Dirac~delta function)}$$

one must show that

$$\lim\limits_{w\to\infty}\int_{-a}^a \delta_w(x)f(x)\, dx={1\over 2}f(0^+) +{1\over 2}f(0^-)$$

One way of doing this is to follow the approach used to obtain an analogous result in the process of establishing the validity of the Fourier series theorem, and then use the result that

$$\int_0^\infty {\sin y\over y}\, dy={\pi \over 2}$$

Exercise 22.2 (DERIVATIVE OF THE DIRAC DELTA)
Consider the integral

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

Assuming that $f(x)$ is nearly linear so that

$$f(-a)=f(0)-af'(0)+\textrm{(neglegible terms)},$$ (224)

show that $I$ can be evaluated by means of the formal equation

$$\boxed{\delta (x+a)=\delta (x)+a\,\delta' (x)}~,$$ (225)

where $\delta (x)$ and $\delta (x+a)$ are defined by Eq.(2.22) and $\delta' (x)$ is defined by

$$\delta' (x):\quad \int_{-\infty}^\infty \delta' (x) f(x)\, dx \equiv \lim_{\alpha \to 0} \int_{-\infty}^\infty \delta'_\alpha (x) f(x)\, dx ~.$$ (226)

Comment: For obvious reasons it is invalid to claim

$$\delta_\alpha (x+a)=\delta_\alpha (x)+a\,\delta'_\alpha (x)$$

without referring to test functions that can be approximated by Eq.(2.24).