Hassani S. Mathematical Methods: For Students of Physics and Related Fields

Подождите немного. Документ загружается.

5.1 One-Variable Case 143

One can think of the last equality as an identity satisﬁed by the Dirac delta

function:

Box 5.1.2. The Dirac delta function is zero everywhere except at the point

which makes its argument zero, in which case the Dirac delta function is

inﬁnite.

Since the Dirac delta function is zero almost everywhere, we can shrink

the region of integration to a smaller interval. In fact,

δ(x −x

) dx =1

as long as x

lies in the interval (a, b). If x

is outside the interval, then the

integral will be zero because the delta function would always be zero in the

region of integration. We summarize these results:

Box 5.1.3. The Dirac delta function satisﬁes the following relation

δ(x −x

) dx =

1 if a<x

<b,

0 otherwise.

(5.7)

Equation (5.4) is a special case of this, because −∞ <x

< +∞ for any

value of x

5.1.1 Linear Densities of Points

Any function λ(x) whose integral over all real numbers is one is called a linear

density function.Theδ

’s deﬁned above are such functions. If we multiply linear density

function

a linear density function by a physical quantity Q, the result will be a linear

density for Q. In fact, this was how we arrived at δ

.Thus,Q

λ(x)isa

Q-linear density with total magnitude Q

. Similarly, if M represents a mass,

then Mλ(x) is a linear mass density with total mass M .Conversely,iff(x)

describes the linear density of a physical quantity with total magnitude Q,

then λ(x) ≡ f(x)/Q is a linear density function.

Because of Equation (5.4) the Dirac delta function is a linear density func-

tion. What kind of a distribution does it describe? To be speciﬁc, consider

δ function and

densities of point

charges and point

masses

mδ(x, x

)withm designating mass. This function is zero everywhere except

at x

. Thus, if it is to be a mass distribution, it has to be a point mass located

at x

. Keep in mind that mδ(x, x

) is a linear mass density,sothatitsintegral

is the total mass m. The linear “density” of a point mass is inﬁnite because

144 Dirac Delta Function

its length is zero, and this is precisely what mδ(x, x

) describes. In fact, the

linear density of a point physical quantity of magnitude Q located at x

can

be written as Qδ(x, x

)=Qδ(x −x

), or generalizing,

Box 5.1.4. The linear density λ(x) of N point physical quantities

,...,Q

located at x

,...,x

, respectively, can be written as

λ(x)=



k=1

δ(x −x

). (5.8)

We see that with the help of the Dirac delta function we can express discrete

charge distributions (collection of point charges) in terms of functions. This

is the most useful property of the Dirac delta function.

Example 5.1.1.

Three charges −q,2q,and−q are located along the x-axis at

−a, the origin, and +a, respectively. How do we write the linear charge density for

such a charge distribution? We use Equation (5.8) with Q replaced by q:

λ(x)=



k=1

δ(x − x

)=−qδ(x − (−a)) + 2qδ(x − 0) − qδ(x − a)

= −qδ(x + a)+2qδ(x) − qδ(x − a).

Note that the Dirac delta functions ensure that no electric charge is present any

where except at x = a, x = −a,andx =0.



Example 5.1.2. A more interesting example of a linear charge distribution using

the Dirac delta function is that of an inﬁnite array of point charges equally spaced

on a straight line having equal magnitudes and alternating in sign. This is a one-

dimensional model of ionic crystals.

Let us assume that the magnitude of each charge is ±q, the spacing between it

and the neighboring charge is a, and that the charges start at −∞ and extend to

+∞ with one positive charge at the origin as shown in Figure 5.4. Then it is easy

to write the density of this distribution. It isdensity of

one-dimensional

ionic crystal

λ(x)=

+∞



k=−∞

(−1)

qδ(x − ka).

Figure 5.4: A one-dimensional ionic crystal. The black circles represent positive charges

and the white circles the negative charges.

5.1 One-Variable Case 145

Note that for odd k thechargeisnegativeandforevenk it is positive. This is

because we placed a positive charge at the origin. Had we chosen the origin to

be the site of a negative charge, the above arrangement would have shifted by one

spacing.



5.1.2 Properties of the Delta Function

From a mathematical point of view, the most important property, which is

sometimes used to deﬁne the Dirac delta function, occurs when it multiplies a

“smooth”

function in an integrand. First look at an integral with a δ

(x−x

)

inside. If the function f (x) multiplying δ

(x − x

)issmoothandn is large

enough, the product f(x)δ

(x − x

) practically vanishes outside a narrow

interval in which δ

(x − x

) is appreciably diﬀerent from zero. For example,

if n =10

, x = x

+0.001, and we use the exponential function of Equation

(5.1), then δ

(x − x

)=0.08, so that f (x)δ

(x − x

) drops to about 8% of

the value it has at x

, assuming that f does not change appreciably in the

small interval of width 0.002 around x

. For larger values of n this drop is

even sharper. In fact, no matter what function we choose, there is always a

large enough n such that the product f (x)δ

(x − x

) will drop to as small

a value as we please in as short an interval as we please. Therefore, we can

approximate the integral over all real numbers to an integral over that small

interval. Let the interval be (x

− , x

+ ). Then, we have

+∞

−∞

f(x)δ

(x − x

) dx ≈

+

−

f(x)δ

(x − x

) dx

≈ f (x

)

+

−

(x − x

) dx

≈ f (x

)

+∞

−∞

(x −x

)=f(x

The approximation in the second line follows from the fact that f (x)isalmost

constant in the small interval (x

− , x

+ ). The third approximation is a

result of the smallness of δ

outside the interval, and the equality follows

because δ

is a linear density function. The approximation above reaches

equality once the limit of n →∞is taken in which case δ

becomes the Dirac

delta function. Thus, we have the important relation

+∞

−∞

f(x)δ(x − x

) dx = f(x

). (5.9)

In the present context, a smooth function is one that does not change abruptly when

its argument changes by a small amount.

146 Dirac Delta Function

This is equivalent to the following statement:integral of product

of δ(x − x

) and

f(x) is simply

f(x

)

Box 5.1.5. The Dirac delta function satisﬁes

f(x)δ(x − x

) dx =

f(x

) if a<x

<b,

0 otherwise.

(5.10)

In words, the result of integration is the value of f at the root of the

argument of the delta function, provided this root is inside the range

of integration.

Example 5.1.3. In this example we illustrate some of the properties of the

Dirac delta function. For instance



∞

f(t)δ(t) dt = 0 because the root of the

argument of the Dirac delta function (the point that makes the argument of the

Dirac delta function zero)—namely t = 0—is outside the range of integration. The

integral



+∞

−∞

xδ(x) dx is zero because the function x vanishes at the point x = 0 (the

root of the argument of the delta function). Also,

−∞

cos yδ(y −π) dy =0

because π—which makes the argument of the delta function vanish—lies outside the

range of integration. However,

+3.2

−∞

cos yδ(y − π) dy =cosπ = −1

because now π lies inside the range of integration.

The reader is urged to check the following results:

+∞

−∞

cos yδ(y −π) dy = −1,

+∞

−∞

sin zδ(z) dz =0,

+∞

cos yδ(y + π) dy =0,

+∞

−∞

cos

δ(y − π) dy =0,

−1

δ(t) dt =1,

+∞

−∞

xf(x)δ(x) dx =0,

2.7

−∞

ln tδ(t −e) dt =0,

2.8

−∞

ln tδ(t − e) dt =1.



As noted earlier, the Dirac delta function is not an ordinary over-the-

counter function. Nevertheless, it is possible to study it, along with many

other “weird” functions called distributions, in a mathematically rigorous

distributions

and systematic way. It turns out that, in all physical applications, distribu-

tions occur inside an integral, and once they do, Equation (5.10) tells us how

to manipulate such integrals. The result of integration is always well deﬁned

because it is simply the value of a “good” function at a point, say x

.Infact,

5.1 One-Variable Case 147

the result of integration is so nice that one can even deﬁne the derivative of

the Dirac delta function by diﬀerentiating (5.10) with respect to x

.Weleave

the details as an exercise and simply quote the result:

+∞

−∞

f(x)δ



(x − x

) dx = −f



) (5.11)

Higher order derivatives of the Dirac delta function can be obtained similarly.

derivatives of

Dirac delta

function

In fact, we have

Box 5.1.6. The nth derivative of the Dirac delta function satisﬁes

f(x)δ

(n)

(x−x

)(x−x

) dx =

(−1)

(n)

) if a<x

<b,

0 otherwise,

(5.12)

where the superscript (n) indicates the nth derivatives.

In many applications the argument of the Dirac delta function is not of

the simple form (x − x

), but may itself be a function g(x) whose deriva-

tive is assumed to be continuous in (a, b). Since by Equation (5.6) the delta

what happens

when the

argument of δ is

itself a function?

function vanishes except when its argument is zero, in such a case, one has

to concentrate on the roots of g(x), i.e., values c for which g(c)=0. For

simplicity, ﬁrst assume that there is only one root c of g in the interval (a, b)

and that g



in the interval (a, b), except at x = c,wecanshrinktheregionofintegration

to (c −, c + ), and write

δ (g(x)) dx =

c+

c−

δ (g(x)) dx.

Now make the change of variable y = g(x), dy = g



(x) dx with the appropriate

transformation of limits of integration to get

δ (g(x)) dx =

g(c+)

g(c− )

δ(y)



(x)

With g(c)=0andg



(c −, c + ), that g(c −) < 0, and that g(c + ) > 0. We can therefore write

δ (g(x)) dx =

g(c+)

g(c− )

δ(y)



(x)



(x)



y=0



(c)

> 0,

because zero is in the region of integration and y = 0 is equivalent to x = c

there.

148 Dirac Delta Function

When g



g(c − ) > 0andg(c + ) < 0. Thus, ﬂipping the limits of integration so that

the smaller number corresponds to the lower limit, we obtain

δ (g(x)) dx = −

g(c− )

g(c+)

δ(y)



(x)

= −



(x)



y=0

= −



(c)

> 0.

We summarize the two results as

δ (g(x)) dx =



(c)|

If there are two roots of g in the interval, say c

and c

with c

,we

break up (a, b):

δ (g(x)) dx =

  

−

δ (g(x)) dx +

+

−

δ (g(x)) dx

  

−

+

δ (g(x)) dx +

+

−

δ (g(x)) dx

  

+

δ (g(x)) dx =



where in the last line we used the result obtained in the previous paragraph.

It should be clear that if g has n roots c

,...,c

in (a, b), there will be a

summation of n terms in the last line of the above equation. In fact, we can

summarize the result of the foregoing discussion as

Box 5.1.7. If g(x) has the roots c

,...,c

,andg



) =0for all k

between 1 and n,then

δ(g(x)) dx =



k=1

1/|g



)| if a<c

<b,

0 otherwise.

(5.13)

When the delta function is multiplied by a smooth function f (x), a similar

argument as above—which is left to the reader—can be used to show that

f(x)δ(g(x)) dx =



k=1

f(c

)/|g



)| if a<c

<b,

0otherwise,

(5.14)

5.1 One-Variable Case 149

provided g



) = 0. These results are sometimes written as an identity among

the delta functions.

a very important

relation

Box 5.1.8. The Dirac delta function satisﬁes the following relation:

δ(g(x)) =



k=1

δ(x −c

)



) =0, (5.15)

where {c

}

k=1

are all the roots of the equation g(x)=0.

The formula analogous to Equation (5.14) involving the derivative of the Dirac

delta function is

f(x)δ



(g(x)) dx =

−



k=1



)/|g



)| if a<c

<b,

0otherwise.

(5.16)

Example 5.1.4.

As a concrete example, let us evaluate the integral

I ≡

+∞

−∞

f(t)δ(t

− a

) dt,

where f is a smooth function and a is a real constant. We can identify g(t)ast

−a

with roots c

= −a, c

= a and derivative g



(t)=2t. Therefore, Equation (5.15)

reduces to

δ(t

− a

δ(t − c

)



δ(t − c

)



δ(t − (−a))

|−2a|

δ(t − a)

|2a|

{δ(t + a)+δ(t − a)}.

Substituting in the integral, we obtain

I =

2|a|

+∞

−∞

f(t) {δ(t + a)+δ(t − a)}

2|a|

(

+∞

−∞

f(t)δ(t + a)+

+∞

−∞

f(t)δ(t − a)

)

2|a|

{f(−a)+f(a)}.

Note that the integral vanishes—as expected—if f is odd.



Example 5.1.5. We illustrate further the foregoing general discussions with some

more concrete examples. To evaluate the integral

∞

sin tδ(t

− π

/4) dt,

150 Dirac Delta Function

we note that g(t)=t

− π

/4 which has two roots c

= π/2andc

= −π/2with

only the positive root lying in the range of integration. Moreover, g



(t)=2t.Thus,

∞

sin tδ(t

− π

/4) dt =

f(c

)



sin(c

)

|2c

sin(π/2)

On the other hand,

∞

−∞

sin tδ(t

− π

/4) dt =0

because the second root c

is also included in the range of integration and its con-

tribution cancels that of c

To evaluate the integral

∞

ln zδ(z

− 4) dz

we note that g(z)=z

− 4 which has two roots c

=2andc

= −2 with only the

positive root lying in the range of integration. Thus, with g



(z)=2z,wehave

∞

ln zδ(z

− 4) dz =

f(c

)



ln(c

)

|2c

ln 2

=0.1733.

The integral

+∞

−∞

f(y) δ(y

+ a

) dy

is zero because there is no point in the range of integration at which the argument

of the Dirac delta function vanishes. In other words, g(y)=y

+ a

has no real

roots at all.

To evaluate the integral

+π/2

−π/2

(t +1)

δ(sin πt) dt

we note that g(t)=sinπt which has three roots c

= −1, c

=0,andc

=+1in

the range of integration. Thus, with g



(t)=π cos πt,wehave

+π/2

−π/2

(t +1)

δ(sin πt) dt =



k=1

f(c

)





k=1

+1)

|π cos(c

π)|

(−1+1)

|π cos(−π)|

(0 + 1)

|π cos(0)|

(1 + 1)

|π cos(π)|

Some other concrete examples are:

+∞

−∞

sin |t| δ(t

− π

/4) dt =2/π,

+∞

−∞

cos xδ(x

− π

) dx = −1/π,

∞

ln zδ(z

− 1) dz =0,

−∞

cos yδ(y

+ π

) dy =0,

+π

−π

(t +1)

δ(sin πt) dt =35/π,

+∞

−∞

f(t) δ(e

− 1) dt = f(0),

∞

ln xδ(10x

+3x −1) dx = −0.23,

+∞

−∞

f(t) δ(e

) dt =0.

The reader is urged to derive all the above relations.



5.1 One-Variable Case 151

Historical Notes

“Physical laws should have mathematical beauty.” This statement was Dirac’s re-

sponse to the question of his philosophy of physics, posed to him in Moscow in 1955.

He wrote it on a blackboard that is still preserved today.

Paul Adrien Maurice Dirac (1902–1984), was born in 1902 in Bristol, Eng-

land, of a Swiss, French-speaking father and an English mother. His father, a

taciturn man who refused to receive friends at home, enforced young Paul’s silence

by requiring that only French be spoken at the dinner table. Perhaps this explains

Dirac’s later disinclination toward collaboration and his general tendency to be a

loner in most aspects of his life. The fundamental nature of his work made the

involvement of students diﬃcult, so perhaps Dirac’s personality was well-suited to

his extraordinary accomplishments.

Dirac went to Merchant Venturer’s School, the public school where his father

taught French, and while there displayed great mathematical abilities. Upon grad-

uation, he followed in his older brother’s footsteps and went to Bristol University to

study electrical engineering. He was 19 when he graduated from Bristol University

in 1921. Unable to ﬁnd a suitable engineering position due to the economic reces- “The amount of

theoretical ground

one has to cover

before being able

to solve problems

of real practical

value is rather

large, but this

circumstance is an

inevitable

consequence of

the fundamental

part played by

transformation

theory and is likely

to become more

pronounced in the

theoretical physics

of the future.”

P.A.M. Dirac

(1930)

sion that gripped post-World War I England, Dirac accepted a fellowship to study

mathematics at Bristol University. This fellowship, together with a grant from the

Department of Scientiﬁc and Industrial Research, made it possible for Dirac to go

to Cambridge as a research student in 1923. At Cambridge Dirac was exposed to

the experimental activities of the Cavendish Laboratory, and he became a member

of the intellectual circle over which Rutherford and Fowler presided. He took his

PhD in 1926 and was elected in 1927 as a fellow. His appointment as university

lecturer came in 1929. He assumed the Lucasian professorship following Joseph

Larmor in 1932 and retired from it in 1969. Two years later he accepted a position

at Florida State University where he lived out his remaining years. The FSU library

now carries his name.

In the late 1920s the relentless march of ideas and discoveries had carried physics

to a generally accepted relativistic theory of the electron. Dirac, however, was dis-

satisﬁed with the prevailing ideas and, somewhat in isolation, sought for a better

formulation. By 1928 he succeeded in ﬁnding an equation, the Dirac equation,that

accorded with his own ideas and also ﬁtted most of the established principles of the

time. Ultimately, this equation, and the physical theory behind it, proved to be

one of the great intellectual achievements of the period. It was particularly remark-

able for the internal beauty of its mathematical structure, which not only clariﬁed

previously mysterious phenomena such as spin and the Fermi–Dirac statistics

associated with it, but also predicted the existence of an electron-like particle of

negative energy, the antielectron, or positron, and, more recently, it has come to

play a role of great importance in modern mathematics, particularly in the inter-

relations between topology, geometry, and analysis. Heisenberg characterized the

discovery of antimatter by Dirac as “the most decisive discovery in connection with

the properties or the nature of elementary particles . . . . This discovery of particles

and antiparticles by Dirac ...changed our whole outlook on atomic physics com-

pletely.” One of the interesting implications of his work that predicted the positron

Paul Adrien

Maurice Dirac

1902–1984

was the prediction of a magnetic monopole. Dirac won the Nobel Prize in 1933 for

this work.

Dirac is not only one of the chief authors of quantum mechanics, but he is

also the creator of quantum electrodynamics and one of the principal architects of

152 Dirac Delta Function

quantum ﬁeld theory. While studying the scattering theory of quantum particles, he

invented the (Dirac) delta function; in his attempt at quantizing the general theory

of relativity, he founded constrained Hamiltonian dynamics, which is one of the most

active areas of theoretical physics research today. One of his greatest contributions

is the invention of the bra | and ket | notation used in quantum theory.

While at Cambridge, Dirac did not accept many research students. Those who

worked with him generally thought that he was a good supervisor, but one who

did not spend much time with his students. A student needed to be extremely

independent to work under Dirac. One such student was Dennis Sciama,wholater

became the supervisor of Stephen Hawking, the current holder of the Lucasian chair.

Salam and Wigner in their Preface to the Festschrift that honors Dirac on his

seventieth birthday and commemorates his contributions to quantum mechanics

succinctly assessed the man:

Dirac is one of the chief creators of quantum mechanics.... Posterity

will rate Dirac as one of the greatest physicists of all time. The present

generation values him as one of its greatest teachers.... On those privi-

leged to know him, Dirac has left his mark ...by his human greatness.

He is modest, aﬀectionate, and sets the highest possible standards of

personal and scientiﬁc integrity. He is a legend in his own lifetime and

rightly so.

(Taken from Schweber, S. S. “Some chapters for a history of quantum ﬁeld theory:

1938–1952,” in Relativity, Groups, and Topology II,vol. 2,B.S.DeWittandR.

Stora, eds., North-Holland, Amsterdam, 1984.)

5.1.3 The Step Function

The step function θ is deﬁned as

θ(x)=

1ifx>0

0ifx<0

(5.17)

The θ function (as it is often called) is useful in writing functions that have

discontinuities or cusps. For instance, absolute values can be written in terms

of the step function:

|x| = xθ(x) − xθ(−x)or|x − y| =(x − y)[θ(x −y) −θ(y − x)]

A piecewise continuous function such as

g(x)=

(x)if0<x<1

(x)ifx>1

(5.18)

can be written as

g(x)=g

(x)θ(x)θ(1 − x)+g

(x)θ(x − 1)