Dirac Delta Function
© Victor Anisimov, 2011
Dirac Delta Function is a peculiar mathematical function introduced by Paul A. M. Dirac. It can be pictured as an infinitely high spike with infinitesimally narrow width having unit area
Figure 1.
The one dimensional delta function δ(x) has the value of infinity for coordinate x = 0 and is zero everywhere else where x ≠ 0
(1)
The definition of delta function having unit area corresponds to its integral being equal to unit
(2)
A more general form of delta function δ(x – a) has the origin of the function shifted to point x = a. Useful in numerical integration schemes is discrete representation of delta function
(3)
where Δ is width of the spike, and δax is Kronecker delta, which is a simple mathematical function of two variables.
(4)
In Eq. 3, for any coordinate x except the one when x = a, Kronecker delta will be zero thus delta function in Eq. 3 will be also equal to zero. For the case when x ≠ a, Kronecker delta will be unit; unit divided by Δ approaching zero will produce infinity. Thus discrete form of delta function (Eq. 3) is fully in agreement with the general definition of delta function. It is possible to obtain the integral in Eq. 2 by using the discrete form of delta function
(5)
For a = const, there is only one x on the entire coordinate space such as x = a, for which δaa = 1. Thus the integral is equal to unit.
Another useful property of delta function is symmetry about the sign of its coordinate. The symmetry of delta function can be guessed from Fig. 1, which means
(6)
More to the discussion of symmetry property will be given in Eq. 9. Unlike regular mathematical functions which are finite, the property of delta function to have infinite value disqualifies it from being an actual function. Accordingly, delta function has little practical value on its own. Its real value comes from the way how it changes the value of other functions it is multiplied to. Therefore delta function should always be expected to be multiplied on a function and their product integrated. Since delta function is non-zero only at the point x = a, from this it follows that its product with a function selects a single point from the entire spectrum of values of the function.
(7)
Eq. 7 shows the way how to take integral of a product of delta function with the regular function
(8)
The integral over delta function does not always have to have the integration limits set from minus infinity to plus infinity. For an integral to be non-zero it is sufficient that its integration limits include the point x = a, whatever the actual limits are.
In the following we will derive some useful relationships involving delta function based on the information we learned so far about delta function. To start from, we will prove that delta function is indeed symmetric instead of taking Eq. 6 on the word. Since delta function will always be present as a product to some function under an integral we need to make sure that any manipulations with delta function we are going to do will be applied consistently to all parts of the equation. Consider integral
(9)
In Eq. 9, first we replace x by –x. Such substitution requires inverting the integration limits. In the second step we replace d(–x) by dx and use the released minus sign to invert once again the integration limits. In the third and last step we substitute x = –a representing the origin of delta function δ(x + a) to f(–x) thus obtaining f(a). Same result was obtained in Eq. 8, thus Eq. 6 is proved.
So far we have dealt with delta function normalized to unit as described by Eq. 2. However, there are delta functions whose integral is not necessarily unit. Example is δ(kx). In the following we will find out how it relates to the simple delta function δ(x). For k > 0
(10)
In the first equation we applied variable substitution y = kx – a, x = (y + a)/k, and dx = dy/k. In the second equation we used z = –kx – a, x = (z + a)/(–k), and dx = –dz/k; we also had to interchange integration limits first after variable substitution and then once again absorbing the minus sign. From Eq. 10 and Eq. 8 it follows that
(11)
Since according to Eq. 10 the result of integration is independent of sign of k, in Eq. 11 in R.H.S. we put k in bars to include the case of k < 0. Eq. 11 leads to an important observation that the outcome of integration of delta function depends on the rate of change of its argument in respect to variable x. The rate of change is a derivative. The derivative of kx in respect to x is k. Therefore k goes to the denominator in R.H.S. of Eq. 11.
To generalize the relationship given by Eq. 11 we need to consider argument of delta function being a function, δ[f(x)], and its integral in the form
(12)
where g(x) is arbitrary function. To solve this integral we use variable substitution
(13)
where f´(x) is derivative of function f(x), and f–1(x) indicates inverse function. The concept of inverse function can be illustrated on the following example. If f(x) = exp(x) then f–1(x) = ln(x), so x ≡ ln(exp(x)). Finally, after applying variable substitution y = f(x) the integration limits of variable x change to integration limits of variable y, i.e. a and b. The integral in Eq. 12 takes the form
(14)
In the last step in Eq. 14, the condition y = 0 is satisfied with x = x0 for which f(x) = 0. Same result as in Eq. 14 can be obtained from
(15)
From comparison of Eq. 14 with Eq. 15 we conclude that
(16)
which is a general result we were seeking. In Eq. 16, x0 is a root of function f(x). If the function has multiple roots xi, Eq. 16 generalizes to
(17)
This completes the brief introduction of delta function. The purpose of this page is to serve as a starting point to those who seek understanding of Delta function. Therefore the page may lack rigour in favor of simplicity. Please let me know your suggestions and comments about this material by leaving your feedback at the "contact us" link below.
