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The Dirac Delta Function
Overview and Motivation : The Dirac delta function is a concept that is useful
throughout physics. For example, the charge density associated with a point charge
can be represented using the delta function. As we will see when we discuss Fourier
transforms (next lecture), the delta function naturally arises in that setting.
Key Mathematics : The Dirac delta function!
I. Introduction
The basic equation associated with the Dirac delta function δ (^ x )is
( ) x f ( ) x dx = f ( ) 0 ∫
∞
−∞
where f ( ) x is any function that is continuous at x = 0. Equation (1) should seem
strange: we have an integral that only depends upon the value of the function f ( x )at
x = 0. Because an integral is "the area under the curve," we expect its value to not
depend only upon one particular value of x. Indeed, there is no function δ ( x )that
satisfies Eq. (1). However, there is another kind of mathematical object, known as a
generalized function (or distribution ), that can be defined that satisfies Eq. (1).
A generalized function can be defined as the limit of a sequence of functions. Let's
see how this works in the case of δ ( x ). Let's start with the normalized Gaussian
functions
( ) nx^2 n e
n g x − =
Here n = 1 σ^2 , where σ is the standard Gaussian width parameter. These functions
are normalized in the sense that their integrals equal 1,
( ) = 1 ∫
∞
−∞
g n x dx (3)
for any value of n (>0). Let's now consider the sequence of functions for
n = 1 , 2 , 3 , K,
( )
1
x g x e − =
, ( ) 22 2
(^2) x g x e − =
, … , ( ) 1002 100
(^100) x g x e − =
What does this sequence of functions look like? We can summarize this sequence as
follows. As n increases
(a) g (^) n ( ) 0 becomes larger;
(b) g (^) n ( x ≠ 0 )eventually becomes smaller;
(c) the width of the center peak becomes smaller;
(d) but ( ) = 1 ∫
∞
−∞
g n x dx remains constant.
The following figure plots some of the functions in this sequence.
2 1 0 1 2
0
5
10
15
20 n = 1 n = 4 n = 16 n = 64 n = 256
n = 1 n = 4 n = 16 n = 64 n = 256
x
gn(x)
Let's now ask ourselves, what does the n =∞ limit of this sequence look like? Based
on (a) through (d) above we would (perhaps simplistically) say
(a) g (^) ∞ ( ) 0 =∞;
(b) g (^) n ( x ≠ 0 ) = 0 ;
(c) the width of the center peak equals zero;
lim ( ) ( ) 0 lim ( ) 0 lim 1 ( ) 0
2 2 e f x dx f e dx f f n
n nx n
n nx n
→∞
∞
−∞
− →∞
∞
−∞
−
→∞ ∫ π^ ∫ π
Thus, the integral on the lhs of Eq. (1) is really shorthand for the integral on the lhs of
Eq. (6), That is, the Dirac delta function is defined via the equation
( ) ( ) ( )
∞
−∞
− →∞
∞
−∞
x f x dx = nenx f x dx n
2
δ lim π (7)
Now often (as physicists) we often get lazy and write
( )
2 lim n nx n
x e − →∞
but this is simply shorthand for Eq. (7). Eq. (8) really has no meaning unless the
function
n nx^2 e
−
π appears^ inside^ an integral and the limit^ n^ lim→∞^ appears^ outside^ the same
integral. However, after you get used to working with the delta function, you will
rarely need to even think about the limit that is used to define it.
One other thing to note. This particular sequence of functions ( ) nx^2 n g (^) n x e −
= π that we
have used here is not unique. There are infinitely many sequences that can be used to
define the delta function. For example, we could also have defined δ ( ) x via
( )
( )
x
nx x n
1 sin lim
→∞
The sequence of functions sin( nx ) ( π x )is illustrated in the figure at the top of the
next page.
Notice that the key features of both of these two difference sequences are expressed
by (a) – (d) at the top of page 5.
II. Delta Function Properties
There are a number of properties of the delta function that are worth committing to
memory. They include the following,
( x − x ′) f ( ) x dx = f ( x ′)
∞
−∞
′ ( ) x f ( ) x dx =− f ′( ) 0 ∫
∞
−∞
δ ( xa ) = a δ( ) x (12)
The proof of Eq. (10) is relatively straightforward. Let's change the integration
variable to y = x − x ′, dy = dx , which gives
( ) ( ) ( ) ( ) ∫ ∫
∞
−∞
∞
−∞
δ x − x ′ f xdx = δ y f y + x ′ dy. (13)
Then using Eq. (1), we see that Eq. (10) is simply equal to f ( x ′). QED.
Let's also prove Eq. (12). We do this in two steps, for a > 0 and then for a < 0.
( i ) First, we assume that a > 0. Then
4 2 0 2 4
0
5
10
15
20
25
n = 1 n = 4 n = 8 n = 16 n = 32
n = 1 n = 4 n = 8 n = 16 n = 32
x
sin(nx)/(pi*x)
III. Fourier Series and the Delta Function
Recall the complex Fourier series representation of a function f ( ) x defined on
− L ≤ x ≤ L ,
( )
∞
=−∞
n
inxL f x cn e π
, (19a)
( )
−
−
L
L
inxL n f x e dx L
c π 2
. (19b)
Let's now substitute cn from Eq. (19b) into Eq. (19a). Before we do this we must
change the variable x in either Eq. (19a) or (19b) to something else because the
variable x in Eq. (19b) is just a (dummy) integration variable. Changing x to x ′^ in Eq.
(19a) and doing the substitution we end up with
∞
=−∞
′
−
−
n
inxL
L
L
f x ein^ xLdx e L
f x π^ π 2
Let's now switch the integration and summation (assuming that this is OK to do).
This produces
( ) (^ )^ ( )
−
∞
=−∞
′−
L
L n
ein^ x x L f x dx L
f x^ π 2
If we now compare Eq. (21) to Eq. (10), we see that we can identify another
representation of the delta function
( )
( )
∞
=−∞
′− − ′ = n
in x x L e L
x x
π
or setting x ′ = 0 we have
( )
∞
=−∞
−
n
inxL e L
x π
So how is this equation related to the delta function being defined as the limit of a
sequence of functions? Well, we can re-express Eq. (23) as a sequence of functions via
( ) (^) ∑ =−
− →∞
m
n m
inxL m
e L
x
π
lim. (24)
The following figure plots
∑ =−
−
m
n m
inxL e L
π 2
for several values of m (for L = 2 ). Notice
that these functions are quite similar to the function sin^ (^ mx ) (^ π^ x )plotted above.
1
2 1 0 1 2
0
10
20
30
40
m = 2 m = 4 m = 8 m = 16 m = 32
m = 2 m = 4 m = 8 m = 16 m = 32
x
Sum[exp(inpi*x/l)]
However, there is one important difference between these two sequences of functions.
Because the functions
inxL e
− π
that appear in the sum in Eqs. (23) and (24) all repeat
with on an interval of length 2 L , Eq. (24) is actually a series of delta functions,
centered at x = 0 , ± 2 L ,± 4 L ,± 6 L ,K. This is illustrated in the next figure, where we
have expanded the x axis beyond the limits of − L to L. Thus, the equality expressed
by Eq. (23) or (24) is only valid on the interval − L ≤ x ≤ L.
1 We have changed the n to m in the sin ( nx ) ( π x )functions because we are now using m to label the
functions in the sequence.
