Physics 5013 Mathematical Methods of Physics

7/31/2019 Physics 5013 Mathematical Methods of Physics

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Chapter 1

Review of Complex

Numbers

Complex numbers are defined in terms of the imaginary unit, i, having theproperty

i2 = 1. (1.1)A general complex number has the form

z = x + iy, (1.2)

where x, y are real numbers. We also often write

z = Re z + iIm z, (1.3)

where Re z is the real part ofz, and Im z is the imaginary part ofz. Complexnumbers are added and multiplied just like real numbers: If

z1 = x1 + iy1, (1.4a)

z2 = x2 + iy2, (1.4b)

then

z1 + z2 = (x1 + x2) + i(y1 + y2), (1.5a)

z1z2 = x1x2 + iy1x2 + ix1y2 + i2y1y2

= x1x2 y1y2 + i(x1y2 + x2y1). (1.5b)The complex conjugate of a number is obtained by reversing the sign of i: If

z = x + iy, we define the complex conjugate of z by

z = x iy. (1.6)

1 Version of August 22, 2011


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2 Version of August 22, 2011CHAPTER 1. REVIEW OF COMPLEX NUMBERS

E

T

y axisImaginary axis

x axis

Real axis

z-plane

x

y

r

Figure 1.1: Geometrical interpretation of a complex number z = x + iy.

(Sometimes the notation z is used for the complex conjugate of z.) Note that

Re z =z + z

2, (1.7a)

Im z =z z

2i. (1.7b)

Note also thatzz = x2 + y2 (1.8)

is purely real and non-negative, so we define the modulus, or magnitude, or

absolute value of z by

|z| =

zz =

(Re z)2 + (Im z)2, (1.9)

where the positive square root is implied.We give a simple geometrical interpretation to complex numbers, by thinking

of them as two-dimensional vectors, as sketched in Fig. 1.1. Here the length ofthe vector is the magnitude of the complex number,

r = |z|, (1.10)and the angle the vector makes with the real axis is , where

tan = y/x; (1.11)

the quadrant lies in is determined by the sign of x and y. We call

= arg z (1.12)

the argument or phase of z. The above geometrical picture is sometimes calledan Argand diagram.


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E

T

y axis

x axis

z-plane

r

z = x + iy

ddddd

r

z = x iy

Figure 1.2: Geometrical interpretation of complex conjugation.

There is an arbitrariness in the choice of the argument of a complex numberz, for one can always add an arbitrary multiple of 2 to without changing z,

+ 2n, n an integer, z z. (1.13)

It is often convenient to define a single-valued argument function arg z. Byconvention, the principal value of arg z is that phase angle which satisfies theinequality

< arg z . (1.14)(Note that radian measure is always employed.) For every z there is a uniquearg z lying in this range.

The geometrical significance of complex conjugation is shown in Fig. 1.2.Complex conjugation corresponds to reflection in the x-axis.

From the Argand diagram we can write down the polar representation ofa complex number,

z = r cos + ir sin

= r(cos + i sin ), (1.15)

so if we have two complex numbers,

z1 = r1(cos 1 + i sin 1), (1.16a)

z2 = r2(cos 2 + i sin 2), (1.16b)

the product is

z1z2 = r1r2 {cos 1 cos 2 sin 1 sin 2+ i [cos 1 sin 2 + cos 2 sin 1]}

= r1r2 [cos(1 + 2) + i sin(1 + 2)] . (1.17)


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4 Version of August 22, 2011CHAPTER 1. REVIEW OF COMPLEX NUMBERS

That is, the moduli of the complex numbers multiply,

|z1z2| = |z1||z2|, (1.18a)while the arguments add,

arg(z1z2) = arg z1 + arg z2. (1.18b)

The latter statement is to be understood as modulo 2, i.e., equality up to theaddition of an arbitrary integer multiple of 2. In particular, note that

1z z =

1z |z| = 1, (1.19a)

while

0 = arg1z

z = arg 1z

+ arg z, (1.19b)

implying that1z = 1|z| , (1.20a)

arg1

z= arg z. (1.20b)

1.1 De Moivres Theorem

From the above, if we choose a unit vector,

z = cos + i sin , (1.21)

successive powers follow a simple pattern:

z2 = cos 2 + i sin2, (1.22a)

z3 = cos 3 + i sin3, (1.22b)

. . . ,

zn = cos n + i sin n, (1.22c)

or(cos + i sin )n = cos n + i sin n, (1.23)

where n is a positive integer. This is called De Moivres theorem.

1.2 Roots

Suppose we wish to find all the nth roots of unity, that is, all solutions to theequation

zn = 1, (1.24)


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1.2. ROOTS 5 Version of August 22, 2011

1

i

1

i

1+i2

1+i2

1i2

1i2

Figure 1.3: The eight 8th roots of unity.

where n is a positive integer. If we take the polar form,

z = (cos + i sin ), (1.25)

this meansn(cos n + i sin n) = 1, (1.26)

which implies

= 1, (1.27a)

n = 2k, (1.27b)

where k is any integer. Thus the nth root of unity has the form

z = cos2k

n+ i sin

2k

n. (1.28)

These are distinct fork = 0, 1, 2, . . . , n 1; (1.29)

outside of these values of k, the roots repeat. Thus there are n distinct nthroots of unity. For example, for n = 8, the roots are as shown in Fig. 1.3, in the

complex plane.


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Chapter 2

Infinite Series

2.1 Sequences

A sequence of complex numbers {zn}n=1 is a countably infinite set of numbers,z1, z2, z3, . . . , zn, . . . . (2.1)

That is, for every positive integer k, there is a number, the kth term of thesequence, zk, in the set {zn}n=1. Mathematically, a sequence is a complex-valued function defined on the positive integers.

We say that the sequence possesses a limit l,

limn

zn = l, or zn l as n , (2.2)

if, for every > 0, no matter how small, there exists a number N for which|zn l| < for all n > N. (2.3)

(The number N will depend on .) That is, {zn}n=N+1 all lie within a circle ofradius centered on the point l in the complex plane.

A necessary and sufficient condition for a sequence {zn}n=1 to converge toa limit is Cauchys criterion: A sequence {zn}n=1 possesses a limit if and onlyif for every > 0, no matter how small, it is possible to find a number N suchthat

|zn zm| < for all n,m > N. (2.4)(Note that the difference |n m| may be arbitrarily large.) Thus, all elementsof the sequence {zn}n=N+1 lie within a disk of radius . Briefly, we say that theCauchy condition is

|zn zm| 0 for all n, m sufficiently large. (2.5)Sequences having this property are called Cauchy sequences. Every Cauchysequence of complex numbers possesses a limit (which is, of course, a complexnumber)this property means that the complex numbers form a complete space.



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8 Version of August 27, 2011 CHAPTER 2. INFINITE SERIES

2.2 Series

Suppose we have a sequence {ak}k=1 from which we construct the finite sumssn =

nk=1

ak, n = 1, 2, 3, . . . . (2.6)

The set of all these sums, {sn}n=1, itself forms a sequence. If this latter sequencehas a limit S,

sn S as n , (2.7)then we say that the infinite series

k=1

ak = limn

nk=1

ak (2.8)

possesses the limit S (or converges to S),

k=1

ak = S. (2.9)

By the Cauchy criterion, this will be true if and only ifn

k=m

ak

< (2.10)for any fixed > 0, whenever n m > N, N a number depending on .

Obviously, a necessary condition fork=1

ak (2.11)

to converge is for ak 0 as k . However, this is not sufficient, as thefollowing example shows.

2.3 Examples

2.3.1 Harmonic series

Consider the sum of the reciprocals of the integers,

1 +1

2+

1

3+ . . . =

n=1

1

n. (2.12)

Note that if the nth term of the series is denoted an = 1/n, we have for the sum

of n adjacent terms

an+1 + . . . + a2n =1

n + 1+ . . . +

1

2n> n

1

2n

=

1

2, (2.13)

no matter how large n is. This violates Cauchys criterion, so the harmonicseries diverges.


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2.4. ABSOLUTE AND CONDITIONAL CONVERGENCE9 Version of August 27, 2011

2.3.2 Geometric series

Consider the series m=0

arm, (2.14)

where a is a constant and r 0. For r = 1, the nth partial sum is

sn =

nm=0

arm = a1 rn+1

1 r , (2.15)

so

S = limn

sn =a

1 r if r < 1, (2.16)

while the series diverges if r 1.

2.4 Absolute and Conditional Convergence

Suppose we have a convergent series

n=1 an. If also

n=1 |an| converges, wesay that the original series converges absolutely. Otherwise, the original seriesis conditionally convergent. (That is, it converges because of sign alternations.)A sufficient condition for (at least) conditional convergence is provided by thefollowing theorem due to Leibnitz:

If the terms of a series are of alternating sign and in addition their absolute

values tend to zero, |an| 0, monotonically, i.e., |an| > |an+1| for sufficientlylarge n, then

n=1 a

n converges. (2.17)

In absolutely convergent series one can rearrange the terms without affectingthe value of the sum. With conditionally convergent series, one cannot rearrangeterms; in fact, such rearrangements can make a conditionally convergent seriesconverge to any desired value, or to diverge!

2.4.1 Example

Consider the conditionally convergent series formed from the divergent harmonicseries by alternating every other sign:

1 1

2 +1

3 1

4 +1

5 1

6 + . . . = ln 2, (2.18)

which converges to the natural logarithm of 2. Multiply this equation term byterm by 1/2:

1

2 1

4+

1

6 1

8+

1

10 . . . = 1

2ln 2. (2.19)


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Add these two series:

1 + 13 12 + 15 + 17 14 + 19 + 111 16 + . . . = 32 ln 2. (2.20)

Since the reciprocal of each integer occurs exactly once in the last series, wewould be tempted to rearrange the series to obtain

1 12

+1

3 1

4+

1

5 1

6+ . . . = ln 2, (2.21)

which is identical to the original series. There is an obvious contradiction here!In order to obtain the rearrangement (2.21), we have to go further and furtherout in the series (2.20), which apparently is not permissible.

2.4.2 A Theorem About Absolutely Convergent Series

Not only can absolutely convergent series be rearranged without changing theirvalue, but they can be multiplied together term by term: If two series

S =

i=1

ui, (2.22a)

T =

i=1

vi (2.22b)

are both absolutely convergent, the series

P =

i=1j=1

uivj (2.23)

formed from the product of their terms written in any order, is absolutely con-vergent, and has a value equal to the product of of the individual series,

P = ST . (2.24)

2.5 Convergence Tests

The following tests can determine whether a given series is absolutely convergentor not.

2.5.1 Comparison test

If bn > 0 for all n and

n=1 bn is convergent, and if |an| bn for all n, thenn=1

an is absolutely convergent. (2.25a)


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2.5. CONVERGENCE TESTS 11 Version of August 27, 2011

Also, if |an| bn > 0 for all n, and

n=1 bn diverges, then

n=1

an is not absolutely convergent. (2.25b)

2.5.2 Root test

The series

n=1 an converges absolutely if from a certain term onward

n

|an| q < 1, (2.26)where q 0 is independent of n.

Proof: If the inequality holds, |an| qn. But

n=1 qn converges for q < 1,

it being the geometric series, so by 2.5.1,

n=1 |an| converges.

2.5.3 Ratio test

The series

n=1 an converges absolutely if from a certain term onwardan+1an q < 1, (2.27)

where q 0 is independent of n.Proof: Without loss of generality, we may assume the inequality holds for

all n; otherwise, we renumber the {an} sequence so that 1 labels the first termfor which the inequality (2.27) holds. Then

ana1

=

anan1

an1an2

an2an3

a2a1

q

n1. (2.28)

Convergence is again assured by comparison with the geometric series. (Whetherthese tests are satisfied by the first few terms of a series is immaterial, since afinite number of terms of an infinite seris has no effect on the convergence.)

Example

When does

n=1 nqn converge? If we use the root test, we examine

limn

n

|an| = |q| limn

n

n = |q|, 1 (2.29a)

while if we use the ratio test, we look at

limnan+1an

= |q| limnn + 1

n = |q|. (2.29b)

In either case, we see that the series is absolutely convergent if |q| < 1, anddivergent otherwise.

1Because ln nn = 1

nlnn, which tends to zero as n, n

n 1.


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The following are refinements of the ratio test, which fails (that is, fails toreveal whether the tested series converges or not) when

limn

an+1an = 1. (2.30)

For example, this indeterminate limit results for the case an = 1/n, whichyields a divergent series, but also for an = 1/(n ln

2 n), which corresponds to aconvergent sum (see Sec. 2.5.8).

2.5.4 Kummers test

Choose a sequence of positive constants bn. If

bn anan+1

bn+1 C > 0, (2.31)for all n N, where N and C are fixed numbers, then

n=1

an converges absolutely. (2.32)

On the other hand, if

bn

anan+1 bn+1 0, (2.33)

and

n=1

b1n diverges, (2.34)

thenn=1

|an| diverges. (2.35)

Proof: If the inequality (2.31) holds, take l N, so that

C|al+1| bl|al| bl+1|al+1|. (2.36)

So we have the inequality

n

l=N+1

|al| bN|aN|C

bn|an|C

bN|aN|C

. (2.37)

Hence, the nth partial sum, for n > N, is

sn =ni=1

|ai| Ni=1

|ai| + bN|aN|C

. (2.38)


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2.5. CONVERGENCE TESTS 13 Version of August 27, 2011

The right-hand side of this inequality is a constant, independent of n. There-fore, the positive sequence of increasing terms

{sn

}is bounded above, and con-

sequently possesses a limit. The series is absolutely convergent.If the inequality (2.33) holds,

|an| |aN|bNbn

, n > N, (2.39)

so since

n=1 b1n diverges, so does

n=1 |an|.

2.5.5 Raabes test

Raabes criterion for absolute convergence is

n

anan+1

1

K > 1, (2.40)

for all n N, where N and K are fixed. And if

n

anan+1 1

1, (2.41)

thenn=1

|an| diverges. (2.42)

Proof: In Kummers test put bn = n.

2.5.6 Gauss test

If anan+1 = 1 + hn + B(n)n2 , (2.43)

where h is a constant and the function B(n) is bounded as n , then

n=1 |an| converges for h > 1 and diverges for h 1.Proof: For h = 1 we can use Raabes test:

limn

n

h

n+

B(n)

n2

= h. (2.44)

For h = 1, Raabes test is indeterminate. In that case use Kummers test withbn = n ln n: for large n,

n ln n

1 +

h

n+

B(n)

n2

(n + 1)ln(n + 1)

n ln n

1 + hn

+ B(n)n2

(n + 1)

ln n + 1

n

h +B(n)

n

ln n ln n 1

(h 1)ln n 1 < 0, if h 1. (2.45)


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f

1 2 3 4 5 6 7 8 9 10Figure 2.1: Bounds on a monotone series provided by an integral.

Because

n=2

1

n ln ndiverges (2.46)

(see homework), the series

n=1 |an| diverges.

2.5.7 Integral test

If f(x) is a continuous, monotonically decreasing real function of x such that

f(n) = |an|, (2.47)then

n=1

|an| converges if

1

dx f(x) < , (2.48)

and diverges otherwise.Proof:

It is geometrically obvious that

1

dx f(x) 1, 1. (2.56)

Thus the series converges if > 1 and diverges for other real .

2.6 Series of Functions

2.6.1 Continuity

A (complex-valued) function f(z) of a complex variable is continuous at z0 if

f(z) f(z0) as z z0 (2.57)

from any direction. That is, given > 0 we may find a > 0 such that

|f(z) f(z0)| < whenever |z z0| < . (2.58)

In other words, z lies within a circle of radius around z0.


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f

x

= f= fn

2

Figure 2.2: Uniform convergence of the partial sum fn(x) to the limit f(x). Forall x, fn(x) is within a band of width 2 about f(x).

2.6.2 Uniform Convergence

Consider the infinite seriesf(z) =

i=1

gi(z) (2.59)

constructed from the sequence of functions {gi}i=1. The condition that thisseries converge is expressed in terms of the partial sums,

fn(z) =

ni=1

gi(z) (2.60)

thusly: given > 0 we can find an integer N so that for n > N

|fn+p(z) fn(z)| < for all p > 0. (2.61)This is Cauchys criterion. In general the N required for this to occur will dependon the point z. If, however, Eq. (2.61) holds for all z if n > N independent ofz, we say that the series converges uniformly throughout the region of interest.Equivalently, there exists a function f(z) such that

|f(z) fn(z)| < for all n > N, N independent of z. (2.62)That is, the partial sum fn is everywhere uniformly close to f, the limitingfunction. This situation is illustrated in Fig. 2.2 for a real function of a realvariable.

Contrast absolute and uniform convergence through the following examples.The series

n=1

(1)nn + z2

(2.63)

is only conditionally convergent, because asymptotically the terms become (1)n/n.On the other hand, for real z it is uniformly convergent because

N+pn=N+1

(1)nn + z2

0, a value of n such that

|fn(z) f(z)| < for all z (2.69)

throughout the domain. Then

|f(z) f(z)| = |f(z) fn(z) + fn(z) f(z) + fn(z) fn(z)| |f(z) fn(z)| + |f(z) fn(z)| + |fn(z) fn(z)|.(2.70)

Since the fns are continuous, we can find a for any given such that

|fn(z) fn(z)| < whenever |z z| < . (2.71)

Therefore, |f(z) f(z)| < 3 whenever |z z| < . (2.72)QED.

Even if the limit function is continuous, convergence to it need not be uni-form, as the following example shows:


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dd

dd

df

x0 1/n 2/nFigure 2.3: Sketch of the function fn(x) given by Eq. 2.73).

Example

Consider the sequence of continuous functions,

fn(x) =

nx, 0 x 1/n,(2/n x)n, 1/n x 2/n,

0, otherwise.(2.73)

This function in sketched in Fig. 2.3. Note that the maximum of the functionfn(x) is 1. On the other hand, for all x,

limn

fn(x) = 0, (2.74)

which is certainly a continuous limit function. But the convergence to this limitis not uniform, for there is always a point, x = 1/n, for which

0 fn

1

n

= 1 (2.75)

no matter how large n is. So the convergence is nonuniform.

Properties of Uniformly Convergent Series

Consider the series of functions of a real variable,

f(x) =

n=1

gn(x). (2.76)

1. If the gn are continuous, we can integrate term by term if

n gn is uni-formly convergent over the domain of integration:b

a

dx f(x) =n=1

ba

dx gn(x). (2.77)

2. If the gn and g

n =ddx

gn are continuous, and

n g

n is uniformly conver-gent, then we can differentiate term by term:

f(x) =

n=1

gn(x). (2.78)


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2.7. POWER SERIES 19 Version of August 27, 2011

Condition for Uniform Convergence

The following condition is sufficient, but not necessary, to ensure that a seriesis uniformly convergent.If |gn(z)| < an, where {an} is a sequence of constants such that

n=1 anconverges, then

n=1 gn(z) converges uniformly and absolutely.Proof: The hypothesis implies

N+pn=N

gn(z)

1 cancels, as may be seen by considering the net phase changewhen both branch points are encircled:

arg

z2 1

arg z=2

arg z=0

= 0 mod 2. (6.22)

This means that we may take

z2 = z, and we can immediately write down theexpansion for large z:

z2 1 = z

1 1

z2

1/2

= z

1 1

2z2 1

8z4 . . .

= n=0

(2n 3)!!2nn!

1

z2n1, (6.23)

where we have used the double factorial notation,

(2k + 1)!! = (2k + 1)(2k 1)(2k 3) 3 1, (6.24)and, from the recursion formula

(2k + 1)!! = (2k + 1)(2k 1)!! (6.25)identify

(1)!! = 1, (3)!! = 1. (6.26)The Laurent expansion (6.23) converges for |z| > 1.


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6.4. CLASSIFICATION OF SINGULARITIES59 Version of October 12, 2011

6.4 Classification of Singularities

Suppose in the neighborhood of z0 a function f(z) may be written as

f(z) = (z) +a1

z z0 +a2

(z z0)2 + . . . +an

(z z0)n , (6.27)

where (z) is analytic in the neighborhood of z0, and a1, a2, . . . , an arecomplex constants. When the above expansion holds true, f is said to have apole of order n at z = z0. When n = 1, the singularity is called a simple pole.When f has a pole of order n at z0,

(z z0)nf(z) (6.28)

is analytic at z = z0. If the function

(z z0)mf(z) (6.29)is not analytic at z = z0 no matter how large the integer m is, we say that fhas an essential singularity at z0. (This definition applies to functions whichare single-valued without the introduction of branch lines.)

If an essential singularity is isolated, that is, in a sufficiently small neigh-borhood ofz0, f is analytic except at z0, f may be expanded in a Laurent seriesconverging in an annulus:

f(z) =

n=

an(z z0)n, > |z z0| > , (6.30)

where is arbitrarily small, and is the distance to the next singularity. (The

proof for this statement is provided in the homework.)

6.4.1 WeierstrassPicard Theorem

In the neighborhood of an essential singularity, f(z) becomes arbitrarily close toevery complex value. This theorem, due to Weierstrass, was greatly sharpenedby Picard.

Picards Theorem

In any neighborhood of an essential singularity, the function assumes every finitevalue, with one possible exception, an infinite number of times.

Example: Consider

e1/z =n=0

1n! zn

, (6.31)

which has an essential singular point at z = 0. Let be any complex numberexcept 0. For what zs is

= e1/z? (6.32)


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60 Version of October 12, 2011CHAPTER 6. TAYLOR AND LAURENT SERIES

Recalling the 2i periodicity of the exponential function, we see

log = 1z + 2in, n = integer, (6.33)

or

z =1

log 2in . (6.34)Thus in any neighborhood of 0 there are an infinite number of these zs.

6.4.2 Branch Points and Cuts

Recall log z was defined in the cut plane shown in Fig. 3.1. The location of thecut line is arbitrary, but the location of the end point, z = 0 is not. This branchpoint is a singular point of log z:

d

dz log z =

1

z , (6.35)

which does not exist at z = 0. This type of singularity is neither a pole nor anessential singularity. Once the cut is specified, thus defining log z, the functionis not analytic on the branch cut or branch line; in fact, it is discontinuous acrossthe cut:

disc(log z) = log ei log ei = 2i. (6.36)The same applies to square roots, and all nonintegral powers, which are

defined in terms of the logarithm,

z = z1/2 = e1

2log z. (6.37)

Here the discontinuity across the branch line is

disc(z) = ei ei=

ei/2 ei/2

= 2i

. (6.38)

6.5 Liouvilles Theorem

First we prove Cauchys inequality. Recall the integral representation for thederivative of an analytic function, Eq. (5.34),

f(n)(z0) =n!

2i

C

f(z)

(z z0)n+1 dz (6.39)

if z0 is inside C and f is analytic on and within C. If C is a circle of radius r

centered about z0,z = z0 + re

i, (6.40)

we write this integral more explicitly as

f(n)(z0) =n!

2

20

f

z0 + rei

rneind (6.41)


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6.5. LIOUVILLES THEOREM 61 Version of October 12, 2011

or

f

(n)

(z0)

n!

220

fz0 + reirn d

n!M

rn , (6.42)

where M is the maximum value attained by |f| on C.Now Liouvilles theorem(also really due to Cauchy) states: An entire bounded

function is constant.

Proof: Since f(z) is entire, the Taylor series converges everywhere,

f(z) =

n=0

1

n!f(n)(0)zn. (6.43)

But from Cauchys inequality,

f(n)(0) M n!

Rn

, (6.44)

where R is the radius of an arbitrarily large circle about the origin, and M maybe taken as the bound on |f|,

|f(z)| M z. (6.45)

Hence by taking R , we see that

f(n)(0) = 0, n > 0, (6.46)

and sof(z) = f(0). (6.47)

QED.

Example:

Although ez is entire, is is certainly not bounded.


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Chapter 7

The Calculus of Residues

If f(z) has a pole of order m at z = z0, it can be written as Eq. (6.27), or

f(z) = (z) =a1

(z z0) +a2

(z z0)2 + . . . +am

(z z0)m , (7.1)

where (z) is analytic in the neighborhood of z = z0. Now we have seen that ifC encircles z0 once in a positive sense,

C

dz1

(z z0)n = 2in,1, (7.2)

where the Kronecker -symbol is defined by

m,n = 0, m = n,1, m = n. . (7.3)

Proof: By Cauchys theorem we may take C to be a circle centered on z0. Onthe circle, write z = z0 + re

i. Then the integral in Eq. (7.2) is

i

rn1

20

d ei(1n), (7.4)

which evidently integrates to zero if n = 1, but is 2i if n = 1. QED.Thus if we integrate the function (7.1) on a contour C which encloses z0,

while (z) is analytic on and within C, we find

C

f(z) dz = 2ia1. (7.5)

Because the coefficient of the (z z0)1 power in the Laurent expansion of fplays a special role, we give it a name, the residue of f(z) at the pole.

If C contains a number of poles of f, replace the contour C by contours ,, , . . . encircling the poles singly, as shown in Fig. 7.1. The contour integral

63 Version of October 26, 2011


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64 Version of October 26, 2011CHAPTER 7. THE CALCULUS OF RESIDUES

'&

$%

C

i ii

Figure 7.1: Integration of a function f around the contour C which containsonly poles of f may be reduced to the integrals around subcontours , , ,etc., each of which contains but a single pole of f.

around C may be distorted to a sum of disjoint ones around , , . . . , soC

f(z) dz =

f(z) dz +

f(z) dz + . . . , (7.6)

and since each small contour integral gives 2i times the reside of the singlepole interior to that contour, we have established the residue theorem: If f beanalytic on and within a contour C except for a number of poles within,

C

f(z) dz = 2i

poleswithinC

residues, (7.7)

where the sum is carried out over all the poles contained within C.This result is very usefully employed in evaluating definite integrals, as the

following examples show.

7.1 Example 1

Consider the following integral over an angle:

I =

20

d

1 2p cos + p2 , 0 < p < 1. (7.8)

Let us introduce a complex variable according to

z = ei, dz = iei d = iz d, (7.9)

so that

cos =1

2z + 1

z . (7.10)

Therefore, we can rewrite the angular integral as an integral around a closedcontour C which is a unit circle about the origin:

I =

C

dz

iz

1

1 p z + 1z

+ p2


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7.2. A FORMULA FOR THE RESIDUE 65 Version of October 26, 2011

=

Cdz

i

1

z

p(z2 + 1) + p2z

= 1i

C

dz 1(1 pz)(z p) . (7.11)

The integrand exhibits two poles, one at z = 1/p > 1 and one at z = p < 1.Only the latter is inside the contour C, so since

1

1 pz1

z p =

1

z p +p

1 pz

1

1 p2 , (7.12)

we have from the residue theorem

I = 2i1

i

1

1 p2 =2

1 p2 . (7.13)

Note that we could have obtained the residue without partial fractioning byevaluating the coefficient of 1/(z p) at z = p:

1

1 pzz=p

=1

1 p2 . (7.14)

This observation is generalized in the following.

7.2 A Formula for the Residue

If f(z) has a pole of order m at z = z0, the residue of that pole is

a1 =1

(m 1)!dm1

dzm1[(z z0)mf(z)]

z=z0

. (7.15)

The proof follows immediately from Eq. (7.1).

7.3 Example 2

This time we consider an integral along the real line,

I =

dx1

(x2 + 1)3= lim

RR

R

dx1

(x2 + 1)3, (7.16)

where we have made explicit the meaning of the upper and lower limits. Werelate this to a contour integral as sketched in Fig. 7.2. Thus we have

C

dz

(z2 + 1)3=

RR

dx

(x2 + 1)3+

dz

(z2 + 1)3, (7.17)


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R R

i

id

dd

dddsR

Figure 7.2: The closed contour C consists of the portion of the real axis betweenR and R, and the semicircle of radius R in the upper half plane. Also shownin the figure are the location of the poles of the integrand in Eq. (7.17).

where we are to understand that the limit R

is to be taken at the end

of the calculation. It is easy to see that the integral over the large semicirclevanishes in this limit:

dz

(z2 + 1)3=

0

R i eid

(R2e2i + 1)3 0, R . (7.18)

Hence the integral desired is just the closed contour integral,

I =

C

dz

(z2 + 1)3= 2i(residue at i). (7.19)

By the formula (7.15) the desired residue is

a1 =1

2!

d2

dz2 (z i)3 1

(z

i)3(z + i)3 z=i

=1

2!

d2

dz21

(z + i)3

z=i

=1

2!

(3)(4)(z + i)5

z=i

=3

16i, (7.20)

so

I =3

8. (7.21)

7.4 Jordans Lemma

The evaluation of a class of integrals depends upon this lemma. If f(z) 0uniformly with respect to arg z as |z| for 0 arg z , and f(z) isanalytic when |z| > c > 0 and 0 arg z , then for > 0,

lim

eizf(z) dz = 0, (7.22)


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pole inside the contour C is at ia.) The second integral on the right-hand sidevanishes as R

by Jordans lemma. (Note carefully that this would not be

true if we replace eiz by cos z in the above.) Because only the even part of eixsurvives symmetric integration,

I =1

2

eix

x2 + a2dx =

1

2

C

eiz

z2 + a2dz

=1

22i

1

2iaei(ia) =

2aea. (7.30)

(Note that if C were closed in the lower half plane, the contribution from theinfinite semicircle would not vanish. Why?)

7.6 Cauchy Principal Value

To this point we have assumed that the path of integration never encounters anysingularities of the integrated function. On the contrary, however, let us nowsuppose that f(x) has simple poles on the real axis, and try to attach meaningto

f(x) dx. (7.31)

For simplicity, suppose f(z) has a simple pole at only one point on the real axis,

f(z) = (z) +a1

z x0 , (7.32)

where (z) is analytic on the entire real axis. Then we define the (Cauchy)principal value of the integral as

P

f(x) dx = lim0+

x0

f(x) dx +

x0+

f(x) dx

, (7.33)

which means that the immediate neighborhood of the singularity is to be omittedsymmetrically. The limit exists because f(x) a1/(xx0) near x = x0, whichis an odd function.

We can apply the residue theorem to such integrals by considering a deformed(indented) contour, as shown in Fig. 7.3. For simplicity, suppose the functionfalls off rapidly enough in the upper half plane so that

f(z) dz = 0, (7.34)

where is the infinite semicircle in the upper half plane. Then the integralaround the closed contour shown in the figure is

C

f(z) dz = P

f(x) dx ia1, (7.35)


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7.6. CAUCHY PRINCIPAL VALUE 69 Version of October 26, 2011

x0

Figure 7.3: Contour which avoids the singularity along the real axis by passingabove the pole.

x0

Figure 7.4: Contour which avoids the singularity along the real axis by passingbelow the pole.

where the second term comes from an explicit calculation in which the simplepole is half encircled in a negative sense (giving 1/2 the result if the pole

were fully encircled in the positive sense). On the other hand, from the residuetheorem, C

f(z) dz = 2i

polesUHP(residues), (7.36)

where UHP stands for upper half plane. Alternatively, we could consider adifferently deformed contour, shown in Fig. 7.4. Now we have

C

f(z) dz = P

f(x) dx + ia1

= 2i

polesUHP

(residues) + a1

, (7.37)

so in either case

P

f(x) dx = 2i

polesUHP(residues) + ia1, (7.38)

where the sum is over the residues of the poles above the real axis, and a1 isthe residue of the simple pole on the real axis.


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R R k + i

k i

Figure 7.5: The closed contour C for the integral in Eq. (7.41).

Equivalently, instead of deforming the contour to avoid the singularity, onecan displace the singularity, x0 x0 i. Then

dx g(x)x x0 i = P

dx g(x)x x0 ig(x0), (7.39)

if g is a regular function on the real axis. [Proof: Homework.]

7.7 Example 4

Consider the integral

I =

eiqx

q2 k2 + i dq, x > 0, (7.40)

which is important in quantum mechanics. We can replace this integral by thecontour integral

C

eiqxq2 k2 + i dq, x > 0, (7.41)

where the closed contour C is shown in Fig. 7.5. The integral over the infinitesemicircle is zero according to Jordans lemma. By redefining , but notchanging its sign, we write the integral as (k = +

k2)

I =

C

dq

1

q (k i)1

q + (k i)

eiqx = 2ieiqx

q (k i)

q=(ki)

= ik

eikx, (7.42)

in the end taking 0.

7.8 Example 5

We will consider two ways of evaluating

I =

0

dx

1 + x3. (7.43)


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7.8. EXAMPLE 5 71 Version of October 26, 2011

Figure 7.6: Contour C used in the evaluation of the integral (7.44). Shownalso is the branch line of the logarithm along the +z axis, and the poles of theintegrand.

The integrand is not even, so we cannot extend the lower limit to . Howcan contour methods be applied?

7.8.1 Method 1

Consider the related integral C

log z

1 + z3dz, (7.44)

over the contour shown in Fig. 7.6. Here we have chosen the branch line of thelogarithm to lie along the +z axis; the discontinuity across it is

disc log z = log log e2i = 2i. (7.45)The integral over the large circle is zero, as is the integral over the little circle:

lim,0

20

log ei

1 + 2e3iei i d = 0. (7.46)

Therefore,

I = 12i

C

log z

1 + z3dz

=

poles insideC

(residues). (7.47)


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2/3

C1

C2

C3

Figure 7.7: Contour used in the evaluation of Eq. (7.50).

The only pole of 1/(z3 + 1) contained within C is at z = ei/3, the residue ofwhich is

1ei/3 ei 1ei/3 ei5/3 = e

2i/3

ei/3 ei/3 e3i/3

e2i/3 e2i/3 , (7.53)so

I = 2ie6i/3

(2i)3

sin 32

sin 23

, (7.54)

or since sin 3 = sin23 =

32 ,

I =

4

2

3

3=

2

3

3, (7.55)

the same result (7.49) as found by method 1.

7.9 Example 6

Consider

I =

0

x1

1 + xdx, 0 < < 1. (7.56)

We may use the contour integralC

(z)11 + z

dz =

0

ei(1)x1 dx

1 + x0

ei(1)x1 dx

1 + x, (7.57)

where C is the same contour shown in Fig. 7.6, and because is between zeroand one it is easily seen that the large circle at infinity and the small circleabout the origin both give vanishing contributions. The pole now is at z = 1,so

C

(z)1 dz1 + z

= 2i, (7.58)

where the phase is measured from the negative real z axis. Thus

2i =

ei(1) ei(1)

I = 2iIsin , (7.59)


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c

T

12

12

R Rx

y

Figure 7.8: Contour C used in integral K, Eq. (7.62). Here the two lines makingan angle of /4 with respect to the real axis are closed with vertical lines atx = R, where we will take the limit R .

or

I =

sin . (7.60)

7.10 Example 7

Here we demonstrate a method of evaluating the Gaussian integral,

J =

ex2

dx. (7.61)

Consider the contour integral

K =

C

eiz2

csc z dz, (7.62)

where C is the contour shown in Fig. 7.8. The equation for the two lines makingangles of /4 with respect to the real axis are

z = 12

+ ei/4, (7.63)

so

z2 =1

4 ei/4 + i2. (7.64)


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Within the contour the only pole of csc z is at z = 0, which has residue 1/,so by the residue theorem

K = 2i 1

= 2i. (7.65)

Directly, however,

K =

ei/4 d exp

i

i2 + ei/4 +

1

4

csc

ei/4 +

1

2

ei/4 d exp

i

i2 ei/4 + 1

4

csc

ei/4 1

2

,(7.66)

since the vertical segments give exponentially vanishing contributions as R . Combining these two integrals, we encounter

exp

ie

i/4csc

e

i/4

+

1

2 exp iei/4 csc ei/4

1

2

= 2exp

iei/4

+ exp

iei/4exp

iei/4

+ exp

iei/4 = 2, (7.67)since ei/2 = i. Hence

K = 2ei/4ei/4

d e2

=2i

dx ex2

, (7.68)

so comparing with Eq. (7.65) we have for the Gaussian integral (7.61)

J =

. (7.69)

7.11 Example 8

Our final example is the integral

I =

0

x dx

1 ex . (7.70)

If we make the substitution ex = t, this is the same as

I =

1

log t

1 tdt

t=

1

dt log t

1

t+

1

1 t

. (7.71)

If we make the further substitution in the first form of Eq. (7.71)

u =1

t,

du

u=

dt

t, (7.72)

we have

I =

10

log 1u

1 1u

du

u=

10

log u

1 u du, (7.73)


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If we average the two forms (7.71) and (7.73) we have

I = 12

1

dt log tt

+ 12

0

dt log t1 t . (7.74)

The two integrals here separately are divergent, but the sum is finite. Weregulate the two integrals by putting in a large t cutoff:

I =1

2lim

1

dtlog t

t+

0

dtlog t

1 t

. (7.75)

The first integral here is elementary,1

dtlog t

t=

1

2log2 t

1

=1

2log2 , (7.76)

while the second is evaluated by considering

K =

C

dzlog2 z

1 z , (7.77)

where again C is the contour shown in Fig. 7.6. Now, however, the sole pole ison the positive real axis, so no singularities are contained within C, and henceby Cauchys theorem K = 0.

This time the contribution of the large circle is not zero:20

ei idlog2 ei

1 ei = i20

d [log + i]2

=

i 2 log2 + 2i 1

2

(2)2 log

1

3

(2)3 .(7.78)The discontinuity of the log2 across the branch line is

log2 x log2 xe2i = log2 x (log x + 2i)2= 4i log x + 42. (7.79)

Finally, notice that there is a contribution from the pole at z = 1 below the realaxis (see Fig. 7.9): Explicitly, the contribution from the small semicircle belowthe pole is

2

diei

ei

4i log

1 + ei 42 = 4i3, (7.80)

as 0. The desired integral is obtained by taking the imaginary part,

Im K = 40

dtlog t

1 t

2 log2 83

3

43 = 0, (7.81)

so 0

dtlog t

1 t = 1

2log2

2

3. (7.82)


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1

Figure 7.9: Portion of integral K, Eq. (7.77), corresponding to the integrationbelow the cut on the real axis. The pole of the integrand at z = 1 contributeshere because log(1 i) = 2i. Thus the contribution of the small semicircle toK is +i(2i)2 = 4i3, in agreement with Eq. (7.80).

Thus averaging this with Eq. (7.76) we obtain

I = 2

6. (7.83)

A slight check of this procedure comes from computing the real part of K:

Re K = 42P

0

dt

1 t + 42 log = 0. (7.84)


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80 Version of November 14, 2011 CHAPTER 8. APPROXIMANTS

8.1.1 Shanks Transformation

The Shanks transformation is good for alternating series, or oscillating partialsums, such as Eq. (8.1). For the series

S =n=1

an, (8.5)

consider the Nth partial sum

SN =

Nn=1

an. (8.6)

Let us suppose that, for sufficiently large N,

SN = S+ AbN, (8.7)

where 1 < b < 0, so that as N, SN S. We will take this as an ansatzfor all N, to obtain an estimate for the limit S. Then, successive partial sumssatisfy

SN1 = S+ AbN1, (8.8a)

SN = S+ AbN, (8.8b)

SN+1 = S+ AbN+1, (8.8c)

so that

b =SN+1 S

SN

S

=SN S

SN1

S

, (8.9)

which may be immediately solved for S,

S(N) =SN+1SN1 S2N

SN+1 + SN1 2SN, (8.10)

where now weve inserted the (N) subscript on the left to indicate this is anestimate for the limit, based on the N, N + 1, and N 1 partial sums.

For the series (8.1) the first 5 partial sums are

S1 = 1, S2 =1

2= 0.5, S3 =

5

6= 0.833, S4 =

7

12= 0.5833,

S5 =47

60= 0.7833, (8.11)

which oscillate around the correct limit ln 2 = 0.693147, but are not good ap-proximations. Using the Shanks transformation (8.10) we obtain much betterapproximants:

S(1) =7

10= 0.700, S(2) =

29

42= 0.690, S(3) =

25

36= 0.6944, (8.12)


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8.1. ACCELERATED CONVERGENCE 81 Version of November 14, 2011

which use only the first 3, 4, and 5 terms in the original series. We can do evenbetter by iterating the Shanks transformation,

S[2](N) =

S(N+1)S(N1) S2(N)S(N+1) + S(N1) 2S(N) , (8.13)

and then we find using the same data (only 5 terms in the series)

S[2](2) =

165

238= 0.693277, (8.14)

an error of only 0.02%! For more detailed comparison of Shanks estimates forthis series, see Table 8.2 on page 373 of Bender and Orzag.

8.1.2 Richardson Extrapolation

For monotone series, Richardson extrapolation is often very useful. In this casewe are considering partial sums SN which approach their limit S monotonically.In this case we assume an asymptotic form for large N

SN S+ aN

+b

N2+

c

N3+ . . . . (8.15)

The first Richardson extrapolation consists of keeping only the first correctionterm,

SN = S+a

N, SN+1 = S+

a

N + 1, (8.16)

which may be solved for the limit

S[1](N) = (N + 1)SN+1 N SN, (8.17)

where again weve inserted on the left a superscript [1] indicating the firstRichardson extrapolation, and a subscript (N) to indicate the approximantcomes from the Nth and N + 1st partial sums.

We consider as an example Eq. (8.2). Here, the first 4 partial sums are

S1 = 1, S2 =5

4= 1.25, S3 =

49

16= 1.361, S4 =

205

144= 1.424, (8.18)

to be compared with 2/6 = 1.644934. The first three Richardson extrapolantsare much better:

S[1](1) =

3

2= 1.5, S

[1](2) =

19

12= 1.58, S

[1](3) =

29

18= 1.611. (8.19)

Iteration of these results by inserting S[1]

(N)in (8.17) yields further improvement:

5/3 = 1.667, but this iteration improves only slowly with N.To do better we keep the first two terms in (8.15). This gives the second

Richardson extrapolant,

S[2](N) =

1

2

(N + 2)2SN+2 2(N + 1)2SN+1 + N2SN

. (8.20)


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When applied to the series (8.2) the first three terms in the series yields nearly1% accuracy:

S[2](1) = 138= 1.625. (8.21)

For further numerical details, see Table 8.4 on page 377 of Bender and Orzag.

8.2 Summing Divergent Series

The series encountered in physics, typically perturbation expansions, are usuallydivergent. How can one extract a meaningful number from such series, whichrepresent physical processes and so reflect real processes?

On the surface, it would seem impossible to attach any meaning to suchobviously divergent series as

1 + 1 + 1 + 1 + 1 + . . . , (8.22a)1 1 + 1 1 + 1 . . . . (8.22b)

However, as we will now see, perfectly finite numbers can be associated withthese series. Again there are various procedures, of which we give a sampling.Throughout, we are considering a divergent series of the form

n=0

an. (8.23)

8.2.1 Euler Summation

Supposen=0

anxn = f(x) (8.24)

converges if |x| < 1. Then we define the limit of the series (8.23) byS = lim

x1f(x). (8.25)

Thus, for the series (8.22b),

S =

n=0

(1)n, (8.26)

f(x) is

f(x) =

n=0

(1)n

xn

=1

1 + x , (8.27)

so S = 1/2. To supply more credence to this result, we note that it is reproducedby the Shanks transformation. The partial sums of the series are

S0 = 1, S1 = 0, S2 = 1, S3 = 0, . . . , (8.28)


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8.2. SUMMING DIVERGENT SERIES 83 Version of November 14, 2011

so

S =SN+1SN1

S2n

SN+1 + SN1 2Sn=

1

2(8.29)

for all N.What if we apply Euler summation to the series

1 + 0 1 + 1 + 0 1 + 1 + 0 1 + 1 + 0 1 + . . .? (8.30)

Now

f(x) = 1 x2 + x3 x5 + x6 x8 + x9 . . .

=

n=0

x3n x2n=0

x3n

=

1

x2

1 x3 =1 + x

1 + x + x2 , (8.31)

so the sum of (8.30) is

S = f(1) =2

3. (8.32)

Thus the process of summation is not (infinitely) associative. In this case theShanks transformation does not work.

8.2.2 Borel Summation

Now we use the Euler representation of the Gamma function, or the factorial,

n! =0 dt t

n

et

. (8.33)

Then we formally interchange summation and integration:

S =n=0

an1

n!

0

dt tnet =

0

dt etn=0

1

n!ant

n, (8.34)

which defines the sum if

g(t) =n=0

1

n!ant

n (8.35)

exists.Thus for (8.22b),

g(t) =n=0

(1)n tnn!

= et, (8.36)

and so

S =

0

dt e2t =1

2, (8.37)


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which coincides with the result found by Euler summation. In general, Borelsummation is more powerful than Euler summation, but if both Euler and Borel

sums exist, they are equal.In fact, we can prove that any summation that is both

1. linear, meaning that if

n=0

an = A,n=0

bn = B, (8.38a)

thenn=0

(an + bn) = A + B, (8.38b)

and

2. satisfies n=0

an = a0 +n=1

an, (8.39)

is unique. In fact, from these two properties alone (which are satisfied by bothEuler and Borel summation) we can find the value of the sum. Thus for example,

1 1 + 1 1 + 1 1 + . . . = S = 1 (1 1 + 1 1 + 1 1 + . . .) = 1S, (8.40)implies S = 1/2. Slightly more complicated is

S = (1 + 0 1 + 1 + 0 1 + 1 + 0 1 + . . .)= 1 + (0 1 + 1 + 0 1 + 1 + 0 1 + . . .)= 1 + 0 + (1 + 1 + 0 1 + 1 + 0 1 + 1 + 0 . . .), (8.41)

where adding the three lines gives

3S = 2 + (0 + 0 + 0 + 0 + 0 + . . .) = 2, (8.42)

or S = 2/3 as before.But there are sums resistant to such schemes. An example is (8.22a), because

the above process leads to

S = 1 + (1 + 1 + 1 + . . .) = 1 + S, (8.43)

which is only satisfied by S = . Yet such a series can be summed.

8.2.3 Zeta-function Summation

Recall that the zeta function is defined by

(s) =n=1

1

ns, Re s > 1. (8.44)


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8.2. SUMMING DIVERGENT SERIES 85 Version of November 14, 2011

In fact, (s) exists for all s = 1, so we can use that function to define the sumalmost everywhere in the complex s plane. In particular, for s = 0:

1 + 1 + 1 + 1 + . . . = (0) = 12

. (8.45)

Even a more divergent sum can be evaluated this way:

n=1

n = (1) = 112

. (8.46)

Note the remarkable fact that these sums are not only finite, but negative, eventhough each term in the sum is positive!

8.2.4 Casimir Effect

Here we give a physical example of the utility of this last mode of summation.The physics is that of a pair of parallel metallic plates, separated by a distancea in the vacuum. Because the plates modify the properties of the vacuum, thereis a change in the zero-point energy of the electromagnetic field, which feels theplates because they are conductors. The result is an attraction between theplates, the famous Casimir effect, predicted by Casimir in 1948 (the same yearthat Schwinger discovered how to renormalize quantum electrodynamics), andnow verified by many experiments at the percent level. The zero-point energy(per unit area) of modes confined by the plane boundaries at z = 0 and z = ais

E =1

2

h =

hc

2

n=1

d2k

(2)2

k2 +

na

, (8.47)

where in the mode sum we have integrated over the two transverse wavenumberskx and ky, and summed over the discrete modes, which, say, must vanish at z = 0and a, that is, be given by an (unnormalized) mode function

(z) = sinn

az. (8.48)

Now we write the square root as integral, putting its argument in the exponen-tial:

k2 +n

a

2=

1

12

0

ds

ss1/2e(k

2+(n/a)2)s, (8.49)

and then interchange the two integrals:

E =

hc

2

n=1

0

ds

s3/2 e

(n/a)2s

dk

2 e

k2s2

1

2 . (8.50)Here we have recognized that the two-dimensional integral over k = (kx, ky) canbe broken into the product of two one-dimensional integrals because

e(k2x+k

2y)s = ek

2xsek

2ys. (8.51)


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These one-dimensional integrals are simply Gaussians, so the squared factor in(8.50) is simply 1/(4s). The remaining s-integral is again a gamma function:

E = hc163/2

n=1

0

ds

s5/2e(n/a)

2s

= hc163/2

3

2

n=0

na

3

= hc2

1440a3, (8.52)

where we have used the facts that

3

2

=

4

3

, (3) = 1

120, (8.53)

together with the zeta-function continuation embodied in Eq. (8.44) Whenmultiplied by 2, for the two polarization states of the photon, this is exactlyCasimirs result, which implies an attractive force per unit area between theplates,

P = a

E = hc2

240a4= 1.30 1027N m2/a4. (8.54)

8.3 Pade Approximants

Consider a partial Taylor sum,

TN+M(z) =N+M

n=0

anzn, (8.55)

which is an N + Mth degree polynomial. Write this in a rational form,

PNM(z) =

Nn=0 Anz

nMm=0 Bmz

m, (8.56)

which is called the [N, M]th Pade approximant. Here the coefficients are de-termined from the Taylor series coefficients as follows: We set B0 = 1, anddetermine the (N+ M+ 1) coefficients A0, A1, . . . , AN and B1, B2, . . . , BM byrequiring that when the rational function (8.56) be expanded in a Taylor seriesabout z = 0 the first N + M + 1 coefficients match those of the original Taylorexpansion (8.55).

Example

Consider the exponential function

ez = 1 + z +1

2z2 + . . . . (8.57)


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8.3. PADE APPROXIMANTS 87 Version of November 14, 2011

The [1, 1] Pade of this is of the form

P11 (z) = A0 + A1z1 + B1z, (8.58)

which, when expanded in a series about z = 0 reads

P11 (z) A0 + (A1 B1A0)z + (B21A0 A1B1)z2. (8.59)

Matching this with Eq. (8.57), we obtain the equations

A0 = 1, (8.60a)

A1 B1A0 = 1, (8.60b)B1(B1A0 A1) = 1

2, (8.60c)

so we learn immediately that

A0 = 1, B1 = 12

, A1 =1

2, (8.61)

so the [1, 1] Pade is

P11 (z) =1 + 12z

1 12z. (8.62)

How good is this? For example, at z = 1,

P11 (1) = 3, (8.63)

which is 10% larger than the exact answer e = 2.718281828 . . ., and is not quite

as good as the result obtained from the first three terms in the Taylor series,

1 + z +1

2z2z=1

= 2.5, (8.64)

about 8% low. However, in higher orders, Pade approximants rapidly outstripTaylor approximants. Table 8.1 compares the numerical accuracy of PMN withTN+M.

Note that typically the Pade approximant, obtained from a partial Taylorsum, is more accurate than the latter. This comes at a price, however; the Pade,being a rational expression, has poles, which are not present in the originalfunction. Thus, ez is an entire function, while the [1, 1] Pade approximant ofthis function has a pole at z = 2.

Example

Heres another example:

1

zlog(1 + z) = 1 z

2+

z2

3 z

3

4+

z4

5 z

5

6+ . . . . (8.65)


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TN+M(1) PNM(1) Relative error of Pade

T3(1) = 2.667 P12 (1) = 2.667 1.9%

T4(1) = 2.708 P22 (1) = 2.71429 0.15%

T5(1) = 2.717 P23 (1) = 2.71875 +0.017%

T6(1) = 2.71806 P33 (1) = 2.71831 +0.00103%

T7(1) = 2.71825 P34 (1) = 2.71827957 0.000083%

Table 8.1: Comparison of partial Taylor series with successive Pade approxi-mants for the exponential function, evaluated at z = 1. Note that precisely thesame data is incorporated in TN+M and in P

NM.

Approximant z = 0.5 z = 1 z = 2

Exact 0.810930216 0.69314718 0.549306P33 0.810930365 0.69315245 0.549403P34 0.810930203 0.69314642 0.549285

Table 8.2: Pade approximations for the function (1/z)log(1+ z) compared withthe exact values. Note that the Taylor series for this function has a radius ofconvergence of unity, yet the Pade approximations converge rapidly even beyondthe circle of convergence.

It is a simple algebraic task to expand the form of an [ N, M] Pade in a Taylorseries and compute the Pade coefficients by matching with the above. Thiscan, of course, be easily implemented in a symbolic program. For example, inMathematica,

PNM(z) = P adeApproximant[f[z], {z, 0, {N, M}}]. (8.66)

Doing so here yields

P33 (z) =1 + 1714z +

13z

2 + 1140z3

1 + 127 z +67z

2 + 435z3

. (8.67)

Table 8.2 shows representative numerical values for P33 and P34 . The Padeapproximants rapidly converge to the correct value even well beyond the circleof convergence of the original series. Note further in this example that

PNN is larger than the function, and decreases monotonically toward it,and


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8.3. PADE APPROXIMANTS 89 Version of November 14, 2011

PNN+1 is smaller than the function, and increases monotonically toward it.

This bounding behavior is typical of a class of functions. For more detail seeC. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientistsand Engineers (McGraw-Hill, New York, 1978), pp. 383ff.

Field Theory Examples

The following function occurs in the field theory of a massless particle in zerodimensions,

Z() =

dx

e(x2)1+

=2

1

(2 + 2)

0

dt

tt1/(2+2)et =

2

1

(2 + 2)

1

2 + 2

=2

3 + 2

2 + 2

, (8.68)

where the gamma function was defined by Euler as

(z) =

0

dt

ttz et, (8.69)

and satisfies the identity

(z + 1) = z(z). (8.70)

The gamma function generalizes the factorial to complex values:

(n + 1) = n!, n = 0, 1, 2, . . . . (8.71)

Because the gamma function (z) has poles when z = N, N = 0, 1, 2, .. . , thisfunction has an infinite number of singularities between = 3/2 and = 1.Thus the radius of convergence of the Taylor series about = 0 is 1. Yet loworder Pades for E() = log Z() give an excellent approximation well outsideof this radius, as Table 8.3 shows.

The partition function for a zero-dimensional field theory with a mass is given by the function

Z() =

2

0

dx e2

2x2(x2)1+ . (8.72)

We consider two cases. If 2

> 0, the power series in again has radius ofconvergence 1, but the Pade approximants are accurate far beyond this radius,as shown in Table 8.4.

If, on the other hand 2 < 0 (which corresponds to the Higgs mechanism inparticle physics), the Taylor series converges nowhere, yet the Pade approximantis still quite good, as seen in Table 8.5.


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T10() T20() P32 () P

54 () E()

2.0 1266.97 2.0 106 0.651267 0.692962 0.6931470.5 0.120055 0.120781 0.120831 0.12078223848 0.120782237640.5 0.00781712 0.00759091 0.00759097 0.0075905958951 0.00759059589491.0 0.367098 0.516940 0.0225167 0.022510401233 0.0225104012132.0 465.821 688611 0.0458145 0.04575620415 0.045756203495.0 5.5 106 7.8 1013 0.0786672 0.078172915 0.078172899

Table 8.3: Approximations to the function (8.68). What is approximated isE() = log Z(). The Pade approximants based on 6 and 10 terms in theTaylor series of this function are far more accurate that the 10 and 20 termtruncated Taylor series, and even are remarkably accurate far outside the circleof convergence, where the Taylor series is meaningless.

T8() P44 () Z()

0.5 1.04631 1.04630 1.046301.0 1.07719 1.07436 1.074362.0 1.81047 1.10647 1.106495.0 745.176 1.14253 1.14285

Table 8.4: Comparison of Z(), Eq. (8.72), 2 > 0, with the 8-term truncated

power series, and the corresponding [4, 4] Pade. Here we have taken 2 = 1, = 1.

T8() P44 () Z()

0.1 0.94808 0.94790 0.947900.5 137.697 0.88388 0.883811.0 40109.3 0.87323 0.872532.0 1.1 107 0.88334 0.879745.0 1.8

1010 0.91830 0.90517

Table 8.5: Comparison of Z(), Eq. (8.72), 2 < 0, with the 8-term truncatedpower series, and the corresponding [4, 4] Pade. Here we have taken 2 = 1, = 1.


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8.4. CONTINUED FRACTIONS 91 Version of November 14, 2011

8.4 Continued Fractions

8.4.1 Number Theory

The most familiar way of representing real numbers is in terms of a decimalfraction, which is nonterminating and nonrepeating if the number is irrational.However, there are other representations which, if less familiar, can be veryuseful. For example, the base of the natural logarithms e can be written in theform of a continued fraction,

e = 2 +1

1 + 12+ 1

1+ 1

1+ 14+...

. (8.73a)

Because this built-up form is cumbersome to write, we could write this as

e = 2 + 1/(1+ 1/(2+ 1/(1+1/(1+1/(4+ 1/(1+ 1/(1+1/(6+ 1/(1+1/(1 + . . . ,(8.73b)

or even more compactly as

e = 2 +1

1+

1

2+

1

1+

1

1+

1

4+

1

1+

1

1+

1

6 + . . .. (8.73c)

The form seen here is the representation of a real number x in the form

x = a0 +1

a1+

1

a2+

1

a3+

1

a4 + . . ., (8.74)

where the numbers an are integers called partial quotients. The rational numberformed by including only the first n + 1 partial quotients a0, a1, . . . , an is called

the n convergent of x. So the continued fraction is given by the set of ans:

e = {2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, . . .}, (8.75)

and the successive convergents, which rapidly approach e = 2.718281828 . . ., are

2, 3,

8

3,

11

4,

19

7,

87

32,

106

39,

193

71,

1264

465,

1457

536,

2721

1001,

23225

8544,

25946

9545,

49171

18089, . . .

= {2, 3, 2.666666667, 2.750000000, 2.714285714, 2.718750000, 2.717948718,2.718309859.2.718279570, 2.718283582, 2.718281718, 2.718281835,

2.718281823, 2.718281829, 2.718281828, . . .} . (8.76)The partial quotients of x are determined by successively determining the

unique integer that provides a bound for x for a given truncation of the partialfraction. Thus in the above example, where in each case 0 < r < 1,

2 < e, (8.77a)

5

2< 2 +

1

1 + r< 3, (8.77b)


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8

3< 2 +

1

1+

1

2 + r

2

or if

2

3

2must be supplied in any case.


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9.1. THE AIRY FUNCTION 99 Version of November 15, 2011

23

23

C

Figure 9.2: Deformed contour C which passes through the saddle point.

Note that in the perpendicular direction, along the real axis, the functionis a minimum at the stationary point. Thus, the stationary point is a saddlepoint, and this method is also referred to as the saddle point method.

The reason for choosing C to be the path of steepest descents is that, forlarge ||, most of the contribution comes from the immediate neighborhood ofthe saddle point. Then we can make use of the approximation above, so thatwe approximate the Airy function by

Ai() 12i

e2

33

2

C

d e2 , (9.8)

where the integral is just a Gaussian one,C

d e2 =

1

4

ii

du eu2

= i1

4

dt et2

= i1

4

. (9.9)

Thus we obtain the leading asymptotic behavior of the Airy function

Ai() 12

1

4 e2

3 32 , . (9.10)

This result is actually valid for complex values of subject to the restriction

| arg | < . (9.11)This asymptotic approximation is really quite good for modest as Fig. 9.3shows.

9.1.1 Asymptotic series

Let us calculate the corrections to this result. We return to Eq. (9.7) and keepthe next term in :

(z) = 23

3/2 + 1/22 13

3, (9.12)

which is exact in this case. Thus the Airy function is exactly represented by theintegral

Ai() =1

2i

ii

d e2

33/2e

1/22e1

33 . (9.13)


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100 Version of November 15, 2011CHAPTER 9. ASYMPTOTIC EXPANSIONS

0.0 1.0 2.0 3.0 4.0 5.0

x

0.00

0.10

0.20

0.30

0.40

0.50

Ai(x)

f(x)

r(x)

Figure 9.3: The Airy function Ai(x) compared with the asymptotic approxima-tion (9.10), denoted f(x), and the relative error of the latter, denoted r(x). Theerror is less than 10% even for x as small as 1.


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9.2. SYNCHROTRON RADIATION 101 Version of November 15, 2011

We approximate this by expanding the last exponential, since for large theintegrand is dominated by small . Expanding out to fourth order, and omitting

odd terms, we have after substituting = iu1/4:

Ai() 1/4

2e

2

33/2

du eu2

1 1

18

u6

3/2+

1

24

1

81

u12

3+ . . .

. (9.14)

The integrals may be evaluated starting from

du eu2

=

, (9.15)

so

du u2keu2

=

d

d

k

du eu2

=

(2k 1)!!

2k1

(2k+1)/2. (9.16)

Thus, the two leading corrections to the asymptotic expression for the Airyfunction given in Eq. (9.10) are

Ai() 12

1/4e

2

33/2

1 5

48

1

3/2+

385

4608

1

3+ . . .

, (9.17)

which is the beginning of an asymptotic series expansion in powers of 3/2.

9.2 Synchrotron Radiation

A charged particle moving in a circular orbit emits electromagnetic radiationcalled (for the machine in which such radiation was first observed) synchrotron

radiation. For details of the theory, see, for example, J. Schwinger, L. L. De-Raad, Jr., K. A. Milton, and W.-y. Tsai, Classical Electrodynamics (Perseus,1998), p. 401 ff. In particular, the power radiated in the mth harmonic of thefrequency of revolution of the charged particle moving in a circle with speedv = c is, in part, proportional to

J2m(2m) = 0

d

sin sin2m(sin ). (9.18)

In the ultrarelativistic limit when 1, most of the radiation occurs for largeharmonic numbers, m 1, and the main contribution comes from the regionnear = 0. Therefore, we may expand the integrand in Eq. (9.18) as follows:

sin sin2m(sin ) sin2m 3

3!

= sin

2m

(1 ) 1

63

sin

m

(1 2) + 1

33


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=

1 2 x sin

m(1 2)3/2

x +1

3x3

,

(9.19)

where we have introduced the change of scale

=

1 2 x. (9.20)As a result, in this limit, Eq. (9.18) can be approximated by2

J2m(2m) (1 2)

0

dx

x sin

m(1 2)3/2

x +

1

3x3

=(1 2)

Im

0

d x x eim(12)3/2(x+x3/3). (9.21)

For m fixed and approaching unity in such a way that m(1 2

)3/2

1, thesignificant contribution to Eq. (9.21) comes from the region where x is large,and Eq. (9.21) reduces to

J2m(2m) (1 2)0

dx

x sin

m3

(1 2)3/2x3

=

0

d

sin

m3

3

, (9.22)

where all reference to the speed of the particle has disappeared. By changingvariables, we may write this as

J2m(2m) Im0

d

eim

3/3

= Im

3

m

2/3ei/3

0

dt

1

3t2/3

t1/3et

= Im

3

m

2/3(2/3)

3ei/3

=31/6

2

(2/3)

m2/3, for m 1. (9.23)

In the above evaluation, we have used Cauchys theorem to perform a change ofcontour, as shown in Fig. 9.4, and have used the definition of the gamma function(8.69). Notice that Eq. (9.23) is valid for m either integer or half-integer.

However, for sufficiently large m, the parameter m(12)3/2 becomes large,

and the integrand in Eq. (9.21) undergoes rapid oscillations in x except nearthe stationary points, which satisfy

d

dx

x +

1

3x3

= 1 + x2 = 0; (9.24)

2Evidently, this integral is related to that defining the Airy function, Eq. (9.1).


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plane:

-

?

-

QQ

QQQk

PPPq /6

Figure 9.4: Change of contour used in evaluating Eq. (9.23).

i

= =

=

Complex x plane

Figure 9.5: Stationary phase contour for evaluation of (9.21).

that is, the stationary phase points are located at

x = i. (9.25)By extending the region of integration from

to +

, we evaluate Eq. (9.21)

asymptotically by following the standard procedure of the saddle point method(or the method of steepest descents). We deform the contour of integration sothat it passes through the stationary point x = i, because then the dominantcontribution comes from the vicinity of that point. (See Fig. 9.5.) In theneighborhood of x = i, we let

x = i + , (9.26)

where is real, to take advantage of the saddle point character. For arbitrary

x +1

3x3 = (i + ) +

1

3(i + )3 = i

2

3+ 2

+

1

33, (9.27)

so that for small , if we drop the cubic term in , the exponential factor inEq. (9.21) becomes

e

2

3m(12)3/2

em(12)3/22

, (9.28)which falls off exponentially on both sides of x = i. The resulting Gaussianintegral in (9.21) leads to the following asymptotic form:

J2m(2m) 1

2

(1 2)1/4m

e2

3m(12)3/2 , m(1 2)3/2 1. (9.29)


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Thus, for very large harmonic numbers, the power spectrum3 decreases expo-nentially in contrast to the behavior for smaller values of m where it increases

like m1/3. The transition between these two regimes occurs near the criticalharmonic number, mc , for which

mc(1 2)3/2 1, (9.31)

or

mc = (1 2)3/2 =

E

c2

3, (9.32)

which uses the relativistic connection between the energy and the rest mass, E = c2(1 2)1/2. The bulk of the radiation is emitted with harmonicnumbers near mc. The qualitative shape of the spectrum is shown in Fig. 9.6.

9.2.1 First correction

Corrections to the formula (9.29) may be computed by retaining the 3 term,but treating it as small, so the correction may be obtained by Taylor expandingthe exponential:

J2m(2m) 1 2

2Im e

2

3m(12)3/2

d em(12)3/22(i + )

1 + im(1 2)3/2 3

3 1

2m2(1 2)3

6

9+ . . .

=1 2

2e

2

3m(12)3/2(1 2)3/4m1/2

dt et2

1 + 13

t4

m(1 2)3/2 118 t6

m(1 2)3/2 + . . .

. (9.33)

Here we noted that the imaginary part only receives the contribution of theeven terms in , which are all that survive symmetric integration. Finally, theGaussian integrals are evaluated according to

dt t2n et2

=

0

dxx

xn ex =

n +

1

2

, (9.34)

where

5

2

=

3

4,

7

2

=

15

8. (9.35)

3The power radiated into the mth harmonic by a particle of charge e moving in a circle ofradius R with angular frequency 0 is given by

Pm =e2

Rm0

22J2m(2m) (1

2)

2m

0

dx J2m(x)

(9.30)

The two terms in the square brackets have similar asymptotic behavior.


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1 10 100 1000 10000

m

0.0

1.0

2.0

3.0

2mJ2

m(2m)

Figure 9.6: Sketch of power emitted into mth harmonic as a function of m.What is actually plotted is 2mJ2m(2m) for = 0.99. In this case mc = 356.


99/136


Thus

J2m(2m) = (1 2

)1/4

2

me 23m(12)3/2

1 +7

48

1

m(1 2)3/2 + O

1

m2(1 2)3

. (9.36)


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Chapter 10

Linear Operators,

Eigenvalues, and Greens

Operator

We begin with a reminder of facts which should be known from previous courses.

10.1 Inner Product Space

A vector space V is a collection of objects {x} for which addition is defined.That is, if x, y V, x + y V, which addition satisfies the usual commutativeand associative properties of addition:

x + y = y + x, x + (y + z) = (x + y) + z. (10.1)

There is a zero vector 0, with the property

0 + x = x + 0 = x, (10.2)

and the inverse of x, denoted x, has the property

x x x + (x) = 0. (10.3)

Vectors may be multiplied by complex numbers (scalars) in the usual way.That is, if is a complex number, and x V, then x V. Multiplication byscalars is distributive over addition:

(x + y) = x + y. (10.4)

Scalar multiplication is also associative: If and are two complex numbers,

(x) = ()x. (10.5)

107 Version of November 16, 2011


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108 Version of November 16, 2011 CHAPTER 10. LINEAR OPERATORS

An inner product space is a vector space possessing an inner product. If xand y are two vectors, the inner product

x, y (10.6)

is a complex number. The inner product has the following properties:

x, y + z = x, y + x, z, (10.7a)x + y,z = x, z + y, z, (10.7b)

x, y = y, x, (10.7c)x, x > 0 if x = 0, (10.7d)

where and are scalars. Because of the properties (10.7a) and (10.7b), wesay that the inner product is linear in the second factor and antilinear in the

first. Because of the last property (10.7d), we define thenorm

of the vector by

x =

x, x. (10.8)

10.2 The Cauchy-Schwarz Inequality

An important result is the Cauchy-Schwarz inequality,1 which has an obviousmeaning for, say, three-dimensional vectors. It reads, for any two vectors x andy

|x, y| xy, (10.9)where equality holds if and only if x and y are linearly dependent.

Proof: For arbitrary we have

0 x y,x y = x2 x, y y, x + ||2y2. (10.10)

Because the inequality is trivial if y = 0, we may assume y = 0, and so we maychoose

=y, xy2 . (10.11)

The the inequality (10.10) read

0 x2 2y2 |x, y|2 +

|y, x|2y2

=

x

2

|x, y|2

y2

, (10.12)

from which Eq. (10.9) follows. Evidently inequality holds in Eq. (10.10) unless

x = y. (10.13)

1The name Bunyakovskii should also be added.


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10.3. HILBERT SPACE 109 Version of November 16, 2011

From the Cauchy-Schwarz inequality, the triangle inequality follows:

x + y x + y. (10.14)

Proof:

x + y2 = x + y, x + y= x2 + y2 + 2Re x, y x2 + y2 + 2|x, y| x2 + y2 + 2xy = (x + y)2. (10.15)

QED

10.3 Hilbert Space

A Hilbert space H is an inner product space that is complete. Recall fromChapter 2 that a complete space is one in which any Cauchy sequence of vectorshas a limit in the space. That is, if we have a Cauchy sequence of vectors, i.e.,for any > 0,

{xn}n=1 : xn xm < n,m > N(), (10.16)

then the sequence has a limit in H, that is, there is an x H for which for any > 0 there is an N() so large that

x xn < n > N(). (10.17)We will mostly be talking about Hilbert spaces in the following.

Suppose we have a countable set of orthonormal vectors {ei}, i = 1, 2, . . .,in H. Orthonormality means

ei, ej = ij. (10.18)

The set is said to be complete if any vector x in H can be expanded in terms ofthe eis:

2

x =i=1

ei, xei. (10.19)

Here convergence is defined in the sense of the norm as described above. Geo-metrically, the inner product ei, x is a kind of direction cosine of the vector x,or a projection of the vector x on the basis vector ei.

2If the space is finite dimensional, then the sum runs up to the dimensionality of the space.


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Example

Consider the space of all functions that are square integrable on the closedinterval [, ]:

|f(x)|2 dx < . (10.20)

The functions (not the values of the functions) are the vectors in the space, andthe inner product is defined by

f, g =

f(x)g(x) dx. (10.21)

It is evident that this definition of the inner product satisfies all the properties(10.7a)(10.7d). This space, called L2(, ), is in fact a Hilbert space. Acomplete set of orthonormal vectors is

{fn} : fn(x) = 12

einx, n = 0, 1, 2, . . . . (10.22)

whose inner products satisfy

fn, fm = n,m. (10.23)

The expansion

f =

n=

fn, ffn (10.24)

is the Fourier expansion of f:

fn, f = 12

f(x)einx dx = an, (10.25)

where in terms of the Fourier coefficient an

f(x) =

n=

an12

einx. (10.26)

This Fourier series does not, in general, converge pointwise, but it does convergein the mean:

f(x) 1

2

N

n=N

aneinx

0 as N

0, (10.27)

that is,

limN

dx

f(x) 12

Nn=N

an einx

2

= 0. (10.28)


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10.4. LINEAR OPERATORS 111 Version of November 16, 2011

10.4 Linear Operators

A linear operator T on a vector space V is a rule assigning to each f V aunique vector T f V. It has the linearity property,

T(f + g) = T f + Tg, (10.29)

where , are scalars. In an inner product space, the adjoint (or Hermitianconjugate) of T is defined by

f , T g = Tf, g, f, g V. (10.30)

T is self-adjoint (or Hermitian) if

T = T. (10.31)

10.4.1 Sturm-Liouville Problem

Consider the space of twice continuously differentiable real functions defined ona segment of the real line

x0 x x1, (10.32)an incomplete subset of the Hilbert space L2(x0, x1). Under what conditions isthe differential operator

L = p(x)d2

dx2+ q(x)

d

dx+ r(x), (10.33)

where p, q, and r are real functions, self-adjoint?

Let u, v be functions in the space. In terms of the L2 inner product

u,Lv =x1x0

dx u(x)Lv(x)

=

x1x0

dx u(x)

p(x)

d2

dx2v(x) + q(x)

d

dxv(x) + r(x)v(x)

= u(x)p(x)v(x)

x1

x0

x1x0

dx [u(x)p(x)]

v(x)

+ u(x)q(x)v(x)

x1

x0

x1x0

dx [u(x)q(x)]

v(x)

+ x1

x0

dx u(x)r(x)v(x)

=

u(x)p(x)v(x) + u(x)q(x)v(x) [u(x)p(x)] v(x)x1

x0

+

x1x0

dx

[u(x)p(x)]

v(x) [u(x)q(x)] v(x) + u(x)r(x)v(x)


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= {p(x) [u(x)v(x) u(x)v(x)] + [q(x) p(x)] u(x)v(x)}x1

x0

+

x1

x0

dx

p(x)u(x) + [2p(x) q(x)] u(x)

+ [p(x) q(x) + r(x)] u(x)v(x). (10.34)The last integral here equals, for all v, v,

Lu,v =x1x0

dx [Lu(x)] v(x) (10.35)

if and only if2p q = q, p q + r = r, (10.36)

which imply the single condition

p(x) = q(x). (10.37)

If this condition holds for all x in the interval [x0, x1], the integrated term is

p(x) [u(x)v(x) u(x)v(x)]x1

x0

. (10.38)

Only if this is zero is L Hermitian:

u,Lv = Lu,v. (10.39)The vanishing of the integrated term may be achieved in various ways:

1. The function p may vanish at both boundaries:

p(x0) = p(x1) = 0, and u, v bounded for x = x0, x1. (10.40)

Thus, for example, the Legendre differential operator

(1 x2) d2

dx2 2x d

dx(10.41)

is self-adjoint on the interval [1, 1].2. The functions in the space satisfy homogeneous boundary conditions:

(a) The functions vanish at the boundaries,

u(x0) = u(x1) = 0, v(x0) = v(x1) = 0. (10.42)

These are called homogeneous Dirichlet boundary conditions.(b) The derivatives of the functions vanish at the boundaries,

u(x0) = u(x1) = 0, v

(x0) = v(x1) = 0. (10.43)

These are called homogeneous Neumann boundary conditions.


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10.5. EIGENVECTORS 113 Version of November 16, 2011

(c) Homogeneous mixed boundary conditions are a linear combinationof these conditions,

u(x0) + (x0)u(x0) = 0, (10.44a)

u(x1) + (x1)u(x1) = 0, (10.44b)

where is some function, the same for all functions u in the space.

3. A third possibility is that the solutions may satisfy periodic boundaryconditions,

u(x0) = u(x1) and u(x0) = u

(x1). (10.45)

This only works when the function p is also periodic,

p(x0) = p(x1). (10.46)

Conditions such as the above, which insure the vanishing of the integratedterm (or, in higher dimensions, surface terms) are called self-adjoint boundaryconditions. When they hold true, the differential equation

d

dx

p(x)

d

dxu(x)

+ r(x)u(x) = 0 (10.47)

is self-adjoint. This equation is called the Sturm-Liouville equation.

10.5 Eigenvectors

If T is a (linear) operator and f = 0 is a vector such that

T f = f, (10.48)

where is a complex number, then we say that f is a eigenvector (characteristicvector) belonging to the operator T, and is the corresponding eigenvalue.

The following theorem is most important. The eigenvalues of a Hermitianoperator are real, and the eigenvectors belonging to distinct eigenvalues are or-

thogonal. The proof is quite simple. If

T f = f, T g = g, (10.49)

theng , T f = g, f = T g , f = g, f. (10.50)

Thus if g and f are the same, we conclude that

= , (10.51)

i.e., the eigenvalue is real, while then if = , we must haveg, f = 0. (10.52)


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10.5.1 Bessel Functions

The Bessel operator isB =

d2

dx2+

1

x

d

dx

2

x2. (10.53)

where is a real number. This is Hermitian in the space of real functionssatisfying homogeneous boundary conditions (Dirichlet, Neumann, or mixed),where the inner product is defined by

u, v =ba

x d x u(x)v(x). (10.54)

Proof: Note that

xB =d

dxx

d

dx

2

x(10.55)

is of the Sturm-Liouville form, (10.47), with p(x) = x, which is Hermitian withthe L2(a, b) inner product. Then

u, Bv =ba

dx u(x)xBv(x) =

ba

dxxBu(x)v(x) = Bu, v. (10.56)

When a = 0, the lower limit of the integrated term is zero automatically ifthe functions are finite at x = 0See Eq. (10.38). Suppose we demand thatDirichlet conditions hold at x = b, i.e., that the functions must vanish there.Then we seek solutions to the following Hermitian eigenvalue problem,

Bn = nn, (10.57)

with the boundary conditions

n(b) = 0, n(0) = finite. (10.58)

Here n enumerates the eigenvalues. The solutions to this problem are the Besselfunctions, which satisfy the differential equation

d2

dz2+

1

z

d

dz+ 1

2

z2

J(z), (10.59)

which are finite at the origin, z = 0.3 This is the same as the eigenvalue equation(10.57) provided we change the variable z =

nx. That is,

n(x) = J(nx). (10.60)

The solutions we seek are Bessel functions of a real variable, so the acceptableeigenvalues satisfy

n < 0, (10.61)

3The second solution to Eq. (10.59), the so-called Neumann function N(z) [it is alsodenoted by Y(z) and is more properly attributed to Weber], is not regular at the origin.


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10.6. DUAL VECTORS. DIRAC NOTATION115 Version of November 16, 2011

so we write

n = k

2

n. (10.62)

Finally, we impose the boundary condition at x = b:

0 = n(b) = J(knb), (10.63)

that is, knb must be a zero ofJ. There are an infinite number of such zeros, asFig. 10.1 illustrates. Let the nth zero of J be denoted by n, n = 1, 2, 3, . . ..For example, the first three zeros of J0 are

01 = 2.404826, 02 = 5.520078, 03 = 8.653728, (10.64)

while the first three zeros of J1 (other than 0) are

11 = 3.83171, 12 = 7.01559, 13 = 10.17347. (10.65)

Then the eigenvalues of the Bessel operator are

n = n

b

2, (10.66)

and the eigenfunctions are

J

n

x

b

. (10.67)

Because of the Hermiticity of B, these have the following orthogonality prop-erty, from Eq. (10.52),

b0

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