Physics 5013 Mathematical Methods of Physics

Chapter 1

Review of Complex Numbers Complex numbers are defined in terms of the imaginary unit, i, having the property 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 of z,” and Im z is the “imaginary part of z.” Complex numbers are added and multiplied just like real numbers: If z1 = x1 + iy1 , z2 = x2 + iy2 ,

(1.4a) (1.4b)

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

(1.5a) 2

z1 z2 = x1 x2 + iy1 x2 + ix1 y2 + i y1 y2 = x1 x2 − y1 y2 + i(x1 y2 + x2 y1 ).

(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 Version of August 22, 2011

(1.6)

2 Version of August 22, 2011CHAPTER 1. REVIEW OF COMPLEX NUMBERS y axis Imaginary axis 6

y

z-plane

 r θ x

x axis Real axis

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 z + z∗ , 2 z − z∗ Im z = . 2i Re z =

(1.7a) (1.7b)

Note also that zz ∗ = x2 + y 2

(1.8)

is purely real and non-negative, so we define the modulus, or magnitude, or absolute value of z by p √ (1.9) |z| = zz ∗ = (Re z)2 + (Im z)2 ,

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 of the vector is the magnitude of the complex number, r = |z|,

(1.10)

and the angle the vector makes with the real axis is θ, where tan θ = y/x;

(1.11)

the quadrant θ lies in is determined by the sign of x and y. We call θ = arg z

(1.12)

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

3 Version of August 22, 2011 z-plane y axis 

6

z = x + iy

r θ @−θ x axis @ @ r @ R @ z ∗ = x − iy Figure 1.2: Geometrical interpretation of complex conjugation. There is an arbitrariness in the choice of the argument θ of a complex number z, for one can always add an arbitrary multiple of 2π to θ without changing z, θ → θ + 2πn,

n an integer,

z → z.

(1.13)

It is often convenient to define a single-valued argument function arg z. By convention, the principal value of arg z is that phase angle which satisfies the inequality −π < arg z ≤ π. (1.14) (Note that radian measure is always employed.) For every z there is a unique arg 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” of a complex number, z = r cos θ + ir sin θ = r(cos θ + i sin θ),

(1.15)

so if we have two complex numbers, z1 = r1 (cos θ1 + i sin θ1 ), z2 = r2 (cos θ2 + i sin θ2 ),

(1.16a) (1.16b)

the product is z1 z2 = r1 r2 {cos θ1 cos θ2 − sin θ1 sin θ2 + i [cos θ1 sin θ2 + cos θ2 sin θ1 ]} = r1 r2 [cos(θ1 + θ2 ) + i sin(θ1 + θ2 )] .

(1.17)

4 Version of August 22, 2011CHAPTER 1. REVIEW OF COMPLEX NUMBERS That is, the moduli of the complex numbers multiply, |z1 z2 | = |z1 ||z2 |,

(1.18a)

arg(z1 z2 ) = arg z1 + arg z2 .

(1.18b)

while the arguments add,

The latter statement is to be understood as modulo 2π, i.e., equality up to the addition of an arbitrary integer multiple of 2π. In particular, note that 1 1 z = |z| = 1, (1.19a) z z while

0 = arg



1 z z



= arg

1 + arg z, z

(1.19b)

implying that 1 = 1 , z |z| 1 arg = −arg z. z

1.1

(1.20a) (1.20b)

De Moivre’s Theorem

From the above, if we choose a unit vector, z = cos θ + i sin θ,

(1.21)

successive powers follow a simple pattern: z 2 = cos 2θ + i sin 2θ, 3

(1.22a)

z = cos 3θ + i sin 3θ, ...,

(1.22b)

z n = cos nθ + i sin nθ,

(1.22c)

(cos θ + i sin θ)n = cos nθ + i sin nθ,

(1.23)

or where n is a positive integer. This is called De Moivre’s theorem.

1.2

Roots

Suppose we wish to find all the nth roots of unity, that is, all solutions to the equation z n = 1, (1.24)

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1.2. ROOTS

−1+i √ 2

• i

•

•

1+i √ 2

−1 • −1−i √ 2

1 • •

• −i

•

1−i √ 2

Figure 1.3: The eight 8th roots of unity. where n is a positive integer. If we take the polar form, z = ρ(cos φ + i sin φ),

(1.25)

ρn (cos nφ + i sin nφ) = 1,

(1.26)

this means which implies ρ = 1,

(1.27a)

nφ = 2πk,

(1.27b)

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

2πk 2πk + i sin . n n

(1.28)

These are distinct for k = 0, 1, 2, . . . , n − 1;

(1.29)

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

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 the sequence, zk , in the set {zn }∞ n=1 . Mathematically, a sequence is a complexvalued function defined on the positive integers. We say that the sequence possesses a limit l, lim zn = l,

n→∞

or zn → l

as

n → ∞,

(2.2)

if, for every ǫ > 0, no matter how small, there exists a number N for which |zn − l| < ǫ

for all n > N.

(2.3)

(The number N will depend on ǫ.) That is, {zn }∞ n=N +1 all lie within a circle of radius ǫ centered on the point l in the complex plane. A necessary and sufficient condition for a sequence {zn }∞ n=1 to converge to a limit is Cauchy’s criterion: A sequence {zn }∞ possesses a limit if and only n=1 if for every ǫ > 0, no matter how small, it is possible to find a number N such that |zn − zm | < ǫ for all n, m > N. (2.4)

(Note that the difference |n − m| may be arbitrarily large.) Thus, all elements of the sequence {zn }∞ n=N +1 lie within a disk of radius ǫ. Briefly, we say that the Cauchy condition is |zn − zm | → 0

for all n, m

sufficiently large.

(2.5)

Sequences having this property are called Cauchy sequences. Every Cauchy sequence of complex numbers possesses a limit (which is, of course, a complex number)—this property means that the complex numbers form a complete space. 7 Version of August 27, 2011

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2.2

CHAPTER 2. INFINITE SERIES

Series

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

n X

ak ,

n = 1, 2, 3, . . . .

(2.6)

k=1

The set of all these sums, {sn }∞ n=1 , itself forms a sequence. If this latter sequence has a limit S, sn → S as n → ∞, (2.7)

then we say that the infinite series ∞ X

ak = lim

n→∞

k=1

n X

ak

(2.8)

k=1

possesses the limit S (or converges to S), ∞ X

ak = S.

(2.9)

k=1

By the Cauchy criterion, this will be true if and only if n X ak < ǫ

(2.10)

k=m

for any fixed ǫ > 0, whenever n ≥ m > N, N a number depending on ǫ. Obviously, a necessary condition for ∞ X

ak

(2.11)

k=1

to converge is for ak → 0 as k → ∞. However, this is not sufficient, as the following example shows.

2.3 2.3.1

Examples Harmonic series

Consider the sum of the reciprocals of the integers, ∞ X 1 1 1 . 1 + + + ... = 2 3 n n=1

(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   1 1 1 1 + ...+ >n (2.13) = , an+1 + . . . + a2n = n+1 2n 2n 2 no matter how large n is. This violates Cauchy’s criterion, so the harmonic series diverges.

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

2.3.2

Geometric series

Consider the series

∞ X

arm ,

(2.14)

m=0

where a is a constant and r ≥ 0. For r 6= 1, the nth partial sum is sn =

n X

arm = a

m=0

so S = lim sn = n→∞

while the series diverges if r ≥ 1.

2.4

a 1−r

1 − rn+1 , 1−r if r < 1,

(2.15)

(2.16)

Absolute and Conditional Convergence

P∞ P∞ Suppose we have a convergent series n=1 an . If also n=1 |an | converges, we say that the original series converges absolutely. Otherwise, the original series is conditionally convergent. (That is, it converges because of sign alternations.) A sufficient condition for (at least) conditional convergence is provided by the following 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 sufficiently large n, then ∞ X an converges. (2.17) n=1

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

2.4.1

Example

Consider the conditionally convergent series formed from the divergent harmonic series by alternating every other sign: 1−

1 1 1 1 1 + − + − + . . . = ln 2, 2 3 4 5 6

(2.18)

which converges to the natural logarithm of 2. Multiply this equation term by term by 1/2: 1 1 1 1 1 1 − + − + − . . . = ln 2. (2.19) 2 4 6 8 10 2

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CHAPTER 2. INFINITE SERIES

Add these two series: 1+

1 1 3 1 1 1 1 1 1 − + + − + + − + . . . = ln 2. 3 2 5 7 4 9 11 6 2

(2.20)

Since the reciprocal of each integer occurs exactly once in the last series, we would be tempted to rearrange the series to obtain 1−

1 1 1 1 1 + − + − + . . . = ln 2, 2 3 4 5 6

(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 further out 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 their value, but they can be multiplied together term by term: If two series ∞ X

S =

ui ,

(2.22a)

vi

(2.22b)

ui vj

(2.23)

i=1 ∞ X

T =

i=1

are both absolutely convergent, the series P =

∞ X i=1 j=1

formed from the product of their terms written in any order, is absolutely convergent, and has a value equal to the product of of the individual series, P = ST.

2.5

(2.24)

Convergence Tests

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

2.5.1

Comparison test

If bn > 0 for all n and

P∞

n=1 bn

∞ X

n=1

an

is convergent, and if |an | ≤ bn for all n, then is absolutely convergent.

(2.25a)

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2.5. CONVERGENCE TESTS Also, if |an | ≥ bn > 0 for all n, and ∞ X

an

P∞

n=1 bn

diverges, then

is not absolutely convergent.

(2.25b)

n=1

2.5.2

Root test

The series

P∞

n=1

an converges absolutely if from a certain term onward p n |an | ≤ q < 1, (2.26)

where q ≥ 0 is independent of n. P∞ n Proof: If the inequality holds, |an | ≤ qP . But n=1 q n 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

P∞

n=1

an converges absolutely if from a certain term onward an+1 (2.27) an ≤ q < 1,

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 term for which the inequality (2.27) holds. Then an an an−1 an−2 a2 n−1 = . (2.28) a1 an−1 an−2 an−3 · · · a1 ≤ q

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

P∞

n=1

nq n converge? If we use the root test, we examine p √ lim n |an | = |q| lim n n = |q|, 1 n→∞

n→∞

while if we use the ratio test, we look at an+1 = |q| lim n + 1 = |q|. lim n→∞ n→∞ an n

(2.29a)

(2.29b)

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

ln

√ n

n=

1 n

ln n, which tends to zero as n → ∞,

√ n

n → 1.

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CHAPTER 2. INFINITE SERIES

The following are refinements of the ratio test, which fails (that is, fails to reveal whether the tested series converges or not) when an+1 = 1. (2.30) lim n→∞ an

For example, this indeterminate limit results for the case an = 1/n, which yields a divergent series, but also for an = 1/(n ln2 n), which corresponds to a convergent sum (see Sec. 2.5.8).

2.5.4

Kummer’s test

Choose a sequence of positive constants bn . If an − bn+1 ≥ C > 0, bn an+1

(2.31)

∞ X

(2.32)

for all n ≥ N , where N and C are fixed numbers, then an

converges absolutely.

n=1

On the other hand, if an − bn+1 ≤ 0, bn an+1

and

(2.33)

∞ X

b−1 n

diverges,

(2.34)

∞ X

|an | diverges.

(2.35)

n=1

then

n=1

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 X

l=N +1

|al | ≤

bN |aN | bN |aN | bn |an | − ≤ . C C C

(2.37)

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

n X i=1

|ai | ≤

N X i=1

|ai | +

bN |aN | . C

(2.38)

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2.5. CONVERGENCE TESTS

The right-hand side of this inequality is a constant, independent of n. Therefore, the positive sequence of increasing terms {sn } is bounded above, and consequently possesses a limit. The series is absolutely convergent. If the inequality (2.33) holds, |aN |bN , n > N, bn P∞ diverges, so does n=1 |an |. |an | ≥

so since

2.5.5

P∞

−1 n=1 bn

(2.39)

Raabe’s test

Raabe’s criterion for absolute convergence is   an − 1 ≥ K > 1, n an+1

for all n ≥ N , where N and K are fixed. And if   an n − 1 ≤ 1, an+1 then

∞ X

n=1

|an | diverges.

(2.40)

(2.41)

(2.42)

Proof: In Kummer’s test put bn = n.

2.5.6 If

Gauss’ test an h B(n) an+1 = 1 + n + n2 ,

(2.43)

where P∞ 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 6= 1 we can use Raabe’s test:   h B(n) lim n = h. (2.44) + n→∞ n n2 For h = 1, Raabe’s test is indeterminate. In that case use Kummer’s test with bn = n ln n: for large n,   h B(n) − (n + 1) ln(n + 1) n ln n 1 + + n n2     1 h B(n) − (n + 1) ln n + ≈ n ln n 1 + + n n2 n   B(n) ≈ h+ ln n − ln n − 1 n ≈ (h − 1) ln n − 1 < 0, if h ≤ 1. (2.45)

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CHAPTER 2. INFINITE SERIES

f

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

∞ X

1 diverges n ln n n=2 P (see homework), the series ∞ n=1 |an | diverges.

2.5.7

(2.46)

Integral test

If f (x) is a continuous, monotonically decreasing real function of x such that f (n) = |an |, then

∞ X

n=1

|an | converges if

∞

Z

1

and diverges otherwise. Proof: It is geometrically obvious that Z ∞ Z ∞ X dx f (x) < f (n) < 1

n=1

(2.47)

dx f (x) < ∞,

(2.48)

dx f (x) + f (1),

(2.49)

∞

1

for this follows merely from the geometrical meaning of the integral as the area under the curve of the function. See Fig. 2.1.

2.5.8

Examples

• The Riemann zeta function is defined by the series ∞ X 1 = ζ(α). α n n=1

We can test for convergence using Gauss’ test, by examining α  α n+1 ≈1+ for large n. n n Thus the series converges if α > 1, and diverges if α ≤ 1.

(2.50)

(2.51)

15 Version of August 27, 2011

2.6. SERIES OF FUNCTIONS • Consider the series

∞ X

1 . (ln n)α n=1

(2.52)

Let’s use Raabe’s test: α  α  α ln n + ln(1 + 1/n) ln(n + 1) = ≈1+ . ln n ln n n ln n Because n

 α  α = →0 n ln n ln n

as n → ∞,

(2.53)

(2.54)

we conclude that the series is divergent. • To test for convergence of ∞ X

1 , n(ln n)α n=2

(2.55)

let us use the integral test: Z

∞

2

dx = x(ln x)α =

=

Z

∞

 

∞ 1 1 , 1−α (ln x)α−1 ∞ x=2 ln(ln x) x=2 ,

ln 2

 

d(ln x) (ln x)α α 6= 1, α=1

finite α > 1, ∞ α ≤ 1.

(2.56)

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

2.6 2.6.1

Series of Functions 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 ǫ > 0 we may find a δ > 0 such that |f (z) − f (z0 )| < ǫ

whenever |z − z0 | < δ.

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

(2.58)

16 Version of August 27, 2011

CHAPTER 2. INFINITE SERIES ↓

2ǫ

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

2.6.2

Uniform Convergence

Consider the infinite series f (z) =

∞ X

gi (z)

(2.59)

i=1

constructed from the sequence of functions {gi }∞ i=1 . The condition that this series converge is expressed in terms of the partial sums, fn (z) =

n X

gi (z)

(2.60)

i=1

thusly: given ǫ > 0 we can find an integer N so that for n > N |fn+p (z) − fn (z)| < ǫ for all p > 0.

(2.61)

This is Cauchy’s criterion. In general the N required for this to occur will depend on the point z. If, however, Eq. (2.61) holds for all z if n > N independent of z, 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)| < ǫ

for all n > N,

N

independent of z.

(2.62)

That is, the partial sum fn is everywhere uniformly close to f , the limiting function. This situation is illustrated in Fig. 2.2 for a real function of a real variable. Contrast absolute and uniform convergence through the following examples. The series ∞ X (−1)n (2.63) n + z2 n=1 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 +p X (−1)n 1 1 ≤ , (2.64) < n + z2 N + z2 N n=N +1

17 Version of August 27, 2011

2.6. SERIES OF FUNCTIONS which is the Cauchy criterion with ǫ = 1/N . In contrast, consider, for real z, the series S(z) =

∞ X

z2 (1 + z 2 )n n=0

(2.65)

which converges absolutely. For z = 0, S(0) = 0; and for z 6= 0, S(z) = z 2

∞ X

1 z2 = 1 + z2. = 1 2 n (1 + z ) 1 − 2 1+z n=0

(2.66)

Thus S(z) is discontinuous at z = 0. The following theorem shows that this series cannot be uniformly convergent there. Theorem If a series of continuous functions of z is uniformly convergent for all values of z in a given closed domain, the sum is continuous throughout the domain. Proof: Let n X gi (z). (2.67) fn (z) = i=1

Since fn (z) → f (z) uniformly,

(2.68)

we can find, for any ǫ > 0, a value of n such that |fn (z) − f (z)| < ǫ

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 fn ’s are continuous, we can find a δ for any given ǫ such that |fn (z) − fn (z ′ )| < ǫ whenever |z − z ′ | < δ.

(2.71)

|f (z) − f (z ′ )| < 3ǫ whenever |z − z ′ | < δ.

(2.72)

Therefore,

QED. Even if the limit function is continuous, convergence to it need not be uniform, as the following example shows:

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CHAPTER 2. INFINITE SERIES

f

@ @ @ @ @ 1/n 2/n x 0 Figure 2.3: Sketch of the function fn (x) given by Eq. 2.73). Example Consider the sequence of continuous functions,  nx, 0 ≤ x ≤ 1/n,  fn (x) = (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 function fn (x) is 1. On the other hand, for all x, lim fn (x) = 0,

(2.74)

n→∞

which is certainly a continuous limit function. But the convergence to this limit is not uniform, for there is always a point, x = 1/n, for which   0 − fn 1 = 1 (2.75) n

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) =

∞ X

gn (x).

(2.76)

n=1

1. If the gn are continuous, we can integrate term by term if formly convergent over the domain of integration: Z b ∞ Z b X dx gn (x). dx f (x) = a

n=1

∞ X

n=1

n gn

is uni-

(2.77)

a

d gn are continuous, and 2. If the gn and gn′ = dx gent, then we can differentiate term by term:

f ′ (x) =

P

gn′ (x).

P

′ n gn

is uniformly conver-

(2.78)

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2.7. POWER SERIES Condition for Uniform Convergence

The following condition is sufficient, but not necessary, to ensure that a series is uniformly convergent. P∞ If |gn (z)| < aP n , where {an } is a sequence of constants such that n=1 an ∞ converges, then n=1 gn (z) converges uniformly and absolutely. Proof: The hypothesis implies N +p N +p X X an , (2.79) gn (z) < n=N

n=N

PN +p

so that if N is chosen so that n=N an < ǫ, then N +p X gn (z) < ǫ ∀z.

(2.80)

n=N

2.7

Power Series

By a power series, we mean a series of the form, ∞ X

cn z n = c0 + c1 z + c2 z 2 + . . . ,

(2.81)

n=0

where the cn s form a sequence of complex constants, and z is a complex variable. If a power series converges for one point, z = z0 , it converges uniformly and absolutely for all z satisfying |z| ≤ η, (2.82) where η is any positive less than |z0 |. P∞ number n Proof: Since c z converges, it must be true that the terms are n 0 n=0 bounded, |cn z0n | < M, (2.83) where M is independent of n (but not of |z0 |). Hence if Eq. (2.82) is satisfied, n ∞ ∞ ∞  X X X η < ∞, (2.84) |cn z n | ≤ |cn |η n < M |z0 | n=0 n=0 n=0 since η/|z0 | < 1. This proves absolute convergence. Uniform convergence follows from the theorem above.

2.7.1

Radius of Convergence

Use the root test to determine where the power series ∞ X

n=0

cn z n

(2.85)

20 Version of August 27, 2011

CHAPTER 2. INFINITE SERIES z-plane '$  ρ &%

Figure 2.4: Circle of convergence of a power series. The series (2.85) converges inside the circle, and diverges outside. The radius of convergence ρ is given by Eq. (2.87). converges. That test says if p lim n |cn ||z| < 1, n→∞

the series converges,

(2.86a)

the series diverges.

(2.86b)

while if lim

n→∞

p n |cn ||z| > 1,

Therefore, the power series converges within a circle of convergence of radius ρ, the radius of convergence, where ρ=

1 limn→∞

p , n |cn |

(2.87)

and diverges outside that circle, as shown in Fig. 2.4. More detailed examination is required to determine whether or not the series converges on the circle of convergence.

2.7.2

Properties of Power Series Within the Circle of Convergence

1. The function defined by the power series is continuous. [This follows from the theorem in Sec. 2.6.2.] 2. It may be differentiated or integrated term by term. follows from P∞ [This n c z converges, so the theorem above, togther with the fact that if n n=0 P n−1 does ∞ nc z , by the ratio test, n n=0 (n + 1)cn+1 z n n + 1 cn+1 = |z|. (2.88) ncn z n−1 n cn Now if z lies within the circle of convergence, cn+1 |z| < 1. lim n→∞ cn

(2.89)

21 Version of August 27, 2011

2.7. POWER SERIES

Since limn→∞ (n + 1)/n = 1, convergence of the differentiated series is assured.] 3. Two such power series may be multiplied together term by term, within the smaller of the two circles of convergence. [This follows from the theorem in Sec. 2.4.2.] 4. The power series is unique. [It suffices to show that if f (z) =

∞ X

cn z n = 0

∀z,

n=0

cn = 0 ∀n.

(2.90)

Indeed, f (0) = c0 = 0,

(2.91a)

′

2.7.3

f (0) = c1 = 0, ...,

(2.91b)

f (n) (0) = n! cn = 0.]

(2.91c)

Taylor Expansion

The Taylor expansion for a real function of a real variable is obtained from the above argument. If we write a function as a power series, f (x) =

∞ X

cn xn ,

(2.92)

n=0

then

1 (n) f (0). (2.93) n! Hence, the power series is the Taylor series of the function it represents, cn =

∞ X 1 (n) f (x) = f (0)xn . n! n=0

2.7.4

(2.94)

Hypergeometric Function

The hypergeometric function F is defined by the power series F (a, b; c; z) =

∞ X

n=0

An z n =

∞ X (a)n (b)n z n . (c)n n! n=0

(2.95)

Here the coefficients are defined in terms of the Pochhammer symbol, (a)n = a(a + 1)(a + 2) · · · (a + n − 1) =

Γ(a + n) . Γ(a)

(2.96)

22 Version of August 27, 2011

CHAPTER 2. INFINITE SERIES

To determine convergence, we examine Γ(c + n + 1) (n + 1)! z n Γ(a + n)Γ(b + n) An z n = An+1 z n+1 Γ(c + n) Γ(a + n + 1)Γ(b + n + 1) n! z n+1

c 1 1+ n 1+ n 1

= 1 + na 1 + nb z    1 1 1 = 1 + (c + 1 − a − b) + O , (2.97) z n n2 where O(1/n2 ) means that the next term goes to zero as n → ∞ at least as fast as 1/n2 . According to the ratio test, the radius of convergence of this series is |z| = 1; that is, the series diverges for |z| > 1, and converges uniformly and absolutely for any z such that |z| ≤ η < 1. The remaining question is what happens on the circle of convergence, |z| = 1. According to Gauss’ test, the series is then absolutely convergent if c > a + b [if the constants are complex, if Re (c − a − b) > 0]. For the point z = 1 the series is certainly divergent if this condition is not satisfied; however, if −1 < Re (c − a − b) ≤ 0 the series is conditionally convergent on the unit circle except for the exceptional point z = 1. On the other hand, if Re (c − a − b) ≤ −1 the series is divergent on the unit circle because the terms in the series increase in magnitude.

Chapter 3

Elementary Transcendental Functions 3.1

Exponential Function

Define, for all complex z, the exponential function by exp(z) = ez =

∞ X 1 n z . n! n=0

(3.1)

By the ratio test, n! |z| |z| = →0 (n + 1)! n+1

∀z,

(3.2)

the series converges everywhere. By the theorem of the Sec. 2.7, that means that the series converges uniformily in any finite closed region. Note that the following property holds: exp(z1 + z2 ) =

∞ X 1 (z1 + z2 )n n! n=0

∞ X n X 1 n! z1m z2n−m n! m! (n − m)! n=0 m=0 ! ∞ ! ∞ X1 X 1 k l z z = k! 1 l! 2

=

l=0

k=0

= exp(z1 ) exp(z2 ).

(3.3)

n

(3.4)

Then, by induction

(ez ) = enz , where n is any positive integer. 23 Version of September 7, 2011

24 Version of September 7, 2011CHAPTER 3. ELEMENTARY TRANSCENDENTAL FUNCTIONS Hyperbolic and trigonometric functions are defined in terms of the exponential function: ez − e−z 2 eiz − e−iz sin z = 2i sinh z =

ez + e−z , 2 eiz + e−iz cos z = , 2

cosh z =

(3.5a) (3.5b)

so that i sin z = sinh iz, cos z = cosh iz,

(3.6a) (3.6b)

eiz = cos z + i sin z.

(3.7)

for all complex z. Note that

Therefore, the polar representation of a complex number, z = r(cos θ + i sin θ) = reiθ ,

(3.8)

becomes a most useful and compact representation. In particular, z n = rn einθ

(3.9)

cos nθ + i sin nθ = (cos θ + i sin θ)n .

(3.10)

implies De Moivre’s formula,

3.1.1

Definition of π

There exists a positive number π such that 1. eπi/2 = i,

and

(3.11a)

2. ez = 1 if and only if z = 2πin,

(3.11b)

where n is an integer. Hence exp(z) is periodic with period 2πi, exp(z + 2πi) = exp(z) exp(2πi) = exp(z).

(3.12)

25 Version of September 7, 2011 z 

3.2. THE NATURAL LOGARITHM iy

θ @ @ @ @ @ “cut” or “branch line”

x

Figure 3.1: Cut plane for defining the logarithm.

3.2

The Natural Logarithm

If z = reiθ , we define ln z ≡ log z ≡ ln r + iθ,

(3.13)

where ln r is defined as the inverse of the exponential function for real positive r, r = eln r . (3.14) Thus we have z = eζ

where

ζ = ln r + iθ = log z.

(3.15)

Recall that θ = arg z is a multivalued function, because θ is only defined up to an arbitrary multiple of 2π. [This is just the periodic property (3.12).] Recall further that we defined the principal value of the argument as that which satisfied −π < arg z ≤ π. (3.16) Correspondingly, we say that the single-valued logarithm function (also denoted log z) is defined in the cut plane shown in Fig. 3.1. In measuring θ from the +x axis, one is not allowed to cross the cut along the −x axis. (Where the cut is placed is an arbitrary convention.) The correspondingly defined single-valued functions arg z and log z = log |z| + i arg z, (3.17) or −π < Im log z ≤ π,

(3.18)

are also referred to as the principal values of the argument and logarithm, respectively. Now we define complex powers of complex numbers as follows: ζ z ≡ ez log ζ ,

(3.19)

where log ζ is defined in the cut plane. Then eξ


z

ξ

= ez log e = ez(Re ξ+iIm ξ) = eξz

(3.20)

26 Version of September 7, 2011CHAPTER 3. ELEMENTARY TRANSCENDENTAL FUNCTIONS when arg eξ = Im ξ

(3.21)

−π < Im ξ ≤ π.

(3.22)

log eξ = ξ + 2πin

(3.23)

−π < Im (ξ + 2πin) ≤ π,

(3.24)

lies between If this is not so, where n is so chosen that

and eξ For example,

√


z

= ez(ξ+2πin) . 1

z = z 1/2 = e 2 log z

(3.25) (3.26)

is defined as a single-valued function only in the cut plane −π < arg z ≤ π.

3.3

(3.27)

Inverse Hyperbolic and Trigonometric Functions

The inverse hyperbolic and trigonometric functions are defined in terms of the logarithm: i h (3.28a) arcsinh z = log z + (z 2 + 1)1/2 , i h (3.28b) arccosh z = log z + (z 2 − 1)1/2 , arctanh z =

1 1+z log , 2 1−z

(3.28c)

which are defined in the cut planes shown in Fig. 3.2. arcsin z = −i arcsinh iz i h = −i log iz + (1 − z 2 )1/2 ,

arccos z = −i arccosh z h i = −i log z + (z 2 − 1)1/2 ,

arctan z = −i arctanh iz 1 − iz i i+z i = log , = log 2 1 + iz 2 i−z

(3.29a) (3.29b)

(3.29c)

which are defined in the cut planes shown in Fig. 3.3. Note that the branch

3.3. INVERSE HYPERBOLIC AND TRIGONOMETRIC FUNCTIONS27 Version of September 7, 2011 arcsinh z:

iy •i •−i

iy

arccosh z: x

+1 •

iy

arctanh z: −1 •

+1 •

x

x

Figure 3.2: Cut planes for defining the inverse hyperbolic functions. The thick lines represent the cuts. iy iy arctan z: arcsin z: arccos z: •+i −1 •

+1 •

x

x •−i

Figure 3.3: Cut plane for defining the inverse trigonometric functions. lines (cuts) are chosen so as not to cross the region where both the range and the domain of the functions are real, because for real x, sin x,

cos x ∈ [−1, 1], tan x ∈ (−∞, ∞),

sinh x ∈ (−∞, ∞), cosh x ∈ [1, ∞),

tanh x ∈ [−1, 1].

(3.30a) (3.30b) (3.30c) (3.30d) (3.30e)

An alternative notation for the inverse functions is provided by the superscript −1, as for example, arcsinh z = sinh−1 z, (3.31) which does not mean 1/ sinh z.

Chapter 4

Bernoulli Polynomials 4.1

Bernoulli Numbers

The “generating function” for the Bernoulli numbers is ∞ X x Bn n = x . ex − 1 n=0 n!

(4.1)

That is, we are to expand the left-hand side of this equation in powers of x, i.e., a Taylor series about x = 0. The coefficient of xn in this expansion is Bn /n!. Note that we can write the left-hand side of this expression in an alternative form ex

x x
= x/2 x/2 −1 e e − e−x/2 x e−x/2 2 sinh x2 x cosh x2 − sinh x2 = 2 sinh x2 x x x = coth − . 2 2 2

=

(4.2)

Note that x2 coth x2 is an even function of x, while x2 is odd. Therefore we conclude that all but one of the Bernoulli numbers of odd order are zero: B1 = −

1 2

B2k+1 = 0,

(4.3a) k = 1, 2, 3, . . . .

(4.3b)

By writing x = iy and noting that coth

iy y = −i cot , 2 2

29 Version of September 16, 2011

(4.4)

30 Version of September 16, 2011CHAPTER 4. BERNOULLI POLYNOMIALS we conclude that ∞ X B2n (iy)2n iy y y iy coth = cot = , 2 2 2 2 n=0 (2n)!

(4.5)

∞ X y B2n 2n y cot = (−1)n y . 2 2 n=0 (2n)!

(4.6)

or

By straightforward expansion in powers of x we can read off the first few Bernoulli numbers: x x x cosh x2 coth = 2 2 2 sinh x2 ≈ ≈

= =





1 x 2 1 x 4 1 x 6 1 x 8 + 4! + 6! + 8! + ... x 1 + 2! 2 2 2 2



3 5 7 9 2 x + 1 x + 1 x + 1 x + 1 x + ... 5! 2 7! 2 9! 2   2 3! 2 1  x 4 1  x 6 1  x 8 1 x 2 + + + 1+ 2! 2 4! 2 6! 2 8! 2      1 x 2 1  x 4 1  x 6 1  x 8 × 1− + + + 3! 2 5! 2 7! 2 9! 2    2     2 4 6 1 x 1 x 1 x + + + 3! 2 5! 2 7! 2 3    4     1  x 4 1 x 2 1 x 2 + O(x10 ) + + − 3! 2 5! 2 3! 2 . . .        x4 1 x6 1 x8 1 x2 1 + − + + − + .... 1+ 2! 6 4! 30 6! 42 8! 30 (4.7)

So by comparison with Eq. (4.1) we find B0 = 1,

B2 =

1 , 6

B4 = −

1 , 30

B6 =

1 , 42

B8 = −

1 . 30

(4.8)

What is the radius of convergence of the series ∞

z z X B2n 2n =− + z ? z e −1 2 n=0 (2n)!

(4.9)

Recall that a power series converges everywhere within its circle of convergence, and diverges outside that circle. Since a uniformly convergent series must converge to a continuous function, the power series must converge to a well-behaved function within the circle of convergence. That is, the limit function must have a singularity somewhere on the circle of convergence, but must be singularityfree within the circle of convergence. The precise theorem, proved in Chapter

31 Version of September 16, 2011

4.1. BERNOULLI NUMBERS n

B2n

Asymptotic value

Relative error

0 1

1

−2

1 6 1 − 30 1 42 1 − 30 5 66 691 − 2730 7 6 − 3617 510 43867 798 − 174611 330

1 π2 − π34 45 2π 6 − 315 π8 14175 2π 10 − 467775 2π 12 42567525 4π 14 638512875 − π16 97692469875 2π 18 − 9280784638125 2π 20

300% 39%

2 3 4 5 6 7 8 9 10

7.6% 1.7% 0.41% 0.099% 0.025% 0.0061% 0.0015% 0.00038% 0.0000095%

Table 4.1: The Bernoulli numbers B2n for n from 0 to 10, compared with the asymptotic values (4.12). The last column shows the relative error of the asymptotic estimate. Note that the later rather rapidly approaches the true value.

5, is that the radius of convergence of a power series is the distance from the origin to the nearest singularity of the function the series represents. In this case, it is clear that the generating function is singular wherever ez = 1, except for z = 0. Thus the closest singularities to the real axis occur at ±2πi, so that the radius of convergence is 2π. On the other hand B2n |B2n | [2(n + 1)]! , (4.10) (2π) = ρ = lim = lim (2n + 2)(2n + 1) n→∞ (2n)! |B2(n+1) | n→∞ B2n+2 2

2

from which we can infer the fact that the Bernoulli numbers grow rapidly with n, |B2n | ∼

(2n)! , (2π)2n

n → ∞.

(4.11)

We cannot deduce the sign or overall constant from this analysis: The true asymptotic behavior of B2n is B2n ∼ 2(−1)n+1

(2n)! . (2π)2n

(4.12)

The table shows the relative accuracy of the asymptotic approximation (4.12).

32 Version of September 16, 2011CHAPTER 4. BERNOULLI POLYNOMIALS

4.2

Bernoulli Polynomials

The Bernoulli polynomials are defined by the generating function F (x, s) =

∞ X xn x exs = Bn (s) , x e − 1 n=0 n!

(4.13)

that is, according to Eq. (2.94), Bn (s) =



∂ ∂x

n

F (x, s)

.

(4.14)

x=0

From the properties of F (x, s) we can deduce all the properties of these polynomials: 1. Note that F (x, 0) =

∞ X xn x = B . n ex − 1 n=0 n!

(4.15)

Therefore, we conclude that the Bernoulli polynomials at zero are equal to the Bernoulli numbers, Bn (0) = Bn . (4.16) 2. Next we notice that F (x, 1) =

−x x x ex = −x = = F (−x, 0), x −x e −1 1−e e −1

(4.17)

so that by comparing corresponding terms in the generating function expansion, we find Bn (1) = (−1)n Bn (0) = (−1)n Bn .

(4.18)

3. If we differentiate the generating function with respect to its second argument, we obtain the relation ∞ X ∂ x2 exs xn F (x, s) = x = Bn′ (s) . ∂s e − 1 n=0 n!

(4.19)

∞ X xn+1 x2 exs = xF (x, s) = B , n ex − 1 n! n=0

(4.20)

But obviously

so equating coefficients of xn /n! we conclude that Bn′ (s) = nBn−1 (s).

(4.21)

4.3. EULER-MACLAURIN SUMMATION FORMULA33 Version of September 16, 2011 (Note that B0′ (s) = 0 is consistent with this if B−1 (s) is finite.) Again, by direct power series expansion of the generating function we can read off the first few Bernoulli polynomials: 1 + xs + 12 (xs)2 + 61 (xs)3 1 3 x x + 21 x2 + 3!  2    s s 1 1 1 + x2 + ..., − − + ≈ 1+x s− 2 2 2 6 4

F (x, s) ≈ x

(4.22)

from which we read off B0 (s) = 1,

(4.23a)

1 B1 (s) = s − , 2

(4.23b)

1 B2 (s) = s2 − s + . 6

(4.23c)

By keeping two more terms in the expansion we find 1 3 B3 (s) = s3 − s2 + s, 2 2 B4 (s) = s4 − 2s3 + s2 −

(4.23d) 1 . 30

(4.23e)

Note that the properties (4.16) and (4.18) are satisfied. Note further we can use the property (4.21) to derive higher Bernoulli polynomials from lower ones. Thus from Eq. (4.23c) we know that 1 B3′ (s) = 3s2 − 3s + . 2

(4.24)

The expression for B3 (s), (4.23d) is recovered, when it is recalled that B3 = 0.

4.3

Euler-Maclaurin Summation Formula

Using the above recursion relation (4.21) we can deduce a very important formula which allows a precise relation between a discrete sum and a continuous integral. First note that since B0 = B0 (s) = 1 we can write Z 1 Z 1 f (x)B0 (x) dx = f (x) dx, (4.25) 0

0

valid for any function f . But now we can integrate by parts using B1′ (x) = B0 (x) : Z

0

1

f (x) dx =

Z

0

1

f (x)B1′ (x) dx

(4.26)

34 Version of September 16, 2011CHAPTER 4. BERNOULLI POLYNOMIALS 1 = f (x)B1 (x)

−

x=0

=

1 [f (1) + f (0)] − 2

Z

1

Z

1

f ′ (x)B1 (x) dx

0

f ′ (x)B1 (x) dx.

(4.27)

0

Here, we have used the facts that 1 B1 (0) = B1 = − , 2 1 B1 (1) = −B1 = . 2

(4.28a) (4.28b)

Now we can continue integrating by parts by noting that B1 (x) =

1 ′ B (x), 2 2

(4.29)

so that Z

0

1

1 1 [f (1) + f (0)] − [f ′ (1)B2 (1) − f ′ (0)B2 (0)] 2 2 Z 1 1 ′′ + f (x)B2 (x) dx 2 0 1 1 = [f (1) + f (0)] − B2 [f ′ (1) − f ′ (0)] 2 2 Z 1 1 + f ′′ (x)B2 (x) dx. 2 0

f (x) dx =

(4.30)

A general pattern is emerging. Let us assume the following formula holds for some integer k (we have just proved it for k = 1): Z 1 1 f (x) dx = [f (1) + f (0)] 2 0 k i X B2m h (2m−1) − f (1) − f (2m−1) (0) (2m)! m=1 Z 1 1 f (2k) (x)B2k (x) dx. (4.31) + (2k)! 0 We shall then prove that the same formula holds for k → k + 1, thereby establishing this formula, the Euler-Maclaurin summation formula, for all k. We proceed as follows. Note that B2k (x) =

′ ′′ B2k+1 (x) B2k+2 (x) = , 2k + 1 (2k + 1)(2k + 2)

so that by integrating by parts, we rewrite the last term in Eq. (4.31) as Z 1 1 1 B ′′ (x) dx f (2k) (x) (2k)! 0 (2k + 1)(2k + 2) 2k+2

(4.32)

4.3. EULER-MACLAURIN SUMMATION FORMULA35 Version of September 16, 2011  1 ′ ′ f (2k) (1)B2k+2 (1) − f (2k) (0)B2k+2 (0) (2k + 2)!  Z 1 ′ − f (2k+1) (x)B2k+2 (x) dx 0  1 − f (2k+1) (1)B2k+2 (1) + f (2k+1) (0)B2k+2 (0) = (2k + 2)!  Z 1 (2k+2) + f (x)B2k+2 (x) dx , =

(4.33)

0

where we have noted that for k > 0 ′ B2k+2 (0) = (2k + 2)B2k+1 (0) = 0,

(4.34a)

′ B2k+2 (1) = (2k + 2)B2k+1 (1) = −(2k + 2)B2k+1 (0) = 0.

(4.34b)

Hence Z

1

f (x) dx =

0

1 [f (1) + f (0)] 2 k+1 X

i B2m h (2m−1) f (1) − f (2m−1) (0) (2m)! m=1 Z 1 1 f (2k+2) (x)B2k+2 (x) dx. + (2k + 2)! 0

−

(4.35)

This is exactly Eq. (4.31) with k replaced by k + 1; so since the formula is true for k = 1 it is true for all integers k ≥ 1. Notice that the last term in this formula, the remainder, can also be written in the form Z 1 1 f (2k+3) (x)B2k+3 (x) dx. (4.36) − (2k + 3)! 0 Now consider the integral (N a positive integer) Z

0

N

f (s) ds =

N −1 Z k+1 X k=0

k

f (s) ds =

N −1 Z 1 X k=0

f (k + t) dt,

(4.37)

0

where we have introduced a local variable t. For the latter integral, we can use the Euler-Maclaurin sum formula, which here reads Z 1 1 f (k + t) dt = [f (k + 1) + f (k)] 2 0 n i X B2m h (2m−1) − f (k + 1) − f (2m−1) (k) (2m)! m=1 Z 1 1 f (2n) (k + t)B2n (t) dt. (4.38) + (2n)! 0

36 Version of September 16, 2011CHAPTER 4. BERNOULLI POLYNOMIALS Now when we sum the first term here on the right-hand side over k we obtain N −1 X k=0

N

X 1 1 f (k) − [f (0) + f (N )], [f (k + 1) + f (k)] = 2 2

(4.39)

k=0

while the second term when summed on k involves N −1 h X

i f (2m−1) (k + 1) − f (2m−1) (k) = f (2m−1) (N ) − f (2m−1) (0).

k=0

(4.40)

Thus we find Z N N X 1 f (s) ds = f (k) − [f (0) + f (N )] 2 0 k=0

n X

h i 1 B2m f (2m−1) (N ) − f (2m−1) (0) (2m)! m=1 Z 1 NX −1 1 + f (2n) (t + k)B2n (t) dt. (2n)! 0

−

(4.41)

k=0

Equivalently, we can write this as a relation between a finite sum and an integral, with a remainder Rn : N X

k=0

f (k) =

Z

N

0

1 f (s) ds + [f (0) + f (N )] 2

+

n X

h i 1 B2m f (2m−1) (N ) − f (2m−1) (0) + Rn , (4.42) (2m)! m=1

where the remainder Rn = −

1 (2n)!

Z

0

−1 1N X

f (2n) (t + k)B2n (t) dt.

(4.43)

k=0

is often assumed to vanish as n → ∞. Note that the remainder can also be written as Z N 1 f (2n) (t)B2n (t − ⌊t⌋) dt, (4.44) Rn = − (2n)! 0 where ⌊t⌋ signifies the greatest integer less than or equal to t.

4.3.1

Examples

1. Use the Euler-Maclaurin formula to evaluate the sum N X

2πn cos = N n=0

Z

0

N

dn cos

PN

n=0

cos(2πn/N ).

2πn 1 + (1 + 1) + 0 = 1, N 2

(4.45)

4.3. EULER-MACLAURIN SUMMATION FORMULA37 Version of September 16, 2011 because f (2m−1) (0) = f (2m−1) (N ) = 0 and Z

N

0

N 2nπ = dn cos N 2π

Z

(4.46)

2π

dx cos x = 0.

(4.47)

0

Of course, the sum may be carried out directly, N X

cos

n=0

N  1 X  i2πn/N 2πn e + e−i2πn/N = N 2 0   1 1 − e2πi(N +1)/N 1 − e−2πi(N +1)/N = + 2 1 − e2πi/N 1 − e−2πi/N 1 = (1 + 1) = 1. 2

(4.48)

2. The following sum occurs, for example, in computing the vacuum energy in a cosmological model: ∞ X

(2l + 1)e−l(l+1)t .

(4.49)

l=0

How does this behave as t → 0? We will answer this question by using the Euler-Maclaurin formula assuming that the remainder Rn tends to zero as n → ∞. Thus we will write the limiting form of that sum formula as ∞ X

f (l) =

Z

0

l=0

∞

1 dl f (l) + [f (∞) + f (0)] 2 +

∞ i X B2k h (2k−1) f (∞) − f (2k−1) (0) . (2k)!

(4.50)

k=1

Here f (l) = (2l + 1)e−l(l+1)t ,

(4.51)

f (∞) = f (2k−1) (∞) = 0,

(4.52)

so that while a very simple calculation shows f (0) = 1, f ′ (0) = 2 − t, f ′′′ (0) = −12t + 12t2 − t3 , f (5) (0) = 120t2 − 180t3 + 30t4 − t5 , f (7) (0) = −1680t3 + 3360t4 − 840t5 + 56t6 − t7 , f (0) = O(t4 ), k ≥ 5. (2k−1)

(4.53a) (4.53b) (4.53c) (4.53d) (4.53e) (4.53f)

38 Version of September 16, 2011CHAPTER 4. BERNOULLI POLYNOMIALS Thus Eq. (4.50) yields ∞ X

(2l + 1)e

−l(l+1)t

=

Z

∞

dl (2l + 1)e−l(l+1)t +

0

l=0

1 2

B4 ′′′ B2 ′ f (0) − f (0) − . . . 2 4!   Z ∞ 1 1 1 1 (t − 2) = du e−u + + t 0 2 2 6   1 1 + − [12t + O(t2 )] + O(t2 ) 4! 30 1 1 t 4 2 1 3 = + + + t + t + .... (4.54) t 3 15 315 315 −

Here the integral was evaluated by making the substitution u = l(l + 1)t, du = (2l + 1)t dl, and in the last line we have displayed the next two terms in this asymptotic expansion for small t. 3. The Riemann zeta function (2.50) is defined by ζ(α) =

∞ X 1 , nα n=1

Re α > 1.

(4.55)

Suppose we approximate this by the first M terms in the sum occurring in the Euler-Maclaurin formula (4.42): ζ(α, M ) =

M 1 X B2m (2m−1) 1 + − f (1), α − 1 2 m=1 (2m)!

(4.56)

where f (n) = n−α , and the first two terms here come from the integral and the 12 f (1) terms in the EM formula. It is easy to see that f (2m−1) (1) = −

Γ(α + 2m − 1) . Γ(α)

(4.57)

Given the asymptotic behavior of the Bernoulli numbers in (4.12), it is apparent that the limit M → ∞ of ζ(α, M ) does not exist. This limit is an example of an asymptotic series. However, in Table 4.2 we compare the sum of the first N terms of the series in (4.55) with the first N terms in the series defined by (4.56), that is ζ(α, N ), for α = 3, where ζ(3) = 1.2020569. The original series converges monotonically to the correct limiting value, but not spectacularly fast. For N = 9 terms, the relative error is about −0.5%. The asymptotic series is divergent; however, the N = 1 term is in error by only 4%, and the average of the N = 1 and N = 2 is larger than the true value by only +0.5%. This illustrates a characteristic feature of asymptotic series: A few terms in the series approximates the function rather well, but as more and more terms are included the series deviates from the true value by an ever increasing amount.

4.3. EULER-MACLAURIN SUMMATION FORMULA39 Version of September 16, 2011

N 1 2 3 4 5 6 7 8 9

PN

n=1

n−3

1 1.125 1.1620 1.1777 1.1857 1.1903 1.1932 1.1952 1.1965

ζ(3, N ) 1.25 1.1667 1.25 1.1 1.5167 −0.1286 8.6214 −51.6619 470.564

1 2 [ζ(3, N )

+ ζ(3, N + 1)] 1.208 1.208 1.175 1.308 0.694

Table 4.2: Two approximations compared for ζ(3) = 1.20206 . . .: N terms in the defining series (4.55) and N terms (without the remainder) in the EulerMaclaurin sum (4.56). The former converges monotonically to the limit from below, while the later diverges, yet approximates the true value to better than 1% for low values of N .

Chapter 5

Analytic Functions 5.1

The Derivative

Let f (z) be a complex-valued function of the complex variable z. The derivative of f is defined as f ′ (z) =

f (z + δz) − f (z) δf df = lim = lim , δz→0 δz dz δz→0 δz

(5.1)

if the limit exists and is independent of the way in which δz approaches zero. This is illustrated in Fig. 5.1

5.1.1

Examples

What is the derivative of z n ? (z + δz)n − z n nz n−1 δz d n z = lim = lim δz→0 δz→0 dz δz δz n−1 = nz .

(5.2)

@ -?  R @ @ I 6 @

Figure 5.1: In the complex plane, δz, as indicated by the arrows in the figure, can approach zero from any direction. 41 Version of October 1, 2011

42 Version of October 1, 2011

CHAPTER 5. ANALYTIC FUNCTIONS

Then, since ez 1s represented by a power series which converges everywhere, and therefore converges uniformly in any finite bounded (compact) region, it is also differentiable everywhere, ∞ ∞ 1 d X 1 n X d z e = z = z n−1 dz dz n=0 n! (n − 1)! n=1

= ez .

(5.3)

The derivative of the exponential function is the function itself.

5.2

Analyticity

Whenever f ′ (z0 ) exists, f is said to be analytic (or regular, or holomorphic) at the point z0 . The function is analytic throughout a region in the complex plane if f ′ exists for every point in that region. Any point at which f ′ does not exist is called a singularity or singular point of the function f . If f (z) is analytic everywhere in the complex plane, it is called entire. Examples • 1/z is analytic except at z = 0, so the function is singular at that point. • The functions z n , n a nonnegative integer, and ez are entire functions.

5.3

The Cauchy-Riemann Conditions

The Cauchy-Riemann conditions are necessary and sufficient conditions for a function to be analytic at a point. Suppose f (z) is analytic at z0 . Then f ′ (z0 ) may be obtained by taking δz to zero through purely real, or through purely imaginary values, for example. If δz = δx, δx real, we have, upon writing f in terms of its real and imaginary parts, f = u + iv,   ∂u ∂v ′ f (z0 ) = . (5.4) +i ∂x ∂x z=z0 On the other hand, if δz = iδy, δy real, we have similarly,   ∂u ∂v f ′ (z0 ) = +i i∂y i∂y z=z0   ∂u ∂v + = −i . ∂y ∂y z=z0

(5.5)

Since the derivative is independent of how the limit is taken, we can equate these two expression, meaning that they must have equal real and imaginary parts, ∂v ∂v ∂u ∂u = , =− . (5.6) ∂x ∂y ∂x ∂y

43 Version of October 1, 2011

5.4. CONTOUR INTEGRALS

These are the Cauchy-Riemann conditions. These conditions are not only necessary, but if the partial derivatives are continuous, they are sufficient to assure analyticity. Write f (z + δz) − f (z) = u(x + δx, y + δy) − u(x, y) + i[v(x + δx, y + δy) − v(x, y)] = u(x + δx, y + δy) − u(x, y + δy) + u(x, y + δy) − u(x, y) + i[v(x + δx, y + δy) − v(x, y + δy) + v(x, y + δy) − v(x, y)]   ∂u ∂v ∂v ∂u + δy + i δx + δy = δx (5.7) ∂x ∂y ∂x ∂y which becomes, if the Cauchy-Riemann conditions hold   ∂u ∂v ∂v ∂u f (z + δz) − f (z) = δx − δy + i δx + δy ∂x ∂x ∂x ∂x   ∂v ∂u , +i = (δx + iδy) ∂x ∂x

(5.8)

so since δz = δx + iδy, we see ∂u ∂v δf → +i δz ∂x ∂x

(5.9)

independently of how δz → 0, so f ′ (z) =

∂u ∂v +i ∂x ∂x

(5.10)

exists. Example Consider the function z ∗ of z; that is, if z = x + iy, z ∗ = x − iy. The CauchyRiemann conditions never hold, ∂(−y) ∂x = 1 6= = −1, ∂x ∂y

(5.11)

so z ∗ is nowhere an analytic function of z.

5.4

Contour Integrals

Suppose we have a smooth path in the complex plane, extending from the point a to the point b. Suppose we choose points z1 , z2 ,. . . , zn − 1 lying on the curve, and connect them by straight-line segments. Likewise connect a = z0 with z1 amd b = zn with zn−1 . See Fig. 5.2. Then the contour integral of a function f is defined by the following limit, Z b n X f (zi )∆zi , ∆zi = zi − zi−1 , (5.12) f (z) dz = lim a C

∆zi →0 n→∞

i=1

44 Version of October 1, 2011

CHAPTER 5. ANALYTIC FUNCTIONS

• •  z2 •  •H • z•3  • z1 • a = z0

• b = zn • • • BB • •  •

Figure 5.2: Path C in the complex plane approximated by a series of straightline segments. and the limit taken is one in which the number n of straight-line segments goes to infinity, while the length of the largest one goes to zero. Whenever this limit exists, independently of how it is taken, the integral exists. Note that in general the integral depends on the path C, as well as on the endpoints. Example Consider

I

K

dz z

(5.13)

where K is a circle about the origin, of radius r. (The circle on the integral sign signifies that the path of integration is closed.) From the polar representation of complex numbers, we may write z = reiθ ,

(5.14a)

dz = reiθ i dθ.

(5.14b)

so since r is fixed on K, we have

Let us assume that the integration is carried out in a positive (counterclockwise) sense, so then Z 2π I dz dθ = 2πi, (5.15) =i 0 K z which is independent of the value of r.

5.5

Cauchy’s Theorem

Chauchy’s theorem states that if f (z) is analytic at all points on and inside a closed contour C, then the integral of the function around that contour vanishes, I f (z) dz = 0. (5.16) C

45 Version of October 1, 2011

5.5. CAUCHY’S THEOREM

MBB C

Ci

Figure 5.3: The integral around the contour C may be replaced by the sum of integrals around the subcontours Ci . Proof: Subdivide the region inside the contour in the manner shown in Fig. 5.3. Obviously I XI f (z) dz, (5.17) f (z) dz = C

i

Ci

where Ci is the closed path around one of the mesh elements, since the contribution from the side common to two adjacent subcontours evidently cancels, leaving only the contribution from the exterior boundary. Now because f is analytic throughout the region, we may write for small δz f (z + δz) = f (z) + δz f ′ (z) + O(δz 2 ),

(5.18)

where O(δz 2 ) means only that the remainder goes to zero faster that δz. We apply this result by assuming that we have a fine mesh subdividing C—we are interested in the limit in which the largest mesh element goes to zero. Let zi be a representative point within the ith mesh element (for example, the center). Then I I I I ′ O((z − zi )2 ) dz. (5.19) (z − zi ) dz + dz + f (zi ) f (z) dz = f (zi ) Ci

Ci

Ci

Ci

Now it is easily seen that for an arbitrary contour Ci I I (z − zi ) dz = 0, dz = Ci

(5.20)

Ci

so if the length of the cell is ε, I f (z) dz = O(ε3 ) = Ai O(ε),

(5.21)

Ci

which is to say that the integral around the ith cell goes to zero faster than the area Ai of the ith cell. Thus the integral required is I X Ai O(ε) = AO(ε), (5.22) f (z) dz = C

i

46 Version of October 1, 2011

CHAPTER 5. ANALYTIC FUNCTIONS @ @ @ @ @ y C @  @ @ @ @

R

Figure 5.4: A multiply connected region R consisting of the area within a triangle but outside of an circular region. The closed contour C cannot be continuously deformed to a point without crossing into the disk, which is outside the region R. where A is the finite area contained within the contour C. As the subdivision becomes finer and finer, ε → 0 and so I f (z) dz = 0. (5.23) C

To state a more general form of Cauchy’s theorem, we need the concept of a simply connected region. A simply connected region R is one in which any closed contour C lying in R may be continuously shrunk to a point without ever leaving R. Fig. 5.4 is an illustration of a multiply connected region. C lies entirely within R, yet it cannot be shrunk to a point because of the excluded region inside it. We can now restate Cauchy’s theorem as follows: If f is analytic in a simply connected region R then I f (z) dz = 0 (5.24) C

for any closed contour C in R. That simple connectivity is required here is seen by the example of the function 1/z, which is analytic in any region excluding the origin. Here is another proof of Cauchy’s theorem, as given in the book by Morse and Feshbach. If the closed contour C lies in a simply-connected region where f ′ (z) exists then I f (z) dz = 0.

(5.25)

C

Proof: Let us choose the origin to lie in the region of analyticity (if it does not, change variables so that z = 0 lies within C). Define I f (λz) dz. (5.26) F (λ) = λ C

Then the derivative of this function of λ is I I zf ′ (λz) dz f (λz) dz + λ F ′ (λ) = C

C

47 Version of October 1, 2011

5.6. CAUCHY’S INTEGRAL FORMULA '

′′

C ↑↓C

z0

&

$

′

•iγ 


↑C %

Figure 5.5: Distortion of a contour C to a small one γ encircling the singularity at z0 .

=

I

C

z=end of C f (λz) dz + zf (λz)

−

I

f (λz) dz = 0, (5.27)

C

z=beginning of C

where we have integrated by parts, because the function f is single valued. Thus F (λ) is constant. But F (0) = lim λ λ→0

I

f (λz) dz = lim

λ→0

C

I

f (z) dz = 0

(5.28)

λC

because f (0) is bounded because f is analytic at the origin. (We have deformed the contour to an infinitesimal one about the origin.) Thus we conclude that F (1) = 0. This proves the theorem.

5.6

Cauchy’s Integral Formula

If f (z) is analytic on and within the closed contour C, and z0 lies within C, then the value of f at z0 is given in terms of its boundary values by f (z0 ) =

1 2πi

I

C

f (z) dz, z − z0

(5.29)

where the contour is traversed in the positive (counterclockwise) sense. Proof: f (z)/(z − z0 ) is not analytic within C, so choose a contour inside of which this function is analytic, as shown in Fig. 5.5. Here we have connected the contour C to the small contour γ by two overlapping lines C ′ , C ′′ which are traversed in opposite senses. Now f (z)/(z − z0 ) is analytic on the inside of the contour C + C ′ + C ′′ + γ. (By inside, we mean that if you follow the path in the direction indicated by the arrows, the inside is only your left, and the outside is on your right.) Thus, by Cauchy’s theorem I

C+C ′ +C ′′ +γ

f (z) dz = 0. z − z0

(5.30)

48 Version of October 1, 2011

CHAPTER 5. ANALYTIC FUNCTIONS

Now because we choose the lines C ′ , C ′′ as overlapping, since f is continuous in the neighborhood of those lines those two integrals cancel, Z f (z) dz = 0. (5.31) C ′ +C ′′ z − z0 And since the circle γ may be chosen arbitrarily small I I f (z) dz dz = f (z0 ) = −2πif (z0 ), z − z z − z0 0 γ γ

(5.32)

since γ is traversed in a negative or clockwise sense. Thus the theorem (5.29) is proved. (Implicit in the above is the assumption that the contour does not cross itself to wind around z0 more than once. If this happens, Cauchy’s formula is modified. See homework.) It is now easily shown from the definition of the derivative that if f is analytic on and within C, we may express the derivative by I 1 f (z) dz, (5.33) f ′ (z0 ) = 2πi C (z − z0 )2 and in fact the nth derivative is given by I f (z) n! (n) dz. f (z0 ) = 2πi C (z − z0 )n+1

(5.34)

That is, if f is analytic, so is its derivative. An analytic function is infinitely differentiable, a property which is not true for a differentiable function of a real variable.

5.7

Morera’s Theorem

The converse to Cauchy’s theorem is the following: If f (z) is continuous in a region R, and for all contours C lying in R I f (z) dz = 0, (5.35) C

then f (z) is analytic throughout R. Proof: If f satisfies the above hypotheses, then the integral Z z2 f (z) dz = F (z2 ) − F (z1 )

(5.36)

z1

is a function of the endpoints only, and not of the path, as is evident from Fig. 5.6. But now the function F has a unique derivative, F ′ (z) = f (z), so that F (z) is analytic. Hence, so is its derivative f (z). QED.

(5.37)

5.8. THE LOGARITHM z1 •

C1 ↑

49 Version of October 1, 2011

↑ C2

• z2 Figure 5.6: point z1 with H Two paths C1 and C2 connecting Rthe R z the point z2 . z Because C1 −C2 f (z) dz = 0, we conclude that z12C1 f (z) dz = z12C2 f (z) dz. t-plane z • •

θ

• 1

|z|

Figure 5.7: Path of integration in the cut t plane used in defining the logarithm in Eq. (5.38).

5.8

The Logarithm

An alternative definition to that given in Sec. 3.2 is given by the path integral Z z dt log z = , (5.38) t 1 over any contour connecting 1 with z which does not cross the cut line shown in Fig. 5.7. The cut is present so the contour cannot encircle the singularity of the integrand at t = 0. Because the arg function must be single-valued, the cut supplies the restriction −π < arg(z) ≤ π. (5.39) The last equality means for negative z we approach the cut from above. Since the integral is path independent, we may chose the path to consist of a segment along the positive z axis and an arc of a circle, as also shown in Fig. 5.7. Then the logarithm may be written as Z θ Z |z| ′ |z| i eiθ dθ′ dt + log z = t |z| eiθ′ 0 1

50 Version of October 1, 2011

CHAPTER 5. ANALYTIC FUNCTIONS

= log |z| + iθ = log |z| + i arg z,

(5.40)

which coincides with the previous definition. The logarithm is analytic in the cut plane, and its derivative is 1 d log z = . dz z

(5.41)

If ξ = log z, define the inverse function by z = exp ξ. Since when z = 1, ξ = 0, we have exp(0) = 1. (5.42) Also we have

dz dz d exp ξ = = = z = exp ξ. dξ dξ d log z

(5.43)

These two properties uniquely define the exponential function.

5.9

A Theorem for Functions Represented by Series

Let us suppose that the function Φ defined by the series ∞ X

Φ(z) =

fn (z)

(5.44)

n=0

converges uniformly on a closed contour C, and that each fn is analytic on and within C. Then, on and within C ∞ X

Φ(z) =

fn (z)

(5.45)

n=0

converges and Φ is analytic. Proof: Since a uniformly convergent series may be integrated term by term, we have for z0 within C 1 2πi

I

C

I ∞ X 1 Φ(z) fn (z) dz = dz z − z0 2πi z − z0 C n=0 =

∞ X

fn (z0 ),

(5.46)

n=0

by Cauchy’s integral formula. So this last sum exists; call it Φ(z0 ) =

∞ X

n=0

fn (z0 ).

(5.47)

5.9. A THEOREM FOR FUNCTIONS REPRESENTED BY SERIES51 Version of October 1, 2011 Now Φ′ (z0 ) exists as well: Φ′ (z0 ) = so Φ is analytic within C.

1 2πi

I

C

∞ X Φ(z) dz = fn′ (z0 ), (z − z0 )2 n=0

(5.48)

Chapter 6

Taylor and Laurent Expansions— Analytic Continuation 6.1

Taylor expansion

Let f (z) be analytic within and on a circle C with center at z0 . Let z be a point within the circle. Then Cauchy’s integral formula can be written as I I 1 1 f (z ′ ) ′ f (z ′ ) f (z) = dz = dz ′ . (6.1) 2πi C z ′ − z 2πi C (z ′ − z0 ) − (z − z0 ) Because z lies inside the circle, |z ′ − z0 | > |z − z0 |,

(6.2)

we can expand the denominator, n ∞  X z − z0 1 1 1 1 = ′ . = ′ 0 (z ′ − z0 ) − (z − z0 ) z − z0 1 − zz−z z − z0 n=0 z ′ − z0 ′ −z 0

(6.3)

This series converges absolutely and uniformly for z ′ on the circle and z fixed inside, so it may be integrated term by term: f (z) = =

∞ X

n=0 ∞ X

n=0

f (z ′ ) dz ′ − z0 )n+1

(z − z0 )n

1 2πi

(z − z0 )n

1 (n) f (z0 ), n!

I

C

(z ′

using the result of Eq. (5.34). 53 Version of October 12, 2011

(6.4)

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

z0 • ξ0 •

ρ

Figure 6.1: Circle of convergence for the Taylor series (6.4). Here z0 is the point about which the Taylor expansion is performed, ξ0 is the closest singularity of f to z0 , and ρ = |ξ0 − z0 | is the radius of convergence. The Taylor series converges within the circle of convergence, and diverges outside the circle of convergence. It may either diverge or converge on the circle of convergence. This Taylor series will converge inside a circle having radius equal to the distance from z0 to the nearest singularity, and diverge outside such a circle, as illustrated in Fig. 6.1. Proof: For |z − z0 | < ρ, we can choose C in the above derivation to have radius r, where |z −z0 | < r < ρ, so the above expansion converges. For |z −z0| > ρ, suppose it were true that the Taylor series converged. Then, according to the theorem in Sec. 2.7, it would converge at z = ξ0 , to an analytic function (Sec. 5.9). This is contrary to the assertion that ξ0 is a singular point. QED. Example Consider the function f (z) =

1 , 1−z

(6.5)

which is analytic except at z = 1. The Taylor series about the origin, 1 = 1 + z + z2 + . . . , 1−z

(6.6)

converges only for |z| < 1. We may obtain a larger circle of convergence by expanding about some other point, say z = −1: 1 1 1 1 = = 1−z 2 − (z + 1) 2 1 − z+1 2 " # 2  z+1 1 z+1 1+ + ... , + = 2 2 2

(6.7)

which converges inside a circle of radius 2, centered about z = −1. In both cases the singularity at z = 1 lies on the circle of convergence.

55 Version of October 12, 2011

6.2. ANALYTIC CONTINUATION

6.2

Analytic Continuation

The process of extending a power series representation of an analytic function is called analytic continuation. It can be done whenever there are only isolated singular points. The general idea is as follows. Suppose we have a power series about z0 f (z) =

∞ X

n=0

an (z − z0 )n ,

(6.8)

which has radius of convergence ρ. (That is, it converges if |z − z0 | < ρ and diverges if |z − z0 | > ρ.) The function f has a singular point somewhere on the circle of convergence. Since this power series represents an analytic function inside its circle of convergence, it can, by the above, be Taylor expanded about any other point lying within the circle of convergence, say z1 , f (z) =

∞ X

n=0

bn (z − z1 )n .

(6.9)

In general,1 the circle of convergence of this series will lie partly outside the original circle. Thus f is now defined in a larger domain. In the new region, f may be expanded once again, and usually the new circle of convergence will lie partly outside both the first two circles, so again the meaning is extended. And so on. The idea is sketched in Fig. 6.2. Entire functions may be represented by power series (Taylor expansions) valid everywhere, since they have no singular points.

6.3

Laurent Expansion

Let f (z) be analytic in the annulus defined by two concentric circles C1 and C2 , both centered on z0 , including the bounding circles. See Fig. 6.3. If z lies in the annulus, Cauchy’s integral formula says (the interior boundary C1 must be traversed in a clockwise sense—hence, the minus sign) I I 1 1 f (z ′ ) ′ f (z ′ ) ′ f (z) = dz − dz 2πi C2 z ′ − z 2πi C1 z ′ − z I I 1 1 f (z ′ ) dz ′ f (z ′ ) dz ′ = − .(6.10) 2πi C2 (z ′ − z0 ) − (z − z0 ) 2πi C1 (z ′ − z0 ) − (z − z0 ) For the C2 integral, |z − z0 | < |z ′ − z0 | so we expand in (z − z0 )/(z ′ − z0 ); for the C1 integral |z − z0 | > |z ′ − z0 |, so we expand in (z ′ − z0 )/(z − z0 ). Thus we have I ∞ X 1 (z − z0 )n ′ f (z) = dz ′ f (z ) ′ − z )n+1 2πi C2 (z 0 n=0 1 But

not always. See Whittaker and Watson, §5.501.

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

ξ0

z0 •  z1  • • z2 ξ2 • •  • ξ1

Figure 6.2: The process of analytic continuation of a function defined by a power series. The original series is a Taylor expansion about the point z0 , which converges inside a circle having radius equal to the distance to the nearest singularity ξ0 . If the function is instead expanded about the point z1 , it converges in a different circle, having radius equal to the distance from z1 to the singular point closest to z1 , namely ξ1 . Instead the function can be expanded about z2 , lying outside the first circle of convergence, but inside the second, which will define the function in a different circle of convergence, with radius of convergence equal to the distance to the singularity closest to z2 , namely, ξ2 . This process may be repeated indefinitely. f is defined in the union of all such circles of convergence.

'$ •z  •z0C1  &% C2 Figure 6.3: Annular region defined by two concentric circles.

57 Version of October 12, 2011

6.3. LAURENT EXPANSION 1 + 2πi

∞ X (z ′ − z0 )n dz ′ . f (z ) (z − z0 )n+1 C1 n=0

I

′

(6.11)

H Now C f (z ′ )(z ′ − z0 )k dz ′ , where k is a positive or negative integer, has the same value for all contours circling z0 once and lying in the annulus, since f (z ′ )(z ′ −z0 )k is analytic there. Therefore the two sums above may be combined to yield ∞ X f (z) = an (z − z0 )n , (6.12) n=−∞

where the expansion coefficients are I 1 f (z ′ ) an = dz ′ . ′ 2πi C (z − z0 )n+1

(6.13)

where C is any contour lying in the annulus. This is called the Laurent expansion. It generalizes the Taylor expansion in the case when there are singularities interior to C1 . (When there are no such singularities, the terms for negative n are identically zero.) Example The function exp

   1 x z− 2 z

(6.14)

is analytic except at z = 0. So it has a Laurent expansion about zero:    ∞ X 1 x z− = an z n , exp 2 z n=−∞ where an =

1 2πi

I

C

e 2 (z x

′

− z1′ )

dz ′ z ′ n+1

.

(6.15)

(6.16)

We make this last integral more explicit by choosing C to be a circle of unit radius, z ′ = eiθ , so Z 2π 1 eix sin θ e−inθ i dθ 2πi 0 Z 2π 1 cos (nθ − x sin θ) dθ, = 2π 0

an =

(6.17)

because Z

0

2π

sin (nθ − x sin θ) dθ = 0,

(6.18)

58 Version of October 12, 2011CHAPTER 6. TAYLOR AND LAURENT SERIES owing to the integrand changing sign under the substitution θ → 2π − θ. This function    ∞ X 1 x z n Jn (x) (6.19) z− = exp 2 z n=−∞

is the generating function for the Bessel functions of integer order, Jn (x). Thus we have derived the following integral representation of the Bessel functions, Z 2π 1 cos (nθ − x sin θ) dθ. (6.20) Jn (x) = 2π 0

6.3.1

Example

Here is another example, which shows that the Laurent expansion holds for functions with branch points and branch lines, provided those are entirely inside the inner annular boundary. Consider the function p z 2 − 1. (6.21)

This function has branch points at z = +1 and at z = −1, and a branch line connecting these two points. Because it is a square-root singularity, the branch line for |z| > 1 cancels, as may be seen by considering the net phase change when both branch points are encircled:  arg z=2π p z 2 − 1 = 0 mod 2π. (6.22) arg arg z=0

This means that we may take expansion for large z:

√

z 2 = z, and we can immediately write down the

 1/2 p 1 2 z −1 = z 1− 2 z   1 1 = z 1 − 2 − 4 − ... 2z 8z ∞ X (2n − 3)!! 1 , = − 2n n! z 2n−1 n=0

(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.

The Laurent expansion (6.23) converges for |z| > 1.

(6.26)

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) = φ(z) +

a−2 a−n a−1 + + ...+ , z − z0 (z − z0 )2 (z − z0 )n

(6.27)

where φ(z) is analytic in the neighborhood of z0 , and a−1 , a−2 , . . . , a−n are complex constants. When the above expansion holds true, f is said to have a pole 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 )n f (z)

(6.28)

is analytic at z = z0 . If the function (z − z0 )m f (z)

(6.29)

is not analytic at z = z0 no matter how large the integer m is, we say that f has an essential singularity at z0 . (This definition applies to functions which are single-valued without the introduction of branch lines.) If an essential singularity is “isolated,” that is, in a sufficiently small neighborhood of z0 , f is analytic except at z0 , f may be expanded in a Laurent series converging in an annulus: f (z) =

∞ X

n=−∞

an (z − z0 )n ,

∆ > |z − z0 | > δ,

(6.30)

where δ is arbitrarily small, and ∆ is the distance to the next singularity. (The proof for this statement is provided in the homework.)

6.4.1

Weierstrass–Picard Theorem

In the neighborhood of an essential singularity, f (z) becomes arbitrarily close to every complex value. This theorem, due to Weierstrass, was greatly sharpened by Picard. Picard’s Theorem In any neighborhood of an essential singularity, the function assumes every finite value, with one possible exception, an infinite number of times. Example: Consider ∞ X 1 e1/z = , (6.31) n! z n n=0

which has an essential singular point at z = 0. Let α be any complex number except 0. For what zs is α = e1/z ? (6.32)

60 Version of October 12, 2011CHAPTER 6. TAYLOR AND LAURENT SERIES Recalling the 2πi periodicity of the exponential function, we see log α =

1 + 2πin, z

n = integer,

(6.33)

or

1 . (6.34) log α − 2πin Thus in any neighborhood of 0 there are an infinite number of these zs. z=

6.4.2

Branch Points and Cuts

Recall log z was defined in the cut plane shown in Fig. 3.1. The location of the cut line is arbitrary, but the location of the end point, z = 0 is not. This branch point is a singular point of log z: d 1 log z = , dz z

(6.35)

which does not exist at z = 0. This type of singularity is neither a pole nor an essential singularity. Once the cut is specified, thus defining log z, the function is not analytic on the branch cut or branch line; in fact, it is discontinuous across the cut: disc(log z) = log ρeiπ − log ρe−iπ = 2iπ. (6.36) The same applies to square roots, and all nonintegral powers, which are defined in terms of the logarithm, √ 1 z = z 1/2 = e 2 log z . (6.37)

Here the discontinuity across the branch line is p p √ disc( z) = ρeiπ − ρe−iπ  √ √  = ρ eiπ/2 − e−iπ/2 = 2i ρ.

6.5

(6.38)

Liouville’s Theorem

First we prove Cauchy’s inequality. Recall the integral representation for the derivative of an analytic function, Eq. (5.34), I f (z) n! dz (6.39) f (n) (z0 ) = 2πi C (z − z0 )n+1 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 + reiθ , (6.40) we write this integral more explicitly as
Z n! 2π f z0 + reiθ dθ f (n) (z0 ) = 2π 0 rn einθ

(6.41)

61 Version of October 12, 2011

6.5. LIOUVILLE’S THEOREM or


Z n!M n! 2π f z0 + reiθ (n) dθ ≤ n , f (z0 ) ≤ n 2π 0 r r

(6.42)

where M is the maximum value attained by |f | on C. Now Liouville’s 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) =

∞ X 1 (n) f (0)z n . n! n=0

(6.43)

But from Cauchy’s inequality, (n) M n! f (0) ≤ n , R

(6.44)

where R is the radius of an arbitrarily large circle about the origin, and M may be taken as the bound on |f |, |f (z)| ≤ M

∀z.

(6.45)

n > 0,

(6.46)

Hence by taking R → ∞, we see that f (n) (0) = 0, and so f (z) = f (0). QED. Example: Although ez is entire, is is certainly not bounded.

(6.47)

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) = φ(z) =

a−1 a−m a−2 + ...+ , + (z − z0 ) (z − z0 )2 (z − z0 )m

(7.1)

where φ(z) is analytic in the neighborhood of z = z0 . Now we have seen that if C encircles z0 once in a positive sense, I 1 dz = 2πiδn,1 , (7.2) (z − z0 )n C where the Kronecker δ-symbol is defined by  0, m 6= n, δm,n = . 1, m = n.

(7.3)

Proof: By Cauchy’s theorem we may take C to be a circle centered on z0 . On the circle, write z = z0 + reiθ . Then the integral in Eq. (7.2) is i rn−1

Z

2π

dθ ei(1−n)θ ,

(7.4)

0

which evidently integrates to zero if n 6= 1, but is 2πi if n = 1. QED. Thus if we integrate the function (7.1) on a contour C which encloses z0 , while φ(z) is analytic on and within C, we find I f (z) dz = 2πia−1 . (7.5) C

Because the coefficient of the (z − z0 )−1 power in the Laurent expansion of f plays 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 α, β, γ, . . . encircling the poles singly, as shown in Fig. 7.1. The contour integral 63 Version of October 26, 2011

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

γ •i

•iα

&

$ β •i

%

→ C

Figure 7.1: Integration of a function f around the contour C which contains only poles of f may be reduced to the integrals around subcontours α, β, γ, etc., each of which contains but a single pole of f . around C may be distorted to a sum of disjoint ones around α, β, . . . , so I I I f (z) dz = f (z) dz + f (z) dz + . . . , (7.6) C

α

β

and since each small contour integral gives 2πi times the reside of the single pole interior to that contour, we have established the residue theorem: If f be analytic on and within a contour C except for a number of poles within, I X f (z) dz = 2πi residues, (7.7) C

poles within C

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: Z 2π dθ I= , 1 − 2p cos θ + p2 0

0 < p < 1.

(7.8)

Let us introduce a complex variable according to z = eiθ , so that

dz = ieiθ dθ = iz dθ,

(7.9)

  1 z+ . z

(7.10)

1 cos θ = 2

Therefore, we can rewrite the angular integral as an integral around a closed contour C which is a unit circle about the origin: I dz 1
I = 1 2 C iz 1 − p z + z + p

65 Version of October 26, 2011

7.2. A FORMULA FOR THE RESIDUE

1 dz 2 2 C i z − p(z + 1) + p z I 1 1 = . dz i C (1 − pz)(z − p) =

I

(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 p 1 1 , (7.12) = + 1 − pz z − p z − p 1 − pz 1 − p2 we have from the residue theorem I = 2πi

2π 1 1 = . i 1 − p2 1 − p2

(7.13)

Note that we could have obtained the residue without partial fractioning by evaluating the coefficient of 1/(z − p) at z = p: 1 1 . (7.14) = 1 − pz z=p 1 − p2 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 1 dm−1 m = [(z − z ) f (z)] . 0 (m − 1)! dz m−1 z=z0

a−1

(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=

Z

∞

−∞

dx

1 = lim R→∞ (x2 + 1)3

Z

R

dx

−R

1 , (x2 + 1)3

(7.16)

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

C

dz = (z 2 + 1)3

Z

R

−R

dx + (x2 + 1)3

Z

Γ

dz , (z 2 + 1)3

(7.17)

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

−R

@ I @ R@ @ @ • i @ → • −i

Γ R

Figure 7.2: The closed contour C consists of the portion of the real axis between −R and R, and the semicircle Γ of radius R in the upper half plane. Also shown in 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 semicircle vanishes in this limit: Z Z π dz R i eiθ dθ = R → ∞. (7.18) 3 → 0, 2 3 Γ (z + 1) 0 (R2 e2iθ + 1) Hence the integral desired is just the closed contour integral, I dz I= = 2πi(residue at i). 2 + 1)3 (z C By the formula (7.15) the desired residue is   1 1 d2 3 (z − i) a−1 = 2 3 3 2! dz (z − i) (z + i) z=i 1 d2 1 = 2 3 2! dz (z + i) z=i 1 (−3)(−4) = 2! (z + i)5 z=i 3 = , 16i so 3π I= . 8

7.4

(7.19)

(7.20) (7.21)

Jordan’s Lemma

The evaluation of a class of integrals depends upon this lemma. If f (z) → 0 uniformly with respect to arg z as |z| → ∞ for 0 ≤ arg z ≤ π, and f (z) is analytic when |z| > c > 0 and 0 ≤ arg z ≤ π, then for α > 0, Z eiαz f (z) dz = 0, (7.22) lim ρ→∞

Γρ

67 Version of October 26, 2011

7.5. EXAMPLE 3

where Γρ is a semicircle of radius ρ above the real axis with center at the origin. (Cf. Fig. 7.2.) Proof: Putting in polar coordinates, Z π Z
iαz (7.23) eiα(ρ cos θ+iρ sin θ) f ρeiθ ρeiθ i dθ. e f (z) dz = 0

Γρ

If we take the absolute value of this equation, we obtain the inequality Z Z π
iαz e f (z) dz ≤ e−αρ sin θ f ρeiθ ρ dθ Γρ 0 Z π < ε e−αρ sin θ ρ dθ, (7.24) 0


if f ρeiθ < ε for all θ when ρ is sufficiently large. (This is what we mean by going to zero uniformly for large ρ.) Now when 0≤θ≤

π , 2

sin θ ≥

2θ , π

(7.25)

which is easily verified geometrically. Therefore, the integral on the right-hand side of Eq. (7.24) is bounded as follows, Z

π

e

−αρ sin θ

ρ dθ < 2ρ

π/2

e−2αρθ/π dθ

0

0

= Hence

Z


π 1 − e−αρ . α

Z επ
iαz 1 − e−αρ e f (z) dz < Γρ α

(7.26)

(7.27)

may be made as small as we like by merely choosing ρ large enough (so ε → 0). QED.

7.5

Example 3

Consider the integral I=

Z

0

∞

cos x dx. x2 + a2

The associated contour integral is Z R Z I eix eiz eiz dz = dx + dz, 2 2 2 2 2 2 −R x + a Γ z +a C z +a

(7.28)

(7.29)

where the contour Γ is a large semicircle of radius R centered on the origin in the upper half plane, as in Fig. 7.2. (The only difference here is that the

68 Version of October 26, 2011CHAPTER 7. THE CALCULUS OF RESIDUES pole inside the contour C is at ia.) The second integral on the right-hand side vanishes as R → ∞ by Jordan’s 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 eix survives symmetric integration, Z I 1 ∞ 1 eix eiz I = dx = dz 2 −∞ x2 + a2 2 C z 2 + a2 1 1 i(ia) π −a = 2πi e = e . (7.30) 2 2ia 2a (Note that if C were closed in the lower half plane, the contribution from the infinite semicircle would not vanish. Why?)

7.6

Cauchy Principal Value

To this point we have assumed that the path of integration never encounters any singularities of the integrated function. On the contrary, however, let us now suppose that f (x) has simple poles on the real axis, and try to attach meaning to Z ∞ f (x) dx. (7.31) −∞

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

a−1 , z − x0

(7.32)

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

∞

x0 −δ

f (x) dx = lim

δ→0+

−∞

f (x) dx +

∞

f (x) dx ,

(7.33)

x0 +δ

−∞

which means that the immediate neighborhood of the singularity is to be omitted symmetrically. The limit exists because f (x) ≈ a−1 /(x−x0 ) near x = x0 , which is 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 function falls off rapidly enough in the upper half plane so that Z f (z) dz = 0, (7.34) Γ

where Γ is the “infinite” semicircle in the upper half plane. Then the integral around the closed contour shown in the figure is Z ∞ I f (z) dz = P f (x) dx − iπa−1 , (7.35) C

−∞

69 Version of October 26, 2011

7.6. CAUCHY PRINCIPAL VALUE ←

Γ •

→

x0 Figure 7.3: Contour which avoids the singularity along the real axis by passing above the pole. ←

Γ x0 •

→

Figure 7.4: Contour which avoids the singularity along the real axis by passing below the pole. where the second term comes from an explicit calculation in which the simple pole 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 residue theorem, I X f (z) dz = 2πi (residues), (7.36) C

poles ∈ UHP

where UHP stands for upper half plane. Alternatively, we could consider a differently deformed contour, shown in Fig. 7.4. Now we have Z ∞ I f (z) dz = P f (x) dx + iπa−1 C −∞   X = 2πi  (7.37) (residues) + a−1  , poles ∈ UHP

so in either case P

Z

∞

−∞

f (x) dx = 2πi

X

poles ∈ UHP

(residues) + πia−1 ,

(7.38)

where the sum is over the residues of the poles above the real axis, and a−1 is the residue of the simple pole on the real axis.

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

−R

Γ

• − k + iǫ

→

•k − iǫ R

Figure 7.5: The closed contour C for the integral in Eq. (7.41). Equivalently, instead of deforming the contour to avoid the singularity, one can displace the singularity, x0 → x0 ± iǫ. Then Z ∞ Z ∞ g(x) g(x) ± iπg(x0 ), (7.39) dx =P dx x − x ∓ iǫ x − x0 0 −∞ −∞ if g is a regular function on the real axis. [Proof: Homework.]

7.7

Example 4

Consider the integral I=

Z

∞

−∞

q2

eiqx dq, − k 2 + iǫ

x > 0,

(7.40)

which is important in quantum mechanics. We can replace this integral by the contour integral I eiqx dq, x > 0, (7.41) 2 2 C q − k + iǫ where the closed contour C is shown in Fig. 7.5. The integral over the “infinite” semicircle Γ is zero according to Jordan’s lemma. √ By redefining ǫ, but not changing its sign, we write the integral as (k = + k 2 )   I eiqx 1 1 dq I= eiqx = 2πi q − (k − iǫ) q + (k − iǫ) q − (k − iǫ) q=−(k−iǫ) C = −

πi −ikx e , k

(7.42)

in the end taking ǫ → 0.

7.8

Example 5

We will consider two ways of evaluating Z ∞ I= 0

dx . 1 + x3

(7.43)

71 Version of October 26, 2011

7.8. EXAMPLE 5

• • •

Figure 7.6: Contour C used in the evaluation of the integral (7.44). Shown also is the branch line of the logarithm along the +z axis, and the poles of the integrand. The integrand is not even, so we cannot extend the lower limit to −∞. How can contour methods be applied?

7.8.1

Method 1

Consider the related integral I

C

log z dz, 1 + z3

(7.44)

over the contour shown in Fig. 7.6. Here we have chosen the branch line of the logarithm to lie along the +z axis; the discontinuity across it is disc log z = log ρ − log ρe2iπ = −2iπ.

(7.45)

The integral over the large circle is zero, as is the integral over the little circle: Z 2π log ρeiθ ρeiθ i dθ = 0. (7.46) lim ρ→∞,0 0 1 + ρ2 e3iθ Therefore, I =− =−

1 2πi

I

C

X

log z dz 1 + z3

poles inside C

(residues).

(7.47)

72 Version of October 26, 2011CHAPTER 7. THE CALCULUS OF RESIDUES To find the sum of the residues, we note that the poles occur at the three cube roots of −1, namely, eiπ/3 , eiπ , and e5iπ/3 , so   X 1 1 iπ/3 (residues) = log e eiπ/3 − eiπ eiπ/3 − ei5π/3   1 1 iπ + log e eiπ − eiπ/3 eiπ − ei5π/3   1 1 5iπ/3 + log e e5iπ/3 − eiπ/3 e5iπ/3 − eiπ 1 π 1  √ = i  √ √  1+ 3i 3 1+ 3i + 1 − 1− 3i 2

2

+ iπ  −1 −

+

2

1

√  1+ 3i 2

1 i5π  √ 3 1− 3i − 2

1



−1 −

√  1+ 3i 2

2 4 iπ 1 √ √ + iπ 3 3i 3 + 3i 9+3  √ π 2 √ (3 − 3i) + 4i − = 12 3 3 2π = − √ , 3 3 =

or

7.8.2

√  1− 3i 2

1



√ 1− 3i 2

+1



2 i5π −1 √ √ 3 3i 3 − 3i  √ 10 √ (3 + 3i) 3 3 +

2π I= √ . 3 3

(7.48)

(7.49)

Method 2

An alternative method which is simpler algebraically is the following. Consider I dz , (7.50) 3+1 z C where the contour C is shown in Fig. 7.7. The integral over the arc of the circle at “infinity,” C2 , evidently vanishes as the radius of that circle goes to infinity. The integral over C1 is the integral I. The integral over C3 is Z Z 0 dz d(xe2iπ/3 ) = = −e2iπ/3 I, (7.51) 3 2iπ/3 )3 + 1 z + 1 (xe C3 ∞ since e2iπ/3


3

= 1. Thus I   π dz = I 1 − e2πi/3 = −Ieiπ/3 2i sin . 3 z + 1 3 C

(7.52)

73 Version of October 26, 2011

7.9. EXAMPLE 6

C2 C3

2π/3

C1 Figure 7.7: Contour used in the evaluation of Eq. (7.50). The only pole of 1/(z 3 + 1) contained within C is at z = eiπ/3 , the residue of which is

so

1 1 e−2πi/3 e−3iπ/3 = , eiπ/3 − eiπ eiπ/3 − ei5π/3 e−iπ/3 − eiπ/3 e−2iπ/3 − e2iπ/3 I =−

or since sin π3 = sin 2π 3 =

√

3 2 ,

I=

2πie−6πi/3 ,
2 (2i)3 sin π3 sin 2π 3 π 4



2 √ 3

3

2π = √ , 3 3

(7.53)

(7.54)

(7.55)

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

7.9

Example 6

Consider I=

Z

0

∞

xµ−1 dx, 1+x

0 < µ < 1.

We may use the contour integral I Z ∞ Z ∞ (−z)µ−1 xµ−1 dx xµ−1 dx e−iπ(µ−1) eiπ(µ−1) dz = − , 1+z 1+x 1+x C 0 0

(7.56)

(7.57)

where C is the same contour shown in Fig. 7.6, and because µ is between zero and one it is easily seen that the large circle at infinity and the small circle about the origin both give vanishing contributions. The pole now is at z = −1, so I (−z)µ−1 dz = 2πi, (7.58) 1+z C where the phase is measured from the negative real z axis. Thus   (7.59) 2πi = e−iπ(µ−1) − eiπ(µ−1) I = 2iI sin πµ,

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

6 

− 12

x 1 2

−R

R

? Figure 7.8: Contour C used in integral K, Eq. (7.62). Here the two lines making an angle of π/4 with respect to the real axis are closed with vertical lines at x = ±R, where we will take the limit R → ∞. or I=

7.10

π . sin πµ

(7.60)

Example 7

Here we demonstrate a method of evaluating the Gaussian integral, Z ∞ 2 J= e−x dx.

(7.61)

−∞

Consider the contour integral K=

I

2

eiπz csc πz dz,

(7.62)

C

where C is the contour shown in Fig. 7.8. The equation for the two lines making angles of π/4 with respect to the real axis are 1 z = ± + ρeiπ/4 , 2 so z2 =

1 ± ρeiπ/4 + iρ2 . 4

(7.63)

(7.64)

75 Version of October 26, 2011

7.11. EXAMPLE 8

Within the contour the only pole of csc πz is at z = 0, which has residue 1/π, so by the residue theorem 1 (7.65) K = 2πi = 2i. π Directly, however,      Z ∞ 1 1 iπ/4 iπ/4 2 iπ/4 csc π ρe + K = e dρ exp iπ iρ + ρe + 4 2 −∞      Z ∞ 1 1 iπ/4 2 iπ/4 iπ/4 − e dρ exp iπ iρ − ρe + csc π ρe − ,(7.66) 4 2 −∞ since the vertical segments give exponentially vanishing contributions as R → ∞. Combining these two integrals, we encounter     i h i h 1 1 − exp −iπρeiπ/4 csc π ρeiπ/4 − exp iπρeiπ/4 csc π ρeiπ/4 + 2 2

iπ/4 iπ/4 exp iπρe + exp −iπρe
= 2,
=2 (7.67) iπ/4 + exp −iπρeiπ/4 exp iπρe since e±iπ/2 = ±i. Hence K = 2e

iπ/4 iπ/4

e

Z

∞

dρ e

2i = √ π

−πρ2

−∞

Z

∞

2

dx e−x ,

so comparing with Eq. (7.65) we have for the Gaussian integral (7.61) √ J = π.

7.11

(7.68)

−∞

(7.69)

Example 8

Our final example is the integral I=

Z

0

∞

x dx . 1 − ex

If we make the substitution ex = t, this is the same as   Z ∞ Z ∞ 1 1 log t dt . = dt log t + I= 1−t t t 1−t 1 1

(7.70)

(7.71)

If we make the further substitution in the first form of Eq. (7.71) u= we have I=

Z

0

1

1 , t

du dt = , u t

log u1 du = 1 − u1 u

Z

0

1

log u du, 1−u

(7.72)

(7.73)

76 Version of October 26, 2011CHAPTER 7. THE CALCULUS OF RESIDUES If we average the two forms (7.71) and (7.73) we have Z Z log t 1 ∞ log t 1 ∞ + . dt dt I= 2 1 t 2 0 1−t

(7.74)

The two integrals here separately are divergent, but the sum is finite. We regulate the two integrals by putting in a large t cutoff: # "Z Z Λ Λ 1 log t log t I= . (7.75) lim + dt dt 2 Λ→∞ 1 t 1−t 0 The first integral here is elementary, Λ Z Λ log t 1 1 dt = log2 t = log2 Λ, t 2 2 1 1 while the second is evaluated by considering I log2 z K= dz , 1−z C

(7.76)

(7.77)

where again C is the contour shown in Fig. 7.6. Now, however, the sole pole is on the positive real axis, so no singularities are contained within C, and hence by Cauchy’s theorem K = 0. This time the contribution of the large circle is not zero: Z 2π Z 2π log2 Λeiθ 2 iθ = −i dθ [log Λ + iθ] Λe idθ iθ 1 − Λe 0 0  1 1 = −i 2π log2 Λ + 2i (2π)2 log Λ − (2π)3 .(7.78) 2 3 The discontinuity of the log2 across the branch line is
2 log2 x − log2 xe2iπ = log2 x − (log x + 2iπ) = −4iπ log x + 4π 2 .

(7.79)

Finally, notice that there is a contribution from the pole at z = 1 below the real axis (see Fig. 7.9): Explicitly, the contribution from the small semicircle below the pole is Z π
 iρeiθ  dθ 4iπ log 1 + ρeiθ − 4π 2 = −4iπ 3 , (7.80) iθ −ρe 2π as ρ → 0. The desired integral is obtained by taking the imaginary part,   Z Λ log t 8π 3 − 4π 3 = 0, (7.81) dt Im K = −4π − 2π log2 Λ − 1 − t 3 0 so

Z

0

Λ

dt

1 π2 log t = − log2 Λ − . 1−t 2 3

(7.82)

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7.11. EXAMPLE 8

1• 



Figure 7.9: Portion of integral K, Eq. (7.77), corresponding to the integration below the cut on the real axis. The pole of the integrand at z = 1 contributes here because log(1 − iǫ) = 2iπ. Thus the contribution of the small semicircle to K is +iπ(2iπ)2 = −4iπ 3 , in agreement with Eq. (7.80). Thus averaging this with Eq. (7.76) we obtain I=−

π2 . 6

(7.83)

A slight check of this procedure comes from computing the real part of K: Re K = 4π 2 P

Z

0

Λ

dt + 4π 2 log Λ = 0. 1−t

(7.84)

Chapter 8

Summation Techniques, Pad´ e Approximants, and Continued Fractions 8.1

Accelerated Convergence

Conditionally convergent series, such as 1−

∞ X 1 1 1 1 1 1 + − + − ... = (−1)n+1 = ln 2, 2 3 4 5 6 n n=1

(8.1)

converge very slowly. The same is true for absolutely convergent series, such as ∞ X 1 π2 = ζ(2) = . n2 6 n=1

(8.2)

If we call the partial sum for the latter N X 1 = SN , n2 n=1

the difference between the limit S and the N th partial sum is Z ∞ ∞ X 1 1 dn S − SN = ≈ = , 2 2 n N N n

(8.3)

(8.4)

n=N +1

which means that it takes 106 terms to get 6-figure accuracy. Thus, to evaluate a convergent series, the last thing you want to do is actually literally carry out the sum. We need a method to accelerate the convergence, and get good accuracy from a few terms in the series. There are several standard methods. 79 Version of November 14, 2011

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CHAPTER 8. APPROXIMANTS

Shanks’ Transformation

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

∞ X

an ,

(8.5)

n=1

consider the N th partial sum SN =

N X

an .

(8.6)

n=1

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 ansatz for all N , to obtain an estimate for the limit S. Then, successive partial sums satisfy

so that b=

SN −1 = S + AbN −1 , SN = S + AbN ,

(8.8a) (8.8b)

SN +1 = S + AbN +1 ,

(8.8c)

SN − S SN +1 − S = , SN − S SN −1 − S

(8.9)

which may be immediately solved for S, S(N ) =

2 SN +1 SN −1 − SN , SN +1 + SN −1 − 2SN

(8.10)

where now we’ve inserted the (N ) subscript on the left to indicate this is an estimate 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,

1 5 = 0.5, S3 = = 0.833, 2 6 47 S5 = = 0.7833, 60 S2 =

S4 =

7 = 0.5833, 12 (8.11)

which oscillate around the correct limit ln 2 = 0.693147, but are not good approximations. Using the Shanks transformation (8.10) we obtain much better approximants: S(1) =

7 = 0.700, 10

S(2) =

29 = 0.690, 42

S(3) =

25 = 0.6944, 36

(8.12)

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 even better by iterating the Shanks transformation, 2 S(N +1) S(N −1) − S(N )

[2]

S(N ) =

S(N +1) + S(N −1) − 2S(N )

,

(8.13)

and then we find using the same data (only 5 terms in the series) [2]

S(2) =

165 = 0.693277, 238

(8.14)

an error of only 0.02%! For more detailed comparison of Shanks estimates for this 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 case we are considering partial sums SN which approach their limit S monotonically. In this case we assume an asymptotic form for large N SN ∼ S +

c b a + 2 + 3 + .... N N N

(8.15)

The first Richardson extrapolation consists of keeping only the first correction term, a a SN = S + , SN +1 = S + , (8.16) N N +1 which may be solved for the limit [1]

S(N ) = (N + 1)SN +1 − N SN ,

(8.17)

where again we’ve inserted on the left a superscript [1] indicating the first Richardson extrapolation, and a subscript (N ) to indicate the approximant comes from the N th and N + 1st partial sums. We consider as an example Eq. (8.2). Here, the first 4 partial sums are S1 = 1,

S2 =

5 = 1.25, 4

S3 =

49 = 1.361, 16

S4 =

205 = 1.424, 144

(8.18)

to be compared with π 2 /6 = 1.644934. The first three Richardson extrapolants are much better: [1]

S(1) =

3 = 1.5, 2

[1]

S(2) =

19 = 1.58, 12

[1]

S(3) =

29 = 1.611. 18

(8.19)

[1]

Iteration of these results by inserting S(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, [2]

S(N ) =

 1 (N + 2)2 SN +2 − 2(N + 1)2 SN +1 + N 2 SN . 2

(8.20)

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When applied to the series (8.2) the first three terms in the series yields nearly 1% accuracy: 13 [2] = 1.625. (8.21) S(1) = 8 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 usually divergent. How can one extract a meaningful number from such series, which represent physical processes and so reflect real processes? On the surface, it would seem impossible to attach any meaning to such obviously divergent series as 1 + 1 + 1 + 1 + 1 + ..., 1 − 1 + 1 − 1 + 1 − ....

(8.22a) (8.22b)

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

an .

(8.23)

an xn = f (x)

(8.24)

n=0

8.2.1

Euler Summation

Suppose ∞ X

n=0

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

(8.25)

x→1

Thus, for the series (8.22b), S=

∞ X

(−1)n ,

(8.26)

n=0

f (x) is f (x) =

∞ X

(−1)n xn =

n=0

1 , 1+x

(8.27)

so S = 1/2. To supply more credence to this result, we note that it is reproduced by 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 so

1 SN +1 SN −1 − Sn2 = SN +1 + SN −1 − 2Sn 2

S=

(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 − . . . ∞ ∞ X X = x3n − x2 x3n n=0

n=0

2

=

1−x 1+x = , 1 − x3 1 + x + x2

(8.31)

2 . 3

(8.32)

so the sum of (8.30) is S = f (1) =

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

8.2.2

Borel Summation

Now we use the Euler representation of the Gamma function, or the factorial, Z ∞ dt tn e−t . (8.33) n! = 0

Then we formally interchange summation and integration: ∞ X

1 S= an n! n=0

Z

∞ n −t

dt t e

=

0

Z

∞

dt e−t

0

which defines the sum if

∞ X 1 a n tn n! n=0

g(t) =

∞ X 1 a n tn , n! n=0

(8.34)

(8.35)

exists. Thus for (8.22b), g(t) =

∞ X

(−1)n

n=0

and so S=

Z

0

tn = e−t , n!

∞

dt e−2t =

1 , 2

(8.36)

(8.37)

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which coincides with the result found by Euler summation. In general, Borel summation 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 ∞ X

an = A,

n=0

then

∞ X

∞ X

bn = B,

(8.38a)

n=0

(αan + βbn ) = αA + βB,

(8.38b)

n=0

and 2. satisfies

∞ X

an = a0 +

n=0

∞ X

an ,

(8.39)

n=1

is unique. In fact, from these two properties alone (which are satisfied by both Euler 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 + . . .) = 1 − S, (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 ∞ X 1 , ζ(s) = s n n=1

Re s > 1.

(8.44)

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In fact, ζ(s) exists for all s 6= 1, so we can use that function to define the sum almost everywhere in the complex s plane. In particular, for s = 0: 1 1 + 1 + 1 + 1 + . . . = ζ(0) = − . 2

(8.45)

Even a more divergent sum can be evaluated this way: ∞ X

n=1

n = ζ(−1) = −

1 . 12

(8.46)

Note the remarkable fact that these sums are not only finite, but negative, even though 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 distance a in the vacuum. Because the plates modify the properties of the vacuum, there is a change in the zero-point energy of the electromagnetic field, which feels the plates because they are conductors. The result is an attraction between the plates, the famous Casimir effect, predicted by Casimir in 1948 (the same year that Schwinger discovered how to renormalize quantum electrodynamics), and now 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 = a is r ∞ Z  nπ  d2 k hc X ¯ 1X 2+ , (8.47) hω = ¯ k E= 2 2 n=1 (2π)2 a where in the mode sum we have integrated over the two transverse wavenumbers kx and ky , and summed over the discrete modes, which, say, must vanish at z = 0 and a, that is, be given by an (unnormalized) mode function φ(z) = sin

nπ z. a

(8.48)

Now we write the square root as integral, putting its argument in the exponential: r Z ∞  nπ 2 ds −1/2 −(k2 +(nπ/a)2 )s 1 2
k + = s e , (8.49) 1 a Γ −2 0 s and then interchange the two integrals: 2 Z ∞ ∞ Z 1 hc X ∞ ds −(nπ/a)2 s ¯ dk −k2 s √ . e e E= 2 n=1 0 s3/2 2π −2 π −∞

(8.50)

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

2

2

2

e−(kx +ky )s = e−kx s e−ky s .

(8.51)

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These one-dimensional integrals are simply Gaussians, so the squared factor in (8.50) is simply 1/(4πs). The remaining s-integral is again a gamma function: ∞ Z ¯ c X ∞ ds −(nπ/a)2 s h e 16π 3/2 n=1 0 s5/2  ∞  ¯hc 3 X  nπ 3 = − Γ − 2 n=0 a 16π 3/2

E = −

= −

¯ cπ 2 h , 1440a3

(8.52)

where we have used the facts that   4√ 3 = Γ − π, 2 3

ζ(−3) =

1 , 120

(8.53)

together with the zeta-function continuation embodied in Eq. (8.44) When multiplied by 2, for the two polarization states of the photon, this is exactly Casimir’s result, which implies an attractive force per unit area between the plates, ∂ ¯hcπ 2 P =− E=− = −1.30 × 10−27 N m2 /a4 . (8.54) ∂a 240a4

8.3

Pad´ e Approximants

Consider a partial Taylor sum, TN +M (z) =

NX +M

an z n ,

(8.55)

n=0

which is an N + M th degree polynomial. Write this in a rational form, PN An z n N , (8.56) PM (z) = PMn=0 m m=0 Bm z

which is called the [N, M ]th Pad´e approximant. Here the coefficients are determined from the Taylor series coefficients as follows: We set B0 = 1, and determine the (N + M + 1) coefficients A0 , A1 , . . . , AN and B1 , B2 , . . . , BM by requiring that when the rational function (8.56) be expanded in a Taylor series about z = 0 the first N + M + 1 coefficients match those of the original Taylor expansion (8.55). Example Consider the exponential function 1 ez = 1 + z + z 2 + . . . . 2

(8.57)

´ APPROXIMANTS 8.3. PADE

87 Version of November 14, 2011

The [1, 1] Pad´e of this is of the form P11 (z) =

A0 + A1 z , 1 + B1 z

(8.58)

which, when expanded in a series about z = 0 reads P11 (z) ≈ A0 + (A1 − B1 A0 )z + (B12 A0 − A1 B1 )z 2 .

(8.59)

Matching this with Eq. (8.57), we obtain the equations A0 = 1, A1 − B1 A0 = 1, 1 B1 (B1 A0 − A1 ) = , 2

(8.60a) (8.60b) (8.60c)

so we learn immediately that A0 = 1,

1 B1 = − , 2

A1 =

1 , 2

(8.61)

so the [1, 1] Pad´e is P11 (z) =

1 + 12 z . 1 − 21 z

(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 1 + z + z 2 = 2.5, (8.64) 2 z=1

about 8% low. However, in higher orders, Pad´e approximants rapidly outstrip Taylor approximants. Table 8.1 compares the numerical accuracy of PNM with TN +M . Note that typically the Pad´e approximant, obtained from a partial Taylor sum, is more accurate than the latter. This comes at a price, however; the Pad´e, being a rational expression, has poles, which are not present in the original function. Thus, ez is an entire function, while the [1, 1] Pad´e approximant of this function has a pole at z = 2. Example Here’s another example: z z2 z3 z4 z5 1 log(1 + z) = 1 − + − + − + .... z 2 3 4 5 6

(8.65)

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CHAPTER 8. APPROXIMANTS

TN +M (1)

N PM (1)

Relative error of Pad´e

T3 (1) = 2.667

P21 (1) = 2.667

T4 (1) = 2.708

P22 (1) = 2.71429 P32 (1) = 2.71875 P33 (1) = 2.71831 P43 (1) = 2.71827957

−1.9%

T5 (1) = 2.717 T6 (1) = 2.71806 T7 (1) = 2.71825

−0.15%

+0.017% +0.00103% −0.000083%

Table 8.1: Comparison of partial Taylor series with successive Pad´e approximants for the exponential function, evaluated at z = 1. Note that precisely the N same data is incorporated in TN +M and in PM . Approximant

z = 0.5

z=1

z=2

Exact P33 P43

0.810930216 0.810930365 0.810930203

0.69314718 0.69315245 0.69314642

0.549306 0.549403 0.549285

Table 8.2: Pad´e approximations for the function (1/z) log(1 + z) compared with the exact values. Note that the Taylor series for this function has a radius of convergence of unity, yet the Pad´e approximations converge rapidly even beyond the circle of convergence. It is a simple algebraic task to expand the form of an [N, M ] Pad´e in a Taylor series and compute the Pad´e coefficients by matching with the above. This can, of course, be easily implemented in a symbolic program. For example, in Mathematica, N PM (z) = P adeApproximant[f [z], {z, 0, {N, M }}].

(8.66)

Doing so here yields P33 (z) =

17 1 3 1 + 14 z + 31 z 2 + 140 z 6 2 4 3 . 12 z 1 + 7 z + 7 z + 35

(8.67)

Table 8.2 shows representative numerical values for P33 and P43 . The Pad´e approximants rapidly converge to the correct value even well beyond the circle of convergence of the original series. Note further in this example that • PNN is larger than the function, and decreases monotonically toward it, and

´ APPROXIMANTS 8.3. PADE

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 see C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers (McGraw-Hill, New York, 1978), pp. 383ff. Field Theory Examples The following function occurs in the field theory of a massless particle in zero dimensions, Z ∞ 2 1+δ dx √ e−(x ) Z(δ) = π −∞   Z ∞ 1 1 2 1 dt 1/(2+2δ) −t 2 t e =√ Γ = √ π (2 + 2δ) 0 t π (2 + 2δ) 2 + 2δ   2 3 + 2δ = √ Γ , (8.68) 2 + 2δ π where the gamma function was defined by Euler as Z ∞ dt z −t t e , Γ(z) = t 0

(8.69)

and satisfies the identity Γ(z + 1) = zΓ(z).

(8.70)

The gamma function generalizes the factorial to complex values: Γ(n + 1) = n!,

n = 0, 1, 2, . . . .

(8.71)

Because the gamma function Γ(z) has poles when z = −N , N = 0, 1, 2, . . . , this function has an infinite number of singularities between δ = −3/2 and δ = −1. Thus the radius of convergence of the Taylor series about δ = 0 is 1. Yet low order Pad´e’s for E(δ) = − log Z(δ) give an excellent approximation well outside of this radius, as Table 8.3 shows. The “partition function” for a zero-dimensional field theory with a mass µ is given by the function r Z ∞ µ2 2 2 1+δ 2 (8.72) dx e− 2 x −λ(x ) . Z(δ) = µ π 0 We consider two cases. If µ2 > 0, the power series in δ again has radius of convergence 1, but the Pad´e approximants are accurate far beyond this radius, as shown in Table 8.4. If, on the other hand µ2 < 0 (which corresponds to the “Higgs mechanism” in particle physics), the Taylor series converges nowhere, yet the Pad´e approximant is still quite good, as seen in Table 8.5.

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CHAPTER 8. APPROXIMANTS

δ

T10 (δ)

T20 (δ)

P23 (δ)

P45 (δ)

E(δ)

−2.0 −0.5 0.5 1.0 2.0 5.0

−1266.97 −0.120055 −0.00781712 −0.367098 −465.821 −5.5 × 106

−2.0 × 106 −0.120781 −0.00759091 −0.516940 −688611 −7.8 × 1013

−0.651267 −0.120831 −0.00759097 −0.0225167 −0.0458145 −0.0786672

−0.692962 −0.12078223848 −0.0075905958951 −0.022510401233 −0.04575620415 −0.078172915

−0.693147 −0.12078223764 −0.0075905958949 −0.022510401213 −0.04575620349 −0.078172899

Table 8.3: Approximations to the function (8.68). What is approximated is E(δ) = − log Z(δ). The Pad´e approximants based on 6 and 10 terms in the Taylor series of this function are far more accurate that the 10 and 20 term truncated Taylor series, and even are remarkably accurate far outside the circle of convergence, where the Taylor series is meaningless.

δ

T8 (δ)

P44 (δ)

Z(δ)

0.5 1.0 2.0 5.0

1.04631 1.07719 1.81047 745.176

1.04630 1.07436 1.10647 1.14253

1.04630 1.07436 1.10649 1.14285

Table 8.4: Comparison of Z(δ), Eq. (8.72), µ2 > 0, with the 8-term truncated power series, and the corresponding [4, 4] Pad´e. Here we have taken µ2 = 1, λ = 1.

δ

T8 (δ)

P44 (δ)

Z(δ)

0.1 0.5 1.0 2.0 5.0

0.94808 137.697 40109.3 1.1 × 107 1.8 × 1010

0.94790 0.88388 0.87323 0.88334 0.91830

0.94790 0.88381 0.87253 0.87974 0.90517

Table 8.5: Comparison of Z(δ), Eq. (8.72), µ2 < 0, with the 8-term truncated power series, and the corresponding [4, 4] Pad´e. Here we have taken µ2 = −1, λ = 1.

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8.4. CONTINUED FRACTIONS

8.4

Continued Fractions

8.4.1

Number Theory

The most familiar way of representing real numbers is in terms of a decimal fraction, which is nonterminating and nonrepeating if the number is irrational. However, there are other representations which, if less familiar, can be very useful. For example, the base of the natural logarithms e can be written in the form of a continued fraction, e=2+

1 1+

.

1 2+

(8.73a)

1 1

1+ 1+

1 4+...

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 1 1 1 1 1 . 1+ 2+ 1+ 1+ 4+ 1+ 1+ 6 + . . .

(8.73c)

The form seen here is the representation of a real number x in the form x = a0 +

1 1 1 1 , a1 + a2 + a3 + a4 + . . .

(8.74)

where the numbers an are integers called partial quotients. The rational number formed 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 an s: 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   8 11 19 87 106 193 1264 1457 2721 23225 25946 49171 2, 3, , , , , , , , , , , , ,... 3 4 7 32 39 71 465 536 1001 8544 9545 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 partial fraction. Thus in the above example, where in each case 0 < r < 1, 2 < e, 1 5 < 2+ < 3, 2 1+r

(8.77a) (8.77b)

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1 1 11 8 < 2+ < , 3 1+ 2 + r 4 19 1 1 1 11 < 2+ < , 7 1+ 2+ 1 + r 4 19 1 1 1 1 30 < 2+ < , 7 1+ 2+ 1+ 1 + r 11

(8.77c) (8.77d) (8.77e)

and so on. The successive convergents are the upper and lower bounds corresponding to r = 0. The partial fraction representation of real numbers can be generated using your favorite symbolic program. For example, in Mathematica the first n partial quotients of x are given by ContinuedF raction[x, n],

(8.78)

and the first n convergents are given by Convergents[x, n].

(8.79)

Let us conclude this subsection with the following comments. • Evidently, a rational number is represented by a terminating continued fraction. For example, 12357 = {0, 99908, 2, 1, 1, 1, 1, 3, 3, 2, 1, 2} 1234567890

(8.80)

exactly. • An algebraic number, that is one which is a solution of an algebraic equation, which is not rational, is represented by a repeating pattern of partial quotients. For example, √ 137 = {11, 1, 2, 2, 1, 1, 2, 2, 1, 22, 1, 2, 2, 1, 1, 2, 2, 1, 22, . . .}. (8.81) • A trancendental number is represented by a nonrepeating pattern. That pattern is simple in the case of e, but not so for the case of π: π = {3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, 2, 1, 1, 15,

3, 13, 1, 4, 2, 6, 6, 99, 1, 2, 2, 6, 3, 5, 1, 1, 6, 8, 1, 7, 1, 2, 3, 7, . . .}.(8.82)

The first few convergents are π = {3, 3.14285714, 3.141509434, 3.141592920, 3.141592653, 3.141592654, . . .}, (8.83) so ten figure accuracy requires 6 terms. However, there are other continued fraction representations for π that have simple patterns: 4 12 32 52 72 , 1+ 2+ 2+ 2+ 2 + . . .

(8.84)

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8.4. CONTINUED FRACTIONS which has rather poor partial sums:

π = {4, 2.6667, 3.4667, 2.89524, 3.33968, . . .};

1 32 52 72 , 6+ 6+ 6+ 6 + . . . where the convergents are somewhat better π =3+

π = {3, 3.16667, 3.13333, 3.14524, 3.13968, 3.14271, . . .}; 2

π=

2

(8.85) (8.86)

(8.87)

2

3 4 1 2 , 1+ 3+ 5+ 7 + . . .

(8.88)

which is comparable, π = {4, 3, 3.16667, 3.13725, 3.14234, . . .}.

(8.89)

All of these are much worse than the rapid convergence of the standard convergents. But it is the existence of simple patterns that is perhaps remarkable.

8.4.2

Continued Fraction Representation of Functions

If a function is represented by a power series about the origin, f (x) =

∞ X

an xn ,

(8.90)

n=0

we can also write it in a continued-fraction form. The standard approach here is to write b0 f (x) = b1 x 1+ b2 x 1+

1+

b3 x b4 x 1+...

1+

b0 b1 x b2 x b3 x b4 x . (8.91) 1+ 1+ 1+ 1+ 1 + . . . Evidently, there is a one-to-one correspondance between the Taylor-series coefficients {an } and the continued-fraction coefficients {bn }, which may be determined by expanding the continued fraction in a power series for small x. The theory of such a representation is discussed also in the book by Bender and Orzag. Let us consider a function with the property f (0) = 1; this is merely a convenient choice of normalization. Then the relation between the continued fraction coefficients and the series coefficients is easily found to be =

a0 = b0 = 1,

(8.92a)

a1 = −b1 , a2 = b1 (b1 + b2 ),

(8.92b) (8.92c)

a3 = −b1 [b2 b3 + (b1 + b2 )2 ], a4 = b1 [b2 b3 (b3 + b4 ) + 2(b1 + b2 )b2 b3 + (b1 + b2 )3 ],

(8.92d) (8.92e)

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and so on. This constitutes a nonlinear mapping from the set of numbers {bn } to the set {an } or vice versa. This mapping seems to be quite remarkable in that the sequence of bn s is typically much simpler than the sequence of an s. Here are some examples: Example 1 Let bn = n, that is, b1 = 1, b2 = 2, etc. Then by computing the first few an s from the above formulæ we find bn = n

⇒ |an | = (2n − 1)!!.

(8.93)

Example 2 Let the continued-fraction sequence be {bn } = {1, 1, 2, 2, 3, 3, 4, 4, . . .}. Then the power series coefficients are given by the factorial, |an | = n!.

(8.94)

Example 3 What if bn = n2 ? The first few an are a1 = −1, a2 = 5,

(8.95a) (8.95b)

a3 = −61, a4 = 1385.

(8.95c) (8.95d)

These are recognized as the first few Euler numbers, defined by the generating function ∞ X tn 1 = En , (8.96) cosh t n=0 n! namely E0 = 1,

(8.97a)

E2 = −1, E4 = 5,

(8.97b) (8.97c)

E6 = −61, E4 = 1385,

(8.97d) (8.97e)

an = E2n .

(8.98)

and we conclude

8.4. CONTINUED FRACTIONS

95 Version of November 14, 2011

Example 4 This suggests that we ask what sequence of bn s corresponds to the Bernoulli numbers. It takes a bit of playing around to find the correct normalization, which matters since the transformation is nonlinear. If we take an = 6B2n+2 ,

(8.99)

we find that the corresponding continued-fraction coefficients are given by bn =

n(n + 1)2 (n + 2) . 4(2n + 1)(2n + 3)

(8.100)

Although the latter seems a bit complicated, it is a closed algebraic expression. It further grows with n as a low power. Neither of these features hold for the Bernoulli numbers, which grow more rapidly than exponentially, and have no closed-form representation. These ideas are provocative, yet the general significance of these results remains elusive. There appears also to be some deep connection to field theory. See C. M. Bender and K. A. Milton, J. Math. Phys. 35, 364 (1994) for more details.

Chapter 9

Asymptotic Expansions We will illustrate the notions with a couple of carefully chosen examples. For more detail, you are referred to C. M. Bender and S. A. Orzag, Advanced Mathematical Methods for Physicists and Engineers: Asymptotic Methods and Perturbation Theory (Springer, 1999).

9.1

The Airy Function

The Airy function, which occurs, for example, in various radiation problems, is defined by the integral   Z ∞ 1 3 dt cos ζt + t πAi(ζ) = 3 0 Z ∞ 3 1 = dt ei(ζt+t /3) . (9.1) 2 −∞ Let z = it; then this integral can also be given as Ai(ζ) =

1 2πi

Z

i∞

dz eζz−z

3

/3

,

(9.2)

−i∞

where the path of integration is along the imaginary axis. Now, to this point, this integral has only a formal existence, since the magnitude of the integrand is unity. However, if we distort the contour to C, as shown in Fig. 9.1, which passes through the origin, but is asymptotic to the lines arg z = ±2π/3, we obtain a convergent integral since 3  z 3 = ρe±i2π/3 = ρ3 > 0.

(9.3)

This deformation of the contour is permissible because the contributions of the arcs at infinity, connecting the ends of C to the imaginary axis, are negligible. 97 Version of November 15, 2011

98 Version of November 15, 2011CHAPTER 9. ASYMPTOTIC EXPANSIONS C π 3

− π3

Figure 9.1: Contour C used to define the Airy function in Eq. 9.5. That is, if z = Reiθ , R → ∞, we have Re z 3 = R3 cos 3θ > 0

if

π 2π ≥θ> 3 2

or if

−

2π π ≤θ π2 must be supplied in any case.

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9.1. THE AIRY FUNCTION C

√ • ζ

2π 3

− 2π 3

Figure 9.2: Deformed contour C which passes through the saddle point. Note that in the perpendicular direction, along the real axis, the function is a minimum at the stationary point. Thus, the stationary point is a saddle point, 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, for large |ζ|, most of the contribution comes from the immediate neighborhood of the saddle point. Then we can make use of the approximation above, so that we approximate the Airy function by Z √ 2 1 − 2 ζ 32 e 3 Ai(ζ) ∼ dξ e ζξ , (9.8) 2πi C where the integral is just a Gaussian one, Z i∞ Z Z √ 2 ζξ − 14 u2 − 41 =ζ du e = iζ dξ e C

−i∞

∞ 2

1

dt e−t = iζ − 4

√

π.

(9.9)

−∞

Thus we obtain the leading asymptotic behavior of the Airy function 3 1 1 2 Ai(ζ) ∼ √ ζ − 4 e− 3 ζ 2 , 2 π

ζ → ∞.

(9.10)

This result is actually valid for complex values of ζ subject to the restriction | arg ζ| < π.

(9.11)

This asymptotic approximation is really quite good for modest ζ as Fig. 9.3 shows.

9.1.1

Asymptotic series

Let us calculate the corrections to this result. We return to Eq. (9.7) and keep the next term in ξ: 1 2 (9.12) φ(z) = − ζ 3/2 + ζ 1/2 ξ 2 − ξ 3 , 3 3 which is exact in this case. Thus the Airy function is exactly represented by the integral Z i∞ 1/2 2 2 3/2 1 1 3 dξ e− 3 ζ eζ ξ e− 3 ξ . Ai(ζ) = (9.13) 2πi −i∞

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

0.50 Ai(x) f(x) r(x)

0.40

0.30

0.20

0.10

0.00 0.0

1.0

2.0

3.0

4.0

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

5.0

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We approximate this by expanding the last exponential, since for large ζ the integrand is dominated by small ξ. Expanding out to fourth order, and omitting odd terms, we have after substituting ξ = iuζ −1/4 :   Z 1 u6 1 1 u12 ζ −1/4 − 2 ζ 3/2 ∞ −u2 3 1− du e + . . . . (9.14) e + Ai(ζ) ∼ 2π 18 ζ 3/2 24 81 ζ 3 −∞ The integrals may be evaluated starting from r Z ∞ 2 π du e−λu = , λ −∞ so Z

∞

−∞

2k −λu2

du u e

=



d − dλ

k Z

∞ 2

du e−λu = −∞

(9.15)

√ (2k − 1)!! 1 . (9.16) π k (2k+1)/2 2 λ

Thus, the two leading corrections to the asymptotic expression for the Airy function given in Eq. (9.10) are   2 3/2 1 5 1 385 1 Ai(ζ) ∼ √ ζ −1/4 e− 3 ζ 1− + . . . , (9.17) + 2 π 48 ζ 3/2 4608 ζ 3 which is the beginning of an asymptotic series expansion in powers of ζ −3/2 .

9.2

Synchrotron Radiation

A charged particle moving in a circular orbit emits electromagnetic radiation called (for the machine in which such radiation was first observed) synchrotron radiation. For details of the theory, see, for example, J. Schwinger, L. L. DeRaad, 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 the frequency of revolution of the charged particle moving in a circle with speed v = βc is, in part, proportional to Z π dφ ′ J2m (2mβ) = − sin φ sin 2m(β sin φ − φ). (9.18) π 0 In the ultrarelativistic limit when β → 1, most of the radiation occurs for large harmonic numbers, m ≫ 1, and the main contribution comes from the region near φ = 0. Therefore, we may expand the integrand in Eq. (9.18) as follows:     φ3 −φ sin φ sin 2m(β sin φ − φ) ≈ φ sin 2m β φ − 3!    1 = φ sin 2m −φ(1 − β) − βφ3 6    1 ≈ −φ sin m (1 − β 2 )φ + φ3 3

102 Version of November 15, 2011CHAPTER 9. ASYMPTOTIC EXPANSIONS    p 1 = − 1 − β 2 x sin m(1 − β 2 )3/2 x + x3 , 3 (9.19) where we have introduced the change of scale p φ = 1 − β 2 x.

(9.20)

As a result, in this limit, Eq. (9.18) can be approximated by2    Z ∞ 1 3 dx 2 3/2 ′ 2 x+ x x sin m(1 − β ) J2m (2mβ) ∼ (1 − β ) π 3 0 Z ∞ 2 3/2 3 (1 − β 2 ) Im dx x eim(1−β ) (x+x /3) . (9.21) = π 0 For m fixed and β approaching unity in such a way that m(1 − β 2 )3/2 ≪ 1, the significant contribution to Eq. (9.21) comes from the region where x is large, and Eq. (9.21) reduces to Z ∞  m dx ′ 2 J2m (2mβ) ∼ (1 − β ) x sin (1 − β 2 )3/2 x3 π 3 0 Z ∞   dφ m 3 = (9.22) φ sin φ , π 3 0 where all reference to the speed of the particle has disappeared. By changing variables, we may write this as Z ∞ dφ −imφ3 /3 ′ φe J2m (2m) ∼ −Im π 0  2/3 −iπ/3 Z ∞   3 e 1 −2/3 1/3 −t = −Im t e t dt m π 3 0  2/3 3 Γ(2/3) −iπ/3 = −Im e m 3π =

31/6 Γ(2/3) , 2π m2/3

for

m ≫ 1.

(9.23)

In the above evaluation, we have used Cauchy’s theorem to perform a change of contour, 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(1 − β 2 )3/2 becomes large, and the integrand in Eq. (9.21) undergoes rapid oscillations in x except near the stationary points, which satisfy   1 d x + x3 = 1 + x2 = 0; (9.24) dx 3 2 Evidently,

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

103 Version of November 15, 2011

9.2. SYNCHROTRON RADIATION φ plane:

- Q π/6 k Q Q PP q∞ Q Q

? Figure 9.4: Change of contour used in evaluating Eq. (9.23). Complex x plane

i •=⇒ =⇒

=⇒

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 so that it passes through the stationary point x = i, because then the dominant contribution comes from the vicinity of that point. (See Fig. 9.5.) In the neighborhood of x = i, we let x = i + ξ, (9.26) where ξ is real, to take advantage of the saddle point character. For arbitrary ξ   1 3 1 1 2 3 2 x + x = (i + ξ) + (i + ξ) = i + ξ + ξ3 , (9.27) 3 3 3 3 so that for small ξ, if we drop the cubic term in ξ, the exponential factor in Eq. (9.21) becomes 2 3/2 2 3/2 2 2 e− 3 m(1−β ) e−m(1−β ) ξ , (9.28) which falls off exponentially on both sides of x = i. The resulting Gaussian integral in (9.21) leads to the following asymptotic form: ′ J2m (2mβ) ∼

1 (1 − β 2 )1/4 − 2 m(1−β 2 )3/2 √ , e 3 2 πm

m(1 − β 2 )3/2 ≫ 1.

(9.29)

104 Version of November 15, 2011CHAPTER 9. ASYMPTOTIC EXPANSIONS Thus, for very large harmonic numbers, the power spectrum3 decreases exponentially 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 critical harmonic number, mc , for which mc (1 − β 2 )3/2 ≡ 1,

(9.31)

or 2 −3/2

mc = (1 − β )

=



E µc2

3

,

(9.32)

which uses the relativistic connection between the energy and the rest mass µ, E = µc2 (1 − β 2 )−1/2 . The bulk of the radiation is emitted with harmonic numbers 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 ξ 3 term, but treating it as small, so the correction may be obtained by Taylor expanding the exponential: Z ∞ 2 3/2 2 1 − β2 ′ − 32 m(1−β 2 )3/2 J2m (2mβ) ∼ Im e dξ e−m(1−β ) ξ (i + ξ) 2π −∞   6 3 1 2 ξ 2 3ξ 2 3/2 − m (1 − β ) + ... × 1 + im(1 − β ) 3 2 9 Z ∞ 2 1 − β 2 − 2 m(1−β 2 )3/2 (1 − β 2 )−3/4 m−1/2 dt e−t e 3 = 2π −∞   t4 t6 1 1 × 1+ − + . . . . (9.33) 3 m(1 − β 2 )3/2 18 m(1 − β 2 )3/2 Here we noted that the imaginary part only receives the contribution of the even terms in ξ, which are all that survive symmetric integration. Finally, the Gaussian integrals are evaluated according to   Z ∞ Z ∞ dx n −x 1 2n −t2 √ x e =Γ n+ , (9.34) dt t e = x 2 −∞ 0 where

  √ 5 3 π Γ = , 2 4

  √ 15 π 7 = Γ . 2 8

(9.35)

3 The power radiated into the mth harmonic by a particle of charge e moving in a circle of radius R with angular frequency ω0 is given by

Pm



e2 ′ mω0 2β 2 J2m = (2mβ) − (1 − β 2 ) R

Z

2mβ



dx J2m (x)

0

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

(9.30)

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9.2. SYNCHROTRON RADIATION

3.0

2mJ′2m(2mβ)

2.0

1.0

0.0

1

10

100

1000

10000

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

106 Version of November 15, 2011CHAPTER 9. ASYMPTOTIC EXPANSIONS Thus ′ J2m (2mβ) =

(1 − β 2 )1/4 − 2 m(1−β 2 )3/2 √ e 3 2 mπ    1 1 7 . (9.36) +O × 1+ 48 m(1 − β 2 )3/2 m2 (1 − β 2 )3

Chapter 10

Linear Operators, Eigenvalues, and Green’s 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 commutative and 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 λ is a complex number, and x ∈ V , then λx ∈ V . Multiplication by scalars is distributive over addition: λ(x + y) = λx + λy.

(10.4)

Scalar multiplication is also associative: If λ and µ are two complex numbers, λ(µx) = (λµ)x. 107 Version of November 16, 2011

(10.5)

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CHAPTER 10. LINEAR OPERATORS

An inner product space is a vector space possessing an inner product. If x and y are two vectors, the inner product hx, yi

(10.6)

is a complex number. The inner product has the following properties: hx, y + αzi = hx, yi + αhx, zi, hx + βy, zi = hx, zi + β ∗ hy, zi, hx, yi = hy, xi∗ , hx, xi > 0 if x 6= 0,

(10.7a) (10.7b) (10.7c) (10.7d)

where α and β are scalars. Because of the properties (10.7a) and (10.7b), we say that the inner product is linear in the second factor and antilinear in the first. Because of the last property (10.7d), we define the norm of the vector by p (10.8) kxk = hx, xi.

10.2

The Cauchy-Schwarz Inequality

An important result is the Cauchy-Schwarz inequality,1 which has an obvious meaning for, say, three-dimensional vectors. It reads, for any two vectors x and y |hx, yi| ≤ kxkkyk, (10.9) where equality holds if and only if x and y are linearly dependent. Proof: For arbitrary λ we have 0 ≤ hx − λy, x − λyi = kxk2 − λhx, yi − λ∗ hy, xi + |λ|2 kyk2 .

(10.10)

Because the inequality is trivial if y = 0, we may assume y 6= 0, and so we may choose hy, xi . (10.11) λ= kyk2 The the inequality (10.10) read 2 |hy, xi|2 |hx, yi|2 + 2 kyk kyk2 |hx, yi|2 , = kxk2 − kyk2

0 ≤ kxk2 −

(10.12)

from which Eq. (10.9) follows. Evidently inequality holds in Eq. (10.10) unless x = λy. 1 The

name Bunyakovskii should also be added.

(10.13)

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10.3. HILBERT SPACE

From the Cauchy-Schwarz inequality, the triangle inequality follows: kx + yk ≤ kxk + kyk.

(10.14)

Proof: kx + yk2 = hx + y, x + yi = kxk2 + kyk2 + 2Re hx, yi

≤ kxk2 + kyk2 + 2|hx, yi| ≤ kxk2 + kyk2 + 2kxkkyk = (kxk + kyk)2 .

(10.15)

QED

10.3

Hilbert Space

A Hilbert space H is an inner product space that is complete. Recall from Chapter 2 that a complete space is one in which any Cauchy sequence of vectors has a limit in the space. That is, if we have a Cauchy sequence of vectors, i.e., for any ǫ > 0, {xn }∞ n=1 :

kxn − xm k < ǫ

∀ n, m > N (ǫ),

(10.16)

then the sequence has a limit in H, that is, there is an x ∈ H for which for any ǫ > 0 there is an N (ǫ) so large that kx − xn k < ǫ

∀ n > N (ǫ).

(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 hei , ej i = δij .

(10.18)

The set is said to be complete if any vector x in H can be expanded in terms of the ei s:2 ∞ X hei , xiei . (10.19) x= i=1

Here convergence is defined in the sense of the norm as described above. Geometrically, the inner product hei , xi is a kind of direction cosine of the vector x, or a projection of the vector x on the basis vector ei . 2 If

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 closed interval [−π, π]: Z π |f (x)|2 dx < ∞. (10.20) −π

The functions (not the values of the functions) are the vectors in the space, and the inner product is defined by Z π f (x)∗ g(x) dx. (10.21) hf, gi = −π

It is evident that this definition of the inner product satisfies all the properties (10.7a)–(10.7d). This space, called L2 (−π, π), is in fact a Hilbert space. A complete set of orthonormal vectors is 1 fn (x) = √ einx , 2π

{fn } :

n = 0, ±1, ±2, . . . .

(10.22)

whose inner products satisfy hfn , fm i = δn,m .

(10.23)

The expansion ∞ X

f=

hfn , f ifn

(10.24)

f (x)e−inx dx = an ,

(10.25)

n=−∞

is the Fourier expansion of f : 1 hfn , f i = √ 2π

Z

∞

−∞

where in terms of the Fourier coefficient an f (x) =

∞ X

1 an √ einx . 2π n=−∞

(10.26)

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

N

X 1

inx an e → 0 as N → 0, (10.27)

f (x) − √

2π n=−N

that is,

lim

N →∞

2 N 1 X inx an e = 0. dx f (x) − √ 2π n=−N −π

Z

π

(10.28)

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10.4

Linear Operators

A linear operator T on a vector space V is a rule assigning to each f ∈ V a unique vector T f ∈ V . It has the linearity property, T (αf + βg) = αT f + βT g,

(10.29)

where α, β are scalars. In an inner product space, the adjoint (or Hermitian conjugate) of T is defined by hf, T gi = hT † f, gi,

∀ f, g ∈ V.

(10.30)

T is self-adjoint (or Hermitian) if T † = T.

10.4.1

(10.31)

Sturm-Liouville Problem

Consider the space of twice continuously differentiable real functions defined on a segment of the real line x0 ≤ x ≤ x1 , (10.32) an incomplete subset of the Hilbert space L2 (x0 , x1 ). Under what conditions is the differential operator L = p(x)

d d2 + q(x) + r(x), 2 dx dx

(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 Z x1 dx u(x)Lv(x) hu, Lvi = x0   Z x1 d2 d dx u(x) p(x) 2 v(x) + q(x) v(x) + r(x)v(x) = dx dx x0 x1 Z x1 ′ dx [u(x)p(x)] v ′ (x) = u(x)p(x)v ′ (x) − x0

x0

x1 Z x1 ′ dx [u(x)q(x)] v(x) + u(x)q(x)v(x) − x0 x0 Z x1 dx u(x)r(x)v(x) + x

0  x1 ′ ′ = u(x)p(x)v (x) + u(x)q(x)v(x) − [u(x)p(x)] v(x)

x0

+

Z

x1

x0

′′ ′ dx [u(x)p(x)] v(x) − [u(x)q(x)] v(x) + u(x)r(x)v(x) 

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x1 = {p(x) [u(x)v ′ (x) − u′ (x)v(x)] + [q(x) − p′ (x)] u(x)v(x)} x0

+

Z

x1



′′

′

′

dx p(x)u (x) + [2p (x) − q(x)] u (x) + [p′′ (x) − q ′ (x) + r(x)] u(x) v(x). x0

The last integral here equals, for all v, v, Z x1 dx [Lu(x)] v(x) hLu, vi =

(10.34)

(10.35)

x0

if and only if 2p′ − 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 x1 (10.38) p(x) [u(x)v ′ (x) − u′ (x)v(x)] . x0

Only if this is zero is L Hermitian:

hu, Lvi = hLu, vi.

(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 )

d d2 − 2x dx2 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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(c) Homogeneous mixed boundary conditions are a linear combination of these conditions, u′ (x0 ) + α(x0 )u(x0 ) = 0, ′

u (x1 ) + α(x1 )u(x1 ) = 0,

(10.44a) (10.44b)

where α is some function, the same for all functions u in the space. 3. A third possibility is that the solutions may satisfy periodic boundary conditions, 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 integrated term (or, in higher dimensions, surface terms) are called self-adjoint boundary conditions. When they hold true, the differential equation   d d p(x) u(x) + r(x)u(x) = 0 (10.47) dx dx is self-adjoint. This equation is called the Sturm-Liouville equation.

10.5

Eigenvectors

If T is a (linear) operator and f 6= 0 is a vector such that T f = λf,

(10.48)

where λ is a complex number, then we say that f is a eigenvector (“characteristic vector”) belonging to the operator T , and λ is the corresponding eigenvalue. The following theorem is most important. The eigenvalues of a Hermitian operator are real, and the eigenvectors belonging to distinct eigenvalues are orthogonal. The proof is quite simple. If T f = λf,

T g = µg,

(10.49)

then hg, T f i = λhg, f i = hT g, f i = µ∗ hg, f i.

(10.50)

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

(10.51)

i.e., the eigenvalue λ is real, while then if λ 6= µ, we must have hg, f i = 0.

(10.52)

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Bessel Functions

The Bessel operator is 1 d ν2 d2 + . (10.53) − dx2 x dx x2 where ν is a real number. This is Hermitian in the space of real functions satisfying homogeneous boundary conditions (Dirichlet, Neumann, or mixed), where the inner product is defined by Z b hu, vi = x dx u(x)v(x). (10.54) Bν =

a

Proof: Note that d d ν2 x − (10.55) dx dx x is of the Sturm-Liouville form, (10.47), with p(x) = x, which is Hermitian with the L2 (a, b) inner product. Then xBν =

hu, Bν vi =

Z

b

dx u(x)xBν v(x) =

a

Z

b

a

dx xBν u(x)v(x) = hBν u, vi.

(10.56)

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

(10.57)

with the boundary conditions ψνn (b) = 0,

ψνn (0) = finite.

(10.58)

Here n enumerates the eigenvalues. The solutions to this problem are the Bessel functions, which satisfy the differential equation   2 1 d ν2 d (10.59) + + 1 − 2 Jν (z), dz 2 z dz z 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 = −λνn x. That is, p ψνn (x) = Jν ( −λνn x). (10.60)

The solutions we seek are Bessel functions of a real variable, so the acceptable eigenvalues satisfy λνn < 0, (10.61) 3 The second solution to Eq. (10.59), the so-called Neumann function N (z) [it is also ν denoted by Yν (z) and is more properly attributed to Weber], is not regular at the origin.

10.6. DUAL VECTORS. DIRAC NOTATION115 Version of November 16, 2011 so we write 2 −λνn = kνn .

(10.62)

Finally, we impose the boundary condition at x = b: 0 = ψνn (b) = Jν (kνn b),

(10.63)

that is, kνn b must be a zero of Jν . There are an infinite number of such zeros, as Fig. 10.1 illustrates. Let the nth zero of Jν be denoted by ανn , n = 1, 2, 3, . . .. For example, the first three zeros of J0 are α01 = 2.404826,

α02 = 5.520078,

α03 = 8.653728,

(10.64)

while the first three zeros of J1 (other than 0) are α11 = 3.83171,

α12 = 7.01559,

α13 = 10.17347.

Then the eigenvalues of the Bessel operator are  α 2 νn λνn = − , b

(10.65)

(10.66)

and the eigenfunctions are

 x . (10.67) Jν ανn b Because of the Hermiticity of Bν , these have the following orthogonality property, from Eq. (10.52), Z b  x x  Jν ανm = 0, n 6= m. (10.68) dx x Jν ανn b b 0

10.6

Dual Vectors. Dirac Notation

It is often convenient to think of the inner product as being composed by the multiplication to two different kinds of vectors. Thus, in 2-dimensional vector space we have column vectors,   v1 , (10.69) v= v2 and row vectors, v† = (v1∗ , v2∗ ).

(10.70)

†

As the notation indicates, the row vector v is the adjoint, the complex conjugate of the transpose of the column vector v. The inner product is then formed by the rules of matrix multiplication, hv, ui = v† u = v1∗ u1 + v2∗ u2 .

(10.71)

We generalize this notion to abstract vectors as follows. Denote a “right” vector (Dirac called it a “ket”) by |λi where λ is a name, or number, or set

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1.0

J0 J1 J2

0.5

0.0

-0.5 0.0

5.0

10.0 x

15.0

Figure 10.1: Plot of the Bessel functions of the first kind, J0 , J1 , and J2 , as functions of x.

20.0

10.6. DUAL VECTORS. DIRAC NOTATION117 Version of November 16, 2011 of numbers, labeling the vector. For example, if |λi is an eigenvector of some operator, λ might be the corresponding eigenvalue. The dual (or “conjugate”) vector to |λi is hλ| = (|λi)† ,

(10.72)

which is a “left” vector or a “bra” vector. For every right vector there is a unique left vector, and vice versa, in an inner product space. The inner product of |αi with hβ| is denoted hβ|αi. Note that the double vertical line has coalesced into a single line. This notation is a bracket notation, hence Dirac’s nomenclature. With row and column vectors there is not only an inner product, but an outer product as well: †

vu =



v1 v2



(u∗1 , u∗2 )

=



v1 u∗1 v1 u∗2 v2 u∗1 v2 u∗2



.

(10.73)

The result is a matrix or operator. So it is with abstract left and right vectors. We may define a dyadic by |αihβ| (10.74) which is an operator. When it acts on the right vector |γi it produces another vector, |αihβ||γi = |αihβ|γi, (10.75) where hβ|γi is a complex number, the inner product of hβ| and |γi; evidently the properties of an operator are satisfied.

10.6.1

Basis Vectors

Let |ni, n = 1, 2, . . . be a complete, orthonormal set of vectors, that is, they satisfy the properties hm|ni = δmn , (10.76a) and if |λi is any vector in the space, |λi =

∞ X

n=1

|nihn|λi.

(10.76b)

This is just a rewriting of the statement in Eq. (10.19). Since |λi is an arbitrary vector, we must have ∞ X |nihn| = I, (10.77) n=1

where I is the identity operator. This operator expression is the completeness relation for the vectors {|ni}.

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CHAPTER 10. LINEAR OPERATORS

L2 (V )

As we have seen, an important example of a Hilbert space is the space of all functions square-integrable in some region. For example, suppose we consider complex-valued functions f (r), where r = (x, y, z), r ∈ V , where V is some volume in 3-dimensional space, such that Z (dr)|f (r)|2 < ∞, (10.78) V

where the volume element (dr) = dx dy dz. We call this Hilbert space L2 (V ). Vectors in this space are functions: The function f corresponds to |f i, which we write as f (r) −→ |f i. (10.79) The inner product is

hf |gi =

Z

(dr)f ∗ (r)g(r).

(10.80)

V

It is most convenient to define the “function” δ(r − r0 ), the Dirac delta function, by the property Z f (r0 ) = (dr) δ(r − r0 )f (r) (10.81) V

for all f provided r0 lies within V . Regarding δ as a function (it is actually a linear functional defined by the integral equation above), we denote the corresponding vector in Hilbert space by |r0 i: δ(r − r0 ) −→ |r0 i.

(10.82)

(Actually, |r0 i is not a vector in L2 (V ), because it is not a square-integrable function.) Pictorially, |r0 i represents a function which is localized at r = r0 , i.e., it vanishes if r 6= r0 , but with the property Z (dr)δ(r − r0 )f (r) = f (r0 ); (10.83) hr0 |f i = V

the number hr0 |f i is the value of f at r0 . Also note that Z (dr)δ(r − r0 )δ(r − r1 ) = δ(r0 − r1 ). hr0 |r1 i =

(10.84)

V

In the above, we always assume that r0 and r1 lie in the volume V . It may be useful to recognize that in quantum mechanics |r0 i is an eigenvector of the position operator. It represents a state in which the particle has a definite position, namely r0 . Now notice that if the completeness relation (10.77) is multiplied on the right by |r′ i and on the left by hr|, it reads ∞ X

hr|nihn|r′ i = δ(r − r′ ).

n=1

(10.85)

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If we define ψn (r) = hr|ni as the values of what is now a complete set of functions, ∞ X ψn∗ (r′ )ψn (r) = δ(r − r′ ). (10.86) n=1

Implicit in what we are saying here is the assumption that the set of vectors |ri, r ∈ V , is complete: Z hg|f i = (dr)g ∗ (r)f (r) V Z (dr)hg|rihr|f i, (10.87) = V

which must mean, since |gi and hf | are arbitrary, Z (dr)|rihr|. I=

(10.88)

V

(Because the vectors are continuously, not discretely, labeled, the sum in Eq. (10.77) is replaced by an integral.) This will not be true if there are other variables in the problem, such as spin, but in that case the inner product is not given in terms of an integral over r alone.

10.8

Green’s Operator

We have now reached the taking-off point for the discussion of Green’s functions. We will in this section sketch the general type of problem we wish to consider. In the next chapter we will fill in the details, by considering physical examples. Let L be a self-adjoint linear operator in a Hilbert space. We wish to find the solutions |ψi to the following vector equation (L − λ)|ψi = |Si,

(10.89)

where |Si is a prescribed vector, the “source,” and λ is a real number not equal to any of the (real) eigenvectors of L. Suppose the eigenvectors of L, which satisfy L|ni = λn |ni, are complete, and are orthonormalized, X |nihn| = I.

(10.90)

(10.91)

n

We may then expand |ψi in terms of these, X |ψi = |nihn|ψi. n

(10.92)

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When we insert this expansion into Eq. (10.89) and use the eigenvalue equation (10.90), we obtain X (λn − λ)|nihn|ψi = |Si. (10.93) n

Now multiply this equation on the left by hn′ |, and use the orthonormality property hn′ |ni = δn′ n , (10.94) to find (n′ → n)

(λn − λ)hn|ψi = hn|Si,

or, provided λ 6= λn , hn|ψi = Then from Eq. (10.92) we deduce |ψi =

hn|Si . λn − λ

X |nihn| |Si, λn − λ n

(10.95)

(10.96)

(10.97)

which means we have solved for |ψi in terms of the presumably known eigenvectors and eigenvalues of L. We write this more compactly as |ψi = G|Si,

(10.98)

where G, the Green’s operator, is G=

X |nihn| ; λn − λ n

(10.99)

the sum ranges over all the eigenvectors of L. We regard Eq. (10.98) as the definition of G: the response of a linear system is linear in the source. Eq. (10.99) is the eigenvector expansion of G. Two properties of G follow immediately from the above: 1. From Eq. (10.99), since both λ and λn are real, we see that G is Hermitian, G† = G.

(10.100)

2. From either of Eqs. (10.98) or (10.99) we see that G satisfies the operator equation (L − λ)G = I. (10.101) The case of functions is the most important one. Then if we use Eq. (10.88), the inhomogeneous equation (10.89) becomes Z (10.102) hr|(L − λ) (dr′ )|r′ ihr′ |ψi = hr|Si. V

121 Version of November 16, 2011

10.8. GREEN’S OPERATOR Suppose

ˆ hr|L|r′ i = Lδ(r − r′ ),

(10.103)

ˆ is a differential operator (the usual case), and let us further write where L hr′ |ψi = ψ(r′ ),

hr|Si = S(r).

(10.104)

Then the inhomogeneous equation (10.102) reads ˆ − λ)ψ(r) = S(r) (L

(10.105)

The solution to Eq. (10.105) is given by hr| times Eq. (10.98): Z hr|ψi = hr|G (dr′ )|r′ ihr′ |Si,

(10.106)

V

or ψ(r) =

Z

(dr′ ) G(r, r′ )S(r′ ),

(10.107)

V

where we have written the Green’s function as G(r, r′ ) = hr|G|r′ i.

(10.108)

The eigenfunction expansion of G(r, r′ ) is G(r, r′ ) =

X ψ ∗ (r′ )ψ(r) n

λn − λ

,

(10.109)

where the eigenfunctions, satisfying Eq. (10.86), are ψn (r) = hr|ni. Now the properties of G(r, r′ ) are 1. The reciprocity relation: G(r, r′ ) = G∗ (r′ , r),

(10.110)

which follows immediately from the eigenfumction expansion (10.109) or from Eq. (10.100): hr|G† |r′ i = hr′ |G|ri∗ = hr|G|r′ i.

(10.111)

2. The differential equation satisfied by the Green’s function is ˆ − λ)G(r, r′ ) = δ(r − r′ ), (L

(10.112)

which follows from Eqs. (10.107), (10.109), or (10.101). 3. Now we have an additional property. If ψn (r) satisfy homogeneous boundary conditions, for example, ψn (r) = 0 on the surface of V , G(r, r′ ) satisfies the same conditions, for example it vanishes when r or r′ lies on the surface of V .

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Note that the eigenfunction expansion of G(r, r′ ), Gλ (r, r′ ) =

X ψ ∗ (r′ )ψn (r) n

n

λn − λ

,

(10.113)

where now the parameter λ has been made explicit in G, says that Gλ has simple poles at each of the eigenvalues λn , and that the residue of the pole of Gλ at λ = λn is Res Gλ (r, r′ ) λ=λn = −ψn∗ (r′ )ψ(r). (10.114) If the eigenvalue is degenerate, that is, there is more than one eigenfunction corresponding to a given eigenvalue, one obtains a sum over all the ψn∗ ψn corresponding to λn . Thus, if G may be determined by other means than by an eigenfunction expansion, such as directly solving the differential equation (10.112), from it ˆ may be determined. We will the eigenvalues and normalized eigenfunctions of L illustrate this eigenfunction decomposition in the next chapter.

Chapter 11

Green’s Functions 11.1

One-dimensional Helmholtz Equation

Suppose we have a string driven by an external force, periodic with frequency ω. The differential equation (here f is some prescribed function) 

∂2 1 ∂2 − 2 2 2 ∂x c ∂t



U (x, t) = f (x) cos ωt

(11.1)

represents the oscillatory motion of the string, with amplitude U , which is tied down at both ends (here l is the length of the string): U (0, t) = U (l, t) = 0.

(11.2)

We seek a solution of the form (thus we are ignoring transients) U (x, t) = u(x) cos ωt,

(11.3)

so u(x) satisfies 

 d2 2 + k u(x) = f (x), dx2

k = ω/c.

(11.4)

The solution to this inhomogeneous Helmholtz equation is expressed in terms of the Green’s function Gk (x, x′ ) as u(x) =

Z

l

dx′ Gk (x, x′ )f (x′ ),

(11.5)

0

where the Green’s function satisfies the differential equation 

 d2 2 Gk (x, x′ ) = δ(x − x′ ). + k dx2

123 Version of December 3, 2011

(11.6)

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As we saw in the previous chapter, the Green’s function can be written down in terms of the eigenfunctions of d2 /dx2 , with the specified boundary conditions,   2 d (11.7a) − λn un (x) = 0, dx2 un (0) = un (l) = 0. (11.7b) The normalized solutions to these equations are r  nπ 2 nπx 2 un (x) = , n = 1, 2, . . . . (11.8) sin , λn = − l l l p The factor 2/l is a normalization factor. From the general theorem about eigenfunctions of a Hermitian operator given in Sec. 10.5, we have Z mπx nπx 2 l sin = δnm . (11.9) dx sin l 0 l l Thus the Green’s function for this problem is given by the eigenfunction expansion ∞ 2 nπx nπx′ X ′ l sin l sin l Gk (x, x ) = (11.10)
2 . k 2 − nπ n=1 l

But this form is not usually very convenient for calculation. Therefore we solve the differential equation (11.6) directly. When x 6= x′ the inhomogeneous term is zero. Since Gk (0, x′ ) = Gk (l, x′ ) = 0,

(11.11)

we must have x < x′ : x > x′ :

Gk (x, x′ ) = a(x′ ) sin kx, Gk (x, x′ ) = b(x′ ) sin k(x − l).

(11.12a) (11.12b)

We determine the unknown functions a and b by noting that the derivative of G must have a discontinuity at x = x′ , which follows from the differential equation (11.6). Integrating that equation just over that discontinuity we find  Z x′ +ǫ  2 d 2 Gk (x, x′ ) = 1, (11.13) + k dx dx2 x′ −ǫ or

x=x′ +ǫ d ′ Gk (x, x ) = 1, dx x=x′ −ǫ

(11.14)

d Gk (x, x′ ) is discontinuous at because 2ǫGk (x′ , x′ ) → 0 as ǫ → 0. Although dx ′ ′ x = x , G(x, x ) is continuous there: Z x′ +ǫ d G(x′ + ǫ, x′ ) − G(x′ − ǫ, x′ ) = dx G(x, x′ ) dx x′ −ǫ

11.1. ONE-DIMENSIONAL HELMHOLTZ EQUATION125 Version of December 3, 2011 Z x′ +ǫ d d ′ dx G(x, x′ ) dx G(x, x ) + = dx dx ′ ′ x x −ǫ   d d ′ ′ =ǫ , G(x, x ) + G(x, x ) dx dx x=x′ −ξ x=x′ +ξ¯ Z

x′

(11.15)

where by the mean value theorem, 0 < ξ ≤ ǫ, 0 < ξ¯ ≤ ǫ. Therefore x=x′ +ǫ ′ G(x, x ) = O(ǫ) → 0 as ǫ → 0. ′

(11.16)

x=x −ǫ

Now using the continuity of G and the discontinuity of G′ , we find two equations for the coefficient functions a and b: ′

a(x′ ) sin kx′ = b(x′ ) sin k(x′ − l), ′

′

′

a(x )k cos kx + 1 = b(x )k cos k(x − l).

(11.17a) (11.17b)

It is easy to solve for a and b. The determinant of the coefficient matrix is sin kx′ − sin k(x′ − l) = −k sin kl, D = (11.18) k cos kx′ −k cos k(x′ − l) independent of x′ . Then the solutions are 1 0 − sin k(x′ − l) sin k(x′ − l) ′ , = a(x ) = D −1 −k cos k(x′ − l) k sin kl sin kx′ 1 sin kx′ 0 = . b(x′ ) = D k cos kx′ −1 k sin kl

(11.19a) (11.19b)

Thus we find a closed form for the Green’s function in the two regions: x < x′ : x > x′ :

sin k(x′ − l) sin kx , k sin kl ′ sin kx sin k(x − l) , Gk (x, x′ ) = k sin kl

Gk (x, x′ ) =

(11.20a) (11.20b)

or compactly,

1 sin kx< sin k(x> − l), k sin kl where we have introduced the notation Gk (x, x′ ) =

(11.21)

x< is the lesser of x, x′ , x> is the greater of x, x′ .

(11.22)

Note that Gk (x, x′ ) = Gk (x′ , x) as is demanded on general grounds, as a consequence of the reciprocity relation (10.110). Let us analyze the analytic structure of Gk (x, x′ ) as a function of k. We see that simple poles occur where kl = nπ,

n = ±1, ±2, . . . .

(11.23)

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There is no pole at k = 0. For k near nπ/l, we have sin kl = sin nπ + (kl − nπ) cos nπ + . . . = (kl − nπ)(−1)n .

(11.24)

If we simply sum over all the poles of Gk , we obtain Gk (x, x′ ) =

∞ X

(−1)n

n=−∞ n6=0

< sin nπ sin nπx l l (x> − l) nπ l (kl − nπ)

∞ nπx′ X sin nπx l sin
l = nπ k − nπ n=−∞ l n6=0

= =

∞ X

n=1 ∞ X

sin

  1 1 nπx′ 1 nπx − sin l l nπ k − nπ k + nπ l l

2 1 nπx nπx′ sin sin
2 . 2 l l l k − nπ n=1

(11.25)

l

This is in fact equal to Gk , as seen in the eigenfunction expansion (11.10), because the difference is an entire function vanishing at infinity, which must be zero by Liouville’s theorem, see Sec. 6.5.

11.2

Types of Boundary Conditions

Three types of second-order, homogeneous differential equations are commonly encountered in physics (the dimensionality of space is not important):   1 ∂2 2 (11.26a) Hyperbolic: ∇ − 2 2 u(r, t) = 0, c ∂t
(11.26b) Elliptic: ∇2 + k 2 u(r) = 0,   1 ∂ Parabolic: ∇2 − T (r, t) = 0. (11.26c) κ ∂t The first of these equations is the wave equation, the second is the Helmholtz equation, which includes Laplace’s equation as a special case (k = 0), and the third is the diffusion equation. The types of boundary conditions, specified on which kind of boundaries, necessary to uniquely specify a solution to these equations are given in Table 11.1. Here by Cauchy boundary conditions we means that both the function u and its normal derivative ∂u/∂n is specified on the boundary. Here ∂u =n ˆ · ∇u, (11.27) ∂n where n ˆ is a(n outwardly directed) normal vector to the surface. As we have seen previously, Dirichlet boundary conditions refer to specifying the function u on the surface, Neumann boundary conditions refer to specifying the normal derivative ∂u/∂n on the surface, and mixed boundary conditions refer to

11.3. EXPRESSION OF FIELD IN TERMS OF GREEN’S FUNCTION127 Version of December 3, 2011 Type of Equation Hyperbolic Elliptic Parabolic

Type of Boundary Condition Cauchy Dirichlet, Neumann, or mixed Dirichlet, Neumann, or mixed

Type of Boundary Open Closed Open

Table 11.1: Boundary conditions required for the three types of second-order differential equations. The boundary conditions referred to in the first and third cases are actually initial conditions. specifying a linear combination, αu + β∂u/∂n, on the surface. If the specified boundary values are zero, we say that the boundary conditions are homogeneous; otherwise, they are inhomogeneous. Example. To determine the vibrations of a string, described by  2  ∂ 1 ∂2 u = 0, − ∂x2 c2 ∂t2

(11.28)

we must specify u(x, 0),

∂u (x, 0) ∂t

(11.29)

at some initial time (t = 0). The line t = 0 is an open surface in the (ct, x) plane.

11.3

Expression of Field in Terms of Green’s Function

Typically, one determines the eigenfunctions of a differential operator subject to homogeneous boundary conditions. That means that the Green’s functions obey the same conditions. See Sec. 10.8. But suppose we seek a solution of (L − λ)ψ = S

(11.30)

subject to inhomogeneous boundary conditions. It cannot then be true that Z (dr′ ) G(r, r′ )S(r′ ). (11.31) ψ(r) = V

To see how to deal with this situation, let us consider the example of the three-dimensional Helmholtz equation, (∇2 + k 2 )ψ(r) = S(r).

(11.32)

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We seek the solution ψ(r) subject to arbitrary inhomogeneous Dirichlet, Neumann, or mixed boundary conditions on a surface Σ enclosing the volume V of interest. The Green’s function G for this problem satisfies (∇2 + k 2 )G(r, r′ ) = δ(r − r′ ),

(11.33)

subject to homogeneous boundary conditions of the same type as ψ satisfies. Now multiply Eq. (11.32) by G, Eq. (11.33) by ψ, subtract, and integrate over the appropriate variables: Z   (dr′ ) G(r, r′ )(∇′2 + k 2 )ψ(r′ ) − ψ(r′ )(∇′2 + k 2 )G(r, r′ ) ZV (dr′ ) [G(r, r′ )S(r′ ) − ψ(r′ )δ(r − r′ )] . (11.34) = V

Here we have interchanged r and r′ in Eqs. (11.32) and (11.33), and have used the reciprocity relation, G(r, r′ ) = G(r′ , r). (11.35) (We have assumed that the eigenfunctions and hence the Green’s function are real.) Now we use Green’s theorem to establish Z   dσ · G(r, r′ )∇′ ψ(r′ ) − ψ(r′ )∇′ G(r, r′ ) −  ZΣ ψ(r), r ∈ V, + (dr′ ) G(r, r′ )S(r′ ) = (11.36) 0, r 6∈ V, V where in the surface integral dσ is the outwardly directed surface element, and r′ lies on the surface Σ. This generalizes the simple relation given in Eq. (11.31). How do we use this result? We always suppose G satisfies homogeneous boundary conditions on Σ. If ψ satisfies the same conditions, then for r ∈ V Eq. (11.31) holds. But suppose ψ satisfies inhomogeneous Dirichlet boundary conditions on Σ, ψ(r′ ) r′ ∈Σ = ψ0 (r′ ), (11.37)

a specified function on the surface. Then we impose homogeneous Dirichlet conditions on G, G(r, r′ ) r′ ∈Σ = 0. (11.38)

Then the first surface term in Eq. (11.36) is zero, but the second contributes. For example if S(r) = 0 inside V , we have for r ∈ V Z ψ(r) = dσ · [∇′ G(r, r′ )]ψ0 (r′ ), (11.39) Σ

which express ψ in terms of its boundary values. If ψ satisfies inhomogeneous Neumann conditions on Σ, ∂ψ ′ (r ) = N (r′ ), ∂n′ ′ r ∈Σ

(11.40)

11.4. HELMHOLTZ EQUATION INSIDE A SPHERE129 Version of December 3, 2011 a specified function, then we use the Green’s function which respects homogeneous Neumann conditions, ∂ ′ G(r, r ) = 0, (11.41) ′ ∂n r′ ∈Σ so again if S = 0 inside V , we have within V Z ψ(r) = − dσ G(r, r′ )N (r′ ).

(11.42)

Σ

Finally, if ψ satisfies inhomogeneous mixed boundary conditions,   ∂ ′ ′ ′ ψ(r ) + α(r )ψ(r ) = F (r′ ), ′ ∂n ′ r ∈Σ

(11.43)

then when G satisfies homogeneous boundary conditions of the same type   ∂ ′ ′ + α(r ) G(r, r ) = 0, (11.44) ′ ∂n r′ ∈Σ

we have for r ∈ V

ψ(r) =

Z

V

11.4

(dr′ ) G(r, r′ )S(r′ ) −

Z

dσ G(r, r′ )F (r′ ).

(11.45)

σ

Helmholtz Equation Inside a Sphere

Here we wish to find the Green’s function for Helmholtz’s equation, which satisfies (∇2 + k 2 )Gk (r, r′ ) = δ(r − r′ ), (11.46) in the interior of a spherical region of radius a, with homogeneous Dirichlet boundary conditions on the surface, ′ Gk (r, r ) = 0. (11.47) |r|=a

We will use two methods.

11.4.1

Eigenfunction Method

We know that the eigenfunctions of the Laplacian are jl (kr)Ylm (θ, φ),

(11.48)

in spherical polar coordinates, r, θ, φ; that is, (∇2 + k 2 )jl (kr)Ylm (θ, φ) = 0.

(11.49)

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Here jl is the spherical Bessel function, r π Jl+1/2 (x), jl (x) = 2x

(11.50)

and the Ylm are the spherical harmonics, Ylm (θ, φ) =



2l + 1 (l − m)! 4π (l + m)!

1/2

Plm (cos θ)eimφ ,

(11.51)

where Plm is the associated Legrendre function. Here l is a nonnegative integer, and m is an integer in the range −l ≤ m ≤ l. For example, the first few spherical Bessel functions (which are simpler than the cylinder functions, the Bessel functions of integer order) are sin x , x sin x cos x j1 (x) = − , 2 x x 3 1 3 sin x − 2 cos x, − j2 (x) = x3 x x

j0 (x) =

(11.52a) (11.52b) (11.52c)

and in general l

jl (x) = x



1 d − x dx

l

sin x . x

The associated Legrendre function is given by  l+m d (cos2 θ − 1)l Plm (cos θ) = (−1)m sinm θ . d cos θ 2l l!

(11.53)

(11.54)

For example, the first few spherical harmonics are 1 Y00 = √ , 4π r 3 Y11 = − sin θ eiφ , 8π r 3 Y10 = cos θ, 4π r 3 1 sin θ e−iφ , Y1 = 8π r 15 2 Y2 = sin2 θ e2iφ , 32π r 15 1 Y2 = − cos θ sin θ eiφ , 8π r 5 0 Y2 = (3 cos2 θ − 1), 16π

(11.55a) (11.55b) (11.55c) (11.55d) (11.55e) (11.55f) (11.55g)

11.4. HELMHOLTZ EQUATION INSIDE A SPHERE131 Version of December 3, 2011 Y2−1 Y2−2

r

15 cos θ sin θ e−iφ , 8π r 15 = sin2 θ e−2iφ . 32π =

(11.55h) (11.55i)

The eigenfunctions must vanish ar r = a, so if βln is the nth zero of jl , jl (βln ) = 0,

n = 1, 2, 3, . . . ,

(11.56)

the desired eigenfunctions are  r m Y (θ, φ), ψnlm (r, θ, φ) = Anl jl βln a l

(11.57)

and the eigenvalues are

λln =

2 −kln

=−



βln a

2

.

(11.58)

The normalization constant Anl is determined by the requirement that Z 2 r2 dr dΩ |ψnlm (r, θ, φ)| = 1, (11.59) where dΩ = sin θ dθ dφ is the element of solid angle. Since the spherical harmonics are normalized so that [Ω = (θ, φ) represents a point on the unit sphere] Z ′ ∗ (11.60) dΩ Ylm (Ω)Ylm (Ω) = δll′ δmm′ , ′ the normalization constant is determined by the requirement Z a h  r i2 |Anl |2 r2 dr jl βln = 1. a 0 Now

Z

0

a

1 2 (βln ), r2 dr jl (βln r/a)jl (βlm r/a) = δnm a3 jl+1 2

which for n 6= m follows from the orthogonality property (10.68). So r 1 2 , |Anl | = 3 a jl+1 (βln )

(11.61)

(11.62)

(11.63)

and the Green’s function has the eigenfunction expansion Gk (r, r′ ) =

X 2 Ylm (Ω)Ylm∗ (Ω′ )jl (βln r/a)jl (βln r′ /a) 1 , 2 a3 jl+1 (βln ) k 2 − (βln /a)2

nlm

where Ω = (θ, φ), Ω′ = (θ′ , φ′ ).

(11.64)

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This result can be simplified by carrying out the sum on m, using the addition theorem for spherical harmonics, l 4π X m∗ ′ m Yl (Ω )Yl (Ω) = Pl (cos γ), 2l + 1

(11.65)

m=−l

where Pl (cos γ) = Pl0 (cos γ) is Legendre’s polynomial, and γ is the angle between the directions represented by Ω and Ω′ , or cos γ = cos θ cos θ′ + sin θ sin θ′ cos(φ − φ′ ).

(11.66)

Then we obtain Gk (r, r′ ) =

2 X 2l + 1 jl (βln r/a)jl (βln r′ /a) 1 . Pl (cos γ) 2 3 a 4π jl+1 (βln ) k 2 − (βln /a)2

(11.67)

nl

This leads us to the second method.

11.4.2

Discontinuity (Direct) Method

Let us adopt the angular dependence found above: Gk (r, r′ ) =

∞ X 2l + 1 l=0

4π

Pl (cos γ)gl (r, r′ ),

(11.68)

where we will call gl the reduced Green’s function. Because Ylm is an eigenfunction of the angular part of the Laplacian operator, ∇2 Ylm (Ω) = −

l(l + 1) m Yl (Ω), r2

(11.69)

and the delta function can be written as δ(r − r′ ) =

1 δ(r − r′ )δ(Ω − Ω′ ), rr′

(11.70)

we see that, because of the orthonormality of the spherical harmonics, Eq. (11.60), the Green’s function equation (11.46) corresponds to the following equation satisfied by the reduced Green’s function, the inhomogeneous “spherical Bessel equation,”  2  d 1 2 d l(l + 1) 2 gl (r, r′ ) = ′ δ(r − r′ ). + + k (11.71) − dr2 r dr r2 rr We solve this equation directly. For (0 < r′ < a) 0 ≤ r < r′ : r′ < r ≤ a :

gl (r, r′ ) = a(r′ )jl (kr), gl (r, r′ ) = b(r′ )jl (kr) + c(r′ )nl (kr).

(11.72a) (11.72b)

11.4. HELMHOLTZ EQUATION INSIDE A SPHERE133 Version of December 3, 2011 Only jl appears in the first form because the solution must be finite at r = 0, and the second solution to the spherical Bessel equation, r π Nl+1/2 (x), (11.73) nl (x) = 2x where Nν is the Neumann function, is singular at x = 0. For example, n0 (x) = −

cos x , x

(11.74)

and in general  l 1 d cos x nl (x) = −xl − . x dx x

(11.75)

To determine the functions a, b, and c, we proceed as follows. The boundary condition at r = a, gl (a, r′ ) = 0, implies

or

0 = b(r′ )jl (ka) + c(r′ )nl (ka),

(11.76)

nl (ka) b(r′ ) =− . c(r′ ) jl (ka)

(11.77)

Thus we can write in the outer region, a ≥ r > r′ :

gl (r, r′ ) = A(r′ )[jl (kr)nl (ka) − nl (kr)jl (ka)].

(11.78)

The next condition we impose is that of the continuity of gl at r = r′ : a(r′ )jl (kr′ ) = A(r′ )[jl (kr′ )nl (ka) − nl (kr′ )jl (ka)].

(11.79)

On the other hand, the derivative of gl is discontinuous at r = r′ , as we may see by integrating Eq. (11.71) over a tiny interval around r = r′ :

which implies

r=r′ +ǫ d 1 ′ gl (r, r ) = ′2 , dr r ′ r=r −ǫ

ka(r′ )jl′ (kr′ ) − kA(r′ )[jl′ (kr′ )nl (ka) − n′l (kr′ )jl (ka)] = −

(11.80)

1 . r′2

(11.81)

Now multiply Eq. (11.79) by kjl′ (kr′ ), and Eq. (11.81) by jl (kr′ ), and subtract: jl (kr′ ) = −kA(r′ )jl (ka)[jl (kr′ )n′l (kr′ ) − nl (kr′ )jl′ (kr′ )]. r′2

(11.82)

Now jl , nl are the independent solutions of the spherical Bessel equation     1 d l(l + 1) 2 d 2 r − u = 0, (11.83) + k r2 dr dr r2

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the Wronskian of which, ∆(r) ≡ jl (kr)n′l (kr) − nl (kr)jl′ (kr)

(11.84)

has the form

const. , (11.85) r2 as we saw in Problem 4 of Assignment 8. We can determine the constant by considering the asymptotic forms of jl , nl , ∆(r) =

sin(kr − lπ/2) , kr ≫ 1, kr cos(kr − lπ/2) nl (kr) ∼ − , kr ≫ 1, kr jl (kr) ∼

(11.86a) (11.86b)

which imply ∆(r) = =

1 k 2 r2

[sin2 (kr − lπ/2) + cos2 (kr − lπ/2)]

1 . (kr)2

(11.87)

Thus since the right-hand side of Eq. (11.82) is proportional to the Wronskian, we find the function A: jl (kr′ ) , (11.88) A(r′ ) = −k jl (ka) and then from Eq. (11.79) we find the function a: a(r′ ) = −

k [jl (kr′ )nl (ka) − nl (kr′ )jl (ka)]. jl (ka)

(11.89)

Hence the Green’s function is explicitly r < r′ : r > r′ : or

jl (kr) [jl (kr′ )nl (ka) − nl (kr′ )jl (ka)], (11.90a) jl (ka) jl (kr′ ) gl (r, r′ ) = −k [jl (kr)nl (ka) − nl (kr)jl (ka)], (11.90b) jl (ka) gl (r, r′ ) = −k

 nl (ka) nl (kr> ) , − gl (r, r ) = −kjl (kr< )jl (kr> ) jl (ka) jl (kr> ) ′



(11.91)

where r< is the lesser of r, r′ , and r> is the greater. From this closed form we may extract the eigenvalues and eigenfunctions of the spherical Bessel differential operator appearing in Eq. (11.83). The poles of gl occur where jl (ka) has zeroes, all of which are real, at ka = βln , the nth zero of jl , or  2 βln k2 = . (11.92) a

11.5. HELMHOLTZ EQUATION IN UNBOUNDED SPACE135 Version of December 3, 2011 In the neighborhood of this zero, jl (ka) = (ka − βln )jl′ (βln ).

(11.93)

But at the zero the Wronskian is 1 = −nl (βln )jl′ (βln ). (βln )2

(11.94)

Now from the recursion relation Jλ′ (z) =

λ Jλ (z) − Jλ+1 (z), z

(11.95)

we see that the derivative of the spherical Bessel function (11.50) satisfies, at the zero, jl′ (βln ) = −jl+1 (βln ). (11.96) Thus the residue of the pole of gl at k = βln /a is 1 a2 β

ln

jl (βln r< /a)jl (βln r> /a) . 2 (β ) jl+1 ln

(11.97)

Now jl is an even or odd function of z depending on whether n is even or odd. So if βln is a zero of jl , so is −βln , and hence if we add the contributions of these two poles, we get the corresponding contribution to gl :   1 1 1 jl (βln r/a)jl (βln r′ /a) ′ . (11.98) − gl (r, r ) ∼ 2 a βln [jl+1 (βln )]2 k − βln /a k + βln /a Summing up the contribution of all such pairs of poles, we obtain gl (r, r′ ) =

∞ 2 X jl (βln r/a)jl (βln r′ /a) 1 , a3 n=1 [jl+1 (βln )]2 k 2 − (βln /a)2

(11.99)

which is the eigenfunction expansion displayed in Eq. (11.67).

11.5

Helmholtz Equation in Unbounded Space

Again we are solving the equation (∇2 + k 2 )Gk (r, r′ ) = δ(r − r′ ),

(11.100)

but now in unbounded space. The solution to this equation is an outgoing spherical wave: ′ 1 eik|r−r | . (11.101) Gk (r, r′ ) = Gk (r − r′ ) = − 4π |r − r′ |

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This may be directly verified. Consider a small sphere S, of radius ǫ, centered on r′ :   Z Z 1 eikρ 2 2 2 ′ (dρ)∇ρ − (dr)(∇ + k )Gk (r − r ) ≈ 4π ρ S S   Z 1 eikρ 2 d − = dΩ ρ dρ 4π ρ ρ=ǫ → 1,

(11.102)

as ǫ → 0. Evidently, for r 6= r′ , Gk satisfies the Helmholtz equation, (∇2 + k 2 )Gk = 0. Alternatively, we may construct Gk from the eigenfunction expansion (10.109), X ψ ∗ (r′ )ψn (r) n Gk (r − r′ ) = (11.103) λn − λ n where λ = −k 2 , λn = −k ′2 , where the eigenfunctions are solutions of (∇2 + k ′2 )ψk′ (r) = 0,

(11.104)

that is, they are plane waves, ψk′ (r) =

′ 1 eik ·r , 3/2 (2π)

Here the (2π)−3/2 factor is for normalization: Z (dk′ ) ψk′ (r)∗ ψk′ (r′ ) = δ(r − r′ ), Z (dr) ψk′ (r)∗ ψk (r) = δ(k − k′ ),

(11.105)

(11.106a) (11.106b)

where we have noted that the spectrum of eigenvalues is continuous, Z X → (dk). (11.107) n

Thus the eigenfunction expansion for the Green’s function has the form Z ′ ′ ′ (dk′ ) e−ik ·r eik ·r . (11.108) Gk (r − r′ ) = (2π)3 k 2 − k ′2 Let us evaluate this integral in spherical coordinates, where we write (dk′ ) = k ′2 dk ′ dφ′ dµ′ ,

µ′ = cos θ′ ,

(11.109)

where we have chosen the z axis to lie along the direction of r − r′ . The integration over the angles is easy: Z ∞ Z 2π Z 1 ik′ |r−r′ |µ′ 1 ′ ′2 ′ ′e dk k dφ dµ Gk (r − r′ ) = (2π)3 0 k 2 − k ′2 0 −1 Z ∞   1 1 dk ′ k ′2 1 ik′ ρ −ik′ ρ , (11.110) e − e = (2π)2 2 −∞ k 2 − k ′2 ik ′ ρ

11.5. HELMHOLTZ EQUATION IN UNBOUNDED SPACE137 Version of December 3, 2011 

k •

 • −k

-

 Figure 11.1: Contour in the k ′ plane used to evaluate the integral (11.110). The integral is closed in the upper (lower) half-plane if the exponent is positive (negative). The poles in the integrand are avoided by passing above the one on the left, and below the one on the right. R∞ R∞ defining ρ = |r − r′ |, where we have replaced 0 by 21 −∞ because the integrand is even in k ′2 . We evaluate this integral by contour methods. Because now k can coincide with an eigenvalue k ′ , we must choose the contour appropriately to define the Green’s function. Suppose we choose the contour as shown in Fig. 11.1, passing below the pole at k and above the pole at −k. We close the contour in the upper half plane for the eikρ and in the lower half plane for the e−ikρ term. Then by Jordan’s lemma, we immediately evaluate the integral:   1 1 2πi k eikρ 2πi k eikρ Gk (r − r ) = − + (2π)2 2 2k iρ −2k iρ ′

= −

1 eikρ , 4π ρ

(11.111)

which coincides with Eq. (11.101). If a different contour defining the integral had been chosen, we would have obtained a different Green’s function, not one corresponding to outgoing spherical waves. Boundary conditions uniquely determine the contour. Note that Gk (r, r′ ) = Gk (r′ , r), (11.112) even though Gk is complex. The self-adjointness property (10.110) implied by the eigenfunction expansion is only formal, and is spoiled by the contour choice.

138 Version of December 3, 2011

11.6

CHAPTER 11. GREEN’S FUNCTIONS

Green’s Function for the Scalar Wave Equation

The inhomogeneous scalar wave equation,   1 ∂2 ∇2 − 2 2 ψ(r, t) = ρ(r, t), c ∂t

(11.113)

requires boundary and initial conditions. The boundary conditions may be Dirichlet, Neumann, or mixed. The initial conditions are Cauchy (see Sec. 11.2). ∂ Thus, we might specify at an initial time t = t0 both ψ(r, t0 ) and ∂t ψ(r, t0 ) at every point r in the region being considered. The corresponding Green’s function G(r, t; r′ , t′ ) satisfies   1 ∂2 2 (11.114) ∇ − 2 2 G(r, t; r′ , t′ ) = δ(r − r′ )δ(t − t′ ). c ∂t It must satisfy the homogeneous form of the boundary conditions satisfied by ψ. Thus, if ψ has a specified value everywhere on the bounding surface, the corresponding Green’s function must vanish on the surface. In classical physics it is customary to adopt as initial conditions  G(r, t; r′ , t′ ) = 0 if t < t′ . (11.115) ∂G ′ ′ (r, t; r , t ) ∂t These then define the so-called retarded Green’s functions. They ensure that an effect occurs after its cause. (In fact, however, this time asymmetry of the Green’s function, which is not present in the wave equation, is not necessary; and in fact it is impossible to maintain in relativistic quantum mechanics.) With such a Green’s function, what takes the place of the self-adjointness property given in Sec. 10.8? Since the second time derivative is invariant under t → −t, we have in addition to the inhomogeneous equation (11.114)   1 ∂2 2 ∇ − 2 2 G(r, −t; r′′ , −t′′ ) = δ(r − r′′ )δ(t − t′′ ). (11.116) c ∂t Multiply Eq. (11.116) by G(r, t; r′ , t), Eq. (11.114) by G(r, −t; r′′ , −t′′ ), subtract, and integrate over the volume being considered, and over t from −∞ to T , where T > t′ , t′′ :  Z T Z dt (dr) G(r, t; r′ , t′ )∇2 G(r, −t; r′′ , −t′′ ) −∞

V

− G(r, −t; r′′ , −t′′ )∇2 G(r, t; r′ , t′ ) 1 ∂2 − G(r, t; r′ , t′ ) 2 2 G(r, −t; r′′ , −t′′ ) c ∂t  1 ∂2 + G(r, −t; r′′ , −t′′ ) 2 2 G(r, t; r′ , t′ ) c ∂t = −G(r′ , −t′ ; r′′ , −t′′ ) + G(r′′ , t′′ ; r′ , t′ ).

(11.117)

11.6. GREEN’S FUNCTION FOR THE SCALAR WAVE EQUATION139 Version of December 3, 2011 Now use Green’s theorem, together with the corresponding identity, ∂ ∂t

  ∂ ∂2 ∂ ∂2 A B − B A = A 2 B − B 2 A, ∂t ∂t ∂t ∂t

(11.118)

to conclude that G(r′′ , t′′ ; r′ , t′ ) − G(r′ , −t′ ; r′′ , −t′′ )  Z T Z dσ · G(r, t; r′ , t′ )∇G(r, −t; r′′ , −t′′ ) = dt Σ −∞  ′′ ′′ ′ ′ − G(r, −t; r , −t )∇G(r, t; r , t )  Z ∂ 1 − (dr) 2 G(r, t; r′ , t′ ) G(r, −t; r′′ , −t′′ ) c ∂t V  t=T ′′ ′′ ∂ ′ ′ − G(r, −t; r , −t ) G(r, t; r , t ) . ∂t t=−∞

(11.119)

The surface integral vanishes, since both Green’s functions satisfy the same homogeneous boundary conditions on Σ. (The boundary conditions are time independent.) The second integral is also zero because from Eq. (11.115) G(r, −∞; r′ , t′ ) ∂G ′ ′ ∂t (r, −∞; r , t ) since −∞ < t′ , and



G(r, −T ; r′′ , −t′′ ) ∂G ′′ ′′ ∂t (r, −T ; r , −t )

= 0,



= 0,

(11.120a)

(11.120b)

since −T < −t′′ . Thus the reciprocity relation here is G(r, t; r′ , t′ ) = G(r′ , −t′ ; r, −t)

(11.121)

How do we express a solution to the wave equation (11.113) in terms of the Green’s function? The procedure is the same as that given earlier. The field, and the Green’s function, satisfy 1 ∂2 ψ(r′ , t′ ) = ρ(r′ , t′ ), (11.122a) c2 ∂t′2 2 1 ∂ ∇′2 G(r, t; r′ , t′ ) − 2 ′2 G(r, t; r′ , t′ ) = δ(r − r′ )δ(t − t′ ). (11.122b) c ∂t ∇′2 ψ(r′ , t) −

Note that the differentiations on G are with respect to the second set of arguments (this equation follows from the reciprocity relation). Again multiply the first equation by G(r, t; r′ , t′ ), the second by ψ(r′ , t′ ), subtract, integrate over the volume, and over t′ from t0 < t to t+ , where t+ means t + ǫ, ǫ → 0 through

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positive values. Then for r ∈ V , t+

 Z dt′ (dr′ ) G(r, t; r′ , t′ )∇′2 ψ(r′ , t′ ) − ψ(r′ , t′ )∇′2 G(r, t; r′ , t′ ) V t0   2 2 1 ′ ′ ∂ ′ ′ ′ ′ ∂ ′ ′ − 2 G(r, t; r , t ) ′2 ψ(r , t ) − ψ(r , t ) ′2 G(r, t; r , t ) c ∂t ∂t Z t+ Z dt′ (dr′ ) G(r, t; r′ , t′ )ρ(r′ , t′ ). (11.123) = −ψ(r, t) + Z

V

t0

Now we again use Green’s theorem and the identity (11.118) to conclude ψ(r, t) =

Z

t+

dt

′

Z

(dr′ ) G(r, t; r′ , t′ )ρ(r′ , t′ )

V

t0

Z

t+

I

  dσ · G(r, t; r′ , t′ )∇′ ψ(r′ , t′ ) − ψ(r′ , t′ )∇′ G(r, t; r′ , t′ ) t0 Σ   Z ∂ ∂ 1 ′ ′ ′ ′ ′ ψ(r , t0 ) − ψ(r , t0 ) G(r, t; r , t0 ) . (dr ) G(r, t; r , t0 ) − 2 c V ∂t0 ∂t0 (11.124) −

dt

′

This is our result. The interpretation is as follows: 1. The first integral represents the effect of the sources ρ distributed throughout the volume V . 2. The second integral represents the boundary conditions. If, for example, ψ satisfies inhomogeneous Neumann boundary conditions on Σ, n ˆ · ∇ψ = f (r′ ) (11.125) Σ

is specified, then we use homogeneous Neumann boundary conditions for G, ′ ′ n ˆ · ∇G(r, t; r , t ) = 0. (11.126) Σ

Then the second integral reads −

Z

t+

t0

dt′

I

Σ

dσ · G(r, t; r′ , t)∇′ ψ(r′ , t′ ).

(11.127)

That is, −ˆ n · ∇′ ψ(r′ , t′ ) represents a surface source distribution. Other types of boundary conditions are as discussed earlier. 3. The third integral represents the effect of the initial conditions, where ψ(r′ , t0 ),

∂ ψ(r′ , t0 ) ∂t0

(11.128)

11.7. WAVE EQUATION IN UNBOUNDED SPACE141 Version of December 3, 2011 are specified. They correspond to impulsive sources at t = t0 :   ∂ 1 ψ(r′ , t0 )δ(t′ − t0 ) + ψ(r′ , t0 )δ ′ (t′ − t0 ) . (11.129) ρinit (r′ , t′ ) = − 2 c ∂t0 We verify this statement by integrating by parts, and letting the lower limit of the t′ integration be t0 − ǫ.

11.7

Wave Equation in Unbounded Space

We now wish to solve Eq. (11.114)   1 ∂2 ∇2 − 2 2 G(r, t; r′ , t′ ) = δ(r − r′ )δ(t − t′ ), c ∂t

(11.130)

in unbounded space by noting that then G is a function of R = r − r′ and T = t − t′ only, G(r, t; r′ , t′ ) = G(r − r′ , t − t′ ) = G(R, T ). Then we can introduce a Fourier transform in space and time, Z g(k, ω) = (dR) dT eik·R e−iωT G(R, T ).

(11.131)

(11.132)

The Fourier transform of the Green’s function equation is (we have set c = 1 temporarily for convenience), (−k 2 + ω 2 )g(k, ω) = 1,

(11.133)

where we write k 2 = k · k, which has the immediate solution g(k, ω) =

ω2

1 . − k2

Thus the Green’s function has the formal representation Z 1 (dk) dω −ik·R iωT . e e G(R, T ) = (2π)3 2π ω2 − k2

(11.134)

(11.135)

The ω integral here is not well defined until we impose the boundary condition (11.115) G(R, T ) = 0 if T < 0. (11.136) This will be true if the poles are located above the real axis, as shown in Fig. 11.2. Here the contour is closed in the upper half plane if T > 0, and in the lower half plane if T < 0. In both cases, by Jordan’s lemma, the infinite semicircle gives no contribution. We have  (  ikT Z ∞ −ikT dω iωT 1 , T > 0, i e2k − e 2k (11.137) e = 2π (ω − k)(ω + k) 0 T < 0. −∞

142 Version of December 3, 2011

CHAPTER 11. GREEN’S FUNCTIONS 

−k + iǫ • -

k + iǫ • -

 Figure 11.2: Contour in the ω plane used to evaluate the integral (11.135). Thus, if T > 0, Z ∞ Z 1
1 i 2 eikT − e−ikT k dk 2π dµ e−ikRµ 3 (2π) 0 2k −1 Z ∞

k dk ikR i e − e−ikR eikT − e−ikT = 2 (2π) 0 2ikR Z   1 1 ∞ 1 ik(R+T ) −ik(R+T ) ik(R−T ) ik(T −R) = dk e + e − e − e (2π)2 2R 2 −∞ 1 1 = [δ(R + T ) − δ(R − T )] . (11.138) 2π 2R

G(R, T ) =

But R and T are both positive, so R + T can never vanish. Thus we are left with 1 1 δ(R − T ), (11.139) G(R, T ) = − 4π R or restoring c,   1 |r − r′ | 1 ′ δ − (t − t ) . (11.140) G(r − r′ , t − t′ ) = − 4π |r − r′ | c The effect at the observation point r at time t is due to the action at the source point r′ at time |r − r′ | t′ = t − . (11.141) c Physically, this means that the “signal” propagates with speed c.

11.7. WAVE EQUATION IN UNBOUNDED SPACE143 Version of December 3, 2011 Let us make this more concrete by considering a simple example, a point d r(t), “charge” moving with velocity v(t) = dt ρ(r, t) = qδ(r − r(t)).

(11.142)

There are no effects from the infinite surface, nor from the infinite past, so we have from Eq. (11.124) ψ(r, t) =

Z

t+

dt

′

−∞

= −

q 4π

Z

V

Z

t+

(dr′ ) G(r − r′ , t − t′ )ρ(r′ , t′ )

dt′

−∞

1 δ |r − r(t′ )|



 |r − r(t′ )| − (t − t′ ) . (11.143) c

If we let R(t′ ) = r − r(t′ ), the distance from the source to the observation point ′ at time t′ = t − R(tc ) , we write this as q ψ(r, t) = − 4π

t+

1 δ dt R(t′ ) −∞

Z

′



 R(t′ ) ′ − (t − t ) . c

(11.144)

Let τ = R(t′ )/c + t′ , where τ = t determines the “retarded time” t′ so   1 dR , (11.145) dτ = dt′ 1 + c dt′ where

that is

dR 1 d R·v = R·R = − , dt′ 2R dt′ R

(11.146)

  R·v dτ = dt 1 − . Rc

(11.147)

′

Thus the field is evaluated as Z t+R(t)/c q 1 dτ
ψ(r, t) = − δ(τ − t) 4π −∞ R(τ ) 1 − R·v Rc (τ ) 1 q = − . 4π R(t) − R(t) · v(t)/c

(11.148)