Mathematical Methods and Physical Insights: An Integrated Approach (Instructor Solution Manual, Solutions) [1 ed.] 1107156416, 9781107156418

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Solution Manual for Mathematical Methods and Physical Insights An Integrated Approach (ISBN 9781107156418) [Updated December 2022]

Alec J. Schramm Occidental College

[Corrections? Please let me know: [email protected]]

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

Contents

2 Coordinating Coordinates 3 Complex Numbers

page 1 8

4 Index Algebra

14

5 Brandishing Binomials

20

6 Series

27

7 Orbits in a Central Potential

46

8 Integration

50

9 Dirac Delta

67

10 Coda: Statistical Mechanics

73

11 Visualizing Vector Fields

75

12 Grad, Div & Curl

78

13 Interlude: Irrotational and Incompressible

85

14 Integrating Scalar and Vector Fields

87

15 The Theorems of Gauss and Stokes

98

16 Mostly Maxwell

109

17 Coda: Simply Connected Regions

113

18 Path Independence in the Complex Plane

115

19 Series, Singularities & Branches

122

2

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20 Interlude: Conformal Mapping

130

21 The Calculus of Residues

136

22 Coda: Analyticity & Causality

152

23 Prelude: Superposition

156

24 Vector Space

157

25 The Inner Product

159

26 Interlude: Rotations

170

27 The Eigenvalue Problem

182

28 Coda: Normal Modes

201

29 Cartesian Tensors

212

30 Beyond Cartesian

221

31 Prelude: 1 2 3 . . . Infinity

237

32 Eponymous Polynomials

239

33 Fourier Series

253

34 Convergence & Completeness

260

35 Interlude: Beyond the Straight & Narrow

264

36 Fourier Transforms

275

37 Coda: Of Time Intervals and Frequency Bands

294

38 First-Order ODEs

298

39 Second-Order ODEs

302

40 The Sturm-Liouville Problem

318

41 Partial Differential Equations

332

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3

42 Green’s Functions

352

43 Coda: Quantum Scattering

371

Appendix B

Rotations in R3

375

Appendix C

The Bessel Family of Functions

384

4

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Coordinating Coordinates

2

2.1 Starting with ~ r = ρ cos φˆı + ρ sin φˆ : ρˆ =

∂~ r/∂ρ = |∂~ r/∂ρ|

cos φˆı + sin φˆ 

p

cos2 φ + sin2 φ

= cos φˆı + sin φˆ  X

and ∂~ r/∂φ φˆ = = |∂~ r/∂φ|

−ρ sin φˆı + ρ cos φˆ 

p

ρ2 sin2 φ + ρ2 cos2 φ

= − sin φˆı + cos φˆ  X

Similar manipulations in spherical coordinates verify Eqn. (2.16).

2.2

⇢ˆ2

ˆ2

@ ⇢ˆ d @ @ˆ d @

⇢2

ˆ1

⇢ˆ1

d

⇢1

2.3 Cylindrical: d~ r=

∂~ r dρ ∂ρ

+

∂~ r dφ ∂φ

+



∂~ r dz: ∂z



∂~ r ∂~r = ρˆ = ρˆ |cos φˆı + sin φˆ | = ρˆ ∂ρ ∂ρ ∂~ r ∂~r ˆ ˆ = φˆ = φ |−ρ sin φˆı + ρ cos φˆ| = φρ ∂φ ∂φ ∂~ r r ˆ ˆ ∂~ =k =k ∂z ∂z Spherical: d~ r=

∂~ r dr ∂~ r

+

∂~ r dθ ∂θ



+

∂~ r dφ: ∂φ



∂~ r ∂~r ˆ = rˆ = rˆ = rˆ sin θ cos φˆı + sin θ sin φˆ  + cos θk ∂r ∂r
∂~ r ∂~ r ˆ ˆ = θˆ r = θ = θˆ r cos θ cos φˆı + cos θ sin φˆ  − sin θk ∂θ ∂θ ∂~ r ∂~r = φˆ |r (− sin θ sin φˆı + sin θ cos φˆ)| = φˆ r sin θ = φˆ ∂φ ∂φ ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

!

cos φ sin φ 0 − sin φ cos φ 0 . 0 0 1 ˆ to {ˆ ˆ φ}. ˆ So the matrix mapping Similarly, Eqn. (2.19) gives the matrix N mapping {ˆı, ˆ, k} r, θ, spherical coordinates into cylindrical coordinates is M N −1 . Since these are rotations, we can ˆ φ} ˆ to {ˆ ˆ k} ˆ save a lot of work invoking N −1 = N T . Then that the mapping from {ˆ r, θ, ρ, φ,

ˆ to {ˆ ˆ k} ˆ is M = 2.4 From Eqn. (2.18), the matrix mapping {ˆı, ˆ, k} ρ, φ,

multiplies out to be

MN

T

=

sin θ 0 cos θ

cos θ 0 − sin θ

!

0 1 0

.

The inverse transformation is N M T — which is just the transpose of M N T .

2.5 Writing out the matrix equation rˆ θˆ φˆ

! =

sin θ cos φ cos θ cos φ − sin φ

sin θ sin φ cos θ sin φ cos φ

!

cos θ − sin θ 0

ˆı ˆ ˆ k

! ,

ˆ etc. it’s straightforward to verify that rˆ · rˆ = θˆ · θˆ = φˆ · φˆ = 1 and that rˆ × θˆ = φ,

2.6 (a) cartesian: x2 + y 2 + z 2 = 1 cylindrical: ρ2 + z 2 = 1 spherical: r = 1 (b) cartesian: x2 + y 2 = 1 cylindrical: ρ = 1 spherical: r sin θ = 1

2.7 The direction cosines are the cartesian components of a unit vector from the origin making ~ and B, ~ we see A ~ ·B ~ = cos θ angles α, β, γ with the axes. Thus, given two different unit vectors A gives the identity in (b); part (a) is just a special case of this result.

2.8 Decomposing the vectors into cartesian components, but using spherical coordinates, ˆ ~ r = r sin θ cos φˆı + r sin θ sin φˆ  + r cos θk ˆ ~ r 0 = r 0 sin θ 0 cos φ 0ˆı + r 0 sin θ 0 sin φ 0 ˆ + r 0 cos θ 0 k Then ~ r·~ r 0 ≡ rr 0 cos γ = rr 0 sin θ sin θ 0 cos φ cos φ 0 + rr 0 sin θ sin θ 0 sin φ sin φ 0 + rr 0 cos θ cos θ 0 . Solving:




cos γ = = sin θ sin θ 0 cos φ cos φ 0 + sin φ sin φ 0 + cos θ cos θ 0 = sin θ sin θ 0 cos(φ − φ 0 ) + cos θ cos θ 0 .

2

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2.9 Executing the steps outlined in Example 2.2:


d ρˆ ˙ ρ + ρφ˙ φˆ dt ˙ = ρ¨ρˆ + ρ˙ ρˆ˙ + ρ˙ φ˙ φˆ + ρφ¨φˆ + ρφ˙ φˆ

~a =

˙ρ = ρ¨ρˆ + ρ˙ φ˙ φˆ + ρ˙ φ˙ φˆ + ρφ¨φˆ − ρφ˙ φˆ ˙ ˆ ¨ ˆ ˙ ˙ = ρ¨ρˆ + 2ρ˙ φφ + ρφφ − ρφφˆ ρ







= ρ¨ − ρφ˙ 2 ρˆ + ρφ¨ + 2ρ˙ φ˙ φˆ




= ρ¨ − ρω 2 ρˆ + (ρα + 2ρω) ˙ φˆ .

2.10 First, |J|2 = |J| |J| = |J T | |J| = |J T J|. Then

  ∂x/∂u ∂y/∂u ∂x/∂u |J| = ∂x/∂v ∂y/∂v ∂y/∂u ∂x 2 ∂y ∂y ∂y ∂x ∂x ( ∂u ) + ( ∂u )2 ∂u + ∂u ∂v ∂v = ∂x ∂x + ∂y ∂y ( ∂x )2 + ( ∂y )2

∂x/∂v ∂y/∂v

2

∂v ∂u

∂v ∂u

∂v

∂v





Before trying to calculate this horrific determinant, note that since u ˆ ∼ ∂~ r/∂u and vˆ ∼ ∂~ r/∂v, the off-diagonal terms are just u ˆ · vˆ — which vanishes for an orthogonal system. Moreover, the diagonal terms are just the scale factors h2u and h2v . Thus

h2u

|J|2 =

0

0 h2v

= h2u h2v . X

ˆ does 2.11 Since θˆ and φˆ span the tangent plane to the sphere, then using little more than ˆı × ˆ = k the trick:




ˆ sin θ × (−ˆı sin φ + ˆcos φ) n ˆ ≡ θˆ × φˆ = ˆı cos θ cos φ + ˆcos θ sin φ − k ˆ cos θ cos2 φ + k ˆ cos θ sin2 φ + ˆsin θ sin φ + ˆı sin θ cos φ =k ˆ cos θ ≡ rˆ X = ˆı sin θ cos φ + ˆsin θ sin φ + k

2.12 Area elements (a) In cylindrical coordinates, the scale factors are hρ = 1, hφ = ρ, hz = 1. Then i. on surface of constant ρ, d~a = ρˆ ρ dφ dz ii. on surface of constant φ, d~a = φˆ dρ dz ˆ ρ dρ dφ iii. on surface of constant z, d~a = k (b) In spherical coordinates, the scale factors are hr = 1, hθ = r, hφ = r sin θ. Then i. on surface of constant r, d~a = rˆ r2 sin θ dθ dφ = rˆ r2 dΩ ii. on surface of constant θ, d~a = θˆ r sin θ dr dφ iii. on surface of constant φ, d~a = φˆ r dr dθ

2.13 (a) Directly leveraging Eqn. (2.12) immediately yields radial equation: m¨ r − mrφ˙ 2 = − rk2 angular equation: rφ¨ + 2r˙ φ˙ = 0
d (b) Simplifying the angular equation as r1 dt r2 φ˙ = mentum conservation: ` ≡ mr2 φ˙ = constant

1 d mr dt

mr2 φ˙




= 0 reveals angular mo-

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3

(c) Using Eqn. (2.9) to express the kinetic energy in polar coordinates gives 1 1 mv 2 = m 2 2



d~ r dt

d~ r dt

  ·



=


1 m r˙ 2 + r2 φ˙ 2 . 2

Using angular momentum conservation in part (b), the kinetic energy is KE =

1 `2 mr˙ 2 + . 2 2mr2

For a circular orbit r˙ = 0, so the entire kinetic energy is due to angular motion.

~ = 2.14 (a) A point charge q has electric field E ~ ·n dΦ = E ˆ dA =

∆A

n /

θ

q r ˆ , 4π0 r 2

so that

q q rˆ · n ˆ dA = cos θ dA 4π0 r2 4π0 r2

r n r θ

r

/

∆Acosθ

AQ

∆A

∆Ω q

q

∆Ω

S

(b) ~ ≡ dA n (c) The component of the vector-valued area dA ˆ perpendicular to the radial line is dA cos θ, and subtends the same solid angle as dA; using the definition of solid angle (which requires r perpendicular to the subtended area), dΩ =

dA cos θ r2

dΦ =

q dΩ . 4π0

we have

Since the total solid angle is 4π, the net flux is

I Φ=

q 4π0

~ ·n E ˆ dA =

S

I dΩ = S

q , 0

where no explicit integration was necessary!

2.15 Starting with x = r sin θ cos φ ,

y = r sin θ sin φ ,

z = r cos θ

we get

|J| ≡ = 4

∂x/∂r ∂y/∂r ∂z/∂r

∂x/∂θ ∂y/∂θ ∂z/∂θ

sin θ cos φ sin θ sin φ cos θ

∂x/∂φ ∂y/∂φ ∂z/∂φ

r cos θ cos φ r cos θ sin φ −r sin θ

−r sin θ sin φ r sin θ cos φ 0

= . . . = r2 sin θ .

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2.16 (a)

y2 x2 + a2 sinh 2u a2 cosh2 u

= cos2 φ+sin2 φ = 1. For constant u, this describes an ellipse with semi-

major and minor axes a cosh u, a sinh u and foci at c = ±

p

a2 cosh2 u − a2 sinh2 u = ±a.

elliptic[u_, φ_] := {a Cosh[u]Cos[φ], a Sinh[u]Sin[φ]}; a = 1; uPlots = ParametricPlot[elliptic[u, φ], {u, 0, 1}, {φ, 0, 2π}, BoundaryStyle → Dashed, Mesh → 9, Frame → False, PlotStyle → Gray, Ticks → None, PlotRange → All]; zeroPlot = ParametricPlot[elliptic[0, φ], {φ, 0, 2π}, PlotStyle → {Red, Thick}, Ticks → None]; Show[uPlots,zeroPlot]

(b) The easiest approach is via Eqn. (2.27b):

h2u =



∂x ∂u

2

+



∂y ∂u

2

= (a sinh u cos φ)2 + (a cosh u sin φ)2











= a2 sinh2 u 1 − sin2 φ + 1 + sinh2 u sin2 φ = a2 sinh2 u + sin2 φ




Similarly,

h2φ =



∂x ∂φ

2

+



∂y ∂φ

2

= (−a cosh u sin φ)2 + (a sinh u cos φ)2 = a2

2.17 (a) –

–






1 + sinh2 u sin2 φ + sinh2 u 1 − sin2 φ

x2 +y 2 a2 cosh2 u




= a2 sinh2 u + sin2 φ




2

z 2 2 + a2 sinh 2 u = sin θ + cos θ = 1. For constant u, this is the equation of an ellipsoidal surface. Note that the surface intersects the xy-plane in circles of radius a cosh u, whereas ellipses in the xz- and yz-planes have semi-major axis a cosh u, and semi-minor axis a sinh u. x2 +y 2 a2 sin2 θ

2

z 2 2 − a2 cos 2 θ = cosh u − sinh u = 1. For constant θ, this is the equation of a hyperbolic surface. Once again, the curves in the xy-plane are circles (this time of radius a sin θ), but those in the xz- and yz-planes are hyperbolas.

(b) Starting with x = a cosh u sin θ cos φ, y = a cosh u sin θ sin φ, z = a sinh u cos θ, the easiest

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5

approach is via Eqn. (2.27b): h2u =



∂x ∂u

2

+



∂y ∂u

2

+



∂z ∂u

2

= (a sinh u sin θ cos φ)2 + (a sinh u sin θ sin φ)2 + (a cosh u cos θ)2 = a2 sinh2 u sin2 θ + cosh2 u cos2 θ














= a2 sinh2 u 1 − cos2 θ + 1 + sinh2 u cos2 θ = a2 sinh2 u + cos2 θ




X

It’s easy to see that hθ gives the same. As for hφ : h2φ =



∂x ∂φ

2

+



∂y ∂φ

2

+



∂z ∂φ

2

= (−a cosh u sin θ sin φ)2 + (a cosh u sin θ cos φ)2 = a2 cosh2 u sin2 θ




(c) Easy: |J| = hu hθ hφ = a3 cosh u sin θ sinh2 u + cos2 θ . But if you prefer the long way:

∂x/∂u ∂x/∂θ ∂x/∂φ |J| ≡ ∂y/∂u ∂y/∂θ ∂y/∂φ ∂z/∂u ∂z/∂θ ∂z/∂φ sinh u sin θ cos φ cosh u cos θ cos φ − cosh u sin θ sin φ = a3 sinh u sin θ sin φ cosh u cos θ sin φ cosh u sin θ cos φ cosh u cos θ − sinh u sin θ 0
3 2 2 = . . . = a cosh u sin θ sinh u + cos θ



.

(d) Surfaces of constant u have da = hθ hφ dθ dφ = a2 cosh u sin θ

p

sinh2 u + cos2 θ dθ dφ .

Since hu = hθ , surfaces of constant θ have the almost-identical da = hu hφ du dφ = a2 cosh u sin θ

p

sinh2 u + cos2 θ du dφ .

2.18 Using the hi for Problem 2.17 in Eqns. (2.27), and with ˆ, ~ r = a cosh u sin θ cos φ ˆı + a cosh u sin θ sin φ ˆ + a sinh u cos θ k we find eˆu =

eˆθ =

1 ∂~ r = hu ∂u

p

1 ∂~ r = hθ ∂θ

p

1 2

sinh u +

ˆ sinh u sin θ cos φ ˆı + sinh u sin θ sin φ ˆ + a cosh u cos θ k cos2




θ

1

ˆ cosh u cos θ cos φ ˆı + cosh u cos θ sin φ ˆ − sinh u sin θ k




sinh2 u + cos2 θ

and eˆφ =

1 ∂~ r 1 = (− cosh u sin θ sin φ ˆı + cosh u sin θ cos φ ˆ) . hφ ∂φ cosh u sin θ

Multiple use of the identities cos2 + sin2 = 1 and cosh2 − sinh2 = 1 readily demonstrates that these vectors form an orthonormal set.

2.19 From Eqns. (2.27), we have hi eˆi = ∂~ r/dui : hr rˆ = (sin ψ sin θ cos φ, sin ψ sin θ sin φ, sin ψ cos θ, cos ψ) hψ ψˆ = r (cos ψ sin θ cos φ, cos ψ sin θ sin φ, cos ψ cos θ, − sin ψ) hθ θˆ = r (sin ψ cos θ cos φ, sin ψ cos θ sin φ, − sin ψ sin θ, 0) hφ φˆ = r (− sin ψ sin θ sin φ, sin ψ sin θ cos φ, 0, 0)

6

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X

Each is indeed orthogonal to the other three, and the magnitude of each gives the scale factor: hr = 1

hψ = r

hθ = r sin ψ

hφ = r sin ψ sin θ ,

which together produce the required line element in R4 . Finally, the product of the four gives the Jacobian, |J| = r3 sin2 ψ sin θ.

2.20 For hyperspherical coordinates x4 = r cos ψ x3 = r sin ψ cos θ x2 = r sin ψ sin θ cos φ x1 = r sin ψ sin θ sin φ , the Jacobian matrix is (in reverse order, as suggested)

∂(x4 , x3 , x2 , x1 ) ∂(r, ψ, θ, φ) cos ψ −r sin ψ 0 0 sin ψ cos θ r cos ψ cos θ −r sin ψ sin θ 0 = sin ψ sin θ cos φ r cos ψ sin θ cos φ r sin ψ cos θ cos φ −r sin ψ sin θ sin φ sin ψ sin θ sin φ r cos ψ sin θ sin φ r sin ψ cos θ sin φ r sin ψ sin θ cos φ r cos ψ cos θ −r sin ψ sin θ 0 = cos ψ r cos ψ sin θ cos φ r sin ψ cos θ cos φ −r sin ψ sin θ sin φ r cos ψ sin θ sin φ r sin ψ cos θ sin φ r sin ψ sin θ cos φ sin ψ cos θ −r sin ψ sin θ 0 +r sin ψ sin ψ sin θ cos φ r sin ψ cos θ cos φ −r sin ψ sin θ sin φ . sin ψ sin θ sin φ r sin ψ cos θ sin φ r sin ψ sin θ cos φ

|J| =

Now each additive element of a determinant has exactly one contribution from each row and column. So a common multiplicative factor of a column or row can be moved outside the determinant. We can do just this with all the factors of sin ψ and cos ψ, giving

cos θ
|J| = r cos2 ψ sin2 ψ + r sin4 ψ sin θ cos φ sin θ sin φ

−r sin θ r cos θ cos φ r cos θ sin φ



0 −r sin θ sin φ . r sin θ cos φ

Of course, we could also have taken out two more factors of r, but leaving them helps us recognize the remaining 3 × 3 determinant as the Jacobian r2 sin θ of standard spherical coordinates in R3 . Thus |J| = (r cos2 ψ sin2 ψ + r sin4 ψ) r2 sin θ = r3 sin2 ψ sin θ .

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7

Complex Numbers

3 3.1 (a) (b) (c) (d)

|e−iπ |2 = eiπ e−iπ = 1 X π ie = eie ln π X (−π)ie = (eiπ eln π )ie = e−eπ eie ln π ×
eπ 2 ieπ = eiπ/2 = eieπ /2 X i

(e) (−i)eπ = (ei3π/2 )e

(1+i ln π)

= ei3eπ/2 e−3π ln(π)/2 ×

3.2 In the cartesian representation z = a + ib, |z|2 = zz ∗ = (a + ib)(a − ib) = a2 + b2 |z 2 | = |(a + ib)2 | = |(a2 − b2 ) + i(2ab)| =

p

(a2 − b2 )2 + (2ab)2 =

p

(a2 + b2 )2 = a2 + b2

In the polar representation z = reiϕ it’s even easier: |z|2 = zz ∗ = reiϕ re−iϕ = r2 |z 2 | = |(reiϕ )2 | = |r2 e2iϕ | = r2 In general, |z n | = |z|n = rn .

3.3 Starting with z = cos θ + i sin θ, we have dz = (− sin θ + i cos θ)dθ = i(cos θ + i sin θ)dθ = izdθ . Thus

Z

dz =i z

Z dθ

z = eiθ ,

=⇒

where the constant of integration is determined from z(θ = 0) = 1.

3.4 Compute the square in two equivalent ways: 1) |(a + ib)(c + id)|2 = |(ac − bd) + i(bc + ad)|2 = (ac − bd)2 + (bc + ad)2 ≡ p2 + q 2 2) |(a + ib)(c + id)|2 = (a2 + b2 )(c2 + d2 ) ≡ M N . Thus M N = p2 + q 2 . For example, M = 13 = 22 + 32 N = 25 = 32 + 42

 M N = 325 = (2 · 3 − 3 · 4)2 + (3 · 3 − 2 · 4)2 = 62 + 172

√
arg(z1 ) = arctan 1/ 3 = π6 =⇒ z1 = 2eiπ/6 √
=⇒ z2 = 2ei5π/6 = 2e−iπ/6 z2 = i − 3 : |z2 | = 2, arg(z1 ) = arctan −1/ 3 = 5π   6 √ √ √ (b) 2i = 2eiπ/2 2i = ± 2eiπ/4 = ± 2 √1 + √i = ±(1 + i)

3.5 (a) z1 = i +

√

√

3 : |z1 | =

√

1 + 3 = 2,

2

2 + 2i

√

3 = 4eiπ/3

p

2 + 2i

√

2

3 = ±2eiπ/6 = ±2

√

3 2

+

i 2



= ±(

√

3 + i)

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3.6 Real: z = z ∗ ; Imaginary z = −z ∗ (a) [(−1)1/i ]∗ = [(−1)−i ]∗ = (−1)i = [1/(−1)]−i = [1/(−1)]1/i = (−1)1/i = z Using −1 = eiπ , (−1)1/i = e+π ≈ 23.1 (b)

=⇒ REAL

∗



(z/z ∗ )i = (z ∗ /z)−i = (z/z ∗ )i =⇒ REAL Using z = reiθ , z/z ∗ = e2iθ so (z/z ∗ )i = e−2θ

(c) [z1 z2∗ − z1∗ z2 ]∗ = z1∗ z2 − z1 z2∗ = −[z1 z2∗ − z1∗ z2 ] =⇒ IMAGINARY
z1 z2∗ − z1∗ z2 = r1 r2 ei(θ1 −θ2 ) − e−i(θ1 −θ2 ) = 2ir1 r2 sin(θ1 − θ2 ) = 2iIm[z1 z2∗ ]

 (d)

N P

∗ einθ

=

P∞ n=0

e−inθ

=⇒ COMPLEX

n=0

 (e)

N P

∗ einθ

=

P∞ n=−∞

e−inθ =

P∞ n=−∞

einθ

=⇒ REAL

n=−N

The imaginary parts for positive n cancel in pairs with the imaginary contributions for negative n, so

N P

einθ = 2

n=−N

3.7 (a) (1 + i)4 = (b)

√

2eiπ/4

4

N P

cos (nθ)

n=0

= 4eiπ = −4

· 2+i = 5i =i 2+i 5 2 (1+4i)(1−4i) 1+4i = (4+i)(4−i) 4+i

1+2i 2−i

17 = 17 =1 (3+4i)4 4 (d) (3−4i)3 = zz∗3 = |z| = 5 (c)

√ 3.8 – a ≡ (2 + i)(3 + i) = 5 + 5i = 5(1 + i) = 5 2eiπ/4 √ −1 – b ≡ (2 + i) = 5ei tan (1/2) √ −1 – c ≡ (3 + i) = √10ei tan (1/3) √ iπ/4 a = b · c =⇒ 5 2e = 5 2 exp{i[tan−1 ( 12 ) + tan−1 ( 13 )]} π = tan−1 ( 12 ) + tan−1 ( 13 ) 4

3.9 (a)



 1  i(a+b) e + e−i(a+b) + ei(a−b) + e−i(a−b) 2

 1  ia ib = e e + e−ib + e−ia eib + e−ib 2  1  ia = e · 2 cos b + e−ia · 2 cos b 2
= cos b eia + e−ia = 2 cos a cos b .

cos (a + b) + cos (a − b) =

(b) Same procedure works here, but for variety, let’s do it a little differently:



sin (a + b) + sin (a − b) = =m ei(a+b) + ei(a−b)



= =m eia eib + e−ib










= =m eia (2 cos b)

= 2 sin a cos b .

3.10 (a) As for any complex number, we can write eiα + eiβ ≡ Reiθ , where (using the cartesian

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9

representation) R2 = [cos α + cos β]2 + [sin α + sin β]2 = 2 + 2 (cos α cos β + sin α sin β)

h

= 2 [1 + cos (α − β)] = 2 2 cos2



α−β 2

i

= 4 cos2



α−β . 2



Alternatively, one can stay in the polar representation:

2



R2 = eiα + eiβ = eiα + eiβ




eiα + eiβ


∗

= 2 + 2 cos(α − β) .

As for the phase (using the results of Problem 3.9 ): θ = tan−1

 "

= tan

−1

sin α + sin β cos α + cos β

α−β 2 α−β 2

α+β 2

i

α+β 2





2 cos = tan−1 tan




#



cos


α+β 2 α+β 2

2 sin

h



cos =

α+β

i




2 (b) Same procedure establishes eiα −eiβ = 2i sin α−β e . (It helps to note that the cot x 2 is odd, and to recall that the cot of an angle is the tan of its complement.)

These results can also be worked out geometrically by drawing eiα and eiβ in the complex plane and adding them with the usual rules of vector addition.

3.11 (a) z = z1 z2 = r1 r2 ei(φ1 +φ2 ) . Thus |z| = r1 r2 , and arg(z) = φ1 + φ2 . So geometrically, multiplication of a complex number by reiθ rotates it by θ and magnifies it by r. (b) z = z1 + z2 : |z|2 = |z1 + z2 |2 = (z1 + z2 )(z1 + z2 )∗ = (z1 + z2 )(z1∗ + z2∗ ) = |z1 |2 + |z2 |2 + z1∗ z2 + z1 z2∗ = |z1 |2 + |z2 |2 + 21 = 0) and B(x) = ex (so that bn = 1/n!) we get

! X

1=

! X

bn xn

n

cm xm

=

m

X

(bn cm ) xn+m

n,m

This is probably easier to do by writing out the middle sums explicitly: 1 = (1 + x + x2 /2 + x3 /3! + x4 /4! + . . .)(c0 + c1 x + c2 x2 + c3 x3 + . . .) Matching like coefficients of integer powers of x allows us to solve for the cm : x0 : 1 = c 0 x1 : 0 = c 0 + c 1

c1 = −1

x2 : 0 = c0 /2! + c1 + c2

c2 = +1/2

x3 : 0 = c0 /3! + c1 /2! + c2 + c3

c3 = −1/6

.. . The emerging pattern is cn = (−1)n /n! (as you can easily check); thus C(x) =

X (−1)n n!

xn = e−x

X

n=0

6.19 (a) 1 1 = a−x (a − k) − (x − k) 1 1 1 = · == (a − k) − x 0 a − k [1 − x 0 /(a − k)]

f (x) =

=

1 a−k

∞  n X x0

a−k n=0

=

∞ X

(x 0 )n (a − k)n+1

n=0

which converges for

32

|x 0 |

< |a − k| — i.e., within a circle centered at k of radius |a − k|.

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(b) Decomposing in partial fractions of the form 1/(a − x) gives 1 1 = 1 − x2 2

g(x) =

h

1 1 − (1 − x) (−1 − x)

i

.

Expanding each around k as in part (a) yields ∞ h X (x − k)n

1 1 = 1 − x2 2 =

(1 − n=0 ∞

1 2

Xh

k)n+1

−

(x − k)n (−1 − k)n+1

i

1 1 − (x − k)n (1 − k)n+1 (−1 − k)n+1

i

n=0

The individual sums converge for |x − k| < |1 − k| and |x − k| < | − 1 − k|, that is, within concentric circles centered at k of radius R1 = |1 − k| and R2 = | − 1 − k|. The full expansion for g(x) converges within a circle whose radius is the smaller of R1 and R2 — i.e., the distance to the nearest singularity (k could be negative). (c) Decomposing in partial fractions of the form 1/(a − z) gives 1 1 = 1 + z2 2i

h(z) =

h

1 1 − (−i − z) (i − z)

i

.

Expanding each around k as before yields 1 1 = 1 + z2 2i

∞ h X

1 1 − (z − k)n . (−i − k)n+1 (i − k)n+1

i

n=0

The individual sums converge for |z −k| < |−i−k| and |z −k| < |i−k|, i.e., within concentric circles centered at k of radius R1 = |i + k|, R2 = |i − k|. The full expansion for h(z) converges within a circle whose radius is the smaller of R1 , R2 . (d) Restricting to the real axis, z = x + iy → x. Then

h(x) =

1 1 = 1 + x2 2i

∞ h X

1 1 − (x − k)n . (−i − k)n+1 (i − k)n+1

i

n=0

Next, since k is real we can write (i − k) = |i − k|eiφ =



φ = arg (i − k) = tan−1 −

Its complex conjugate (−i − k) = 1 1 = 1 + x2 2i

√

∞ X 

1 k



 , with

√

1 + k2 eiφ , where

0 ≤ φ < π/2 , π/2 < φ ≤ π ,

1 + k2 e−iφ . Then

e+i(n+1)φ − e−i(n+1)φ



n=0

=

∞ X sin [(n + 1)φ] n=0

(

√

k0

1 + k2 )n+1

(x − k)n

(x − k)n √ ( 1 + k2 )n+1

X

The figure compares the exact expression for k = 2 with the first six terms in the expansion. Note the near-perfect agreement within the circle of convergence.

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33

2.0

1.5

1.0

0.5

1

-1

2

3

4

5

6.20

" C(x) = A(x)B(x) =

∞ X

#" an x

n

n=0 ∞

=

X

∞ X

# bm x

m

m=0

an bm xn+m .

n,m

Let ` = n + m, and substitute it for m. Then ` ranges from 0 to ∞ — but for any fixed ` value, n ranges from 0 (where m = `) up to n = ` (where m = 0). Thus

C(x) =

∞ X

an bm xn+m =

n,m

∞ ` X X `=0

6.21 Starting only from sin ϕ =

! an b`−n

x` ≡

n=0

2n+1

φ (−1)n (2n+1)! and cos ϕ =

P

n=0

∞ X

∞ X

`=0

`=0

(a ∗ b)` x` =

P

c` x` .

2n

φ (−1)n (2n)! :

n=0

(a) Clearly sin 0 = 0 and cos 0 = 1. As for the derivatives: (sin ϕ)0 =

X

(cos ϕ)0 =

X

(−1)n (2n + 1)

X φ2n φ2n = (−1)n = cos ϕ (2n + 1)! (2n)!

n=0

n=0

X φ2n−1 φ2n−1 (−1)n 2n = (−1)n = − sin ϕ (2n)! (2n − 1)!

n=1

(b) With an = bn =

c` (ϕ) =

` X

P

(−1)n

n

n=1

(−1)n ϕ2n+1 /(2n + 1)!, Eqn. (6.52) gives

ϕ2n+1 ϕ2(`−n)+1 (−1)`−n (2n + 1)! [2(` − n) + 1]!

n=0

= (−1)` ϕ2`+2

` X

1 1 = (−1)` ϕ2`+2 (2n + 1)!(2` − 2n + 1)! (2` + 2)!

n=0

 `  X 2` + 2 2n + 1

,

n=0




where ab is a binonial coefficient. From Problem 5.2c, the sum over these coefficients is 22`+1 , giving the concise expression c` (ϕ) =

34

(−1)` (2ϕ)2`+2 . 2 (2` + 2)!

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Thus sin2 ϕ =

∞ X

c`0 (ϕ) =

`0 =0

∞ 0 0 X (−1)` (2ϕ)2` +2 `0 =0 ∞

=−

(2`0 + 2)!

2

X (−1)` (2ϕ)2` 2

(2`)!

` = `0 + 1

,

`=1

1 = 2

" 1−

∞ X (−1)` (2ϕ)2`

2

#

(2`)!

,

`=0

where we shifted the index in the second step, and then added and removed the ` = 0 contribution to get the desired result. The derivation of the series expansion for cos2 ϕ follows similarly — though without the need of an index shift. However, note that in the sum over the binomial coefficients, ` = 0 contributes 1, not 22`−1 = 1/2. (c) The series solutions in (b) reveal that sin2 ϕ + cos2 ϕ = 1 for all ϕ. So both sin ϕ and cos ϕ must bounded. This together with results from (a) immediately establish that | cos ϕ| ≤ 1. Now for small ϕ, (sin ϕ)0 = cos ϕ > 0 and (cos ϕ)0 = − sin ϕ < 0 — so sin ϕ increases from 0 and (since bounded) must reach a maximum when cos ϕ reaches 0 at some ϕ = ξ. Thus there must be a ξ > 0 for which sin ξ = 1, cos ξ = 0. (d)

sin ϕ cos α + cos ϕ sin α =

`  XX

(−1)n

ϕ2n+1 α2(`−n) (−1)`−n (2n + 1)! (2` − 2n)!

`=0 n=0

+(−1)n

=

 `  X (−1)` X 2` + 1



 ϕ2n+1 α2`−2n +

2` + 1 2n



 ϕ2n α2`−2n+1

n=0

 `  X (−1)` X 2` + 1 n

(2` + 1)! `=0

=

2n + 1

(2` + 1)! `=0

=

ϕ2n α2(`−n)+1 (−1)`−n (2n)! (2` − 2n + 1)!

ϕn α(2`+1)−n

n=0

X (−1)` (2` + 1)!

(ϕ + α)2`+1 = sin(ϕ + α)

X

`=0

A similar calculation verifies the identity for cos(ϕ + α). Then, with sin ξ = 1 and cos ξ = 0, we can verify a 4ξ periodicity: sin(ϕ + 4ξ) = sin(ϕ + 3ξ) cos(ξ) + cos(ϕ + 3ξ) sin(ξ) = cos(ϕ + 3ξ) = cos(ϕ + 2ξ) cos(ξ) − sin(ϕ + 2ξ) sin(ξ) = − sin(ϕ + 2ξ) = − sin(ϕ + ξ) cos(ξ) − cos(ϕ + ξ) sin(ξ) = − cos(ϕ + ξ) = − cos(ϕ) cos(ξ) + sin(ϕ) sin(ξ) = sin ϕ , (e) Since sin2 ϕ + cos2 ϕ = 1 is the unit circle, the periodicity established in (d) gives 4ξ = 2π.

6.22 (a) Singularities generally occur when the denominator goes to zero, ez = 1 =⇒ z = 2πin for integer n. Note, however, that the numerator also vanishes at z = 0; indeed, L’Hôpital’s

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35

rule shows that n = 0 is not actually a singularity:

lim z→0

z 1 = lim z = 1. ez − 1 z→0 e

Thus all the singularities occur along the imaginary axis at z = 2πin, n = 1, 2, 3, · · · (b) The radius of convergence is the distance to the nearest singularity in the complex plane. Thus the radius of convergence is 2π — even though 2π has no special significance along the real axis! 1.0

x ãx - 1 â Bn xn n! ¥

0.5

n=0

Π

3Π

Π

2

2Π

2

-0.5

6.23 (a) This series is clearly absolutely convergent; comparison with the expansion for ex shows that it sums to e. (b) This series is also absolutely convergent; comparison again with the expansion for ex shows that it sums to 1/e (c) This is the alternating harmonic series — which converges since each successive term decreases and limn→∞ 1/n = 0. Comparison with the series for ln(1 + x) shows that it sums to ln(2). It is not absolutely convergent since (as we’ve seen) the harmonic series diverges. (d) Again, the terms diminish to zero for large n. Comparison with the series for arctan x shows that this alternating series converges to π/4. Perhaps the simplest way to see that it is not absolutely convergent is to split the harmonic series into the sum of two separate infinite series, one over even, the other of odd integers (which is our series). The convergence of these “sub-sums" should be identical. But since they combine to give a divergent series, each individually must also be divergent. If this argument is a bit too glib for you, we can also be more formal. Comparing term by term, we see

1+

1 1 1 1 1 + + ··· > + + + ··· . 3 5 2 4 6

Factoring 1/2 from the right-hand side,

1+

1 1 1 + + ··· > 3 5 2



1+

1 1 + + ··· 2 3



.

That, of course, is the harmonic series in parentheses — which is divergent. So the sum over 1/n for odd n is also divergent, and thus the original series is not absolutely convergent.

36

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6.24 Using

d dx

sinh x = cosh x,

d dx

sinh x =

cosh x = sinh x, sinh (0) = 0, and cosh (0) = 1, we get

∞ X xn

dn sinh x dxn x=0



n! n=0

=0+x+0+ =

1 3 1 5 x +0+ x + ... 3! 5!

∞ X x2n+1

(note that 2n+1 is always odd)

(2n + 1)! n=0

cosh x =

∞ X xn

dn cosh x dxn x=0



n! n=0

=1+0+ =

1 2 1 4 x +0+ x + ... 2! 4!

∞ X x2n

(2n)!

(note that 2n is always even)

n=0

6.25 Expand ex with x = iθ: (iθ)2 (iθ)3 (iθ)4 (iθ)5 + + + + ··· 2! 3! 4! 5! iθ3 θ4 iθ5 θ2 − + + + ··· = 1 + iθ − 2! 3! 4! 5!

eiθ = 1 + (iθ) +

 =

θ2 θ4 1− + − ··· 2! 4!

= cos θ + i sin θ





θ3 θ5 θ− + − ··· 3! 5!

+i



X

6.26 Take derivatives of the function for use in a Taylor series: f (x) =

sin x cosh x + 2

f (0) = 0 f 0 (0) =

1 cos x(cosh x + 2) − sin x(sinh x) = (cosh x + 2)2 3 x=0



f 0 0 (0) = 0 (since all terms are proportional to either sin or sinh) 2 f 0 0 0 (0) = − 3 Then 1 1 x+0+ 3 3! 1 1 = x − x3 + . . . 3 9

f (x) = 0 +



−

2 3



x3 + . . .

Note that f is an odd function, so only odd powers of x can appear in its expansion. How good is the series through x3 ? Well, the series gives f (0.1) = 0.0332222 while the exact expression yields f (0.1) = 0.0332224. Not bad!

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37

6.27 Begin by calculating the first few derivatives of f (x) = ln(x): 1 x 1 f 0 0 (x) = − 2 x 2 f 0 0 0 (x) = + 3 x .. . f 0 (x) = +

The emerging pattern is f (n>0) (x) = (−1)n+1 (n − 1)!/xn . Expanding f about x = 1 gives (x − 1)3 0 0 0 (x − 1)2 0 0 f (1) + f (1) + . . . 2! 3! 2 3 (x − 1) (x − 1) (x − 1)4 = 0 + (x − 1) + (−1) + (2) + (−6) + . . . 2! 3! 4! 2 3 4 (x − 1) (x − 1) (x − 1) = (x − 1) − + − + ... 2! 3 4

f (x) = f (1) + (x − 1)f 0 (1) +

Making the substitution x → x + 1 then gives the well-known result ln(1 + x) = x −

x2 x3 x4 + − + ... . 2 3 4

Note that a Maclauren series is an expansion around x = 0, where ln(x) is undefined. For z = i, √ iπ/4
all even powers in the series are real, all odd powers imaginary. Using ln(1+i) = ln 2e = √ ln( 2) + iπ/4, we find ln(

√

1 1 1 1 − + − + ··· 2 4 6 8 1 1 1 π/4 = 1 − + − + · · · 3 5 7 2) =

6.28 (a) 1

1+

1




−1 ln(1 + x) 1 n n ≈ (1 + x)1/n = 1 + x + x2 + n n 2!

1 1 −1 −1 n n 2 =⇒ ln(1 + x) ≈ x + x + 2! 3!

1 n 1 n

1 n




−1

1 n


3!

−2




−2

x3 + · · ·

x3 + · · ·

In the n → ∞ limit, this becomes exact, ln(1 + x) = x −

x3 x4 x2 + − + ··· 2 3 4

(b) With f (x) = c1 x + c2 x2 + c3 x3 + c4 x4 + · · · , f2 f3 + + ··· 2! 3! = 1 + (c1 x + c2 x2 + c3 x3 + c4 x4 + · · · ) + 1 (c1 x + c2 x2 + c3 x3 + c4 x4 + · · · )2 + 2! 1 (c1 x + c2 x2 + c3 x3 + c4 x4 + · · · )3 + · · · 3!     1 1 = 1 + c1 x + c2 + c21 x2 + c3 + c1 c2 + c31 x3 + 2 6   1 2 1 4 1 2 c4 + c1 c3 + c2 + c1 + c1 c2 x4 + O(x5 ) 2 24 2

1 + x = ef (x) = 1 + f +

38

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Clearly c1 = 1; all the other terms in parentheses must vanish, giving the other cn .

6.29 L’Hôpital’s rule: lim x→x0

f (x) = lim g(x) x→x0





g(x0 ) + (x − x0 )g 0 (x0 ) + 21 (x − x0 )2 g 0 0 (x0 ) + . . .

 = lim

(x − x0 )f 0 (x0 ) + 12 (x − x0 )2 f 0 0 (x0 ) + . . .



(x − x0 )g 0 (x0 ) + 21 (x − x0 )2 g 0 0 (x0 ) + . . .

x→x0

 = lim

f 0 (x0 ) + 21 (x − x0 )f 0 0 (x0 ) + . . .

, since f (x0 ) = g(x0 ) = 0

 , canceling common factor (x − x0 )

g 0 (x0 ) + 21 (x − x0 )g 0 0 (x0 ) + . . .

x→x0

=

f (x0 ) + (x − x0 )f 0 (x0 ) + 21 (x − x0 )2 f 0 0 (x0 ) + . . .

f 0 (x0 ) f 0 (x) = lim g 0 (x0 ) x→x0 g 0 (x)

6.30 (a) Find ymin : dy/dx = sinh(x/a) ≡ 0 −→ x = 0 =⇒ ymin = y(0) = a cosh(0) = a.
1 1 (x/a)2 + 4! (x/a)4 + · · · . (b) Use the Taylor expansion of cosh around x = 0: y(x) = a 1 + 2! For small sag, a  x — so to lowest non-trivial order in x/a, y(x) ≈ a +

1 2 x , 2a

a parabola.

6.31 From the text, we know f (~ r + ~a) =

∞ X 1

n!

~ ~a · ∇


n

f (~ r) ,

n=0

~ = (∂/∂x1 , . . . , ∂/∂xN ) is the gradient operator in cartesian coordinates. In two diwhere ∇ ~ = (∂x , ∂y ), this gives (through third order) mensions, with f (~ r) = f (x, y), ~a = (ax , ay ), and ∇ f (x + ax , y + ay ) ≈ f (x, y) + (ax ∂x + ay ∂y )f (x, y) +

h

= 1 + (ax ∂x + ay ∂y ) +

1 (ax ∂x + ay ∂y )2 f (x, y) + 2 1 (ax ∂x + ay ∂y )3 f (x, y) 3!


1 2 2 ax ∂x + 2ax ay ∂x ∂y + a2y ∂y2 + 2
i 1 3 3 ax ∂x + 3a2x ay ∂x2 ∂y + 3ax a2y ∂x ∂y2 + a3y ∂y3 f (x, y) 6

The pattern is that of a binomial (c + d)n , where c ≡ ax ∂x and d ≡ ay ∂y .

6.32 (a) For a function given by its values fn on a lattice of spacing a, the Taylor expansion gives fn+1 − fn = a

fn − fn−1 = a

 

∂f ∂x



∂f ∂x



+ n

− n

a2 2



a2 2



∂2f ∂x2



∂2f ∂x2



+ n

+ n

a3 6



a3 6



∂3f ∂x3



∂3f ∂x3



+ ··· n

− ··· n

Dividing by a shows both forward and backward difference schemes approximate (∂f /∂x)n to O(a). The central difference scheme is just the average of the previous two, fn+1 − fn−1 = 2a



∂f ∂x



+ n

a2 6



∂3f ∂x3

 + ··· n

leaving only O(a2 ) errors.

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39

(b) The second derivative can be isolated by subtracting the two Taylor series above a2 fn00 = (fn+1 − fn ) − (fn − fn−1 ) = fn+1 − 2fn + fn−1 + O(a4 ) . The same result can be obtained by taking the first derivative using forward differences and the second using backward differences,



d2 f dx2



fn+1 − fn a

=

d dx

=

fn+1 − 2fn + fn−1 . a2

n





=

1 a

h

fn+1 − fn a





−

fn − fn−1 a

i

(c) Using central differences,



∂2f ∂x∂y

 m,n

1 = 2a



∂f ∂y



−



m+1,n

∂f ∂y





. m−1,n

The first derivatives are



∂f ∂y



= m±1,n

1 2a



∂f ∂y



−



m±1,n+1

∂f ∂y





. m±1,n−1

Inserting into the previous expression yields the desired result.

6.33 Expanding the cosines, 1−

1 2!

 2 a r

+

1 4!

 4 a r

 − ··· = 1 − =1−

1 2!

 2 b r

+

1 4!

 4 b r

 − ···

1−

1 2!

 2 c r

+

1 4!

 4 c r

 − ···



1 1 b 2 + c2 + b4 + 6b2 c2 + c4 + O(1/r6 ), 2 4 2r 24r

or, through O(1/r2 ), a2 −

r2 2

 4 a r

= b 2 + c2 −


1 b4 + 6b2 c2 + c4 , 12r2

which, in the r → ∞ limit, reduces to the familar Pythagorean theorem. Which is reasonable: in this limit, the geometry of a sphere flattens to that of the plane.

6.34 (a) From Problem 2.8, we see that γ is the central angle separating points on the spherical surface. Then the arc length is simply s = aγ. As the segment of a circle whose radius is the same as the sphere’s, it lies on a great circle — which, requiring only the specification of a unique direction from one point to the next is generally what one thinks of as the shortest and straightest distance between them. (We’re discounting arcs that go the long way around.)




(b) The line element in spherical coordinates is ds2 = dr2 + a2 dΩ2 = dr2 + a2 dθ2 + sin2 θ dφ . For points on the surface of the sphere, dr ≡ 0 — and the desired result emerges. (c) For θ0 = θ + dθ, φ0 = φ + dφ: cos(dγ) = cos θ cos(θ + dθ) + sin θ sin(θ + dθ) cos(dφ) = cos θ [cos θ cos(dθ) − sin θ sin(dθ)] + sin θ cos(dφ) [sin θ cos(dθ) + cos θ sin(dθ)] .

40

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Expanding to 2nd order, keeping only through quadratic terms in dθ and dφ, 1−

1 1 1 2 1 sin θ 1 − dθ2 + cos θdθ dγ = cos θ cos θ 1 − dθ2 − sin θdθ + sin θ 1 − dφ2 2 2 2 2 h   i  1 1 1 = cos2 θ + 1 − dφ2 sin2 θ 1 − dθ2 − sin θ cos θ dθ dφ2 2 2 2    
1 2 1 1 1 2 2 2 2 1 − dθ ≈ cos θ + sin θ − sin θ dφ = 1 − dθ2 − sin2 θ dφ2 . 2 2 2 2

h





i



h






Thus dγ 2 = dθ2 + sin2 θ dφ2 and ds2 = a2 dγ 2 = a2 dθ2 + sin2 θ dφ2 .



X

6.35 With r1 = |~ r1 | and r2 = |~ r2 |, then for r1  r2 , |~ r1 − ~ r2 | =

p

r12 − 2~ r1 · ~ r2 + r22

 = r1

~ r1 · ~ r2 1−2 + r12

 = r1



1 ~ r1 · ~ r2 + 1− 2 r12

r2 r1



2 1/2

r2 r1

2

 + . . . = r1 + rˆ1 · ~ r2 + O(1/r1 )

6.36 To first order, the tangent vector ~vi along ui at P = (u1 , u1 , · · · , un ) is the difference in position ~vi (P ) = ~ r(P + dui ) − ~ r(P ) ∂~ r ∂~ r dui − ~ r(P ) = dui . ≈~ r(P ) + ∂ui ∂ui So the set {∂~ r/∂ui } is a basis for the tangent space at P . Note there’s no guarantee the basis is orthonormal.

6.37 To use the given condition h  RE , factor out RE as follows: F =

GmME GmME = 2 (RE + h)2 RE (1 + h/RE )2

Now the expression is in a form ripe for a binomial expansion, F =

GmME 1 2 (1 + h/RE )2 RE

≈

 GmME  1 − 2h/RE + O(h/Re )2 2 RE

≈

GmME . 2 RE

Plugging in ME = 6 × 1024 kg, RE = 6.4 × 106 m, and G = 6.67 × 10−11 N-m2 /kg2 gives an acceleration of a=

GME = 9.8 m/s2 ≡ g . 2 RE

A 10% correction to this requires 2h/RE = 0.1 h = RE /20; this results in a smaller gravitational acceleration since the first-order correction is negative.

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41

i

6.38 For x  a,

"

Q E(x) = 2π0 a2

#

x

1− x

p

1+

= a2 /x2

  Q 1 − (1 + a2 /x2 )−1/2 2π0 a2

Q 1 1 − 1 − a2 /x2 + O(a4 /x4 ) 2π0 a2 2 i h Q Q 1 1 2 2 ≈ a /x = 2 2π0 a 2 4π0 x2

h

=





i

So for large x, E looks like the field of a point charge!

6.39 For small t (or, more accurately, for small gt/c), c2 x= g =

c2 g

( 1+

gt c

    2 gt c

1 2

)

 2 1/2

=

−1

c2 ≈ g



1 1+ 2

 2  gt c

 −1

1 2 gt , 2

which is the non-relativistic expression for motion in a constant gravitational field.

6.40 The kinetic energy T is T = E − mc2 = c

p

m2 c2 + p2 − mc2

" = mc

p2 1+ 2 2 m c

2

 ≈ mc2

1+

#

1/2

−1

p2 p4 − 2m2 c2 8m4 c4



 −1 =

p2 p4 − 2m 8m3 c2

4

So the first-order correction is ∆T = − 8mp3 c2 . A 25% error results when ∆T p2 = − = 0.25 =⇒ pc = mc2 . p2 /2m 4m2 c2

6.41 The precise identity is E 2 − p 2 c2 =

m2 c4 1 − β2




1 − β 2 = m2 c4 .

Now we’ll expand E and p up to 4th order in β and verify that the identity still holds: 2 1 2 3 β + β4 2 8   1 2 1 3 2 4 = m c 1 + 2 · β + β 4 + 2 · β 4 + O(β 6 ) 2 4 8



E 2 = m 2 c4 1 +






= m2 c4 1 + β 2 + β 4 + O(β 6 ) Similarly

p2 = m2 c2 β 2 1 + β 2 + β 4 + O(β 6 )

42




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So






E 2 − p2 c2 = m2 c4 1 + β 2 + β 4 + O(β 6 ) − β 2 1 + β 2 + β 4 + O(β 6 ) = m 2 c4

6.42 The elliptic integral can be approximated for small k2 sin2 φ using a binomial expansion: 1 − k2 sin2 φ


−1/2

=1+

1 2 3 k sin2 φ + k4 sin4 φ + O(k6 ) 2 8

The so-called “zeroth order” result uses the first term in this series,

r Z L g

T =4

π/2

0

dφ −→ T0 = 4 1 − k2 sin2 φ

r Z L g

π/2

0

dφ = 2π 1

r

L , g

a result we recognize from intro physics. The lowest-order correction (first-order in the expansion, second-order in k) is given by π/2

r Z L g

T1 = 4

=4

h

dφ 1 +

r 0 L g

π 1 π + k2 · 2 2 4

1 2 k sin2 φ 2



i

r = 2π

L g



1+

1 2 k 4



The fractional change is T1 − T0 1 δT 1 = sin2 (α/2) = k2 = T T0 4 4 so that for α = π/3, δT /T = 1/16, i.e.,the first-order term makes a 6.25% correction.

6.43 (a) The easiest thing to do is demonstrate the equality directly:

 a b

e e =

a2 a3 1+a+ + + ··· 2 6



b2 b3 1+b+ + +· 2 6



1 2 1 (a + b2 + 2ab) + (a3 + 3a2 b + 3ab2 + b3 ) + · · · 2 6 1 1 2 = 1 + (a + b) + (a + b) + (a + b)3 + · · · 2 6 = ea+b . = 1 + (a + b) +

For this to actually constitute a proof, one must show that the terms implied by the ellipses have the general form (a + b)n /n!. Another approach is to start with exact sums and use the binomial theorem, e(a+b) =

∞ X (a + b)n

n! n=0 ∞

=

n X 1 X n


n! n=0

=

k k=0

∞ n X 1 X

n! n=0

ak bn−k

k=0

∞

X n! ak bn−k = k!(n − k)! n=0

n X ak bn−k

k! (n − k)! k=0

which looks an awful lot like the desired result. Just as we can usually change the order of

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43

integration, so too we can usually flip the order of the sums, ∞ n X X n=0

=

k=0

∞ ∞ X X k=0

n=k

Then if we define ` ≡ n − k, we have ea+b =

∞ ∞ X X ak bn−k

k! (n − k)! k=0

=

n=k

∞ ∞ X X ak b`

k! `! k=0

∞ X ak

=

∞ X b`

k!

!

`!

k=0

`=0

= ea eb

!

`=0

X

(b) Both methods in (a) rely on ab = ba. Thus if AB , BA, the simple result does not hold. One way to work out the result is to use the expansion for ln(1 + x):





ln eA eB = ln

h

1+A+

1 2 A + ... 2



1+B+

1 2 B + ... 2

i

1 1 1 2 A + AB + B 2 − (A + B + ...)2 2 2 2 1 = A + B + [A, B] + ... 2

=A+B+

If you  have the intestinal fortitude, 1you can show that the next two terms in the expansion 1 are 12 [A, [A, B]] + [[A, B], B] + 24 [A, [B, [A, B]]]. (c) Probably the simplest way to do this is to use the Taylor expansion of the exponential together with (6.51) from Problem 6.15 — discarding terms which vanish in the n → ∞ limit: eiAt/n eiBt/n

lim


n

n→∞

= lim






1 + iAt/n + O(1/n2 )


n

1 + iBt/n + O(1/n2 )

n→∞

= lim [(1 + iAt/n) (1 + iBt/n)]n n→∞

= lim



1 + i(A + B)t/n + O(1/n2 )

n

= ei(A+B)t .

n→∞

Note that the i’s are irrelevant to this formula.

6.44 First, calculate powers of M :

 M =



0 1 0 0

0 0 2 0

0 0 0 3

0 0 0 0





 , M2 =   

0 0 2 0

0 0 0 6

0 0 0 0

0 0 0 0





 , M3 =   

0 0 0 6

0 0 0 0

0 0 0 0

0 0 0 0

 . 

All higher powers of M are zero. Thus

 eM = 11 + M +

1 1 2  M + M3 =  2! 3!

1 1 1 1

0 1 2 3

0 0 1 3

0 0 0 1

  

Pascal’s triangle!

6.45 Calculating powers of σ1 and σ2 , one discovers that both square to the identity matrix, σi2 = 11.

44

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As a result, all odd powers of σi are equal to σi , and all even powers give the identity. Thus eiσ1 θ = 11 + iσ1 θ −

 1−

=

θ2 iθ3 θ4 iθ5 11 − σ1 + 11 + σ1 − · · · 2! 3! 4! 5!

θ2 θ4 + − ··· 2! 4!



 11 + iσ1

θ−

θ3 θ5 + − ··· 3! 5!



= 11 cos θ + iσ1 sin θ

 =

cos θ i sin θ

i sin θ cos θ

 .

Similarly,

 e

iσ2 θ

= 11 cos θ + iσ2 sin θ =

cos θ − sin θ

sin θ cos θ

 .

d

6.46 Applying the translation operator eτ dt to f (t) = sin ωt: d

f (t + τ ) = eτ dt f (t) =

 1+τ

τ 2 d2 τ 3 d3 τ 4 d4 τ 5 d5 d + + + + + ··· 2 3 4 dt 2! dt 3! dt 4! dt 5! dt5

= sin ωt + ωτ cos ωt −

 =

1−

 sin ωt

ω2 τ 2 ω3 τ 3 ω4 τ 4 ω5 τ 5 sin ωt − cos ωt + sin ωt + cos ωt − · · · 2! 3! 4! 5!

ω2 τ 2 ω4 τ 4 + − ··· 2! 4!



 sin ωt +

ωτ cos ωt −

ω3 τ 3 ω5 τ 5 + − ··· 3! 5!

 cos ωt

= cos ωτ sin ωt + sin ωτ cos ωt ≡ sin[ω(t + τ )]

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45

Orbits in a Central Potential

7

7.1 Recasting Eqn. (7.1) in cartesian coordinates x = r cos φ, y = r sin φ, gives r0 = r(1 +  cos φ) = r + x so that r2 = (r0 − x)2 = r02 − 2r0 x + 2 x2 or, using r2 = x2 + y 2 , r02 = (1 − 2 )x2 + 2r0 x + y 2 For  = 1, this is clearly a parabola. For 0 <  < 1, completing the square gives



r02 = (1 − 2 ) x +

r0  1 − 2

2

−

r02 2 1 − 2

+ y2 .

Letting c ≡ r0 /(1 − 2 ), a little algebra reveals 1= ≡

(x + c)2 y2 + √ [r0 /(1 − 2 )]2 [r0 / 1 − 2 ]2 (x + c)2 y2 + 2 , a2 b

which describes an ellipse with semi-major and semi-minor axes a=

r0 1 − 2

b= √

r0 1−

2

=a

p

1 − 2 .

Note the a is in fact the average of perihelion and aphelion, 2a = r(0) + r(φ). The center of the ellipse is on the x-axis at c = −a, to the left of the coordinate origin at the focus. For  > 1, the algebra is identical except that b2 is negative, resulting in the equation for a hyperbola — unbound orbits of, e.g., comets which swing by the sun once and never return.)

7.2 (a) Inserting Eqn. (7.14) for z˙ in Eqn. (7.21), then with φ˙ = `/mr2 we get Ae−i(φ−φ0 ) = −i`r˙ + `2 /r − mk




.

Taking the real part gives A cos(φ − φ0 ) = `2 /mr − k . (b) The imaginary part is −A sin(φ − φ0 ) = −`r. ˙ Multiplying by

mr2 /`

= dt/dφ and doing a bit of rearranging gives A sin(φ − φ0 ) dφ =

`2 dr . m r2

Integrating reveals the orbit equation A cos(φ − φ0 ) =

`2 +C , mr

where the integration constant C must be −k.

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7.3 (a) For a central force f (r) = −k/r2 , the assignment rˆ → eiφ gives f (z) = −

k iφ k z kz e =− 2 =− 3 . |z|2 |z| |z| |z|

(b) z ∗ z¨ − z z¨∗ = −

k (z ∗ z − zz ∗ ) = 0 . |z|3

To find the associated conserved quantity, we need to write this as a derivative, 0 = z ∗ z¨ − z z¨∗ =

d ∗ (z z˙ − z z˙ ∗ ) dt

Thus J ≡ z ∗ z˙ − z z˙ ∗ is a constant of the motion. Its form looks suspiciously like a cross product (see Problem 7.4). This, together with the identification of the complex momentum mz, ˙ leads us to expect the constant value J is actually the statement of angular momentum conservation. We can confirm this using the polar representation z = eiφ and Eqn. (7.14): ˙ − r(r˙ − irφ) ˙ = 2i`/m . J = z ∗ z˙ − z z˙ ∗ = r(r˙ + irφ) (c) By its very form, we can see that Q ≡ part, then, must be P =

1 J 2

= i`/m is the imaginary part of W = z ∗ z. ˙ Its real

1 d 1 d 2 d 1 ∗ (z z˙ + z˙ ∗ z) = (zz ∗ ) = |z| = |z| |z| . 2 2 dt 2 dt dt

d |z| + i`/m. Thus z ∗ z˙ = |z| dt (d) Multiplying Q by m¨ z , W and P by −kz/|z|3 , gives

−im¨ z

` z d = −k 3 |z| |z| − z ∗ z˙ m |z| dt   z˙ d z z d = −k |z| − = +k , |z|2 dt |z| dt |z|





or d dt



i`z˙ + k

d z dt |z|



=0.

Thus the complex expression A ≡ i`z˙ + k

z |z|

is a constant of the motion.

7.4 (a) Using a the cartesian representation ~a = (a1 , a2 ) → A = a1 +ia2 , ~b = (b1 , b2 ) → B = b1 +ib2 , A∗ B = (a1 + ia2 )∗ (b1 + ib2 ) = (a1 b1 + a2 b2 ) + i(a1 b2 − a2 b1 ) = ~a · ~b + i(~a × ~b) . The right-hand rule is reflected by the choosing A∗ B rather than its complex conjugate AB ∗ . (b) Using the polar representation z = reiφ , ~ r×p ~ = =m [z ∗ mz] ˙ =

1 m (z z˙ ∗ − z ∗ z) ˙ = mrφ˙ = ` . 2i

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47

~ = A ~ + mkˆ (c) From Eqns. (7.8) and (7.21), we know that p ~×L r → −i`z. ˙ Let’s see what BAC − CAB gives: ~ =p p ~×L ~ × (~ r×p ~) = ~ r (~ p·p ~) − p ~ (~ r·p ~)




= z 0. (We also discover that 0 0! = Γ(1) = 1.) (c) 1 z z(z + 1) z(z + 1)(z + 2) z(z + 1) . . . (z + n) = = = = ... = , Γ(z) Γ(z + 1) Γ(z + 2) Γ(z + 3) Γ(z + n + 1) which is zero for all integer z ≤ 0.

8.13 Γ

1 2

 

∞

Z

t−1/2 e−t dt

0

Z0 ∞

−y 2

(1/y) e

=

∞

Z

t1/2−1 e−t dt =

=

2ydy , where t = y 2 , dt = 2ydy

0 ∞

Z

√

2

e−y dy =

=2

π

0

Then using Γ(n + 1) = nΓ(n): Γ

3 2

 

1 Γ 2

=

√

1 2

 

π 2

=

Similarly,



Γ −

1 2



1 1 Γ − +1 −1/2 2



=



= −2 Γ

1 2

 

= −2

√

π

8.14 I3 can be calculated explicitly, using substitution and integration by parts: ∞

Z

3 −ax2

I3 =

x e 0

1 = 2



ue−au ∞ − −a 0

Z



∞

Z

1 dx = 2

e−au u du ,

u = x2 , du = 2xdx

0

∞



0

1 − e−au a



 du =

1 2a2

But this isn’t what the problem asked. We can find I3 from I1 as follows: ∞

Z

2

x3 e−ax dx =

I3 =



−

0

d d = − I1 (a) = − da da 1 =+ 2 2a



1 2a

d da

Z

∞ 2

xe−ax dx

0



Similarly,

Z

∞

I4 =

4 −ax2

x e

dx =



0

d2 =+ 2 da

Z

d2 da2

1 2

8.15 (a) J0 =

56

R∞ −∞

e−x

2

/2a2 dx

2 Z

∞ 2

e−ax dx

0

∞ 2

e−ax dx

0

 q  π a

=

= 2I0 (1/2a2 ) =

p

=

d − da

3 8

q

π 1/2a2

π a5

=

√

2π a.

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

(b) The result directly follows from n derivatives of J0 with respect to β = 1/2a2 ,



J2n =

−

n

∂ ∂β

J0 (β),

to give J2n /J0 = a2n (2n − 1)(2n − 3) · · · 5 · 3 · 1 ≡ a2n (2n − 1)!! A nice way to envision the double factorial is as the pairing of 2n points into n pairs: The first point can be paired with (2n − 1) others, the second with the still-available (2n − 3) points, etc.

8.16 (a) By the fundamental theorem of calculus, d(erf)/dx =

Z

Z 

erf(x) dx =



Z

∞

Z

erf( 0

√

d(erf) dx dx

x

Z

2 = x erf(x) − √ π



then

d x erf(x) dx dx

= x erf(x) −

(b) By the chain rule, d erf(

2 2 √ e−x ; π



1 √ π

qx) /dx =

2 2 1 x e−x dx = x erf(x) + √ e−x . π

pq

e−qx ; then

x

1 √ √ ∞ qx) e−px dx = − erf( qx) e−px − p 0

∞

q Z



h

q π

1 √ e−(p+q)x dx . x

i

0

Since erf(0) = 0 and erf(∞) = 1, the so-called “surface term" vanishes, leaving only (with u2 ≡ x)

Z

∞

erf(

√

qx)e−px dx = +

0

2 = p

1 p

∞

q Z q π

q

2 dx e−(p+q)x √ = x p

0

r

q 1 · π 2

1 π = p+q p

r

∞

q Z q π

2

e−(p+q)u du

0

q . p+q

2

8.17 Integrating by parts once, with u = 1/t and dv = e−t t dt:

Z

∞

2

2

e−t dt =

x

1 e−x − 2x 2

∞

Z

2

e−t dt t2

x

2

2

e−x e−x 3 = − 2 3 + 2 2x 2 x 2

Z

∞

2

e−t dt , t4

x

(8.1)

where in the second integration by parts we used u = 1/t3 ; for the nth , take u = t−(2n−1) .

R∞

2

R∞

2

2

e−(x−1/x) dx. Then 8.18 Multiplying out the quadratic, we easily find e−(x +1/x ) dx = e−2 0 0 note that the substitution x = 1/y yields the same exponent, but a modified integrand:

Z

∞ −(x+1/x)2

e

∞

Z

0

dy y2

2

e−(y+1/y)

dx = 0

Putting all this together, we have

Z I=

∞

e 0

−(x2 +1/x2 )

1 dx = 2



1 e2

Z 0

∞

1 1+ 2 x





−(x−1/x)2

e

 dx

.

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57

∞

Z

e−(x

2

+1/x2 )

dx =

0




1 x2

Finally, letting u = x − 1/x gives du/dx = 1 +

. Then since u ranges between ±∞, we find

1 2e2

√

∞

Z

2

e−u du =

−∞

π . 2e2

8.19 (a) ∞

Z

t(n+1)−1 e−t dt =

Γ(n + 1) =

∞

Z

tn e−t dt

0

0 ∞

Z

2

x2n e−x · 2x dx

=

(t = x2 )

0 ∞

Z

2

x2n+1 e−x dx

=2 0



= 2I2n+1 (1) = 2 − = =

 

−

n

d da

d da

n



I1 (a)

a=1



a−1

a=1



a−(n+1)

1 · 2 · 3···n

= n!

X

a=1

(b)



Γ n+

1 2



∞

Z

t(n+1/2)−1 e−t dt =

=

Z0 ∞

1

tn− 2 e−t dt

0

x

=

∞

Z

2n−1 −x2

e

(t = x2 )

· 2x dx

0 ∞

Z

2

x2n e−x dx

=2 0



= 2I2n (1) = 2 − = =

√

d π − da



n

d da

n

−1/2

a





I0 (a)

a=1

a=1

h √ 1 3 5

√ (2n − 1) −(2n+1)/2 π · · ··· a = π (2n − 1)!!/2n 2 2 2 2 a=1



i

X

where (2n − 1)!! = (2n − 1) · (2n − 3) · (2n − 5) · · · 3 · 1 (2n)! = (2n)(2n − 2)(2n − 4) · · · 4 · 2 (2n)! = n 2 n(n − 1)(n − 2) · · · 2 · 1 (2n)! = n 2 n! Note that this also shows that (2n)!! = 2n n!.

8.20

Z

v

Γ(r)Γ(s) = 0

58

∞ r−1 −v

e

 Z

∞

w

dv 0

s−1 −w

e

 dw

Z

∞

dv v

= 0

r−1

Z

∞

dw ws−1 e−(v+w) .

0

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

Let u = v + w. For fixed u, v = u implies w = 0, and v = 0 goes with w = u. Thus ∞

Z

e−u du

Γ(r)Γ(s) =

Z

u

dw ws−1 (u − w)r−1 .

0

0

Then with w = ut,

Z

∞ r+s−1 −u

u

Γ(r)Γ(s) =

e

1

 Z

 s−1

t

du

r−1

(1 − t)

dt

= Γ(r + s)B(r, s) ,

0

0

with 1

Z

tr−1 (1 − t)s−1 dt =

B(r, s) =

∞

Z 0

0

ur−1 du , (1 + u)r+s

where the second form emerges from the variable change t =

u . u+1

8.21 (a) Since (as shown in the Problem 8.20) ∞

Z

α−1 −t

t

Γ(α) =

e

Z

∞ 2

y 2α−1 e−y dx

dt = 2

0

0

we have ∞

Z

∞

Z

y 2α−1 x2β−1 e−(x

Γ(α)Γ(β) = 4 0

2

+y 2 )

dxdy .

0

To adapt the procedure used to compute the gaussian I0 , we need to transform to polar coordinates. However, since the parity of the integrand is unspecified (it depends on the particular forms of α and β), we cannot extend the integral to the entire plane. But we can rewrite our rectangular integral over the entire first quadrant as a sum over a quarter circle in polar coordinates — that is, ∞

Z

π/2

Z

2

(r sin θ)2α−1 (r cos θ)2β−1 e−r rdrdθ

Γ(α)Γ(β) = 4 0

0

Z

∞ 2

r2α+2β−1 e−r dr

=4

 Z

π/2

 sin2α−1 θ cos2β−1 θ dθ

0

0

1 = 4 · Γ(α + β) · 2

Z = 2Γ(α + β)

π/2

Z

 sin2α−1 θ cos2β−1 θ dθ

0 π/2

 sin

2α−1

θ cos

2β−1

θ dθ

,

0

which is the desired result. (b) i. For α = n + 1/2, β = 1/2, we get

Z 0

π/2

1 1 1 Γ(n + 2 )Γ( 2 ) 2 Γ(n + 1) √ √ 1 [ π(2n − 1)!!/2n ] π = 2 n! π (2n − 1)!! π (2n − 1)!! = = 2 2n n! 2 (2n)!!

sin2n θ dθ =

since 2n n! = [2n] · [2(n − 1)] · [2(n − 2)] · · · [2(1)] = (2n) · (2n − 2) · (2n − 4) · · · 4 · 2 = (2n)!!

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59

An identical calculation for α=1/2, β=n+1/2 gives the same result for

R π/2 0

cos2n θ dθ.

ii. For α = n + 1, β = 1/2, we get π/2

Z

1

1 Γ(n + 1)Γ( 2 ) 2 Γ(n + 32 ) √ 1 n! π = √ 2 π(2n + 1)!!/2n+1 2n n! (2n)!! = = , (2n + 1)!! (2n + 1)!!

sin2n+1 θ dθ =

0

An identical calculation for α=1/2, β=n+1 gives the same result for

R π/2 0

cos2n+1 θ dθ.

8.22 (a) In cartesian coordinates, “all space" means each xi ranges from −∞ to +∞:

Z

2

d3 x e−r =

Z

2

2

2

dx1 dx2 dx3 e−(x1 +x2 +x3 )

Z

∞

−∞

Z

 Z

2

dx2 e−x2

−∞

∞

dx e−x

=

∞

 Z

2

dx1 e−x1

=

2

3

√ =

∞ 2

dx3 e−x3



−∞

π3

−∞

(b) In spherical coordinates, “all space” requires that r go from 0 to ∞:

Z

3

d xe

−r 2

Z =

2

e−r r2 drdΩ

Z

Z

∞

dΩ

=

2

e−r r2 dr .

0

The r integral is just the Gaussian

1 I 2 2

=

√

π/4.

(c) Putting it all together: √

 √

Z π3 =

dΩ

π ⇒ 4

Z dΩ = 4π .

8.23 (a) As we did in the previous problem,

Z Jn =

2

dn+1 x e−r =

Z

∞ 2

dx1 e−x1

=

Z

 Z

−∞

Z

∞

dx e−x

2

2

dx2 e−x2

−∞

∞

=

2

2

2

dx1 dx2 . . . dxn+1 e−(x1 +x2 +...+xn+1 )



Z

∞

...

2

dxn+1 e−xn+1

−∞

n+1 = π (n+1)/2

−∞

60

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(b) In spherical coordinates,

Z

∞

r drdΩn =

e

Jn =

 dΩn

=

Z

 dΩn

=

Z

1 · 2

Z

1 · 2

Z

 ·

dΩn

=

R

Z

∞ 2

rn e−r dr

dΩn 0

0 Z

(c)

Z

−r 2 n

∞

t(n−1)/2 e−t dt

(t = r2 )

0 ∞

n+1 −1 2

t

e−t dt

0

1 Γ 2

n+1 2





.

dΩn is the surface area of the unit n-sphere, so the unit sphere has area 1 Γ 2



Z

n+1 2

dΩn = π (n+1)/2 ⇒

Z dΩn =

2π (n+1)/2 Γ

n+1 2




Then

Z

S1 :

dΩ1 =

2π = 2π X Γ(1)

dΩ2 =

2π 3/2 = 4π X Γ(3/2)

dΩ3 =

2π 2 = 2π 2 Γ(2)

Z

2

S :

Z

S3 :

(d) The infinitesimal volume in (n + 1)-dimensions is the n-dimensional surface area of the shell sn Ωn times its thickness ds. So a ball of radius R has volume

Z

R

Vn+1 ≡ Ωn

sn ds =

0

2π (n+1)/2 (n + 1)Γ( n+1 ) 2

Rn+1 .

Then V2 = πR2 X 4π 3 V3 = R X 3 2 π V4 = R4 2

R1R1

n exp[− a (x1 +x2 +···xn )]

R1

n 8.24 Let J(a) = limn→∞ ... x1 +x2 +···+xn 0 0 0 ferentiating with respect to the parameter a,

dJ = − lim da n→∞

Z

1

1

Z

1

Z

a e[− n (x1 +x2 +···xn )] dx1 dx1 · · · dxn

... 0

0

Z = − lim

0

1

e

n→∞

dx1 dx1 · · · dxn ; our goal is J(0). Dif-

−ax/n

n dx

0


n n −a/n e −1 a n→∞ h in 2n −a/2n e sinh(a/2n) = − lim a n→∞ h i sinh(a/2n) n −a/2 = −e lim = −e−a/2 . a/2n n→∞ = − lim

h

−

i

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61

Then

Z J(a) = −

e−a/2 da = +2e−a/2 + C ,

where, since J(∞) = 0, the integration constant C = 0. Thus J(0) = 2. Note that it’s a Rbit simpler to build that final a integral into the first step by simply replacing the ∞ da exp[−a(x1 +x2 +· · · xn )/n], and then proceeding with the x integrations. integrand with 0

8.25 Integrating by parts: π

Z

Z 0 : π   sin(x) cos2n (x) dx = cos2n−1 (x) + (2n − 1)   0



0

π

cos2n (x) dx =

0

cos2n−2 (x) sin2 (x) dx .

0

Rπ

Then sin2 = 1 − cos2 and a little algebra to collect all the

Z

π

cos2n (x) dx’s yields

π

Z

2n − 1 2n

0

cos2n−2 (x) dx .

0

Repeating the procedure gives

Z

π

cos

2n

(x) dx =



0

2n − 1 2n



2n − 1 2n



2n − 3 2n − 2

Z

2n − 3 2n − 2



π

cos2n−4 (x) dx

0

.. . =



···

π

  Z 3 4

1 2

dx .

0

Thus:

Z

1 π

π

cos2n (x) dx =

0

(2n − 1)!! (2n)! = n 2 . (2n)!! (2 n!)

(The relationships between factorial and double factorial are derived in the solution to Problem 8.21.) It’s not hard to see that the same procedure for

Rπ 0

sin2n (x) dx yields the exact same result.

8.26 (a) Let u = sin kx, dv = e−ax dx; then

Z 0

∞

1 ∞ e−ax sin kx dx = − e−ax sin kx − a 0



=+

k a

Z

∞

Z



+k −a

 Z

∞

0



e−ax cos kx dx



−k −a

∞

e−ax cos kx dx

0

Now integrate by parts again, this time with u = cos kx: k a

Z

∞

e−ax cos kx dx = −

0

k −ax ∞ e cos kx − a2 0

k =+ 2 − a



 2 Z k a

R∞

0

62

∞

e−ax sin kx dx =

0



e−ax sin kx dx

∞

e−ax sin kx dx

0

Putting it all together, we find e−ax sin kx dx = 0 can solve algebraically for the desired integral,

Z

k a

k a2

−

k 2 a


R∞ 0

e−ax sin kx dx,which we

k/a2 k = 2 [1 + (k/a)2 ] a + k2

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(b) ∞

Z

1 sin kx dx = 2i

−ax

e 0

1 = 2i 1 2i 1 = 2i =

∞

Z

e−ax eikx − e−ikx

Z0 ∞




e(ik−a)x − e−(ik+a)x




0

1 1 e(ik−a)x − e−(ik+a)x ik − a −(ik + a)   −1 −1 k + = 2 ik − a ik + a k + a2



 ∞ 0

(c) ∞

Z

e−ax sin kx dx = =m

0

∞

Z

e−ax eikx

Z0 ∞



 e(ik−a)x

= =m

= =m

h

0

= =m

h

ik + a k2 + a2

i

=

−1 ik − a

i

k k2 + a2

8.27 Differentiation with respect to a parameter: ∞

Z

x e−ax sin kx dx = −

d da

=−

d da

0

∞

Z

x e−ax cos kx dx = +

0

=

e−ax sin kx dx

h0

d dk

d dk

∞

Z

a2

k + k2

Z

∞

i

=+

2ak (a2 + k2 )2

e−ax sin kx dx

0

h a2

k + k2

i

=

a2 − k2 (a2 + k2 )2

8.28 (a) Letting u = x − π/2,

Z

π

I=

Z

π/2

ln (sin x) dx = π/2

Z ln [sin(u + π/2)] du dx =

0

π/2

ln (cos u) du 0

(b)

Z

π/2

Z [ln (sin x) + ln (cos u)] dx =

2I = 0

Z =

ln (sin x cos x) dx 0

π/2

ln 0

π/2



1 sin 2x 2



dx

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63

(c) Since | sin x| ≤ 1, we expect I to be negative; let’s see: π/2

Z

ln

2I = 0



1 sin 2x 2

dx =

1 2

   ln

0

+ ln(sin 2x)



dx

π

Z

π 1 ln 2 + 2 2 π = − ln 2 + I 2

π/2

Z



=−

ln (sin u) du 0

I=−

=⇒

π ln 2 . 2

8.29 Since x4 − 1 = (x2 − 1)(x2 + 1), the partial fraction decomposition is 1 ax + b cx + d = 2 + 2 x4 − 1 x −1 x +1 Multiplying both sides by (x2 − 1)(x2 + 1) gives 1 = (a + c)x3 + (b + d)x2 + (a − c)x + b − d . Equating the coefficients of like powers on both sides, we have a+c=0 , a−c=0 , b+d=0 , b−d=1 or simply b = −d = 1/2. Thus

Z

1 1 = x4 − 1 2 =−

Z

1 − x2 − 1

Z

1 x2 + 1



 1 tanh−1 (x) + tan−1 (x) 2

8.30 (a) Since 1 + u2 is even, parity gives I(α) =

1 π

Z

∞

−∞

1 eiαu du = 1 + u2 π

Z

∞

2 cos αu + i sin αu du = 1 + u2 π

−∞

Z 0

∞

cos αu du 1 + u2

Then 2 I(α) = π

Z

∞

0

∞ 2 du −1 = tan u =1. 1 + u2 π 0



(b) Taking a derivative with respect to α, 2 I (α) = − π 0

∞

Z 0

u sin αu du . 1 + u2

Without the sin αu, this would be logarithmically divergent for large u; since | sin | ≤ 1, the integral converges. However the log divergence gives rise to an ambiguous result in the α → 0 limit, so we need to massage it into a different form. Using 1 1 u = − u(1 + u2 ) u 1 + u2 we get I 0 (α) =

2 π

Z 0

∞

sin αu 2 − u(1 + u2 ) π

Z 0

∞

sin αu 2 = u π

Z 0

∞

sin αu −1 . u(1 + u2 )

Now the α → 0 limit is unambiguous: I 0 (0) = −1.

64

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(c) Taking a second derivative, I 00 (α) =

d dα

 Z 2 π

∞



0

∞

Z

sin αu 2 −1 = u(1 + u2 ) π

cos αu = I(α) . (1 + u2 )

0

I 00

(Calculating directly yields a divergent integral.) This familiar differential equation has solution I(α) = c1 e−α + c2 e+α . With I(0) = −I 0 (0) = 1, we get c1 = 1, c2 = 0. Thus I(α) = e−α ,

α≥0.

(d) The only change for α < 0 is that, since sin αu is odd, I 0 (0) = +1. Then c1 = 0, c2 = 1 to give solution e+α . For all α, then, I(α) = e−|α| .

8.31 Using the substitution u = tan φ/2 of Example 8.6, Eqn. (7.26) gives:

Z

Z

dφ 2

(1 +  cos φ)

= =

2(u2 + 1) [(1 − )u2 + (1 + )]2

2 (1 + )2

Z

du

u2 + 1 du , + 1)2

(β 2 u2

where β 2 ≡ (1 − )/(1 + ) is the ratio of perihelion to aphelion. A partial fraction expansion yields 2 (1 − 2 )

Z 

1 (β 2 u2 + 1)

+

β2 − 1 2 (β u2 + 1)2

 du .

(8.2)

(Note that we factored out a 1/β 2 .) The first integral should be familiar: it’s just the inverse tangent,

Z I1 =

du 1 = tan−1 (βu) . (β 2 u2 + 1) β

(8.3)

The second integral will benefit from the trig substitution tan v ≡ βu:

Z I2 =

du 1 = (β 2 u2 + 1)2 β =

Z

1 2β

sec2 v dv 1 = (tan2 v + 1)2 β



v+

1 sin 2v 2



=

Z

cos2 v dv

1 (v + sin v cos v) . 2β

(8.4)

In terms of u, we find sin v cos v = βu/(β 2 u2 + 1); thus I2 =

1 1 βu (v + sin v cos v) = tan−1 (βu) + 2 2 2β 2β β u +1

h

i

.

(8.5)

Putting this all together,

Z

dφ (1 +  cos φ)2

=


 2 I1 + β 2 − 1 I2 (1 − 2 )

1 = (1 − 2 ) =

"



1 β+ β




!#

tan

−1

β2 − 1 u (βu) +

β 2 u2 + 1

2 2 tan−1 (βu) − (1 − 2 )(1 + ) (1 − 2 )3/2



u β 2 u2 + 1

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(8.6)

65

It’s fairly simple to rewrite the first term in terms of φ; as for the second: −2 (1 − 2 )(1 + )



u β 2 u2 + 1

(1 + ) tan(φ/2) 2 (1 − 2 )(1 + ) (1 − ) tan2 (φ/2) + (1 + ) h i  2 sin(φ/2) cos(φ/2) =− (1 − 2 ) (1 − ) sin2 (φ/2) + (1 + ) cos2 (φ/2)  sin(φ) =− . (1 − 2 ) 1 +  cos φ



h

=−

i

(8.7)

and so — at last! — Eqn. (7.26) gives µ = 1 − 2

Z


3/2

dφ

= 2 tan−1 (βu) −  = 2 tan

−1

(8.8)

(1 +  cos φ)2

r

p

1 − 2

sin(φ) 1 +  cos φ



p 1− sin φ tan (φ/2) −  1 − 2 1+ 1 +  cos φ

(8.9)

Phew!

8.32 Steepest Descent and The Stirling approximation: (a) n! = (b, c)

R∞ 0

xn e−x dx: a simple max/min exercise verifies that xn e−x peaks at x = n. ∞

Z

n −x

n! =

x e

Z

∞

en ln x−x dx .

dx =

0

0

Then, using f (x) = n ln x − x; f 0 (x) = n/x − 1; f 00 (x) = −n/x2 , we expand f (x) around x = n: f (x) = (n ln n − n) + 0 −

1 (x − n)2 + ··· . n 2!

Thus

Z n! ≈

∞ 2 1 e[(n ln n−n)− 2n (x−n) ] dx

0

= e(n ln n−n) (n ln n−n)

Z

∞

Z0 ∞

≈e

1

2

1

2

e− 2n (x−n) dx e− 2n (x−n) dx ,

−∞

since in the large-n the integrand is very small for negative x. Evaluating the remaining gaussian integral yields the desired result.

66

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Dirac Delta

9 9.1 (a)

R3

(b)

R3

(c)

R1

−2 0

[x3 − (2x + 5)2 ] δ(x − 1) dx = (1)3 − [2(1) + 5]2 = −48

(5x2 − 3x + 4) δ(x + 2) dx = 0, since range of integration does not include x = −2 √ cos x δ(x − π/6) dx = cos(π/6) = 3/2

(d)

R0π

(e)

R1

(f)

R1

(g)

R

−π −1 −1 π

ln (sin x + 2) δ(x + π/2) dx = ln(1) = 0

2

4x2 cos−1 x δ(2x − 1) dx =



(h)

R∞

(i)

R 2b

(j)

R π/2

p

0

δ(x + q) dx =

R π/2

ex δ(tan x) dx =

−π/2

2

R

−q > p −q < p

1, 0,

ex

(x3 − 3x2 + 2) δ(x) dx = 7 · 2 = 14

(x − 1) ex δ(x) dx = −1/3

−1 π 1 2 −π

x δ[(x2 − b2 )(x − b/2)] dx =

−π/2

−1

R1 1

(x − 1) ex δ(−3x) dx = + 3

−π

R1

1 1/7

(x3 − 3x2 + 2) δ(x/7) dx =

4x2 cos−1 x δ(x − 21 ) =

1 2

· 4 · ( 12 )2 ·

π 3

=

π 6

= Θ(−p − q) .

R 2b 0

x [ b12 δ(x − b) +

P

1 δ(x 3b2

+ b) +

cos2 xn δ(x − xn ) dx = 1 ,

n

4 δ(x 3b2

− b/2)] dx =

5 3b

xn = nπ

9.2 Since strictly speaking a delta function is only defined under an integral, we need to prove this using integration:

Z

∞

f (x) −∞

dΘ ∞ − dx = f (x)Θ(x) dx −∞



Z

Z

∞

f 0 (x)Θ(x)dx

−∞

0

= f (∞) −

∞

Z

0

0

f (x)Θ(x)dx + −∞





f (x)Θ(x)dx 0

∞

Z



f 0 (x) dx

= f (∞) − 0 + 0





= f (∞) − f (∞) − f (0) = f (0)

X

9.3 Limit of Gaussians First note that

Z

∞

n φn (x) dx = √ π −∞

Z

∞ 2 2

e−n

x

dx = 1 ,

−∞

independent of n. So far so good. But to demonstrate that the sequence of φn ’s provides a representation of δ(x) in the n → ∞ limit, we need to show

Z

∞

f (x)φn (x) dx = f (0) .

lim n→∞

−∞

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(9.1)

A suggestive, though non-rigorous approach is with a simple change of variables, y = nx: ∞

Z lim n→∞

n f (x)φn (x) dx = lim √ π n→∞ −∞

Z

1 = lim √ π n→∞

Z

∞

f (x) e−n

2 2

x

dx

−∞ ∞ 2

f (y/n) e−y dy

Z −∞ ∞

1 = f (0) √ π

2

e−y dy = f (0) .

−∞

The problem with this argument is that while f (y/n) → f (0) as n → ∞ is fine for finite y, our integral has infinite limits. This wouldn’t be a concern had we taken the limit after integrating — as we’re supposed to. For more rigor, we split the integration in (9.1) into three intervals,

Z

∞

Z

−1/

√ n

= −∞

Z

1/

√

+ −∞

−1/

n

√

Z

∞

+ n

1/

√ n

.

Making the variable change y = nx in the outer two integrals shows that they each vanish in √ the limit. As for the middle integral, the reason we chose limits of ±1/ n is so that we can squeeze f (x) to f (0) the same way we squeezed f (ck ) in Example 9.2 using the mean value theorem. Once f (cn ) is removed from the integral, the change y = nx will yield the error √ function erf ( n) — which in the limit is just 1. Poof.

9.4 As was worked it out in the text, for a > 0 we can make the variable change y = ax:

Z

∞

Z

∞

f (x) δ(ax) dx = −∞

f (y/a) δ(y) −∞

1 dy a

=

1 f (0) . a

For a < 0, the manipulations are almost identical — the only difference is an overall sign,

Z

∞

Z

−∞

f (x) δ(ax) dx = −∞

∞

1 f (y/a) δ(y) dy = − a

Z

∞

f (y/a) δ(y) −∞

1 dy a

1 = − f (0) . a

Both cases can be succinctly stated as δ(ax) =

R

9.5 Using

δ(ax)dx =

1 |a|

R

1 δ(x) . |a|

δ(x)dx, we find that δ is even:

R

δ(−x)dx = +

R

δ(x)dx.

9.6 First, the only contributions to the integral are at x = ±a, so

Z

∞

f (x) δ(x2 − a2 ) dx =

−∞

Z

−a+

−a−

f (x) δ(x2 − a2 ) dx +

Z

a+

f (x) δ(x2 − a2 ) dx .

a−

Now in the vicinity of x = a, x2 − a2 = (x + a)(x − a) ≈ 2a(x − a) ; similarly, close to x = −a, x2 − a2 = (x + a)(x − a) ≈ −2a(x + a) .

68

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So the integral becomes

Z

∞ 2

Z

2

−a+

f (x) δ(x − a ) dx =

a+

Z







f (x) δ 2a(x − a) dx

f (x) δ –2a(x + a) dx + −a− −a+

−∞

Z =

−a−

1 f (x) δ x + a) dx + 2a

Z

a− a+

f (x) a−

1 δ(x − a) dx 2a

 1  = f (−a) + f (a) . 2a Thus δ(x2 − a2 ) =

1 2a





δ(x + a) + δ(x − a) .

p

p2 c2 + m2 c4 . Then —

9.7 For convenience, define Ep ≡

Z

d3 p

Z

∞

dE δ(E 2 −p2 c2 −m2 c4 ) =

Z

d3 p

Z

∞

dE

0

h

0

1 1 δ(E − Ep ) + δ(E + Ep ) = 2E 2E

i

Z

d3 p . 2Ep

9.8 In all three cases, we change the integration variable to y ≡ nx2 : (a) δ(x) =

1 √ π

2 2

limn→∞ n e−n

x

:

∞

1 |x| δ(x2 ) dx = √ lim n π n→∞ −∞

Z

2 = √ lim n π n→∞

Z

2 = √ lim n π n→∞

Z

Z

1 = √ π

(b) δ(x) =

1 π

limn→∞

∞

e−n

x

|x| dx

−∞ ∞

e−n

2 4

x

x dx

0 ∞

e−y

0

∞

Z

2 4

2

e−y dy =

0

2

dy 2n

1 . 2

n : 1+n2 x2

Z

∞

1 |x| δ(x ) dx = lim π n→∞ −∞ 2

2 = lim π n→∞ 1 = π =

Z 0

∞

Z

∞

Z−∞ ∞ 0

n |x| dx 1 + n2 x4 n x dx 1 + n 2 x4

dy 1 + y2

1 ∞ 1 π 1 tan−1 y = · = . π π 2 2 0



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69

(c) δ(x) = limn→∞

sin2 (nx) : nπx2

Z

∞ n→∞

−∞

∞

Z

|x| δ(x2 ) dx = lim

−∞ ∞

sin2 (nx2 ) |x| dx nπx4

Z

= 2 lim n→∞

2 = lim π n→∞ 1 = π

Z

∞

Z0

∞

0

sin2 (nx2 ) x dx nπx4 sin2 y dy (y 2 /n) 2n

sin2 y dy . y2

0

Comparing this last integral to the original sequence of functions — and knowing that each function in the sequence is normalized, independent of n — we easily see that the integral is π/2, once again yielding the correct answer 1/2. [The function sin(y)/y is often called the “sinc” function which, amongst other things, gives the diffraction pattern for the canonical double-slit interference experiment. It has the curious property that its oscillations precisely cancel so that integrals from 0 to ∞ of sinc(y) and sinc2 (y) are both π/2.]

9.9

Z

Z ∞  : 0

∞

−∞

 xn δ (n) (x) dx = xn δ (n−1) (x) 

∞

−n

−∞

xn−1 δ (n−1) (x) dx

−∞

Z ∞ ∞ :0    = −nx δ  (x) + n(n − 1) xn−2 δ (n−3) (x) dx −∞  −∞ n−1 (n−2)

= . . . = (−1)n n! δ(x) .

9.10 (a) Cylindrical coordinates: We confine the ring to the xy-plane with a factor of δ(z). The term δ(r − R) restricts contributions to a circle of radius R. (We denote the radial coordinate by r rather than ρ to avoid confusion with the density.) With no restriction on φ, we have

ρ(~ r) =

M δ(r − R)δ(z) . 2πR

(b) Spherical coordinates: We confine the ring to the xy-plane with a factor of δ(cos θ). The term δ(r − R) again restricts contributions to a circle of radius R. With no restriction on φ, we have

ρ(~ r) =

70

M δ(r − R)δ(cos θ) . 2πR2

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9.11 We set up the defining relation and reverse the order of the integrals: b

Z

b

Z

1 2π

Z

1 = 2π

Z

=

1 2πi

∞

Z

−∞

∞

Z

−∞ ∞ −∞ ∞

1 2πi

Z



dx

a b

Z

dx f (x)

dk

Z

dk ikx e 2π

dx f (x)eikx

dk

1 = 2πi =

=

−∞ b

a

a

=

∞

Z

f (x)

f (x)δ(x)dx =

−∞ ∞ −∞

a

dk k

Z

dk k




1 d eikx ik dx

b

dx f (x) a


d eikx dx

a

Z a



b

dx a

i

Z

Z

df dx

b

dx a

∞

−∞



dk ikx e k

df · iπ sgn(x) dx

df |x| dx dx x

1 1 f (b) sgn(b) − f (a) sgn(a) − sgn(b) = 2 2

h

b

Z

1 1 [f (b) · iπ sgn(b) − f (a) · iπ sgn(a)] − 2πi 2πi

1 1 = [f (b) sgn(b) − f (a) sgn(a)] − 2 2

df ikx e dx

dx

a

 dk  1 f (b)eikb − f (a)eika − k 2πi

b

Z

b

f (x)eikx −

Z

b

dx 0

df + sgn(a) dx

Z

0

dx a

df dx



1 1 = f (b) sgn(b) − f (a) sgn(a) − sgn(b) f (b) − f (0) + sgn(a) f (0) − f (a) 2 2 h i 1 = f (0) sgn(b) − sgn(a) 2

h

(

i

f (0) , −f (0) , 0,

=

a 0 G(x, t) = −

c c [iπ sgn(ct + x) + iπ sgn(ct − x)] = − Θ(ct − |x|) , 4iπ 4

We can include the fact that I is non-zero only for t > 0 with an additional step function in t, resulting in Eqn. (22.28), GR (x, t) = −

c Θ(ct − |x|) Θ(t) . 2

The derivation of the advanced Green’s function GA (x, t) in Eqn. (22.29) proceeds almost identically; the only fundamental difference is that the poles of I need to be pushed into the UHP to ensure that GA vanishes for t > 0.

22.6 As with the wave on a string, the denominator in the integrand for G can be found by inserting U (x, t) = Cei(kx−ωt) into the differential equation — in this case, the Schrödinger equation (22.30): ∞

Z

Z

dk 2π

G(x, t) = i −∞

∞

dω −∞

ei(kx−ωt) . ω − ~k2 /2m

The e−iωt term requires closing the contour with a semicircle in the UHP for t < 0, so by pushing the pole at ~k2 /2m into the LHP we easily achieve GR (t < 0) = 0. For t > 0 the contour must be closed with a semicircle in the LHP; that’s where the effect of the pole at ~k2 /2m − i will be seen: G(x, t > 0) =

i 2π 2

Z

∞

dk eikx

−∞

i = −2πi · 2π 2 =

1 2π

Z

∞

Z

∞

dω −∞

Z

e−iωt ω − ~k2 /2m + i

∞

dk eikx e−i(~k

2

/2m)t −t

e

−∞ 2 −i ~k t−kx 2m




dke −∞

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153

where in that last step we took the  → 0 limit. What remains is a gaussian integral in k, which can be evaluated by completing the square. The result is (22.33):

q

G(x, t > 0) =

2 m eimx /2~t . 2πi~t

22.7 (a) With G(~ r) in Eqn. (22.43), 2

∇ +k

2




2

G(~ r) = ∇ + k

Z

2




Z

d3 q ei~q·~r (2π)3 k2 − q 2


ei~q·~r

d3 q (2π)3

∇2 + k 2

Z

d3 q (2π)3

−q 2 + k2

Z

d3 q

= = =

k2 − q 2


ei~q·~r k2 − q 2

ei~q·~r = δ(~ r)

(2π)3

X

(b)

Z G(~ r) =

d3 q 1 ei~q·~r = 3 2 (2π) k − q 2 (2π)2

Z

1 = (2π)2

Z

∞

q 2 dq

−1

0 ∞

 2

q dq 0

i =+ 2 4π r

Z

i 8π 2 r

Z

=+

1

Z

∞



∞



0

−∞

eiqru du k2 − q 2

1 eiqr − e−iqr iqr k2 − q 2

eiqr − e−iqr q 2 − k2 eiqr − e−iqr q 2 − k2



 q dq

 q dq .

(Anticipating the q integral, we elect not to write this in terms of sin qr.) (c) Note that the integral over eiqr must be closed in the UHP, while that over e−iqr must be closed in the LHP. The poles at q = ±k can be moved off the real axis in four different ways:

(i)

(ii)

AAAB6XicbVBNS8NAEJ3Ur1q/qh69LBahXkqigh4LXjxWtB/QhrLZbtqlm03YnQgl9Cd48KLi1V/k0X/jts1BWx8MPN6bYWZekEhh0HW/ncLa+sbmVnG7tLO7t39QPjxqmTjVjDdZLGPdCajhUijeRIGSdxLNaRRI3g7GtzO//cS1EbF6xEnC/YgOlQgFo2ilh6o475crbs2dg6wSLycVyNHol796g5ilEVfIJDWm67kJ+hnVKJjk01IvNTyhbEyHvGupohE3fjY/dUrOrDIgYaxtKSRz9fdERiNjJlFgOyOKI7PszcT/vG6K4Y2fCZWkyBVbLApTSTAms7/JQGjOUE4soUwLeythI6opQ5uOzcBb/niVtC5q3mXNvb+q1Gt5GkU4gVOoggfXUIc7aEATGAzhGV7hzRk7L86787FoLTj5zDH8gfP5AwxfjTA=

(iii)

AAAB6nicbVBNS8NAEJ34WetX1aOXxSLUS0lU0GPBi8cK9gPaUDbbSbt0swm7G6GE/gUPXlS8+oc8+m/ctDlo64OBx3szzMwLEsG1cd1vZ219Y3Nru7RT3t3bPzisHB23dZwqhi0Wi1h1A6pRcIktw43AbqKQRoHATjC5y/3OEyrNY/lopgn6ER1JHnJGTS7VOL8YVKpu3Z2DrBKvIFUo0BxUvvrDmKURSsME1brnuYnxM6oMZwJn5X6qMaFsQkfYs1TSCLWfzW+dkXOrDEkYK1vSkLn6eyKjkdbTKLCdETVjvezl4n9eLzXhrZ9xmaQGJVssClNBTEzyx8mQK2RGTC2hTHF7K2FjqigzNh6bgbf88SppX9a9q7r7cF1t1Is0SnAKZ1ADD26gAffQhBYwGMMzvMKbI5wX5935WLSuOcXMCfyB8/kD0PuNow==

(iv)

AAAB63icbVBNS8NAEJ3Ur1q/qh69LBahXkJiBT0WvHisYNpCG8pmu2mX7m7C7kYoob/BgxcVr/4gj/4bt20O2vpg4PHeDDPzopQzbTzv2yltbG5t75R3K3v7B4dH1eOTtk4yRWhAEp6oboQ15UzSwDDDaTdVFIuI0040uZv7nSeqNEvko5mmNBR4JFnMCDZWCuqMsctBtea53gJonfgFqUGB1qD61R8mJBNUGsKx1j3fS02YY2UY4XRW6WeapphM8Ij2LJVYUB3mi2Nn6MIqQxQnypY0aKH+nsix0HoqItspsBnrVW8u/uf1MhPfhjmTaWaoJMtFccaRSdD8czRkihLDp5Zgopi9FZExVpgYm4/NwF/9eJ20r1y/4XoP17WmW6RRhjM4hzr4cANNuIcWBECAwTO8wpsjnBfn3flYtpacYuYU/sD5/AGWEY4W

AAAB6nicbVBNS8NAEJ3Ur1q/qh69LBahXkKigh4LXjxWsB/QhrLZbtqlu5uwuymU0L/gwYuKV/+QR/+NmzYHbX0w8Hhvhpl5YcKZNp737ZQ2Nre2d8q7lb39g8Oj6vFJW8epIrRFYh6rbog15UzSlmGG026iKBYhp51wcp/7nSlVmsXyycwSGgg8kixiBJtcqrPp5aBa81xvAbRO/ILUoEBzUP3qD2OSCioN4Vjrnu8lJsiwMoxwOq/0U00TTCZ4RHuWSiyoDrLFrXN0YZUhimJlSxq0UH9PZFhoPROh7RTYjPWql4v/eb3URHdBxmSSGirJclGUcmRilD+OhkxRYvjMEkwUs7ciMsYKE2PjsRn4qx+vk/aV61+73uNNreEWaZThDM6hDj7cQgMeoAktIDCGZ3iFN4c7L86787FsLTnFzCn8gfP5A+SvjbA=

(a) Move both poles into the UHP, ±k −→ ±k + i. Thus only the eiqr term contributes — G++ (~ r) =

i 8π 2 r

Z

∞

−∞

i = · 2πi 8π 2 r

q eiqr dq (q − k − i)(q + k − i)



k eikr −k e−ikr + 2k −2k

 =−

1 cos kr . 4π r

(b) Move both poles into the LHP, ±k −→ ±k − i. This time, only the integral over e−iqr

154

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contributes — G−− (~ r) =

=

i 8π 2 r

Z

∞

−∞

−q e−iqr dq (q − k + i)(q + k + i)

i · −2πi 8π 2 r



−k e−ikr k eikr + 2k −2k

 =−

1 cos kr . 4π r

(c) Move the pole at +k into the UHP, the other into the LHP, ±k −→ ±k ± i. For this choice, each term has a contribution from one pole. From the UHP:

Z

∞

−∞

q eiqr dq keikr = 2πi · = iπ eikr (q + k + i)(q − k − i) 2k

and from the LHP:

Z

∞

−∞

ke−i(−k)r −q e−iqr dq = −2πi · = iπ eikr (q + k + i)(q − k − i) −2k

All together, G+− (~ r) =

1 ikr i · 2iπ eikr = − e . 8π 2 r 4πr

(d) The complement of the previous one, ±k −→ ±k ∓ i. From the UHP:

Z

∞

−∞

q eiqr dq −kei(−k)r = 2πi · = iπ e−ikr (q + k − i)(q − k + i) −2k

and from the LHP:

Z

∞

−∞

−ke−ikr −q e−iqr dq = −2πi · = iπ e−ikr (q + k − i)(q − k + i) 2k

All together, G−+ (~ r) =

i 1 −ikr · 2iπ e−ikr = − e . 8π 2 r 4πr

(d) Appending the time-dependence e−iωt = e−ikct to the G(r)’s shows that both G++ and G−− describe standing wave, whereas G+− and G−+ describe outgoing and incoming spherical waves, respectively. Note incidentally, that in the k → 0 limit, the Helmholtz equation reduces the Poisson equation — and sure enough, the Coulomb potential emerges from all four G’s in that limit.

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155

Prelude: Superposition

23

 23.1 (a) The transformation can be achieved with the matrix M = (b) (c) (d) (e)

a c



b . Hence it’s linear. d

Tˆ: Non-linear unless ~b = 0. ˆ Non-linear due to cross terms S: ˆ : Non-linear, since scalars need not be positive N
ˆ Linear, since ∇ ~ × aA ~ + bB ~ = a∇ ~ ×A ~ + b∇ ~ ×B ~ C:

ˆ Linear D: ˆ 2 : Linear D ˆ 2 + x2 : Linear D ˆ D2 [f (x)] + sin2 [f (x)]: Non-linear R P2 R P2 R P2 ˆ K(x, t)g(t) dt. K(x, t)f (t) dt+b K(x, t) (af (t) + bg(t)) dt = a (j) Linear: K[af +bg] =

(f) (g) (h) (i)

P1

P1

P1

23.2 Since operators ∂x2 , ∂t and multiplication by a function V (x) are all linear, the Schrödinger equation is linear. Therefore its solutions are superposable. [Hence the appellation “the Schrödinger wave equation.”]

23.3 Let’s prove it two ways: (a) L(0) = L(0) + L(0) = 2L(0) =⇒ L(0) = 0. (b) L(0) = L(a0) = aL(0) must be true for all a. Hence L(0) = 0.

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

Vector Space

24 24.1 (a) Yes

(b) No: since scalars can be negative, space not closed. (c) No: not closed under scalar multiplication. (d) Yes: this is a plane in Rn through the origin. (e) Yes (f) No: matrix multiplication is not commutative, AB , BA. (g) Yes (h) No: not closed (i) Yes (j) Yes

24.2 Notice that the definition of U does not mix the vectors from spaces V and W . Thus if V and W are vector spaces, so is U . In ket notation, one often writes |v, wi = |vi|wi.

24.3 (a) Independent (though not orthogonal!)

  (b) Not independent:

1 0

  +

0 1

  =

1 1

(c) Independent: as in Example 24.7, the determinant of the matrix of columns vectors is nonzero. (d) Independent (e) Independent: no linear combination axn + bxm can yield cx` for arbitrary `, m, n

24.4 The space C is equivalent to R2 , as is easily appreciated by considering a complex number z as an arrow in the plane. So the set of complex numbers forms a vector space. And then it becomes obvious that z = a + ib = (a, b) and its complex conjugate z ∗ = a − ib = (a, −b) are linearly independent for non-zero a and b.

24.5 Taking the vectors on the standard basis

!

|1i →

1 1 1

!

, |2i →

1 −2 1

!

, |3i →

1 0 −1

we form the determinant

1 1 1 −21 0 1 1 −1

=6,

which, being non-zero, shows that the three vectors |ii are linearly independent.

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(24.1)

~ both as 24.6 If the components are not unique on a given basis, we can expand a vector A ~= A

n X

ci ~ei

~= A

and

i=1

n X

ci0 ~ei .

i=1

Subtracting the two expansions gives 0=

n X

(ci − ci0 )~ei .

i=1

But since the ~ei ’s form a basis, they are linearly independent vectors — and so this last expansion can have only the trivial solution in which the coefficients di ≡ ci − ci0 are zero. Thus ci = ci0 , and the expansion on a given basis is unique.

a1 a2 a3

!

!

b1 24.7 (a) |vi → |wi → b2 b3 ! ! ! b1 a1 + b1 a1 (b) a2 + b2 = a2 + b2 a3 b3 3 ! ! a3 + b! b1 a1 ka1 (c) b2 = k a2 = ka2 =⇒ bi = kai b3 a3 ka3 a1 + b1 a2 + b2 a3 + b3

(d) If |vi and |wi are inverses, then bi = −ai Then

! =0

24.8 The simplest is to assign them to the standard basis,

!

|Ri →

1 0 0

!

|Gi →

0 1 0

!

0 0 1

|Bi →

!

1 White is equal admixture of all the primary colors, |W i → 1 . Black is given by the zero 1 vector. Orange is an equal admixture of yellow !and red, and yellow is equal parts red and green 2 — so |Oi = |Y i + |Ri = 2|Ri + |Gi → 1 . Note that only the ratios of the components 0 matter here, so the overall norm of the vector is not relevant.

24.9 The line in French translates as “This is not a pipe.” And indeed, it isn’t: the painting is a representation of a pipe, not a physical pipe. [Of course, you can’t use the painting to smoke tobacco — so I guess it’s not a faithful representation!]

158

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The Inner Product

25 25.1 (a) h1|1i =

1


∗  1 i

i

1

=2

−2i

−i

h2|2i =

1


∗

1 −2i 1

h4|4i =


∗  −i 2i

2i

=5

!

h3|3i =

i

1

i

h5|5i =

=6


∗

eiϕ

−1


∗

eiϕ −1



!

i 1 i

=3

Then 1 |10 i = √ |1i 2

1 |20 i = √ |2i 5

1 |30 i = √ |3i 2

1 |40 i = √ |4i 6

1 |50 i = √ |5i 3

(b) Similar matrix multiplication yield: h1|2i = 2 − i

h2|1i = 2 + i

h1|3i = eiϕ + i

X

h3|1i = eiϕ − i

X

h3|2i = −ie−iϕ − 2i

h2|3i = ieiϕ + 2i h4|5i = 4i

h5|4i = −4i

X

X

25.2 With |Bi = |iAi, hB|Bi = hiA|iAi = ihiA|Ai = ihA|iAi∗ = (i)(i)∗ hA|Ai = hA|Ai ≥ 0 . So despite the factor of i, the positivity of |Ai guarantees the positivity of |Bi. With hiA| = −ihA| we can skip the conjugate symmetry step: hB|Bi = hiA|iAi = (i)(i)∗ hA|Ai = hA|Ai ≥ 0 .

25.3 (a) hA ± B|A ± Bi = [hA| ± hB|] [|Ai ± |Bi] = hA|Ai ± hA|Bi ± hB|Ai + hB|Bi Thus hA + B|A + Bi + hA − B|A − Bi = 2 [hA|Ai + hB|Bi]. (b) This one is similar, except this time |Ai and |Bi are orthogonal: hA + B|A + Bi = hA|Ai + hA|Bi + hB|Ai + hB|Bi = hA|Ai + hB|Bi .

25.4 (a) Direct calculation gives

 |ui = |Mvi = M|vi →

2 2−i



and †

hu| = hMv| = hv|M →

1

i




1 −i



2 1

= (2

2 + i) ,

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

=2

which are clearly adjoints of one another. (b) −1

hw|M|vi =

1




1 −i

2 1



i 1

1 2

1 −i

 = −i ,

and 1

†

hv|M |wi =

+i






−1 1



= +i = hw|T |vi∗ .

25.5 From Eqn. (25.23): |ui = T |wi = |T wi and hT † v| = hv|T ; then starting with hv|ui — hv|T wi = hv|T |wi = hT † v|wi. We can just as easily start with hu|vi = hv|ui∗ : hu|vi = hv|ui∗ hT w|vi = hv|T wi∗ hw|T † |vi = hv|T |wi∗ .

25.6 (a) hw|T |vi = hw|T |vi∗∗ = hv|T † |wi∗ = hw|T †† |vi −→ T †† = T , which is the obvious generalization of z ∗∗ = z for complex numbers. (b) Defining |ui = T |vi: hw|ST |vi∗ = hw|S|ui∗ = hu|S † |wi = hv|T † S † |wi , It’s perhaps more elegant, however, to just use the T |vi ≡ |T vi notation: hw|ST |vi∗ = hw|S|T vi∗ = hT v|S † |wi = hv|T † S † |wi . Either way, Eqn. (25.24)
then gives (ST
)† = T † S † . 1 2 1 i For matrices S = −2 1 and T = 2 1 : (ST )† =



1 2



i 1

1 −2

†

2 1

 =

1 − 2i 0

2+i 5

†

 =

1 + 2i 2−i



0 5

and T †S† =



1 2



−2 1

1 −i



2 1

 =

1 + 2i 2−i



0 5

.

’Nuff said.

25.7 Schwarz: cos θ ≡

|hv|wi| : kvkkwk

√ 1 (a) hv|wi = 1, kvk = 5, kwk = 2 −→ cos ϕ = √ . 5 2 p √ √ (b) hv|wi = −4 + 5i, kvk = 7, kwk = 10 −→ cos ϕ = 41/70. √ √ √ (c) hv|wi = 4, kvk = 7, kwk = 5 −→ cos ϕ = 4/ 35.

160

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25.8 1

h1|1i =

0


∗  1 0

1

i


∗ 

1 h↑ | ↑i = 2

1

1


∗ 

hI|Ii =

1 9

i−

√

1 i

3

h2|2i =

=1

1 2

h+|+i =

1 1

0


∗  0 1

1

=1

h−|−i =

1 2

1

−i

=1

1 h↓ | ↓i = 2

1

−1


∗

1 + 2i

√  i− 3 =1 1 + 2i

1

h1|2i =

=1


∗ 

0


∗  1 −i


∗

h+|−i =

=1



1 −1

1 2

0 1

=0

1

i

1 h↑ | ↓i = 2

=1

1


∗  1 −i


∗

1

=0



1 −1

=0

 √
∗ 1 1 − 2i i + 3 1 −√ 2i =1 i+ 3 9 
∗ √ 1 −√ 2i i − 3 1 + 2i =0 i+ 3

hII|IIi =

hI|IIi =

1 9

The elements of these column vectors are all given on the standard basis — e.g., 1 i 1 |+i = √ |1i + √ |2i → √ 2 2 2

  1 i

.

25.9 |Ai = a1 |1i + a2 |2i = a1 |1i + a2 |2i = a+ |+i + a− |−i = a↑ | ↑i + a↓ | ↓i = aI |Ii + aII |IIi where a1 = h1|Ai =

1


∗

0

1−i 2 + 2i

 =1−i

a2 = h2|Ai =

0

1


∗

1−i 2 + 2i

 = 2 + 2i

Similarly, 1 = h+|Ai = √ 2

1

i


∗

1 a↑ = h↑ |Ai = √ 2

1

1


∗

a+

1−i 2 + 2i

1 aI = hI|Ai = 3 aII

1 = hII|Ai = 3

1−i 2 + 2i

i−

√





3

1 − 2i

3 − 3i = √ 2

3+i = √ 2

1 + 2i

i+

√

3

1 a↓ = h↓ |Ai = √ 2


∗


∗

a−

1 = h−|Ai = √ 2

1−i 2 + 2i

1−i 2 + 2i

 =

 =

−i

1

1

−1


∗

1−i 2 + 2i


∗

1−i 2 + 2i





=

1 + 3i =− √ 2

√ √ 1 [(5 − 3) + i( 3 − 3)] 3

√ √ 1 [(5 + 2 3) + i(2 3 − 1)] 3

A straightforward calculation verifies that the norm is independent of basis: |a1 |2 + |a2 |2 = |a+ |2 + |a− |2 = |a↑ |2 + |a↓ |2 = |aI |2 + |aII |2 = 10 .

25.10 With inner product hA|Bi =

1 Tr(A† B), 2

 A=

a c

b d

the decomposition of

 = m0 11 + m1 σ1 + m2 σ2 + m3 σ3

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−1 + i √ 2

161

on the basis of Pauli matrices is m0 =

1 1 1 Tr(11A) = Tr(A) = (a + d) , 2 2 2 1 1 Tr(σ1 A) = Tr 2 2



m1 =

1 i Tr(σ2 A) = Tr 2 2



m2 =

1 1 Tr(σ3 A) = Tr 2 2



m3 =

c a

d b

 =

1 (b + c) 2



−c a

−d b

a −c

b −d

=

i (b − c) 2

=

1 (a − d) 2



which is just Eqn. (24.23).

~ ·u 25.11 From the directional derivative df /du ≡ ∇f ˆ, the Schwarz inequality states

~ u ~ . df /du ≤ ∇f ˆ = ∇f



~ is the maximum value of the change in f , occurring for u ~ (i.e., cos ϕ = 1). So ∇f ˆ parallel to ∇f

25.12 This is just the triangle inequality ~ + Bk ~ ≤ kAk ~ + kBk ~ . kA ~= With A

P

~ + Bk ~ = kA

i

~ = ai eˆi and B

sX

P

b eˆ i i i

(ai + bi )ˆ ei ·

i

and eˆi · eˆj = δij , we have:

X

(aj + bj )ˆ ej =

sX

j

(ai + bi )(aj + bj )δij =

sX

ij

(ai + bi )2

i

and, similarly, ~ + kBk ~ = kAk

sX

a2i +

sX

i

b2i

i

Then, by the triangle inequality,

2 i1/2 h X i1/2 h X i1/2 hX ≤ |ai |2 + |bj |2 . (ai + bi ) i

i

j

25.13 hA|Ai =



hˆ e1 | + hˆ e2 | + hˆ e3 |





|ˆ e1 i + |ˆ e2 i + |ˆ e3 i

= hˆ e1 |ˆ e1 i + hˆ e2 |ˆ e2 i + hˆ e3 |ˆ e3 i + hˆ e1 |ˆ e2 i + hˆ e1 |ˆ e3 i + hˆ e2 |ˆ e1 i + hˆ e2 |ˆ e3 i + hˆ e3 |ˆ e1 i + hˆ e3 |ˆ e2 i = 3 , ˆ = where we’ve used the orthonormality, hˆ ei |ˆ ej i = δij . Thus the normalized vector is |Ai

25.14 Using only the basis-independent orthonormality relation, eˆi · eˆj = δij : v10 ≡ hˆ ex0 |V i =



v20 ≡ hˆ ey0 |V i =

162

ˆ ı√ −ˆ  2





ˆ ı√ +ˆ  2

· (−ˆı + 2ˆ ) = −



· (−ˆı + 2ˆ ) =

3 √ 2

1 √ 2

X X

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1 √ |Ai. 3

and w10 ≡ hˆ ex0 |W i =



w20 ≡ hˆ ey0 |W i =



ˆ ı√ −ˆ  2

ˆ ı√ +ˆ  2





· (ˆı − iˆ ) =

1+i √ 2

X

· (ˆı − iˆ ) =

1−i √ 2

X

25.15 It’s easier to work with the norm-squared, N ≡ k|Ai −

X

ai |ˆ ei ik2 = hA|Ai −

X

i



ei |Ai + ai hA|ˆ ei i + a∗i hˆ

X i,j

i

= hA|Ai − 2

ej |ˆ ei i a∗j ai hˆ

X

ai hA|ˆ ei i +

X

i

a2i

,

i

since the vector space is real and the basis is orthonormal. Then dN = −2hA|ˆ ek i + 2ak = 0 dak

ak = hA|ˆ ek i .

=⇒

Noting that d2 N /da2k > 0, we see that indeed ak = hA|ˆ ek i minimizes the norm.

25.16 The vectors are  obviously normalized;  as for their orthogonality: hˆ e1 |e2 i = hˆ e1 | |φ2 i − hˆ e1 |φ2 i |ˆ e1 i

= hˆ e1 |φ2 i − hˆ e1 |φ2 i = 0









hˆ e1 |e3 i = hˆ e1 | |φ3 i − hˆ e1 |φ3 i |ˆ e1 i − hˆ e2 |φ3 i |ˆ e2 i hˆ e2 |e3 i = hˆ e2 | |φ3 i − hˆ e1 |φ3 i |ˆ e1 i − hˆ e2 |φ3 i |ˆ e2 i

= hˆ e1 |φ3 i − hˆ e1 |φ3 i − 0 = 0 = hˆ e2 |φ3 i − 0 − hˆ e2 |φ3 i = 0

25.17 The questions asks to verify the last two orthonormal vectors in a 4-dimensional space. So we already have 1 |ˆ e1 i → √

2



1 1

i |e2 i → √

1 1

 2



−1 1

1 −1

.





1 0

Then with |M3 i →



0 , we find i

|e3 i ≡ |M3 i − hˆ e1 |M3 i |ˆ e1 i − hˆ e2 |M3 i |ˆ e2 i

 →



1 0

0 i

1 − Tr 2

"

1 − Tr 2

 =

=

1 2



1 0

0 i

−



1−i 0

Then, with he3 |e3 i =

1 Tr 2

1 (1 + i) 4 0 −1 + i

"



1 1

† 

1 1

1 1

−i √ 2



−1 1



1 1

−

#

1 0

0 i

† 



1 √ 2

1 −1

1 0

1 (−1 − i) 4





1 1

1 1

#

i √

0 i

−1 1

 2

−1 1



1 −1



1 −1

 .

e†3 e3 =

|ˆ e3 i =



1 √ 2




1 , 2

the normalized basis vector is

|e3 i

p

1 → √ 2 he3 |e3 i



1−i 0

0 −1 + i

 .

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163

 Similarly, |M4 i →

0 1



−1 0

gives

|e4 i ≡ |M4 i − hˆ e1 |M4 i |ˆ e1 i − hˆ e2 |M4 i |ˆ e2 i − hˆ e3 |M4 i |ˆ e3 i



0 1

→



−1 0

1 − Tr 2 1 − Tr 2

"

"

1 √ 2



−i √ 2



† 

1 1

1 1

1 −1



0 1

"

1 √ 2

−1 0

0 1



1−i 0

i √

0 −1 + i



1 1

1 1

 2

−1 1

† 

0 1



1 −1 −1 0

#

1 √ 2



1−i 0

0 −1 + i





−1 0

,

since all the traces vanish. Then, since he4 |e4 i =

P

c |φ i m m m

e†4 e4 = 1,




1 Tr 2

 |ˆ e4 i = |M4 i →

25.18 (a) For |vi =



1 √ 2

#

† 

−1 1

1 − Tr 2 =

#

−1 0

0 1



−1 0

0 1

.

we have

! X

hφn |vi = hφn |

cm |φm i

=

X

m

=

X

cm hφn |φm i

m

cm km δmn = cn kn .

m

Thus cn =

1 hφn |vi . kn

(b) Using the result above for cn ,

|vi =

X

cm |φm i =

m

X 1 m

km



hφm |vi |φm i =

X 1 m

km

|φm ihφm |vi =

X 1 m

km

! |φm ihφm |

which requires

X 1 m

km

|φm ihφm | = 11 .

25.19 It’s clearly true when |vi and |wi are basis vectors, since Mij ≡ hˆ ei |M|ˆ ej i . For the more general case, insert complete sets of states on either side of the operators in Eqn. (25.24) in

164

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|vi

order to extract the elements Mij of the operator M on the basis: hv|M† |wi = hw|M|vi∗

!∗ X

X

†

hv|ˆ ei ihˆ ei |M |ˆ ej ihˆ ej |wi =

ij

hw|ˆ ei ihˆ ei |M|ˆ ej ihˆ ej |vi

ij

!∗ X

† wj vi∗ Mij

X

=

wi∗ Mij vj

ij

ij

=

X

X

∗ ∗ vj = wi Mij

∗ wj . vi∗ Mji

ij

ij † ∗ =⇒ Mij = Mji

25.20 Tr(A) =

X

hˆ ei |A|ˆ ei i =

X

i

hˆ ei |

" X

i

=

# " |ˆ a ihˆ a | A

a

X

# X

|ˆ b ihˆ b | |ˆ ei i

b

hˆ ei |ˆ a ihˆ a |A|ˆ b ihˆ b |ˆ ei i

iab

=

X

hˆ b |ˆ ei ihˆ ei |ˆ a ihˆ a |A|ˆ b i

iab

=

X

hˆ b |ˆ a ihˆ a |A|ˆ b i =

ab

X

hˆ a |A|ˆ a i = Tr(A) X

a

25.21 (a) Once a vector has been projected down to a subspace, repeating the same projection has no effect. (b) (|ˆ ei ihˆ ej |)2 = |ˆ ei ihˆ ej |ˆ ei ihˆ ej | = |ˆ ei ihˆ ej |δij = |ˆ ei ihˆ ei |. So it’s only a projection for i = j. Thus P must be hermitian, P † = P. (c) Yes, it’s a projection:



(m) P⊥

2

" =

#"

m−1

11 −

X

11 −

|ˆ e` ihˆ e` |

#

m−1

`=1

X

|ˆ ek ihˆ ek |

k=1

m−1

= 11 − 2

X

|ˆ e` ih+|

m−1

m−1

X

|ˆ e` ih+|

`=1 (m)

(d) hA|Bi = hV |P⊥



(m)



|V i−P⊥ |V i

|ˆ e` ihˆ e` ||ˆ ek ihˆ ek |

`,k

`=1

= 11 − 2

X

X

(m)

|ˆ e` ih= |P⊥

X

`=1 (m)

(m) 2

= hV |P⊥ |V i−hV |P⊥



(m)

|V i = hV | P⊥

(m) 2

− P⊥



|V i =

0X

25.22 No: The ai and σi are different P mathematical objects. In particular, ~σ = (σ1 , σ2 , σ3 ) is not a vector in R3 . And indeed, a · σi is not a scalar. i i 25.23 Declaring that an arbitrary 2 × 2 complex matrix M can be expanded on the basis of Pauli

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165

matrices plus the identity is itself a statement of completeness — but a completeness relation should utilize only the basis, without reference to an arbitrary vector. That’s the goal of this problem:

M =

3 X

cα σα ,

α=0

where projection gives cα = hσα |M i =


1 1 † Tr σα M = Tr (σα M ) , 2 2

† since σα = σα . Thus

M =

3 X

cα σα =

1 2

a=0

X

Tr (σα M ) σα .

α=03

In terms of the elements Mk` for k, ` = 1, 2, 3:

Mk` =

1 2

3 X

Tr (σα M ) (σα )k`

α=0

1 = 2

" 3 X X α=0

=

# (σα )ij Mji (σα )k`

ij

" 3 X 1X 2 ij

# (σα )ij (σα )k` Mji

α=0

This can only be true if the sum over the σα ’s collapses the i and j sums — specifically, 1 2

3 X

(σα )ij (σα )k` = δi` δjk .

α=0

25.24 (a) Tr(σi ) = 0, and σi† = σi for all i. (b) The simplest is to just verify by direct multiplication of all possible pairs of σ’s. (c) (~a · ~ σ )(~b · ~ σ) =

X

a` bm (σ` σm )

`m

=

X

a` bm (11δ`m + i`mn σn )

`m

=

X

a` b` 11 + i

X

`

`mn a` bm σn = ~a · ~b 11 + i~ σ · (~a × ~b) .X

`m

(d) Direct calculation: hA|Bi =

 1 1  Tr(A† B) = Tr (~a · ~ σ )(~b · ~ σ) 2 2 1X 1 = ai bj Tr (σi σj ) = 2 2 ij

166

X

ai bj 2δij = ~a · ~b .

ij

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Or, using results of parts (a) and (c):

 1 1  Tr(A† B) = Tr (~a · ~ σ )(~b · ~ σ) 2 2  1  σ · (~a × ~b) = ~a · ~b . = Tr ~a · ~b 11 + i~ 2

hA|Bi =

25.25 Taking the basis

!

!

1 −i 1

|φ1 i →

i 0 i

|φ2 i →

1 i −1

|φ3 i →

!

in reverse order:

!

1 i −1

|φ3 i

1 |ˆ e30 i = p → √ 3 hφ3 |φ3 i

.

Then, since |ˆ e30 i is orthogonal to |φ2 i,

!

|e2 i = |φ2 i −

hˆ e30 |φ2 i|ˆ e30 i

hˆ e30 |φ1 i|ˆ e30 i

hˆ e20 |φ1 i|ˆ e20 i

→

i 0 i

→

1 −i 1

=⇒

|ˆ e20 i

i → √

−

1 0 1

!

1 0 1

2

.

And finally, |e1 i = |φ1 i −

−

!

1 + 3

!

1 i −1

!

!

1/3 −2i/3 −1/3

=

=⇒

|ˆ e10 i

1 → √

25.26 The determinant of the 4 × 4 matrix of these column vectors is −4 − 4i , 0 — so the vectors are linearly independent. Applying Gram-Schmidt yields:

 |ˆ e1 i →

1 2

 

1 1 1 1

  

 |ˆ e2 i →

1 2

 

1 −1 1 −1

  

 1 |ˆ e3 i → √ 2

 

0 −i 0 i



 |ˆ e4 i →

 

1 2

 

1−i 0 −1 + i 0

25.27 The probabilities are the sum of the magnitude-squared of the expansion coefficients on the basis. So the probability of finding spin up along y is



1 |c + |2 = |h+|ψi|2 = √ 2

1 |c− |2 = |h−|ψi|2 = √

2


∗  2
1 1 a = |a − ib|2 = |a|2 + |b|2 b 2 2
∗  2
1 1 a 1 −i |a|2 + |b|2 = |a + ib|2 = b 2 2 1

i

If |ψi is normalized, then we know that |a|2 + |b|2 = 1. Thus the total probability of finding the particle either spin up or down along y is 1 — that is, 100%.

25.28 Using the orthonormality of | ↑i, | ↓i: ~2 | ↑ ih ↓ | + | ↓ ih ↑ | | ↓ ih ↑ | − | ↑ ih ↓ | 4   2 ~ ~ =i | ↑ ih ↑ | − | ↓ ih ↓ | = i Sz 4 2

Sx Sy = i







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167

  

6

1 −2i −1

! .

Similarly, Sx Sy = −i ~2 Sz . The remainder of the i , j cases follow the same way. As for i = j, ~2 | ↑ ih ↓ | + | ↓ ih ↑ | | ↑ ih ↓ | + | ↓ ih ↑ | 4   2 ~ ~2 = | ↑ ih ↑ | + | ↓ ih ↓ | = 11 . 4 4



Sx Sx =





On the standard matrix basis, these operators are represented by the Pauli matrices, Sx →



~ ~ σ1 = 2 2

~ ~ Sy → i σ2 = 2 2 Sz →

1 0





0 −1



~ ~ σ3 = 2 2



0 1

1 0



1 0

0 −1

as is easily confirmed. And indeed, σi σj = δij + iijk σk .

25.29 (a) T = |1ih2| + |2ih1| (b) Three possible choices of distinct bases:

 – On the standard matrix basis, T =



0 1

1 . 0

  – On the orthonormal basis |1i → 1 T → 2





1 −1

1

1


∗

1 + 2

1 , |2i → 1

1 √ 2

  1 1

−1

1





1 = 2



1 , −1

1 √ 2


∗

  – On the orthonormal basis |1i → 1 T → 5

  1 2i

2

−i


∗

1 , |2i → 2i

1 √ 5

 

1 + 5

2 −i

1

2i

1 −1

1 −1



 +

1 1



−1 −1


∗

1 = 5



2 4i

+i −2



 +

2 −i

−4i −2

On the standard matrix basis

S→S=

T = |3ih1| + |2ih2| + |1ih3|

!

1 0 , 0

0 1 , 0

1 0 0

0 0 −1

0 0 0

!

0 0 , this gives 1

! T →T =

0 0 1

0 1 0

!

1 0 0

25.31 Using the bases from Eqns. (25.128) and (25.131), 1 |ˆ 1 i → √ 2

168

  1 1

1 |ˆ 2 i → √

 2



1 −1

0 −1



2 , −i

25.30 The outer product sum can be read directly from the operators’ actions on the basis,

!

=

1 0

  1 √ 5

(Note that T 2 = 11 — and indeed, all three matrices square to the identity.)

S = |1ih1| − |3ih3|



1 |ˆ ε1 i → √ 2

  1 i

1 |ˆ ε2 i → √ 2

 

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1 −i



1 = 5



4 3i

−3i −4



gives h1 |ε1 i = h2 |ε2 i =

1 (1 + i) 2

h1 |ε2 i = h2 |ε1 i =

1 (1 − i) . 2

Then Eqn. (25.126) yields M

00

1 = 4



1−i 1+i

1+i 1−i



3 0



0 −1

1+i 1−i

1−i 1+i



 =

1 2i

−2i 1

 .

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169

Interlude: Rotations

26

26.1 The goal is to verify the passive expression Eqn. (26.14), |vi = the components from Eqn. (26.2), we get |vi = v10 |ˆ e10 i + v20 |ˆ e20 i =





P i

vi |ˆ ei i =

P i



vi0 |ˆ ei0 i. Taking



v1 cos ϕ − v2 sin ϕ |ˆ e10 i + v1 sin ϕ + v2 cos ϕ |ˆ e20 i





= v1 cos ϕ|ˆ e10 i + sin ϕ|ˆ e20 i + v2





− sin ϕ|ˆ e10 i + cos ϕ|ˆ e20 i

≡ v1 |ˆ e1 i + v2 |ˆ e2 i , where |ˆ e1 i = cos ϕ|ˆ e10 i + sin ϕ|ˆ e20 i

|ˆ e2 i = − sin ϕ|ˆ e10 i + cos ϕ|ˆ e20 i

is (the inverse of) the passive basis transformation, Eqn. (26.8).

26.2 Starting with Eqn. (26.13), |v 0 i = v10 |ˆ e1 i + v20 |ˆ e2 i







= v1 cos ϕ − v2 sin ϕ |ˆ e1 i + v1 sin ϕ + v2 cos ϕ |ˆ e2 i







= v1 cos ϕ|ˆ e1 i + sin ϕ|ˆ e2 i + v2 − sin ϕ|ˆ e1 i + cos ϕ|ˆ e2 i ≡ v1 |ˆ e10 i + v2 |ˆ e20 i .

This primed basis is a counter-clockwise rotation. Note that we started with a positive active rotation using Eqn. (26.13), not (26.14). So the new vector |v 0 i expanded on the identicallyrotated basis |ˆ ei0 i has components vi because it’s as if no rotation has occurred at all!

26.3 Dotting both sides of ˆi = sought-after result,

P j

Rij eˆj with ˆk and using orthonormality quickly yields the

! ˆi · eˆk =

X

X

· eˆk =

Rij eˆj

j

Rij (ˆ ej · eˆk ) = Rik .

j

Applied to the given orthonormal bases gives

 R=

ˆ1 · eˆ1 ˆ2 · eˆ1

which is a rotation by ϕ = cos−1 (1.4/

ˆ1 · eˆ2 ˆ2 · eˆ2 √



1 = √

 2

2) = sin−1 (0.2/

1.4 −0.2 √

0.2 1.4

 ,

2) = 8.13◦

m (ϕ) = R (mϕ). For rotations in the plane with m = 3, matrix 26.4 Rn ˆ (ϕ)Rn ˆ (ϕ) · · · Rn ˆ (ϕ) = Rn n ˆ ˆ multiplication



cos 3ϕ sin 3ϕ



− sin 3ϕ cos 3ϕ

 =

cos ϕ sin ϕ

3

− sin ϕ cos ϕ

reveals cos 3ϕ = cos3 ϕ − 3 sin2 ϕ cos ϕ and sin 3ϕ = − sin3 ϕ + 3 cos2 ϕ sin ϕ.

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26.5 (a) Orthogonal (but with det = −1, its a reflection). Since real and orthogonal, it’s also unitary. It’s not hermitian. (b) Hermitian. Neither orthogonal nor unitary. (c) Neither orthogonal, unitary, or hermitian. (The columns are complex-orthogonal, but they’re not orthonormal) (d) The exponential of a real anti-symmetric matrix is orthogonal and unitary. (e) Hermitian and unitary.

26.6 (a) symmetric: there are n diagonal elements and 12 n(n − 1) elements above the diagonal. All together then 21 n(n + 1) independent parameters are required. (b) anti-symmetric: the diagonal elements are zero, leaving only the 12 n(n − 1) off-diagonal elements to be specified. [Note that sym+antisym = n2 ] (c) orthogonal: 12 n(n − 1) [See BTW 26.4.] (d) unitary: There are n real elements on the diagonal plus 12 n(n − 1) complex = n(n − 1) real numbers in the off-diagonals slots. So a total of n2 independent parameters are needed. (e) hermitian: since complex, there are 2n2 real numbers. But the diagonal elements must be real, and the off-diagonals complex conjugates — n2

26.7 Referring  to Example 26.5 and Eqn. (26.34), i can be represented by the orthogonal matrix 0 −1 , which is a 90◦ -rotation in the plane. Then 1 0

 2

i →

0 1

2

−1 0



−1 0

=



0 −1

= −11 ←→ −1 .

Similarly,

 i3 →

0 1

3

−1 0

 =

−1 0



0 −1

0 1



−1 0



0 1

=−



−1 0

←→ −i .

26.8 Using U † = U −1 , 1 = det(11) = det U U † = det (U ) det U † = det (U ) det (U )∗ = | det(U )|2 ⇒ det U = eiα ,







for α real.

26.9 The transformation from the cartesian basis {ˆ ex , eˆy , eˆz }, to the spherical basis {ˆ e+ , eˆ− , eˆ0 } is essentially a π/4 rotation in the complex xy-plane, eˆ+ eˆ− eˆz

!

eˆx eˆy eˆz

=T

! ,

where 1 T = √ 2

−1 +1 0

−i −i 0

!

0 √0 2

.

Not surprisingly, this is a unitary transformation, T T † = 11. The spherical basis provides a complex representation for R3 vectors, ~ = a+ eˆ+ + a− eˆ− + a0 eˆ0 , A

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171

where a0 ≡ az , and

a±

±i

1

1 ≡ hˆ e± |Ai = ∓ √ 2

0


∗

!

ax ay az

1 = √ (∓ax + iay ) . 2

A little algebra suffices to verify that hA|Bi = ax bx + ay by + az bz = a∗+ b+ + a∗− b− + a0 b0 = −a− b+ − a+ b− + a0 b0 .

26.10 The key is to appreciate that the rotation



~0 E ~0 cB





~ E ~ cB

= R(ϕ)





− sin ϕ cos ϕ

cos ϕ sin ϕ

R(ϕ) ≡

,



is in “field space”, and is not a function of either ~ r or t. (a) Taking the divergence of the rotated fields and using the source-free Maxwell equations (26.6):



~ ·E ~0 ∇ ~ ·B ~0 c∇



 = R(ϕ)

~ ·E ~ ∇ ~ ·B ~ c∇



1 ∂t c



=0.

Similarly taking the curls (recall µ0 0 = 1/c2 ),



~ ×E ~0 ∇ ~ ×B ~0 c∇





~ ×E ~ ∇ ~ ×B ~ c∇



~ −∂t B ~ cµ0 0 ∂t E

= R(ϕ)

= R(ϕ)



=

1 ∂t R(ϕ) c

 



~ −cB ~ E

 =

~0 −cB ~ E0



 =

~0 −∂t B 0 ~ E /c

 .

(b) It’s clear that the duality symmetry is spoiled if we include the electric sources ρ and J~ — so instance, we’d get ~ 0 = ρ cos ϕ , ρ ~ ·E 0 ∇

~ 0 = ρ sin ϕ , 0 . ~ ·B c∇

(c) With electric and magnetic sources transforming as







ρ0e ρ0m /c

= R(ϕ)











J~e0 J~0 /c

ρe ρm /c

 = R(ϕ)

m



J~e J~m /c

,

we now get ~ ·E ~0 ∇ ~ ·B ~0 c∇

 = R(ϕ)

~ ·E ~ ∇ ~ ·B ~ c∇



 = R(ϕ)

ρe /0 c µ0 ρm

=

1 R(ϕ) 0

=

1 0









ρe ρm /c



ρ0e ρ0m /c

 =

ρ0e /0 c µ0 ρ0m



The curls follow similarly.

172

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P

26.11 As a rotation, the magnitude of a vector, k~v k2 = v ∗ v , must be invariant. For a transfori i i mation U which takes a complex vector ~v into another complex vector ~v 0 ,

!∗ hv|vi = ||~v 0 k2 =

X

vi0∗ vi0 =

X X i

i

=

X

=

X X

Uik vk

k

! X

Ui` v`

`

∗ Uik Ui` vk∗ v`

ik`

! k`

where, recall, that adjoint

U†

† Uki Ui`

vk∗ v` ,

i

is the transpose complex-conjugate, U† ≡ UT∗ .

Thus in order for the length of the vector to be constant, we must have

X

† Uki Ui` = δk` ⇐⇒ U † U = I ,

i

or simply U † = U −1 . We get the same result by insisting that the relative orientation of vectors is invariant. Using bra-ket notation, hA 0 |B 0 i = hU A|U Bi = hA|U † U |Bi which only yields hA|Bi for U † U = 11.

26.12 Just as picking an axis in R3 requires two parameters, so too picking an axis in Rn requires n − 1 parameters. So then all we need is to determine the number of additional parameters required to rotate in the (n − 1)-dimensional hyperplane. To do this rotation requires n − 2 parameters for an axis, leaving a rotation in the (n − 2)-dimensional subspace. We continue in this way until we get to R3 , which needs two parameters for an axis, and one more (the rotation angle) to perform the rotation in the plane. All told then, the number of parameters required to specify rotations in Rn is (n − 1) + (n − 2) + (n − 3) + · · · 2 + 1 =

n−1 X

m=

1 n(n − 1) . 2

m=1

26.13 Combining vectors must be a commutative operation, |v1 i + |v2 i = |v2 i + |v1 i. Group elements need not obey this. (Groups that do are said to be abelian.) Also, groups do not have scalar multiplication defined. For example, for |a| , 1, aR is not a rotation, and so is not an element of the group!





 

 

1 0 1 0 easily reveals that v1 = and v2 = are invariant. 0 −1 0 1 The first is the x-axis — i.e., that plane of the mirror, so reflection leaves it completely unscathed, B0 v1 = v1 . The other is the y-axis, which the mirror reflects, sending all of y > 0 to negative values and vice-versa, B0 v2 = −v2 .

26.14 (a) Inspection of B0 =

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173

(b) Rotating v1 and v2 gives



 

− sin ϕ cos ϕ

cos ϕ sin ϕ

1 0

 =



cos ϕ sin ϕ

and



− sin ϕ cos ϕ

cos ϕ sin ϕ

  0 1

 =

− sin ϕ cos ϕ

 .

Check:



cos 2ϕ sin 2ϕ



sin 2ϕ − cos 2ϕ





cos ϕ sin ϕ

=







cos 2ϕ cos ϕ + sin 2ϕ sin ϕ sin 2ϕ cos ϕ − cos 2ϕ sin ϕ

 =



cos ϕ sin ϕ

X





and



cos 2ϕ sin 2ϕ



− sin ϕ cos ϕ

sin 2ϕ − cos 2ϕ



− cos 2ϕ sin ϕ + sin 2ϕ cos ϕ sin 2ϕ cos ϕ − cos 2ϕ sin ϕ

=

=−

− sin ϕ cos ϕ

X

26.15 Using the venerable conversion factor 1 picture = 1000 words: y

y

y

Bx

P z x

x

x

z

x z

z

y

ˆ is mapped into a left-handed one (ˆı × ˆ = −k) ˆ . In each case, a right-handed system (ˆı × ˆ = k)

26.16 Since ~ r · (δ ϕ ~ ×~ r) = 0, ~ r · δ~ r=

1 1 δ (~ r·~ r) = δ(r2 ) = 0 . 2 2

Thus the magnitude of ~ r stays constant.

26.17 To first order, successive infinitesimal rotations by δ ϕ ~ 1 and δ ϕ ~ 2 gives Rn2 (δϕ2 )Rn1 (δϕ1 )~v = ~v + (δ ϕ ~ 1 × ~v ) + δ ϕ ~ 2 × [~v + (δ ϕ ~ 1 × ~v )] = ~v + (δ ϕ ~ 1 × ~v ) + (δ ϕ ~ 2 × ~v ) = ~v + (δ ϕ ~ 1 + δϕ ~ 2 ) × ~v . Similarly, Rn1 (δϕ1 )Rn2 (δϕ2 )~v = ~v + (δ ϕ ~ 2 + δϕ ~ 1 ) × ~v . Thus not only do infinitesimal rotations commute, but we see that two successive infinitesimal rotations add vectorially. √

!

−1 √ 26.18 (a) A = − 6 √ 2 − 6 has orthonormal columns (including the 41 factor in front) 6 3 −1 and it has det A = 1. So it’s a proper rotation with ! angle given by cos ϕ = (TrA − 1) /2 = 1 1 → ϕ = π/3. The axis is given by n ˆ = √1 0 , since Aˆ n=n ˆ. 2 2 −1 1 4

174

3√

6

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 (b) B =

0 1



1 . Clearly orthogonal — but since det B = −1, it’s a reflection. The reflection 0

 

1 axis is at ϕ such that cos 2ϕ = 0, sin 2ϕ = 1, or ϕ = π/4 — that is, the line . 1 √ ! 3 0 1 (c) C = 21 0 −2 √ 0 . Orthogonal, but with det C = −1. So it can be factored into −1 0 3 a reflection and a ! rotation. If we factor out a reflection across y, we are left with C 0 = √ 3 0 1 1 0 2 0 , which is a rotation around y by π/6. We could also factor out a parity 2 √ −1 0 3 operator P = −11, so that C is a reflection though the origin and a 5π/6 rotation around the y-axis. √ ! ! 0√ 2 0 0 1 (d) D = − √ 0 1 1 . This is a π rotation around 1 − 2 . 2 0 1 −1 1

26.19 0 1 0

0

~v = Rx (π/2)Ry (π/2)~v =

!

0 0 1

!

1 0 0

1 1 0

!

=

0 1 1

=

1 0 −1

whereas 0 0 −1

!

!

0 ~v 00 = Ry (π/2)Rx (π/2)v = −1 0 √ The angle between them is (note that each has length 2) ϕ = cos−1



1 0 ~v · ~v 00 2



1 0 0



= cos−1 −

1 2



1 1 0

! .

= 2π/3 = 120◦ .

26.20 The matrix in Eqn. (26.99) is 1 Rq = 3

!

−1 2 −2

2 −1 −2

2 2 1

.

As an R3 rotation, we know the rotation angle is ϕ = cos−1



TrRq − 1 2



= cos−1 (1/3) = 70.5◦ .

To verify that qˆ is the axis, just show that it is unscathed by the rotation:

3

1 √

2

2 −1 −2

−1 2 −2

!

2 2 1

−1 1 0

! =

3

1 √

!

−3 3 0

2

1 = √ 2

!

−1 1 0

.

26.21 Since n ˆ is at 45◦ in the yz-plane, the two simplest ways are: 1. Rotate Ry counter-clockwise around x to align rotation axis with n ˆ —



cos α

1 T Rx (π/4)Ry (α)Rx (π/4) =  √2 sin α − √1 sin α 2

− 1 2 1 2

1 √ 2

sin α

(1 + cos α) (1 − cos α)

1 2 1 2

1 √ 2

sin α



(1 − cos α) (1 + cos α)

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175

2. Rotate Rz clockwise around x to align rotation axis with n ˆ – T Rx (−π/4)Rz (α)Rx (−π/4)

which yields the same result: a rotation by α around n ˆ=

1 √ (0, 1, 1). 2

26.22 A vector transforms as vi0 =

X

Rij vj .

ij

A matrix transforms by similarity, M 0 = RM RT — which in index notation is 0 = Mij

X

T Rik Mk` R`j =

k`

X

Rik Rj` Mk` .

k`

In both examples, one rotation matrix R is required for each vector index. Thus we expect a three-index object to transform as 0 Tijk =

X

Ri` Rjm Rkn T`mn .

`mn

Note that unlike the previous two cases, there’s no way to express this using conventional two-dimensional arrays.

26.23 (a) The simplest 1-1 correspondence between index a and permutations of {ij} is









w1 u1 v1 w2  = u1 v2  w3  u2 v1  w4 u2 v2 (b)



cos2 ϕ cos2 ϕ T → 2 cos ϕ cos2 ϕ

− sin ϕ cos ϕ sin ϕ cos ϕ − sin ϕ cos ϕ sin ϕ cos ϕ

− sin ϕ cos ϕ − sin ϕ cos ϕ sin ϕ cos ϕ sin ϕ cos ϕ



sin2 ϕ − sin2 ϕ − sin2 ϕ sin2 ϕ

(c)







w10 cos2 ϕ w20  cos2 ϕ w0  = cos2 ϕ 3 w40 cos2 ϕ

− sin ϕ cos ϕ sin ϕ cos ϕ − sin ϕ cos ϕ sin ϕ cos ϕ

− sin ϕ cos ϕ − sin ϕ cos ϕ sin ϕ cos ϕ sin ϕ cos ϕ





sin2 ϕ u1 v1 − sin2 ϕ u1 v2  − sin2 ϕ u2 v1  sin2 ϕ u2 v2





(u1 cos ϕ − u2 sin ϕ) (v1 cos ϕ − v2 sin ϕ)  (u cos ϕ − u2 sin ϕ) (v1 cos ϕ + v2 sin ϕ) = 1 (u1 cos ϕ + u2 sin ϕ) (v1 cos ϕ − v2 sin ϕ) (u1 cos ϕ + u2 sin ϕ) (v1 cos ϕ + v2 sin ϕ)

26.24 In index notation — using little more than RT = R−1 ,

! R~v · Rw ~ =

X X i

=

X ik`

176

Rik vk

! X

k

Rik Ri` vk w` =

Ri` w`

`

X ik`

T Rki Ri` vk w` =

X k`

δk` vk w` =

X

vk wk = ~v · w. ~

k

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The cross product requires more serious index gymnastics:

! R~v × Rw ~ →

X

X

pmn

mn

! X

Rmj vj

j

Rnk wk

k

! =

X

pmn Rmj Rnk vj wk =

X X

jkmn

jk

Now use the orthogonality of rotations to insert

P i

pmn Rmj Rnk

vj wk

mn

T =δ , R`i Rip `p

! X

pmn Rmj Rnk =

mn

X

X

`mn Rmj Rnk

`mn

T R`i Rip

! =

i

X X i

`mn R`i Rmj Rnk

Rpi =

`mn

i

where we’ve used Eqn. (4.42), ijk det[M ] =

X

`mn Mi` Mjm Mkn .

`mn

As a proper rotation, det[R] = +1. Thus

! R~v × Rw ~ →

X

ijk Rpi vj wk =

ijk

X

Rpi

X

i

ijk vj wk

→ R (~v × w) ~ .

jk

26.25 You can work this out in detail using the explicit forms of the Pauli matrices; we’ll take an   1 0 ˆ so that we only need consider σ3 = easier approach and choose n ˆ = k, . The other 0 −1 directions work the same way. Then eiϕ σ3 /2 = 11 + iϕσ3 /2 +

1 1 1 (iϕ σ3 /2)2 + (iϕ σ3 )3 + + (iϕ σ3 )4 + · · · 2! 3! 4!

Since all the Pauli matrices have σi2 = 11, so the infinite series separates into even terms proportional to the identity, and odd terms proportional to σ3 : 1 1 1 1 (ϕ/2)2 + (ϕ/2)4 + · · · + iσ3 ϕ/2 − (ϕ/2)3 + (ϕ/2)5 + · · · 2! 4! 3! 5! = 11 cos(ϕ/2) + iσ3 sin(ϕ/2) .





eiϕ σ3 /2 = 11 1 −





The most general approach uses Eqn. (25.138) to show that (ˆ n·~ σ )m = 11 for even m, n ˆ ·~ σ for odd. Then: ˆ ·~ σ /2 eiϕ n =

X (iϕ/2)m m!

(ˆ n·~ σ )m

m=0

= 11

X (iϕ/2)m m! m=even

+n ˆ ·~ σ

X

X (iϕ/2)m m! m=odd

= 11 cos(ϕ/2) + iˆ n·~ σ sin(ϕ/2) .

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177

ijk det[R]Rpi ,

26.26 Using Eqns. (26.60) and (26.65),



cosh Φ − sinh Φ

− sinh Φ cosh Φ



0 −1

cosh Φ − sinh Φ



cosh Φ − sinh Φ

− sinh Φ cosh Φ



cosh Φ + sinh Φ

− sinh Φ − cosh Φ



cosh2 Φ − sinh2 Φ 0

ΛT ηΛ =

=

=



1 0

0 sinh2 Φ − cosh2 Φ









1 0

=



− sinh Φ cosh Φ

0 −1

=η

26.27 From the matrix generators in Eqns. (26.137), we find 0 i 0

J3 =

!

−i 0 0

0 0 0

1 0 0

J32 =

,

0 1 0

0 0 0

! ≡ 1112 .

So all odd powers of J3 equal J3 , all even powers equal 1112 . Defining 11z ≡

0 0 0

0 0 0

0 0 1

! to

simplify the notation, Taylor expanding reveals 1 1 (−iϕJ3 )2 + + (−iϕJ3 )3 + . . . 2! 3!     1 1 1 1 = 11z + 1112 1 − ϕ2 + ϕ4 − . . . − iJ3 ϕ − ϕ3 + ϕ5 − . . . 2! 4! 3! 5!

e−iϕJ3 = 11 − iϕJ3 +

− sin ϕ cos ϕ 0

cos ϕ sin ϕ 0

= 11z + 1112 cos ϕ + (−iJ3 ) sin ϕ =

!

0 0 1

= Rz (ϕ)

X

All-but-identical calculations verify that −iϕJ1

e

=

1 0 0

0 cos ϕ sin ϕ

0 − sin ϕ cos ϕ

! −iϕJ2

= Rx (ϕ),

e

cos ϕ 0 − sin ϕ

=

0 1 0

sin ϕ 0 cos ϕ

! = Ry (ϕ).

26.28 A real orthogonal symmetric matrix has SS T = S 2 = 11. Thus it can only represent rotation by ϕ = 0 or π.

26.29

11 = Rz† ()Rz () = 11 + iJ3†




11 − iJ3



= 11 + i J3† − J3 + O(2 ) −→ J3† = J3 .




26.30 As a succession of N rotations by ϕ/N , we have Rz (ϕ) ≈



1−i

ϕ J3 N



1−i

ϕ ϕ J3 · · · 1 − i J3 N N







=



1−i

ϕ J3 N

N

In the limit, the approximation becomes exact, and we get Rz (ϕ) = lim

N →∞

178



1−i

ϕ J3 N

N

= eiϕJ3 .

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.

26.31 [A, [B, C]] = [A, BC − CB] = [A, BC] − [A, CB] = ABC − BCA − ACB + CBA We can then cyclically permute to get [B, [C, A]] = BCA − CAB − BAC + ACB [C, [A, B]] = CAB − ABC − CBA + BAC Thus we arrive at the Jacobi identity [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0 .

26.32 The generator for R2 rotations must be a hermitian 2 × 2 matrix J such that

 R = exp[−iJϕ] =



− sin ϕ cos ϕ

cos ϕ sin ϕ

.

Since R must be orthogonal, −iJ must be real, antisymmetric; since det R = 1, J must be traceless. (See Problem  26.33.)  In addition, since cos ϕ appears in R with matrix coefficient 11, 0 −1 and sin ϕ with  ≡ , Taylor expansion of exp[−iJϕ] requires even powers of iJ to be 1 0

 ±11, and odd ±. Straightforward manipulation shows that J =


T

0 −i



i . 0

T

26.33 (a) For A antisymmetric, RT = eA = eA = e−A . Then RRT = eA e−A = 11, so it’s orthogonal. Using (26.54), we can show it’s a proper rotation: det R = det eA = eTrA = 1, since antisymmetric matrices are traceless. In fact, they have all zeros along the diagonal — so the total number of parameters required to specify A is n(n − 1)/2. Thus this is the number needed to specify a rotation in Rn .


†

†

(b) For H hermitian, U † = eiH = e−iH = e−iH . Then U U † = eiH e−iH = 11, so it’s unitary. Now an n × n hermitian matrix has n(n − 1) free parameters in the upper diagonal, and n more along the diagonal, for a total of n2 . Since specification of the generator gives U , an n × n unitary transformation must also require n2 free parameters. For proper complex rotations, det U = eiTrH = 1, so H must be traceless. Unlike the antisymmetric generator in part (a), hermitian matrices are not automatically traceless. Thus a restriction to proper complex rotations imposes an additional constraint, leaving n2 − 1 free parameters.

26.34 (a) Direct matrix multiplication of

J1 ≡

0 0 0

0 0 i

0 −i 0

! ,

J2 ≡

0 0 −i

0 0 0

i 0 0

! ,

J3 ≡

0 i 0

−i 0 0

!

0 0 0

.

(26.1)

verifies the commutation relation [Ji , Jj ] = iijk Jk . (b) Again, direct matrix multiplication verifies [ 21 σi , 12 σj ] = iijk 12 σk . ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

179

(c) ~ m , (~ ~ n] [Lm , Ln ] = (−i)2 [(~ r × ∇) r × ∇) =−

X

ijm k`n [ri ∂j , rk ∂` ]

ijk`

=−

X

=−

X

=−

X




ijm k`n ri ∂j (rk ∂` ) − rk ∂` (ri ∂j )

ijk`

ijm k`n δjk ri ∂` + ri rk ∂j ∂` − δi` rk ∂j − rk ri ∂` ∂j




ijk`

X

ijm j`n ri ∂` +

ij`

=−

ijm kin rk ∂j

ijk

X

(δm` δin − δmn δi` )ri ∂` +

X

i`

(δjn δmk − δjk δmn )rk ∂j

jk




~ + rm ∂n − δmn ~ ~ = − rn ∂m − δmn ~ r·∇ r·∇




= rm ∂n − rn ∂m ~ = imnp −i~ r×∇


p

= imnp Lp .

X

26.35 To verify (Lk )`m = −ik`m , let k = 1; then 111 121 131

i(L1 )`m = −i

112 122 132

113 123 133

! =

0 0 0

0 0 i

0 −i 0

! .

X

L2 and L3 follow the same way. As for the commutator:

X

(Li )`k (Lj )km =

X

k

(−ii`k ) −ijkm




k

=−

X

i`k mjk

k

= − δim δ`j − δij δ`m




Then [Li , Lj ]`m =

X

(Li )`k (Lj )km −

X

k

(Lj )`k (Li )km

k




= − δim δ`j − δij δ`m + (δjm δ`i − δji δ`m ) = δjm δ`i − δim δ`j =

X

jik m`k = +i

X

ijk (−ik`m ) = i

X

k

k

ijk (Lk )`m .

k

26.36 ~ ×L ~ L


i

=

X

ijk LjLk

jk

=

1 2

X

ijk (LjLk − Lk Lj ) =

1 2

jk

1 =i 2

X

ijk [Lj , Lk ]

jk

! X

ijk jk`

L` = iLi ,

jk

180

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where we’ve used Eqn. (4.24),

P

  jk ijk `jk

= 2δi` .

26.37 (a) [A, BC] = ABC − BCA = ABC + (BAC − BAC) − BCA = (BAC − BCA) + (ABC − BAC) = B[A, C] + [A, B]C (b) For j, k , i: [Ji , J 2 ] = [Ji , Jj2 + Jk2 ] = [Ji , Jj2 ] + [Ji Jk2 ] = Jj [Ji , Jj ] + [Ji , Jj ]Jj + Jk [Ji , Jk ] + [Ji , Jk ]Jk













= Jj iijk Jk + iijk Jk Jj + Jk iikj Jj + iikj Jj Jk = iijk (Jj Jk + Jk Jj − Jk Jj − Jj Jk ) = 0 . (c) From Eqns. (26.137), we get L2 = L21 + L22 + L23 = 211 — and everything commutes with the identity.

26.38 (a) Applying Lx = −i (y ∂z − z ∂y ) = a ∂r + b ∂θ + c ∂φ to r gives



a = −i (y ∂z r − z ∂y ) = −i y

z z −y r r



=0.

Applied to x and z (using a = 0), gives




0 = b ∂θ + c ∂φ r sin θ cos φ = b r cos θ cos φ − c r sin θ sin φ and




−iy = −ir sin θ sin φ = b ∂θ + c ∂φ r cos θ = −b r sin θ . So b = i sin φ and c = b cot θ cot φ = i cot θ cos φ — and the desired result emerges. The same approach gives the claimed results for Ly and Lz in spherical coordinates. (b) Since each of the Li is proportional to i, L2 must be real — so the imaginary parts sum to zero. Straightforward (if tedious) algebra then gives the desired result L2 = −

h

1 1 ∂θ (sin θ ∂θ ) + ∂2 sin θ sin2 θ φ

i

which is r2 times the angular portion of the Laplacian.

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181

The Eigenvalue Problem

27

  1 1

27.1 (a) Real, symmetric. λ1 = 3, ~v1 =

 ;



1 −1

λ2 = −1, ~v1 =

  1 i

(b) Real, antisymmetric. λ1 = 1 + 2i, ~v1 =

  ;

λ2 = 1 − 2i, ~v1 =

  (c) Complex, symmetric. λ1 = 1 + 2i, ~v1 =

1 1

 ;

λ2 = 1 − 2i, ~v1 =

  (d) Hermitian. λ1 = 3, ~v1 =

1 −i

1 −i



1 −1

  ;

1 i

λ2 = −1, ~v1 =

The matrices with real eigenvalues are both hermitian, H = H † .





a b , TrA = λ1 + λ2 = a + d, det A = λ1 λ2 = ad − bc. This gives two independent c d conditions to find the two eigenvalues. With TrA = 2, det A = −3, we find λ1 = −1, λ2 = 3.

27.2 For A =

27.3 The new eigenvalues are now 6, 6 (doubly degenerate) and 0. That these are all reals is due to the fact that D0 is hermitian.

27.4 The eigenvalues are λ1 = −2, λ2 = λ3 ! = 1, so the system 1 eigenvector is straightforward, ~v1 = −2 . The other two, 1

is!doubly degenerate. The first a b must satisfy a + c = 2b. Any c ! 1 orthogonal pair that respects this works. For simplicity, we can choose b = 0 to get ~v2 = 0 . −1 ! 1 The orthogonal degenerate eigenvector must satisfy a − c = 0, so that ~v3 = 1 . 1

27.5 Solving the secular determinant for M =

λ1 = −1

|1i →

1 −1 0

1 2 1

2 1 1

!

1 2 1

!

gives

!

λ2 = 0

|2i →

1 0 −1

!

λ3 = 4

|3i →

The eigenvectors are not orthogonal, which is expected since M M T , M T M . ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

7 8 5

27.6 Direct multiplication of −2 3 −2

E=

!

−4 5 −2

−3 3 −1

with

!

!

1 0 −1

!

1 −1 1

1 −1 0

verifies that they are eigenvectors with eigenvalues 1, −1, and +2, respectively. Similar multiplication verifies that they are also eigenvectors of 4 −1 −3

F =

!

5 −2 −3

1 −1 0

with eigenvalues 3, 0, −1. Since E and F share an eigenbasis, they must commute:

!

EF − F E =

−2 3 −2

−4 5 −2

−3 3 −1

=

5 −2 −3

7 −4 −3

2 −2 0

4 −1 −3

!

 27.7 The eigensystem of R =

cos ϕ sin ϕ

5 −2 −3

−

 whereas that of S =

 |1i →

1 −1 0

7 −4 −3

4 −1 −3

−

5 −2 −3

!

1 −1 0

−2 3 −2

−4 5 −2

!

−3 3 −1

!

2 −2 0

=0

X



− sin ϕ cos ϕ

 

is

 

1 i

|1i →

!

5 −2 −3

|2i →

1 −i

with λ1,2 = e∓iϕ ,







cos ϕ sin ϕ

sin ϕ − cos ϕ

cos ϕ + 1 sin ϕ

is



cos ϕ − 1 sin ϕ

|2i →

with λ1,2 = ±1 .

By Euler’s rotation theorem, in two dimensions proper, det = +1 rotations like R do not have a rotation axis — and indeed, the eignvectors do not lie in R2 . By contrast, improper, det = −1 rotations like S do have real eigenvectors which give the reflection axes: one for the “mirror", the other orthogonal to it.

 27.8 The eigensystem of R =

cos ϕ sin ϕ



− sin ϕ cos ϕ

  |1i →

1 i

is

  |2i →

1 −i

with λ1,2 = e∓iϕ .

These are C2 eigenvectors. In the R2 subspace, however, they take the from |10 i =

1 (|1i + |2i) → 2

  1 0

|20 i =

1 (|1i − |2i) → 2i

  0 1

which gives the plane of rotation. The rotation angle of rotation is the phase of the eigenvalues.

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183

27.9 (a) Expanding over the eigenspace |φˆn i of A, |vi = a |φˆ i, makes clear that A − λ11 is only n n n invertible for λ , λn . With this condition in mind, we can verify the claim for the an :

P

(A − λ11)

X

X

an |φˆn i =

n

an (λn − λ)|φˆn i

n

X hφˆn |ρi

=

X

=

(λn − λ)|φˆn i

λn − λ

n

hφˆn |ρi|φˆn i =

X

n

|φˆn ihφˆn |ρi = |ρi.

n

So indeed, |vi =

X hφˆn |ρi X |φˆn ihφˆn | |φˆn i = |ρi ≡ G|ρi . n

λn − λ

λn − λ

n

Clearly, the operator G blows up if λ = λn . (b) A − λ11 is not invertible and G does not exist if λ = λm is an eigenvalue of A. But projecting the inhomogeneous equation onto that eigenmode gives

X

an hφˆm |A − λ11|φˆn i = am (λm − λ) = hφˆm |ρi ,

n

ˆ where we’ve used the orthonormality of the |φi’s. So if hφˆm |ρi = 0, then the mth mode is not excited and so makes no contribution to the defining sum for G. Hence a divergent sum is dodged, and a solution exists.

27.10 The eigensystem of cos ϕ sin ϕ 0

Rz =

− sin ϕ cos ϕ 0

!

0 0 1

is

!

|1i →

0 0 1

!

,

λ1 = 1

1 i 0

|2i →

,

−iϕ

λ2 = e

|3i →

1 −i 0

! λ3 = e+iϕ

,

The first gives the axis of rotation; the other two can be combined to give their real subspace, 1 |10 i = (|2i + |3i) → 2

!

1 0 0

1 |20 i = (|2i − |3i) → 2

0 1 0

! ,

which together give the plane of rotation. The sum of the eigenvalues is 1 + e−iϕ + e+iϕ = 1 + 2 cos ϕ — as expected.

27.11 (a)(a) λhv|vi = hv|λ|vi: λ is a scalar (b) = hv|H|vi: Eigenket equation, H|vi = λ|vi (c) = hv|Hvi: By definition, |Hvi ≡ H|vi (d) = hH † v|vi: hHv| ≡ hv|H † , so hH † v| ≡ hv|H (e) = hHv|vi: Hermitian H = H † (f) = hv|λ∗ |vi: Eigenbra equation, hv|H = hv|λ∗ (g) = λ∗ hv|vi: λ∗ is a scalar

184

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(b) If λ’s not real, we’d get λ2 hv1 |v2 i = hv1 |Hv2 i = hH † v1 |v2 i = hHv1 |v2 i = λ∗1 hv1 |v2 i , so that (λ∗1 −λ2 )hv1 |v2 i = 0. Thus non-degenerate states with complex conjugate eigenvalues need not be orthogonal.

27.12 (a) λhv|vi = hv|λ|vi = hv|Q|vi = hv|Qvi = hQ† v|vi = −hQv|vi = −hv|λ∗ |vi = −λ∗ hv|vi. So λ = −λ∗ . In other words, the eigenvalues must be imaginary. (b) λ2 hv1 |v2 i = hv1 |Av2 i = −hAv1 |v2 i = −λ∗1 hv1 |v2 i = +λ1 hv1 |v2 i =⇒ (λ2 − λ1 ) hv1 |v2 i = 0 — so non-degenerate eigenvectors are orthogonal.

27.13 (a) M |vi = λ|vi =⇒ |vi = λM −1 |vi =⇒ M −1 |vi =

1 |vi λ

(b) For M |vi = λ|vi, and using det[A† ] = det[A]∗ , 0 = det[M − λ11] = det[(M − λ11)† ]∗ = det[M † − λ∗ 11]∗ . Since this is set to zero, the overall complex-conjugation is of no import. This leaves us with the secular determinant for M † — and thus we see that the eigenvalues of M † are λ∗ . (c) A unitary matrix has U −1 = U † — so combining the previous two results gives: 1/λ = λ∗ =⇒ λλ∗ = |λ|2 = 1 . So indeed, the eigenvalues are pure phases, λ = eiα .

27.14 Using the condition U † U = 11 for unitary matrices, hv1 |v2 i = hv1 |U † U |v2 i = hv1 |λ∗1 λ2 |v2 i = λ∗1 λ2 hv1 |v2 i . Now the eigenvalues of unitary operators are pure phases, λ = eiα . So if |v1,2 i are degenerate, then λ∗1 λ2 = |λ1 |2 = 1 — giving no condition on hv1 |v2 i. But if they’re non-degenerate, then λ∗1 λ2 , 1 — and this requires orthogonal eigenvectors, hv1 |v2 i = 0.

(`)

27.15 The columns of the diagonalizing matrix U are the eigenvectors vi Then, since

(`) T v k ik k

P

(T U )ij =

X

=

(`)

— that is, Ui` = vi .

(`) λ(`) vk ,

Tik Ukj =

X

k

(j)

Tik vk

(j)

= λ(j) vi

k

= λ(j) Uij =

X

Uik δkj λ(j) = (U Λ)ij .

k

Note that this is not the same as ΛU unless all the eigenvalues are equal.

4√ −√ 2 i 6

27.16 The matrix D =

λ1 = 6, |1i →

−

1 √ 0

− i

√

5 √

2

3

√ i √6 i 3 3

! has eigensystem √

! 2

λ2 = 3 + 3i, |2i →

√

!

2 1 √ 3

λ3 = 3 − 3i, |3i →

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

!

2 1√ − 3

185

Normalizing these and assembling into the unitary matrix

 U = −

1 √ √3 √2 3

1 √ 3 1 √ 6 1 √ 2

0

1 √ 3 1 √ 6 1 √ 2

 

so that

 D0 = U † DU = 

√ √2 3 1 √ 6 1 √ 6

1 √ 3 1 √ 3 1 √ 3

= ··· =

−

6 0 0

0 1 √ 2 1 √ 2

0 3 + 3i 0



−

4√ −√ 2 i 6

 0 0 3 − 3i

i

√

5 √

√ i √6 i 3 3

2

3

! −

1 √ √3 √2 3

1 √ 3 1 √ 6 1 √ 2

0

1 √ 3 1 √ 6 1 √ 2

 

! .

27.17 Using the orthogonal matrices constructed from the basis of (27.1), − sin φ cos φ 0

R=

− cos θ cos φ − cos θ sin φ sin θ

sin θ cos φ sin θ sin φ cos θ

!

a similarity transformation takes the moment of inertia tensor in (4.61) and (4.62), I = 2mr2

1 − sin2 θ cos2 φ − sin2 θ sin φ cos φ − sin θ cos θ cos φ

− sin2 θ sin φ cos φ 1 − sin2 θ sin2 φ − sin θ cos θ sin φ

− sin θ cos θ cos φ − sin θ cos θ sin φ sin2 θ

!

to (after a somewhat tedious calculation) 1 0 0

I 0 = U T IU = 2mr2

!

0 1 0

0 0 0

.

The two principal axes perpendicular to the rod have identical moments due to the inherent symmetry of the configuration. The third coincides with the rod itself — and since we’ve treated the m’s as point masses, the dumbbell has zero moment of inertia around this axis.

P

27.18 Start with the probability-weighted sum of H’s eigenvalues, hHi = |c |2 λi . In the superpoi i P sition over an orthonormal basis |ψi = c |φ i, where c = hφ |ψi, we get i i i i i hHi =

X

|ci |2 λi =

X

=

X

i

|hφi |ψi|2 λi

i

hφi |ψi∗ hφi |ψiλi =

X

i

"

# X

= hψ|

"

#

"

|φi ihφi |λi |ψi = hψ|

i

= hψ|

hψ|φi ihφi |ψiλi

i

X

λi |φi ihφi | |ψi

i

# X

H|φi ihφi | |ψi ,

i

where we’ve used the fact that the |φi i’s are eigenvectors of H. Thus

" hHi = hψ|

# X i

186

H|φi ihφi | |ψi = hψ|H

"

# X

|φi ihφi | |ψi = hψ|H|ψi ,

i

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

where

P i

|φi ihφi | = 11 for a complete orthonormal basis.

27.19 If a matrix M is normal, it can be unitarily diagonalized. Then the diagonal matrix of complexconjugate eigenvalues, Λ† , is Λ† = (U † M U )† = U † M † U




.

But if all the eigenvalues are real, then Λ† = Λ. And if Λ is hermitian, then so too is M .

27.20 (a) Orthonormal:

h1|2i =

3

√

−1

1 h1|1i = 3

1 √




−1

√

√

!

−1 √ 2 0

2 0

1 h2|2i = 6

=1


√ !

√

2

2 0

1 √

2

2 1

1 h2|3i = 6

=0 3

√


√ ! 2 1 √ 3

3

√

2 1

3




=1

1 h2|2i = 6

√ ! − 2 =0 −1 √ 3

−

√

1 h3|1i = 6

2 −1

−

√

√




3

2 −1

√ ! − 2 −1 =1 √ 3

√




3

!

−1 √ 2 0

Complete:

X

|iihi| −→

1 3

i

1 = 3

!

−1 √ 2 0

−

1√

√

−1

0

+

− 2

√

!

2

0 0 0

2 0

0

√



2

1 6

! √

2 1 √ 3

2 √ √2 6

1 + 6

2

1

√



3

+ √ 1 √

√ ! √6 1 3 = 6 3

2 3

1 6

2 √ √2 − 6

√ ! √ √  − 2 − 2 −1 3 −1 √ 3 √ √ ! ! 2 − √6 1 0 0 . 1√ − 3 = 0 1 0 0 0 1 − 3 3

(b) Spectral decomposition:

X

6 λi |iihi| −→ 3

i

=

−

1√

− 2

2 0

0 4√ −√ 2 i 6

√

− i

√

5 √

2

3

2

!

0 3 + 3i 0 + 6 0 √ ! i √6 i 3 ≡M . 3

√ 2 1 √ 3

2 √ √2 6

√ ! √6 3 − 3i 3 + 6 3

2 √ √2 − 6

√

2 1√ − 3

Straightforward multiplication verifies the eigensystem, M |ii = λi |ii.

27.21 From Problem 4.16b, the moment of inertia tensor is

M a2 I= 12

8 −3 −3

−3 8 −3

!

−3 −3 8

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187

√ ! − √6 − 3 3

=0

The principal moments and axes are I’s eigenvalues and eigenvectors:

!

1 λ1 = 6

~v1 =

1 1 1

~v2 =

1 −1 0

~v3 =

1 0 −1

!

11 λ2 = 12

!

11 λ3 = 12

The first is an axis passing diagonally through the center of the cube; around this axis, the ~ and angular velocity ω angular momentum L ~ are parallel. The orthogonal degenerate eigensubspace defined by the other two solutions is a plane perpendicular to ~v1 . The symmetry of the cube underlies the result that any axis in that plane — i.e., any linear combination of ~v2 and ~v3 — is a principal axis.

27.22 M =

X

λi |ˆ vi ihˆ vi |

i

1 2 1

1 −→ 2 · 6 1 = 3

1 2 1

1 = 3

−1 1 5

!

1




2

1

!

2 4 2

1 2 1 1 5 1

−

!

1 −2· 2

1 0 −1

1 0 −1

−1 0 1

0 0 0

!

1




−1

0

1 −1 1

1 + 3

1 +1· 3

!

1 −1 1

1

−1




1

!

−1 1 −1

1 −1 1

!

5 1 −1

.

27.23 The surface z = f (x, y) = 2x2 + y 2 − 2xy 2 has equilibria determined by ~ = (4x − 2y 2 , 2y − 4xy) = 0 . ∇f So there are three equilibrium points are the origin and (1/2, ±1). Using (27.107), the Hessian at these points is



4 −4y

−4y 2 − 4x



 = (0,0)

4 0



0 2

.

and



4 −4y

−4y 2 − 4x



 = 1 ,±1) (2

4 ∓4



∓4 0

.

Thus the origin, with two positive eigenvalues, is a miminum, whereas the other two, with one positive and one negative eigenvalues, are saddle points. [Note that the determinant in the latter case is negative, so the two eigenvalues must be of opposite signs.]

188

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-2

1

2

2

1

0

-1

y

0.0

-0.5

-1.0

-2

-1

0

z

1.0

0.5

x

27.24 Applying

R

1

T

dn x e− 2 x

q

Ax

=



(2π)n det A

to the two-dimensional matrix A =

2

5 √

2 2

√



2

3

gives

Z

1

d2 x e− 2 x

T

r Ax

=

√ (2π)2 = 2π/ 7. det A

27.25 Since Q = xT Ax is a scalar, Q = QT = (xT Ax)T = xT AT x. Therefore Q=



1 T
1 1 Q + QT = x Ax + xT AT x = xT A + AT x , 2 2 2

so that only the symmetric part of A actually contributes.

27.26 Since M is normal, it can be diagonalized by the unitary matrix of its eigenvectors, Λ = U † M U :







(a) TrM = Tr M U U † = Tr U † M U = Tr(Λ) = (b) det M = det M det U U


†

P i

λi

= det M det U det U †




= det U † det M det U = det U † M U = det Λ = Πi λi

27.27 – Since TrM1 = 0 and det M1 = 0, one eigenvalue must be zero, and the other two negatives of one another. – Now TrM2 = 0, but det M2 = acd > 0, one eigenvalue must be positive, the other two negatives of one another.

27.28 (a) Using the matrix Taylor expansion of eT and the unitarity of U : 1 2 1 T + T3 + ··· U 2! 3! i h


1 1 = U † 11 + T + T U U † T + T U U † T U U † T + · · · U 2! 3!
2 1 †
3 1 † † = 11 + U T U + U TU + U T U + · · · = exp[U † T U ] . 2! 3!

h

i

U † eT U = U † 11 + T +

(b) If T is a normal matrix, then it can be diagonalized by a unitary matrix U , Λ = U † T U . Using the results of part (a):








 

det eT = det U † eT U = det exp(U † T U ) = det eΛ

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189

But for the diagonal matrix Λ with elements λi , eΛ

 

det eT = det eΛ =

Y




= eλi δij . Thus

ij

eλi = exp [Tr T ] .

i

(c) As shown in Example 26.19, the matrix T =

3 0 −1

!

−1 0 3

0 −2 0

can be diagonalized with 1 U = √ 2

1 1 0

0 0 √

!

2

1 −1 0

Then U† .MatrixExp[T].U − MatrixExp[U† .T.U] yields zero, and Det[MatrixExp[T]] both return

Exp[Tr[T]]

e4 .

27.29 “If and only if” requires a two-way proof: (a) If hT v|T wi = hT † v|T † wi then hv|T † T |wi = hv|T T † |wi =⇒ [T, T † ] = 0 (b) If [T, T † ] = 0, then hT v|T wi = hv|T † T |wi = hv|T T † |wi = hT † v|T † wi

27.30 (a) Since the identity is hermitian, then for hermitian H:

11 = H −1 H = H −1 H


†

= H † H −1† = HH −1†

−→

H −1† = H −1

X

(b) For unitary U1 and U2 : (U1 U2 )† U1 U2 = U2† U1† U1 U2 = U2† U1† U1 U2 = U2† 11U2 = U2† U2 = 11




X

For hermitian H1 and H2 : (H1 H2 )† = H2† H1† = H2 H1 , H1 H2 So the product is of two hermitian matrices is hermitian only if the two commute: H1 H2 = H2 H1 — or equivalently, [H1 , H2 ] = 0.

27.31 Hermitian operators are diagonalized by the unitary matrix of their eigenvectors — so this problem is tantamount to showing hermitian A and B with [A, B] = 0 have the same eigenbasis. For |vi i the eigenvectors of A with eigenvalues ai , we have 0 = hvi |[A, B]|vj i = hvi |AB − BA|vj i = (ai − aj )hvi |B|vj i , where we’ve used that the ai are real. Now if i , j, then Bij = hvi |B|vj i = 0 (assuming A is non-degenerate); if i = j, then Bii = hvi |B|vi i = bi , the eigenvalues of B. So B is diagonal on the same eigenbasis as A.

27.32 (a) AB = (AB)† = B † A† = BA = AB only for [A, B] = 0

190

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(b) To share an eigenbasis with A, we need [A, AB] = 0: [A, AB] = AAB − ABA = A[A, B] = 0. So if A and B share an eigenbasis, [A, B] = 0, then AB shares the same one.

27.33 [A, B]† = (AB − BA)† = B † A† − A† B † = BA − AB = [B, A] = −[A, B]

X

27.34 Invertible operators must have non-zero determinant. But the determinant is the product of the eigenvalues. Thus an invertible operator can only have non-zero eigenvalues.

 27.35 For vn =

fn

 , the next term in the Fibonacci sequence is

fn−1

 vn+1 =

fn+1 fn





1 1

=



1 0

fn

fn−1

 ≡ F vn = F 2 vn−1 = F 3 vn−2 = . . . = F n v1 .

  where v1 =

1 . F is a normal matrix, with eigenvalues λ1,2 = 0

1 (1 2

√

±

5) and orthogonal

eigenvectors

 |1i →

1 (1 2

+ 1

√

5)



 



λ1 1

=

|2i →

1 (1 2

√

− 1

5)



  =

λ2 1

.

Given their simplicity in terms of the eigenvalues, rather than normalizing the eigenvectors and constructing a unitary (orthogonal) matrix, it’s easier just to use the inverse when diagonalizing:

 S=

λ1 1

λ2 1

 S

−1

1 = λ1 − λ2



−λ2 λ1

1 −1

 .

It’s not hard to verify that S −1 F S = Λ. Then vn+1 = F n v1 = SΛn S −1 v1 =

1 λ1 − λ2



λ1 1

1 λ1 − λ2



=

− λn+1 λn+1 2 1 n λ1 − λn 2




λ2 1



λn 1 0

0 λn 2





 ≡

−λ2 λ1

1 −1

fn+1 fn

  1 0

 .

Then, noting that λ2 /λ1 < 1, the limit we want is lim n→∞

√ λn+1 − λn+1 fn+1 1 − (λ2 /λ1 )n+1 1 2 = lim 1 n = λ1 lim = λ1 = (1 + 5) n fn λ1 − λn 1 − (λ /λ ) 2 n→∞ n→∞ 2 1 2

X

1/2 — defined by D = D 1/2 D 1/2 — 27.36 (a) For diagonal matrix D with elements √ di , the matrix D is a diagonal matrix with elements di . (b) If M is diagonalizable, D = U † M U , then M 1/2 can be found by diagonaling M , taking the square root, and then rotating back to the original basis:

M 1/2 = U D1/2 U † . Check: M 1/2 M 1/2 = U D1/2 U †







U D1/2 U † = U D1/2 D1/2 U † = U DU † = M

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X

191

 (c) M1 =

1 −2i

2i 1





1 2



−1 0

has D =

0 3



  1 i

with eigenvectors |1i →

 

1 . −i

and |2i →

So 1/2

M1

=

−i i

1 1



i 0

0 √



1 i

3

1 −i

 =





ie−iπ/3 −e−iπ/3

= 3 0 −1

M2 =

!

0 1 0

−1 0 3

has D =

4 0 0

0 0 1

√ ! −1/ √ 2 1/ 2 0

2 0 0

!

0 2 0

√ i(1 − i √3) −1 − i 3

1 2

eiπ/3 ie−iπ/3

!

0 0 , |1i → 1

√  1 + i √3 i(1 − i 3)



!

1 0 , |2i → −1

!

1 0 , and |3i → 1

0 1 . 0

So 1/2 M1

√ 1/ √2 1/ 2 0

=

0 √

!

0 0 1

2

0

1/

√

√ 1/ 2 0√ 1/ 2

2

0√ −1/ 2

√ 1 + 1/ 2 0 √ −1 + 1/ 2

!

0 1 0

=

 

1 , |ˆ 2i → i

1 √ 2

27.37 M1 has eigenvalues λ1 = −1 and λ2 = 3 with eigenvectors |ˆ 1i →

0 1 0

√ ! −1 + 1/ 2 0 √ 1 + 1/ 2

 

1 . −i

1 √ 2

Then 1/2

M1

p

=

λ1 |ˆ 1ihˆ 1| +

 

1 = i 2

p

λ 2 |ˆ 2ihˆ 2|



√

−i 1

1 i

+

 3



i 1

1 −i



1 = 2

√ i+ √ 3 −1 − i 3

√  1 + i√ 3 i+ 3

With M2 ’s eigenvalues λ1 = 1, λ2 = 2, λ3 = 4 and eigenvectors |ˆ 1i → 1 √ 2

|ˆ 3i → 1/2

M2



1 0 −1



= |ˆ 1ihˆ 1| + 0 0 0

=

  0 1 0

X

, |ˆ 2i →

1 √ 2

  1 0 1

,

, we get: √

2|ˆ 2ihˆ 2| + 2|ˆ 3ihˆ 3|

!

0 1 0

0 0 0

√

+

1 2

2

1 2

0 0 0

0 1 2

1 2

!

0

+2

0 − 12

1 2

− 12 0

0 0 0

!

1 2



1 + √1 2  = 0 −1 + √1

2

0 1 0



−1 + √1 2  0 1 + √1

X

2

Similarly



exp(M1 ) =

1 −1 e 2



1 i

−i 1



 + e3

1 −i



i 1

=

e 2



 =e



e−2 + e2 −i(−e−2 + e2 )

i(−e−2 + e2 ) e−2 + e2

i sinh 2 cosh 2

cosh 2 −i sinh 2

 .

and exp(M2 ) = e

27.38

H 1/2


†

27.39 M =

192

!

0 1 0

0 0 0

= U D1/2 U †

 1 2

0 0 0

−1 √ 3

√ 1


†

2

+e

1 2

0 1 2

= U D1/2

0 0 0


†

1 2

0 1 2

! +e

4

1 2

0 − 12

0 0 0

− 12 0 1 2

1 = 2

e2 + e4 0 e2 − e4

e2 − e4 0 e2 + e4

0 2e 0

U † = U D1/2 U † = H 1/2



3

!

 has eigenvalues ±1 with eigenvectors |1i →

1 √ 3

 √ 

 and |2i →

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

−

3

1

,

!

 so it can be diagonalized with U =

M = U ΛU † =

1 2



1 2

1 √ 3

−

1 √ 3 −

√



3

such that

1 √



3

 

1 0

1

1 2

0 −1

−

1 √

√  3 1

3

So (reading right-to-left) M is a −π/3 rotation followed by reflection across y-axis, followed by π/3 rotation.

27.40 With M T = ±M , we have [M, M T ] = ±[M, M ] = 0. So both real symmetric and antisymmetric matrices are normal — and hence diagonalizable.

27.41 For M = H1 + iH2 : [M, M † ] = [H1 + iH2 , H1 − iH2 ] = [H1 , H1 ] + i[H2 , H1 ] − i[H1 , H2 ] + [H2 , H2 ] = −2i[H1 , H2 ] which vanishes only if H1 and H2 commute.

27.42 (a) Solving det(A − λB) = 0 yields eigenvalues λ1,2 = ±4, λ3 = 12 with generalized eigenvectors

!

1 −2 0

1 ~v1 = 2

~v2 =

1 √

4

!

!

7 −2 √ −4 3

3

~v3 =

1 0 0

(b) Generalized orthonormality: 1 v1T Bv2 = √ 32 3 1 = √ 32 3




−2

1

0




−2

1

0

2 1 √

√

1 7/2 0

3

!

!

7 −2 √ −4 3

3

0 4

!

0 0√ −9 3

X

=0

Similarly, v2T Bv3 = v3T Bv1 = 0. (c) First verify the square root: √ B 1/2 B 1/2

1 = 8

=

1 4

3 0 1

2 1 √ 3

−

0√

3

2 1 7/2 0

1 2 √ √

! 3

√

3 0 1

−

0√

3

2

1 2 √

! 3

!

3 0 4

=B

X

As for the eigenvectors, ~ u1 · ~ u2 = ~v1 B 1/2 · B 1/2~v2 = ~v1 · B~v2 = 0, as shown in (b). Similarly, ~ u2 · ~ u3 = ~ u3 · ~ u1 = 0.

27.43 (a) Labeling the eigenbasis by their eigenvalues, eiϕ1 |1i → √ 2

!

1 i 0

|0i → e

iϕ2

0 0 1

!

eiϕ3 | − 1i → √ 2

!

1 −i 0

,

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193

(b) On this basis, L3 becomes e−iϕ1 0 e−iϕ3

1 J3 ≡ U L3 U = 2 †

=

1 0 0

−ie−iϕ1 0 ie−iϕ3

√ 0−iϕ 2 2e 0

!

0 i 0

−i 0 0

!

0 0 0

eiϕ1 ieiϕ1 0

eiϕ3 −ieiϕ3 0

0 √ 0 iϕ 2e 2

!

!

0 0 0

0 0 −1

(diagonal, of course), while the other two become −e−i(ϕ1 −ϕ2 ) 0 ei(ϕ2 −ϕ3 )

0

1 J1 ≡ U L1 U = √ 2 †

−ei(ϕ1 −ϕ2 ) 0

!

0 e−i(ϕ2 −ϕ3 ) 0

and 1 J2 ≡ U L2 U = √ 2 †

ie−i(ϕ1 −ϕ2 ) 0 iei(ϕ2 −ϕ3 )

0 −iei(ϕ1 −ϕ2 ) 0

!

0 −ie−i(ϕ2 −ϕ3 )

.

0

All three are clearly hermitian. Straightforward matrix multiplication confirms the commutation relation [Ji , Jj ] = iijk Jk . (c) We can make J1 real and positive by choosing ϕ1 = π, ϕ2 = ϕ3 = 0. This same choice also makes J2 purely imaginary — 1 J1 = √ 2

0 1 0

1 0 1

!

0 1 0

1 J2 = √

,

2

0 i 0

−i 0 i

0 −i 0

! .

With these phase choices, L3 ’s eigenbasis is identical to the spherical basis of Problem 26.9, 1 |1i → − √ 2

!

1 i 0

!

|0i →

0 0 1

1 | − 1i → √ 2

1 −i 0

! ,

27.44 Mathematica code: (a) L1 = {{0, 0, 0}, {0, 0, -1}, {0, 1, 0}}; L2 = {{0, 0, 1}, {0, 0, 0}, {-1, 0, 0}}; L3 = {{0, -1, 0}, {1, 0, 0}, {0, 0, 0}}; Ry[φ_] := MatrixExp[-i φ L2] Rz[φ_] := MatrixExp[-i φ L3] Then eulerR=Rz[α].Ry[β].Rz[γ] returns cos(α) cos(β) cos(γ) − sin(α) sin(γ) sin(α)(− cos(γ)) − cos(α) cos(β) sin(γ) sin(α) cos(β) cos(γ) + cos(α) sin(γ) cos(α) cos(γ) − sin(α) cos(β) sin(γ) sin(β)(− cos(γ)) sin(β) sin(γ) (b) On the spherical basis, the generators Li and rotations R become J1 = √1 {{0, 1, 0}, {1, 0, 1}, {0, 1, 0}}; 2

J2 = √1 {{0, -i, 0}, {i, 0, -i}, {0, i, 0}}; 2 J3 = {{1, 0, 0}, {0, 0, 0}, {0, 0, -1}}; Uy[φ_] := MatrixExp[-i φ J2] Uz[φ_] := MatrixExp[-i φ J3] Then wignerD=Uz[α].Uy[β].Uz[γ] returns the requested Wigner D matrix.

194

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cos(α) sin(β) sin(α) sin(β) cos(β)

!

(c) Using the spherical basis to construct the diagonalizing rotation, √ SphRot = √1 {{{-1, 0, 1}, {-i, 0, -i}, {0, 2, 0}}; 2 Then SphRot† .eulerR.SphRot returns wignerD.

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195

27.45 (a) Mathematica code: ��� ��� ���� ��

��� ��� ���� ��

��� ��� ���� ��

��� ���

M = {{u1 v1, u1 v2}, {u2 v1, u2 v2}} 

�� �� �� ��  �� �� �� ��

w = Flatten[M] {�� ��� �� ��� �� ��� �� ��} R = {{Cos[ϕ], -Sin[ϕ]}, {Sin[ϕ], Cos[ϕ]}} 

���(ϕ) -���(ϕ)  ���(ϕ) ���(ϕ)

TensorProduct[R, R] ���� (ϕ) ���(ϕ) (-���(ϕ)) ���(ϕ) ���(ϕ) ���� (ϕ) ���(ϕ) ���(ϕ) -���� (ϕ) ���� (ϕ) ���(ϕ) ���(ϕ)

���� ��

���(ϕ) (-���(ϕ)) ���� (ϕ) -���� (ϕ) ���(ϕ) (-���(ϕ)) ���� (ϕ) ���(ϕ) (-���(ϕ)) ���(ϕ) ���(ϕ) ���� (ϕ)

T = ArrayFlatten[R ⊗ R] ���� (ϕ) ���(ϕ) (-���(ϕ)) ���(ϕ) (-���(ϕ)) ��� � (ϕ) ���(ϕ) ���(ϕ) ���� (ϕ) -���� (ϕ) ���(ϕ) (-���(ϕ)) ���(ϕ) ���(ϕ) -���� (ϕ) ���� (ϕ) ���(ϕ) (-���(ϕ)) ���� (ϕ) ���(ϕ) ���(ϕ) ���(ϕ) ���(ϕ) ���� (ϕ)

���� ��

��� ��� ���� ��

��� ��� ���� ��

R.M.R // Simplify 

(�� ���(ϕ) - �� ���(ϕ)) (�� ���(ϕ) - �� ���(ϕ)) (�� ���(ϕ) - �� ���(ϕ)) (�� ���(ϕ) + �� ���(ϕ))  (�� ���(ϕ) + �� ���(ϕ)) (�� ���(ϕ) - �� ���(ϕ)) (�� ���(ϕ) + �� ���(ϕ)) (�� ���(ϕ) + �� ���(ϕ))

U.w // Simplify {(�� ���(ϕ) - �� ���(ϕ)) (�� ���(ϕ) - �� ���(ϕ))� (�� ���(ϕ) - �� ���(ϕ)) (�� ���(ϕ) + �� ���(ϕ))� (�� ���(ϕ) + �� ���(ϕ)) (�� ���(ϕ) - �� ���(ϕ))� (�� ���(ϕ) + �� ���(ϕ)) (�� ���(ϕ) + �� ���(ϕ))}

(b) Define rotation matrices R: ��� ���

L1 = {{0, 0, 0}, {0, 0, -ⅈ}, {0, ⅈ, 0}}; L2 = {{0, 0, ⅈ}, {0, 0, 0}, {-ⅈ, 0, 0}}; L3 = {{0, -ⅈ, 0}, {ⅈ, 0, 0}, {0, 0, 0}};

��� ���

���� ��

��� ���

L1.L1 + L2.L2 + L3.L3 � � � � � � � � � rotx = MatrixExp[-ⅈ α L1] roty = MatrixExp[-ⅈ β L2] rotz = MatrixExp[-ⅈ γ L3]

196

���� ��

� � � � ���(α) -���(α) � ���(α) ���(α)

���� ��

���(β) � ���(β) � � � -���(β) � ���(β)

���� ��

���(γ) -���(γ) � ���(γ) ���(γ) � � � �

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

(c) Mathematica code: ��� ���

Tx[α] = ArrayFlatten[rotx ⊗ rotx]; Ty[β] = ArrayFlatten[roty ⊗ roty]; Tz[γ] = ArrayFlatten[rotz ⊗ rotz];

��� ���

Series[Tx[α], {α, 0, 1}] // Normal Series[Ty[β], {β, 0, 1}] // Normal Series[Tz[γ], {γ, 0, 1}] // Normal

���� ��

� � � � � � � � �

���� ��

� � β � � � � � -β � � � � � � � � � � � � � � -β -β � � � � -β � � � � -β �

���� ��

� -γ � -γ � � � � � γ � � � -γ � � � � � � � � � -γ � � � γ � � � -γ � � � � � γ � γ � � � � � � � γ � � � � � � � � � � � � � -γ � � � � � � � γ � � � � � � � � � � �

��� ���

� � � � � � � � � -α � � � � � � α � � � � � � � � � � � � -α � � � � � � -α � -α � � � � α � � � -α � � α � � � � � � � � α � � � -α � � � � α � α �

genx = geny = genz =

���� ��

� � � � � � � � �

� � ⅈ � � � � � �

���� ��

� � -ⅈ � � � -ⅈ � �

���� ��

� ⅈ � ⅈ � � � � �

1 α 1 β 1 γ

� -ⅈ � � � � � � � � � � � � � � -ⅈ �

-ⅈ � � � ⅈ � � � �

� β � � � � β � � � � � � � � � � -β

� β � � � � � � �

� � β � � � β � �

ⅈ (Normal[Series[Tx[α], {α, 0, 1}]] - IdentityMatrix[9]) ⅈ (Normal[Series[Ty[β], {β, 0, 1}]] - IdentityMatrix[9]) ⅈ (Normal[Series[Tz[γ], {γ, 0, 1}]] - IdentityMatrix[9]) � � � � � � ⅈ � �

ⅈ � � � � � � � -ⅈ � � � � � ⅈ � � �

� � � � � � � � �

� � � � � ⅈ � ⅈ �

� � � � -ⅈ � � � ⅈ

� � � � � -ⅈ � � �

� � � � � � � � �

-ⅈ � � � ⅈ � � � �

� -ⅈ � -ⅈ � � � � �

� � � -ⅈ � � � � �

� � � � -ⅈ � � � ⅈ

� � � � � -ⅈ � -ⅈ �

� � � ⅈ � � � � �

ⅈ � � � � � � � -ⅈ

� ⅈ � � � � � � �

� � ⅈ � � � ⅈ � �

� � -ⅈ � � � � � �

� � � � � � � ⅈ �

� � � � � � -ⅈ � �

� � � � � � � � �

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

197

(d) K3’s eigensystem is degenerate: ��� ��� ���� ��

Eigenvalues[K3] {-�� �� -�� -�� �� �� �� �� �}

But since Kz commutes with Ksq (which also has a degenerate spectrum)... ��� ���

Ksq = K1.K1 + K2.K2 + K3.K3 � � � � -� � � � -�

���� ��

��� ���

� � � � � � � � �

� � � � � � � � �

� -� � � � -� � � � � � � � � � � � � � � � � � � � � � � � -� � � � � � � � � � � � � � � � � � � � -� � � � �

Ksq.K3 - K3.Ksq � � � � � � � � �

���� ��

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

��� ���

Eigenvalues[Ksq]

���� ��

{�� �� �� �� �� �� �� �� �}

� � � � � � � � �

...we can simultaneously diagonalize both to li� all the degeneracies: ��� ���

{evals, evecs} = Eigensystem[a Ksq + b K3] � �� �� ��- � � (� � - �) ��- � ��+ � � (� � + �) ��+ � � � {�� �� �� �� �� �� �� �� �} {�� -�� �� �� �� �� �� �� �} - � �� �� �� - � �� �� �� � {�� �� -ⅈ� �� �� -�� ⅈ� �� �} {-�� ⅈ� �� ⅈ� �� �� �� �� �} {�� �� ⅈ� �� �� �� ⅈ� �� �} {�� �� ⅈ� �� �� -�� -ⅈ� �� �} {-�� -ⅈ� �� -ⅈ� �� �� �� �� �} {�� �� -ⅈ� �� �� �� -ⅈ� �� �} � �

���� ��

The eigenvalue pairs -- Ksq’s with coefficient a, K3’s with coefficient b -- are unique: ��� ��� ���� ��

evals {�� � �� � �� � � - �� � (� � - �)� � � - �� � � + �� � (� � + �)� � � + �}

(e) Separating the eigenvectors into sets from highest to lowest Ksq eigenvalue, and then within each set from lowest to highest K3 eigenvalue, we get the matrix U (with overall phase chosen to give ϕ1=0): U = evecs[[1]] 

3 , Exp[ⅈ ϕ2] evecs[[7]] / 2, Exp[ⅈ ϕ3] evecs[[2]] 

Exp[ⅈ ϕ5] evecs[[8]] / 2,

2 , Exp[ⅈ ϕ4] evecs[[4]] / 2,

Exp[ⅈ ϕ6] evecs[[9]] / 2, Exp[ⅈ ϕ7] evecs[[3]] 

3/2 ,

Exp[ⅈ ϕ8] evecs[[6]] / 2, Exp[ⅈ ϕ9] evecs[[5]] / 2 � �

�

�

� � �

�

ⅈ ⅇⅈ ϕ�

�

�

�

� �

�

�

�

ⅇⅈ ϕ� �

ⅈ ϕ�

�

�

�

- � ⅈ ⅇⅈ ϕ�

�

�

ⅈ ϕ�

�

� � � �

-ⅇ

�

�

�

� � ⅈ ⅇⅈ ϕ� � ⅈ ϕ� ⅇ

�

ⅇⅈ ϕ�

-ⅇ

- � ⅈ ⅇⅈ ϕ�

ⅈ ⅇⅈ ϕ�

-ⅇ

�

�

�

�

-� �

�

ⅈ ϕ�

- � ⅈ ⅇⅈ ϕ�

�

�

�

�

� -ⅇ

�

�

ⅈ ϕ�

ⅇⅈ ϕ� �

�

� �

���� ��

� -ⅇ

-� �

ⅈ ϕ�

�

�

ⅈ ⅇⅈ ϕ�

�

�

�

�

-ⅇ

ⅇⅈ ϕ� � - � ⅈ ⅇⅈ ϕ� � ⅈ ⅇ ϕ�

ⅈ ϕ�

�

� �

ⅈ ⅇⅈ ϕ�

� �

ⅈ ⅇⅈ ϕ�

ⅈ ⅇⅈ ϕ�

� ⅇⅈ ϕ� � � ⅈ ⅇⅈ ϕ� � ⅈ ⅇ ϕ�

� � � �

-ⅇ

�

�

�

�

�

� �

�

ⅇⅈ ϕ�

ⅈ ϕ�

�

�

�

�

ⅇⅈ ϕ� �

� � � �

Verify that U is unitary: ��� ���

���� ��

198

U.U // ComplexExpand � � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

(f) On the new basis, the matrices have block diagonal form: ��� ���

ϕ2 = π; ϕ3 = 0; ϕ4 = 0; ϕ5 = 0; ϕ6 = π / 2; ϕ7 = π; ϕ8 = π / 2; ϕ9 = 0;

��� ���

J1 = U.K1.U // Simplify J2 = U.K2.U // Simplify J3 = U.K3.U // Simplify � �

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ⅈ

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ⅈ

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Jsq = U.Ksq.U // Simplify � � � � � � � � �

���� ��

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

(g) The blocks are distinguished by the Jsq eigenvalue j3(j3 + 1), where j3 is the largest eigenvalue of J3 in the block; each matrix has dimension 2j3 + 1. The three blocks have, respectively, j3 = 0, 1, 2. Notice that the j3 = 1 matrix is the same as found in Problem 27.44. ��� ���

wigner[α_ , β_ , γ_ ] = MatrixExp[-ⅈ α J3].MatrixExp[-ⅈ β J2].MatrixExp[-ⅈ γ J3] // FullSimplify �

� � -ⅈ α ���( β) � � ⅇ -ⅈ (α+γ) (���(β) + �) - ⅇ �

�

�

ⅇ-ⅈ γ ���( β)

���(β)

�

� ⅇ -ⅈ (α-γ) ��� �  β  �

-

ⅇⅈ γ ���( β) �

�

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ⅇ ⅈ (α-γ) ��� �  β 

ⅇⅈ α ���( β) �

� ⅈ (α+γ) ⅇ (���(β) + �) �

�

�

�

�

�

�

� � + ⅇ ⅈ β � ⅇ -� ⅈ (α+β+γ) ��

- � ⅇ -ⅈ (� α+γ) (� ���(β) + ���(� β))

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ⅇ ⅈ (α-� γ) ��� �  β  ���(β)

� ⅈ (α-γ) ⅇ (���(β) - ���(� β)) �

�

�

�

�

ⅇ � ⅈ (α-γ) ��� �  β 

ⅇ � ⅈ α-ⅈ γ ��� �  β  ���(β)

�

���� ��

�

�

� �

� -ⅈ (α+� γ) ⅇ ���(β) (���(β) + �) ��� �  β  (� ���(β) - �) (���(α + γ) - ⅈ ���(α + γ)) � � � �

� -� ⅈ γ ⅇ ��� � (β) �

�

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� -ⅈ γ ⅇ ���(β) ���(β) �

�

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� -� ⅈ α ⅇ ��� � (β) �

-ⅇ ⅈ γ-� ⅈ α ��� �  β  ���(β)

� -� + ⅇ ⅈ β � ⅇ -� ⅈ (α+β-γ) ��

� -ⅈ α ⅇ ���(β) ���(β) �

� -ⅈ (α-γ) ⅇ (���(β) - ���(� β)) �

-ⅇ -ⅈ (α-� γ) ��� �  β  ���(β)

�

� (� ���(� β) + �) � � ⅈα ⅇ ���(β) ���(β) � � �

�

� �ⅈα ⅇ ��� � (β) �

-

�

� ⅈγ ⅇ ���(β) ���(β) �

� �

� �ⅈγ ⅇ ��� � (β) �

��� �  β  (� ���(β) - �) (���(α + γ) + ⅈ ���(α + γ)) - � ⅇ ⅈ (α+� γ) ��� � (β) ����  β  �

�

� ⅈ (� α+γ) ⅇ ��� � (β) ����  β  � �

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

�

ⅇ � ⅈ (α+γ) ����  β  �

199

27.46 Start with expressing the Li ’s in the L3 eigenbasis, which we did in Problem 27.43: Here are the L matrices on L3’s eigenbasis: 1

��� ��� K1 =

{{0, 1, 0}, {1, 0, 1}, {0, 1, 0}};

2 1

K2 =

{{0, -ⅈ, 0}, {ⅈ, 0, -ⅈ}, {0, ⅈ, 0}};

2

K3 = {{1, 0, 0}, {0, 0, 0}, {0, 0, -1}};

The tensor product generators are: ��� ���

J1 = ArrayFlatten[K1⊗ IdentityMatrix[3] + IdentityMatrix[3] ⊗ K1] J2 = ArrayFlatten[K2⊗ IdentityMatrix[3] + IdentityMatrix[3] ⊗ K2] J3 = ArrayFlatten[K3⊗ IdentityMatrix[3] + IdentityMatrix[3] ⊗ K3]

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Jsq = J1.J1 + J2.J2 + J3.J3 � � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

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{evals, evecs} = Eigensystem[a Jsq + b J3] �

��

��

��-�

� (� � - �)

��-�

��+�

� (� � + �)

��+�

���� ��  {�� �� �� �� - �� �� �� �� �} {�� �� - �� �� �� �� �� �� �} {�� �� �� �� �� �� �� �� �} {�� �� �� �� �� - �� �� �� �} {�� �� �� �� �� �� �� �� �} {�� �� �� �� �� �� �� �� �} {�� - �� �� �� �� �� �� �� �} {�� �� �� �� �� �� �� �� �} {�� �� �� �� �� �� �� �� �} 

Setting all the phases to zero, we get the diagonalizing matrix ��� ���

U = evecs[[1]] 

3 , evecs[[7]] 

evecs[[9]]  �

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2 , evecs[[2]]  6,

evecs[[6]] 

2 , evecs[[4]] 

2 , evecs[[8]],

2 , evecs[[5]]

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- � �

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2 , evecs[[3]] 

� �

U.J1.U // Simplify U.J2.U // Simplify U.J3.U // Simplify U.Jsq.U // Simplify

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Since these results are identical to those of Problem 22.27c, the rest follows.

200

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

Coda: Normal Modes

28

28.1 Adding and subtracting the equations of motion for equal m’s and k’s, m¨ q1 = −2kq1 + kq2 m¨ q2 = −2kq2 + kq1 yields the decoupled equations m (¨ q1 + q¨2 ) = −k(q1 + q2 ) m (¨ q1 − q¨2 ) = −3k(q1 − q2 ) which describe two decoupled oscillators φ1,2 ≡ q1 ±q1 with frequencies ω12 = k/m, ω22 = 3k/m.



1 k 2m



1 1



1 −1



−1 2

2 −1

k m

28.2 The coupled system is described by

−1 2

2 −1

 with eigenmodes



1 1

1 −1



k = m



1 √ 2



1 . Then ±1



1 0

0 3

.

28.3 The general solution is

  q1 q2

 



1 1

= A1 cos(ω1 t + δ1 )

+ A2 cos(ω2 t + δ2 )



1 −1

.

Releasing the masses from rest means q˙i (0) = 0:

  0 0

  = −A1 ω1 sin(δ1 )

1 1

 − A2 ω2 sin(δ2 )



1 −1

.

Since the eigenvectors are linearly independent, this requires δi = 0. Then for initial positions q1 (0) = a, q2 (0) = b,

  a b

  = A1

1 1

 + A2



1 −1

,

or simply q1,2 (0) = A1 ± A2

=⇒

A1,2 =

1 (a ± b) . 2

Altogether, then, we find q1,2 (t) =

a b [cos (ω1 t) ± cos (ω2 t)] + [cos (ω1 t) ∓ cos (ω2 t)] . 2 2

28.4 For k1 = k2 = k, the K matrix in (28.6),

 K −→

k+κ −κ



−κ k+κ

,

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

and with equal masses, M = m11. Solving the secular determinant (see (28.12)) det(K − ω 2 M ) = 0 yields ω12 =

k m

ω22 =

k + 2κ , m

which reduces to the eigenfrequencies found in Example 28.1 for κ = k. Notice that, as expected, the coupling is clearly irrelevant for the symmetric mode. [Indeed, the modes are the same.] Also as expected, for weak coupling κ  k the frequencies differ little — see Problem 28.5 .

28.5 (a) The initial conditions just required setting a = 1, b = 0 in (28.19) — so q1 (t) =

1 [cos (ω1 t) + cos (ω2 t)] 2

q2 (t) =

1 [cos (ω1 t) − cos (ω2 t)] . 2

(b) With ω1,2 = ωav ± ∆ω and the identity cos(A ± B) = cos A cos B ∓ sin A sin B, the positions q1,2 (t) simplify to q2 (t) = − sin ωav t sin ∆ωt .

q1 (t) = cos ωav t cos ∆ωt

The plots reveals a beat phenomenon, in which each mass goes through maxima and minima corresponding to constructive and destructive interference of the two underlying frequencies. 1 cos !1 t 2 1 ⇠2 = cos !2 t 2 AAACA3icbVDLSgMxFM3UV62vUZfdBIvgasgMtdWFUHDjsoJ9QKcMmTTThmYmQ5IRS+nChd/iwo2KWz/CpX9j+kDUeuDC4Zx7ufeeMOVMaYQ+rdzK6tr6Rn6zsLW9s7tn7x80lcgkoQ0iuJDtECvKWUIbmmlO26mkOA45bYXDy6nfuqVSMZHc6FFKuzHuJyxiBGsjBXbRv2OBe+FHEhPX84lQvohpHwcu1IFdQo6HUNVDEDnueaV66kHXQTN8kxJYoB7YH35PkCymiSYcK9VxUaq7Yyw1I5xOCn6maIrJEPdpx9AEx1R1x7MnJvDYKD0YCWkq0XCm/pwY41ipURyazhjrgfrrTcX/vE6mo7PumCVppmlC5ouijEMt4DQR2GOSEs1HhmAimbkVkgE2eWiTm8lg6eNl0vQct+KUr8ulWmWRRh4UwRE4AS6oghq4AnXQAATcg0fwDF6sB+vJerXe5q05azFzCH7Bev8CFqiXXw==

⇠1 = AAACA3icdZDLSgMxGIUz9VbrbdRlN8EiuCqTUnpZCAU3LivYC3SGIZNm2tDMZEgyYhm6cOGzuHCj4taHcOnbmF4EFT0QOJzz/yT5goQzpR3nw8qtrW9sbuW3Czu7e/sH9uFRV4lUEtohggvZD7CinMW0o5nmtJ9IiqOA014wuZj3vRsqFRPxtZ4m1IvwKGYhI1ibyLeL7i3zK+duKDFBFZcI5YqIjrBfgdq3S6jsLASdcrWOUKNpDKo1EarCr6oEVmr79rs7FCSNaKwJx0oNkJNoL8NSM8LprOCmiiaYTPCIDoyNcUSVly0+MYOnJhnCUEhzYg0X6feNDEdKTaPATEZYj9Xvbh7+1Q1SHTa8jMVJqmlMlheFKYdawDkROGSSEs2nxmAimXkrJGNseGjD7TuD/023YqiUq1fVUqu2opEHRXACzgACddACl6ANOoCAO/AAnsCzdW89Wi/W63I0Z612jsEPWW+fKISXbA==

⇠1 + ⇠2 = cos !av t cos AAAB8nicbVBNS8NAEJ34WetX1aOXYBEEoSSlVC9CwYvHCvYDmlA22027dLMJuxOxlP4ND15UvPpnPPpv3LQ5aOuDGR7vzbCzL0gE1+g439ba+sbm1nZhp7i7t39wWDo6bus4VZS1aCxi1Q2IZoJL1kKOgnUTxUgUCNYJxreZ33lkSvNYPuAkYX5EhpKHnBI0kuc98b57mfXqTb9UdirOHPYqcXNShhzNfunLG8Q0jZhEKojWPddJ0J8ShZwKNit6qWYJoWMyZD1DJYmY9qfzm2f2uVEGdhgrUxLtufp7Y0oirSdRYCYjgiO97GXif14vxfDan3KZpMgkXTwUpsLG2M4CsAdcMYpiYgihiptbbToiilA0MZkM3OUfr5J2teLWK7X7WrlRz9MowCmcwQW4cAUNuIMmtIBCAs/wCm8WWi/Wu/WxGF2z8p0T+APr8wdr65Di

AAACE3icbVDLSgNBEJz1/Tbq0ctgEAQh7EpQj4IePCoYFbIh9E46cXBmZ5npFcKyH+HBb/HgRcWrB4/+jZPHwVdBQ1HVTXdXkinpKAw/g4nJqemZ2bn5hcWl5ZXVytr6pTO5FdgQRhl7nYBDJVNskCSF15lF0InCq+T2eOBf3aF10qQX1M+wpaGXyq4UQF5qV3ZjYVxsNPagXcRWF3BXlsQHahGfoCIYmZxK3q5Uw1o4BP9LojGpsjHO2pWPuGNErjElocC5ZhRm1CrAkhQKy4U4d5iBuIUeNj1NQaNrFcOnSr7tlQ7vGusrJT5Uv08UoJ3r68R3aqAb99sbiP95zZy6h61CpllOmIrRom6uOBk+SIh3pEVBqu8JCCv9rVzcgAVBPkefQfT747/kcq8W7dfq5/Xq0f44jTm2ybbYDovYATtip+yMNZhg9+yRPbOX4CF4Cl6Dt1HrRDCe2WA/ELx/AUu5nzk=

⇠1 AAAB8nicbVBNS8NAEJ34WetX1aOXYBG8WJJSqheh4MVjBfsBTSib7aZdutmE3YlYSv+GBy8qXv0zHv03btoctPXBDI/3ZtjZFySCa3Scb2ttfWNza7uwU9zd2z84LB0dt3WcKspaNBax6gZEM8ElayFHwbqJYiQKBOsE49vM7zwypXksH3CSMD8iQ8lDTgkayfOeeN+9zHr1pl8qOxVnDnuVuDkpQ45mv/TlDWKaRkwiFUTrnusk6E+JQk4FmxW9VLOE0DEZsp6hkkRM+9P5zTP73CgDO4yVKYn2XP29MSWR1pMoMJMRwZFe9jLxP6+XYnjtT7lMUmSSLh4KU2FjbGcB2AOuGEUxMYRQxc2tNh0RRSiamEwG7vKPV0m7WnHrldp9rdyo52kU4BTO4AJcuIIG3EETWkAhgWd4hTcLrRfr3fpYjK5Z+c4J/IH1+QNu/ZDk

AAACE3icbVC7SgRBEJz1/fbU0GTwEATx2BVRQ0EDQwVPhdvj6J3rOwdnZpeZXuFY9iMM/BYDExVTA0P/xrlH4KugoajqprsryZR0FIafwdj4xOTU9Mzs3PzC4tJyZWX10qW5FVgXqUrtdQIOlTRYJ0kKrzOLoBOFV8ntcd+/ukPrZGouqJdhU0PXyI4UQF5qVbZ3YidNnGrsQquIrS7griyJ99UiPkFFMDQ5la1KNayFA/C/JBqRKhvhrFX5iNupyDUaEgqca0RhRs0CLEmhsJyLc4cZiFvoYsNTAxpdsxg8VfJNr7R5J7W+DPGB+n2iAO1cTye+UwPduN9eX/zPa+TUOWwW0mQ5oRHDRZ1ccUp5PyHelhYFqZ4nIKz0t3JxAxYE+Rx9BtHvj/+Sy91atF/bO9+rHu2P0phh62yDbbGIHbAjdsrOWJ0Jds8e2TN7CR6Cp+A1eBu2jgWjmTX2A8H7F2ZPn1A=

!t

⇠2 =

sin !av t sin

!t

¨ = −K|Qi. Using |Qi ¨ = −ω 2 |Qi, the eigensolutions 28.6 We want to verify our solution to M |Qi |Φ1,2 i give −ω 2 M |Φ1,2 i → (2 ±

√

2)

g `



2 1



1 1

∓

1 √

 2

= (2 ±

=

g `

√

 ∓

g `

2)

2 √

 2

→ −K|Φ1,2 i

202

√  2 ∓ √2 1∓ 2



=

g `



2 0



0 1

X

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

∓

1 √

 2

28.7 Equations of motion: m¨ y1 = −k1 y1 − k2 (y1 − y2 ) = −5ky1 + 2ky2 m¨ y2 = −k2 (y2 − y1 ) = 2ky1 − 2ky2 . or

  y¨1 y¨2



k = m

 

−5 2

y1 y2

2 −2

.

With the usual ansatz y¨i = −ω 2 yi we find eigensolutions

 ω12 = 6 k/m,

|1i →



 

−2 1

ω22 = 1 k/m,

1 2

|2i →

.

28.8 Since we chose the cosine for our ansatz, releasing the blocks from rest sets all the phases to zero, δi = 0. Then the general solution for q1 (0) = a, q2 (0) = b, q3 (0) = c is q1 (t) q2 (t) q3 (t)

! = A1 cos ω1 t

1 √

!

!

+ A2 cos ω2 t

2 1

1 0 −1

−

+ A3 cos ω3 t

1 √

! 2

,

1

where



√ √ 1 1 1 A2 = (a − c) A3 = a + 2b + c a − 2b + c . 4 2 4 √ For: i) Initial conditions a = b/ 2 = c gives A2 = A3 = 0, putting the system is in mode 1; √ ii) a = −c, b = 0 gives A1 = A3 = 0, placing the system is in mode 2; and iii) a = −b/ 2 = c gives A1 = A2 = 0, and the system is in mode 3 A1 =

28.9 The general solution is





q1 (t) q2 (t)

 = A1 cos(ω1 t + δ1 )

1 −2



  + A2 cos(ω2 t + δ2 )

3 2

,

where ω1 = π sec−1 and ω2 = 2π sec−1 . So all we need do is apply initial conditions to find the four unknowns A1,2 and δ1,2 (a) from rest, q1 (0) = 4 cm, q2 (0) = 8 cm: Since from rest, δ1 = δ2 = 0, leaving





q1 (0) q2 (0)

  4 8

=





 

1 −2

= A1

+ A2

3 2

=⇒ A1 = 2, A2 = −2 .

Therefore, at t = 1 sec,





q1 (1) q2 (1)

 = −2 cos(π)



1 −2

  + 2 cos(2π)

3 2

  =

8 0

cm .

(b) from equilibrium, q˙1 (0) = 4 cm/s, q˙2 (0) = 8 cm/s: Since from equilibirum δ1 = δ2 = π/2 (which essentially turns time dependence into sine’s — so we could’ve just started with sines), leaving





q˙1 (0) q˙2 (0)

  4 8

=

 = −A1 ω1

1 −2



  − A2 ω2

3 2

=⇒ A1 ω1 = 2, A2 ω2 = −2 .

Therefore, at t = 1 sec,





q1 (1) q2 (1)

2 = − cos(3π/2) π





1 −2

1 + cos(5π/2) π

  3 2

  =

0 0

cm .

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

203

28.10 For two masses connected by a spring, we expect one vibrational mode and one zero mode. The equations of motion are (equivalent to setting k1 = k2 = 0 in Eqn. (28.1)) m1 q¨1 = −κ (q1 − q2 )

m2 q¨2 = −κ (q2 − q1 ) .

 So we need to solve for the eigenmodes of

−κ/m1 κ/m2

κ/m1 −κ/m2

 (we’ve flipped the overall sign

to cancel the implicit sign in −ω 2 in the other side of the matrix equation). Straightforward — and by now familiar! — calculation yields the expected zero mode

  1 1

|0i →

,

ω0 = 0

and the sought-after vibrational mode

 |1i →

m2 −m1

 ω2 = κ

,

m1 + m2 = κ/µ , m1 m2

where µ is the reduced mass. For CO, with atomic masses m1 ≡ mC = 1.99 m2 ≡ mO = 2.66µkg — both in 10−26 kg — the reduced mass is µ = 1.14 × 10−26 kg. Thus the effective spring constant is




κ = µω 2 = 1.14 × 10−26 kg

2π 6.42 × 1013 s−1


2

= 1855N/m

28.11 This is best set up as a generalized eigenvalue problem, det(K − ω 2 M) = 0, with

K=k

!

−1 2 −1

2 −1 0

0 −1 2

,

M=

m 0 0

0 M 0

0 0 m

!

adapted from Example 28.4. Direct calculation yields

k ω12 = m

"

m 1+ − M

r

m2 1+ M2



#

|Φ1 i −→ 1 −

m M

+

1q

 1+

m2 M2



1 ω22

k ω32 = m

"

m 1+ + M

!

2k = m

r

m2 1+ M2

|Φ2 i −→

1 0 −1



#

|Φ3 i −→ 1 −

m M

−

1 q

1+

 m2 M2



1 — all of which reduces to the results of Example 28.4 for m = M .

28.12 Since one of the eigenvalues in zero, the system is free to translate freely in a zero mode with eigenvector 1 vˆ3 = √ 3

!

1 1 1

.

So this system has the configuration of Example 28.5. Only the modes with λ , 0 contribute in

204

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P

the spectral decomposition A = λ |ˆ v ihˆ vj |. There are two possibilities [using Mathematica’s j j j TensorProduct command removes the tedium]: √ ! √ ! √ √ √
√ √ √
√ √ 3 + √3 3+ 3 −3 + 3 −2 3 3 − √3 3 − 3 −3 − 3 2 3 3± 3 3∓ 3 + −3 +√ 3 −3 − √ 3 6 6 −2 3 2 3 where the factors of

1 6

are included to normalize the eigenvectors. The upper sign yields

!

1 −1 0

−1 3 2

0 −2 2

3 −1 −2

−1 1 0

−2 0 2

,

and the lower sign

! .

Comparison with the matrix in (28.33) show that the first has the correct form for this configuration — so as expected, the lower frequency corresponds to the mode with higher symmetry. It also reveals that n = 2.

28.13 Rather than a linear string of atoms, H2 O is a triangular molecule with the H atoms forming “Mickey Mouse ears” at a 104◦ angle to one another. Still, the three vibrational normal modes have a direct correspondence to three masses in a line connected by springs. Indeed, the equivalence of the two H atoms allow the identification of a symmetric and antisymmetric mode. (Notice that each remains unchanged by a rotation around the vertical axis through the oxygen atom. These configurations are eigenstates of this π rotation with eigenvalue +1. )

H

H

H

O

H

O

H

H

O

The third mode requires a bit more thought, but it too is a direct analog of a linear chain mode. (This configuration is an eigenstate of the same π rotation with eigenvalue −1. )

28.14 (a) Kirchhoff’s law for voltage around an LC circuit is V =

dI Q +L . C dt

Since there is no battery in this circuit, V = 0; then with I = dQ/dt, we get a 2nd-order equation for Q: L

d2 Q Q =− . dt2 C

√ So charge oscillates harmonically around the circuit with frequency ω0 = 1/ LC. (b) The mutual inductance adds a term to Kirchhoff’s law — leading to the coupled equations for the two circuits d2 Q1 d2 Q2 Q1 +M =− 2 dt dt2 C d2 Q2 d2 Q1 Q2 L +M =− dt2 dt2 C

L

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

205

,

These can be decoupled simply by adding and subtracting them — as can the two-mass/threespring system discussed in the chapter. But as was done there, we prefer to cast this explicitly as an eigenvalue problem. Following the template of (28.3)–(28.7), we introduce to vector

 |Qi →

Q1 Q2



where of course here Q denotes charge, not position. The coupled equations can be written in the unified form



L M

M L



¨1 Q ¨2 Q

 =−

1 C



Q1 Q2



or equivalently



1 M/L

M/L 1



¨1 Q ¨2 Q





Q1 Q2

= −ω02

 ,

which not only has the advantage of a dimensionless matrix, but highlights the role of ω0 . Note that unlike the two-mass system, the coupling occurs amongst the second-derivative terms. Even so, the usual cos(ωt + δ) ansatz gives



1 M/L

M/L 1



Q1 Q2

 =



ω0 ω

2 

Q1 Q2

 ,

which is an eigenvalue equation for λ = ω02 /ω 2 . Solving yields the eigensystem λ1,2

M =1± L

ξ~1,2 =





1 ±1

.

As you may have anticipated, there’s a symmetric and antisymmetric mode with frequencies 2 ω1,2 =

ω02 1 ± M/L

=

1 . C(L ± M )

(c) The general expression for the charges Q1 and Q2 is just (28.17),

 |Qi →

Q1 Q2



  = A1 cos(ω1 t + δ1 )

1 1

 + A2 cos(ω2 t + δ2 )



1 −1

.





So the general expression for the currents is the derivative

  |Ii →

I1 I2

  = −A1 ω1 sin(ω1 t + δ1 )

1 1

− A2 ω2 sin(ω2 t + δ2 )

1 −1

.

28.15 Though the forces here are different than for a spring system, once linearized to describe small oscillations, this system is equivalent to the three-mass problem of Example 28.5 — but with m1 = m3 = m and m2 = M (see Problem 28.10). So there are three normal modes: a zero mode, a mode in which the pendulums swing out of phase such that the straw remains stationary, and a mode in which the pendulums swing in phase and the straw out of phase in order that the center of mass remains at rest (conservation of momentum). Thus the normal modes must

206

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be essentially those of Example 28.5 with m2 = M — scaled not by k/m, but instead by g/`:

!

ω12

|Φ1 i −→

=0

!

g ω22 = `

1 0 −1

|Φ2 i −→

g = α `

ω32

1 1 1

1 −2β 1

|Φ3 i −→

! .

for dimensionless parameters α, β. Perhaps the most straightforward way to find α and β is to recast the matrix in Example 28.5 with the replacements m2 → M and k/m → g/`, −1 2m/M −1

1 −m/M 0

g `

0 −m/M 1

! .

Direct multiplication then gives the parameters. Alternatively, taking the trace gives g g (2 + 2m/M ) = ω12 + ω22 + ω32 = (1 + α) . ` ` Thus α = 1 + 2m/M , so that ω32

g = (1 + 2m/M ) `

1 −2M/m 1

|Φ3 i −→

!

Note that as M → ∞, the straw should not move at all — as can readily be seen from |Φ3 i — and the frequency reduces to that of the uncoupled pendulums.

`π ) N +1 2`π sin( N +1 )

sin( 28.16 Using ω`2 =

2k m



`π N +1

1 − cos




=

4k m

sin2

`π 2(N +1)






 

and |Φ` i → 

  

.. .

N `π ) sin( N +1

(a) N = 2: ω12 =

4k sin2 m

π 6

 

=

k m

ω22 =

4k sin2 m



2π 6



3k m

=

X

with

 |Φ1 i →



sin(π/3) sin(2π/3)

=

1 2

 



1 1

|Φ2 i →

sin(2π/3) sin(4π/3)

 =

1 2





1 −1

X

N = 3: ω12 =

4k sin2 m

π 8

 

=

√
k 2− 2 m

ω22 =

4k sin2 m

π 4

 

=

2k m

ω32 =

4k sin2 m



2π 8



√
k 2+ 2 m

=

X

with |Φ1 i →

sin π/4 sin π/2 sin 3π/4

!

1 = √ 2

!

1 √ 2 1

!

|Φ2 i →

sin π/2 sin π sin 3π/2

1 = 2

!

1 0 −1

!

|Φ3 i →

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

sin 3π/4 sin 3π/2 sin 9π/4

207

1 = √ 2

−

1 √ 1

! 2

X

(b) The N = 4 chain has eigenmodes

 





c− sin π/5 sin 2π/5 = c+  |Φ1 i →  sin 3π/5 c+  c− sin 4π/5









c+ sin 3π/5  sin 6π/5  −c−  = |Φ3 i →  sin 9π/5  −c−  c+ sin 12π/5

















c+ sin 2π/5 sin 4π/5 =  c−  |Φ2 i →  sin 3π/5 −c−  −c+ sin 8π/5 c− sin 4π/5  sin 8π/5  −c+  = |Φ4 i →  sin 12π/5  c+  −c− sin 16π/5

p √ √ where c± = (5 ± 5)/8. (Problem 5.5 gives sin2 π/5 = 18 (5 − 5) and sin2 3π/5 = √ 1 (5 + 5).) The modes are in accord — both in relative magnitude and phase — with the 8 depiction in Figures 28.4 and 28.5. The eigenfrequencies are √
k π 4k sin2 = 3− 5 m 10 2m   √
k 4k 3π ω32 = sin2 = 3+ 5 m 10 2m



ω12 =



√
k 4k 2π sin2 = 5− 5 m 10 2m   √
k 4k 4π ω42 = sin2 = 5+ 5 m 10 2m



ω22 =



which verifies the asserted hierarchy ω1 < ω2 < ω3 < ω4 .

28.17 The eigenfrequencies are given by Eqn. (28.43) with N = 6: ω`2 =

4k sin2 (π`/6) =⇒ ω02 = 0, m

ω12 = ω52 =

k , m

ω22 = ω42 =

3k , m

ω32 =

4k . m

where ` = 0 − 5. The eigenvectors are the real and imaginary parts from Eqn. (28.40b) hqn |s` i = e2πi`n/6 = eπi`n/3 , giving 6 normal modes (` = 0-5), each a six-component vector (n = 0-5). The real eigenvectors (which, not coincidentally, are non-degenerate) are

 

1 1 1  |` = 0i →  1 1 1





1 −1  1  |` = 3i →  −1  1 −1

while the degenerate and complex eigenvectors are





1





1

eiπ/3

eiπ/3

e e5iπ/3

−e2iπ/3

    e2iπ/3   e2iπ/3      |` = 1i →  3iπ/3  =  −1  e  4iπ/3   −eiπ/3   

1 e4iπ/3

 

 

1 −eiπ/3

 

 e8iπ/3   e2iπ/3     |` = 4i →  e12iπ/3  =  1   16iπ/3  −eiπ/3  e e2iπ/3 e20iπ/3





1



e2iπ/3



1 e2iπ/3

     e4iπ/3  −eiπ/3      |` = 2i →  6iπ/3  =  1  e  8iπ/3   e2iπ/3  e e10iπ/3

 

1 e5iπ/3

 

−eiπ/3

 

1 −e2iπ/3

 

e10iπ/3   −eiπ/3     |` = 5i →  e15iπ/3  =  −1  .  20iπ/3   e2iπ/3  e eiπ/3 e25iπ/3

Note that each column correctly has a common factor ei`π/3 between adjacent elements, and

208

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

that all 6 are mutually orthogonal The vectors |0i ≡ |φ0 i and |3i ≡ |φ3 i are normal modes; the other 4 require taking the real and imaginary parts of the 2-4 and 1-5 degeneracies:

















√0  √3   3  |φ5 i = |1i − |5i →   √0  − 3 √ − 3

2  1 −1 |φ1 i = |1i + |5i →   −2 −1 1 and

√0  √3  − 3  |φ4 i = |2i − |4i →   √0   3 √ − 3

2 −1 −1  |φ2 i = |2i + |4i →   2 −1 −1

You can readily verfiy that the |φn i form an orthogonal set of eigenmodes of the bracelet.

28.18 Since S only has non-zero elements are 1’s above the diagonal, we can express it in terms of kronecker delta’s, Sij = δi,j−1 = δi+1,j . T =

M −1 K

has 2’s along the diagonal, and -1’s above and below the diagonal, Tij = 2δij − δi,j−1 − δi,j+1 = 2δij − δi+1,j − δi−1,j .

Then [S, T ]ik =

X

Sij Tjk − Tij Sjk

k

=

X

δi+1,j (2δjk − δj,k−1 − δj,k+1 ) −

j

X

(2δij − δi+1,j − δi−1,j )δj,k−1

j

= 2δi+1,k − δi+1.k−1 − δi+1,k+1 − 2δi,k−1 − δi+1.k−1 − δi−1,k−1




.

Now if integers i, j range over ±∞, this gives zero; if they don’t, terms at either end of the range will not cancel, and the operators do not commute.

28.19 (a) Straightforward calculation yields eigenvalues 1 ω1 + ω2 ± 2

h

ω± =

p

= ωav ±

p

(ω12 − ω22 ) + 4|A|2

∆ω 2 + |A|2 = ωav ± ∆ω

i

p

1 + tan2 θ ,

where tan θ ≡ |A|/∆ω. Using this last form to solve for the eigenvectors, and simplifying with half-angle trig identities, gives cos(θ/2)e−iϕ/2 |1i + sin(θ/2)eiϕ/2 |2i

|+i =

|−i = − sin(θ/2)e−iϕ/2 |1i + cos(θ/2)eiϕ/2 |2i or





|+i |−i

 =

cos(θ/2) e−iϕ − sin(θ/2) e−iϕ

sin(θ/2) eiϕ cos(θ/2) eiϕ

  |1i |2i

.

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209

(b) Weak coupling: |A|  ∆ω ←→ θ & 0, tan θ ≈ θ — so the binomial expansion gives



ω±

1 = ωav ± ∆ω 1 + 2



|A| ∆ω

2

 + ···

and |A| |2i + · · · 2∆ω i h |A| |1i + · · · . |−i = e+iϕ/2 |2i − e−iϕ 2∆ω

h

i

|+i = e−iϕ/2 |1i + e+iϕ

So weak coupling doesn’t mix the states |1i and |2i very much, and only affects the eigenfrequencies in second order. Strong coupling: |A|  ∆ω ←→ θ . π/2, tan θ  1: ω± ≈ ωav ± ∆ω tan θ = ωav ± |A| and

 1  |±i ≈ √ ±e−iϕ/2 |1i + e+iϕ/2 |2i . 2 So strong coupling leads to maximal change in the eigenfrequencies, the splitting from the unperturbed average directly proportional to the coupling A. The perturbed normal modes are maximally mixed, with equal contributions of |1i and |2i. (c) The expressions for ω± − ωav are hyperbolas, with asymptotes at ω1,2 :

!± AAACA3icbZC7SgNBFIZn4y3G26plmsEg2Bh2JaiFRcDGMoK5QHZZZieTZMhclpnZQFhSWPgsFjYqtj6EpW/jJNlCE38Y+PjPOZw5f5wwqo3nfTuFtfWNza3idmlnd2//wD08ammZKkyaWDKpOjHShFFBmoYaRjqJIojHjLTj0e2s3h4TpakUD2aSkJCjgaB9ipGxVuSWA8nJAEVBwuF5zlmgOETjaeRWvKo3F1wFP4cKyNWI3K+gJ3HKiTCYIa27vpeYMEPKUMzItBSkmiQIj9CAdC0KxIkOs/kRU3hqnR7sS2WfMHDu/p7IENd6wmPbyZEZ6uXazPyv1k1N/zrMqEhSQwReLOqnDBoJZ4nAHlUEGzaxgLCi9q8QD5FC2NjcbAb+8sWr0Lqo+pfV2n2tUr/J0yiCMjgBZ8AHV6AO7kADNAEGj+AZvII358l5cd6dj0VrwclnjsEfOZ8/4VuX5A==

!av AAAB+XicbZDLSgMxFIYzXmu9TXXpJlgEQSgzUqoboeDGZQV7gXYYMulpG5pMhiSjlKGP4sKNilufxKVvY9rOQlt/CHz85xzOyR8lnGnjed/O2vrG5tZ2Yae4u7d/cOiWjlpapopCk0ouVSciGjiLoWmY4dBJFBARcWhH49tZvf0ISjMZP5hJAoEgw5gNGCXGWqFb6kkBQxJe3OTgh27Zq3hz4VXwcyijXI3Q/er1JU0FxIZyonXX9xITZEQZRjlMi71UQ0LomAyhazEmAnSQzU+f4jPr9PFAKvtig+fu74mMCK0nIrKdgpiRXq7NzP9q3dQMroOMxUlqIKaLRYOUYyPxLAfcZwqo4RMLhCpmb8V0RBShxqZlM/CXf7wKrcuKX6tU76vlei1Po4BO0Ck6Rz66QnV0hxqoiSh6Qs/oFb05mfPivDsfi9Y1J585Rn/kfP4A62qTYw==

!+ = !1

|A| AAAB6XicbVDLTgJBEOzFF+IL9ehlIjHxRHaNQY8YLx4xyiOBDZkdBpgwO7uZ6TUhC5/gwYsar36RR//GAfagYCWdVKq6090VxFIYdN1vJ7e2vrG5ld8u7Ozu7R8UD48aJko043UWyUi3Amq4FIrXUaDkrVhzGgaSN4PR7cxvPnFtRKQecRxzP6QDJfqCUbTSw+Rm0i2W3LI7B1klXkZKkKHWLX51ehFLQq6QSWpM23Nj9FOqUTDJp4VOYnhM2YgOeNtSRUNu/HR+6pScWaVH+pG2pZDM1d8TKQ2NGYeB7QwpDs2yNxP/89oJ9q/9VKg4Qa7YYlE/kQQjMvub9ITmDOXYEsq0sLcSNqSaMrTp2Ay85Y9XSeOi7FXKl/eXpWolSyMPJ3AK5+DBFVThDmpQBwYDeIZXeHNGzovz7nwsWnNONnMMf+B8/gDRiI2+

AAAB8nicbVA9SwNBEN2LXzF+RS1tFoNgFe4kqIVFQAvLCOYDckfY28wlS/Zuj905IYT8DQsbFVv/jKX/xk1yhSY+GHi8N8PMvDCVwqDrfjuFtfWNza3idmlnd2//oHx41DIq0xyaXEmlOyEzIEUCTRQooZNqYHEooR2Obmd++wm0ESp5xHEKQcwGiYgEZ2gl378DicxXMQxYr1xxq+4cdJV4OamQHI1e+cvvK57FkCCXzJiu56YYTJhGwSVMS35mIGV8xAbQtTRhMZhgMr95Ss+s0qeR0rYSpHP198SExcaM49B2xgyHZtmbif953Qyj62AikjRDSPhiUZRJiorOAqB9oYGjHFvCuBb2VsqHTDOONiabgbf88SppXVS9y2rtoVap3+RpFMkJOSXnxCNXpE7uSYM0CScpeSav5M1B58V5dz4WrQUnnzkmf+B8/gBtPpGQ

AAAB6nicbVBNSwMxEJ34WetX1aOXYBG8WHalVI8VLx4r2A9ol5JNs21oNrskWaFs+xc8eFHx6h/y6L8x2+5BWx8MPN6bYWaeHwuujeN8o7X1jc2t7cJOcXdv/+CwdHTc0lGiKGvSSESq4xPNBJesabgRrBMrRkJfsLY/vsv89hNTmkfy0Uxi5oVkKHnAKTGZdDm9nfZLZafizIFXiZuTMuRo9EtfvUFEk5BJQwXRuus6sfFSogyngs2KvUSzmNAxGbKupZKETHvp/NYZPrfKAAeRsiUNnqu/J1ISaj0JfdsZEjPSy14m/ud1ExPceCmXcWKYpItFQSKwiXD2OB5wxagRE0sIVdzeiumIKEKNjcdm4C5/vEpaVxW3Vqk+VMv1Wp5GAU7hDC7AhWuowz00oAkURvAMr/CGBHpB7+hj0bqG8pkT+AP0+QM7DI31

!

|A| AAAB+XicbZDLSgMxFIYz9VbrbapLN8EiuLHMlFLdCAU3LivYC7TDkEkzbWguQ5JRytBHceFGxa1P4tK3MW1noa0/BD7+cw7n5I8SRrXxvG+nsLG5tb1T3C3t7R8cHrnl446WqcKkjSWTqhchTRgVpG2oYaSXKIJ4xEg3mtzO691HojSV4sFMExJwNBI0phgZa4VueSA5GaHw8iaHWuhWvKq3EFwHP4cKyNUK3a/BUOKUE2EwQ1r3fS8xQYaUoZiRWWmQapIgPEEj0rcoECc6yBanz+C5dYYwlso+YeDC/T2RIa71lEe2kyMz1qu1uflfrZ+a+DrIqEhSQwReLopTBo2E8xzgkCqCDZtaQFhReyvEY6QQNjYtm4G/+uN16NSqfqNav69Xmo08jSI4BWfgAvjgCjTBHWiBNsDgCTyDV/DmZM6L8+58LFsLTj5zAv7I+fwB8AWTZg==

! = !2

This shows that the coupling (perturbation) creates a gap between the eigenfrequencies; this is most stark with the system is degenerate, ∆ω = 0. But no matter the value of ∆ω, the difference between ω± is always greater than that between ω1,2 . However the average value of the eigenfrequencies is the same for all |A|. (d) First, use the unitarity of U to trivially invert the expressions for |±i: |1i = e+iϕ/2 [cos(θ/2)|+i − sin(θ/2)|−i] |2i = e−iϕ/2 [sin(θ/2)|+i + cos(θ/2)|−i] Now the general time dependence of the system is a linear combination of normal modes, |ψ(t)i = aei(ω+ t+δ+ ) |+i + bei(ω− t+δ− ) |−i. If the system is initially in the state |ψ(0)i = |1i, then a = cos(θ/2), b = − sin(θ/2) and δ+ = δ− = eiϕ/2 . Thus



|ψ(t)i = e+iϕ/2 cos(θ/2)eiω+ t |+i − sin(θ/2)eiω− t |−i

210



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The system will have completely evolved into |2i when h1|ψ(t)i = 0. It’s easiest to compute the magnitude-squared (which not incidentally corresponds to probability in quantum mechanics):

2 h1|ψ(t)i 2 = cos(θ/2)eiω+ t h1|+i − sin(θ/2)eiω− t h1|−i 2 = cos2 (θ/2)eiω+ t + sin2 (θ/2)eiω− t = cos4 (θ/2) + sin4 (θ/2) + 2 sin2 (θ/2) cos2 (θ/2) cos(ω+ − ω− )t = 1 − 2 sin2 (θ/2) cos2 (θ/2) [1 − cos(ω+ − ω− )t] [As a quick consistency check, note that this equals 1 at t = 0.] For strong coupling, θ ≈ π/2 so sin2 (θ/2) cos2 (θ/2) ≈ 1/4. Is this limit,

h1|ψ(t)i 2

θ=π/2

=

1 [1 + cos(ω+ − ω− )t] , 2

which goes to zero at t=

π π = . ω+ − ω− 2A

At this time, what started in state |1i has evolved to be purely in state |2i. [Indeed, in this limit, |h2|ψ(t)i|2 = 1.] And the stronger the coupling, the sooner this happens. For weak coupling, however,

h1|ψ(t)i 2

θ&0

≈ 1 − θ [1 + cos(ω+ − ω− )t] .

So |ψ(t)i stays almost exclusively in the state |1i for all t.

28.20 By symmetry, the two states |1i and |2i are degenerate, so Eav = E0 . And since ∆E = 0, this is a strong coupling system — so that the perturbed energies are E± = E0 ± A . Note that as the distance between the atom and the proton decreases, the proton will start to penetrate hydrogen’s electron cloud and thus start to feel a coulomb repulsion from the atom’s proton. Thus E0 must grow with decreasing separation of H and p. However, so too does the coupling A, which increases the energy gap between E± — and in particular, works to decrease E− . Now a bound state will occur when the system energy is less than that of the individual components — in this case, one hydrogen atom and a proton. As the distance decreases, it’s reasonable to expect that these two effects — growing E0 and growing A — collude to form a stable molecule at some distance r0 . A detailed calculation gives the correct size of the molecule, r0 ≈ 1.3 angstroms.

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211

Cartesian Tensors

29 29.1 (a) (b) (c) (d)

No Yes Yes (scalar) ~ and P ~ are vectors Yes: both E

29.2 A matrix of temperatures is not a tensor, since the individual elements are scalar. The temperature Tij is a discretized scalar field.

29.3 Since the numerical value of vector components change under rotation, they are not scalars in the tensorial sense.

~ has components 29.4 In the primed frame, the force F Fy0 = N − mg cos ϕ

Fx0 = mg sin ϕ

where N is the normal force. Rotating into the unprimed frame gives



Fx Fy



 =



cos ϕ − sin ϕ

sin ϕ cos ϕ



mg sin ϕ N − mg cos ϕ

 =

N sin ϕ N cos ϕ − mg

 .

Since in the primed frame ay0 = 0, we have a = ax0 = g sin ϕ so the block has speed v(t) = g sin ϕ t. On the unprimed basis, may = N cos ϕ − mg

max = N sin ϕ so that vx =

N sin ϕ t m

vy =



N cos ϕ − g t . m



The speed is v(t) =

p

r

N2 sin2 ϕ + m2

r

N N2 − 2g + g 2 = t g sin ϕ , m2 m

vx2 + vy2 = t =t



N cos ϕ − g m

2

where we’ve used N = mg cos ϕ.

↔

29.5 (a) w ~ · T −→

P i

↔

(b) w ~ · T · ~v −→ ↔

↔

(c) T · U −→

wi Tij — vector

P ij

P j

wi Tij vj — scalar

Tij Ujk —2nd-rank tensor

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

↔

(d) T ~v −→ Tij vk — 3rd-rank tensor ↔↔

(e) T U −→ Tij Uk` — 4th-rank tensor

29.6 (a) (b) (c) (d) (e) (f) (g) (h)

axial vector (cross product of polar vectors) axial vector (cross product of polar vectors) axial vector (cross products of axial vectors) polar vector axial vector (cross product of polar vectors) polar vector (vector + cross product of polar vector and axial vector) axial vector scalar

29.7 A 4th-rank tensor in 2 dimensions has 24 = 16 components.

↔

↔

↔

29.8 For W = a S + b T , 0 Wij =

X

Rik Rj` Wk` =

X

k`

Rik Rj` (aSk` + bTk` )

k`

=a

X

Rik Rj` Sk` + b

k`

X

0 0 Rik Rj` Tk` = aSij + bTij

X

k`

29.9 Tr(T 0 ) =

X

Tij δij =

ij

X

Tii =

i

X

Rik Ri` Tk`

ik`

=

X

T Rki Ri` Tk` =

X

δk` Tk` = Tr(T )

k`

ik`

0 − T 0 . In the unprimed: 29.10 First, note that in the primed frame, ij Tij0 = T12 21

ij Tij0 = ij Rik Rj` Tk` = (R1k R2` − R1` R2k ) Tk` = (R11 R22 − R12 R21 ) T12 + (R12 R21 − R11 R22 ) T21







= cos2 ϕ + sin2 ϕ T12 + − sin2 ϕ − cos2 ϕ T21 = T12 − T21

X

29.11 (a) Tij =

1 1 (Tij + Tji ) + (Tij − Tji ) ≡ Sij + Aij 2 2

(b) Index notation: 0 Sij =

X

Rik Rj` Sk` =

k`

X k`

Rik Rj` S`k =

X

0 Rj` Rik S`k = Sji

k`

and 0 Aij =

X k`

Rik Rj` Ak` = −

X k`

Rik Rj` A`k = −

X

0 Rj` Rik A`k = −Aji

k`

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213

Matrix notation: S 0 T = RSRT


T

= RS T RT = RSRT = S 0

and A 0 T = RART


T

= RAT RT = −RART = −A 0

~ · ~v is a scalar. Then 29.12 (a) For a vector ~v , we know that ∇ ∂x0 vx0 + ∂y0 vy0 = ∂x vx + ∂y vy




= ∂x cos ϕ vx0 + sin ϕ vy0 + ∂y − sin ϕ vx0 + cos ϕ vy0




= (cos ϕ ∂x − sin ϕ ∂y ) vx0 + (sin ϕ ∂x + cos ϕ ∂y ) vy0 Thus ∂x0 = cos ϕ ∂x − sin ϕ ∂y and ∂y0 = sin ϕ ∂x + cos ϕ ∂y . (b) We essentially repeat the previous calculation, but without limiting ourselves to n = 2:

!

! ~ 0 · ~v 0 = ∇ ~ · ~v = ∇

X

∂i v i =

X

i

∂i

X

i

T 0 Rij vj

j

=

X

T Rij ∂i vj0

=

X X

ij

j

Rji ∂i

vj0

i

~ is a vector. Note that fundamental to this result is that R be R ∂ . So ∇ Thus ∂j0 = i ji i independent of position — otherwise it would not commute with ∂i .

P

(c) By the chain rule,

X ∂xj ∂ ∂ = . 0 ∂xi ∂x0i ∂xj j

Now since xi is a vector and R is not a function of position, we can write

X X ∂x0k ∂xj T T T = Rjk = Rjk δki = Rji = Rij . ∂x0i ∂x0i k

k

Thus

X ∂ ∂ = Rij . 0 ∂xi ∂xj j

29.13 Applying a rotation to dFi =

P j

σij daj gives dFi0 :

! dFi0

=

X k

Rik

X j

σkj daj

=

X kjm`

T Rik σkj Rjm Rm` da` ≡

X

0 σim da0m ,

m

where in the second step we inserted 11 = RT R. So σ 0 = RσRT is a 2nd-rank tensor.

214

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29.14 Starting with Eqn. (29.7), 0 = σij

X

Rik Rj` σk`

k`

X

=

Rik Rj` Yk`mn εmn

k`mn

! X

=

X

Rik Rj` Yk`mn

k`mn

pq

X

=

−1 −1 0 Rnq εpq Rmp

T T 0 Rik Rj` Rmp Rnq Yk`mn εpq

k`mnpq

! =

X

X

pq

k`mn

≡

X

Rik Rj` Rpm Rqn Yk`mn

0 εpq

0 0 Yijpq εpq

pq

~ · ~v = 29.15 (a) ∇

P j

~ × ~v )i = (b) (∇

∂j v j =

P ij

~ v ). ∂j vi δij = Tr(∇~

P

 ∂ v jk ijk j k

1 2

= 2Ai , where Ai =

↔

P

 T jk ijk jk

~ v. is the dual of T ≡ ∇~

29.16 The flux of a vector is a scalar, whereas the flux of a 2nd-rank tensor is a vector,

Z

Z X

↔

T · d~a

~v =

←→

vi =

S

S

Tij daj ,

j

so the divergence theorem becomes vi =

I X S

Tij daj =

Z X V

j

∂j Tij dτ .

j

Applied to the definition of stress in Eqn. (29.2): dFi =

X

σij daj =

X

j

←→

∂j σij dτ

↔

~ =∇ ~ · σ dτ , dF

j

allowing the identification of the force density f , ~ ·↔ f~ ≡ ∇ σ .

29.17 Start with (29.2) dFi = −

X

σij daj

↔

~ = − σ · d~a , dF

←→

j

where, since d~a is the outward pointing normal, the negative sign renders this as the inward force due to the fluid. Integrating over the area S of the body gives the full buoyancy force ~b = − F

I

↔

Z

σ · d~a = −

S

~ ·↔ ∇ σ dτ ,

V

where we’ve used the divergence theorem to convert to an integral over the volume V of the

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215

~ ·↔ body. This reveals ∇ σ to be the force density f~. But the force density of the fluid on the ˆ Thus body is simply ρ~g = −ρg k. ˆ ~b = +g k F

Z

ˆ, ρ dτ = +M g k

V

which is Archimedes’ Principle. ~ ·↔ 29.18 (a) We want need to verify that f~ = ∇ σ − tensor:

X

h 

∂i 0 Ei Ej −

1 2 E δij 2

i

= 0

Xh

i

1 ~ ∂ S. c2 t

Start with the electric part of the stress


1 ~ ·E ~ (∂i Ei ) Ej + Ei (∂i Ej ) − δij ∂i E 2

i

i

X




~ ·∇ ~ Ej − 0 = ρEj + 0 E

δij






~ ·∇ ~ Ei + E ~× ∇ ~ ×E ~ E



 i

i



~ ~ × ∂B E ∂t

= ρEj + 0

 j

~ ·E ~ and the vector identity for where in the second line we used both Gauss’ law ρ = 0 ∇ ~ Similar manipulation ~ A ~ · B). ~ In the third line we invoked Faraday’s law ∇ ~ ×E ~ = −∂t B. ∇( ↔ for the divergence of the magnetic part of σ gives:

X i

∂i

h

1 µ0



Bi Bj −

1 2 B δij 2

i

=

1 µ0

Xh


1 ~ ·B ~ (∂i Bi ) Bj + Bi (∂i Bj ) − δij ∂i B 2

i

i

1 ~ ~
1 =0+ B · ∇ Bj − µ0 µ0

X

δij






~ ·∇ ~ Bi + B ~× ∇ ~ ×B ~ B

i

~ ~ × J~ − 0 (B ~ × ∂ E )j , = −B ∂t ~ All together then ~ ·B ~ = 0 and Ampère’s law ∇ ~ ×B ~ = µ0 J~ + 0 µ0 ∂t E. using ∇ ~ ·↔ ~ + J~ × B ~ + 0 ∇ σ = ρE






~ ×B ~ ~ + ∂t E ~ × ∂t B E




~ , = f~ + µ0 0 S

~= 1 E ~ × B. ~ where the Poynting vector S µ0 (b) The total force on charges within the volume is p ~ = d~ F = dt

Z V

d ~ ·↔ ∇ σ dτ − 0 µ0 dt

Z S dτ . V

We can recast the left-hand side as a volume integral by introducing the momentum density ~ charge of the charges themselves, P

Z p ~=

~ charge dτ . P

V

The result is a continuity equation,



∂ ~ ~ · −↔ ~ fields , ∇ σ =− Pcharge + P ∂t ~ fields = µ0 0 S. ~ Thus we find the where the momentum density contained in the fields in P statement of momentum conservation: any change of momentum — whether in the charges (“mechanical”) or stored in the fields — must manifest as a momentum flux. This not ↔ only demonstrates that electromagnetic fields contain momentum, but identifies − σ as the momentum flux density.

216

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 i

ˆ and B ~ = −α/0 k ~ = 0, so direct application of (29.53) gives 29.19 (a) The capacitor has electric field E −1 0 0

α σ −→ − 20

↔

!

0 −1 0

0 0 1

~ = 0, so the force on the upper plate with area A is (b) Since there’s no magnetic field, S

I

~ = F

↔

ˆ σ · d~a = −k

Z

ˆ σzz dx dy = −k

S

α A, 20

ˆ so the force per unit area is − 2α k. 0

↔

(c) Problem 29.18 identified − σ and the momentum flux density. Thus −σzz = +α/20 is the momentum crossing the xy plane per area per time.

29.20 (a) dτi =

P

 r dFk jk ijk j

=

P

 r σ da` . jk` ijk j k`

P

P

(b) By the divergence theorem,  r σ da` =  ∂ (rj σk` ) d3 x, so that the torque jk` ijk j k` jk` ijk ` per volume — which of course is the change of angular momentum per volume — is

X




∂` ijk rj σk` =

jk`

X

ijk (∂` rj ) σk` +

jk`

=

X

ijk rj ∂` (σk` )

jk`

X

ijk σkj +

jk

X

ijk rj fk ,

jk`

where we’ve used ∂` rj = δ`j , and Eqn. P(29.52) for the force density fk . (c) Choosing the origin to be at ~ r leaves  σ , which must vanish since angular momenjk ijk kj tum is constant. (See Newton’s 3rd law!) The only way for this to be generally true is for σ to be symmetric — since then its contraction against the totally antisymmetric  vanishes,

X

ijk σkj = −

X

jk

ikj σkj = −

jk

X

ikj σjk = −

jk

X

ijk σkj ,

jk

where we’ve relabeled dummy indices in the last step.

29.21 Expanding ~ = [~ ∆ r + d~ r+~ u(~ r + d~ r)] − [~ r+~ u(~ r)] to first order gives ∆i = dxi +

X ∂ui ∂xj

j

dxj + O(∂u/∂x)2 .

Then, again through 1st order, 2

∆ ≈

X

dxi +

X ∂ui

i

=

X

j

dx2i + 2

i

=

X i

+

dxj

X ∂ui i,j

dx2i

∂xj

!2

∂xj

X  ∂ui i,j

dxi dxj

∂uj + ∂xj ∂xi

 dxi dxj

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217

Defining the strain tensor εij ≡

1 ∂ui ( 2 ∂xj

∆2 =

+

∂uj ∂xi

X

) gives the desired result,



δij + 2εij dxi dxj .

ij

Taking the square root, the change or deformation is the part added to δij ,

X

εij dxi dxj .

ij

And what happened to the rotation we found in Eqn. (29.9)? Since that’s rigid body motion, it was subtracted away in the definition of ∆.

29.22 The short answer is that, with only the 2 invariant and , the only way to contract away Ptensors δP a vector’s single index to get a scalar is v 2 = vv = v v δ — valid in any number i i i ij i j ij of dimensions. The problem with  is that it’s totally antisymmetric, so in two dimensions P P  v v = 0, and in three  v v v = 0. So we’d need a pair of epsilons. But this ij i j i j ijk k ij ijk

P

gives nothing new: In two dimensions, dimensions,

P

  v v ijkm ijk ijm k m

=

  v v ijk ij ik j k

P jkm

=

P δ v v = v 2 , while in three
jk jk j k 2

δjj δkm − δjm δjk vk vm = 2v .

29.23 (a) The decomposition of Eqn. (29.50), Tij =

1 1 1 Tr(T )δij + (Tij − Tji ) + 3 2 2



Tij + Tji −

2 Tr(T )δij 3



.

is already down to the irreducible parts of T . The easiest way to see this is the note that nothing new can be found be P deploying more δ’s or ’s. 0 = (b) We need to check that Dij R R D for each irreducible piece D, k` ik j` k` Dij =

1 Tr(T )δij : 3

X X 1 1 1 T Tr(T ) Rik Rj` δk` = Tr(T ) Rik Rkj = Tr(T )δij X 3 3 3 k`

k

Dij = Aij : 0 Aij =

X

Rik Rj` Ak` = −

k`

Dij = Sij −

X

So

X

0 Rj` Rik A`k = −Aji X

k`

First, since D is symmetric:

Rik Rj` Dk` =

X

k`

D0

Rik Rj` A`k = −

k`

1 Tr(T )δij . 3

0 Dij =

X

Rik Rj` D`k =

k`

X

0 Rj` Rik D`k = Dji

k`

is symmetric. It’s also traceless:

Tr(D0 ) =

X j

0 Djj =

X

Rjk Rj` Dk` =

jk`

X

T Rkj Rj` Dk` =

jk`

X

δk` Dk` = Tr(D) = 0 X

k`

~ =A ~ × B, ~ ijk an invariant (pseudo) tensor, and using the summation convention: 29.24 (a) With C




Ci0 = ijk Aj0 Bk0 = ijk Rj` A` (Rkm Bm ) = ijk Rj` Rkm A` Bm




T = njk Rj` Rkm Rip Rpn A` Bm = Rip p`m A` Bm

= Rip Cp X

218

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↔

~ but C, (b) Using the invariance of ijk , define not C

Cij ≡

X

~ × B) ~ k= ijk (A

k

X

ijk k`m A` Bm

k`m

=

X




δi` δjm − δim δj` A` Bm

`m

= Ai Bj − Aj Bi ,

(29.1)

which is an antisymmetric tensor, Cij = −Cji whose 3 independent elements are the individual components of the cross product. That the cross product can be represented either as a vector or antisymmetric tensor is an artifact three dimensional space. The number of indices on , recall, is equal to the dimension of the space N . Thus contracting it against two vectors leaves N − 2 indices — which in three dimensions leaves a single-index object. Generalizing the cross product’s anti-symmetric combination to higher dimensions, however, does not return a vector. For instance, in four dimensions,

Cij =

X

ijk` Ak B`

(29.2)

k`

yields an antisymmetric 2nd rank tensor with N (N − 1)/2 = 6 independent elements. There is no way to package this as a four-dimensional vector.

29.25 Tij =

P

 B : k ijk k

(a) Multiplying both sides by `ij ,

X

`ij Tij =

ij

X

`ij ijk Bk

ijk

=

X

2δk` Ak = 2B`

−→

B` =

1 2

X

`ij Tij X

ij

k

~ is a vector, T is a pseudotensor; if A ~ is a pseudovector, then T is a Due to the factor of , if B ! 0 B3 −B2 ↔ tensor. Stepping through component-by-component easily confirms T −→ −B3 0 B1 . B2 −B1 0 (b) If T is not antisymmetric, then write Tij =P Aij + Sij . TheP contraction of the symmetric S with the antisymmetric  vanishes, so that  T =  A . jk ijk jk jk ijk jk (c) Rotating the dual tensor of Bk , Tij =

P

 B , k ijk k

requires an orthogonal similarity trans-

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219

formation: Tij0 = RT RT


ij

=

X

T Rik Tk` R`j =

X

Rik Rj` Tk`

k`

k`

! =

X

X

Rik Rj`

k`

k`m Bm

m

! =

X X m

k`m Rik Rj`

Bm

k`

=

X

=

X

ˆi × R ˆ j ) Bm = (R m

X

m

ˆ n ) Bm (ijn R m

m

! ijn Rnm Bm =

mn

=

X

X n

ijn

X

Rnm Bm

m

ijn Bn0 ,

(29.3)

n

~i = which is the dual of the rotated vector A

220

P j

Rij Bj .

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Beyond Cartesian

30 30.1 (a) (b) (c) (d) (e) (f) (g) (h)

scalar vector illegal: superscript j cannot be balanced by subscript j on the other side illegal: sums must be over one subscript and one superscript scalar (trace) 2nd-rank tensor illegal: j unbalanced on the two sides illegal: more than two k’s on the right-hand side; also unbalanced j

30.2 As a 2nd-rank mixed tensor, δdc transforms as 0

0

0

0 ∂ua ∂ud c ∂ua ∂uc ∂ua δ = = = δba0 , ∂uc ∂ub0 d ∂uc ∂ub0 ∂ub0

so it’s invariant. On the other hand ∂uc ∂uc ∂uc ∂ud δcd = , δa0 b0 . ∂ua0 ∂ub0 ∂ua0 ∂ub0 30.3 0

0

Aj 0 B j =

∂uk ∂uk ∂uj Ak B ` = Ak B ` = δ`k Ak B ` = Ak B k ∂u` ∂uj 0 ∂u`

0

and similarly for Aj Bj 0 . On the other hand, 0

0

0

Aj B j =

0

∂uj ∂uj k ` A B , Ak B k ∂uk ∂u`

30.4 Using the definition gab = ~ea · ~eb and g ab = ~e a · ~e b in Eqn. (30.12) gives: δca = g ab gbc = (~e a · ~e b )(~eb · ~ec ) = ~e a · (~e b~eb ) · ~ec = ~e a · ~ec , where we’ve used the completeness of the basis, ~e b~eb = 11. (Note the implied sum.) So indeed, the upstairs and downstairs bases are mutually orthonormal.

30.5 The upstairs basis vectors in sheared coordinates are ~e

u

1 → cos (α + β)



cos β − sin β

 ~e

v

1 → cos (α + β)



− sin α cos α

 .

and the metric is

 gab →

1 cos ϕ



cos ϕ 1

,

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where α + β = π/2 − ϕ. Then, invoking appropriate trig identities: eˆu = gua~e a = guu~e u + guv ~e v 1 cos (α + β)

=



cos β − sin β



 + cos ϕ

− sin α cos α



 =

cos α sin α



and eˆv = gva~e a = gvu~e u + gvv ~e v



=



1 cos ϕ cos (α + β)

cos β − sin β



 +



− sin α cos α

 =

sin β cos β

 .

30.6 (a) The line element in spherical coordinates is d~s = rˆ dr + θˆ r dθ + φˆ r sin θ dφ . ˆ φ} ˆ do not form a tensor basis because they don’t satisfy But {ˆ r, θ, d~s = ~er dr + ~eθ dθ + ~eφ dφ . Comparing the two expansions, though, reveals ~eθ = r θˆ

~er = rˆ

~eφ = r sin θ φˆ .

(b) The upstairs basis is given by ~e i = g ij ~ej , with g ij the inverse of gij :

~e r  ~e θ

 1

 1/r2

=

1/r2 sin2 θ

~e φ

!

rˆ rθˆ r sin θφˆ

.

ˆ ˆ sin θ. So ~e r = rˆ, ~e θ = θ/r, and ~e φ = φ/r (c) Given the upstairs vector components Ai , the downstairs components Ai = gij Aj are



Ar Aθ Aφ

  Ar 

 1



r2

=

Aθ Aφ

r2 sin2 θ

,

so that Ar = Ar , Aθ = r2 Aθ , and Aφ = r2 sin2 θ Aφ . (d) The reciprocal distribution of scale factors amongst bases and components allows straight~ = Ai~ei = Ai~e i . forward verification of A

~ to find 30.7 (a) With ~eb · ~e c = δbc , we use the projection ~e c · A (~e c · ~eb ) Ab = (~e c · ~ea0 ) Aa

0

0

0

= (~e c · ~ea0 ) T ab Ab = ~e c · ~ea0 T ab




Ab .

0

Thus the basis vectors transform as ~eb = ~ea0 T ab . (b) This was worked out in Example 30.4,

 T ib =

cos α sin α

sin β cos β

 .

(c) The cartesian components are



222

Ax Ay



 =

cos α sin α

sin β cos β

  1 1

 =

cos α + sin β sin α + cos β

 .

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Then, simplicity of the cartesian metric gives ~ 2 = δij Ai Aj = (Ax )2 + (Ay )2 = (cos α + sin β)2 + (sin α + cos β)2 |A| = 2 (1 + sin α cos β + cos α sin β) = 2 [1 + sin(α + β)] . or in sheared coordinates, ~ 2 = gab Aa Ab = 2 (1 + cos ϕ) = 2 [1 + sin(α + β)] . |A| (d) In terms of the g ab , the inverse of gab , g ab =

1 sin2 ϕ



1 − cos ϕ



− cos ϕ 1

,

the transformation of the metric laid out in (30.48) becomes a similarity transformation g ij = T ia T jb g ab → T gT T . Straightforward matrix multiplication then reveals g ij = δ ij . Thus since the shear metric can be transformed into the cartesian metric, it is an R2 metric.

30.8 Under a Lorentz transformation Λ, the rank-two tensor ηµν transforms as (sum implied) ηα0 β 0 = Λµα0 Λνβ 0 ηµν = Λµα0 ηµν Λνβ 0 . or in matrix notation, η 0 = ΛT ηΛ . The requirement that η be invariant is actually a physical constraint imposed on what form the Lorentz transformations can take — in other words, η = ΛT ηΛ is effectively the defining condition for Λ. But since we already have Λ in (30.52), we need only do the matrix multiplication to confirm that η = ΛT ηΛ. [One could also construe the calculation as confirming that the parametrization of Λ in (30.52) results in the condition γ 2 (1 − β 2 ) = 1.]

30.9 (a) Perhaps the simplest way to see this is to note that even though ∂a ≡ ∂/∂ua is a downstairs vector, the derivative is respect to an upstairs vector. So it must be that in Minkowski space the four-gradient components are ~ ∂µ → ∂0 , +∇




~ ∂ µ → ∂0 , −∇




(b) ∂µ ∂ µ = ∂02 − ∇2 , or in terms of t, ∂µ ∂ µ =

1 2 ∂ − ∇2 . c2 t

Operating on a scalar field, ∂µ ∂ µ Φ(~ r, t) = 0 is the wave equation. Thus  ≡ ∂µ ∂ µ is the wave operator — also called the d’Alembertian. (It’s often denoted 2 to resemble the Laplacian ∇2 . In fact, it’s because it’s a four-dimensional version of the Laplacian that the d’Alembertian has four rather then three sides!) (c) The four-divergence is ~ ·A ~ . ∂µ A µ = ∂ µ A µ = ∂0 A 0 + ∇ So a vanishing four-divergence is the continuity equation.

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223

30.10 Using the product rule, (30.63), and the antisymmetry of F µν , q µν d(m2 c2 ) d dP µ = (P µ Pµ ) = 2 Pµ = 2 F Pν Pµ dτ dτ dτ m 2q µν 2q = F Pµ Pν = − F νµ Pµ Pν . m m





But Pµ Pν is symmetric, so relabeling its dummy indices µ ↔ ν shows that the sum equals its own negative — and thus must vanish.

30.11 The spatial part of Eqn. (30.63) is


dt dP i dP i q q dP i = =γ = Pν F iν = P0 F i0 + Pj F ij dτ dτ dt dt m m

 q  = P0 E i /c + Pj −ijk Bk . m Denoting energy as E, we substitute P0 = E/c, and write Pj = −pj to find q dP i = dt γm



EE i + ijk pj Bk c2

 .

Then, using E = γmc2 and p ~ = γm~v ,


dP i ~ + ~v × B) ~ i . = q E i + ijk vj Bk = q(E dt A similar calculation shows that the time component of the equation reduces to dE ~ · ~v , = qE dt which is the work done by the electric field.

λ ~ = E0 ρˆ = E0 (cos φˆı + sin φˆ 30.12 (a) Let E0 = 2π . Then in cartesian components, E ). Together 0ρ ~ with B = 0, we easily get

 F µν =

E0 c

0 cos φ sin φ 0

 

− cos φ 0 0 0

− sin φ 0 0 0

0 0 0 0

  . 

ˆ most elements of Eqn. (30.52) vanish, leaving only (b) Since the relative velocity is along k,

 0

Λµν = 



γ 0 0 −γv/c

0 1 0 0

0 0 1 0

−γv/c 0 0 γ

  

(c) In Bob’s frame, the electromagnetic field tensor is given by Fµ

224

0 0

ν

0

= Λµµ Λνν 0 F µ

0 0

ν

,

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or in matrix form F 0 = ΛF ΛT

 =

E0 c

  

=γ

E0 c

 

γ 0 0 −γv/c 0 cos φ sin φ 0

0 1 0 0

−γv/c 0 0 γ

0 0 1 0

− cos φ 0 0 cos φ v/c

  

− sin φ 0 0 sin φ v/c

0 cos φ sin φ 0

− cos φ 0 0 0

0 − cos φ v/c − sin φ v/c 0

− sin φ 0 0 0

0 0 0 0

  

γ 0 0 −γv/c

0 1 0 0

0 0 1 0

  . 

~ 0 = γE ~ = γE0 (cos φˆı + sin φˆ So Bob measures E ) (larger than Alice’s by a factor of γ) and v 0 ~ B = c (sin φˆı − cos φˆ ). (d) Alice: E 2 /c2 − B 2 = E02 /c2 .
Bob: E 02 /c2 − B 02 = (γE0 /c)2 1 − v 2 /c2 = E02 /c2 since γ = (1 − v 2 /c2 )−1/2 . ~ = 0, clearly E· ~ B ~ = 0. But despite B ~ 0 , 0, we still have E ~ 0 ·B ~ 0 = 0. Similarly, since Alice has B

30.13 (a) Fµν = ηµα ηνβ F αβ −→ ηF η T , or

 Fµν = 

 

=



1 0 0 0

0 −1 0 0

0 −Ex /c −Ey /c −Ez /c

0 0 −1 0

0 0 0 −1

Ex /c 0 Bz −By

  

Ey /c −Bz 0 Bx

− Ecx 0 Bz −By

0 Ex c Ey c Ez c

Ez /c By −Bx 0

E

− cy −Bz 0 Bx

− Ecz By −Bx 0

  

1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1

  

  . 

Seems like a lot of work for just that! But notice that since the B field components all have two spatial indices, each η flips the sign. But the E components have only one spatial index, and so get only one sign flip. (b) The double sum in Gµν = 12 µναβ Fαβ is not directly amenable to matrix multiplication since µνρσ is rank 4. But we can choose specific values for µ and ν to reduce  to a 4 × 4 matrix and then multiply. Moreover, given the inherent antisymmetry, we need only do this 6 times (rather than 16) to map out all the elements of Gµν . But after doing it once or twice, the pattern becomes readily apparent. (c) Tr(F 2 ) = Fµν F νµ

 = Tr 



0 −Ex /c −Ey /c −Ez /c

Ex /c 0 Bz −By

Ey /c −Bz 0 Bx

Ez /c By −Bx 0

  

0 Ex c Ey c Ez c

− Ecx 0 Bz −By

E

− cy −Bz 0 Bx

− Ecz By −Bx 0

  

Though tedious, we can use vector notation to clean things up — and note that we only

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225

−γv/c 0 0 γ

  

need the diagonal elements:

 Tr(F 2 ) = Tr 





←− 0 Bz −By

0 ↑ ~ −E/c ↓

~ E/c −Bz 0 Bx

−→ By −Bx 0

  

←− 0 Bz −By

0 ↑ ~ E/c ↓

~ −E/c −Bz 0 Bx

−→ By −Bx 0

   

E 2 /c2 Ex2 /c − By2 − Bz2

= Tr 



 

Ey2 /c − Bz2 − Bx2 Ez2 /c − Bx2 − By2

~2

2

2

= 2(E /c − B ) . Similarly, the second invariant is Gµν Fµν = −Gµν Fνµ = −Tr(GF ):

 −Tr(GF ) = −Tr 



0 ↑ ~ −E/c ↓



←− 0 Bz −By

~ E/c −Bz 0 Bx

−→ By −Bx 0

  

0 ↑ ~ −B ↓

~ ·B ~ −E

1  = − Tr  c

←− 0 Ez /c −Ey /c

~ B −Ez /c 0 Ex /c

−→ Ey /c −Ex /c 0

 ~ ·B ~ −E

 

~ ·B ~ −E ~ ·B ~ −E

4~ ~ =− E ·B . c 30.14 The four-cuurent is U µ = γ(c, ~ u), where ~ u is the conventional three-velocity of a particle, and p 2 γ = 1/ 1 − u /c2 . Then the scalar invariant associated with the four-current J µ = ρ0 U µ =
ρc, J~ is J µ Jµ = ρ20 U µ Uµ = ρ20 c2 = ρ2 c2 − J 2 . In the charges’ rest frame, ~ u = J~ = 0 — so ρ0 is the charge density in the rest frame. Charge conservation is given by the continuity equation ∂ρ ~ · J~ . = −∇ ∂t Recasting this in terms of the four-current and four-gradient, this becomes (remember ∂0 =

1 ∂ ) c t

∂0 J 0 = −∂i J i , which can be rendered most-simply as the vanishing four-divergence of the four-current, ∂µ J µ = 0 . .

30.15 (a) Notice there are no sums — implied or otherwise — so we’ll have to verify the claim piecemeal. Also, due to the cyclic permutation of indices and the antisymmetry of F µν , the only nontrivial results have all different indices. With µ = i, α = j, β = k, we get ∂i Fjk + ∂j Fki + ∂k Fij = 0 . Reading the components from (30.64) and Problem 30.13 (no implied sums),










∂i −jki Bi + ∂j −kij Bj + ∂k −ijk Bk = 0

226

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  

~ ·B ~ = 0. The only other non-trivial form has one time derivative. With µ = 0, or simply ∇ α = i, β = j, we get ∂0 Fij + ∂i Fj0 + ∂j F0i = 0 . Reading the components from (30.64),




∂0 −ijk Bk + ∂i (−Ej /c) + ∂j (Ei /c) = 0. In conventional form (∂0 =

1 ∂ ), c t

 1 −ijk ∂t Bk − (∂i Ej − ∂j Ei ) = 0 , c ~ ×E ~ = −∂ B/∂t. ~ revealing Faraday’s law ∇ (b) For β = 0 ∂α F α0 = µ0 J 0 ,




or, since J β = ρc, J~ , ∂i F i0 + ∂0 F 00 = µ0 cρ . Reading the components from (30.64) gives 1 ~ · (E/c) ~ ∇ + ∂t (0) = µ0 ρc. c ~ ·E ~ = ρ/0 . For β = i, Recalling that c2 = 1/µ0 0 reveals Gauss’ law, ∇ ∂0 F 0i + ∂j F ji = µ0 J i , or


1 ∂t (−Ei /c) + ∂j −jik Bk = µ0 J i . c ~ ×B ~ = µ0 J~ + µ0 0 ∂ E/∂t. ~ which is Ampère’s law, ∇ Due to the antisymmetry of F αβ , charge conservation is automatic: ∂β J β = ∂β ∂α F αβ = ∂α ∂β F αβ = −∂α ∂β F βα = −∂β ∂α F αβ ≡ 0 , where the last step swapped the dummy indices α ↔ β.

30.16 (a) The tensor transformation is ∂ua ∂ua0 ∂r = ∂ua0

gacyl 0 b0 =

∂ub sph g ∂ub0 ab ∂r ∂θ ∂θ ∂φ ∂φ grr + gθθ + gφφ , ∂ub0 ∂ua0 ∂ub0 ∂ua0 ∂ub0

sph where we’ve used that gab is diagonal. Solving for spherical coordinates (r, θ, φ) in terms of cylindrical (ρ, ϕ, z) gives

θ = tan−1 (ρ/z)

r2 = ρ2 + z 2

φ=ϕ

From this, we can calculate the partials to find gρρ =

∂r ∂r ∂θ ∂θ ∂φ ∂φ grr + gθθ + gφφ ∂ρ ∂ρ ∂ρ ∂ρ ∂ρ ∂ρ

= sin2 θ · 1 +



cos θ r

2

· r2 + 0 · r2 sin2 θ = 1.

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227

gϕϕ =

∂θ ∂θ ∂φ ∂φ ∂r ∂r grr + gθθ + gφφ ∂ϕ ∂ϕ ∂ϕ ∂ϕ ∂ϕ ∂ϕ

= 0 · 1 + ·r2 + 1 · r2 sin2 θ = r2 sin2 θ = ρ2 ∂r ∂r ∂θ ∂θ ∂φ ∂φ grr + gθθ + gφφ ∂z ∂z ∂z ∂z ∂z ∂z  2 sin θ = cos2 θ · 1 + · r2 + 0 · r2 sin2 θ = 1. r

gzz =

Similar manipulations reveal that all the off-diagonals vanish. (b) We embed the second transformation within the first and use the chain rule, gacyl 0 b0 = =

∂ua ∂ub sph g ∂ua0 ∂ub0 ab ∂ua ∂ub ∂ua0 ∂ub0





∂xi ∂xj δij ∂ua ∂ub



=

∂ua ∂xi ∂ua0 ∂ua

=

∂xi ∂xj δij . ∂ua0 ∂ub0



∂ub ∂xj ∂ub0 ∂ub

 δij

This is the tensor transformation from cartesian to cylindrical — so all we need are those partials. ∂x ∂x ∂y ∂y ∂z ∂z + + = cos2 ϕ + sin2 ϕ + 0 = 1 ∂ρ ∂ρ ∂ρ ∂ρ ∂ρ ∂ρ ∂x ∂x ∂y ∂y ∂z ∂z = + + = ρ2 sin2 ϕ + ρ2 cos2 ϕ + 0 = ρ2 ∂ϕ ∂ϕ ∂ϕ ∂ϕ ∂ϕ ∂ϕ ∂y ∂y ∂z ∂z ∂x ∂x + + =0+0+1=1 = ∂z ∂z ∂ϕ ∂z ∂z ∂z

gρρ = gϕϕ gzz

with again vanishing off-diagonal elements.

30.17 This requires little more than the chain rule:

 1 = det 11 =

det(δca )

= det

0

∂ua ∂uc ∂uc0 ∂ub

 = det



∂u ∂u0



det



∂u0 ∂u



.

30.18 (a) The volume element dτ transforms with a factor of the Jacobian determinant |J|,

∂u dτ, ∂u0

dτ = |J|dτ 0 =

and so is not a scalar; in fact, it’s a scalar density of weight +1. But the metric determinant g is a scalar density of weight -2,

0 0 ∂u ∂u g0 . ∂u ∂u

g= So the product

√

g dτ should be a scalar density of weight 0 — with is to say, a scalar: √

0 p p ∂u ∂u g 0 dτ 0 = g 0 dτ 0 . 0 ∂u ∂u

g dτ =

In cartesian coordinates, the metric is the identity, so its determinant is 1. Thus the covariant √ volume element is 1dτ = dx dy dz. In cylindrical g = r2 , so gdτ = r dr dφ dz; in spherical, √ 4 2 2 g = r sin θ, giving the familiar volume element gdτ = r sin θ dr dθ dφ.

228

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(b) It may be simpler to consider this in three dimensions. According to the transformation rules — and recalling (4.42) in the role of the Levi-Civita epsilon in the determinant, 0

0

0

∂ua ∂ub ∂uc ∂u0 = abc a b c ∂u ∂u ∂u ∂u

abc = a0 b0 c0





Since the epsilon is invariant, this shows that it is a tensor density of weight -1. This directly generalizes to higher dimensions.

30.19 (a) Since in cartesian coordinates gij = δij , Eqn. (30.80) gives the path length along C from t = 0 to 1 as the familiar

Z

r

1

Z ds = 0

dx dt

2

+



dy dt

2

dt =

√

2.

~ =~ (b) The transformation T −1 from x, y to u, v can be obtained from Eqns. (30.47) with A r: 1 1 (x cos β − y sin β) = (cos β − sin β)t sin ϕ sin ϕ 1 1 (−x sin α + y cos α) = (− sin α + cos α)t , v= sin ϕ sin ϕ

u=

where cos (α + β) = sin ϕ. The symmetry of the shear coordinates means that the curve C is specified by u = v — or α = β. Then α + β + ϕ = π/2 gives √ 2 sin(ϕ/2) u=v= t. sin ϕ With this, we’re ready to apply Eqn. (30.19):

Z

1

Z ds =

r

0

= =

√

du dt

2

+ 2 cos ϕ

du dv + dt dt

p 2

Z

2(1 + cos ϕ)

dv dt

2

dt

1

dt

sin(ϕ/2)

sin ϕ



0

√ √ 2 cos(ϕ/2) sin(ϕ/2) 2 = 2. sin ϕ

So the path length is the same in either coordinate system. Of course.

30.20 From Eqn. (30.83), the area contained between u = v = 0 and u = v = 1 is

Z

a

Z du

0

b

dv

p

Z |g| da = α sin ϕ

0

a

Z

b

du 0

dv = α ab sin ϕ , 0

the area of a parallelogram with sides length a and b (multiplied by α, which controls the overall scaling of the (x, y) → (u, v) transformation).

30.21 (a) In spherical coordinates, ds2 = dr2 + r2 dθ2 + r2 sin2 θ dφ. On the surface of a sphere, dr = 0 — so that with the radius fixed at r = R, the metric reduces to the two-dimensional

 gab =

R2 0

0 R2 sin2 θ

 .

(b) Reading the scale factors from the metric, the physical distance s along a curve of constant φ from θ = 0 and the north pole to the circle at θ0 is s = Rθ0 .

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229

(c) Again reading the scale factors from the metric — this time with θ = θ0 constant and φ sweeping through 2π, the measured circumference C is C = 2πR sin θ0 . You can verify this geometrically by noting that the actual radius of the circle, which passes into the sphere, is R sin θ0 . The discrepancy is due to the curvature of the sphere. The circumference as measured on the sphere is less than 2πs. Only for θ0 small is C ≈ 2πs. (d) The simplest approach is to leverage the symmetry of the sphere by noting that all points are effectively equivalent. So if when can demonstrate that close-enough to any one point the space is approximately flat, it must be true at all points. Consider the point P : θ = π/2, φ = 0. The distance ds from P is ds2 = R2 (dθ2 + sin2 θdφ2 ) = R2 (dθ2 + dφ2 ) = du2 + dv 2 , where u ≡ Rθ and v = Rφ. Expressed in these orthogonal coordinates, the space looks flat. Using (29.31) to expand the metric in a Taylor series around the point P ,

h




gab (π/2 + δθ, δφ) = 1 + δθ∂θ + δφ∂φ +




2 1 δθ∂θ + δφ∂φ + . . . gab . 2! P i

The first term is the flat metric R2 δij , and the first-derivative terms all vanish at P . It’s only in second order, since ∂θ2 gab

  → 0 0 , 0 −1 P

that the effect of curvature begins to manifest.

30.22 (a) The radial distance (dθ = dφ = 0) at fixed time (dt = 0) between r1 and r2 (both > rs ) is

Z

r2

∆s = r1

dr

p

.

1 − rs /r

This is a challenging integral — which we don’t actually need to evaluate in order to compare it to ∆r = |r2 − r1 |. That’s because for finite r1,2 , the integrand’s denominator is less than 1, resulting in a larger contribution to the integral. And the difference is larger for the smaller of r1,2 . Thus ∆s > ∆r.

"

Z

For the record:

dr

p

=

p

r(r − 1) + tanh−1

q

1 − rs /r

r r−1



# .

(b) The angular scale factors in the Schwarzschild metric have the usual form for spherical coordinates. So at a fixed instant of time, the surface area of a sphere centered at the origin is the familiar

Z A=

r2 sin θdθdφ = 4πr2 .

So even though the coordinate r is not the distance from the origin, it is the distance measurement upon which the area is based. (c) For a clock at rest (dr = dθ = dφ = 0), the proper time dτ = ds/c, ∆τ =

Z p

1 − rs /r dt =

p

1 − rs /r ∆t .

These times only agree as r → ∞. So the coordinate time interval ∆t is the time measured by an observer at infinity. (d) According to Alice, Bob approaches rs in a time ∆t → ∞. Thus she never sees him cross the horizon at rs .

230

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30.23 Since Φ is a scalar, Dβ Φ = ∂β Φ ≡ Aβ is a vector; then Dα Aβ = ∂α Aβ − Aµ Γµ = ∂α ∂β Φ − Γ µ ∂ Φ. αβ αβ µ Similarly, Dβ Dα Φ = ∂β ∂α Φ − Γµ ∂ Φ. βα µ Now for general curved spaces, Dα Dβ , Dβ Dα . However, in Minkowski space we can in invoke the principal of general covariance to work in a cartesian basis at a point P , which reveals Dα Dβ Φ = ∂α ∂β Φ = ∂β ∂α Φ = Dβ Dα Φ . Thus we find

Γµ αβ

=

Γµ . βα

30.24 (a) Start by taking the inner product of Eqn. (30.86) with a basis vector: ∂~ea · ~ed = Γcab ~ec · ~ed . ∂ub Then Eqn. (30.12) gives ∂b gad − ~ea ·

∂~ed = Γcab gcd ∂ub

or ∂b gad = Γcba gcd + Γcbd gca ,

(∗)

where we’ve used the symmetry of Γ in its lower two indices to “pretty this up" a bit. Now the result we’re trying to confirm combines ∂b gad with its cyclic permutations:







∂b gad + ∂a gdb − ∂d gba = Γcba gcd + Γcbd gca + Γcad gcb + Γcab gcd − Γcdb gca + Γcda gcb




= Γcba gcd + Γcab gcd = 2Γcab gcd . Finally, multiplying both sides by g df and using gcd g df = δcf yields the desired result, Γcab =

1 dc g (∂b gad + ∂a gdb − ∂d gba ) . 2

(b) The vanishing of the covariant derivative of the metric, Eqn. (30.99), is just the starred equation above; the rest is the same!

30.25 Apply Γcab =



1 dc g (∂b gad + ∂a gdb − ∂d gba ) . 2



1 0 polar coordinates, g → 0 r2 Because the metric is diagonal, and the only non-trivial element is gφφ , most of the Christoffel elements vanish: 1 dr 1 g (∂r grd + ∂r gdr − ∂d grr ) = g rr (∂r grr + ∂r grr − ∂r grr ) = 0 2 2
1 1 = g dφ (∂r grd + ∂r gdr − ∂d grr ) = g φφ ∂r grφ + ∂r gφr − ∂φ grr = 0 2 2
1
1 = g dφ ∂φ gφd + ∂φ gdφ − ∂d gφφ = g φφ ∂φ gφφ + ∂φ gφφ − ∂φ gφφ = 0 2 2
1
1 = g dr ∂r gφd + ∂φ gdr − ∂d grφ = g rr ∂r gφr + ∂φ grr − ∂r grφ = 0 = Γrrφ . 2 2

Γrrr = Γφ rr Γφ φφ Γrφr

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231

This leaves as the only non-zero elements,


1 1 dr g ∂φ gφd + ∂φ gdφ − ∂d gφφ = g rr 2 2 1 = 2 2r
1 1 = g dφ ∂r gφd + ∂φ gdr − ∂d grφ = g φφ 2 2 1 = 2 2r

Γrφφ =

Γφ φr

∂φ gφr + ∂φ grφ − ∂r gφφ







∂φ 0 + ∂φ 0 − ∂r r2 = −r ∂r gφφ + ∂φ gφr − ∂φ grφ




∂r r 2 + ∂φ 0 − ∂φ 0 =




1 = Γφ . rφ r

!

1 0 0 spherical coordinates:g → 0 r2 0 0 0 r2 sin2 θ Similarly, the only non-vanishing elements are: Γrθθ = Γrφφ = Γθφφ = Γθrθ = Γφ = rφ = Γφ θφ

1 1 1 dr g (∂θ gθd + ∂θ gdθ − ∂d gθθ ) = g rr (∂θ gθr + ∂θ grθ − ∂r gθθ ) = − ∂r (r2 ) = −r 2 2 2
1
1 dr 1 g ∂φ gφd + ∂φ gdφ − ∂d gφφ = g rr ∂φ gφr + ∂φ grφ − ∂r gφφ = − ∂r (r2 sin2 θ) = −r sin2 θ 2 2 2
1
1 dθ 1 g ∂φ gφd + ∂φ gdφ − ∂d gφφ = g θθ ∂φ gφθ + ∂φ gθφ − ∂θ gφφ = − 2 ∂θ (r2 sin2 θ) = − sin θ cos θ 2 2 2r 1 dθ 1 1 1 g (∂θ grd + ∂r gdθ − ∂d gθr ) = g θθ (∂θ grθ + ∂r gθθ − ∂θ gθr ) = 2 ∂r (r2 ) = = Γθθr 2 2 2r r
1 φφ
1 dφ 1 1 g ∂φ grd + ∂r gdφ − ∂d gφr = g ∂φ grφ + ∂r gφφ − ∂φ gφr = 2 ∂r (r2 sin2 θ) = = Γφ φr 2 2 2r sin2 θ r

1 dc 1 1 g (∂b gad + ∂a gdb − ∂d gba ) = g dφ ∂φ gθd + ∂θ gdφ − ∂d gφθ = g φφ ∂φ gθφ + ∂θ gφφ − ∂φ gφθ 2 2 2 1 2 2 ∂θ (r sin θ) = cot θ = Γφ . = 2 φθ 2r sin2 θ

30.26 Since the shear metric

 gab →

cos ϕ 1

1 cos ϕ



is not a function of either u or v, all the derivatives in Γcab =

1 dc g (∂b gad + ∂a gdb − ∂d gba ) . 2

ˆ vanish. So Γcab ≡ 0. Indeed, both u ˆ and vˆ are constant (unlike, say, cylindrical rˆ, φ).

30.27 (a) The tensorial behavior of Γ can be determined by transforming its definition Eqn. (30.86), 0

Γa ea0 = b0 c0 ~

∂~eb0 ∂uc0

into unprimed coordinates: 0

Γa b0 c0



∂ud ~ed ∂ua0

 =

∂ ∂uc0



∂ud ~ed ∂ub0



∂ 2 ud ∂ud ∂~ed ~ed + ∂ub0 ∂uc0 ∂ub0 ∂uc0  e  ∂ 2 ud ∂ud ∂u ∂~ed = ed + 0 0 ~ 0 0 e b c b c ∂u ∂u ∂u ∂u ∂u =

 =

232

∂ 2 uf ∂ud ∂ue f + Γ ∂ub0 ∂uc0 ∂ub0 ∂uc0 de

 ~ef .

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Relabeling the dummy index d → f on the left allows us to equate the coefficients of ~ef on 0

∂ua ∂uf

both sides. Multiplying this by

0

isolates Γa : b0 c0 0

0

∂ua ∂ud ∂ue f ∂ 2 uf ∂ua + Γ . ∂uf ∂ub0 ∂uc0 ∂uf ∂ub0 ∂uc0 de

0

Γa b0 c0 =

The first term on the right-hand side is precisely what is expected of a third-rank mixed tensor. But the additional so-called inhomogeneous term prevents Γa bc from being a tensor. 0

(b) Taking the derivative ∂b0 of

∂ua ∂uf ∂uf ∂uc0

0

= δca0 , the product rule yields

0

0

∂ua ∂ 2 ua ∂uf ∂ 2 uf 0 0 = − f b c ∂u ∂u ∂u ∂ub0 ∂uf ∂uc0 0

=−

∂ 2 ua ∂ud ∂uf . ∂ud ∂uf ∂ub0 ∂uc0

(c) 0

0

Db0 Aa =

0 0 ∂Aa + Ac Γ a b0 c0 ∂ub0

 =

0

0

∂Ac ∂ud ∂ua ∂ 2 ua ∂ud + Ac 0 d c b ∂u ∂u ∂u ∂uc ∂ud ∂ub0 0

 Ab

+

∂uc ∂ub



 0

0

∂ua ∂ud ∂ue f ∂ 2 ua ∂ud ∂uf 0 0 Γde − f b c ∂u ∂u ∂u ∂ud ∂uf ∂ub0 ∂uc0

 .

The first term of each line combine to give 0

∂Ac ∂ud ∂ua + Ab ∂ud ∂ub0 ∂uc



0

∂ua ∂ud f Γ ∂uf ∂ub0 db

0

 =

∂ud ∂ua ∂ub0 ∂uc



∂Ac + Ab Γcdb ∂ud



,

which is the correct transformation for a 2nd-rank mixed tensor. The remaining terms, then, must vanish: 0

Ac

0

0

0

0

c 2 a 2 a ∂ 2 ua ∂ud ∂ 2 ua ∂ud ∂uf ∂ud ∂ud b ∂u c ∂ u b ∂ u 0 −A 0 0 = A 0 −A c d b d f c d d b b b c b ∂u ∂u ∂u ∂u ∂u ∂u ∂u ∂u ∂u ∂u ∂u ∂u ∂u ∂ub0 = 0. X

30.28 (a) To find the geodesic differential equations, we need the Christoffel symbols, which can be found by taking derivatives of the metric, Eqn. (30.100). But since we already know the Γ’s for spherical coordinates in R3 , Eqn. (30.92), we need not work quite that hard. In the two-dimensional space defined by θ and φ, none of the Γijk ’s will have any of i, j, or k = r. Thus we are left with only Γφ = Γφ = cot θ . φθ φθ

Γθφφ = − sin θ cos θ

Direct substitution into the geodesic equation leads to the desired result, d2 θ − sin θ cos θ dt2



dφ dt

2

=0

d dt



sin2 θ

dφ dt



=0. 2

(b) We chose initial position θ0 = π/2 so that the θ equation reduces to ddt2θ = 0 at t = 0. Together with θ˙0 = 0, this tells us there is no change in θ — so the solution is simply θ(t) = π/2. Since this gives sin θ(t) = 1, the solution to the φ equation must be linear in t — i.e., φ(t) = ωt, consistent with φ0 = 0. (c) The solution we’ve found is the equator — which we know is a great circle. By “undoing" the rotations which took us to the equator, we can generate any and all geodesics. So they must all be great circles.

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233

30.29 (a) Assuming the product rule applies, we get Dc (Aa B b ) = (Dc Aa )B b + Aa (Dc B b )



=

∂Aa b a + Ad Γa cd B + A ∂uc





∂B b + B d Γbcd ∂uc



∂(Aa B b ) a d b + Ad B b Γ a cd + A B Γcd , ∂uc

=

which is Eqn. (30.96) for the dyadic T ab = Aa B b . (b) Using the product rule, we have Db (Aa Aa ) = Aa (Db Aa ) + (Db Aa )Aa  a  ∂A = Aa (Db Aa ) + + Ac Γa bc Aa . b ∂u But Aa Aa is a scalar, and the covariant derivative of a scalar is the normal derivative: Db (Aa Aa ) = ∂b (Aa Aa ) = Aa

∂Aa ∂Aa + Aa . b ∂u ∂ub

Comparison of these two expressions for Db (Aa Aa ) gives ∂Aa − Ac Aa Γa bc ∂ub   ∂Aa = Aa − Ac Γcba , ∂ub

Aa Db Aa = Aa

confirming Eqn. (30.97) for the covariant derivative of a downstairs vector, Db Aa =

∂Aa − Ac Γcba . ∂ub

30.30 (a) Together, (30.12) and (30.86) imply ~ed ·

∂~ea = ~ec · ~ed Γcab = gcd Γcab . ∂ub

Then, using (30.97), ∂gab − gdb Γdac − gad Γdbc ∂uc = ∂c (~ea · ~eb ) − ~eb · ∂c~ea − ~ea · ∂c~eb = 0. X

Dc gab =

!

1 0 0 (b) cylindrical, g → 0 r2 0 : Since the z coordinate is cartesian, all Γ’s with a z element 0 0 1 vanish — and then so must all covariant derivatives of gzz Then as a diagonal metric, with only Γφ = 1/r and Γφ = −r non-zero, we need only check a few cases: φr φφ Dr gφφ = ∂r gφφ − gdφ Γdφr − gφd Γdφr = ∂r (r2 ) − 2r2 (1/r) = 0 Dr grφ = ∂r grφ − gdφ Γdrr − grd Γdφr = 0 Dr grφ = ∂r grφ − gdr Γdrφ − grd Γdφφ = 0 − r2 (1/r) − (−r) = 0 .

spherical, g →

234

1 0 0

0 r2 0

0 0 r2 sin2 θ

! : The metric is diagonal, and the only non-zero Γ’s are

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listed in (30.92). So of the 27 permutations, only 6 are non trivial: Dr gθθ = ∂r gθθ − gdθ Γdrθ − gθd Γdθr = ∂r (r2 ) − 2r2 (1/r) = 0 Dr gφφ = ∂r gφφ − gdφ Γdφr − gφd Γdrφ = ∂r (r2 sin2 θ) − 2(r2 sin2 θ)(1/r) = 0 Dθ grθ = ∂θ grθ − gdθ Γdrθ − grd Γdθθ = 0 − r2 (1/r) − (−r) = 0 Dθ gφφ = ∂θ gφφ − gdφ Γdφθ − gφd Γdφθ = ∂θ (r2 sin2 θ) − 2r2 sin2 θ cot θ = 0 Dφ gθφ = ∂φ gθφ − gdθ Γdφφ − gφd Γdθφ = 0 − r2 (− sin θ cos θ) − r2 sin2 θ(cot θ) = 0 Dφ grφ = ∂φ grφ − gdφ Γdrφ − grd Γdφφ = 0 − r2 sin2 θ(1/r) − (−r sin2 θ) = 0

 (c) – On the surface of a cylinder of radius R, the metric reduces to the 2 × 2 matrix

R2 0



0 1

— which is constant, and so all derivatives ∂c gab ≡ 0.



R2 0

– On the surface of a sphere of radius R, the metric reduces to

0 R2 sin2 θ

 which is

not constant. The results of part (b) show that ∂θ gφφ , 0. The difference between these two results reflects the difference in the types of curvature: The cylinder can be “unrolled” into a plane, and so has no intrinsic curvature; by contrast, the sphere’s curvature is intrinsic (see the discussion at the beginning of BTW 20.1).

30.31 (a) Using ~e a = ˆa / ~ = ∇Φ

√

gaa ,

X

X ˆa

~e a ∂a Φ =

√

a

a

ˆ1 ˆ2 ˆ3 ∂a Φ = √ ∂1 Φ + √ ∂2 Φ + √ ∂3 Φ, g11 g22 g33

gaa

Noting that for orthogonal coordinates the elements gaa are the square of the scale factors ha , this is just (12.49a). ~ on the conventional orthonormal basis ˆa : (b) First use (30.41) to find the component Aaˆ of A ~ = ~ea Aa = √gaa ˆa Aa ≡ ˆa Aaˆ A

Aaˆ Aa = √ . gaa

=⇒

√ √ Second, orthogonal systems have diagonal metrics with gaa = ha and g = h1 h2 h3 , where that ha ’s are the R3 scale factors. So (including explicit summations for clarity): b a b a Da Aa = ∂a Aa + Γa ab A = ∂a A + Γba A

=

X



Aaˆ √ gaa



Aaˆ ha

∂a

a

=

X

∂a

a

=

1 h1 h2 h3

1 = h1 h2 h3



 +

X

1 √ + √ (∂a g) g



a

a

∂a

h

Aaˆ √ gaa

1 ∂a (h1 h2 h3 ) h1 h2 h3

h1 h2 h3 ∂a

X



Aaˆ ha

h1 h2 h3 aˆ A ha

i





Aaˆ ha





 + ∂a (h1 h2 h3 )

Aaˆ ha



,

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235

P

which is (12.49c). [To verify

X

Γa ab =

1X 2

a

a

1 √ ∂ g b

Γa ab =

√

g:

g ac [∂a gbc + ∂b gca − ∂c gab ]

a

1 = 2

X

1 = 2

X

= ∂b

X

g aa [∂a gba + ∂b gaa − ∂a gab ]

a

g aa ∂b gaa =

1 2

1

X

a

=

h2a ∂b h2a

a


1 X ∂b ln h2a 2 a

ln (ha ) = ∂b ln (h1 h2 h3 ) = ∂b ln (

√

1 √ g) = √ ∂b g .] g

a

(c) Together, the results of (a) and (b) yield the covariant expression for the Laplacian: D2 Φ = Db Db Φ =

X



Db

b

1 = h1 h2 h3

ˆb ∂b Φ hb



X b

 ∂b



h1 h2 h3 ∂b Φ , h2b

which is (12.49b). (d) The conventional cartesian form of the curl ~ ×A ~ ∇


i

=

X

ijk ∂j Ak

jk

~ with the covariant is not in covariant tensor form. To do that, we need not only to replace ∇ derivative, the tensor density ijk must be recast as a proper third-rank tensor. From Problem √ 30.18, we know that ijk has weight -1, whereas g has weight +1. Thus √

g abc

and similarly

1 abc √  g

are tensors. So the covariant form of the curl is ~ ×A ~ D


a

1 = √ g

X

abc Db Ac .

bc

Since the antisymmetry of  colludes with the symmetry of Γ to give zero, on the right-hand side Db → ∂b . In addition, on the orthonormal basis (no sum) ~ ×A ~ D


a


1 ~ ×A ~ aˆ D = √ gaa

and

∂b A c = ∂b (

√

gcc Acˆ) .

All together then, in terms of the scale factors ha = gaa , ~ ×A ~ D


a

=

ha h1 h2 h3

X

abc ∂b (hc Acˆ) ,

bc

which is the determinant in (12.49d).

236

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Prelude: 1 2 3 . . . Infinity

31

31.1 In the expression hφn |φm i = kn δmn , the value of kn only matters for n = m — in which case √ the inner product just gives the norm-squared of |φn i. But by definition, the norm kn must be real and positive. Hence kn must be real and positive.

31.2 Start with the observation that

Z bZ a

b

|ψ(x)φ(y) − ψ(y)φ(x)|2 dx dy ≥ 0 .

a

Multiplying out the integrand, we find (re-labeling dummy variables, as appropriate)

Z

b

 Z |ψ(x)|2 dx

2

b

 |φ(y)|2 dy

a

a

b

Z

− 2

a

2

φ∗ (x)ψ(x) dx ≥ 0,

from which the Schwarz inequality directly emerges.

31.3 Using the Schwarz inequality,

Z

b

|φ(x) + ψ(x)|2 dx =

a

b

Z

|φ|2 dx +

a

|φ|2 dx +

≤



φ∗ ψ dx

a

sZ

b

Z

a

b

Z |ψ|2 dx + 21

Note that, other than n = 1 only even n actually contribute. (d) An odd extension with period 2π gives a sine series basis with functions sin nx: bn =

2 π

Z

sin(x + π/4) sin(nx) dx 0

 =

256

π

1 √ , 2 √ n 2 n[1+(−1) ] π n2 −1

n=1 ,

n>1

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Once again, other than n = 1 only even n actually contribute. As the plots show, only the cosine series doesn’t ring: it’s the only one of the of three which is continuous at x = 0.

1

1

-π

-π

π

2

π

2

-π

-π

1

π

2

π

2

(a)

-π

-π

π

2

2

(b)

π

(c)

Problem 33.11: Fourier series of sin(x + π/4) (through n = 16), (a) with period π, (b) cosine series, period 2π, (c) sine series, period 2π

33.12 (a) Period π requires basis functions sin(2nx) and cos(2nx). Now you can certainly integrate to find the Fourier coefficients any number of ways (including invoking orthogonality), but the familiar trig identity sin2 x = 21 (1 − cos 2x) is itself the sought-after Fourier series! In other words, the only non-vanishing Fourier coefficients are a0 = 1, a1 = −1/2. (Indeed, sin2 x is even, so bn ≡ 0.) Similarly, since cos2 x = 12 (1 + cos 2x) the only non-vanishing Fourier coefficient for cos2 x are a0 = 1, a1 = +1/2. (b) An even extension with period 2π requires the cosine series basis functions cos nx. The same 1 (1 ± cos 2x) trig identities once again provide the sought-after Fourier series. So as in (a), 2 we again have only one non-zero frequency — but this time it’s associated with n = 2 rather that n = 1. (c) An odd extension with period 2π requires the sine series basis functions sin nx. But now the integration is not over an entire period, so orthogonality only pertains for n even:

bn =

2 π

Z

π

sin(nx) sin2 (x) dx =

0

Thus sin2 x = − π8

sin nx . n odd n(n2 −4)

P

1 π

Z

π

sin(nx)(1 − cos 2x) dx = − 0

Similarly, cos2 x = + π4

P

4 [1 − (−1)n ] πn(n2 − 4)

n2 −2 n odd n(n2 −4)

sin nx. As the

plots show, only the expansion of cos2 x rings due to the discontinuity of its odd extension at x = 0.

1 1

-π

-π

π

2

2

(a)

π

-π

-π

π

2

2

π

(b)

Problem 33.12: Fourier sine series with period 2π, (a) sin2 x and (b) cos2 x. The solid curves are the Fourier sums through n = 15.

33.13 This merely requires direct application of formulas in Table 33.1 with 2L = 4π:

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257

(a) 2 2π

Z

2 an = 2π

Z

a0 =

So sin x =

8 π

1 3

2π

sin(x) dx = 0 0 2π

 sin(x) cos(nx/2) dx =

1 5

cos(3x/2) −

1 21

n even

8 π(4−n2 )

0

cos(x/2) −

0,

cos(5x/2) − · · ·

,

n odd

,

n even n odd



(b) an = So cos x =

4 π

2 2π

2π

Z

 cos(x) sin(nx/2) dx =

4n π(n2 −4)

0



− 31 cos(x/2) +

3 5

cos(3x/2) +

5 21

cos(5x/2) + · · ·

sin x

-2

0,



cos x

2

-

-2

2

-

33.14 One can redo the example, altering the boundary condition (33.32) apply at ±L/2 rather than at 0 and L. But far simpler — and arguably more intuitive, is to take the solution (33.35) and translate it, letting x → x − L/2: φn (x) = sin



nπ(x − L/2) L



= sin



nπx L



cos(nπ/2) − cos



nπx L



sin(nπ/2)




sin nπx , n even L
= . nπx cos

,

L

n odd

(The sign in front of cos(nπx/L) is irrelevant, since the full solution is a superposition of normal modes.) These are in fact the same normal modes — and application of Schrödinger’s equation (33.30) confirms that they have the same energies. As they had to.

P∞

33.15 Using the Fourier expansion ln(sin x) = − ln 2 −

Z

n=1

π

ln (sin x) dx = −π ln 2 −

S1 = 0

∞ Z X n=1 ∞

= −π ln 2 −

cos 2nx n π

0

X 1 n2

from Problem 6.14:

cos 2nx dx n

sin(2nπ) = −π ln 2 ,

n=1

which should be compared with the approach of Problem 8.28.

Z S2 =

π 2

Z

[ln (sin x)] dx = 0

π

" ln 2 +

0

∞ X cos 2nx

dx

n n=1

2

= π (ln 2) + 2 ln 2

X1Z n n

258

#2

0

π

cos 2nx dx +

X 1 Z nm n,m

π

cos 2nx cos 2nx dx

0

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The second term vanishes by orthogonality (or trivial integration); orthogonality also reduces the integral in the third term to π2 δnm . Thus

Z

π

[ln (sin x)]2 dx = π (ln 2)2 +

0

33.16 f (x) =

an =

1 π 1 π

bn =

1 π

(a) a0 =

0, x,

−π

n2

π π3 ζ(2) = π (ln 2)2 + . 2 12

−π < x < 0 0

` X

∗ Y`m (θ0 , φ0 )Y`m (θ, φ) cos θ0 δ(r0 − a)r02 dr0 dΩ0

m=−`

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For the potential outside the sphere r > a, r> = r and r< = r0 : q Φ(r > a) = 2π0 a2

X

q = 2π0 a2

X

=

=

=

q 2π0

1 2` + 1

`=0

Z ` X Y`m (θ, φ) m=−`

1 Y`m (θ, φ) 2` + 1 r`+1

qa 6π0

Z

∗ a` Y`m (θ0 , φ0 ) cos θ0 a2 dΩ0

`,m

X a`

Y`m (θ, φ) 2` + 1 r`+1

Z

∗ Y`m (θ0 , φ0 ) cos θ0 dΩ0

`,m

q X a` 2π0

∗ r0` Y`m (θ0 , φ0 ) cos θ0 δ(r0 − a)r02 dr0 dΩ0

r`+1

Y`m (θ, φ) 2` + 1 r`+1

Z

r ∗ Y`m (θ0 , φ0 )

4π Y10 (θ0 , φ0 )dΩ0 3

`,m

r

qa cos θ 4π Y10 (θ, φ) = . 3 r2 6π0 r2

The calculation is similar inside the sphere, where r> = r0 and r< = r: Φ(r < a) =

=

q 2π0 a2 q 2π0

X

1 r` Y`m (θ, φ) 2` + 1

Z

∗ (θ 0 , φ0 ) Y`m

r0`+1

cos θ0 δ(r0 − a)r02 dr0 dΩ0

`,m

X

1 r` Y`m (θ, φ) `+1 2` + 1 a

Z

∗ Y`m (θ0 , φ0 ) cos θ0 dr0 dΩ0

`,m

q = r cos θ . 6π0 a2 Incidentally, note that the two solutions agree at r = a — as they must, since the electric field ~ = −∇Φ ~ must be finite there. E

35.19 (a) The addition theorem (35.50) shows that the sum only depends on the angle γ between the unit vectors (θ, φ) and (θ0 , φ0 ), which is invariant under rotation of the coordinates. More intuitively, π/` characterizes the angular separation of extrema of the normal modes Y`m , |m| ≤ `, which is clearly a rotationally-invariant quantity. (b) For φ → φ + α, Y`m (θ, φ) → eimα Y`m (θ, φ), independent of `. Indeed, for any change dφ, dY`m (θ, φ) = imY`m (θ, φ)dφ. So since φ is the angle around the z-axis, the choice of that axis will affect m. For instance, we can choose the z-axis to lie along any given direction p 2`+1 n ˆ = (θ, φ); such an assignment yields m = 0, Y`m (ˆ z) = δm0 . 4π 35.20 (a) Inserting (35.61) into (35.63) — and noting that δT /T¯ is real, C(ϕ) =

D

δT (ˆ n) δT (ˆ n0 ) T¯ T¯

E

=

XX `m

=

ha`m a`0 m0 i Y`m (ˆ n)Y`0 m0 (ˆ n0 )

`0 m0

XX



a`m a∗`0 m0 Y`m (ˆ n)Y`∗0 m0 (ˆ n0 ).

`m `0 m0

(b) Homogeneity requires that measured quantities be independent of position. Thus C depends only on the angle ϕ between the temperature measurements, not their locations. But C(ϕ) can be expanded in Legendre polynomials P` (cos ϕ) — and so it must be that the ensemble average is proportional to δ ``0 . Moreover,  ` is unchanged by a rotation (Problem 35.19), so isotropy is consistent with a`m a∗`0 m0 ∼ δ``0 . As for m, under rotation in φ, Y`m (θ, φ + α) = eimα Y`m (θ, φ) ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

271

which implies that a`m → eimα a`m . So a rotation gives



a`m a∗`0 m0



→ ei(m−m

0

)α



a`m a∗`0 m0



.

Thus isotropy requires m = m0 . Finally,  unlike m, ` corresponds to wavelength λ ∼ 2π/`, invariant under rotation. So a`m a∗`0 m0 can depend only on `. (c) C(ϕ) =

XX



a`m a∗`0 m0 Y`m (ˆ n)Y`∗0 m0 (ˆ n0 ) =

`m `0 m0

XX

δ``0 δmm0 C` Y`m (ˆ n)Y`∗0 m0 (ˆ n0 )

`m `0 m0

=

∞ X

` X

C`

`=0

=

∗ Y`m (ˆ n)Y`m (ˆ n0 )

m=−`

∞ 1 X

(2` + 1)C` P` (cos ϕ) .

4π `=0

35.21 (a) Since the φ-dependence of spherical harmonics comes solely from the factor eimφ , Lz Y`m = −i∂φ Y`m = mY`m .

X

Similarly, all the θ-dependence of Y`m resides in the associated Legendre function P`m (cos θ). Thus the θ derivatives in the Laplacian only act on P`m (cos θ):

  1 ∂θ sin θ ∂θ P`m (cos θ) . sin θ √ √ In terms of x = cos θ, sin θ = 1 − x2 and ∂θ = − 1 − x2 ∂x , this simplifies to r2 ∇2θφ P`m (cos θ) =





r2 ∇2θφ P`m (x) = ∂x (1 − x2 )∂x P`m (x)





= (1 − x2 )∂x2 − 2x ∂x P`m (x) =

m2 − `(` + 1) , 1 − x2

where we’ve used (32.67) in the last step. The φ derivatives in ∇2 only act in eimφ — so all together we get L2 Y`m = −r2 ∇2θφ Y`m = −

h

1 1 ∂θ (sin θ ∂θ ) + ∂ 2 Y`m sin θ sin2 θ φ



m2 m2 − `(` + 1) − 2 1−x 1 − x2

=−

i

 Y`m = `(` + 1)Y`m

X

(b) First square the individual component operators: L2x = −(sin φ∂θ + cot θ cos φ∂φ )2




= − sin φ∂θ (sin φ∂θ ) − cot θ cos φ∂φ cot θ cos φ∂φ − sin φ∂θ (cot θ cos φ∂φ ) − cot θ cos φ∂φ (sin φ∂θ ) = − sin2 φ ∂θ2 − cot2 θ cos2 φ ∂φ2 + cot2 θ sin φ cos φ ∂φ + csc2 θ sin φ cos φ∂φ − cot θ sin φ cos φ∂θ ∂φ − cot θ cos2 φ∂θ − cot θ sin φ cos φ∂θ ∂φ L2y = −(cos φ∂θ − cot θ sin φ∂φ )2




= − cos φ∂θ (cos φ∂θ ) − cot θ sin φ∂φ cot θ sin φ∂φ + cos φ∂θ (cot θ sin φ∂φ ) + cot θ sin φ∂φ (cos φ∂θ ) = − cos2 φ ∂θ2 − cot2 θ sin2 φ ∂φ2 − cot2 θ sin φ cos φ ∂φ − csc2 θ sin φ cos φ∂φ + cot θ sin φ cos φ∂θ ∂φ − cot θ sin2 φ∂θ + cot θ sin φ cos φ∂θ ∂φ

272

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and L2z = −∂φ2 . Then L2x + L2y + L2z = −∂θ2 − cot2 θ∂φ2 − cot θ∂θ − ∂φ2 = −∂θ2 − cot θ∂θ −

1 1 1 ∂2 = − ∂θ (sin θ∂θ ) − ∂ 2 ≡ −r2 ∇2θφ . sin2 θ φ sin θ sin2 θ φ

(c) Since the Y`m ’s are simultaneous eigenfunctions of hermitian operators L2 and Lz , it must be that [L2 , Lz ] = 0. In fact, since physically there’s nothing inherently special about the z-direction, L2 must commute with all the components, [L2 , Li ] = 0. Then since the Y`m are not also eigenvectors of either Lx or Ly , it must be that neither commutes with Lz . And indeed, direct calculation reveals [Lz , Lx ] = iLy , [Lz , Ly ] = −iLx . (d) Direct calculation yields L± = Lx ± iLy = e±iφ ±∂θ + i cot θ ∂φ




.

Working directly with associated Legendre functions P`m is tedious; better to extrapolate from Table 35.2 that Y` ±` = a` sin` θe±i`φ , for constant a` . Then




L± Y`` = a` e±iφ ±∂θ + i cot θ ∂φ sin` θe±i`



= a` e±iφ ±

` cos θ ∓ ` cot θ sin` θe±i`φ = 0 , sin θ



as expected when attempting to climb one rung above (below) the top (bottom) of the ladder. (e) Y22 =

p

15 32π

sin2 θ e2iφ :

r L− Y22 =

r =


15 −iφ e −∂θ + i cot θ ∂φ sin2 θ e2iφ 32π 15 −iφ e (−2 sin θ cos θ − 2 sin θ cos θ) e2iφ 32π

r = −4

p

where, with ` = m = 2, verifies

p

15 sin θ cos θeiφ = −2 32π

r

15 sin θ cos θeiφ = 2Y21 , 8π

(` + m)(` − m + 1) = 2. Repeated application of L− similarly

(` + m)(` − m + 1) all the way down to Y2 −2 .

35.22 Let’s do this in steps: ~

i. Since eik·~r = eikr cos γ , a ˜(k, r) = a ˜(kr), ii. The identity cos γ = cos θ cos θ 0 +sin θ sin θ 0 cos (φ − φ 0 ) shows that the sum can only depend on the combination φ − φ0 — and thus it must be that m0 = −m. Then we can use Y`,−m = ∗ . (−1)m Y`m iii. Similarly, if there’s only one m sum — does it run over ` or `0 ? Since there can’t be any ambiguity, it must be that `0 = `. So far, then, we have a ˜``0 mm0 (kr) → (−1)m a ˜`m (kr) ˆ iv. If we choose zˆ to align with k, then γ = θ and Y`m is proportional to δm0 . But since a ˜(kr) has no knowledge of the choice of coordinate system, it cannot depend on m. Thus (−1)m a ˜`m (kr) → c` (kr). All together then, the sum collapses to ~

eik·~r =

` X X `

ˆ `m (ˆ c` (kr)Y`∗0 m (k)Y r) X

m=−`

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273

~ 35.23 (a) The symmetry between ~k and ~ r in eik·~r is sufficient to conclude that either Y can be starred — but let’s demonstrate this explictly:

~

eik·~r =

∞ ` X X

∞ ` X X

∗ ˆ c` (kr)Y`m (k)Y`m (ˆ r) =

`=0 m=−`

∗ ˆ `,−m (ˆ c` (kr)Y`,−m (k)Y r)

`=0 m=−` ∞ ` X X

=

m ∗ ˆ c` (kr)(−1)m Y`m (k)(−1) Y`,m (ˆ r)

`=0 m=−` ∞ ` X X

=

ˆ ∗ (ˆ c` (kr)Y`m (k)Y `,m r ) .

`=0 m=−` P

~

~

(b) Under parity, eik·~r −→ eik·(−~r) =



~

eik·~r

∗

. We’ll examine the two expressions separately

and then compare: ~

eik·(−~r) =

X

∗ ˆ c` (kr)Y`m (k)Y`m (−ˆ r) =

`m

X

∗ ˆ (−1)` c` (kr)Y`m (k)Y`m (ˆ r) .

`m

and



~

eik·~r

∗

=

X

ˆ ∗ (ˆ c∗` (kr)Y`m (k)Y `m r ) =

`m

X

∗ ˆ c∗` (kr)Y`m (k)Y`m (ˆ r) ,

`m

where we’ve used the result of part (a). Comparison of the two reveals c∗` = (−1)` c` =⇒ c` ∼ i` X

ˆ gives Y`m (ˆ 35.24 Orienting zˆ along k z) = coordinates, e

i~ k·~ r

= eikr cos γ = 4π

` ∞ X X

p 2`+1 4π

δm0 . This makes γ the polar angle in spherical

"r i` j` (kr)

#

2` + 1 δm0 Y`m (ˆ r) 4π

`=0 m=−`

= 4π

∞ X

r `

i j` (kr)

2` + 1 Y`0 (ˆ r) 4π

`=0

= 4π

∞ X

r `

i j` (kr)

2` + 1 4π

!2 P` (cos γ) =

`=0

∞ X

i` (2` + 1)j` (kr)P` (cos γ).

`=0

35.25 From Eqn. (35.71), we get (−i)` j` (0) = 2

Z

1

P` (u) du . −1

But the Legendre polynomials all integrate to zero except for P0 (u) = 1. (This can be seen from the Rodrigues formula for Legendre polynomials, which shows that due to the factor of (1 − x2 ), the integrals vanish between ±1.)

274

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Fourier Transforms

36 36.1

F −1 [F [f ] ] = F −1



1 = √ 2π

Z

∞



Z

=

∞



−∞

∞

 e+ikx

1 2π

0

e−ikx f (x0 ) dx0

−∞

=

Z−∞ ∞

Z

1 √ 2π

Z

1 √ 2π

∞

Z

∞



0

e−ikx f (x0 ) dx0 dk

−∞



0

eik(x−x ) dk f (x0 ) dx0

−∞

δ(x − x0 ) f (x0 ) dx0 = f (x) ,

−∞ F −1

F

So indeed, f −−−−−→ f˜ −−−−−−−→ f .

36.2 For gaussian f (x) = exp(−ax2 ): 1 f˜(k) = √ 2π 1 = √ 2π

Z

∞

Z−∞ ∞

2

e−ax e−ikx dx 2

e−ax

−ikx

dx

−∞

2 1 = √ e−k /4a 2π

Z

∞ 2

e−a(x+ik/2a) dx

(completing the square)

−∞

2 1 = √ e−k /4a 2a

which is also a gaussian.

36.3 1 f˜(k) = √ 2π

Z

∞

1 f (x) e−ikx dx = √ 2π −∞

Z

r

a

e−ikx dx =

−a

2 sin(ka) . π k

As a → ∞,

r f˜(k) =

√ 2 sin(ka) lim = 2π δ(k) . π a→∞ k

So a large distribution in x (i.e., a large slit) requires a narrow distribution in k.

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36.4 For a > 0, we can all-but read the result from Table 36.1:

Z

∞

e−a|x| cos bx dx =

∞

Z





F e−a|x| F [cos bx ] dk

−∞

−∞

∞

Z

"r

2 a π a2 + k2

= −∞

=

#

hq

π (δ(k + b) + δ(k − b)) dk 2

i

2a , a2 + b2

as you can readily verify by directly integrating in x.

36.5

Z

∞

f ∗ (x)g(x) dx =

hf |gi = −∞

Z

∞



−∞ ∞

Z

∞

Z

1 √ 2π

f˜(k) e−ikx dk



−∞ ∞

Z

1 √ 2π

−∞

f˜∗ (k)˜ g (k0 )

=

∗ 

1 2π

Z

∞

ei(k−k

0

 )x

dx

Z

∞

0

g˜(k0 ) e−ik x dk0

 dx

−∞

dk dk0

−∞

f˜∗ (k)˜ g (k0 )δ(k − k0 ) dk dk0

= −∞ ∞

Z

f˜∗ (k)˜ g (k) dk .

= −∞

36.6 The different signs in (36.5) are irrelevant to these identities — after all, the integrals are over all positive and negative k and x. But the difference is consistent for our Fourier transform convention:



f (x) = F −1 [F [f ] ] = F −1

1 = √ 2π

Z

1 √ 2π ∞

Z

Z

which holds only if

1 2π

R∞ −∞



−∞

1 √ 2π



Z

1 = f (x ) 2π −∞ 0

0

f (x0 ) e−ikx dx0



−∞

∞

∞

Z

∞

0



f (x0 ) e−ikx dx0 eikx dx

−∞ ∞ ik(x−x0 )

e



dk dx0

−∞

0

eik(x−x ) dk = δ(x − x0 ).

36.7

Z

∞

hf |f i = hf |



Z

−∞ ∞

Z = hf |



Z

Z

−∞ ∞

dk|kihk| |f i =

Z

36.8 From Table 36.1 lists F (x2 + a2 )−1



=

π 2a2

2

Z Z

∞

|f (x)|2 dx

−∞ ∞

|hk|f i| dx = −∞

p

|hx|f i|2 dx =

−∞ ∞

hf |kihk|f idk = −∞



∞

hf |xihx|f idx =

−∞

276

∞

dx|xihx| |f i =

|f˜(k)|2 dk

−∞

e−|k|a for a > 0, allowing straightforward

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

application of Plancherel’s theorem:

Z

∞

−∞

dx = (x2 + a2 )2

∞

Z

Z−∞ ∞

1 1 dx (x2 + a2 ) (x2 + a2 )

h

F

= −∞ ∞

Z =

(x2

q

−∞

=

π 2a2

Z

1 + a2 )

i h F

π −|k|a e 2a2

(x2

q

∞

π a2

e−2|k|a dk =

−∞

∞

π e−2kx =− 2 a 2a

1 + a2 )

i

dk

π −|k|a e 2a2



dk

∞

Z

e−2ka dk

0

π = . 2a3

0

36.9 With F [f (x)] = f˜(k) — (a)

  1 F e−bx f (ax) = √ 2π

Z

1 = √ 2π

Z

(b) F

[f ∗ (x) ]

∞

f (ax)e−bx e−ikx dx

−∞ ∞ (−ik−b)y/a

f (y)e −∞

dy 1 1 = √ a a 2π

Z

∞




f (y)e

−i k−ib y a

dy =

−∞

1 ˜ k − ib f a a



, f˜∗ (k); rather:

1 F [f (x) ] = √ 2π ∗

Z

∞ ∗

−ikx

f (x)e

 dx =

−∞

 =

1 √ 2π

Z

1 √ 2π

Z

∗

∞ +ikx

f (x)e −∞ ∞

dx

∗

f (x)e−i(−k)x dx

= f˜∗ (−k)

−∞

Note that for real f (x), F [f ∗ (x) ] = f˜(k), confirming the symmetry listed in Table 36.2. (c) 1 f˜(k) = F [f (x) ] = √ 2π

Z

∞

1 f (x)e−ikx dx = √ 2π −∞

Z

1 = √ 2π

Z

∞

f (−x)e−ik(−x) dx

−∞ ∞

f (x)e−i(−k)x dx = f˜(−k)

−∞

Note that for f (x) either real or imaginary, this result together with the previous one confirm the symmetries listed in Table 36.2.

36.10 Integrating by parts, and noting that the surface term vanishes, F −1



df˜ dk



1 = √ 2π

Z

∞

−∞

df˜ ikx 1 e dk = − √ dk 2π

Z


d f˜(k) eikx dk dk −∞

1 = −ix √ 2π



36.11 (a) F −1 e−ak

2



h

= F −1 exp



−k2 2/(2a)

i

=

√1 2a

e−x

2

∞

Z

∞

f˜(k)eikx dk = −ixf (x) .

−∞

/4a

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277





(b) F x e−x (c) F

h

2

/2



1 2 x2 e− 2 a x

(d) F −1

h

e2ik 1+k2

i

2



d = −i dk F e−x

i 2

=

d 2 −i dk




= F −1



1 1+k2

/2

h



d −k = −i dk e

−1 a2 x 2

F e



i 2

pπ

=

/2

= ike−k

d2 − dk 2

 1 a

2

/2

2 − k e 2a2

 1 a3

=



1−

k2 a2



2 − k2 2a

e

e−|x−2|

2

x+2

=

2

2

d 2 is an operator expression, and so should be considered as acting on an (e) Hx = dx 2 − x arbitrary function f (x):

 F [Hx f (x) ] = F

d2 − x2 dx2





 =F

f (x)

d2 f dx2







− F x2 f (x) = −k2 f˜(k) +

d2 f˜ = Hk f˜(k) dk2

So the operator H is its own Fourier transform.

36.12 (a) Since θ0 (x) = δ(x), and F [f 0 (x) ] = ikF [f (x) ], F [θ(x) ] =

1 1 1 F [δ(x) ] = √ , ik 2π ik

valid for all k , 0. (b) The DC average of the step function is 1/2. So the contribution from this zero-frequency component is F [1/2 ] =

1 √ 1 F [1 ] = 2π δ(k) = 2 2

q

π δ(k) . 2

36.13 Using the convolution theorem, 1 ˜ ˜ ˜ ψ(ω) = F [θ(t)φ(t) ] = √ Θ(ω) ∗ φ(ω) 2π

Z

1 = √ 2π 1 = 2πi

Z

d2 −x dx2



∞

1 1 + √ 2π iω 0

h

−∞ ∞ ˜

−∞

q

π ˜ − ω 0 )dω 0 δ(ω 0 ) φ(ω 2

i

1˜ φ(ω − ω 0 ) dω 0 + φ(ω) ω0 2

36.14 Solving Airy’s equation:

 F

 f (x)

= (ik)2 f˜(k) − i

df˜ =0 dk

So f˜0 = ik2 f˜ — or, up to an overall multiplicative constant, df˜ = ik2 dk f˜

−→

3 f˜(k) = eik /3 .

Thus

  1 1 1 Ai(x) = √ f (x) = √ F −1 f˜(k) = 2π 2π 2π

Z

∞

ei(k

3

/3+kx)

dk

−∞

36.15 (a) The Fourier transform of the Dirac comb sa (x) =

∞ X

δ(x − na) ,

n=−∞

278

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is: 1 s˜(k) = √ 2π 1 = √ 2π 1 = √ 2π

Z

∞ X

∞

δ(x − na) e−ikx dx

−∞ n=−∞

Z ∞ X n=−∞ ∞ X

∞

δ(x − na) e−ikx dx

−∞

e−inka =

√

2π δ (ka) ,

n=−∞

where in the last step we identified the Fourier series of the delta function. Fourier series of course are inherent periodic; this one is periodic in k with period 2π/a. (b) The results of (a) reveal s˜ to be a periodic string of delta functions over all k, √

s˜(k) =

∞ X

2π

δ (ka − 2πm)

m=−∞

√

∞ X

2π a

=



δ k−

2πm a



,

m=−∞

which is (36.46).

36.16 ∞ X

Z f (αn) =

f (x)

n=−∞

X

δ(x − αn) dx

n

Z =

Z

f˜(k) s˜α (k)dk

f (x)sα (x)dx =

Z

f˜(k)

= √

2π α

=

√

Z



2π s2π/a (k) α

dk √

f˜(k)

X

δ(k − 2πm/α) =

m

2π α

X

f˜(2πm/α) ,

m

where we’ve used Plancherel’s theorem in the second line.

36.17 For f (x) = e−b|x| , b > 0, f˜(k) = gives

p2

∞ X

b , π k2 +b2

e−b|n| = 2

n=−∞

the Poisson summation formula in Problem 36.16 ∞ X

b (2πm)2 + b2

m=−∞

Manipulating the left-hand side into a geometric series reveals the hyperbolic cotangent: ∞ X n=−∞

e−b|n| = 1 + 2

∞ X

e−bn

n=1

= −1 + 2

∞ X

e−bn = −1 +

2 1 + e−b eb/2 + e−b/2 = = b/2 = coth(b/2) . 1 − e−b 1 − e−b e − e−b/2

n=0

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279

Thus ∞ X

coth(b/2) = 2b

∞ X

1 b = (2πm)2 + b2 2

m=−∞

1 (πm)2 + (b/2)2

m=−∞

So, relabeling b/2 → x,

coth(x) = x

∞ X

1 x2 + π 2 m2

m=−∞ ∞

=

X 1 1 + 2x x x2 + π 2 m2 m=1

=

∞ X

1 + x

2x 2 1 − =− + x2 + π 2 m 2 x x

m=0

∞ X

2x x2 + π 2 m2

m=0

36.18 ~ dψ ~ d − (xψ) i dx i dx h i dψ dψ = −i~ x −ψ−x = i~ . dx dx

[X, P ]ψ(x) = XP ψ − P Xψ = x





36.19 A straightforward lowest-order Taylor expansion gives




~ r) = 1 + ˆ ~ ψ(~ ψ(~ r + ˆ n) = ψ(~ r) + ˆ n · ∇ψ(~ n·∇ r) . Direct comparison with Tn ()ψ(~ r) = (1 + iˆ n·p ~/~) ψ(~ r) ~ A finite translation by a shows that the generator of translation is the operator p ~ = −i~∇. can be accomplished by m successive translations of a/m; in the large-m limit, each is an infinitesimal translation. Thus Tn (a) = lim [Tn (a/m)]m = lim m→∞

h

1+

m→∞

iaˆ n·p ~/~ m

im

n·~ p/~ = eiaˆ .

36.20

Z

∞

f˜(u) g˜(w − u) du

F [f ] ∗ F [g] = −∞ ∞

Z

Z

∞

= −∞

1 = 2π =

1 2π

Z

−∞

Z

dx f (x) e−iux √ 2π

 Z

∞ −iwy

Z

∞

−∞ ∞

dx dy f (x) g(y)e

Z−∞ ∞

e

−iu(x−y)

 du

−∞

dx dy f (x) g(y)e−iwy · 2πδ(x − y)

−∞

∞

=



dy g(y) e−i(w−u)y du √ 2π

dx f (x) g(x)e−iwx =

√

2π F [f g]

−∞

280

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36.21 Since the product of sin a and cos b is the sum of sin(a + b) and sin(a − b), the Fourier transform of the product should reveal only these values in the frequency spectrum. Using

q

π [δ(k + 8π) − δ(k − 8π)] 2 q √ π g˜(k) = F [8 + 6 cos(2πx) ] = 8 2π δ(k) + 6 [δ(k + 2π) + δ(k − 2π)] 2

f˜(k) = F [4 sin(8πx) ] = 4i

the Fourier spectrum of the product is F [f (x)g(x) ] = f˜ ∗ g˜ 1 = √ 2π =i

q

Z

f˜(k)˜ g (w − k)dk

π [12 (δ(w + 10π) − δ(w − 10π) + δ(w + 6π) − δ(w − 6π)) + 32 (δ(w + 8π) − δ(w − 8π))] 2

revealing the expected frequencies.

p

2

2

p

a −ax /2 b −bx /2 e and g = e are both gaussians, their Fourier transforms 36.22 Since f = 2π 2π are also gaussians — making the Fourier transform of their convolution relatively simple:

F [f ∗ g ] = where

1 c

=

1 a

√

2 2 1 2 1 1 2πF [f ] F [g ] = √ e−k /2a e−k /2b = √ e− 2 k /2c , 2π 2π

+ 1b . Thus 1 2 1 f ∗ g = √ F −1 e− 2 k /2c 2π

h

i

=

q

c − 1 cx2 e 2 . 2π

The convolution of gaussians is a gaussian.

36.23 – commutative:

Z

∞

[f ∗ g](t) =

f (u)g(t − u) du −∞ ∞

Z

f (t − w)g(w) dw,

=

w =t−u

−∞

= [g ∗ f ](t) – associative:

Z

∞

[(f ∗ g) ∗ h](t) =

[f ∗ g](u)h(t − u)du −∞

Z Z

 f (v)g(u − v)dv h(t − u) du

=

Z

Z f (v)

=

 g(u − v)h(t − u)du

dv

Z =

f (v)[g ∗ h](t − v) dv = f ∗ [(g ∗ h)](t)

36.24 The convolution of top hat functions

 ha (u) ≡

1, 0,

|u| < a/2 |u| > a/2

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281

is

Z

∞

ha (u)ha (t − u)du .

[ha ∗ ha ](t) = −∞

For t < −a and t > a, the two top hats in the integrand do not overlap — so their product makes no contribution to the integral: the convolution for |t| > a vanishes. As t grows just beyond t = −a, the functions begin to overlap, the area under the curve of the product growing linearly with t until they completely overlap at t = 0. The convolution at t = 0 is a, the area under the curve of ha . As t grows past zero, the overlap decreases linearly as the same rate until it vanishes at t = a. Thus the convolution returns the triangle function (as direct calculation verifies)

( [ha ∗ ha ](t) =

a+t , a−t , 0,

−a < t < 0 0 0 f (x) =

1 2

0

Z

ey e−(x−y) dy +

x

Z

−∞

e−y e−(x−y) dy +

∞

Z



e−y e−(y−x) dy =

x

0

1 (1 + x) e−x . 2

Similarly, for x < 0. f (x) =

1 2

Z

x

ey e−(x−y) dy +

Z

−∞

0

ey e−(y−x) dy +

Z

x

∞



e−y e−(y−x) dy =

0

1 (1 − x) e−x , 2

confirming (36.64).

36.27 Using the convolution theorem, F [f ∗ f ] =

√

h

2

2π f˜2 (k) = F e−x

i

2 1 = √ e−k /4 2

−→

f˜(k) = ±

2 1 e−k /8 (4π)1/4

So





f (x) = F −1 f˜(k) = ±

282

2 1 F −1 e−k /8 (4π)1/4

h

i

=±

 1/4 4 π

e−2x

2

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36.28 This is largely the derivation of (36.87) in reverse — but with variable change w = −u + v:

Z

∞

F



∗

f (u)f (w + u) du −∞

1 = √ 2π 1 = √ 2π = =

where the factor of

√

√ √

Z

∞ −ikw

Z

∞

f (u)f (w + u) du dw

e

Z−∞ ∞



∗

−∞

eik(u−v) f ∗ (u)f (v) du dv

−∞

 2π

1 √ 2π

Z

∗ 

∞ −iku

e

f (u) du

−∞

1 √ 2π

Z

∞ −ikv

e

 f (v) dv

−∞

2π |f˜(k)|2 ,

2π arises from our 2π-convention.

36.29 f (u)g(t − u)du is the fraction of the amount f (u)du left at time t > u. So if the hole is plugged, it must be that g(t) → 1. And indeed,

Z A(t) = [f ∗ g](t) = [f ∗ 1](t) =

t

f (u)du, 0

as it should.

36.30 Coulomb’s law is the convolution of the charge density ρ(~ r) with the 1/r Coulomb potential of a point charge, Φ(~ r) =

1 4π0

Z V

ρ(~ r 0) 3 0 d r . |~ r−~ r 0|

Rather than the diminution of water over a time t−u, this convolution accounts for the decrease in potential due to the distance |~ r−~ r 0 |.

36.31 Q(t) = I ∗ e−t/τ : For 0 < t < τ , in a time du charge I(u)du is added to the capacitor as a fraction e−(t−u) leaks off — so Q(t ≤ τ ) =

1 V0 Vin (t) ∗ e−t/τ = R R

Z

t

e−(t−u)/τ du =

0

V0 τ (1 − e−t/τ ) R

= CV0 (1 − e−t/τ ) = Qmax

1 − e−t/τ . 1 − 1/e

where Qmax = Q(τ ). Though no more charge is added for t > τ , it does continue to leak off — so Q(t > τ ) =

V0 R

Z 0

τ

e−(t−u)/τ du =

V0 τ (e − 1)e−t/τ x = Qmax e−(t−τ )/τ . R

By the convolution theorem, the convolution of I with a damped exponential is the (inverse) Fourier transform of the product of I˜ and the Fourier transform of the damped exponential, which is proportion to (1 + ω 2 )−1 . Thus any large frequencies in the signal are filtered out in the product; only low frequencies pass through. √ 36.32 For a constant force f0 = F/m, f˜ = F [f0 ] = 2πδ(t). Using this in (36.70) gives   √ 2πδ(t) 1 −1 u(t) = F = 2 . ω02 − ω 02 + 2iβω 0 ω0 ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

283

This result is not unfamilar: When a mass m on a spring of stiffness k is suspended vertically, it experiences a constant force mf0 ≡ mg, stretching to a new steady-state equilibrium position mg/k = f0 /ω02 .

36.33 (a) By the convolution theorem [f ∗ ha ](t) =

√



˜ a (ω) 2πF −1 f˜(ω)h



.

˜ a (ω) is proportional to sinc(aω/2), which is dominated by small frequencies — or more But h generally, by a small bandwidth |ω − ω0 |. In the product with f˜, it thus acts as a window, preferentially allowing low-ω components through. So only these low frequencies are passed along when Fourier transforming back to t. (b) From the property of Fourier transforms, a derivative introduces a factor of ω. This essentially enhances the high-frequency noise at the expense of the low-frequency signal. [Note that a smooth signal will not have arbitrarily high frequency components.] So it’s wiser to smooth a signal with a low pass filter before taking a derivative.

36.34 (a) The Fourier transform from x → k, Fx [T (x, t)] = T˜(k, t), takes the derivative ∂x2 → −k2 . So in k and t, the heat equation becomes ∂t T˜(k, t) = −αk2 T˜(k, t) . (b) With T˜(k, 0) = f˜(k), the solution to the Fourier-transformed differential equation is a simple damped exponential, 2 T˜(k, t) = f˜(k)e−αk t .

(c) Rather than directly take the inverse Fourier transform, a simpler expression can be found using the convolution theorem. That’s because the exponential in t is a gaussian in k — so its inverse-Fourier x transform is a gaussian in x, 2 2 1 e−x /4αt . Fk−1 [e−αk t ] = √ 2at

Thus 2

T (x, t) = Fk−1 [T˜(k, t)] = Fk−1 [f˜(k)e−αk t ] 2 1 1 1 = √ f (x) ∗ √ e−x /4αt = √ 2π 2at 4παt

Z

∞ 2

T (u, 0) e−(x−u)

/4αt

.

−∞

As we’ve seen, convolving with a frequency-limited function like a gaussian smooths out a signal. So this integral essentially spreads the initial temperature distribution in x to get a smoother distribution over time. As expected of heat flow.

36.35 If

Z

∞

c(t) = [f ∗ g](t) =

f (u)g(u − t)du , −∞

then shifting t in the argument of f gives

Z

∞

Z

∞

f (u − τ )g(t − u)du = −∞

f (v)g(t − τ − v)dv = c(t − τ ) . −∞

36.36 F [φ ] = λφ.

284

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(a) To find the eigenvalue, start with F 2 :

Z 

1 F [F [f (x) ] ] = √ 2π

Z  =

1 2π



Z

1 √ 2π

f (x) e−ikx dx e−iky dk



Z

e−ik(x+y) dk f (x) dx

Z f (x) δ(x + y) dx = f (−y) .

=

So then F 4 = F [F [f (−y) ] ] = f (x), and λ4 = 1. So λ = ±1, ±i — all pure phases, as expected of a unitary operator. (b) Differentiation and multiplication are interchanged by F . Specifically, (d/dx)2 ↔ −k2 and x2 ↔ −(d/dk)2 . So:



d2 + x2 dx2

F [H ] = F −



d2 ≡H. dk2

= k2 −

2

2

n

2

i

2




d (c) The functions φ(x) = e−x /2 Hn (x), where Hn (x) = (−1)n ex dx e−x , are eigenfuncn tions of H. To show that they’re also eigenfunctions of F , integrate by parts and complete the square to swap the x derivatives for k derivatives:

1 F [φn ] = √ 2π

Z

e−x

1 = (−1) √ 2π n

1 = √ 2π =e

Z

= ek

2

/2

k2 /2

=e

dn dxn

= ek

2

/2

h

d dk

in

= ek

2

/2

h

d dk

in

i

2

1

e2x

Z

in

(−1)n ex

1



d i dk

h

h

e2x

2

−ikx



1 √ 2π

Z 

2

e−k

2

/2







e−x

e−x



2



e−x dx



e−x dx

1

2

1

2

e 2 (x−ik)

e 2 (x−ik)

Z

dn dkn

2

e−x

/2

h

= (−i)n e−k

2

/2

Hn (k) = (−i)n φn (k) .





e−k

2

2

/2 −ikx

2

2

dx

e−x dx

= (−i)n e−k

(−1)n ek



2

2

1 √ 2π

2

e−ikx dx

e−x dx

1

in 

e−k

dn dxn

e 2 (x−ik)

d dk

i

2

dn dxn

−ikx

dn dxn

Z h

1 √ 2π

i

/2

Z

1 √ 2π

+k2 /2

2

2

e

dx

i



(d) From the analysis in (a), we know that F f˜(x) = f (−k). Thus





F f (x) + f˜(x) = f˜(k) + f (−k) = f˜(k) + f (k) for f even.

36.37 (a) For N slits of negligible width separated by a distance d, the mask function is just a series

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285

of delta functions md (x) =

N X

δ(x − nd) ,

n=1

whose Fourier transform is 1 m ˜ d (k) = √ 2π

N X n=1

e−ikd e−inkd = √ 2π e−ikd = √ 2π

N −1

X

e−inkd

n=0



1 − e−iN kd 1 − e−ikd

1

e−i(N + 2 )kd = √ 2π

"

sin sin


#
.

1 N kd 2 1 kd 2

So the intensity of the diffraction pattern is proportional to |m| ˜ 2,

sin I∼

sin


2
.

1 N kd 2 1 kd 2

The intensity of the central peak emerges in the k → 0 limit, I(0) ∼ N 2 . (b) For one large slit of fixed width a = N d, the large-N limit is lim

N →∞ N d fixed

I(k) = I(0) =


2


1 N kd 2 1 kd 2

lim

1 sin N 2 sin

lim

1 sin (ka/2) 2 sin (ka/2) 2 = ka/2 = sinc2 (ka/2). N 2 sin (ka/2N )

N →∞ N d fixed

N →∞ N d fixed









(c) For a grating with a uniform array of d slits of width a < d, the intensity is given by the convolution of the mask md with the top hat function ha :

2



2



˜ (k)h˜ (k) I(k) ∼ F [md (x) ∗ ha (x) ] = m a d

-3

-2

-1

1

2

3

-15

-10



sinc2 (ka/2)

1 N kd 2 1 2 sin 2 kd

sin2 =

-5

5

10

15

36.38 (a) For g(t) = Θ(t)f (t)e−σt ,

286

1 g˜(ω) = F [g(t) ] = √ 2π

Z

1 = √ 2π

Z

∞

Θ(t)f (t)e−σt e−iωt dt

−∞ ∞

f (t)e−st dt ,

0

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where s = σ + iω. Since |f (t)| ∼ eβt , the integral converges for √ σ > β — i.e., β. Thus the Laplace transform of f (t) is proportional to g˜: F (s) = 2πF [g(t) ]. (b) From the inverse Fourier transform Θ(t)f (t)e

∞

Z

1 = √ 2π

−σt

g˜(ω) eiωt dω

−∞

we get 1 Θ(t)f (t) = √ 2πi

Z

1 2πi

=

Z

∞

g˜(ω) e(σ+iω)t idω

−∞ σ+i∞

F (s) est ds

σ−i∞

36.39 (a) ∞

Z

L [cos ωt] =

cos(ωt) e−st dt

0

1 = 2 1 = 2

∞

Z




eiωt + e−iωt e−st dt

h0

1 1 + s − iω s + iω

i

=

s , s2 + ω 2

0 .

(b) Using the result of part (a): L [sin2 ωt] =

∞

Z

sin2 (ωt) e−st dt

0 ∞

1 = 2

Z

=

1 2

h

h

1 √

i

[1 − cos(2ωt)] e−st dt

0

s 1 − 2 s s + 4ω 2

i

=

s3

2ω 2 , + 4ω 2 s2

0 .

(c) L

t

∞

Z

t−1/2 e−st dt

= 0

1 = √ =

Z s

q

∞

u−1/2 e−u du ,

u = st

0

π , s

0 .

Note that this agrees with the entry in Table 36.4, with n! → Γ(α + 1) for non-integer α.

36.40 (a) Using a partial fraction decomposition 1 1 1 1 = − + , s(s2 + 4s + 3) 3s 2(s + 1) 6(s + 3) direct appeal to Table 36.4 yields L −1 One can also use L −1

1 s

h

1 s(s2 + 4s + 3)



F (s) =

Rt 0

i

=

1 1 1 − e−t + e−3t . 3 2 6

f (τ )dτ (Table 36.5).

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287

(b) Using a partial fraction decomposition 1 1 = (s − 1)(s2 + 1) 2

h

1 1 s − 2 − 2 s−1 s +1 s +1

i

,

Table 36.4 reveals

h

L −1

1 (s − 1)(s2 + 1)

i

=

 1 t e − sin t − cos t . 2

(c) Using entries of both Tables 36.4 and 36.5, L

−1



e−2s (s − 2)2



= L −1

h

1 (s − 2)2



i

= (t − 2)e−2(t−2) .

= te−2t

t−2

t−2

Or by convolution (last entry in Table 36.5): L −1



e−2s (s − 2)2







= L −1 e−2s ∗ L −1

h

1 (s − 2)2

Z

i

∞

=

δ(t − τ − 2)τ e−2τ dτ = (t − 2)e−2(t−2) .

0

(d) Using the results of Problem 3.12, Table 36.4 immediately reveals

h

L −1 tan−1

a s

 i

h

= L −1

1 ln 2i



s + ia s − ia

i

 1 1 1  iat e − e−iat = sin at . 2i t t

=

36.41 L [f ]L [g] =

∞

Z

−su

f (u) e

du

g(v) e

0 ∞

Z

∞

 Z

−sv

 dv

0

f (u)g(v) e−s(u+v) du dv .

= 0

Substituting v with w = u + v, L [f ]L [g] =

∞

Z

∞

Z

 f (u)g(w − u) du

0

e−sw dw = L [f ∗ g] .

0

36.42

Z L

t

 f (τ )dτ

= L [f ∗ 1] = F (s) ∗ L [1] =

0

1 F (s) . s

36.43 Given a real function f (t) = eat : (a) For a < 0, the Fourier transform does not exist for t < 0. Appending a step function Θ(t), however, yields f˜(ω) = F (s = iω). (b) For a > 0, the Fourier transform does not exist for t > 0. Appending a step function Θ(−t), however, yields f˜(ω) = F (s = −iω). (c) A step function is not needed for a = 0, but the Fourier transform has a δ function contribution at ω = 0 not found in the Laplace transform.

36.44 (a) With L [cos ωt] = L −1



s , s2 +ω 2

Mathematica gives

s (s2 + ω 2 )(s2 + 2βs + ω02 )

 =

sω 2 − ω02 (2β + s) 4β 2 ω 2

+

ω4

− 2ω02 ω 2 + ω04




s(2β + s) + ω02




Choosing simple parameter values returns the follow plot — revealing that it’s not precisely

288

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a steady-state solution: Plot[InverseLaplaceTransform[Ust[s], s, t] /. {β  1, ω0  2, ω  1.5}, {t, 0, 20}, PlotRange  All] 0.3 0.2 0.1

Out[]=

5

10

15

20

-0.1 -0.2 -0.3

(b) In[]:=

g[s] = Apart[Ust[s], s] [[1]] // Simplify s ω20 - ω2 (s - 2 β)

Out[]=

In[]:=

4 β ω + ω4 - 2 ω20 ω2 + ω40  s2 + ω2  2

h[s] = Apart[Ust[s], s] [[2]] // Simplify s ω2 - ω20 (2 β + s)

Out[]= In[]:=

2

4 β ω + ω - 2 ω20 ω2 + ω40  s (2 β + s) + ω20  2

2

4

Plot[InverseLaplaceTransform[g[s], s, t] /. {β  1, ω0  2, ω  1.5}, {t, 0, 10}, PlotRange  All] 0.3 0.2 0.1

Out[]=

2

4

6

8

10

-0.1 -0.2 -0.3 In[]:=

Out[]=

InverseLaplaceTransform[h[s], s, t] // Simplify 

-t 

β2 -ω2 +β

0

ω20 2

In[]:=

β

2

- ω20

ω2 β + β

2

- ω20 2

2

β2 - ω20  

2t

4

4 β ω + ω

β2 -ω2

0

- 2 ω20

2t

β2 -ω2

0

+1 -β

+1 +β  2

ω

2t

2t

β2 -ω2

0

β2 -ω2

-1

0



+ ω40 

Plot[InverseLaplaceTransform[h[s], s, t] /. {β  1, ω0  2, ω  1.5}, {t, 0, 10}, PlotRange  All]

2

4

6

8

10

-0.05 Out[]=

-0.10

-0.15

-0.20

Clearly, the transient piece of Ust comes not from the harmonic driving force, but from the damped oscillator itself. As one would expect.

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289

(c) The Fourier solution in Example 36.13 does not pre-multiply the driving force by a step function, thereby omitting an actual start time.

36.45 Integrating by parts: L

h

df dt

i

∞

Z = 0

Z

df −st ∞ e dt = f (t)e−st − dt 0



Z

∞

f (t) 0

d −st e dt dt

∞

f (t)e−st dt = sF (s) − f (0) .

= −f (0) + s 0

Using this result and integrating by parts again:

 L

d2 f dt2



∞

Z = 0

d2 f −st e dt = −f˙(0) + s dt2

∞

Z

df −st e dt dt

0

∞

= −f˙(0) + sf (t)e−st

+s

2

Z

0

∞

f (t)e−st dt = s2 F (s) − sf (0) − f˙(0) .

0

The emerging pattern suggests the general formula is L

h

dnf dtn

i

= sn F (s) − sn−1 f (0) − sn−2

dn−1 df d2 f − sn−3 dt2 − . . . − dtn−1 f . dt 0 0 0







36.46 Using the expressions in (36.120) [see also Problem 36.45 ],

 L

d2 f df −3 + 2f dt2 dt



= s2 F (s) − sf (0) − f˙(0) − 3 (sF (s) − f (0)) + 2F (s)




= s2 − 3s + 2 F (s) − (s − 3)f (0) − f˙(0) . (a) g(t) = 0 ↔ G[s] = 0. Then with f (0) = 1, f˙(0) = 0, s−3 ←→ f (t) = 2et − e2t s2 − 3s + 2

F (s) =

(b) g(t) = t ↔ G[s] = 1/s2 . Then with f (0) = 1, f˙(0) = −1, F (s) =

1 s2 − 3s + 2

h


1 1 + s − 4 ←→ f (t) = 8et − 7e2t + 2t + 3 s2 4 i

(c) g(t) = e−t ↔ G[s] = 1/(s + 1). Then with f (0) = 0, f˙(0) = 1, F (s) =

1 s2 − 3s + 2

h


1 1 + 1 =←→ f (t) = 8e3t − 9e2t + 1 e−t s+1 6 i

36.47 F (s) = L [f (t)] =

Z

∞

f (t)e−st dt

0

Z

T

f (t)e

= 0 ∞

=

Z dt +

2T −st

f (t)e T

XZ n=0

290

−st

Z

3T

dt +

f (t)e−st dt + . . .

2T

(n+1)T

f (t)e−st dt

nT

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Since each if these integrals is over a single period of f , we can recast the integral over u = t−nT and extract a geometric series:

F (s) =

∞ Z X n=0

∞ X

T −s(u+nT )

f (u)e

du =

0

−snT

!Z

Z sF (s) = s

R∞ 0

f (u)e−su du =

0

n=0

36.48 (a) lims→∞ F (s) = lims→∞ (b)

T

e

Z

1 1 − e−sT

T

f (u)e−su du .

0

f (t)e−st dt = 0.

∞

f (t)e−st dt = −

∞

Z

f (t) 0

0

d −st e dt dt ∞

∞ Z +

= −f (t)e−st

0

0

df −st e dt = f (0) + dt

Z 0

∞

df −st e dt dt

For non-divergent df /dt — which, from Table 36.5, requires that sF (s) be finite in the limit — the integral vanishes as s goes to infinity, leaving lim sF (s) = f (0) = lim f (t) . s→∞

t→0

(c) If instead s goes to zero, lim sF (s) = f (0) + f (∞) − f (0) = f (∞) s→0

The identity holds as long as f has a finite limit (so, e.g, it doesn’t hold for et and sin t).

36.49 For F (s) =

s , s2 −a2

the inverse Laplace transform 1 f (s) = 2πi

Z C

sest ds , s2 − a2

where the vertical portion of C parallel to the imaginary s axis must lie to the right of the poles at ±a — that is, we must have a. For t > 0, the path can be closed along a semicircle Cr in the left half s-plane,

Z

Z F (s) e ds ≤ |F (s) est ds| Cr Cr Z −π/2 st

|F (reiϕ )| er cos ϕt rdϕ .

= π/2

Since cos ϕ ≤ 0 for ϕ in the left half-plane between ±π/2, the contribution of Cr in the r → ∞ limit vanishes as long as |F (s)| → 0

as

|s| → ∞ .

For t < 0, we’ll need to close the contour in the right half-plane so that cos ϕ ≥ 0 yields a decaying exponential.

36.50 The transformed equations (36.122) follow directly from (36.120b). (a) The conditions q1 (0) = q2 (0) = q˙1 (0) = 0 and q˙2 (0) = v0 , lead directly to (36.124), Q+ (s) =

v0 s2 + k/m

Q− (s) = −

v0 . s2 + 3k/m

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291

Then, with ω12 = k/m and ω22 = 3k/m, q± (t) = L −1 [Q± (s)] =

v0 sin (ω1,2 t) . ω1,2

Inverting gives (36.125), q1,2 (t) =

1 1 1 v0 sin ω1 t ∓ sin ω2 t 2 ω1 ω2

h

i

.

(b) The conditions q1 (0) = a, q2 (0) = b, and q˙1 (0) = q˙2 (0) = 0 give 2k Q1 (s) + m 2k s2 Q2 (s) − bs = − Q2 (s) + m

s2 Q1 (s) − as = −

k Q2 (s) m k Q1 (s) . m

so that (a + b)s s2 + k/m

Q+ (s) =

Q− (s) =

(a − b)s . s2 + 3k/m

Then q± (t) = (a ± b) cos ω1,2 t . Inverting, q1,2 (s) =

1 [(a + b) cos ω1 t ± (a − b) cos ω2 t] . 2

36.51





d2 d + 2β + ω02 u(t) = f0 δ(t − t0 ) dt2 dt

where a quiescent oscillator has u(0) = u(0) ˙ = 0. Then with L [δ(t − t0 )] = e−st0 , (36.128) becomes U (s) = where s± = −β ± iΩ, with Ω =

f0 e−st0 f0 e−st0 = . (s − s+ )(s − s− ) s2 + 2βs + ω02

p

ω02 − β 2 . We’ll solve for u(t) two ways:

– Convolution: For U (s) = F (s)G(s), with F (s) = f0 e−st0 and G(s) =

1 , (s−s+ )(s−s− )

u(t) = f (t) ∗ g(t) where f (t) = f0 δ(t − t0 ) and g(t) =

1 1 1 L −1 − (s+ − s− ) s − s+ s − s−

h

i

 s t s t 1 e + −e − (s+ − s− )  1 −βt 1 −βt  iΩt = e e − e−iΩt = e sin (Ωt) . 2iΩ Ω =

Then u(t) = f0 δ(t − t0 ) ∗

f0 −β(t−t0 ) 1 −βt e sin (Ωt) = e sin [Ω(t − t0 )] Ω Ω

– Residues: The Bromwich integral is 1 u(t) = 2πi

292

Z

σ+i∞

σ−i∞

f0 es(t−t0 ) 1 ds = 2 2πi s + 2βs + ω02

Z

σ+i∞

σ−i∞

f0 es(t−t0 ) ds . (s − s+ )(s − s− )

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For t − t0 > 0,we can close the contour in the left half-plane; this encloses both poles, with residues Res[s+ ] =

f0 es+ (t−t0 ) f0 −β(t−t0 ) iΩ(t−t0 ) = e e s+ − s− 2iΩ

Res[s− ] =

f0 es− (t−t0 ) f0 −β(t−t0 ) −iΩ(t−t0 ) =− e e . s− − s+ 2iΩ

The integral is 2πi times the sum of these residues: u(t) =

f0 −β(t−t0 ) e sin [Ω(t − t0 )] . Ω

Closing the contour in the right half-plane encloses no poles, so u(t < 0) = 0 — as it should. [That u(t ˙ 0 ) , 0 reflects the effect of the sudden impulse at t0 .] 36.52 The Laplace transform of f¨ + ω 2 f = t, with f (0) = a, f˙(0) = b is




s2 + ω 2 F (s) = as + b +

1 , s

or, using Table 36.4 for the first term and the convolution theorem for the second: as + b 1 + s2 + ω 2 s(s2 + ω 2 ) 1 b = a cos (ωt) + sin (ωt) + 1 ∗ sin (ωt) ω ω
 1  = 2 aω 2 − 1 cos(tω) + bω sin(tω) + 1 ω

f (t) = L −1

h

i

36.53 In the form presented in (36.129), Utr (s) =

u0 (s + β) u0 β + v0 + , (s + β)2 + Ω2 (s + β)2 + Ω2

it’s clear that translating s → s + β will simplify the expression, Utr (s) = F (s + β), F (s) =

u0 s u0 β + v0 + 2 s2 + Ω2 s + Ω2

so that u(t) = e−βt f (s). And f (s) can be found directly from Table 36.4, f (t) = u0 cos (Ωt) + (u0 β + v0 )

1 sin (Ωt) Ω

Thus we arrive at (36.130),



utr (t) = e−βt u0 cos Ωt +

1 (βu0 + v0 ) sin Ωt Ω



,

which is equivalent to (3.28) for u0 = C cos δ and βu0 + v0 = Ω C sin δ.

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293

Coda: Of Time Intervals and Frequency Bands

37

37.1 (a) Using the conventions of (37.1),

Z

∞

∞

Z

2

∞

Z

|f (t)| dt =

∗ Z

Z

f˜∗ (ν 0 )f˜(ν)

=

Z−∞ ∞

∞

f˜(ν) e2πiνt dν

 dt

−∞

−∞

−∞ ∞

−∞

0 f˜(ν 0 ) e2πiν t dν 0

Z

∞

0

e2πi(ν−ν



)t

dt dν dν 0

−∞

f˜∗ (ν 0 )f˜(ν) δ(ν − ν 0 ) dν dν 0 =

=

∞

Z

−∞

|f˜(ν)|2 dν .

−∞

(b)

Z

∞

 Z

0

f (t0 ) e−2πiνt dt0

F [f ] F [g ] = −∞ ∞

Z

∞

g(t) e−2πiνt dt



−∞ 0

f (t0 )g(t) e−2πiν(t+t ) dt0 dt .

= −∞

Substituting t0 with u = t + t0 ,

Z

∞

e−2πiuν

F [f ] F [g ] =

Z

−∞

∞

 f (u − t)g(t) dt

du = F [f ∗ g ] .

−∞

37.2 Starting with (37.10) and consulting with Table 37.1, 2µf (t) = F −1 [h2µ ] ∗ f · s1/2µ




"

∞ X

= 2µ sincπ (2µt) ∗ f (t)

# δ(t − n/2µ)

n=−∞

" = 2µ sincπ (2µt) ∗

∞ X

# f (n/2µ)δ(t − n/2µ)

n=−∞

=

∞ X

f (n/2µ) [sincπ (2µt) ∗ δ(2µt − n)] =

∞ X

f (n/2µ) sincπ (2µt − n) ,

n=−∞

n=−∞

where we’ve used (36.96).

37.3 (a) Noting that F −1 [1 ] = δ(t), lim F [n sincπ (nt) ] = lim hn (ν) = 1 .

n→∞

n→∞

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(b) ∞ X

sincπ (t − n) =

n=−∞

∞ X

sincπ (t) ∗ δ(t − n)

n=−∞ ∞ X

= sincπ (t) ∗

δ(t − n)

n=−∞

= sincπ (t) ∗ s1 (t) . Then, by convolution theorem, F [sincπ (t) ∗ s1 (t) ] = F [sincπ (t) ] F [s1 (t) ] = h1 (ν)

∞ X

δ(ν − n) = δ(ν) ,

n=−∞

the inverse Fourier transform  ofwhich is 1. [Can also use periodizing property of the Dirac comb together with F −1 sinc2 = Λ, the triangle function.] (c) Just use the interpolation formula with f (t) = sincπ (t − a) and 2µ = 1: ∞ X

sincπ (t − a) =

sincπ (n − a) sincπ (t − n) .

n=−∞

(d) Take the limit of the identity in part (c) as a → t, noting that sinc is an even function, ∞ X

sinc2π (t − n) = lim sincπ (t − n) sincπ (a − n) a→t

n=−∞

= lim sincπ (t − n) sincπ (n − a) a→t

= lim sincπ (t − a) = lim a→t

a→t

sin [π(t − a)] = 1. π(t − a)

(e) Convolution:

Z

∞



sincπ (τ − t) sincπ (t) dt = sincπ ∗ sincπ = F −1 h1 (ν)2



= F −1 [h1 (ν) ] = sincπ (τ ) .

−∞

37.4 (a) Orthonormality:

Z

∞

hφn |φm i = 2µ

Z

sincπ (2µt − n) sincπ (2µt − m) dt −∞ ∞

sincπ (ξ − n) sincπ (ξ − m) dξ

= −∞ ∞

Z



e−2πinν h2µ (ν)

=

∗ 



e−2πimν h2µ (ν) dν

−∞

Z

1/2

=

e−2πi(m−n)ν dν = δmn .

−1/2

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295

(b) Projection of components: hφn |f i =

p

Z

1 = √ 2µ

Z

∞

sinc(2µt − n)f (t) dt

2µ

Z

−∞ ∞



e−2πinν/2µ h2µ (ν)

∗

f˜(ν) dν

−∞

µ

e2πiν(n/2µ) f˜(ν) dν

= −µ ∞

Z

e2πiν(n/2µ) f˜(ν) dν = f (n/2µ) ,

= −∞

where in the last line we recalled that the interpolation formula only applies to band-limited signals.

37.5 An aliased signal is one that is sampled at a rate below 2µ. Sampling f (t) at a rate ρ yields the measured signal g(t) =

∞ X

f (n/ρ) sinc(ρt − n) =

n=−∞

∞ X

f (tn ) sinc [ρ(t − tn )] ,

n=−∞

where tn = n/ρ. Individual data points are ∞ X

g(tm ) =

f (tn ) sinc [ρ(tm − tn )] .

n=−∞

From the familiar behavior of sin x/x as x goes to zero, we know sincπ [ρ(tm − tn )] = δmn . As a result, g(tm ) =

∞ X

f (tn ) sinc [ρ(tm − tn )] = f (tn ) .

n=−∞

So g is constructed from values of f — and if the sampling rate ρ > 2µ, then f will be wellrepresented by g. But a low sampling rate ρ < 2µ will result in g being an aliased signal, indistinguishable from f based on the measured data set.

37.6 When sampling at a rate νs , there’s a signal with ν > νs /2 which is indistinguishable from a frequency νalias ≤ νs /2. The aliased frequency is determined by integer k for which νalias = |ν − kνs | ≤ νs /2 so that νalias is over-sampled. Thus the integer k must satisfy ν/νs − 1/2 ≤ k ≤ ν/νs + 1/2 . For ν = 150 Hz and νs = 40, 13/4 ≤ k ≤ 17/4 → k = 4. So νalias = 10 Hz.

37.7 A signal frequency νcl can only be fully resolved with a sample frequency νs at the Nyquist rate νs > 2νcl . What happens when the sample frequency is lower than this?

296

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(a) τ = 75 s: It’s not hard to appreciate that the clock appears to advance 15s each flash, completing one period in 75 × 4 = 300s = 5 min. And in fact, νs = 54 min−1 < νcl → νalias = 1 |1 − 4/5| = 51 min−1 = 300 Hz. (b) τ = 60: The clock appears stopped, the second hand always in the same position at the time of each flash. In this case, νs = 1 min−1 = νcl → νalias = |1 − 1| = 0. (c) τ = 45 s: The clock appears to run backwards by 15s each flash, completing one reverse period in 45 × 4 =180 s = 3 min. Mathematically, νs = 34 min−1 , so that νcl < νs < 2νcl : 1 νalias = |1 − 4/3| = 13 min−1 = 180 Hz. (d) τ = 30 s: The clock hand appears to alternate between two fixed, opposite positions, with no sense of forward or backward. One period is 1 min = 60 s. And indeed, νs = 2 min−1 = 1 2νcl → νalias = |1 − 2| = 1 min−1 = 60 Hz. (e) τ = 15 s: In this case, νs = 4 min−1 → νs > 2νcl . The clock appears to run forward 15s each flash, keeping time correctly. There is no aliasing.

37.8 (a) A typical four-blade fan has π/4 symmetry, which raises the effective frequency by a factor of 4. The Nyquist rate is thus νs ≡ 2νeff = 8ν. So for νs = 24 FPS, the largest fan frequency ν that can be resolved without aliasing is ν = 3 FPS. (b) The fan will appear to rotate backwards for 12 νs < νeff < νs — or fan frequency ν between 3 and 6 FPS.

37.9 With six uniformly-distributed blades, the effective Nyquist frequency is 6νrot . So if νrot in an integer multiple of 20 Hz, lights flickering at 120 Hz will make the blades appear stationary, leading unwary factory workers to think the rotor is not spinning, posing a danger to life — and especially limbs.

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297

First-Order ODEs

38

38.1 (a) If the equation is linear, then for solutions u1 (x) and u2 (x), the combination au1 + bu2 must also solve. But for n , 1,










au01 + bu02 + f (x) (au1 + bu2 ) = a u01 + f (x)u1 + b u02 + f (x)u2



n = ag(x)un 1 + bg(x)u2 n n = g(x) (aun 1 + bu2 ) , g(x) (au1 + bu2 ) .

(b) Dividing the equation by un , f (x) u0 + n−1 = g(x), un u or →

1 v 0 + f (x)v = g(x) , 1−n

where v ≡ u1−n .

38.2 Newton’s law of cooling is given by the differential equation dT = b (Tenv − T (t)) . dt Separating the variables gives dT = −b dt T − Tenv Then for constant Tenv and initial temperature T0 , T (t) = ke−bt + Tenv = (T0 − Tenv ) e−bt + Tenv . Thus as time passes, the object asymptotically comes into thermal equilibrium with the environment — regardless of whether T0 is greater and less than Tenv .

38.3 (a) The change dN2 in the number of 222 Rn nuclei is determined by both their creation due to the α decay of 226 Ra and their loss by their own α decay. The probability/time of α decay of a 226 Ra nucleus is λ1 ; given N1 226 Ra nuclei, the number of decays in time dt is λ1 N1 dt. Similarly, given N2 222 Rn nuclei, the probability of a loss of 226 Ra nuclei iss λ2 N2 dt. Thus dN2 = λ1 N1 dt − λ2 N2 dt . (b) Using N1 (t) = N1 (0)e−λ1 t , the differential equation becomes dN2 + λ2 N2 = λ1 N1 (0)e−λ1 t . dt The integrating factor w(x) = eλ2 t results in


d eλ2 t N2 = N1 (0)λ1 e(λ2 −λ1 )t , dt ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

and so



N2 (t) = e−λ2 t N1 (0)λ1 = N1 (0)

Z

e(λ2 −λ1 )t dt + C



λ1 e−λ1 t + Ce−λ2 t λ2 − λ1

The initial condition N2 (0) = 0 gives C = −N1 (0)λ1 / (λ2 − λ1 ), and thus N2 (t) = N1 (0)


λ1 e−λ1 t − e−λ2 t . λ2 − λ1

(c) With λ2  λ1 , for sufficiently large time t we get N2 (t) ≈ N1 (0)


λ1 −λ1 t λ1 −λ1 t − e−λ2 t −→ N1 (0) = (λ1 /λ2 )N1 (t) . e e λ2 λ2

Thus the equilibrium balance of nuclei is given by λ2 N2 = λ1 N1 .

38.4 (a) With the substitution u(x) = xm v(x), the differential equation becomes 2x2m+2 v

dv + x2m+1 (2m + 1)v 2 + 2 = 0 . dx

The first two terms have weight 2m + 1 (the derivative d/dx contributes -1), and the last has weight 0. Thus the entire equation is isobaric — and v is rendered dimensionless — for m = −1/2. (b) With the choice m = −1/2, the differential equation greatly simplifies, 2xv

dv +2=0 . dx

This is now a separable equation: vdv = −

dx x

v 2 (x) = −2 ln(x) + k2 ,

−→

where u(1) = v(1) = k. Since u2 = v 2 /x, the solution to the original equation is u2 (x) =


1 −2 ln(x) + k2 . x

Verifying this solution is easiest in terms of u2 : 2x2 u


2 1 du = x2 − 2 −2 ln(x) + k2 − 2 dx x x h

and xu2 = −2 ln(x) + k2 . Summing these gives −2.

i

= 2 ln(x) − k2 − 2

X

38.5 For high speeds, atmospheric drag goes like v 2 , so the acceleration is dv = g − bv 2 . dt

(38.1)

dv = dt . g − bv 2

(38.2)

Separating v and t gives

Integrating, we find −1

tanh

r

b v g

 =

p

bg t + c0

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(38.3)

299

or

q

v(t) = where c0 = tanh−1

q

b g

g tanh b

p

bg t + c0



,

(38.4)



v0 . So the terminal velocity — the speed at which the drag force

exactly cancels the downward pull of gravity so that the body ceases to accelerate — is vterm ≡ v(t → ∞) =

p

g/b .

(38.5)

38.6 We can do most of the work for both cases by starting with the differential equation dp = mv˙ + mv ˙ = mg − bm2/3 v α dt where α = 1 or 2. Using v˙ = (dm/dt)(dv/dt) ≡ mv ˙ 0 (m) and m ˙ = km2/3 v α−1 , this can be manipulated to give



v α−1 v 0 +

vα m



=

g b vα − , 2/3 k m km

which should be compared with Eqn. (38.22). Changing to u = v α gives u0 = αv α−1 v 0 and so u0 +

α u αg −2/3 (k + b) = m . k m k

To solve, use the integrating factor

 w(m) = exp

α(k + b) k

Z

dm m

 = mα(k+b)/k

Then


αg α(k+b)/k−2/3 d m mα(k+b)/k u = dm k resulting in αg m1/3 αg m1/3 = , k α(k + b)/k + 1/3 k(α + 1/3) + bα

u(m) =

where the constant of integration vanishes for u(m = 0) = 0. Then we can get the acceleration fairly simply: u˙ = u0 m ˙ =

αg m−2/3 αgk v α−1 1 m ˙ = . 3 k(α + 1/3) + bα 3 k(α + 1/3) + bα

With u0 m ˙ = αv α−1 v, ˙ we finally get the acceleration v˙ =

gk 1 g 1 = , 3 k(α + 1/3) + bα 3 (α + 1/3) + bα/k

where, since b and k have the same units, is clearly just some fraction of g. (a) linear drag, α = 1: v˙ =

g 4+3b/k

(b) quadratic drag, α = 2: v˙ =

g 7+3b/k

Note that in the absence of drag, these reduce to Eqns. (38.25) and (38.46) with m0 = 0.

38.7 Consider a rocket which ejects mass at a constant rate, m = m0 − kt. If the rocket takes off

300

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vertical m rest, find its speed as a function of time if the only external force is (a) gravity; (b) linear drag. Start with Eqn. (38.47) m

dv dm = −u + Fext = ku + Fext , dt dt

with m = m0 − kt. Note that, unlike the raindrop, we take up to be the positive direction. (a) Fext = −mg: v

Z

dv ku = −g dt m0 − kt

−→

dv =

Z t 0

0

= u ln



ku − g dt m0 − kt



m0 m0 − kt



− gt = u ln



m0 m



− gt ,

which is just the rocket equation of Eqn. (38.48) reduced by the pull of gravity. Note that if the rocket’s weight is greater than its thrust, mg > ku, the rocket will just sit on the launch pad. (At least until enough fuel is burned to reduce the weight below the thrust.) (b) Fext = −bv: m

dv = ku − bv dt

Z

v

−→ 0

dv = ku − bv

t

Z

1 dt =− m k

0

Z

m

m0

dm , m

since dt = −dm/k. Then −

1 ln b



ku − bv ku



=−

1 ln k



m m0





−→

v=

ku 1− b



m0 m

b/k 

.

In the b → 0 limit:



lim b→0

ku 1− b



m0 m

b/k 

= lim b→0

= lim b→0

ku 1 − exp b



b ln k

m0 m

i

h

ku b

h

b ln k





m0 m

i

= u ln



m0 m



.X

38.8 If w = w(y), the Eqn. (38.34), ∂ ∂ (wα) = (wβ) , ∂y ∂x becomes w0 (y)α + w(y)∂y α = w(y)∂x β or w0 (y) = (∂x β − ∂y α) w(y) ≡ p(y)w(y), which is Eqn. (38.38).

38.9 Probably the least-tedious (non-Mathematica) approach is to write Eqn. (38.44) as v=c where c =

p 6g 7k

1 m

q

7/3

m7/3 − m0

,

, and calculate v˙ using m ˙ = km2/3 v.

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301

Second-Order ODEs

39

39.1 A second-order ODE with constant coefficients has the general form of damped harmonic motion — so the oscillatory ansatz u(t) = Aeiξt is appropriate: Inserting the ansatz into (39.3) gives


d2 u du + 2β + ω02 u = −ξ 2 + 2iβξ + ω02 u = 0 . 2 dt dt Thus ξ = −iβ ±



p

−β 2 + ω02 = −i β ∓

p

β 2 − ω02



,

which gives (39.10).

39.2 (a) Linearly dependent:

sin kx cos kx e−ikx =0 −ikx W = k cos kx −k sin kx −ike −k2 sin kx −k2 cos kx −k2 e−ikx (b) Linearly independent:

x W = 1 0

1+x 1 0



x2 2x = −2 2

(c) Linearly independent (x , 0):

x x 2 x3 2 W = 1 2x 3x = 2x3 0 2 6x (d) Linearly independent:

eαx W = αx αe



xeαx = e2αx (1 + αx) eαx

39.3 With



u W = 10 u1



u2 = u1 u02 − u2 u01 , u02

and using the differential equation to replace the second derivatives:


d d 00 W (x) = u1 u02 − u01 u2 = u1 u00 2 − u1 u2 dx dx


= u1 −pu02 − qu2 − u2 −pu01 − qu1







= −p u1 u02 − u2 u01 = −p W .

39.4 Using (39.18), up to an overall constant:

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(a) Bessel, p(x) = 1/x:

 Z W (x) = exp −



 Z

p(x) dx = exp −

dx x

 =

1 . x

2x (b) Associated Legendre: p(x) = − 1−x 2:

 Z



2x 1 dx = 2 . 1 − x2 x −1

W (x) = exp + (c) Mathieu: p(x) = 0: W (x) = 1.

39.5 Taking a derivative of Eqn. (39.31) gives 0 0 0 0 00 00 u00 p = a1 u1 + a2 u2 + a1 u1 + a2 u2 .

Using the differential equation to replace all u00 ’s




r − (pu0p + qup ) = a01 u01 + a02 u02 − a1 pu01 + qu1 − a2 pu02 + qu2




which, since up = a1 u1 + a2 u2 , simplifies to a01 u01 + a02 u02 = r . This, together with Eqn. (39.32), a01 u1 + a02 u2 = 0 gives the system



u1 u01

u2 u02

  a01 a02

  0 r

=

.

Thus

  a01 a02

 =

u1 u01

u2 u02

−1   0 r

1 = W



u02 −u01

−u2 u1

  0 r

,

from which the result immediately follows.

39.6 We merely need to show that the derivative of p(x)W (x) vanishes:

 d d  [pW ] = p(u1 u02 − u01 u2 ) dx dx d  0 d  0 = u1 pu2 − u2 pu1 = 0 , dx dx where u1,2 (x) are Sturm-Liouville solutions,

d dx





d p(x) dx u1,2 = −qu1,2 .

39.7 (a) u00 + 2u0 + u = e−x : The homogeneous solutions are most-easily found with the ansatz eiξx — or just by transcribing (39.10) — to get u1 (t) = e−x

u2 (t) = xe−x .

For the particular solution, we need

 Z W (x) = exp −



2 dx = e−2x .

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303

Then by using (39.33) with r = e−x

Z a1 (x) = −

Z

1 x dx = − x2 2

a2 (x) = +

dx = x

we get u(x) =

1 2 −x x e + c1 e−x + c2 xe−x 2

as substitution into the differential equation readily confirms. (b) u00 − u = ex : The homogeneous solutions are u2 (x) = e−x .

u1 (x) = e+x The Wronskian is W = −2, so

Z a1 (x) =

dx =

x 2

a2 (x) = −

1 2

Z

1 e2x dx = − e2x . 4

Thus u(x) =

x x 1 1 e − ex + c1 ex + c2 e−x = (2x − 1) ex + c1 ex + c2 e−x . 2 4 4

(c) u00 + u = sin x: The homogeneous solutions are u1 (x) = cos x

u2 (x) = sin x

with W = 1. Then a1 (x) =

1 (sin x cos x − x) 2

a2 (x) = −

1 cos2 x 2

and 1 1 (sin x cos x − x) cos x − cos2 x sin x + c1 cos x + c2 sin x 2 2 1 = − x cos x + c1 cos x + c2 sin x . 2

u(x) =

39.8 Use the exponential form of the homogeneous solutions u1 (t) = e−βt e−iΩt

u2 (t) = e−βt eiΩt ,

where Ω2 = β 2 − ω02 . This gives Wronskian W (t) = 2iΩe−2βt . Then, using variation of parameters with r(t) = δ(t − t0 ), a1 (t) = −2iΩe+βt

0

a2 (t) = 2iΩe+βt

0

to arrive at the particular solution up (t) =

 1 0 1 −β(t−t0 )  −iΩt e −e + eiΩt = e−β(t−t ) sin Ωt. 2iΩ Ω

39.9 The freedom to add a complementary solution uc to up u(x) = up (x) + [c1 u1 (x) + c2 u2 (x)] in order to satisfy conditions u(x0 ) = a, u0 (x0 ) = b requires that c1,2 exist. In other words c1 u1 (x0 ) + c2 u2 (x0 ) = a − up (x0 ) c1 u01 (x0 ) + c2 u02 (x0 ) = b − u0p (x0 )

304

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must always have a solution. Expressed as a matrix equation,



u1 (x0 ) u01 (x0 )

u2 (x0 ) u02 (x0 )

  c1 c2

 =

a − up (x0 ) b − u0p (x0 )

 .

So c1,2 exist as long as the matrix has an inverse — which requires its determinant be non-zero. But that determinant is the Wronskian, which is non-zero for linear independent solutions u1,2 .

39.10 Substituting v = u0 and solving the resulting first-order equation: (a) xu00 − 2u0 = 0 → xv 0 − 2v = 0 → v(x) = ax2 , for some constant a. Then

Z u=

1 3 ax + b . 3

v(x)dx =

(b) 2

u00 + 2xu0 = 0 → v 0 + 2xv = 0 → v(x) = ae−x . This gives the error function (8.68),

Z u=

Z v(x)dx = c

√ 2

e−x dx = a

π erf(x) + b . 2

(c) v = ax . v+1

xu00 − u0 − u02 = 0 → xv 0 − v − v 2 = 0 → Then

Z u(x) =

Z v(x)dx =

ax 1 dx = − [ax + log(1 − ax)] + b 1 − ax a

39.11 (a) With p(u) = u and q(u) = 0, setting v = u0 so that u00 = dv/du, u00 + uu0 = 0 → v

dv + uv = 0. du

So one solution is v = 0 — which is constant u(x). For the other: dv 1 = −u → v(u) = − u2 + a du 2 Then returning to u0 (x) = v(x),

Z

1 1 du = − dx → √ tanh−1 u2 − 2a 2 2a



u √ 2a



=

1 (x + b) 2

Thus u(x) =

√

2a tanh

hq

a (x + b) . 2

i

(b) With p(u) = 0 and q(u) = −u02 /2u, setting v = u0 gives 2uu00 − u02 = 0 → v

dv v2 − = 0. du 2u

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305

Again, one solution is constant u(x). For the other: √ dv v = → v(u) = a u . du 2u Returning to u0 (x) = v(x),

Z

√ 1 du = a u → u(x) = (ax + b)2 . dx 4

39.12 Hsol = u00 [t] + 2βu0 [t] + Ω20 u[t] == 0; IVPsolution = DSolve[Hsol, u[0] == u0, u0 [0] == up0, u, t]; cdamp = Plot[Evaluate[Limit[u[t]/.IVPsolution[[1]], β → Ω0 /.Ω0 → 10], {t, 0, 2.}, PlotRange → All, PlotStyle → {Thick}]; udamp = Plot[Evaluate[u[t]/.IVPsolution[[1]]/.β → Ω0 /5/.Ω0 → 10], {t, 0, 2.}, PlotRange → All, PlotStyle → {Thick, Dotted}]; odamp = Plot[Evaluate[u[t]/.IVPsolution[[1]]/.β → 5Ω0 /.Ω0 → 10], {t, 0, 2.}, PlotRange → All, PlotStyle → {Thick, Dashed}]; Show[cdamp, udamp, odamp, Ticks → None, AxesLabel → “t”, “u(t)”, AxesStyle → 22] u(t)

u(t)

t

t

(a)

(b)

Problem 39.12: (a) u(0) = 1, u0 (0) = 0; (b) u(0) = 0, u0 (0) = 1

39.13 u(t) = up (t) + uc (t) where, from Eqn. (39.42), the particular solution is




up (t) =

2 2 F0 ω0 − ω cos(ωt) + 2βω sin(ωt)
2 m ω 2 − ω 2 + 4β 2 ω 2 0

and the complementary

h

uc (t) = e−βt c1 cos

p



ω02 − β 2 t + c2 sin

p

i

ω02 − β 2 t

(a) u(0) = x0 , u0 (0) = 0:




ω02 − ω 2 F0 c1 = 1 −
m ω 2 − ω 2 2 + 4β 2 ω 2 0

306

c2 =

β

p

ω02 − β 2

ω02 + ω 2 F0 x0 −
m ω 2 − ω 2 2 + 4β 2 ω 2 0

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!

(b) u(0) = 0, u0 (0) = v0 :




ω02 − ω 2 F0 c1 = −
m ω 2 − ω 2 2 + 4β 2 ω 2 0

c2 =

β(ω02 + ω 2 ) F0 v0 −
m ω 2 − ω 2 2 + 4β 2 ω 2 0

1

p

ω02 − β 2

!

39.14 For constant force F (t) = f0 /m, the differential equation becomes


du d2 u + 2β + ω02 u − f0 /ω02 = 0 . dt2 dt This is clearly satisfied by a constant up (t) = f0 /ω02 , which is thus the particular solution. (So despite the constant force, the oscillator is not in fact moving — which makes sense: as can be seen from the differential equation, applying the external force must increase u until F exactly balances the oscillator’s restoring force mω02 u.) The most general solution is thus this constant up plus a linear combination of the homogenous solutions u(t) =

f0 + e−βt [a cos Ωt + a sin Ωt] , ω02

where Ω2 = ω02 − β 2 . So all the constant force does is change the equilibrium position from u = 0 to u = f0 /ω02 . There are no dynamic consequences.

39.15 Using the homogeneous solutions u1,2 (t) = e−βt e

∓i

p

2 −β 2 t ω0

and the coefficients from Eqn. (39.41) (omitting the F0 /m for cleanlienss),

p

 a1 (t) =

−i e

β+i ω+



p

β+i ω+

 ie

a2 (t) =

β+i ω−



β+i ω−

2 −β 2 ω0

ω02 − β 2

p

2 −β 2 ω0

p




t






t

ω02 − β 2

 ,

the particular solution is the real part of



 −1

a1 (t)u1 (t) + a2 (t)u2 (t) = ieiωt 



β+i ω+ = −ieiωt h

p

ω02

−

2i



β+i ω+

p

+

1



p

p ω2 − β 2 i 0h 

p

β2

β+i ω−

ω02 − β 2

=2

p

ω02 − β 2

eiωt (β + iω)2 + (ω02 − β 2 )

=2

p

ω02 − β 2

eiωt (ω02 − ω 2 ) + 2iβω

=2

p

ω02



−

β2

β+i ω−

eiωt (ω02 − ω 2 ) − 2iβω

ω02 − β 2



ω02 − β 2

i



(ω02 − ω 2 )2 + 4β 2 ω 2

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307

The real part (discarding the overall real constant 2

p

ω02 − β 2 ) gives the particular solution

 up (t) =



(ω02 − ω 2 ) cos ωt + 2βω sin ωt

X

(ω02 − ω 2 )2 + 4β 2 ω 2

39.16 The ansatz up = ξ(t) cos(ω0 t−δ) has a time-dependent amplitude, which is physically motivated by the absence of damping. It also has a parameter δ which we can adjust to find the particular solution. So: plugging this into the β = 0 differential equation, −2ω0 ξ˙ sin(ω0 t − δ) + ξ¨ cos(ω0 t − δ) = f0 cos(ω0 t) . With δ = π/2, the equation can be satisfied if ξ˙ = f0 /2ω0 . Thus ξ = ξ0 + f0 t/2ω0 , and up (t) =



ξ0 +

f0 t 2ω0





cos ω0 t −

π 2



=



ξ0 +

f0 t 2ω0



sin(ω0 t) .

The homogeneous solutions are cos ω0 t and sin ω0 t — the latter of which already appears within up . Thus the general solution is



u(t) = c1 cos ω0 t + c2 +

f0 t 2ω0



sin ω0 t

where c2 includes ξ0 . Defining ϕ such that cos ϕ =

c1

q

sin ϕ =


f0 t 2

c21 + c2 +

c2 +

f0 t 2ω0

q

c21 + c2 +

2ω


,

f0 t 2 2ω

the solution can be rendered u(t) = A cos(ω0 t − ϕ) with amplitude and phase

r A=



c21 + c2 +

f0 t 2ω

2

tan ϕ =

1 f0 t c2 + c1 2ω0

h

i

.

√ 39.17 (a) This is damped oscillation with β = 3/2 and ω0 = 2. Since β 2 > ω02 , the homogeneous solutions are best expressed as damped exponentials, u1 (t) = e−2t

u2 (t) = e−t

(b) With Wronskian W = e−3t and source r(t) = cos t, (39.33) gives

Z a1 (t) = −

Z a2 (t) = +

e2t cos t dt = − et cos t dt =

1 (sin t + 2 cos t) e2t 5

1 (sin t + cos t) et . 2

Thus up (t) = a1 u1 + a2 u2 = −

1 1 1 (sin t + 2 cos t) + (sin t + cos t) = (3 sin t + cos t) . 5 2 10

(c) The full solution is u(t) =

308

1 (3 sin t + cos t) + c1 e−2t + c2 e−t . 10

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Applying initial conditions 9 10 3 u0 (0) = 0 −→ 2c1 + c2 = 10 u(0) = 1 −→ c1 + c2 =

gives c1 = −3/5, c2 = 3/2. Thus u(t) =

 1  3 sin t + cos t − 6e−2t + 15e−t . 10

39.18 With ansatz u = xα , the homogeneous equation becomes α(α − 1)xα − 2xα = 0 −→ α = 2, −1 giving homogeneous solutions u1 = x2 , u2 = 1/x. One can continue to solve using variation of parameters — but in this case, it’s simpler to notice that with a source proportional to x, both x2 u00 and u must have order 1 — which is only possible if u00 = 0. With this observation, we easily find up (x) = − 12 x. Thus the general solution is 1 c2 u(x) = − x + c1 x2 + . 2 x Applying boundary conditions u(1) = 0 −→ c1 + c2 =

1 2

u0 (1) = 0 −→ 2c1 − c2 =

1 , 2

gives c1 = 1/3, c2 = 1/6. Thus 1 1 1 u(x) = − x + x2 + . 2 3 6x 39.19 (a) Inverting the simple pendulum by taking ϕ = π − θ flips a crucial sign in the equation of motion, θ¨ = −ω02 sin θ

ϕ ¨ = +ω02 sin ϕ .

−→

So although ϕ = 0 is an equilibrium point, any slight deviation causes ϕ to increase further, demonstrating that ϕ = 0 is an unstable equilibrium (as expected of an inverted pendulum.) (b) For small ϕ, the equation of motion for the horizontally-driven pendulum becomes a linear inhomgeneous ODE, ϕ ¨ − ω02 ϕ = αω 2 cos ωt . From the homogeneous solutions ϕ1 (t) = e−ω0 t

ϕ2 (t) = eω0 t

and their Wronskian W = 2ω0 , (39.33) gives a1 (t) = − a2 (t) =

1 2 ω α 2ω0

1 2 ω α 2ω0

Z

Z

eω0 t cos ωt dt = −

e−ω0 t cos ωt dt =

αω 2 [ω sin(ωt) + ω0 cos(ωt)] eω0 t 2ω0 (ω 2 + ω02 )

αω 2 [ω sin(ωt) − ω0 cos(ωt)] e−ω0 t . 2ω0 (ω 2 + ω02 )

A little algebra returns the particular solution ϕp (t) = a1 ϕ1 + a2 ϕ2 = −

αω 2 cos ωt . ω 2 + ω02

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309

[You should take a moment to appreciate how you could have found this merely from inspection of the differential equation.] The general solution is thus

ϕ(t) = c1 e−ω0 t + c2 eω0 t −

αω 2 cos ωt . ω 2 + ω02

So with appropriate initial conditions to render c2 = 0, once the transient ϕ1 decays, the motion is that of steady oscillation around equilibrium. With initial conditions ϕ(0) = a, ϕ(0) ˙ = b, we get c2 = 0 for

 b = −ω0 a −

αω 2 2 ω + ω02

 .

The motion of the normal pendulum is found by replacing ϕ with π − θ in the original differential equation; this has the effect of flipping the sign of ω02 . Thus replacing ω0 with iω0 gives the solution for the normal pendulum,

θ(t) = c1 e−iω0 t + c2 eiω0 t −

αω 2 cos ωt − ω02

ω2

This exhibits a resonance at ω = ω0 , and so at this frequency the motion becomes unstable.

39.20 (a) We want to average the differential equation over period T = 2π/ω,

hϕi ¨ =−






αω 2 cos ωt − ω02 ϕ

,

using the ansatz ϕ(t) = ξ(t) (1 + α cos ωt):

ξ¨ = −










αω 2 cos ωt − ω02 ξ (1 + α cos ωt)



= α2 ω 2 cos ωt − ω02 ξ =



1 2 2 α ω − ω02 ξ , 2



where, since ξ varies p 1 slowly over time scale 1/ω, hξi = ξ. So indeed, ξ oscillates harmonically with frequency α2 ω 2 − ω02 . 2

310

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(b) As the plots verify, the ansatz is valid for αω 2  ω02 : In[]:=

sol[t_, ω_] =

ϕ[t] /. DSolve[{ϕ ''[t] + (α ω ^ 2 Cos[ω t] - ω0 ^ 2) ϕ[t]  0, ϕ[0]  0.1, ϕ '[0]  0}, ϕ[t], t] 2

Out[]=



0.1 MathieuC- 4 ω20 , - 2 α, ω

2

tω  2



MathieuC- 4. ω20 , - 2. α, 0 ω

In[]:=

10 ; α = .1;

ω0 =

p[ω_] := Plotsol[t, ω], {t, 0, 1.5}, PlotLabel  StringTemplate"αω 2 =`1`"[α ω ^ 2]; In[]:=

Tablep[ω], ω, ω0 

α , 10 ω0 

αω2 =10.0

αω2 =40.0

4 3 Out[]=



,

2 1

3.5 3.0 2.5 2.0 1.5 1.0 0.5

,

0.2 0.4 0.6 0.8 1.0 1.2 1.4

0.2 0.4 0.6 0.8 1.0 1.2 1.4

αω2 =90.0

αω2 =160.0

1.5 1.0

α , 10

,

0.5

0.40 0.35 0.30 0.25 0.20 0.15

0.2 0.4 0.6 0.8 1.0 1.2 1.4

,

0.2 0.4 0.6 0.8 1.0 1.2 1.4

αω2 =250.0

αω2 =360.0

0.10

0.10 0.05

0.05

,

, 0.2 0.4 0.6 0.8 1.0 1.2 1.4

0.2 0.4 0.6 0.8 1.0 1.2 1.4

-0.05

-0.05 -0.10

αω2 =490.0

αω2 =640.0

0.10

0.10

0.05

0.05

,

,

0.2 0.4 0.6 0.8 1.0 1.2 1.4

0.2 0.4 0.6 0.8 1.0 1.2 1.4

-0.05

-0.05

-0.10

-0.10

αω2 =810.0

αω2 =1000.0

0.10

0.10

0.05

0.05

,



0.2 0.4 0.6 0.8 1.0 1.2 1.4

0.2 0.4 0.6 0.8 1.0 1.2 1.4

-0.05

-0.05

-0.10

-0.10

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311

39.21 For u(x) a homogeneous solution the differential equation, if the operator has definite parity, L−x = ±Lx , then 0 = Lx u(x) = L−x u(−x) = ±Lx u(−x) showing that u(−x) is also a homogeneous solutions. We can then use the linearity of L to construct even and odd solution u± (x),

u+ (x) =

39.22 (a) Insert the ansatz u =

x2

X

1 [u(x) + u(−x)] 2

P

c x n=0 n

α+n

u− (x) =

1 [u(x) − u(−x)] . 2

in Eqn. (39.54), x2 u00 + 2xu0 + x2 u = 0:

cn (α + n)(α + n − 1)xα+n−2 + 2x

X

n=0

cn (α + n)xα+n−1 + x2

n

X

cn xα+n = 0

n=0

or 0=

X

=

X

[cn (α + n)(α + n − 1) + 2(α + n)] xα+n +

n=0

X

cn xα+n+2

n=0 α+n

cn (α + n + 1)(α + n)x

+

n=0

X

cn−2 x

α+n

n=2

= c0 α(α + 1)xα + c1 (α + 1)(α + 2)xα+1 +

X

[cn (α + n)(α + n + 1) + cn−2 ] xα+n

n=2

(b) For equation, x2 u 00 + xu 0 + (x2 − λ2 )u = 0, we used the modified ansatz u(x) = P Bessel’s 1 α+n : c x n n n! 0 = x2

X cn n!

(α + n)(α + n − 1)xα+n−2 + x

X cn 

=

X cn 

n!

n!

(α + n)xα+n−1 + (x2 − λ2 )

n=0

n=0

=

X cn

X cn n!



(α + n)(α + n − 1) + (α + n) − λ2 xα+n +

X cn n!

xα+n+2

n=0

n=0

n!

2

(α + n) − λ

n=0

2



x

α+n

+

X cn n!

x

α+n+2

n=0 ∞




= c0 α2 − λ2 xα +

X cn 
c1  cn−2 (α + 1)2 − λ2 xα+1 + (α + n)2 − λ2 + xα+n 1! n! (n − 2)! h

i

n=2

(c) For Hermite’s equation u00 − 2xu0 + 2λu = 0 and u =

0=

X

=

X

cn (α + n) (α + n − 1)xα+n−2 − 2

n=0

n=0

312

xα+n

n=0

X

P

c xα+n : n=0 n

cn (α + n) xα+n + 2λ

n=0

cn (α + n) (α + n − 1)x

α+n−2

−2

X

X

cn xα+n

n=0

cn (α + n − λ) x

α+n

n=0

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39.23 (a) For Legendre’s equation (1 − x2 )u 00 − 2xu 0 + λu = 0, the usual ansatz gives: 0=

X

=

X



cn (1 − x2 ) (α + n) (α + n − 1)xα+n−2 − 2x (α + n) xα+n−1 + λxα+n



n=0

cn (α + n) (α + n − 1)xα+n−2 −

n=0

X

cn [(α + n)(α + n + 1) − λ] xα+n

n=0

= c0 α(α − 1)xα−2 + c1 α(α + 1)xα−1 +

X

[cn (α + n)(α + n − 1) − cn−2 ((α + n − 2)(α + n − 1) − λ)] xα+n ,

n=2

from which one can read off Eqns. (39.81). (b) For Hermite’s equation u00 − 2xu0 + 2λu = 0 and the usual ansatz, the solution to Problem 39.22c

X

cn (α + n) (α + n − 1)xα+n−2 − 2

n=0

X

cn (α + n − λ) xα+n = 0 .

n=0

Extracting the first two terms, 0 = c0 α(α − 1)xα−2 + c1 α(α + 1)xα−1 +

X

[cn (α + n) (α + n − 1) − 2cn−2 (α + n − 2 − λ)] xα+n ,

n=2

from which one can read off Eqns. (39.93).

39.24 A differential operator in the form x2 D2 + P (x)xD + Q(x) leaves each term in the series ansatz proportional to xm+α ; the indicial equation P is thenmgiven p x and by setting m = 0. This means that in the convergent power series P = m=0 m P m , only p and q contribute. Thus the indicial equation is Q= q x m 0 0 m=0 α(α − 1) + p0 α + q0 = 0 . – Example 39.5: 0 = 4xu00 + 2u0 − u = x2 u00 + 12 xu0 − 41 xu. So p0 = is α(α − 1) + 12 α = α(α − 12 ) = 0 X.

1 , 2

q0 = 0. Thus the indicial equation

– Example 39.6: 0 = x2 u00 + 2xu0 + x2 u: p0 = 2, q0 = 0 =⇒ α(α − 1) + 2α = α(α + 1) = 0 X – Example 39.7: 0 = x2 u 00 + xu 0 + (x2 − λ2 )u: p0 = 1, q0 = −λ2 =⇒ α(α − 1) + α − λ2 = α2 − λ2 = 0 X

39.25 With u1 (x) =

sin x , x

Eqn. (39.23) with p = 2/x gives

Z u2 (x) = u1 (x) =

sin x x

Z

 Rx

exp −

2 t



dt

u21 (x) csc2 x dx = −

dx =

sin x x

Z

1/x2 (sin x/x)2

dx

sin x cos x cot x = − , x x

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313

which differs from the series result only by an overall constant (-1). To find u1 from u2 :

 Rx

Z

exp −

u1 (x) = u2 (x) =



dt

dx =

u22 (x)

Z

cos x x

2 t

sec2 x dx =

cos x x

Z

1/x2 (cos x/x)2

dx

cos x sin x tan x = . x x

39.26 With the ansatz u2 (x) = v(x) + Au1 ln x, the Wronskian is



u1 (x)

W =

Au1 ln x + v(x) = (u1 v0 − u01 v) + A u21 /x , Au01 ln x + Au1 /x + v 0

u01 (x)

which differs from Eqn. (39.67) only in the u21 /x term. Then the second solution is

Z

W (x) dx u21 (x)

u2 (x) = u1 (x)

Z

(u1 v 0 − u01 v)

= u1 (x)

u21

Z

d dx

= u1 (x)



v u1



Z

dx x

dx x



dx + A

Z dx + A



= v(x) + Au1 ln x ,

P

consistent with the ansatz. For α1 = α2 , the sum v = xα bm xm must start at m = 1: starting at m = 0 will give the same recursion relation and thus render v ≡ u1 — so it would add nothing linearly independent to u2 .

39.27 Let z = 1/x in the canonical form Lx =



dz d dx dz

d2 dx2

2

d + q(x) and use the chain rule: + p(x) dx

+ p˜(z)

dz d + q˜(z) , dx dz

where the functions p˜(z) = p(x) and q˜(z) = q(x). With dz/dx = −1/x2 = −z 2 ,



dz d dx dz

2

= +z 2

d dz



z2

d dz



 = z2

z2

d d2 + 2z dz 2 dz

 = z4

d2 d + 2z 3 . dz 2 dz

Thus d2 d d + 2z 3 − z 2 p˜(z) + q˜(z) dz 2 dz dz 2 d d = z 4 2 + z 2 (2z − p˜) + q˜(z) dz dz

Lz = z 4

or, in canonical form d2 1 d 1 + 2 (2z − p˜) + 4 q˜(z) dz 2 z dz z (a) For z = 0 to be an ordinary point in z requires lim z→0

2z − p˜ = lim z2 z→0



p˜ 2 − 2 z z



and

lim z→0

1 q˜(z) z4

to be finite — or equivalently, lim 2x − x2 p(x) x→∞

and

lim x4 q(x) x→∞

must be finite.

314

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(b) For z = 0 to be an regular singular point in z requires lim z→0

2z − p˜ z

−→

lim z→0

1 p˜(z) z

and

lim z→0

1 q˜(z) z2

to be finite — or equivalently, lim xp(x)

lim x2 q(x)

and

x→∞

x→∞

must be finite. – For Legendre’s equation, (1 − x2 )u 00 − 2xu 0 + λu = 0: lim xp(x) = −2 lim x→∞

x→∞

x2 =2 1 − x2

and

lim x2 q(x) = 2λ lim x→∞

x→∞

x2 = −2λ , 1 − x2

so infinity is a regular singular point. – For Hermite’s equation, u00 − 2xu0 + 2λu = 0: lim xp(x) = −2 lim x2 = ∞ x→∞

lim x2 q(x) = 2λ lim x2 = ∞ ,

and

x→∞

x→∞

x→∞

so the point at infinity is an irregular singularity.

39.28 Substitute u(x) =

P

c x n=0 n

n

X

into Airy’s equation u00 − xu = 0:

cn n(n − 1)xn−2 −

n=2

X

cn xn+1 = 0 ,

n=0

where we’ve implicatly noted that there’s no point starting the first sum at n = 0. Shifting that first sum n → n + 2 and the second n → n − 1 gives

X

cn+2 (n + 2)(n + 1)xn−2 −

n=0

X

cn−1 xn = 0 .

n=1

Pulling out the n = 0 term allows us to combine the sums, 2c2 +

X

[(n + 2)(n + 1)cn+2 − cn−1 ] xn = 0 .

n=1

Thus we find c2 = 0, and obtain the recursion relation cn+2 =

cn−1 , (n + 2)(n + 1)

n ≥ 1.

Examining the coefficients and looking for patterns reveals three families: – The c3n are all anchored by c0 : c3n =

c0 (2 · 3)(5 · 6) · · · (3n − 1)(3n)

– The c3n+1 are all anchored by c1 : c3n+1 =

c1 (3 · 4)(6 · 7) · · · (3n)(3n + 1)

– c2 = c5 = c8 = . . . = 0 =⇒ c3n+2 = 0 All together then, the solutions are

" u1 (x) = c0 1 +

X

x3k (2 · 3)(5 · 6) · · · (3n − 1)(3n)

#

h

= c0 1 +

1 3 1 6 x + x + ... 6 180

i

n=1

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315

and

" u2 (x) = c1 x +

x3k+1 (3 · 4)(6 · 7) · · · (3n)(3n + 1)

X

#

h

= c1 x +

1 4 1 7 x + x + ... 12 504

i

n=1

39.29 (a) Examining Laguerre’s equation xu00 + (1 − x)u0 + λu = 0 . we see that it does not have definite parity. Its “power structure" implies that we’ll get a two-term recursion relation between cm and cm+1 . Although x = 0 is not an ordinary point (both p and q diverge in the x → 0 limit), it is a regular singular point since lim xp = x x→0

1−x =1 x

lim x2 q = x2 x→0

λ =0 x

are both finite. So at least one series solution exists around x = 0 — and with p and q having no finite singularities other than the origin, it’s convergent for all finite x. However, x = ∞ is an essential singularity: from Problem 39.27, the conditions for a regular singularity there are that both xp(x) and x2 q(x) be finite as x → ∞. Clearly, neither is true in this case. (b) We can use the result of Problem 39.24 to quickly find the indicial equation α(α − 1) + p0 α + q0 = α(α − 1) + α + 0 = α2 = 0 , 2 where p0 = limx→0 xp, and q0 = lim Px→0 x q. mWith α1 = α2 = 0 a repeated root, we’ll be able to find a series solution u1 = c x , but the second solution is given by m=0 m

u2 (x) = u1 (x) ln x + x0

X

c m xm ,

m=1

where for clarity I’ve left in the extraneous factor of xα1 . (c) Plugging in u =

P

X

c x m m

m+α



and collecting like terms, we get



cm (m + α)2 xm+α−1 −

m

X

cm [(m + α) − λ] xm+α = 0

m

The indicial equation is clearly α2 = 0. So the recursion relation is cm+1 =



m−λ (m + 1)2



cm

yielding solution (−λ)(1 − λ) 2 (−λ)(1 − λ)(2 − λ) 3 x + x + ··· (2!)2 (3!)2 X xn [(−λ)(1 − λ)(2 − λ) · · · (n − 1 − λ)] . =1+ (n!)2

u1 (x) = 1 + −λ x +

n=1

Calculating the radius of convergence

(m + 1)2 cm = lim = lim m , m→0 cm + 1 m→∞ m − λ m→∞

|x| < lim

show that the seriess converges for all finite x ≥ 0. For large x, however, it behaves like ex ,

316

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and so grows without bound. The only way to avoid this is for integer λ ≡ n, which makes the series terminate. Then we get a family of solutions for each n: n=0:

u1 −→ 1

n=1:

u1 −→ 1 − x

n=2: n=3:

1 2 x 2 3 2 1 u1 −→ 1 − 3x + x − x3 2 6 u1 −→ 1 − 2x +

which you can identify as the Laguerre polynomials Ln (x). Of course.

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317

40

The Sturm-Liouville Problem 40.1 The eigenfunction of iD = id/dx is φλ (x) = Ae−iλx . And from Eqn. (40.4), we have

Z hψ|iDφi = i

b

ψ∗

a

b dφ dx = iψ ∗ (x)φ(x) − i a dx b = iψ ∗ (x)φ(x) + a

Z

b

a

dψ ∗ φ(x) dx dx

Z b a

i

dψ dx

∗

φ(x) dx

≡ hiDψ|φi

b

if the surface term vanishes. So id/dx is hermitian as long as e−i(λn −λm )x = 0 for distinct a

eigensolutions φn , φm .

(a) For x ∈ [−π, π], the surface term only vanishes for real integer eigenvalues, λn = n. (b) For x ∈ [0, π], the surface term only vanishes for real even-integer eigenvalues, λn = 2n. Note that in both cases, vanishing surface term requires/guarantees both real eigenvalues and orthogonal solutions. (Unlike D2 , sin and cos are not eigenfunctions of D.)

40.2 If the surface term doesn’t vanish, we get (see Problem 40.1)

Z hψ|iDφi = i a

b

b dφ dx = iψ ∗ (x)φ(x) + ψ (x) a dx ∗

=

Z bh

Z b

i

a

dψ dx



∗

φ(x) dx

−i δ(x − b) − δ(x − a) +

a

d dx

 i∗ ψ

φ(x) dx

≡ h(iD)† ψ|φi Thus (iD)† = −i δ(x − b) − δ(x − a) + not hermitian.

d dx




. So without a vanishing surface term, iD is clearly

40.3 This time, we’ll have to integrate by parts twice: hψ|D2 φi =

Z a

= ψ∗

b

ψ∗

d2 φ dx dx2

dφ b − dx a

b

Z a

dψ ∗ dφ dx dx dx

dφ b dψ ∗ b = ψ∗ − φ + a dx a dx

Z a

b

d2 ψ ∗ φ dx dx2

b

= ψ ∗ φ0 − ψ 0∗ φ + hD2 ψ|φi . a

So hermiticity requires that the surface term vanish — which, for x ∈ [−π, π], it does if the eigenfunctions sin λx and cos λx have real, integer eigenvalues λ = n. The result, of course, is Fourier series, for which we already know that the eigenfunctions are orthogonal.

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

40.4 Start with Eqn. (40.3) with L = L† , b

Z

ψ ∗ (x)L[φ(x)] dx −

b

Z

[Lψ(x)]∗ φ(x) dx = 0 .

a

a

(a) For ψ = φ and Lφ = λwφ, we have (λ − λ∗ )

Z

b

w(x)|φ(x)|2 dx = 0 .

a

Since only the trivial vector has vanishing norm, we conclude λ = λ∗ : the eigenvalue must be real. In bra-ket notation: hLφ|φi − hφ|Lφi = (λ − λ∗ )hφ|φi = 0

=⇒

λ = λ∗ .

(b) For Lφ = λ1 wφ and Lψ = λ2 wψ, this time we get (λ1 − λ∗2 )

b

Z

w(x)ψ ∗ (x)φ(x) dx = (λ1 − λ2 )

Z

a

b

w(x)ψ ∗ (x)φ(x) dx = 0 ,

a

where we’ve used that the eigenvalues are real. So either λ1 = λ2 , or the eigenvectors are orthogonal. So eigenfunctions with non-degenerate eigenvalues must be orthogonal. (And as we know from Part 3, eigenfunctions in the degenerate subspace can always be chosen to be orthogonal.) In bra-ket notation: hψ|Lφi − hLψ|φi = (λ1 − λ2 )hψ|φi = 0

=⇒

hψ|φi = 0 for λ1 , λ2 .

40.5 (a) Integrating by parts twice, hψ|D2 φi =

b

Z

ψ ∗ (x)



a

d2 φ dx2

b

= ψ ∗ (x)φ0 (x) −

 dx

Z

a



∗

0

b

a

dψ ∗ dφ dx dx dx

 b +

0∗

Z b

= ψ (x)φ (x) − ψ (x)φ(x)

a

a

d2 ψ dx2

∗ φ(x) dx .

The boundary terms vanish both for Dirichlet and Neumann conditions; hence hψ|D2 φi = hD2 ψ|φi, establishing D2 as a hermitian operator. (b) For D2 φ = −λφ and Neumann conditions: Integral form:

Z λ a

b

|φ(x)|2 dx = −

Z

b

φ∗ (x)

a





d2 φ dx dx2

0Z Z b 2 *  b  b  dφ 2 dx = dφ dx ≥ 0. (x) + = −φ (x)φ dx dx a a a  ∗

0

So λ ≥ 0. Bra-ket notation: since Neumann conditions renders D anti-hermitian, D† = −D, λhφ|φi = hφ|λ|φi = −hφ|D2 φi = +hDφ|Dφi ≥ 0 .

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319

(c) For D2 φ = −λφ and Dirichlet conditions, the argument is the same — but now we can exclude the zero eigenvalue. That’s because λ = 0 implies φ0 (x) = 0, and thus φ(x) is a constant, including on the boundary — where the Dirichlet condition fixes the constant to be zero. But φ = 0 is not an eigenfunction and hence λ = 0 is ruled out: λ > 0.

40.6 (a) For real x,

Z hψ|Xφi =

b

Z

∗

ψ (x) [xφ(x)] dx =

b

[xψ(x)]∗ φ(x) dx = hXψ|φi .

a

a

So X is hermitian. And if X is hermitian, then so too is X n . Similarly, X n +X m is hermitian. Thus any function f (X) which can be expanded in powers of X is also hermitian. (b) In general, Xf (x) = xf (x) is not an eigenvalue equation since the function xf (x) , f (x). The only solution to Xf (x) = λf (x) is f (x) = δ(x − a), for which Xδ(x − a) = aδ(x − a). (c) Assuming vanishing surface term, hψ|XPφi = hXψ|Pφi = hPX|φi. So (XP)† = PX. Then the product XP is hermitian only if X and P commute. Do they? [X, P]f (x) = (XP − PX) f = (−XiD + iDX) f = −Xi∂x f + iD(xf ) = −ix∂x f (x) + if (x) + ix∂x f = if (x) . So they don’t commute, and XP is not hermitian.

40.7 For α(x), β(x), and γ(x) real: b

Z

ψ ∗ (x)

hψ|Lφi = a

 α

d2 d +β +γ dx2 dx

= αψ ∗ φ0 + βψ ∗ φ


b + a

Z bh a

 φ(x) dx

−

d(αψ ∗ ) dφ (dβψ ∗ ) − φ + γψ ∗ φ dx dx dx dx

i


b = αψ φ + βψ φ − (αψ ) φ + ∗ 0

∗

∗ 0

a

Z b a

d2 (αψ ∗ ) (dβψ ∗ ) + φ− φ + γψ ∗ φ 2 dx dx

  b = α(ψ ∗ φ0 − ψ 0∗ φ) + (β − α0 )ψ ∗ φ a Z b  00 ∗ 0




 dx






α ψ + 2α ψ 0∗ + αψ 00∗ φ − β 0 ψ ∗ + βψ 0∗ φ + γψ ∗ φ dx

+

a



∗ 0

0∗

= α(ψ φ − ψ φ) + (β − α0 )ψ ∗ φ

 b

a

Z b  + a

∗



d2 d + (α00 − β 0 + γ) α 2 + (2α0 − β) dx dx

ψ(x)

φ(x) dx ,

For α0 = β, the differential operator reduces to L, and the surface term becomes

b

α(ψ ∗ φ0 − ψ 0∗ φ) , a

as advertised in Eqn. (40.13) with p ≡ −α.

320

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40.8 Expanding |ψi of A’s eigenbasis, hAi = hψ|A|ψi

! X

=

c∗m ham |

! X

A

m

=

X

=

X

=

X

cn |an i

n

cm c∗n ham |A|an i

m,n

cm c∗n an ham |an i

m,n

cm c∗n an δnm =

X

|cn |2 an ,

n

m,n

where |cn |2 is the probability of measuring an .

40.9 (a) Integrating by parts (Problem 40.1) gives

b

hψ|iDφi = iψ ∗ (x)φ(x) + hiDψ|φi . a

Since square well solutions must vanish at the endpoints, the surface term goes to zero. However, φ = 0 at both a and b is sufficient to achieve this — there is no condition of ψ. Thus the operator is hermitian, but not self-adjoint. As a result there is no eigenbasis of P and hence momentum is not an observable quantity of the infinite square well. (Indeed, the eigenfunctions of P are e±ikx , which is never zero.) (b) Integrating by parts twice (Problem 40.3) gives

b

hψ|D2 φi = ψ ∗ φ0 − ψ 0∗ φ + hD2 ψ|φi . a

Now both φ and ψ must vanish at the well boundaries, rendering P 2 (and hence, the energy P 2 /2m) both hermitian and self-adjoint. So the well energies are observable, with a complete basis of eigenfunctions. If this seem strange, remember that even in this one-dimensional well, momentum is a signed quantity: the eigenstates are standing waves, (sin kx, cos kx), not traveling waves (e±ikx ) — and the energy only tells us the magnitude of the momentum. (c) To be self-adjoint, we need the same condition to apply to both φ and ψ. Examining the surface term reveals:

b

ψ ∗ (x)φ(x) = 0 a

=⇒

[ψ(b)/ψ(a)]∗ = φ(a)/φ(b) .

Since φ and ψ are independent functions, this ratio must be a constant independent of a and b. The only constant whose reciprocal is its complex conjugate is a pure phase, eiθ for some constant θ. Thus the most general boundary condition which yields a self-adjoint P is χ(a) = eiθ χ(b) for any eigenfunction χ(x) of P . Note that this is incompatible with φ(a) = φ(b) = 0. The case θ = 0 is familiar as periodic boundary conditions; for non-zero θ, the solution is phase shifted at the boundary.

40.10 (a) Straightforward integration gives

r r cn =

2 L

30 L3

Z 0

L

x sin (nπx/L) x 1 − L





 8 √15 =

π 3 n3

0,

,

n odd . n even

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321

(b) hH 2 i =

X

X  8 √15 2  n2 π2 ~2 2

2 = |cn |2 En

π 3 n3

2mL2

n odd

n

240~4 π 2 m2 L4

=

X 1

=

n2

240~4 π 2 30~4 = 2 4 . π 2 m2 L4 8 m L

n odd

(c) Since H 2 =



2

~ − 2m

d2 dx2

2

and ψ is quadratic, H 2 ψ = 0 and thus hH 2 i = hψ|H 2 |ψi = 0.

(d) For the square well boundary conditions ψ(0) = ψ(L) = 0, two integrations by parts shows ~2 d2 that H = − 2m is self adjoint (Problem 40.3). However for H 2 to be self-adjoint requires dx2 four integrations by parts — revealing the surface term hψ|H 2 φi = ψ ∗ φ000 − ψ 0∗ φ00 + ψ 00∗ φ0 − ψ 000∗ φ


b + hH 2 ψ|φi . a

Thus to be self adjoint requires not only ψ(0) = ψ(L) = 0, but either ψ 0 (0) = ψ 0 (L) = 0 or ψ 00 (0) = ψ 00 (L) = 0 as well. And our wavefunction ψ doesn’t satisfy either of these additional conditions. In other words, it’s simply not true that hH 2 i = hψ|H 2 ψi for this function — rather, since ψ is in the domain of H, we can show hH 2 i = hHψ|Hψi = 30~4 /m2 L4 . X 00 − 2xH 0 + 2nH = 0 in Sturm-Liouville form, we need 40.11 To put Hermite’s equation Hn n n

Z exp



2 −2x dx = e−x . 1

Applying this from the left on Hermite’s equation yields −

2 2 d 0 e−x Hn = 2n e−x Hn , dx

h

i

2

2

which is a Sturm-Liouville equation with p = e−x , q = 0, λ = 2n, and w = e−x .

40.12 To put Chebyshev’s equation (1 − x2 )Tn00 − xTn0 + n2 Tn = 0 in Sturm-Liouville from, we need w(x) =

1 exp 1 − x2

Z

−x 1 − x2

 =

1 1 1 exp − log(1 − x2 ) = √ . 1 − x2 2 1 − x2

h

i

Applying this from the left on Chebyshev’s equation yields −

d dx

hp

n2

i

1 − x2 Tn0 = √

which is a Sturm-Liouville equation with p =

√

1 − x2

Tn ,

1 − x2 , q = 0, λ = n2 , and w = √

1 . 1−x2

40.13 (a) Reading off from xLkn 00 + (k + 1 − x)Lkn 0 + nLkn = 0, 1 w(x) = exp x

Z



k+1−x 1 dx = exp [(k + 1) ln x − x] = xk e−x . x x

Unlike the associated Legendre functions (Example 40.2), there’s only one eigenvalue, n, and hence orthogonality only amongst associated Laguerre polynomials Lkn and Lkm with n , m. So different k’s merely define a family of Sturm-Liouville differential equations, each of which has its own complete basis of orthogonal eigenfunctions.

322

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(b) From the orthogonality of the associated Laguerre polynomials, we have (the normalization was found in Problem 32.24 — though it’s not relevant here) (k + n)! δnm = n!

∞

Z

xk e−x Lkn (x)Lkm (x) dx

0 ∞

Z



xk/2 e−x/2 Lkn (x)

=

Z0 ∞ ≡





xk/2 e−x/2 Lkm (x) dx

φkn (x)φkm (x) dx .

0

Then with Lkm (x) ≡ x−k/2 e+x/2 φkm (x),


d d k d d xk+1 e−x Lm (x) = xk+1 e−x x−k/2 e+x/2 φkm (x) dx dx dx dx h
i 1 +k/2 −x/2 =x e xφkm 00 + φkm 0 − x2 − 2x − 2kx + k2 φkm . 4x i

h

i

h

so that the associated Laguerre equation becomes


1 x2 − 2(k + 1)x + k2 φkm = −nφkm , 4x

xφkm 00 + φkm 0 −

which is in Sturm-Liouville form with weight 1 and eigenvalue m.

40.14 Recall that the Sturm-Liouville operator L has real p, q and w. Thus 0 = (Lφ − λwφ)∗ = Lφ∗ − λwφ∗ , since λ = λ∗ . Now for separated boundary conditions, the eigenvalue spectrum is non-degenerate. So it must be that the eigenfunction is real, φ = φ∗ . With periodic conditions, the eigenfunctions can be doubly-degenerate. The above equations shows they can be expressed as complex conjugates of one another. Combining those, L (φ ± φ∗ ) = λw (φ ± φ∗ ) , we choose as independent solutions the two real functions φ1 =

1 2

(φ + φ∗ ) and φ2 =

1 2i

(φ − φ∗ ).

40.15 With separated boundary conditions, c1 χ(a) + d1 χ0 (a) = 0

c2 χ(b) + d2 χ0 (b) = 0 ,

b

the surface term p(ψ 0∗ φ − ψ ∗ φ0 ) goes to zero at a and b separately. And with a vanishing a

surface term, the operator is hermitian. Similarly, for periodic boundary conditions, p(a) = p(b) and χ(a) = χ(b)

χ0 (a) = χ0 (b) ,

the value of surface term is

b







p(ψ 0∗ φ − ψ ∗ φ0 ) = p(b) ψ 0∗ (b)φ(b) − ψ ∗ (b)φ0 (b) − p(a) ψ 0∗ (a)φ(a) − ψ ∗ (a)φ0 (a) a





= [p(b) − p(a)] ψ 0∗ (b)φ(b) − ψ ∗ (b)φ0 (b) = 0

e 2 r , 40.16 (a) Multiplying both sides by − 2m ~2

d dr



r

2 dR

dr



 −



2me e2 −2me E 2 r − `(` + 1) R = r R, ~2 4π0 ~2

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323

or, with κ =

√

−2me E/~ > 0 and z ≡ κr, d dz



z2

dR dz



+

h

2z − `(` + 1) R = z 2 R , κa0

i

where a0 = ~/me cα. With  = 2/κa0 , we move the terms proportional to z and z 2 to their opposite sides to get d dz



z2

dR dz







− z 2 + `(` + 1) R = −zR.

(b) To get orthogonal eigenfunctions with weight 1, define φ(z) = d dz



z2

dR dz



=

d d z2 dz dz

h



φ √ z

i

=

√

p

w(z)R =



z zφ00 + φ0 −

1 φ 4z



√

zR. Then

.

Using this in the radial equation gives zφ00 + φ0 −

 1  2 4z + (2` + 1)2 φ = −φ . 4z

(c) Using x = 2z in Eqn. (40.55) gives zφkm 00 + φkm 0 −

 1  2 4z − 4(k + 1)z + k2 φkm = −2mφkm , 4z

where φ0 ≡ dφ/dz and m is an integer index (not to be confused with the electron mass me ). Comparing this with the our differential equation reveals k = 2` + 1 and then, since the numerical coefficients of φkm must sum to : (k + 1) + 2m = 

=⇒

 = 2(m + ` + 1) ,

or 1/κa0 = /2 = m + ` + 1 ≡ n for integer n ≥ 0. Using κ2 a20 = 1/n2 = −2E/ne c2 α2 , we find eigenenergies En = −

1 me c2 α2 13.6 eV =− . 2 n2 n2

The radial wavefunctions are found by tracing our steps in reverse: with x = 2z, Eqn. (40.55) has solutions φ(z) = (2z)k/2 e−z Lkm (2z) . With k = 2` + 1 and m = n − ` − 1, this becomes 1

(2z) , φ(z) = (2z)`+ 2 e−z L2`+1 n−`−1 √ Now R = φ/ z, and we know from the energies En that z = κr scales like r/n. Altogether then, we get (up to normalization) Rn` (r) = (2r/na0 )` e−r/na0 L2`+1 (2r/na0 ) , n−`−1 where we label the solutions with the physically meaningful indices n (energy) and ` (angular momentum). [Note that since the energies only depend on n, the system is highly degenerate.]

2

d 40.17 (a) With periodic boundary conditions on ψ, the operator − dθ 2 + (1 − cos θ) is hermitian. Hence the eigenfunctions form a complete orthogonal basis. (b) The Schrödinger equation for the potential (1 − cos θ) can be converted into the form of 2

2

3

Mathieu’s equation with a = 8m` (E − mg`), q = 4m~2g` , and 2η = θ − π. [Placing the ~2 zero of potential energy at θ = π/2 rather than θ = 0 would simplify things a bit.]

324

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(c)

(d) First, Sturm-Liouville orthogonality applies only to eigenfunctions of the same operator. Thus the orthogonality of cen (q, η) and sen (q, η) only holds for the same value of q; orthogonality must be over η. Second, as expected from their names and can be seen from their plots, the elliptic cosine and sine functions cen (q, η) and sen (q, η) have even and odd party, respectively. So they are certainly orthogonal to one another,

Z

π

cen (q, η) sen (q, η)dη = 0 . −π

Furthermore, for q = 0 these functions reduce to cosine and sine; a little playing with Mathematica reveals cen (0, η) = cos(nη) and sen (0, η) = sin(nη). And indeed, for q = 0 the eigenvalues are an = bn = n2 . Thus we know that integrating pairs over a 2π-period vanishes hn|miq=0 = 0 for n , m. One is then motivated to propose

Z

π

hn|miq =

Z

π

cen (q, η) cem (q, η)dη = −π

sen (q, η) sem (q, η)dη = πδnm , −π

as Mathematica readily confirms. (e) The boundary conditions in θ are periodic over 2π, so the solutions must be periodic in η with period π. Inspection of the plots shows that π periodicity only occurs for even n. So the ~2 a energy eigenfunctions are ce2n and se2n . The energy eigenvalues are E = 2m` 2 4 + mg`. The five lowest energies for q = 4 are determined by the five lowest values of the characteristic functions a2n (q) and b2n (q). Mathematica gives a0 (4) = −4.281 b2 (4) = 2.747 a2 (4) = 6.829 b4 (4) = 16.452 a4 (4) = 16.650

40.18 (a) The Schrödinger equation for the square well is φ00 = − 2mE φ. So — ~2 E = 0 : φ(x) = ax + b and E < 0 : φ(x) = aeκx + be−κx ,

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

325

where κ =

p

2m|E|/~2 . In both cases, boundary conditions φ(0) = φ(L) = 0 give φ(x) = 0. 2m(E−V )

(b) We only need examine the Schrödinger equation φ00 = − φ: If E > Vmin for all x, ~2 then there will be intervals in which φ and φ00 have opposite signs, allowing φ to curve back toward the x-axis — and so have the possibility of being square integrable.

40.19 For

d2 φ dξ2

= − [ε − v(ξ)] φ(ξ), we take v0 = 25 and define the Mathematica function v[ξ_] := Piecewise[{{0, Abs[ξ] < 0.5}, {25, Abs[ξ] > 0.5}}] ,

since the well sits between ξ = ±1/2. We can solve and plot even-parity solutions using evensolution = NDSolve[{φ00 [ξ] == −( − v[ξ])φ[ξ], φ[0] == 1, φ0 [0] == 0}, φ, {ξ, −3, 3}]; Plot[φ[ξ]/.evensolution, {ξ, −xrange, xrange}] varying  until a well-behaved solution emerges. We’ll also need to vary xrange upward in order to find  to 5 significant figures. I find  = 4.9329. We can do the same for odd solutions with only a change of the initial conditions, oddsolution = NDSolve[{φ00 [ξ] == −( − v[ξ])φ[ξ], φ[0] == 0, φ0 [0] == 1}, φ, {ξ, −2, 2}]; Plot[φ[ξ]/.oddsolution, {ξ, −xrange, xrange}] This time, I get  = 18.068. 1

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Problem 40.19: Zeroing in on the eigenvalue: (a) ground state; (b) 1st excited state

40.20 Choosing L = 1 for simplicity, we can use Mathematica’s Reduce command to find the roots of tan(k) = −1/k. Since tan x passes through zero every π, the first three roots should be between 0 and 3π — so we need only ask Reduce to search k in this range: kn = Reduce[Tan[k]==-1/k && 0 0, −

Z

∞

 v[ξ]ψ1(ξ)2 dξ

ψ1(ξ)D[ψ1(ξ), ξ, 2]dξ + −∞

−∞

A plot of e1[b] shows that its minimum is around 3 — so that’s the vicinity we’ll tell FindRoot to look: e1[b]/.FindRoot[D[e1[b], b] == 0, {b, 3}] which returns 1 = 5.03 — a much better estimate! (c) Repeating for the first excited state, we use the odd trial function ψ2 (ξ) =



32b3 π

1/4

2

x e−bξ .

Then



Z

∞

e2[b_] = Assuming b > 0, −

Z

∞

−∞

 v[ξ]ψ2(ξ)2 dξ

ψ2(ξ)D[ψ2(ξ), ξ, 2]dξ + −∞

and e2[b]/.FindRoot[D[e2[b], b] == 0, {b, 3}] returns 2 = 18.49 — which compares quite favorably to 18.07 found in Problem 40.19.

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331

Partial Differential Equations

41

41.1 Inserting the ansatz T (x, t) ≡ ψ(x)τ (t) into the heat equation gives ∂ 2 [ψ(x)τ (t)] 1 ∂[ψ(x)τ (t)] = ∂x2 α ∂t

→

τ

d2 ψ 1 dτ = ψ . dx2 α dt

Dividing both sides by ψτ , 1 d2 ψ 1 dτ = . ψ dx2 ατ dt Since the left is solely a function of x, the right of t, both sides must equal the same constant −k2 . Thus we get the separated ODEs d2 ψ = −k2 ψ(x) dx2

dτ = −αk2 τ (t) . dt

41.2 Inserting the ansatz Ψ(~ r, t) ≡ ψ(~ r)τ t into the Schrödinger equation





−

~2 2 ∇ + V (~ r) 2m

Ψ(~ r, t) = i~

∂Ψ(~ r, t) , ∂t

gives

 τ (t)



~2 2 − ∇ + V (~ r) 2m

ψ(~ r, t) = i~ψ(~ r)

dτ (t) , dt

Dividing both sides by ψτ gives 1 ψ(~ r)





−

~2 2 ∇ + V (~ r) 2m

ψ(~ r, t) = i~

1 dτ (τ ) . τ (t) dt

The left is solely a function of ~ r, the right of t, so both sides must equal the same constant. Since the operators on both left and right have units of energy, we’ll denote that constant as E. For τ this gives dτ (τ ) −iE = τ → τ (t) = Ce−iEt/~ = Ce−iωt , dt ~

E ≡ ~ω

and

 −

~2 + V (~ r) 2m

 ψ(~ r, t) = Eψ(~ r)

which is the time-independent Schrödinger equation.

41.3 If the end at x = 0 is free rather than fixed, a Neumann condition applies there, ∂x Ψ(0, t) = 0. This picks out the cosine rather than the sine, replacing (41.23) with Ψk (x, t) = cos(nπx/2L) [ck cos ωk t + dk sin ωk t] , ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

where the Dirichlet condition at x = L requires n odd. The normal mode frequencies are thus ωn = nπv/2L for odd n — or equivalently, ωn = (2n − 1)πv/2L for n = 1, 2, 3, . . .

41.4 For h(y, t) = u(y)τ (t), the modified wave equation in (41.29) becomes d2 τ v(y) 2 = τ (t)g dt



d2 v dv y 2 + dy dy



or, introducing separation constant −ω 2 , d2 τ = −ω 2 τ dt2

−→

τ (t) = Ce±iωt .

and y

d2 v dv 1 + = − ω 2 v(y) . dy 2 dy g

For z 2 = 4y/g, d dz d 2 1 d = = dy dy dz g z dz and d2 dz d = dy 2 dy dz



2 1 d g z dz



=



4 1 g2 z

1 d 1 d2 − 2 z dz 2 z dz

 .

Inserting these into the separated spatial differential equation, with v(y) → u(z), gives z2

du d2 u +z + z2 ω2 u = 0 , dz 2 dz

which is Bessel’s equation with solutions J0 (ωz) and N0 (ωz).

41.5 The matrix of (41.6) for the operator ∂x ∂y ,

 A=

0 1/2



1/2 0

,

has negative determinant. Hence the is hyperbolic. The matrix A has eigenvalues ±1/2  operator  1 with normalized eigenvectors √1 . So the linear combinations on the diagonal basis are 2 ±1 √ √ u = (x + y)/ 2 and v = (x − y)/ 2. Then ∂ ∂u ∂ ∂v ∂ 1 = + = √ ∂x ∂x ∂u ∂x ∂v 2



∂ ∂ + ∂u ∂v

∂ ∂u ∂ ∂v ∂ 1 = + = √ ∂y ∂y ∂u ∂y ∂v 2





∂ ∂ − ∂u ∂v



to give the canonical form in (41.7) for a hyperbolic operator, ∂x ∂ y =


1 ∂u2 − ∂v2 . 2

41.6 (a) A function u(ζ) = u(x ± vt) has derivatives ∂x2 u(x ± vt) = u00 (x ± vt)

∂t2 u(x ± vt) = v 2 u00 (x ± vt)

where u00 ≡ d2 u/dζ 2 . Such a function clearly satisfies the wave equation. The general solution is thus a superposition of both directions of propagation, ψ(x, t) = f (x − vt) + g(x + vt) .

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333

(b) The initial conditions are evaluated at ζ = x; in terms of f and g, we have ψ(x, 0) = α(x) = f (x) + g(x) and

Z

∂t ψ(x, 0) = β(x) = −vf 0 (x) + vg 0 (x)

x

−→

β(ξ)dξ = −vf (x) + vg(x) .

Inverting yields 1 1 f (x) = α(x) − 2 2v

x

Z

1 1 g(x) = α(x) + 2 2v

β(ξ)dξ a

Z

x

β(x)dξ , a

for some reference point a. Thus the full solution is ψ(x, t) = f (x − vt) + g(x + vt) =

1 1 [α(x − vt) + α(x + vt)] + 2 2v

Z

x+vt

β(x)dξ , x−vt

where the lower limit accounts of the sign in front of the integral in f (x). Since this expression is the sum of functions of x ± vt, it’s necessarily a solution to the wave equation. It’s not hard to verify that it also satisfies the initial conditions. (c) The solution found in (b) reveals that at a given point (x0 , t0 ), ψ requires data only from points x in the closed interval [x0 − vt0 , x0 + vt0 ] — the interior for the integral over ∂t ψ(x, 0), and the endpoints in order to specify ψ(x, 0). This creates a cone with slope determined by v, as shown. (Note that time is on the vertical axis, so the slope is 1/v.) Any point x on and within the cone can affect the value of ψ(x0 , t0 ). The reverse must then also be true: any point x within a distance |x − x0 | = vt can be affected by x0 . In other words, signals propagate with finite speed no greater than v (e.g., about 350 m/s for sound, 3 × 108 m/s for light), and so can influence only points within a distance vt away.

t

t

AAAB53icbVBNS8NAEJ3Ur1q/qh69LBbBU0mkVI8FLx5bsB/QhrLZbtqlm03YnQgl9Bd48KLi1Z/k0X/jts1BWx8MPN6bYWZekEhh0HW/ncLW9s7uXnG/dHB4dHxSPj3rmDjVjLdZLGPdC6jhUijeRoGS9xLNaRRI3g2m9wu/+8S1EbF6xFnC/YiOlQgFo2ilFg7LFbfqLkE2iZeTCuRoDstfg1HM0ogrZJIa0/fcBP2MahRM8nlpkBqeUDalY963VNGIGz9bHjonV1YZkTDWthSSpfp7IqORMbMosJ0RxYlZ9xbif14/xfDOz4RKUuSKrRaFqSQYk8XXZCQ0ZyhnllCmhb2VsAnVlKHNxmbgrX+8STo3Va9erbVqlUY9T6MIF3AJ1+DBLTTgAZrQBgYcnuEV3hzhvDjvzseqteDkM+fwB87nD1ypjOU=

AAAB53icbVBNS8NAEJ3Ur1q/qh69LBbBU0mkVI8FLx5bsB/QhrLZbtqlm03YnQgl9Bd48KLi1Z/k0X/jts1BWx8MPN6bYWZekEhh0HW/ncLW9s7uXnG/dHB4dHxSPj3rmDjVjLdZLGPdC6jhUijeRoGS9xLNaRRI3g2m9wu/+8S1EbF6xFnC/YiOlQgFo2ilFg7LFbfqLkE2iZeTCuRoDstfg1HM0ogrZJIa0/fcBP2MahRM8nlpkBqeUDalY963VNGIGz9bHjonV1YZkTDWthSSpfp7IqORMbMosJ0RxYlZ9xbif14/xfDOz4RKUuSKrRaFqSQYk8XXZCQ0ZyhnllCmhb2VsAnVlKHNxmbgrX+8STo3Va9erbVqlUY9T6MIF3AJ1+DBLTTgAZrQBgYcnuEV3hzhvDjvzseqteDkM+fwB87nD1ypjOU=

AAAB73icbVBNSwMxEM36WetX1aOXYBEqSNktRT0WvHisYD+kXZZsmrahSXZNZsWy9Fd48KLi1Z/j0X9j2u5BWx8MPN6bYWZeGAtuwHW/nZXVtfWNzdxWfntnd2+/cHDYNFGiKWvQSES6HRLDBFesARwEa8eaERkK1gpH11O/9ci04ZG6g3HMfEkGivc5JWCl+9JT4J5D4J4FhaJbdmfAy8TLSBFlqAeFr24voolkCqggxnQ8NwY/JRo4FWyS7yaGxYSOyIB1LFVEMuOns4Mn+NQqPdyPtC0FeKb+nkiJNGYsQ9spCQzNojcV//M6CfSv/JSrOAGm6HxRPxEYIjz9Hve4ZhTE2BJCNbe3YjokmlCwGdkMvMWPl0mzUvYuytXbarFWydLIoWN0gkrIQ5eohm5QHTUQRRI9o1f05jw4L8678zFvXXGymSP0B87nD7Bmj0Q=

(x0 , t0 )

x

x0

vt0

x0 + vt0

AAAB7HicbVBNS8NAEJ3Ur1q/qh69BIvgxZBIUY8FLx4r2A9oQ9lsN+3azSbsTool9D948KLi1f/j0X/jts1BWx8MPN6bYWZekAiu0XW/rcLa+sbmVnG7tLO7t39QPjxq6jhVlDVoLGLVDohmgkvWQI6CtRPFSBQI1gpGtzO/NWZK81g+4CRhfkQGkoecEjRS86nnXoyxV664jjuHvUq8nFQgR71X/ur2Y5pGTCIVROuO5yboZ0Qhp4JNS91Us4TQERmwjqGSREz72fzaqX1mlL4dxsqURHuu/p7ISKT1JApMZ0RwqJe9mfif10kxvPEzLpMUmaSLRWEqbIzt2et2nytGUUwMIVRxc6tNh0QRiiYgk4G3/PEqaV463pVTva9Wak6eRhFO4BTOwYNrqMEd1KEBFB7hGV7hzZLWi/VufSxaC1Y+cwx/YH3+AJwjjrk=

AAAB8XicbVBNS8NAEJ3Ur1q/qh69LBZBEEIiRT0WvHisYGuhDWGz3bRLN7thd1MsoT/DgxcVr/4aj/4bt20O2vpg4PHeDDPzopQzbTzv2ymtrW9sbpW3Kzu7e/sH1cOjtpaZIrRFJJeqE2FNORO0ZZjhtJMqipOI08dodDvzH8dUaSbFg5mkNEjwQLCYEWys1H0KPXSBxsiEXlitea43B1olfkFqUKAZVr96fUmyhApDONa663upCXKsDCOcTiu9TNMUkxEe0K6lAidUB/n85Ck6s0ofxVLZEgbN1d8TOU60niSR7UywGeplbyb+53UzE98EORNpZqggi0VxxpGRaPY/6jNFieETSzBRzN6KyBArTIxNyWbgL3+8StqXrn/l1u/rtYZbpFGGEziFc/DhGhpwB01oAQEJz/AKb452Xpx352PRWnKKmWP4A+fzB8WRj9g=

x0

vt

AAAB7HicbVBNS8NAEJ3Ur1q/qh69BIsgCCGRoh4LXjxWsB/QhrLZbtq1m03YnRRL6H/w4EXFq//Ho//GbZuDtj4YeLw3w8y8IBFco+t+W4W19Y3NreJ2aWd3b/+gfHjU1HGqKGvQWMSqHRDNBJesgRwFayeKkSgQrBWMbmd+a8yU5rF8wEnC/IgMJA85JWik5lPPvRhjr1xxHXcOe5V4OalAjnqv/NXtxzSNmEQqiNYdz03Qz4hCTgWblrqpZgmhIzJgHUMliZj2s/m1U/vMKH07jJUpifZc/T2RkUjrSRSYzojgUC97M/E/r5NieONnXCYpMkkXi8JU2Bjbs9ftPleMopgYQqji5labDokiFE1AJgNv+eNV0rx0vCunel+t1Jw8jSKcwCmcgwfXUIM7qEMDKDzCM7zCmyWtF+vd+li0Fqx85hj+wPr8AZkZjrc=

x0 AAAB6XicbVBNT8JAEJ3iF+IX6tHLRmLiibSEqEcSLx4xCphAQ7bLFDZst83u1kgafoIHL2q8+os8+m9coAcFXzLJy3szmZkXJIJr47rfTmFtfWNzq7hd2tnd2z8oHx61dZwqhi0Wi1g9BFSj4BJbhhuBD4lCGgUCO8H4euZ3HlFpHst7M0nQj+hQ8pAzaqx099R3++WKW3XnIKvEy0kFcjT75a/eIGZphNIwQbXuem5i/Iwqw5nAaamXakwoG9Mhdi2VNELtZ/NTp+TMKgMSxsqWNGSu/p7IaKT1JApsZ0TNSC97M/E/r5ua8MrPuExSg5ItFoWpICYms7/JgCtkRkwsoUxxeythI6ooMzYdm4G3/PEqadeq3kW1fluvNGp5GkU4gVM4Bw8uoQE30IQWMBjCM7zCmzN2Xpx352PRWnDymWP4A+fzB4TUjYg=

41.7 For x = cos θ, 1 d 1 dx d d = =− sin θ dθ sin θ dθ dx dx and sin θ

d d = −(1 − x2 ) . dθ dx

Then with χ(x) = Θ(θ): 1 d 0= sin θ dθ =+

334

d dx

h



dΘ sin θ dθ

1 − x2



 +


dχ i dx

x AAAB53icbVBNS8NAEJ3Ur1q/qh69LBbBU0lEqseCF48t2A9oQ9lsJ+3SzSbsbsQS+gs8eFHx6k/y6L9x2+agrQ8GHu/NMDMvSATXxnW/ncLG5tb2TnG3tLd/cHhUPj5p6zhVDFssFrHqBlSj4BJbhhuB3UQhjQKBnWByN/c7j6g0j+WDmSboR3QkecgZNVZqPg3KFbfqLkDWiZeTCuRoDMpf/WHM0gilYYJq3fPcxPgZVYYzgbNSP9WYUDahI+xZKmmE2s8Wh87IhVWGJIyVLWnIQv09kdFI62kU2M6ImrFe9ebif14vNeGtn3GZpAYlWy4KU0FMTOZfkyFXyIyYWkKZ4vZWwsZUUWZsNjYDb/XjddK+qnq16nXzulKv5WkU4QzO4RI8uIE63EMDWsAA4Rle4c3hzovz7nwsWwtOPnMKf+B8/gBitYzp

AAAB53icbVBNS8NAEJ3Ur1q/qh69LBbBU0lEqseCF48t2A9oQ9lsJ+3SzSbsbsQS+gs8eFHx6k/y6L9x2+agrQ8GHu/NMDMvSATXxnW/ncLG5tb2TnG3tLd/cHhUPj5p6zhVDFssFrHqBlSj4BJbhhuB3UQhjQKBnWByN/c7j6g0j+WDmSboR3QkecgZNVZqPg3KFbfqLkDWiZeTCuRoDMpf/WHM0gilYYJq3fPcxPgZVYYzgbNSP9WYUDahI+xZKmmE2s8Wh87IhVWGJIyVLWnIQv09kdFI62kU2M6ImrFe9ebif14vNeGtn3GZpAYlWy4KU0FMTOZfkyFXyIyYWkKZ4vZWwsZUUWZsNjYDb/XjddK+qnq16nXzulKv5WkU4QzO4RI8uIE63EMDWsAA4Rle4c3hzovz7nwsWwtOPnMKf+B8/gBitYzp

AAAB8XicbVBNS8NAEJ3Ur1q/qh69LBbBiyGRoh4LXjxWsLXQhrDZbtqlm92wuymW0J/hwYuKV3+NR/+N2zYHbX0w8Hhvhpl5UcqZNp737ZTW1jc2t8rblZ3dvf2D6uFRW8tMEdoikkvVibCmnAnaMsxw2kkVxUnE6WM0up35j2OqNJPiwUxSGiR4IFjMCDZW6j6FHrpAY2RCL6zWPNebA60SvyA1KNAMq1+9viRZQoUhHGvd9b3UBDlWhhFOp5VepmmKyQgPaNdSgROqg3x+8hSdWaWPYqlsCYPm6u+JHCdaT5LIdibYDPWyNxP/87qZiW+CnIk0M1SQxaI448hINPsf9ZmixPCJJZgoZm9FZIgVJsamZDPwlz9eJe1L179y6/f1WsMt0ijDCZzCOfhwDQ24gya0gICEZ3iFN0c7L86787FoLTnFzDH8gfP5A8ijj9o=

m2 λ− sin2 θ

 +

λ−



m2 1 − x2

Θ(θ)

 χ

X

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

x0 + vt

41.8 In spherical coordinates, Helmholtz equation is 1 1 1 ∂r (r2 ∂r ψ) + 2 ∂θ (sin θ ∂θ ψ) + 2 ∂ 2 ψ + k2 ψ = 0 . r2 r sin θ r sin2 θ φ Inserting the ansatz ψ(~ r) = R(r)Θ(θ)Φ(φ) and dividing by ψ gives 1 1 d R r2 dr



r2

dR dr



1 1 d Θ r2 sin θ dθ

+



sin θ

dΘ dr



1 1 d2 Φ + k2 = 0 . 2 2 Φ r sin θ dφ2

+

Multiplying through by r2 sin2 θ separates the Φ equation, −

1 d2 Φ 1 d = sin2 θ Φ dφ2 R dr



r2

dR dr



+

1 d sin θ Θ dθ



sin θ

dΘ dr



+ k2 r2 sin2 θ

Since the left only depends of φ, the right r and θ, both sides must equal the same constant; denoting it as m2 gives the first separated ODE, d2 Φ = −m2 Φ . dφ2 Dividing the remaining PDE by sin2 θ leads to the second separation, 1 1 d Θ sin θ dθ



sin θ

dΘ dr



−

m2 1 d =− sin2 θ R dr



r2

dR dr



− k2 r2 .

Setting both sides equal to −λ gives the ODEs 1 d sin θ dθ



d dr



sin θ

dΘ dr



 +

λ−

m2 sin2 θ

 Θ(θ) = 0

and r2

dR dr






+ k2 r2 − λ R(r) = 0 .

41.9 (a) Affixing a mass m at the end of the string simply adds weight mg to the tension T . This addition subtly alters (41.29) to become ∂2h ∂2h ∂h = g (µ + y) +g , 2 ∂t ∂y 2 ∂y where µ = m/ρ. The effect is just a constant translation of y, so we still have Bessel’s equation — but with the replacement z 2 → z 2 = 4(y + µ)/g. So the solution still has the general form u(z) = AJ0 (ωz) + BN0 (ωz). (b) The ends at y = 0 and L now respectively correspond to

where cbot = becomes

p

1+

µ , L

zbot = 2

p

(L + µ)/g = 2cbot /ω0

ztop = 2

p

µ/g = 2ctop /ω0 ,

ctop =

pµ L

, and ω02 = g/L. At the top, the boundary condition

AJ0 (2ctop ω/ω0 ) + BN0 (2ctop ω/ω0 ) = 0 . A large mass attached at the bottom end inhibits the string from swinging freely, making it effectively fixed AJ0 (2cbot ω/ω0 ) + BN0 (2cbot ω/ω0 ) = 0 . Although the differential equation is the same, the boundary conditions are not. In particular, since z = 0 no longer describes the bottom of the string, we cannot discard the Neumann

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

335

function. Thus we have two simultaneous equations for A and B, their non-trivial solution guaranteed by the vanishing of their coefficients, J0



2ctop ω ω0



N0



2cbot ω ω0



− J0



2cbot ω ω0



N0



2ctop ω ω0



=0

This is the condition we need to solve to find the normal mode frequencies. (c) For general m, applying the boundary conditions necessitates numerical calculation. But for large m, the tension is dominated by the weight at the end, T ≈ mg. So this configuration should have normal mode frequencies that approach those of a horizontal string fixed at both ends. In this large m (large µ) limit, the problem greatly simplifies with the application of the asymptotic forms (C.8) and (C.13) x large

r

J0 (x) −−−−→

2 π cos x − πx 4





x large

r

N0 (x) −−−−→

2 π sin x − πx 4





apply. (d) Inserting the asymptotic forms into the condition derived in (b) yields

h

sin 2

ω (cbot − ctop ) = 0 ω0

i

−→

2

ω (cbot − ctop ) = nπ ω0

for integer n > 0. To first order, cbot − ctop =

q

µ 1+ − L

q

µ = L

q r



L 1L 1 L2 1+ −1 ≈ − . µ 2µ 4 µ2

µ L

and thus we obtain normal mode frequencies ωn =

1 ω0 nπ = nπ 2 cbot − ctop

q h

µ 1L nπ 1+ ω0 = L 4µ L

i

r

mg 1 ρL 1+ ρ 4 m

h

i

.

Note that the zeroth order term is precisely the normal modes of the horizontal string with tension T = mg. As anticipated.

41.10 In cartesian coordinates in two dimensions, Laplace’s equation for V = χΥ separates into 1 d2 χ = −α2 χ dx2

1 d2 Υ = −β 2 Υ dy 2

with α2 + β 2 = 0. Though the boundary conditions V (0, y) = V (a, y) = 0

V (x, 0) = V0

V (x, y → ∞) = 0

imply α2 > 0, here we’ll assume β 2 > 0. With this choice, the solutions acquire the form χ(x) = Aeβx + Be−βx Υ(y) = C cos(βy) + D sin(βy) , The boundary condition at x = 0 requires B = −A, leaving χ(x) proportional to sinh βx. But the condition at x = a cannot be satisfied unless β = inπ/a for integer n > 0. This flips the sign of β 2 , ultimately leading to the same solution as before, (41.91).

41.11 In Example 41.5, the Laplacian problem is formulated in two space dimensions, whereas a heat diffusion problem along a one-dimensional rod is in one space and one time. Still, both are effectively two-dimensional configurations. Moreover, in both cases one of the dimensions is unbound: y in the conductor, t in the rod. So the boundary conditions for a rod with ends held at T = 0 are identical to those of the conductor. The only difference is the separated y and t equations: the former is second order in y, the latter first order in t. So in place of 2 Υ(y) = e−nπy/a , we get τ (t) = e−α(nπ/a) t — see Example 41.3.

336

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

41.12 (a) V (x, y) = V0 is clearly a solution — and since Laplacian solutions with Dirichlet boundary conditions are unique, this must be the solution. Indeed, extremal values of Laplacian solutions only occur on the boundary — so if the entire boundary sports the same value, V (x, y) must be a constant. (b) Consultation with Figure 41.2Boundary condition in superposition.figure.5645.41.2 shows that the solution we need is V (x, y) + V (a − x, b − y) , where V (x, y) is the solution (41.98) in Example 41.6. The figure gives the Mathematica plot for the sum over odd n up to 9.

41.13 (a) For the given boundary conditions, (41.54) suggests a solution of the form Vnm (x, y, z) = [Anm cosh `x + Bnm sinh `x] sin(nπy/a) sin(mπz/a) , where

`2

= n2 + m2 . We can satisfy the condition at x = a with Bnm = −Anm coth `a,

h

Vnm (x, y, z) = Anm 1 −

coth `a cosh(`x) sin(nπy/a) sin(mπz/a) . coth `x

i

The coefficients Anm are determined by orthogonality in y and z at x = 0 (note that the term in square brackets goes to 1 at x = 0): Anm =

4V0 a2

Z

a

sin(nπy/a) sin(mπz/a)dy dz , 0

to get Anm =

16V0 1 , π 2 nm

n, m odd.

Thus the solution is V (x, y, z) =

16V0 π2

X n,m odd

  √ 1 coth( n2 + m2 a) 1− √ nm coth( n2 + m2 x) × cosh(

p

n2 + m2 x) sin(nπy/a) sin(mπz/a)

(b) For the cube with all faces grounded except for V = V0 at z = a, a similar analysis gives V (x, y, z) =

16V0 π2

X n,m odd

p 1 1 sin(nπx/a) sin(nπy/a) sinh( n2 + m2 z). √ nm sinh( n2 + m2 a)

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

337

(This case is a bit simpler because vanishing potential at z = 0 gives Anm = 0.) The full solution for faces x = 0 and z = a held at V0 is then the superposition of this with the solution in part (a).

41.14 For the given boundary conditions, Table 41.1 suggests a solution of the form Vnm (r, φ, z) = Jm (αnm r/R) [Amn cos(mφ) + Bmn sin(mφ)] sinh(αnm z) where αnm is the nth zero of Jm . Using the orthogonality of Bessel functions (41.130) gives the coefficients which yield V (r, φ, L) = V˜ (r, φ): Anm =

1 π sinh(αnm L)

Z

1 1 2 2 R Jm+1 (αnm R) 2

2π

Z

R

r dr V˜ (r, φ)Jm (αnm r/R) cos mφ ,

dφ 0

0

and the same expression for Bmn with cos mφ → sin mφ. For constant V˜ (r, φ) = V0 , the φ integrals vanish — which can’t be right, since this is a physically acceptable configuration. And in fact, the integrals only vanish for m , 0. (Of course, we could’ve deduced the independence of φ from the axial symmetry of the configuration.) So then there are no B’s, and An ≡ An0 =

2V0 1 sinh(αn0 L) R2 J12 (αn0 R)

Z

R

J0 (αn0 r/R)r dr . 0

This integral can be evaluated using (C.18b) with ν = 1: d [rJ1 (r)] = rJ0 (r) −→ dr

Z

R

R2 J1 (αn0 ) . αn0

J0 (αn0 r/R)r dr = 0

All together then, V (r, φ, z) = 2V0

X n

J1 (αn0 ) 1 J0 sinh(αn0 L) αn0 J12 (αn0 R)



αn0 r R



sinh(αn0 z)

41.15 Set up the chessboard in the z = 0 plane as

 black squares: V (x, y, 0) = V0 =

 white squares: V (x, y, 0) = 0 =

0 < x < a, 0 < y < a a < x < 2a, a < y < 2a

0 < x < a, a < y < 2a a < x < 2a, 0 < y < a

with periodic boundary conditions in the plane V (x + 2a, y, z) = V (x, y, z)

V (x, y + 2a, z) = V (x, y, z)

and V (x, y, |z| → ∞) = 0. Then (41.54) suggests a solution of the form A1 eiαx + B1 e−iαx




A2 eiβy + B2 e−iβy




A3 eγz + B3 e−γz




where γ 2 = α2 + β 2 and A3 = 0 for z > 0, B3 = 0 for z < 0. Since the configuration is clearly symmetric across the plane of the chess board, V (x, y, z) = V (x, y, −z), we can simplify this to A1 eiαx + B1 e−iαx







A2 eiβy + B2 e−iβy e−γ|z| .

Applying the periodic boundary condition in x reveals e±iα2a = 1

338

−→

α = nπ/a, n = 0, 1, 2, . . . .

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

Similarly, in y, e±iβ2a = 1

−→

β = mπ/a, m = 0, 1, 2, . . . .

So as expected, the voltage is periodic in both x and y with period 2a. However unlike Dirichlet or Neumann conditions, this does nothing to reduce the number of coefficients — which we need to do before using orthogonality. But consider a camouflaged symmetry in the z = 0 plane: were the chessboard set up so that the voltage alternated between ± 12 V0 , the solution would be odd in both x and y from square to square. That is to say, we’d have Fourier sine series with period 2a. But 21 V0 is just a DC offset — that “a0 ” term in a Fourier expansion. So uniqueness allows us to express the solution as (see Table 33.1) Vnm (x, y, z) =

1 V0 + Cnm sin 2



nπx a



sin



mπy a



e−γ|z| ,

for integer n, m > 0. The remaining coefficients are obtained via orthogonality, Cnm = =

4 a2

Z

2V0 a2

a

h

Vnm (x, y, 0) −

0

Z

a

sin



0

nπx a



sin

1 V0 sin 2

i



mπy a





nπx a



dx dy =

sin



mπy a



8V0 1 , π 2 nm

dx dy n, m odd.

Thus V (x, y, z) =

8V0 1 V0 + 2 2 π

1 sin nm

X



nπx a



sin



mπy a



√ e−π

n2 +m2 |z|/a

n,m odd

The plots display the potential through n, m = 7 at three values of z.

41.16 The configuration is periodic in z, and therefore so too the solution. Consulting Table 41.1 shows that for this to be possible, β 2 must be negative. This suggests a solution of the form





Im (βr) Km (βr)



cos mφ sin mφ

cos βz sin βz

 ,

where now, having recast the solution in terms of modified Bessel functions, β > 0. In fact, the periodic boundary condition effectively “quantizes” β, −→

V (a, φ, z) = V (a, φ, z + L)

βn = 2πn/L , n = 0, 1, 2, . . .

Since the solution must be finite at r = 0, we can discard Km . In addition, since the configuration has axial symmetry, the solution has no φ dependence — so only m = 0 contributes. Moreover, the cylinder is manifestly odd around z = L/2, so only sin βz contributes (so n = 0 is excluded). All together then, V (r, φ, z) =

∞ X

An I0 (βn r) sin(βn z) .

n=1

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339

Although modified Bessel function are not orthogonal, the An can be extracted from the orthogonality of sine applied at r = a: An =

1 I0

L

V0

= I0

( =


2πna
2πna L

L

2 L

Z

2 L

Z

V (a, φ, z) sin



0 L/2

sin



0

0, 4V0

nπI0 ( 2πna L )

2πnz L

2πnz L



Z

 L

−

sin



L/2

2πnz L



n even . n odd

,

41.17 This is a heat equation problem in 1+1 dimensions with inhomogeneous boundary conditions, 1 ∂T ∂2T = ∂x2 α ∂t

T (0, t) = T1 ,

with

T (L, t) = T2 ,

T (x, 0) = T0 x/L .

The challenge posed by the inhomogeneous conditions is the same as in Example 41.8; though here we’re only in one spatial dimension, the path to the solution is the same. As t increases, the temperature distribution should approach an equilibrium Teq (x), independent of time — and so therefore a solution of the (one-dimensional) Laplace equation: d2 Teq =0 dx2

−→

Teq = a + bx .

Applying the inhomogeneous boundary conditions at x = 0 and L gives Teq = T1 +

T2 − T1 x. L

Then T˜, define by the superposition T (x, t) = Teq (x) + T˜(x, t) , is a solution to the heat equation with homogeneous boundary conditions, T˜(x, t) =

∞ X

An sin



nπx L



e−α(nπ/L)

2

t

,

n=1

where sin survives the condition at x = 0, and nπ/L is required to satisfy the condition at x = L. There remains then
only the initial condition on T˜ (not T ) to find the coefficients An . Orthogonality in sin nπx gives: L An =

=

2 L

Z

2 L

Z

L

T˜(x, 0) sin

0

0

L

h



nπx L



dx

T0 x − Teq (x) sin L

i



nπx L



dx =

 2  (−1)n+1 (T0 + T2 ) + T1 . nπ

41.18 In polar coordinates, the heat equation becomes 1 ∂ r ∂r



r

∂T ∂r



+

1 ∂2T 1 ∂T = . r2 ∂φ2 α ∂t

Since the ring has fixed radius R, this reduces to a 1+1 dimensional problem for T (φ, t), 1 ∂2T 1 ∂T = R2 ∂φ2 α ∂t

340

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the solution of which must satisfy the periodic boundary conditions T (0, t) = T (2π, t)

∂t T (0, t) = ∂t T (2π, t) .

and

Thus we get ∞ X

T (φ, t) =

(An cos nφ + Bn sin nφ) e−αn

2

t/R2

,

n=1

where, with T (φ, 0) = f (φ) = T0 cos2 φ,

Z

T0 An = π

2π

cos2 φ cos nφ dφ = T0

0

 1 , 1 2



n=0 n=2. else

,

0

and Bn =

T0 π

2π

Z

cos2 φ sin nφ dφ = 0 .

0

41.19 The solution in Example 41.8 is 4T∗ π

T (x, y, t) =

X 1 h sin (nπx/a) sinh (nπy/a) i n

sinh (nπb/a)

n odd

+

X

cnm sin(nπx/a) sin(mπy/b) e

−α



2 2 +( mπ ( nπ a ) b )



t

.

nm

This clearly satisfies the homogeneous conditions T (0, y, t) = T (x, 0, t) = T (a, y, t) = 0. As for y = b, T (x, b, t) = Teq (x, b) =

4T∗ π

X 1 h sin (nπx/a) sinh (nπb/a) i n

sinh (nπb/a)

n odd

=

4T∗ π

X sin (nπx/a) n

.

n odd 1 But the Fourier sine series sin (nπx/a) converges to π/4 for all finite a (as can be n odd n verified directly, and as a Mathematica plot reveals) — so the condition at y = b is satisfied.

P

41.20 (a) Dirichlet conditions, u(0) = a, u(b) = b: v(x) = a +

x b

(b − a) 2

(b) Neumann conditions, = a, = b: v(x) = ax + x2b (b − a) 0 (c) Mixed conditions, u (0) = a, u(b) = b: v(x) = a(x − b) + b u0 (0)

u0 (b)

41.21 With homogeneous Dirichlet conditions on all sides, the solution has the form

T (x, y, t) =

∞ X

Cnm sin



nπx a



sin



mπy b



e

−απ 2

n2 a2

2

+ m2 b




t

,

n,m=1

where the sine’s survive to satisfy the conditions at x = 0 and y = 0, and the integers n and m are required to satisfy the conditions at x = a and y = b, respectively. Solving for the expansion

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341

coefficients requires leveraging orthogonality in both x and y: 4 ab

Cnm =

Z

Z

b

T (x, y, 0) sin a

Z

sin



0

8T0 cos π 2 nm

h

=



0

0

4T0 ab

=

a



nπx a

mπ 2





Z

nπx a



b

dx

sin

sin



b/2

− (−1)m

i

,



mπy b

mπy b





dx dy

dy

n odd

The figures show how the temperature distribution becomes uniform over time, from αt = 0.001 to 0.1 (n, m ≤ 21).

41.22 The is Example 41.8, with T∗ = 3T0 . The solution is given by the sum of the solution to Laplace’s equation Teq in (41.112) 12T0 π

Teq (x, y) =

X 1 h sin (nπx/a) sinh (nπy/a) i n

sinh (nπb/a)

n odd

and T˜ in (41.114), T˜(x, y, t) =

X

cnm sin(nπx/a) sin(mπy/b) e

−α



2 2 +( mπ ( nπ a ) b )



t

.

nm

The coefficients are

cnm

4 = ab

Z

a

Z

b

[T (x, y, 0) − Teq (x, y)] sin(nπx/a) sin(mπy/b) dx dy . 0

0

The contribution from T (x, y, 0) gives the homogeneous coefficients found in the previous problem. From these must be subtracted the Teq contribution — which is less tedious to calculate than may appear at first blush, due to the orthogonality of sin nπx/a. But there’s always Mathematica. Either way:

cnm

342

T0 = 2 π n

("

8 cos

πm 2




m

− (−1)m


#

 + (−1)

m

3 · 8a2 m (b2 n2 + a2 m2 )

) ,

n odd

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41.23 For all three sides held at T = 0 except for T (a, y, t) = y, the only substantive difference from the previous problem is the Laplace equation solution Teq . Consultation with Example 41.6 leads to

Teq =

∞ X

Bn sinh (nπx/b) sin (nπy/b) ,

n=1

with 2 Bn sinh (nπa/b) = b

Z 0

b

y sin (nπy/b) dy = (−1)n+1

2b . nπ

41.24 The Laplace equation solution Teq is the sum of those in the previous two problems, plus the contribution given by the new condition at y = 0 — which can be adapted from Example 41.6 with y → b − y and a Fourier sine series for x2 .

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343

41.25 As a circular membrane with Dirichlet boundary conditions, Example 41.11 shows that the normal mode frequencies of a timpani are given by the Bessel function zeros αjm , ψjm = Jm (αjm r) cos mφ , where for simplicity we’ve taken the radius of the drumhead to be R = 1. (a) Striking the drum in the center, the excited modes must all be axially symmetric — that is, they must be independent of φ. So m = 0, making J0 the only relevant Bessel function. (In fact, only J0 is non-zero at the origin — precisely where the drum is being struck.) With none of the other Jm ’s excited, the eigenfrequency spectrum is severely limited, leading to a “hollow” sound. In addition, the ratios of zeros of J0 α20 /α10 = 2.29542

α30 /α10 = 3.59848

α40 /α10 = 4.90328

do not match well with any of the ratios in the twelve-tone equal temperament tuning. So a timpani hit in the center does not sound musically harmonic. (b) Hitting the timpani off-center excites axially asymmetric nodes, thereby suppressing the symmetric J0 modes. Indeed, all Jm>0 modes have nodes at the center. More careful analysis shows that hitting the drum about 2/3 from the center primarily excites the (j, m) = (1, 1) mode. (c) The frequency ratios explains why the (1, m) modes are preferred: In[]:= Table[BesselJZero[n, 1] / BesselJZero[1, 1], {n, 1, 6}] // N Out[]=

{1., 1.3403, 1.6651, 1.98041, 2.28918, 2.59313}

In[]:=

Table[(2 ^ (1 / 12.)) ^ n, {n, 0, 16}]

Out[]=

{1., 1.05946, 1.12246, 1.18921, 1.25992, 1.33484, 1.41421, 1.49831,

1.5874, 1.68179, 1.7818, 1.88775, 2., 2.11893, 2.24492, 2.37841, 2.51984}

The appearance of the timpani ratios amongst the well tempered ratios is explains why the timpani (played properly) sounds musically resonant.

41.26 (a) The Laplacian in curvilinear coordinates given in (12.49b) and (A.12) is ∇2 =

1 ∂1 h1 h2 h3

h



h2 h3 ∂1 f h1



+ ∂2



h1 h3 ∂2 f h2



+ ∂3



h1 h2 ∂3 f h3

i

.

Using the elliptic coordinate scale factors in Table A.1, the Helmholtz equation in the plane becomes

"

#

1


2

c2 sinh2 ρ + sin2 φ


∂ρ2 + ∂φ + k2 u(ρ, φ) = 0 .

Inserting the ansatz u(x, y) = R(ρ)Φ(φ),


1 d2 R 1 d2 Φ + = −k2 c2 sinh2 ρ + sin2 φ , 2 R dρ Φ dΦ2 or, introducing the separation constant λ, 1 d2 R + k2 c2 sinh2 ρ = − R dρ2 (b) With sin2 φ =

1 (1 2

− cos 2φ) and sinh2 φ = d2 Φ + dφ2

344

h

λ+

1 2 2 k c 2



1 d2 Φ + k2 c2 sinh2 φ Φ dΦ2

1 (1 2



−

 ≡λ

− cosh 2φ), the separated equations become 1 2 2 c k cos 2φ Φ = 0 , 2

i

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and d2 R − dρ2

h

1 2 2 k c 2

λ+



−

1 2 2 c k cosh 2ρ R = 0 2

i

which are Mathieu and modified Mathieu equations, respectively, with a = λ + 21 k2 c2 , and q = 14 c2 k2 = 41 c2 ω 2 /v 2 , where v is the speed of the wave appearing in the wave equation. So solving for the normal mode frequencies ω is tantamount to finding the corresponding q’s.

41.27 (a) For semi-major axis x = c cosh ρ cos 0 = 5, and semi-minor axis y = c sinh ρ sin π/2 = 3, ρ0 = tanh−1 (3/5) = 0.69 . The foci are on the x-axis at c2 (cosh2 ρ0 − sinh2 ρ0 ) = 16

→

c = ±4 ,

and ρ = 0 is the line connecting them. (b) Notice that the cen are even, the sen odd. In addition, those with even n have period π, those with odd n have period 2π ce0 (4,ϕ)

ce1 (4,ϕ)

ce2 (4,ϕ) 1.0

1.0

1.2 1.0

0.5

0.5

0.8 0.6

-π

π

0.4

ϕ -π

-0.5

π

0.2 -π

π

-0.5

-1.0

ϕ

ϕ

se1 (4,ϕ)

se2 (4,ϕ) 1.0

1.0

0.5

0.5 -π

π

ϕ

-π

π

-0.5

ϕ

-0.5

-1.0

-1.0

(c) Notice that all cross the q axis; these zeros will give the normal mode frequencies: ce0 (q,iρ0 )

ce1 (q,iρ0 )

0.06

0.5

0.04

0.4

0.02

0.3 2

4

6

8

10

12

ce2 (q,iρ0 ) 2.0 1.5

0.2

14

1.0

0.1

-0.02 -0.04

0.5 2

4

6

8

10

12

14

-0.1

-0.06

-ise1 (q,iρ0 )

2

4

12

14

6

8

10

12

14

-ise2 (q,iρ0 )

0.25

1.2

0.20

1.0 0.8

0.15

0.6

0.10

0.4

0.05

0.2 2

4

6

8

10

12

14

2

4

6

8

10

(d) To satisfy the Dirichlet conditions, the solutions cen (q, iρ) and sen (q, iρ) must vanish at the boundary ρ = ρ0 . FindRoot gives the following values of q as the zeros: Denoting the zeros of cen and sen as cαjn and sαjn , respectively, the first five are cα10 , cα11 , sα11 , cα12 , sα12 . (e) Drumhead solutions must be continuous across the ρ = 0 line between the foci.

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345

ce0 j=1 2 3

ce1

ce2

1.73531 3.35224 5.65301 11.3564 14.6274 18.487 29.7953 34.8439 40.4577

ce0 (4,iρ) 0.3

0.04

0.2

0.02

-0.02

0.4

se2

5.43008 19.4785 42.3061

7.8191 23.6357 48.2484

ce1 (4,iρ)

0.06

0.2

se1

0.6

0.8

1.0

ce2 (4,iρ) 0.8 0.6 0.4

0.1

ρ

0.2

0.4

0.6

0.8

1.0

0.2

ρ

-0.1

-0.04

0.2

-0.2

-0.06

-ise1 (4,iρ)

0.8

1.0

ρ

0.3

0.04

0.2

0.02 0.2 -0.04

0.6

-ise2 (4,iρ)

0.06

-0.02

0.4

-0.2

0.4

0.6

0.8

1.0

ρ

0.1

0.2

0.4

0.6

0.8

1.0

ρ

Comparison of the plots of cen (4, iρ) and −isen (4, iρ) with those of cen (4, φ) and sen (4, φ) in part (b), together with the even and odd parity in ξ of cen (q, ξ) and sen (q, ξ), shows that such continuity can only be maintained by solutions ψ formed by the products cen (q, φ)cen (q, iρ) and sen (q, φ)sen (q, iρ). (f)

41.28 (a) From Table 41.1, the general solutions to the Helmholtz equation in the plane is ψ(r, φ) = R(r)Φ(φ) = [AJm (kr) + BNm (kr)] cos mφ , where k = ω/v for wave speed v. (So specification of k in a normal mode gives that mode’s frequencies.) Since the annular drum does not include r = 0, we cannot discard the Neumann function; in fact, it’s needed in order to satisfy the boundary conditions at r = a and b: AJm (ka) + BNm (ka) = 0 AJm (kb) + BNm (kb) = 0 . Non-trivial A, B require that the determinant of their coefficients vanishes, Jm (ka)Nm (kb) − Jm (kb)Nm (ka) = 0 .

346

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Thus in terms of u = r/a and γ = b/a, the solution is the antisymmetric linear combination

Rm (u) = Jm (kjm u)Nm (kjm γ) − Jm (kjm γ)Nm (kjm u)

where kjm is the jth zero of Rm (u). Specifically, the kjm ensure that R(1) = 0, since the antisymmetry guarantees R(γ) = 0.

(b) First, plot the antisymmetric form of the boundary condition to get a rough location of the zeros; this will make FindRoot more effective: In[]:= bc[k_] := BesselJ[0, k ] BesselY[0, γ k ] - BesselJ[0, γ k ] BesselY[0, k ] In[]:=

Plot[bc[k], {k, 0, 10}] 0.4 0.3 0.2

Out[]=

0.1

2

4

6

8

10

-0.1

So the zeros are k[0, j_] := z /. Table[FindRoot[bc[0, z], {z, i}], {i, 3, 9, 3}]〚j〛; k[1, j_] := z /. Table[FindRoot[bc[1, z], {z, i}], {i, 3, 9, 3}]〚j〛; In[]:=

Table[k[n, j], {j, 1, 3}, {n, 0, 1}]

Out[]=

3.12303 3.19658 6.27344 6.31235 9.41821 9.44446

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347

In[]:=

R[n_, j_, u_] := BesselJ[n, k[n, j] u] BesselY[n, k[n, j] γ] BesselJ[n, k[n, j] γ] BesselY[n, k[n, j] u]

In[]:=

Plot[{R[0, 1, u], R[0, 2, u], R[0, 3, u]}, {u, 1, 2}]

Out[]=

0.10

0.05

1.2

1.4

1.6

1.8

2.0

-0.05

In[]:=

Plot[{R[1, 1, u], R[1, 2, u], R[1, 3, u]}, {u, 1, 2}]

Out[]=

0.10

0.05

1.2

1.4

1.6

1.8

2.0

-0.05

Verifying orthogonality: 2

In[]:=

mag[n_, i_, j_] :=  R[n, i, u] × R[n, j, u] u u // Chop 1

In[]:=

Table[mag[0, i, j], {i, 1, 3}, {j, 1, 3}]

Out[]=

0.0103026 0 0 0 0.00256862 0 0 0 0.00114108 In[]:=

Table[mag[1, i, j], {i, 1, 3}, {j, 1, 3}]

Out[]=

0.0101573 0 0 0 0.00256015 0 0 0 0.00113944

(c) Note that the J0 modes are spherically symmetry, whereas the J1 modes have a factor of

348

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cos φ from the separation ansatz:

41.29 (a) Due to the spherical symmetry, this is heat diffusion in 1+1 dimension – radius r and time t. Unlike the following problem, the container in this problem is not an insulator: heat can flow such that at r = R the ice cream is held at the fixedtemperature of the freezer. Thus T (R, t) = 0, making this is a homogeneous Dirichlet problem. The spatial part of the solution is a combination of spherical Bessel and Neumann functions, jm and nm . The spherical symmetry reduces the contributions to m = 0 only, and continuity at the center requires that we discard the Neumann function. So the general solution is

T (r, t) =

X

cm j0 (nπr/R) e−αm

2

π 2 t/R2

,

m

where nπ are the zeros of j0 , where

cm =

2m2 π 2 R3

Z

R

T (r, 0)j0 (mπr/R)r2 dr ,

0

with the normalization in the projection integral comes from (C.41). The initial condition is r2 T (r, 0) = T0 − (T0 − T1 ) R 2. (b) In terms of u = r/R and τ = αt/R2 ,

T (u, t) =

X

cm j0 (nπu) e−m

2

π2 τ

,

cm = 2m2 π 2

Z

1

T (u, 0)j0 (mπu)u2 du

0

m

with T (u, 0) = T0 − (T0 − T1 )u2 . 1

In[]:=

c[m_] = 2 π ^ 2 m ^ 2  (T0 - (T0 - T1) u ^ 2) SphericalBesselJ[0, m π u] u ^ 2  u;

In[]:=

solDirichlet[u_, τ_, nmax_] :=

0

Sum[c[n] SphericalBesselJ[0, n π u] Exp[- n ^ 2 π ^ 2 τ], {n, 1, nmax}];

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

349

In[]:=

T0 = -15; T1 = 0; Plot[solDirichlet[u, 0, 15], {u, 0, 1}] Plot3DsolDirichlet[u, τ, 15], {u, 0, 1}, {τ, 0, .5}, PlotRange  All, AxesLabel  "r/R", "α t/R2 ", "T(r,t)", LabelStyle  Directive[Bold], AxesEdge  {Automatic, {1, -1}, Automatic}

In[]:=

T0 = -15; T1 = 0; Plot[solDirichlet[u, 0, 15], {u, 0, 1}] 0.2

0.4

0.6

0.8

1.0

-5 Out[]=

-10

-15 In[]:=

Plot3DsolDirichlet[u, τ, 15], {u, 0, 1}, {τ, 0, .35}, PlotRange  All, AxesLabel  "r/R", "α t/R2 ", "T(r,t)", LabelStyle  Directive[Bold], AxesEdge  {Automatic, {1, -1}, Automatic}

Out[]=

41.30 With the surface insulated, no heat can flow across r = R,



dT =0, dr r=R

so this is a homogeneous Neumann problem. The basic structure of the solution is the same; the only substantive difference is that rather than zeros of j0 , we need the zeros of its derivative — which is −j1 . So we need the zeros of j1 = J3/2 : Zeros = Table[BesselJZero[3 / 2, j ], {j, 1, 3}] // N Out[]=

{4.49341, 7.72525, 10.9041}

The zero however are so large, only the first one need be retained in the calculation — so we only need the expansion coefficients c1 and c2 : The expansion coefficients are then

350

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1

c[1] =  (T0 - (T0 - T1) u ^ 2) SphericalBesselJ[0, Zeros〚1〛 u] u ^ 2  u  0

1

 SphericalBesselJ[0, Zeros〚1〛 u] ^ 2 u ^ 2  u ; 0

1

c[0] =  (T0 - (T0 - T1) u ^ 2) u ^ 2  u  0

In[]:=

1

 u^2 u ; 0

solNeumann1[u_, τ_] = c[0] + c[1] SphericalBesselJ[0, Zeros〚1〛 u] Exp[- Zeros〚1〛 ^ 2 τ];

T0 = -15; T1 = 0; Plot[{T0 - (T0 - T1) u ^ 2, solNeumann1[u, 0, 25]}, {u, 0, 1}] 0.2

0.4

0.6

0.8

1.0

-5 Out[]=

-10

-15

This time, the ice cream equilibrates faster — and to a lower value: In[]:=

Plot3DsolNeumann1[u, τ], {u, 0, 1}, {τ, 0, .1}, PlotRange  All, AxesLabel  "r/R", "α t/R2 ", "T(r,t)", LabelStyle  Directive[Bold]

Out[]=

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351

Green’s Functions

42

d2 ψ(x) dx2

42.1 For the Poisson equation in one dimension, function is linear, G(x, x0 ) =



a1 x + b1 , a2 x + b2 ,

= ρ(x), the general form of the Green’s

x < x0 , x0 < x

This must be continuous at x = x0 , and its first derivative must have a jump discontinuity there,



∂ =1. G(x, x0 ) ∂x x=x0 It must also satisfy the boundary conditions. (a) ψ(0) = 0, ψ(L) = 0: This is the case explore in Example 42.1, where we found 1 1 G(x, x ) = x< (x> − L) = L L 0



x (x0 − L) , x0 (x − L) ,

x < x0 , x0 < x

(b) ψ(0) = 0, ψ 0 (L) = 0: From the boundary conditions at x = x0 , we get b1 = a2 = 0. Continuity of G and the discontinuity of ∂x G yield a1 x0 = b2

0 − a1 = 1 .

So a1 = −1 and b2 = −x0 . Thus G(x, x0 ) =



−x , −x0 ,

x < x0 . x0 < x

(c) ψ 0 (0) = 0, ψ(L) = 0: The boundary conditions give a1 = 0 and b2 = −a2 L. Continuity of G and the discontinuity of ∂x G yield b1 = a2 x0 + b2

a2 = 1 .

So



0

G(x, x ) =

x0 − L , x−L ,

x < x0 . x0 < x

42.2 For ρ(x0 ) = x02 : (a)

Z ψ(x) = 0

L

1 G(x, x )ρ(x ) dx = L 0

0

0

Z

x



0



x (x − L) x

0

02

1 dx + L 0

Z

L





x (x0 − L) x02 dx0

x

1 = x(x3 − L3 ) . 12 ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

(b) L

Z

G(x, x0 )ρ(x0 ) dx0 = −

ψ(x) =

x

Z

x0 x02 dx0 −

Z

xx02 dx0

x

0

0

L

1 1 1 = − x4 − x(L3 − x3 ) = x(x3 − 4L3 ) . 4 3 12 (c) L

Z

0

0

Z

0

x

(x − L) x

G(x, x )ρ(x ) dx =

ψ(x) =

02

=

(x0 − L) x02 dx0

dx + x

0

0

L

Z

0

1 4 1 1 1 1 4 x − Lx3 + (L4 − x4 ) − L(L3 − x3 ) = (x − L4 ) . 3 3 4 3 12

42.3 The simplest approach is to first find G with homogeneous conditions. Then note that any linear form c1 x + c2 is annihilated by the Poisson operator d2 /dx2 — so adding it to G will have no effect on its continuity properties. Then we simply choose the values of c1 and c2 which satisfy the inhomogeneous boundary conditions. For ψ(0) = a, ψ 0 (L) = b (Problem 42.1b): G(x, x0 ) =



−x + (bx + a) , −x0 + (bx + a) ,

x < x0 . x0 < x

42.4 For a string fixed at both ends, ψ(0) = ψ(L) = 0. Thus the Green’s function must have the general form G(x, x0 ) =



x < x0 . x0 < x

a sin kx , b sin k(x − L) ,

The continuity and discontinuity conditions on G and ∂x G at x0 = x give a sin kx0 = b sin k(x0 − L)

bk cos k(x0 − L) − ak cos kx0 = 1 .

Solving, we find a=

sin k(x0 − L) k sin kL

b=

sin kx0 . k sin kL

Thus G(x, x0 ) = =

sin kx< sin k(x> − L) k sin kL 1 k sin kL



sin kx sin k(x0 − L) , sin kx0 sin k(x − L) ,

x < x0 . x0 < x

The singularity at k = nπ/L, manifest in (42.76), is fairly obvious here as well. Moreoever, lim k→0

sin kx< sin k(x> − L) 1 = x< (x> − L) , k sin kL L

which is the Poisson equation Green’s function (Example 42.1). As for (42.71), we can construe it as the Fourier series over sin(nπx0 /L) of G in the k → 0 limit, with expansion coefficients 2 bn = L =

2 L

Z

L

G(x, x0 ) sin(nπx0 /L) dx0

0

Z 0

x

x0 (x − L) sin(nπx0 /L) dx0 + L

Z 0

x

x(x0 − L) sin(nπx0 /L) dx0 L



2L = − 2 2 sin (nπx/L) . n π ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

353

42.5 For the Helmholtz equation in one dimension,



d2 dx2



+ k2 ψ(x) = 0, we’ll express the general

form of the Green’s function as



G(x, x0 ) =

x < x0 . x0 < x

a1 cos(kx − δ1 ) , a2 cos(kx − δ2 ) ,

(a) ψ(0) = 0, ψ 0 (L) = 0: The condition at x = 0 gives δ1 = π/2. At x = L, we get sin(kL − δ2 ) = 0, so δ2 = kL. So thus far, we have G(x, x0 ) =



x < x0 . x0 < x

a1 sin kx , a2 cos k(x − L) ,

The (dis)continuity conditions at x = x0 give



sin kx0 k cos kx0

 

cos k(x0 − L) −k sin k(x0 − L)

a1 a2





0 −1

=

which solution a1 = −

cos k(x0 − L) k cos kL

a2 =

sin kx0 k cos kL

Thus: G(x, x0 ) = −



1 k cos kL

sin kx cos k(x0 − L) , sin kx0 cos k(x − L) ,

x < x0 . x0 < x

(b) ψ(0) = a, ψ(L) = b: Since the Helmholtz operator annihilates the quadratic form − 21 k2 x2 + c1 x + c0 , we merely need to find the values of ci which satisfy the inhomogeneous boundary conditions — c0 = a and c1 = 12 k2 L + (b − a)/L. Then add this to the Green’s function found in 42.1a to get 1 1 1 G(x, x0 ) = − k2 x(x − L) + (b − a)x + a − x< (x> − L) . 2 L L 42.6 Starting with (42.25),

 ` ∞ X X a`m r` Y`m (θ, φ) ,

0

G(~ r, ~ r )=

02`+1 a`m rr`+1 Y`m (θ, φ)

`=0 m=−`

0 < r < r0 ,

r0 < r < ∞

.

where we’ve used (42.26). Then from (42.29), the discontinuity at r = r0 read

X





a`m − (` + 1) r0`−1 − `r0`−1 Y`m (θ, φ) = −

1 r02

X

∗ Y`m (θ0 , φ0 ) Y`m (θ, φ) .

`m

`m

Since the Y`m ’s form an orthonormal basis, a`m =

1 1 Y ∗ (θ0 , φ0 ) . 2` + 1 r0`+1 `m

(1)

42.7 With G(r) = cH0 (kr), integrate (42.34) over a small circle of radius ,

Z

Z δ(~ r) da = c

S

354




(1)

∇2 + k2 H0 (kr) da .

S

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Using the divergence theorem in the plane, 1 = c

I

~ (1) · d~ ∇H ` + k2 0

Z S

(1)

Z

dH0 (kr)

=

(1)

H0 da

C

dr



2π

 dφ + 2πk2

Z

 (1)

H0 (kr)rdr . 0

0

Consulting Table C.3 we find d (1) (1) H (x) = −H1 (x) dx 0 so the first term is (1)

2π

Z

dH0 (kr) dr



(1)

 dφ = −2πkH1 (k) = −2πk [J1 (k) + iN1 (k)] .

0

Table C.3 also reveals (1)

xH0 (x) =

d (1) xH1 dx

h

i

so the second term becomes 2πk2

Z

 (1)

(1)

H0 (kr) rdr = 2πkH1 (k) = 2πkJ1 (k) . 0

[The disappearance of the Neumann function here is a bit tricky. Whereas J1 (0) = 0, for small x, xN1 (x) ≈ −2/π. So if we carefully take the lower limit of the integral to zero, then together with  → 0, the two Neumann functions cancel.] Putting everything together gives 1 (1) = 2πk −H1 (x) + J1 (k) = −2πikN1 (k). c

h

i

Then in the  → 0 limit, c = −i/4.

42.8 One can find the Green’s function starting with (41.71) or (41.72) for 0 < r < a and a < r < R, applying the various conditions (continuity and discontinuity at r = a, vanishing at r = R). Or one can leverage the fundamental solution (42.31) and find the homogeneous solution F in (42.44) to give the required Dirichlet Green’s function, GD (~ r, ~ r 0) =

X `,m

` r< 1 ∗ Y`m (θ0 , φ0 )Y`m (θ, φ) + F (~ r, ~ r 0) , `+1 2` + 1 r>



GD (r, a)

= 0. r=R

With r> = R, r< = a, G(R, a) =

X



a` 1 ∗ + F (R, a) Y`m (θ0 , φ0 )Y`m (θ, φ) = 0 2` + 1 R`+1

`,m

we find — recalling that G should be symmetric in r, r0 , F (r, r0 ) = −

` r` r< 1 > Y ∗ (θ0 , φ0 )Y`m (θ, φ) . 2` + 1 R2`+1 `m

The density of charge on the ring is ρ(~ r) =

q δ(r − a)δ(cos θ) . 2πa2

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355

Inserting this and GD into (42.3), the integral over cos θ0 sends Y (θ0 , φ0 ) → the m sum. The φ integral is then trivial, leaving V (~ r) =

q 4πa2

X

r

Z P` (cos θ) 0

`

r0` r0` r` − 2`+1 `+1 r R R

Z +

r` r0`+1

r

collapsing



r` r0` R2`+1

−

2`+1 δm0 , 4π



δ(r0 − a) r02 dr0 ,

where only the first integral contributes for r > a, the second for r < a. Thus q V (~ r) = 4π

X

P` (cos θ)

 

`



`

r` a`+1 a r `+1

h

a 2`+1 R

h

i
r 2`+1




1−

1−

i

R

,

ra

.

42.9 The separated spatial part of the wave equation is




∇2 + k2 ψ(r) = 0 ,

k = ω/c .

From Table 41.1, the general solution is a linear combination of spherical Bessel and Neumann functions. But spherical symmetry collapses the sums to ` = m = 0, leaving the spherical waveforms 1 [a sin(kr) + b cos(kr)] . r

ψ(r) = aj0 (kr) + bn0 (kr) =

For an outgoing wave with time dependence e−iωt , we choose the constants so that the spatial part has the form eikr . r

ψ(r) = c

To find c, integrate the Green’s function equation over of sphere of radius  centered at the origin,

Z

2

∇ +k

2




Z

3

δr d3 r = −1 .

Gd r = −

V

V

The second integral vanishes in the  → 0 limit,

Z

k 2 G d3 r = k

V

Z



0

eikr 2 r dr dΩ = 4πk r

Z



reikr dr → 0 .

0

Evaluate the first integral using the divergence theorem:

Z V

I



~ · d~a = dG ∇ Gd r = ∇G dr r= S 2

3



I

eikr eikr − 2 da = 4π c ik r r S 2



→ −4πc . r=

Putting it all together, G(r) = −

1 eikr 4π r

or, noting the translational symmetry, 0

G(r, r0 ) = −

42.10

R V

356




Φ ∇2 G − G ∇2 Φ d3 r =

H S

1 eik|r−r . 4π |r − r0 |




~ − G ∇Φ ~ Φ∇G · d~a

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

(a) With ∇2 G = −δ(~ r−~ r 0 ) and ∇2 Φ(~ r ) = −ρ(~ r ), (42.51) yields:

Z

I

G(~ r, ~ r 0 ) ρ(~ r 0 ) d3 r0 +

Φ(~ r) = V





~ 0 Φ(~ ~ 0 G(~ G(~ r, ~ r 0 )∇ r 0 ) − Φ(~ r 0 )∇ r, ~ r 0 ) · d~a0 .

S

(b) Starting with ∇2 G = −δ(~ r−~ r 0 ),

Z

I

∇2 G(~ r−~ r 0 ) d3 r0 =

−1 = V

~ r−~ ∇G(~ r 0 ) · d~a0 =

I

S

S

which is inconsistent with the Neumann condition

I S

1 ∂G 0 da = − ∂n A

I

∂G ∂n

∂G 0 da ∂n

= 0. But for ∂G/∂n = −1/A

da0 = −1.

S

(c)

Z Φ(~ r) =

ZV =

ZV = V

G(~ r, ~ r 0 ) ρ(~ r 0 ) d3 r0 + G(~ r, ~ r 0 ) ρ(~ r 0 ) d3 r0 +

I





~ 0 Φ(~ ~ 0 G(~ G(~ r, ~ r 0 )∇ r 0 ) − Φ(~ r 0 )∇ r, ~ r 0 ) · d~a0

IS

G(~ r, ~ r 0)

IS

∂Φ(~ r 0) 0 da − ∂n0

I

Φ(~ r 0)

S

∂G 0 da ∂n

∂Φ(~ r 0) 0 G(~ r, ~ r ) ρ(~ r )d r + G(~ r, ~ r ) da + hΦiS . ∂n0 S 0

0

3 0

0

42.11 (a) The homogeneous equation has solution

 ψ=

A cos 3x + B sin 3x , C cos 3x + D sin 3x ,

x < x0 . x > x0

ψ 0 (0) = 0 gives B = 0; ψ 0 (L) = 0 yields D = C tan 3L . Imposing continuity at x =

x0

gives

A cos 3x0 = C cos 3x0 1 + tan 3L tan 3x0




and discontinuity at x = x0 ,




3C sin 3x0 −1 + tan 3L cot 3x0 + 3A sin 3x0 = 1 . Together these yield the Green’s function 1 G(x, x ) = 3 sin 3L 0



cos 3(x0 − L) cos 3x , cos 3x0 cos 3(x − L) ,

x < x0 . x > x0

(b) The eigenfunctions of the eigenvalue equation



are ψ =

p2 L

d2 +9 dx2

 ψ = −λψ,

ψ 0 (0) = ψ 0 (L) = 0

cos kn x, where k2 = 9 + λ and kn = nπ/L for integer n. So λ=

nπ − 9, L

n = 0, 1, 2, . . .

and the Green’s function is G(x, x0 ) =

2 L

∞ X cos(kn x) cos(kn x0 )

9 − n2 π 2 n=0

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357

(c) A closed form solution emerges directly from the form in (a): L

Z

G(x, x0 )dx0

ψ(x) = 0

=

Z

1 3 sin 3L

x

cos 3x0 cos 3(x − L)x02 dx0 +

L

Z

cos 3x cos 3(x0 − L)x02 dx0



x

0

1 (9x2 − 2) sin 3L + 6L cos 3x = . 81 sin L 2 cos 2L + 1

42.12 (a) The homogeneous equation tions at ±∞,

d2 ψ dx2

− ψ = 0 has solutions y = e±y . Accounting for the condi-



G(x, x0 ) =

x < x0 . x0 < x

Aex , Be−x ,

The (dis)continuity conditions are 0

Aex = Be−x

0

0

0

− Bex − Aex = 1 .

Thus 0 1 G(x, x0 ) = − e−|x−x | . 2

(b) Taking the Fourier transform of the differential equation: ˜ −k2 ψ(k) − ψ˜ = ρ˜(k)

ρ˜(k) ˜ . ψ(k) =− 1 + k2

−→

Thus 1 ˜ = − √1 G 2π 1 + k2 (c) For ρ(x) = e−|x| , ρ˜(k) =

p2

1 ; π 1+k2

r ψ(x) = −

˜ = − 1 e−|x−x0 | . G(x, x0 ) = F −1 G 2

 

−→ then

2 −1 1 F π (1 + k2 )2

h

i

=

1 2



−(x + 1)e−x , (x − 1)ex ,

x>0 . x>0

or, equivalently, 1 ψ(x) = − 2

Z

x

−(x−x0 ) −|x0 |

e

e

0

Z

∞

dx +

−∞

+(x−x0 ) −|x0 |

e

e

0



dx

x

1 = 2



−(x + 1)e−x , (x − 1)ex ,

x>0 . x>0

42.13 (a) The solutions to the homogeneous equation u00 − 2u0 + 1 = 0 are ex and xex ; after applying the boundary conditions u(0) = u(1) = 0, this yields G(x, x0 ) =



x > x0

aex b(1 −

x0 > x

x)ex ,

.

The (dis)continuity conditions give ax0 − b(1 − x0 ) = 0

0

0

bx0 ex − a(a + x0 )ex = 1.

Solving, 0

G(x, x ) =

358



0

x(x0 − 1)ex−x , 0 x0 (x − 1)ex−x

x > x0 . x0 > x

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Then with source f (x) = x,

Z

x

u(x) = (x − 1)

1

Z

0

x02 ex−x dx0 + x

0

x0 (x0 + 1)ex−x dx0

x

0

= 2(x − 1)ex − 3xex−1 + x + 2 . (b) The Green’s function for the same equation, but with initial conditions u(0) = u0 (0) = 0, is 0

t > t0 .

G(t, t0 ) = (t − t0 )et−t , Then with source f (t) = t,

Z

t

u(t) =

0

t0 (t − t0 )et−t dt0 = (t − 2)et + t + 2 .

0

42.14 The Green’s function for the damped harmonic oscillator is the solution to d2 G dG + 2β + ω02 G = δ(t − t0 ) . dt2 dt For the retarded Green’s function, the solution to the homogeneous equation is



0

G(t, t ) =




e−βt ae−Ωt + beΩt ,

t > t0

0,

t < t0 ,

where Ω2 = β 2 − ω02 . Together, the (dis)continuity conditions at t0 give A=

0 1 cos(Ωt0 )eβt Ω

B=−

0 1 sin(Ωt0 )eβt . Ω

Thus 0

G(t, t0 ) =

e−β(t−t Ω

)

0




cos Ωt0 sin Ωt − sin Ωt0 cos Ωt =

e−β(t−t Ω

)

sin Ω(t − t0 )Θ(t − t0 ) .

This is the retarded Green’s function for underdamping (Ω < 0) and overdamping (Ω > 0). For critical damping (Ω = 0),



0

G(t, t ) =

(a + bt)e−βt , 0,

t > t0 t < t0 ,

In this case, the (dis)continuity conditions at t0 give A = −t0 eβt , B = eβt , so 0

0 0 −β(t−t ) Gcrit Θ(t − t0 ) . R (t, t ) = (t − t )e 1.0

Underdamped, β=.05ω0

Overdamped, β=1.1ω0

0.30

Critically damped, β=1 0.3

0.25

0.5

0.20 0.2 10

20

30

40

50

0.15

t

0.10 -0.5

0.1

0.05

2

4

6

8

10

t

2

4

6

8

10

t

42.15 For G(x, x0 , t, t0 ) the wave equation retarded Green’s function (with τ ≡ t − t0 ), ∂ ∂ ψ(x, t) = ∂t ∂t

Z tZ

∞ 0

0

0

0

0

0

f (x , t ) G(x, x ; t, t ) dx dt = 0

Z

f (x0 , t) G(x, x0 ; τ = 0) dx0

−∞

Z tZ +

f (x0 , t0 ) ∂t G(x, x0 ; t, t0 ) dx0 dt0 .

0

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359

The first term vanishes due to the initial condition; then ∂2 ∂t2

Z tZ

0

0

0

0

0

Z

0

f (x0 , t) ∂t G(x, x0 ; τ = 0) dx0

f (x , t ) G(x, x ; t, t ) dx dt = 0

Z tZ +

f (x0 , t0 ) ∂t2 G(x, x0 ; t, t0 ) dx0 dt0

0

Z

0

0

0

Z tZ

f (x , t) δ(x − x ) dx +

=

f (x0 , t0 ) ∂t2 G(x, x0 ; t, t0 ) dx0 dt0

0

Z tZ

f (x0 , t0 ) ∂t2 G(x, x0 ; t, t0 ) dx0 dt0 .

= f (x, t) + 0

Since G satisfies the homogeneous wave equation, c2

∂2 ∂x2

Z tZ

f (x0 , t0 ) G(x, x0 ; t, t0 ) dx0 dt0 =

0

Z tZ

f (x0 , t0 ) c2 ∂x2 G(x, x0 ; t, t0 ) dx0 dt0

0

Z tZ

f (x0 , t0 ) ∂t2 G(x, x0 ; t, t0 ) dx0 dt0 .

=− 0 2

2

2∂ ψ 0 0 Thus ∂∂tψ 2 − c ∂x2 = f (x, t). For G(x, x , t, t ) = (22.28) only by the sign and placement of c):

1 2c

Z tZ

1 = 2c

Z tZ

ψ(x, t) =

1 Θ [c(t 2c

− t0 ) − (x − x0 )] (which differs from

∞





Θ c(t − t0 ) − (x − x0 ) f (x0 , t0 )dx0 dt0

−∞

0

x+c(t−t0 )

f (x0 , t0 )dx0 dt0

x−c(t−t0 )

0

This has the form of d’Alembert’s solution, with a sequence of staggered initial velocities f (x0 , t0 )dt0 — see Problem 41.6.

42.16 (a) The eigenfunction expansion is G(x, x0 , τ ) =

X

an (τ )

n

p2

with Helmholtz eigensystem φˆn = operator:




∂t2 − c2 ∂x2 G =

L

sin(nπx/L), λn = n2 π 2 /L2 . Applying the wave

X φˆn (x0 ) 

a ¨n (τ )φˆn (x) − c2 an (τ )φˆ00 n (x)

−λn

n

=

φˆn (x0 )φˆn (x0 ) , −λn

X φˆn (x0 )φˆn (x)  −λn

n

X

a˙ n (τ )

n



a ¨n (τ ) − c2 λn an (τ ) = 0 .

So an (τ ) = An sin ωn τ + Bn cos ωn τ , where ωn = c G(τ = 0) = 0 eliminates Bn . Then since ∂t G(τ = 0) =



√

λn = nπc/L. The initial condition

φˆn (x0 )φˆn (x0 ) = δ(x − x0 ) −λn

the completeness of the eigenfunctions implies a˙ n = −λn

360

−→

An =

nπ λn = . ωn cL

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All together:

X nπ

G(x, x0 , t, t0 ) =

cL

sin(ωn τ )

n

φˆn (x0 )φˆn (x0 ) 2 = λn c

X 1 nπ

sin(ωn τ ) sin(nπx/L) sin(nπx0 /L) .

n

(b) For L = π, L

Z tZ

f (x0 , t0 )G(x, x0 , t, t0 )dx0 dt0

ψ(x, t) = 0

0

2 c

=

X sin(nπx/L) Z nπ

t




cos 2c(t − t0 ) sin(nπct0 /L) dt0

L

sin(x0 ) sin(nπx0 /L) dx0

0

0

n

Z

1 = 2 sin x [cos(ct) − cos(2ct)] . 3c 42.17 The analysis is the same as in BTW 42.1, but since a and b are now initial conditions, there’s no sign flip in the second integral of (42.21).

42.18 Multiply the unit-source Sturm-Liouville equation for G(~ r, ~ r1 ) by G(~ r, ~ r2 ), subtract from it the same with ~ r1 ↔ ~ r2 and integrate

Z

Z

G(~ r, ~ r1 )δ(~ r−~ r2 ) dτ

G(~ r, ~ r2 )δ(~ r−~ r1 ) dτ −

G(~ r1 , ~ r2 ) − G(~ r2 , ~ r1 ) =

V

V

Z









~ · p(~ ~ r, ~ ~ · p(~ ~ r, ~ G(~ r, ~ r2 )∇ r)∇G(~ r1 ) − G(~ r, ~ r1 )∇ r)∇G(~ r2 ) dτ

=

IV =





~ r, ~ ~ r, ~ p(r) G(~ r, ~ r2 )∇G(~ r1 ) − G(~ r, ~ r1 )∇G(~ r1 ) · d~a ,

S

where we’ve used the divergence theorem. This expression vanishes for either Dirichlet (G|S = 0) ~ ·n or Neumann (∇G ˆ |S = 0) conditions.

42.19 The general solution is given in (42.108) L

Z

0

T (x, t) =

0

0

Z tZ

T0 (x ) G(x, x ; t, 0) dx + 0

0

L

q(x0 , t0 ) G(x, x0 ; t, t0 ) dx0 dt0 ,

0

where the Dirichlet Green’s function is that of (42.102), GD (x, x0 ; t − t0 ) =

2 L

X

2

e−α(nπ/L)

(t−t0 )

sin(nπx0 /L) sin(nπx/L).

n=1

With T0 =

1 2

cos x, the homogeneous contribution is given by a sum over n of

Z hn (x, t) =

L

T0 (x)GD (x, x0 ; t) dx0 =

0

2 2 2 nπL [1 − (−1)n cos L] sin(nπx/L)e−αn π t/L . 2(n2 π 2 − L2 )

The contribution from the source q(x, t) = sin x cos 2t is the sum over n of

Z tZ

L

q(x0 , t0 ) G(x, x0 ; t, t0 ) dx0 dt0

sn (x, t) = 0

0

= (−1)n

2 2 2 L3 sin L sin(nπx/L) cos(2t) e−αn π t/L − 1 nπ(n2 π 2 − L2 )





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.

361

The plot show the solution for α = 1, L = π:

42.20 The procedure is much the same in each case; we’ll focus of (42.108), L

Z

T0 (x0 ) G(x, x0 ; t, 0) dx0 +

T (x, t) =

Z tZ

0

0

L

q(x0 , t0 ) G(x, x0 ; t, t0 ) dx0 dt0 .

0

ˆ ≡ ∂t − α∂ 2 to the first term gives zeros, since it’s the solution Applying the heat operator H x to the homogeneous problem. Applied to the second term gives gives

Z

ˆ (x, t) = HT

L

q(x0 , t) G(x, x0 ; 0) dx0 +

Z tZ

0

Z

0

L

ˆ q(x0 , t0 ) HG(x, x0 ; t, t0 ) dx0 dt0

0

L

q(x0 , t) G(x, x0 ; 0) dx0 ,

= 0

ˆ since HG(x, x0 ; t, t0 ) = 0 for t > t0 . The remaining integral is the initial impulse at t = 0 — see (42.105).

42.21 (a) Expand the solution T (x, t) on the complete basis of homogeneous eigenfunctions — which for these boundary conditions is a Fourier sine series, T (x, t) =

X

an (t) sin



nπx L



,

an (t) =

L

2 L

Z



nπx L

T (x, t) sin 0

n



nπx L



dx

where the initial condition T (x, 0) = f (x) gives 2 an (0) = L

L

Z

f (x) sin 0



dx .

Similarly, q(x, t) =

X

bn (t) sin



nπx L



,

n

with a similar projection integral for the coefficients bn . (b) Inserting these expansion into the differential equation gives

X

a˙ n + α



nπx L

2

 an sin



nπx L

n



=

X

bn sin



nπx L



.

n

Since the basis is complete, the coefficients for each n must vanish separately, a˙ n (t) + α

362



nπx L

2

an (t) = bn (t) .

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

From (38.15) we thus obtain 2

an (t) = e−α(nπ/L)

 t

t

Z

2 0

t

e+α(nπ/L)

an (0) +

bn (t0 )dt0

 .

0

(c) Plugging the solution for an (t) into the expansion for T (x, y), T (x, t) =

X

=

X

an (t) sin



nπx L



n −α(nπ/L)2 t

2 L

e

2 = L

L 0

f (x )

" X

0

0

−α(nπ/L)2 t

e

sin



nπx0 L



t

Z

0

+α(nπ/L)2 t0

0

e

dx +

bn (t )dt

0

 sin



0

0

nπx L

t

XZ

sin



2

bn (t0 )e−α(nπ/L)

(t−t0 )

0

n

=



nπx L



L

f (x0 )GD (x, x0 , t, 0)dx0 +

Z tZ

0

0

GD (x, x0 ; t, t0 ) =

dx0

 Z 2 L

L

sin



0

nπx L



sin



nπx0 L



dx0

L

bn (t0 )GD (x, x0 , t, t0 )dx0 dt0 ,

0

2 L

X

2

e−α(nπ/L)

(t−t0 )

sin(nπx0 /L) sin(nπx/L).

n

42.22 (a) In the limit as a → ∞, the problem acquires complete symmetry in x, thereby reducing to an effective one-dimensional problem in y. (b) Start with (41.112), Teq (x, y) =

4T∗ π

X 1 h sin (nπx/a) sinh (nπy/a) i n

sinh (nπb/a)

.

n odd

Before taking the limit, choose any finite, positive x < a — for specificity, x = a/2. Then

X 1 h sin (nπ/2) (nπy/a) i

Teq (a/2, y) = lim

n

a→∞

nπb/a

n odd

=

4T∗ y π b

X sin (nπ/2) n n odd

= T∗ y/b, which is (42.112) with L ↔ b, y ↔ y, and inhomogeneous conditions T (0, t) = c1 (t) = 0 and T (L, t) = c2 (t) = T∗ , (c) Together, (41.113) – (41.115) give T˜(x, y, t) =

X

−α

cnm sin(nπx/a) sin(mπy/b) e



2 2 +( mπ ( nπ a ) b )



t

,

nm

with cnm



#

where

4 = ab

nπx L

n

+

Z

0

f (x ) sin

n

Z

L

 Z

Z 0

a

Z

b





T0 (x0 , y 0 ) − Teq (x0 , y 0 ) sin(nπx0 /a) sin(mπy 0 /b) dx0 dy 0 .

0

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363



dt0

This has the basic form of (42.117) with T˜0 = T0 − Teq , T (x, t) = Teq (x, y) + T˜(x, t) =

T∗ y+ L

a

Z

Z

b

T˜0 (x0 ) GD (x, x0 ; t, 0) dx0 dy 0 ,

0

0

where GD (x, x0 , y, y 0 ; t, t0 ) =

4 ab

X

e

−α



2 2 +( nπ ( nπ a ) b )



(t−t0 )

nm

× sin(nπx0 /a) sin(nπx/a) sin(nπy 0 /b) sin(nπy/b). Note that since c2 = T∗ is constant, q˜ = 0 in (42.114).

42.23 With infinite limits in x, we can take the spatial Fourier transform of the heat equation: F










∂t − α∂x2 G(x, t) = F δ(x − x0 )δ(t − t0 )



˜ which yields the first-order equation in G(k, t),


˜ = √1 eik(x−x0 ) δ(t − t0 ) ∂t + αk2 G 2π 2

Referring to (38.13) gives an integrating factor e+αk t , and thence the solution 0 2 1 ˜ G(k, t) = √ e−ikx e−αk t Θ(t − t0 ) 2π

˜ with satisfies the initial condition G(k, 0) = 0 for positive t0 . Then the Green’s function is (invoking translational symmetry)



1



˜ G(x − x0 , t − t0 ) = F −1 G(k, t) = Θ(t − t0 )

p

4π(t − t0 )

−

e

(x−x0 )2 4(t−t0 )

.

With source q(x, t), the solution is thus

Z tZ

∞

T (x, t) = 0

q(x0 , t0 )G(x − x0 , t − t0 )dx0 dt0

−∞

The plot shows the temperature distribution for q(x, t) = cos x e−t

42.24 The source at x = x0 can be “mirrored" by an equivalent image sink at x = −x0 , so that the Green’s function source for the half-bar becomes





δ(x − x0 ) − δ(x + x0 ) δ(t) .

364

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(a) For Dirichlet conditions, the Green’s function is the difference between those of the source and sink. Using the G found in Problem 42.23, (x−x0 )2 (x+x0 )2 1 GD (x, x0 , t) = Θ(t) √ e− 4t − e− 4t 4πt

h

i

,

which clearly vanishes at x = 0. (b) Neumann conditions are more subtle — see the discussion following Example 42.6 and BTW 42.2. But since GD = 0 at t = 0, Neumann conditions can be satisfied simply by adding to the integral over GD a homogeneous solution h(x, t) satisfying ∂x h(0, t) = 0.

42.25 Using the method of images, the semi-infinite string Green’s function can be constructed from a superposition of the infinite string Green’s function and its image at x = −x0 , G(x, x0 , t) = −

 c Θ(ct − |x − x0 |) − Θ(ct − |x + x0 |) . 2

42.26 Building off Problem 42.24, superpose alternating sources and sinks every L to construct the Green’s function of the heat equation which satisfies Dirichlet conditions at x = 0 and L — and which exhibits translation invariance, ∞ X

δ(x − x0 − 2nL) − δ(x + x0 + 2nL).

n=−∞

This leads to the infinite sum of fundamental solutions G(x, x0 ; t) = √

∞ h X

1 4αt

e−(x−x

0

−2nL)2 /4αt

− e−(x+x

0

+2nL)2 /4αt

i

.

n=−∞

This clearly satisfies the boundary condition at x = 0 — and translation invariance guarantees that the condition at x = L is also satisfied: G(L, x0 ; t) = √

= √

∞ h X

1 4αt

e−(L−x

−2nL)2 /4αt

− e−(L+x

0

+2nL)2 /4αt

i

n=−∞ ∞ h X

1 4αt

0

e−(x

0

2

+(2n−1)L) /4αt

− e−(x

0

2

−(2n+1)L) /4αt

i

=0,

n=−∞

where we’ve implicitly let n → n + 1 in the first sum — or equivalently, n → n − 1 in the second.

42.27 (a) Given that G(χ, τ ) satisfies the heat equation, define χ ˜ ≡ λχ, t˜ ≡ λ2 t:



∂ ∂2 −α 2 ∂t ∂χ



 G(χ, ˜ t˜) = λ2

∂G(χ, ˜ t˜) ∂ 2 G(χ, ˜ t˜) −α ∂χ ˜2 ∂ t˜

 = 0,

t>0.

At t = 0, however,

 λ2

∂G(χ, ˜ t˜) ∂ 2 G(χ, ˜ t˜) −α ˜ ∂χ ˜2 ∂t

 = λ2 δ(χ) ˜ = λδ(χ) ,

verifying that λG(λχ, λ2 t) = G(χ, τ ). With this revelation, we wish to choose λ to recast G as a function of a single dimensionless variable ξ. The only dimensional parameter in the 2 equation is α, with units of distance2 /time — so that the parameter ξ = xαt is dimensionless. ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

365

But the initial condition reveals that G has units of 1/distance. So setting λ2 = Green’s function can be expressed as

1 , αt

the

1 G(x, t) = √ g(ξ) , αt for dimensionless function g. (b) With ∂2 2ξ = 2 ∂x2 x

1 ∂ ∂ =− ∂t t ∂ξ plugging G(x, t) =

√1 αt



∂2 ∂ ξ 2 + ∂ξ ∂ξ

 ,

g(ξ) into the heat equation gives 4ξ

∂g 1 ∂2g + (ξ + 2) + g=0. ∂ξ 2 ∂ξ 2

We can solve this 2nd-order ODE by factoring it into two successive 1st-order ODEs,



4

d + 1 g(ξ) = f (ξ) dξ





ξ

d 1 + dξ 2



f (ξ) = 0 .

Straightforward calculation shows that f is a constant times ξ −1/2 , which blows up at the origin — so the only acceptable solution is the constant f (ξ) = 0. This leaves a homogeneous equation for g, with solution 2

g(ξ) = Ae−ξ/4 = Ae−χ

/4αt

= Ae−(x−x

0 2

) /4αt

so that 0 2 A G(x − x0 , t) = √ e−(x−x ) /4αt . αt

In the t → 0 limit, this must reduce to the initial value δ(x − x0 ); consultation with the representations inside the back cover fixes A, to get the Green’s function G(x − x0 , t) = √

1 4παt

e−(x−x

0 2

) /4αt

.

42.28 With homogenous Dirichlet conditions, the only problematic solution would be the trivial one ψ = 0 — which is not actually a problem. Physically, fixed T at the ends of the rod allows for heat flow out — so there’s no need to modify the Green’s function with a sink.

42.29 For



d2 dx2



+ λ G(x, x0 ) = δ(x − x0 ) with homogeneous Neumann conditions ∂x G(0, x0 ) = 0 and

∂x G(L, x0 ) = 0, the (dis)continuity conditions lead to the Green’s function

 √  √  λ(1−x0 ) λx cos cos , x < x0 L L  √ 0  √  G(x, x0 ) = √ . √ λ(1−x) λx λ sin( λ) cos cos , x > x0 L L 1

Inspection shows this to be valid as long as λ , n2 π 2 /L2 for integer n ≥ 0 — that is, for all λ √ d2 not an eigenvalue of the operator dx λ — so 2 . As λ → 0, the cosines go to 1 and the sine to G diverges like 1/λ. Subtracting off this divergence, define ˜ =G− 1 , G λ

366

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which goes to zero as λ → 0. Plugging this into the original Neumann problem reveals





d2 +λ dx2

1 ˜ G(x, x0 ) = δ(x − x0 ) − L

p

with the same Neumann boundary conditions as G. Moreover, since φˆλ=0 = satisfied, ˜ = hφˆ0 |Gi

2/L, (42.88) is

L

Z

˜ φˆ0 (x) G(x, x0 ) dx = 0 ,

0

which is readily confirmed by integrating the differential equation, L

Z




˜ ˜ ∂x2 G(x, x0 ) + λG(x, x0 ) dx =

0

Z

L



δ(x − x0 ) −

0

1 L



dx ,

since the right hand side integrates to zero, while the Neumann conditions cause the integral ˜ 00 to vanish. In the λ → 0 limit, G ˜ becomes equivalent to the modified Green’s function over G of Example 42.7

42.30 For ap solution to exist, the source x/L + c must be orthogonal to the homogeneous solution φˆ = 2/L sin (3πx/L), L

Z

sin 0



3πx L



x +c L



dx =

L (1 + 2c) = 0 3π

−→

c = −1/2 .

Only for this value of c is there a solution to the inhomogeneous equation. ˜ we need to find the eigenfunctions of the operator L = d22 + To construct G, dx



d2 9π 2 + 2 2 dx L

9π 2 , L2

 φn (x) = −λn φn (x) ,

where we’ve taken λ → −λ in (42.59). Thus for integer n > 0,

r φˆn (x) =

2 sin L



nπx L



,

λn = (n2 − 9)π 2 /L2 ,

where qthe Dirichlet condition at x = 0 produced the sine, and the one at x = L yields eigenvalues via

9π 2 L2

+ λn L = nπ. Notice that λ3 = 0 — as it must for this L. As indicated in (42.78),

the modified Green’s function must remove this n = 3 eigensubspace in order to satisfy the orthogonality condition (42.82). Thus the differential equation for the modified Green’s function is



d2 9π 2 + 2 2 dx L



2 ˜ G(x, x0 ) = δ(x − x0 ) − sin L



3πx L



sin



3πx0 L



Inserting the eigenfunction expansion ˜ G(x, x0 ) =

X

bn (x0 ) sin



nπx L



n=1

into the differential equation gives −

X

λn bn (x0 ) sin



nπx L



= δ(x − x0 ) −

2 sin L



3πx L



sin



3πx0 L



.

n

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367

The Fourier coefficients are found by projection, L

Z

2 L

λn bn (x0 ) = −

δ(x − x0 ) −

0

2 sin L

=−

2 sin L

h



nπx0 L



+

2 sin L





3πx L

3πx0 L





sin



3πx0 L

i

sin



nπx L



dx

δn3 .

Thus 2 ˜ G(x, x0 ) = − L

X sin(nπx/L) sin(nπx0 /L) λn

n,3

.

At last, we’re ready to find the solution to the inhomogeneous problem with source x/L − 1/2: L

Z

˜ G(x, x0 )

u(x) =



0

1 x0 − L 2



2L2 π3

dx0 =

X sin(nπx/L) n(n2 − 9)

.

n even

42.31 (a) The modified Green’s function satisfying (42.87), ˜ d2 G 1 = δ(x − x0 ) − . 2 dx L has the general form

 ˜= G

a + bx − c + dx −

1 2 x , 2L 1 2 x , 2L

x < x0 . x > x0

Applying the boundary conditions, ˜ x0 ) = 0 ∂x G(0,

−→

˜ x0 ) = 0 ∂x G(L,

b=0

−→

d=1.

Continuity at x = x0 then gives a = x0 + c so that

 ˜= G

1 2 x , c + x0 − 2L 1 2 c + x − 2L x ,

x < x0 x > x0

.

As expected, the discontinuity condition is automatically satisfied. So we need another constraint to determine c. For that we use (42.88), ˜ = hφˆ0 |Gi

r Z 2 L

L

˜ G(x, x0 ) dx = 0

0

which yields c = − 31 L − x02 /2. Thus we obtain (42.89),

 ˜= G

1 − 13 L + x0 − 2L (x2 + x02 ), 1 − 13 L + x − 2L (x02 + x2 ),

x < x0 . x > x0

(b) A solution with a heat source q(x) = sin(2πx/L) exists since it’s orthogonal to φ0 — see (42.86). The particular solution is L

Z

G(x, x0 ) sin(2πx0 /L)dx0

T (x) = 0 x

Z

G(x, x0 ) sin(2πx0 /L)dx0 +

= 0

=−

Z

L

G(x, x0 ) sin(2πx0 /L)dx0

x

L [sin(2πx/L) + π(1 − 2x/L)] . 4π 2

This clearly satisfies the Neumann boundary conditions — and its 2nd derivative is q(x).

368

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42.32 Substituting c1 (t) = cos t, c2 (t) = sin 2t, and T0 (x) = ex into (42.117) using the Dirichlet Green’s function (42.102) with α = 0.01 gives the following plot:

42.33 (a) From Table 41.1, the Green’s function must have of the general form G(~ r, ~ r 0) =





Jm (βr) Nm (βr)

e±imφ e±βz .

Since this must be finite for all z, β must be imaginary — so we set β = ik for real k to get



0

G(~ r, ~ r )=

Im (kr) Km (kr)



e±imφ e±ikz .

A finite solution requires that we discard K for 0 ≤ rr < r0 , whereas I must be discarded for r0 < r < ∞. Thus

Z ∞ X

0

G(~ r, ~ r )=

∞

 imφ ikz

dk e

e

−∞

m=−∞

am Im (kr) , bm Km (kr) ,

0 ≤ r < r0 , r0 < r < ∞

where the lack of finite boundary conditions on z means there is no quantization condition on k — and hence the sum must be over a continuum of k values. (b) Continuity at r = r0 requires am Im (kr0 ) = bm Km (r0 )

−→

bm =

Im (kr0 ) am . Km (kr0 )

(c) In cylindrical coordinates, (42.24) is 1 ∇2 G(~ r, ~ r 0 ) = −δ(~ r−~ r 0 ) = − δ(r − r0 )δ(φ − φ0 )δ(z − z 0 ) r 1 = − δ(r − r0 ) r

∞ 0 Z X eim(φ−φ )

2π m=−∞

∞

−∞

dk −ik(z−z0 ) e . 2π

So integrating by analogy with (42.28) gives

Z

2

0

3

I

∇ G(~ r, ~ r )d r = V

S

~ ∇G(x, x0 ) · d~a =

Z 





∂G ∂G − ∂r r0 + ∂r r0 −

 rdφ dz ,

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

369

which can be satisfied by −

Z ∞ X

1 r0

m=−∞

∞

−∞





dk im(φ−φ0 ) −ik(z−z0 ) ∂G ∂G e e = − 2π ∂r r0 + ∂r r0 −

Z ∞ X

=

m=−∞

is the Wronskian of Im am (k) = −

dk eimφ eikz am k

m=−∞

−∞

∞

∞

m=−∞

−∞

∞

∞

m=−∞

where

∞

X Z

= (kr0 )

−∞

X Z

=





0 0 (kr0 ) (kr0 ) − am Im dk eimφ eikz k bm Km

Z ∞ X

=

W (kr0 )

∞

h

Im (kr0 ) 0 0 (kr0 ) K (kr0 ) − Im Km (kr0 ) m

dk eimφ eikz

 am k  0 0 (kr0 ) (kr0 ) − Km (kr0 )Im Im (kr0 )Km 0 Km (kr )

dk eimφ eikz

am kW (kr0 ) , Km (kr0 )

−∞

and Km (kr0 ). Thus

1 1 Km (kr0 ) −imφ0 −ikz0 e e . 4π 2 kr0 W (kr0 )

(d) For large ξ,

r Im (ξ) ≈

r

1 ξ e 2πξ

Km (ξ) ≈

π −ξ e 2ξ

Then a brief calculation reveals 0 0 W (ξ) = Im (ξ)Km (ξ) − Km (ξ)Im (ξ) = −

1 , ξ

so that am (k) =

0 0 1 Km (kr0 )e−imφ e−ikz . 4π 2

(e) Putting it all together, G(~ r, ~ r 0) =

=

1 4π 2 1 2π 2

Z ∞ X

−∞

∞

∞

X Z

0

dk eim(φ−φ ) eik(z−z

m=−∞

m=−∞

370

∞

0

 )

i

Im (kr)Km (kr0 ) , Km (kr)Im (kr0 ) ,

0 ≤ r < r0 r0 < r < ∞

0

dk Im (kr< ) Km (kr> ) eim(φ−φ ) cos[k(z − z 0 )] .

−∞

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Coda: Quantum Scattering

43

43.1 Applying the Helmholtz operator L = ∇2 + k2 to (43.5), recalling that φ0 is a homogeneous solution, gives



0 2m :  Lφ(~ r ) = Lφ r ) +L 0 (~ ~2 2m ~2

Z

2m = 2 ~

Z

=



Z

G(~ r−~ r 0 )V (~ r 0 ) φ(~ r 0 ) d3 r 0





L G(~ r−~ r 0 ) V (~ r 0 ) φ(~ r 0 ) d3 r 0 δ(~ r−~ r 0 )V (~ r 0 ) φ(~ r 0 ) d3 r 0 =

2m V (~ r ) φ(~ r) , ~2

which is the source ρ(~ r ) of (43.3).

43.2 Since we’re considering a limited-range potential, we use the first-order approximation |~ r −~ r 0| ≈ r − rˆ · ~ r 0 for r  r 0 to write ~0

0

eikr e−ik ·~r eik|~r−~r | ≈ 0 |~ r−~ r | r − rˆ · ~ r0

0

=

eikr −i~k0 ·~r 0 e + O(1/r2 ) r

where ~k0 ≡ kˆ r. Inserting this into the Lippman-Schwinger equation (43.12) produces (43.14), m eikr φ(~ r) = φ0 (~ r)− 2π~2 r

Z

~ 0 ·~ r0

e−ik

V (~ r 0 ) φ(~ r 0 ) d3 r 0 + O(1/r2 ) .

43.3 We’ll do this two ways: – ∇2

1 r




= −4πδ(~ r ): 1 r

h  

~ · ∇V ~ (r) = ∇ ~ · ∇ ~ ∇ = ∇2

1 r

 

1 ~ −µr
∇ e r  

~ 1 ·∇ ~ e−µr + 1 ∇2 e−µr + 2∇ r r

e−µr

h

= −4πδ(~ r)+2 − = −4πδ(~ r)+

i

e−µr +


rˆ · −µe−µr rˆ + 2 r i



µ2 2µ − 2 r r



e−µr

µ2 −µr e = −4πδ(~ r ) + µ2 V (r) . r

Thus




∇2 − µ2 V (r) = −4πδ(r) . – Fourier space: From (43.19), we know that

 F

e−µr r

 =

4π . q 2 + µ2

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Then

 2

F

2

∇ −µ


e−µr

 = (−q 2 − µ2 )

r

4π = −4π = −4πF [δ(r) ] . q 2 + µ2

43.4 With f (θ) =

X

(2` + 1)f` (k)P` (cos θ) .

`

and the orthogonality of Legendre polynomials

Z

1

P` (x)Pn (x)dx = −1

2 δ`n , 2` + 1

the integral in (43.37) becomes a sum over partial waves:

Z

|f (θ)|2 dΩ

σ=

X

= 2π

(2` + 1)(2n + 1)|f` (k)|

2

Z

X

P` (x)Pn (x)dx −1

`,n

= 4π

1

(2` + 1)|f` (k)|2 .

`

43.5 (a) For large r, Ψ(~ r) →

X

(2` + 1)P` (cos θ)

1 2ik

 (1 + 2ikf` )

eikr e−ikr + (−1)` r r

 .

`

Conservation of probability requires that the incoming and outgoing fluxes are the same — which means the coefficients of e±ikr /r must have the same magnitude. Thus it must be that 1 + 2ikf` is a pure phase. Defining the angle δ` via e2iδ` ≡ 1 + 2ikf` , the scattering amplitude in (43.33) becomes f (θ) =

X

(2` + 1)f` (k)P` (cos θ)

`

=

X

=

X

(2` + 1)

 1  2iδ` e − 1 P` (cos θ) 2ik

(2` + 1)

 eiδ`  iδ` 1 e − e−iδ` P` (cos θ) = 2ik k

`

`

X

(2` + 1)eiδ` sin δ` P` (cos θ)

`

(b) From (43.37), σ = 4π

X

= 4π

X

(2` + 1)|f` (k)|2

`

`

372

1

(2` + 1)

2ik


2 4π X (2` + 1) sin2 δ` . = k2

e2iδ` − 1

`

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X

43.6 (a) Since we’re asked to compare with a form of Ψ expressed in terms of incoming and outgoing waves, it’s sensible to write the general solution to the radial equation in terms of spherical Hankel functions, (1)

(2)

R` (r) = c1 h` (kr) + c2 h` (kr) . For large r, (1,2)

h`

1 ±i(kr−`π/2) e . ikr

(kr) →

Comparison with (41.69) then reveals i` 2iδ` e 2

c1 =

c2 =

i` . 2

Thus i` 2iδ` (1) (2) e h` (kr) + h` (kr) 2 h i i` iδ` iδ` (1) (2) = e e h` (kr) + e−iδ` h` (kr) 2 i` iδ` e [(cos δ` + i sin δ` ) (j` (kr) + in` (kr)) = 2 + (cos δ` − i sin δ` ) (j` (kr) − in` (kr))]

R` (r > a) =

h

i

= i` eiδ` [cos δ` j` (kr) − sin δ` n` (kr)] . As a hard, impenetrable sphere, the wave function must vanish at R = a: cos δ` j` (ka) − sin δ` n` (ka) = 0

−→

tan δ` =

j` (ka) . n` (ka)

In particular, tan δ0 =

sin(ka)/ka = − tan(ka) . − cos(ka)/ka

(b) Low-energy, large-wavelength scattering has small ka. The Taylor expansions for Jn and Nn be adapted to j` and n` to find that for small argument (see Problem C.16) j` (x) ≈

x` 2` `! x` = (2` + 1)! (2` + 1)!!

n` (x) ≈ −

(2`)! −(`+1) (2` − 1)!! x =− . 2` `! x`+1

Thus tan δ` ≈ −

(ka)2`+1 (ka)2`+1 =− . (2` + 1)!!(2` − 1)!! (2` + 1)[(2` − 1)!!]2

So only δ0 is relevant. Moreover, for small ka, δ0 = −ka. Then the total scattering cross section is σ=

4π k2

X

(2` + 1) sin2 δ`

`

4π ≈ 2 sin2 δ0 = 4πa2 . k That this is 4 times the classical expectation is due to diffraction of the wavefunction around the sharp surface at r = a.

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373

Appendix B Rotations in R3

B

B.1 We decompose along and perpendicular to the rotation axis using n ˆ

~v = ~vk + ~v⊥ = (~v · n ˆ) n ˆ+n ˆ × (~v × n ˆ) . As a rotation, |~v | = |~v 0 |. But since the parallel component is unchanged under rotation, so too is the perpendicular: vk = vk0 ⇒ |~v 0⊥ | = |~v⊥ |. So in the plane of the perps, we can decompose ~v 0⊥ along the orthogonal directions ~v⊥ and n ˆ ×~v determined by the axis and the unrotated vector, ~v 0⊥

'

~v ~v 0

= ~v⊥ cos ϕ + (ˆ n × ~v ) sin ϕ ,

where we’ve implicitly used |ˆ n × ~v | = |~v⊥ |. Putting it all together, we get (applying BAC − CAB to the 2nd line) ~v 0 = ~vk0 + ~v 0⊥ = (~v · n ˆ) n ˆ + ~v⊥ cos ϕ + (ˆ n × ~v ) sin ϕ = (~v · n ˆ) n ˆ+n ˆ × (~v × n ˆ ) cos ϕ + (ˆ n × ~v ) sin ϕ = ~v cos ϕ + (1 − cos ϕ) (~v · n ˆ) n ˆ + (ˆ n × ~v ) sin ϕ = ~v cos ϕ + (1 − cos ϕ) [~v − n ˆ × (~v × n ˆ )] + (ˆ n × ~v ) sin ϕ = ~v + (1 − cos ϕ) n ˆ × (ˆ n × ~v ) + (ˆ n × ~v ) sin ϕ .

B.2 Using Eqn. (B.3), Nij ≡ −

P

 n ˆ , k ijk k

! N

2


ij

=

X

Ni` N`j =

`

X X `

=

X

i`r n ˆr

r

! X

`js n ˆs

s

i`r `js n ˆr n ˆs

`rs

=

X

(δjr δis − δrs δij ) n ˆr n ˆs

rs

=n ˆin ˆ j − δij n ˆ·n ˆ=n ˆin ˆ j − δij . So

X

2 Nij vj = n ˆ i (ˆ n · ~v ) − vi −→ N 2~v = n ˆ (ˆ n · ~v ) − ~v = n ˆ × (ˆ n × ~v ) ,

j

where we’ve used BAC − CAB in the last step.

~ ·n B.3 Direct comparison of N with the matrices Li show that N = −iL ˆ . Thus we can write ~

ˆ ·L Rn (ϕ) = e−iϕ n = eϕN .

We can demonstrate N n ˆ = 0 in multiple ways. First: Rn (ϕ)ˆ n = eϕN n ˆ≡n ˆ .

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Each term after 11 in the Taylor expansion of eϕN has powers of N . But Rn (ϕ)ˆ n=n ˆ means only the 11 survives. Another way is to look at Eqn. (B.6): since both Rˆ n=n ˆP and RT n ˆ =n ˆ, their difference vanishes. And finally, in the definition Eqn. (B.3), Nij ≡ − k ijk n ˆ k , the P P anti-symmetry of the  makes Nij n ˆj =  n ˆ n ˆ vanish. j jk ijk k j Although it may not be immediate obvious that eϕN is orthogonal, it clearly must be — the secret is N ’s antisymmetry: RRT = eϕN eϕN

T

= eϕN e−ϕN = 11 .

So if one insists that R be manifestly orthogonal rather than unitary, the generators must be real, antisymmetric matrices (rather than hermitian).

B.4 (a) Working out

2

Rn (ϕ) = 11 + (1 − c) N + s N = 11 + (1 − c)

−ˆ nz 0 n ˆx

0 n ˆz −ˆ ny

n ˆy −ˆ nx 0

!2 +s

−ˆ nz 0 n ˆx

0 n ˆz −ˆ ny

n ˆy −ˆ nx 0

gives the expected result, Eqn. (B.5). (b) The trace of the Rodrigues matrix is 3c + (1 − c)ˆ n2x + (1 − c)ˆ n2y + (1 − c)ˆ n2z = 2c + 1 X Similarly direct matrix multiplication confirms Rn n ˆ=n ˆ.

B.5 (a) Starting with 26.89, d~ r =ω ~ ×~ r = ωN~ r, dt where we’ve used Eqn. (B.2)-(B.3). This first-order differential equation has solution (note that as a matrix equation, ~ r(0) must sit to the right of the exponential) ~ r(t) = eωN t ~ r(0) , so that the rotation matrix is R = eωN t . (b) Recalling that for a matrix M , exp[M ] is defined by its Taylor expansion, we need to calculate powers of N . Direct multiplication easily reveals N 2k = (−1)k−1 N 2

N 2k−1 = (−1)k−1 N ,

for integer k ≥ 1. Thus

eωN t = 11 + ωtN +

 = 11 +

ω 2 t2 2 ω 3 t3 3 N + N + ··· 2! 3!

ω 3 t3 ω 5 t5 ωt − + − ··· 3! 5!



 N+

ω 2 t2 ω 4 t4 ω 6 t6 − + − ··· 2! 4! 6!

= 11 + sin(ωt) N + [1 − cos(ωt)] N 2 which is Eqn. (B.4). X

376

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 N2

! ,

B.6 (a)

! N

2


ij

=

X

Ni` N`j =

X X

`

`

=

! X

i`r n ˆr

r

X

`js n ˆs

s

i`r `js n ˆr n ˆs

`rs

=

X

(δjr δis − δrs δij ) n ˆr n ˆs

rs

=n ˆin ˆ j − δij , where we’ve used n ˆ·n ˆ = 1. And it’s clearly symmetric. Then N4


ij

=

X

N2


ik

N2


kj

=

X

=

X

k

(ˆ ni n ˆ k − δik ) n ˆk n ˆ j − δkj




k

n ˆin ˆk n ˆk n ˆ j − δik n ˆk n ˆ j − δkj n ˆin ˆ k + δik δkj




k

=n ˆin ˆj − n ˆin ˆj − n ˆin ˆ j + δij = −ˆ ni n ˆ j + δij = − N 2


ij

.

(b) Test for orthogonality (remembering that N is antisymmetric):





T







RRT = 11 + (1 − cos ϕ) N 2 + sin ϕ N = 11 + (1 − cos ϕ) N 2 + sin ϕ N



= 11 + (1 − cos ϕ) N 2

2

11 + (1 − cos ϕ) N 2 + sin ϕ N 11 + (1 − cos ϕ) N 2 − sin ϕ N

− sin2 ϕ N 2

= 11 + 2(1 − cos ϕ) N 2 + (1 − cos ϕ)2 N 4 − sin2 ϕ N 2 = 11 + 2(1 − cos ϕ) N 2 − (1 − cos ϕ)2 N 2 − sin2 ϕ N 2 = 11 + (1 − cos2 ϕ − sin2 ϕ)N 2 = 11

X

(c) For an anti-symmetric matrix, det(N ) = det(N T ) = det(−N ) = (−1)n det(N ) = 0 for n odd. Then det N = 0 =⇒ det(N 2 ) = (det N )2 = 0, and therefore R is a proper rotation:





det R = det 11 + (1 − cos ϕ) N 2 + sin ϕ N = det 11 = 1.

B.7 To first order in δϕ, Rn (δϕ) = 11 + (1 − cos δϕ) N 2 + sin δϕ N = 11 + δϕ N . Applied to an arbitrary vector ~v — and using Eqn. (B.2) — we get Rn (δϕ) ~v = [11 + δϕ N ] ~v = ~v + δϕ (ˆ n × ~v ) = ~v + δ ϕ ~ × ~v .

B.8 For rotation around z, n ˆx = n ˆ y = 0, n ˆ z = 1. Then

N =

0 1 0

−1 0 0

!

0 0 0

2

N =

−1 0 0

0 −1 0

!

0 0 0

.

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377

So Rz (ϕ) = 11 + sin ϕ N + (1 − cos ϕ) N 2 =

1 0 0

0 1 0

=

cos ϕ sin ϕ 0

!

0 0 1

+ sin ϕ

− sin ϕ cos ϕ 0

0 1 0

−1 0 0

!

0 0 0

+ (1 − cos ϕ)

−1 0 0

!

0 −1 0

0 0 0

!

0 0 1

X

B.9 Although you can do this by hand, using Mathematica is much easier. It’s best to verify that R is in fact a rotation — using Chop to remove any numerical detritus: R.RT returns 1. 0. 0.

0. 1. −1.1102230246251565`*∧ -16

0. −1.1102230246251565`*∧ -16 1.

!

and R.RT //Chop gives the identity. Then from Eqn. (B.7) the vector 2 sin ϕ n ˆ has components v = {R[[3, 2] − R[[3, 2]], R[[1, 3]] − R[[3, 1]], R[[2, 1]] − R[[1, 2]]} = {0.126826, −0.856688, −1.5} Then ϕ = sin−1

h √ i 1 2

~v 2 = π/3

and nhat =

v = {0.0732233, −0.494609, −0.866025} 2 sin ϕ

Using TrR = 2 cos θ + 1 gives the supplementary angle 2π/3, which is the rotation around −ˆ n.

B.10 The canonical Rz matrix operator emerges simply by setting β = 0: R(α, 0, γ) = Rz (α + γ). Similarly, R(0, β, 0) = Ry (β). Rx can be found by a similarity transformation rotating y → z. Or squinting at Figure B.2: R(−π/2, β, π/2) = Rx (β).

B.11 To find the angle of rotation ϕ, use the invariance of the trace and then various trig identities: 1 [Tr (R(α, β, γ)) − 1] 2 1 = [cos α cos β cos γ − sin α sin γ + cos α cos γ − sin α cos β sin γ + cos β − 1] 2 1 = [(1 + cos β) (cos α cos γ − sin α sin γ) + cos β − 1] 2 1 = [(1 + cos β) cos (α − β) + cos β − 1] 2 = cos2 (β/2) cos (α − β) − sin2 (β/2)  i h  α−β 1 = cos2 (β/2) 2 cos2 − − sin2 (β/2) 2 2   α−β = 2 cos2 (β/2) cos2 −1 X 2

cos ϕ =

To find the spherical coordinates of the axis n ˆ = (sin θ cos φ, sin θ sin φ, cos θ), we can use two

378

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of the three equations in Rˆ n=n ˆ (since n ˆ is unit length, the third is not linearly independent). The equation for n ˆ z is simplest, so start there: cos θ = (− sin β cos γ) sin θ cos φ + (sin β sin γ) sin θ sin φ + (cos β) cos θ = sin θ sin β (− cos φ cos γ + sin φ sin γ) + cos θ cos β = − sin θ sin β cos (φ + γ) + cos θ cos β So tan θ =

−1 + cos β . sin β cos (φ + γ)

OK, so now we need another equation relating θ and φ. Either one will do — let’s take n ˆx: sin θ cos φ = (cos α cos β cos γ − sin α sin γ) sin θ cos φ − (cos α cos β sin γ + sin α cos γ) sin θ sin φ + (cos α sin β) cos θ Since our expression for tan θ has cos (φ + γ) that’s our goal here as we look to deploy trig id’s. And indeed, we find sin θ cos φ = sin θ [cos α cos β cos (φ + γ) − sin α sin (φ + γ)] + cos θ cos α cos β . Dividing out sin θ and using our expression for tan θ ultimately reveals cos φ = − cos (α − γ − φ) = cos [(α − γ + π) − φ] so that φ=

1 (α − γ + π) . 2

X

Plugging this back into tan θ and using half-angle identities gives tan θ = tan

β /sin 2

 



α+γ 2



.

X

B.12 A coordinate axis in the body frame is given by the standard matrix basis. Thus in the fixed frame, each is represented by a column of the Euler matrix. For instance, zˆ0 = (cos α sin β, sin α sin β, cos β), which is the third column of R(α, β, γ). The rows, then, give the inverse transformation — i.e., the representation of the fixed coordinate axes in the body frame.

B.13 Substituting Rz0 (γ) = Ry0 (β)Rz (γ)Ry−1 0 (β) Ry0 (β) = Rz (α)Ry (β)Rz−1 (α) one at a time into Eqn. (B.16) yields R(α, β, γ) = Ry0 (β)Rz (γ)Ry−1 0 (β) Ry 0 (β)Rz (α)





= Ry0 (β)Rz (γ)Rz (α)





= Rz (α)Ry (β)Rz−1 (α) Rz (γ)Rz (α) = Rz (α)Ry (β)Rz (α) , where in the last line we’ve used Rz−1 (α)Rz (γ)Rz (α) = Rz (−α + γ + α) = Rz (γ) — or simply that rotations around the same axis commute.

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379

B.14 Uz is largely worked out in the text: setting Eqn. (B.28) and (B.30) equal,



vz0 vx0 + ivy0

vx0 − ivy0 −vz0



 =

vz eiϕ (vx + ivy )



e−iϕ (vx − ivy ) −vz

gives vz0 = vz , vx0 = vx cos ϕ − vy sin ϕ, and vy0 = vx sin ϕ + vy cos ϕ.

 For Ux , we set

Ux (ϕ) V

Ux† (ϕ)

vz0 vx0 + ivy0

vx0 − ivy0 −vz0

 equal to



cos ϕ/2 −i sin ϕ/2



vz cos ϕ + vy sin ϕ vx + ivy cos ϕ − ivz sin ϕ

=

=



−i sin ϕ/2 cos ϕ/2

vz vx + ivy

vx − ivy −vz





cos ϕ/2 +i sin ϕ/2

+i sin ϕ/2 cos ϕ/2



vx − ivy cos ϕ + ivz sin ϕ −vz cos ϕ − vy sin ϕ

,

confirming vx = vx0 , vy0 = vy cos ϕ − vz sin ϕ, and vz0 = vy sin ϕ + vz cos ϕ. Finally, Uy (ϕ) V

Uy† (ϕ)

 =

cos ϕ/2 sin ϕ/2



− sin ϕ/2 cos ϕ/2

 =

vz vx + ivy

vz cos ϕ − vx sin ϕ ivy + vx cos ϕ + vz sin ϕ

vx − ivy −vz



cos ϕ/2 − sin ϕ/2

−ivy + vx cos ϕ + vz sin ϕ −vz cos ϕ + vx sin ϕ



sin ϕ/2 cos ϕ/2



confirms vy0 = vy , vz0 = vz cos ϕ − vx sin ϕ, and vx0 = vx cos ϕ + vz sin ϕ.

B.15 (a) Two consecutive rotations around the same axis is given by the double similarity transformation







U U V U † U † = (U U ) V U † U † ≡ W V W † . To verify all is well, we need only calculate W using the property (25.138) of Pauli matrices: W = U U = [11 cos(ϕ/2) − iˆ n·~ σ sin(ϕ/2)]2 = 11 cos2 (ϕ/2) − (ˆ n·~ σ )2 sin2 (ϕ/2) − 2iˆ n·~ σ sin(ϕ/2) cos(ϕ/2)





= 11 cos2 (ϕ/2) − sin2 (ϕ/2) − iˆ n·~ σ [2 sin(ϕ/2) cos(ϕ/2)] = 11 cos ϕ − iˆ n·~ σ sin ϕ for a net rotation by twice the angle. (b) Once again using (25.138): U2 (ϕ2 )U1 (ϕ1 ) = [cos(ϕ2 /2) − iˆ n2 · ~ σ sin(ϕ2 /2)] [cos(ϕ1 /2) − iˆ n1 · ~ σ sin(ϕ1 /2)] = cos(ϕ2 /2) cos(ϕ1 /2) − (ˆ n2 · ~ σ ) (ˆ n1 · ~ σ ) sin(ϕ2 /2) sin(ϕ1 /2) −i (ˆ n2 · ~ σ ) sin(ϕ2 /2) cos(ϕ1 /2) − i (ˆ n1 · ~ σ ) sin(ϕ1 /2) cos(ϕ2 /2) = cos(ϕ2 /2) cos(ϕ1 /2) − n ˆ1 · n ˆ 2 sin(ϕ2 /2) sin(ϕ1 /2) − iσ · (ˆ n1 × n ˆ 2 ) sin(ϕ2 /2) sin(ϕ1 /2) −i (ˆ n2 · ~ σ ) sin(ϕ2 /2) cos(ϕ1 /2) − i (ˆ n1 · ~ σ ) sin(ϕ1 /2) cos(ϕ2 /2) = [cos(ϕ2 /2) cos(ϕ1 /2) − n ˆ1 · n ˆ 2 sin(ϕ2 /2)] −iσ · [ˆ n1 sin(ϕ1 /2) cos(ϕ2 /2) + n ˆ 2 sin(ϕ2 /2) cos(ϕ1 /2) + (ˆ n1 × n ˆ 2 ) sin(ϕ2 /2) sin(ϕ1 /2)] ≡ cos(θ/2) − iˆ n·~ σ sin(θ/2) .

B.16 (a) Unitarity can be verified without writing out matrix components: U U † = [11 cos(ϕ/2) − iˆ n·~ σ sin(ϕ/2)] [11 cos(ϕ/2) + iˆ n·~ σ sin(ϕ/2)] = 11 cos2 (ϕ/2) + (ˆ n·~ σ ) (ˆ n·~ σ ) sin2 (ϕ/2) = 11 cos2 (ϕ/2) + 11 sin2 (ϕ/2) + i~ σ · (ˆ n×n ˆ ) = 11 .

380

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(b) V 0 ≡ ~v 0 · ~ σ = U (ϕ) ~v · ~ σ U † (ϕ) = [11 cos ϕ/2 − iˆ n·~ σ sin ϕ/2] ~v · ~ σ [11 cos ϕ/2 + iˆ n·~ σ sin ϕ/2] = ~v · ~ σ cos2 ϕ/2 + (ˆ n·~ σ ) (~v · ~ σ ) (ˆ n·~ σ ) sin2 ϕ/2 +i [(~v · ~ σ ) (ˆ n·~ σ ) − (ˆ n·~ σ ) (~v · ~ σ )] cos ϕ/2 sin ϕ/2 . Repeated use of Eqn. (25.138) — and a little BAC-CAB — leads to ~v 0 = ~v cos2 (ϕ/2) + [2(ˆ n · ~v )ˆ n − ~v ] sin2 (ϕ/2) + 2(ˆ n × ~v ) sin(ϕ/2) cos(ϕ/2) = ~v cos ϕ + (1 − cos ϕ) (ˆ n · ~v )ˆ n + (ˆ n × ~v ) sin ϕ = ~v + (1 − cos ϕ) [ˆ n × (ˆ n × ~v )] + (ˆ n × ~v ) sin ϕ which is just the Rodrigues angle-axis formula, Eqn. (B.1).

B.17 (a) Hit Eqn. (B.36),

X

U † σj U =

Rj` σ` ,

`

from the right with σk and take the trace:




Tr U † σj U σk =

X

Rj` Tr (σ` σk ) =

X

`

Rj` 2δ`k = 2Rjk

`




Thus Rjk = 12 Tr U † σj U σk .X (b) Perhaps the simplest way to do this is by explicit matrix multiplication. With

 Uz (ϕ) = 11 cos (ϕ/2) − iσ3 sin (ϕ/2) =

e−iϕ/2 0



0 eiϕ/2

from Eqn. (B.29), we have (Rz )ij


1 1 = Tr Un† σj Un σk = Tr 2 2



e+iϕ/2 0





0 e−iϕ/2

σj

e−iϕ/2 0



0

 σk

e+iϕ/2

So the diagonal elements of Rz are (Rz )11 = =

(Rz )22 = =

1 Tr 2



1 Tr 2



e+iϕ/2 0

e+iϕ 0

1 Tr 2



1 Tr 2



e+iϕ 0

1 0

e−iϕ/2 0

0 e+iϕ/2



1 0



−i 0



0 −1

0 1



 = cos ϕ

X



−i 0

0

0 i

e−iϕ/2

0 e−iϕ



0 1

e−iϕ/2

0 e−iϕ

e+iϕ/2 0



0



e−iϕ/2 0

0 e+iϕ/2

0 i



 = cos ϕ

X

and (Rz )33

1 = Tr 2 =



e+iϕ/2 0

1 Tr11 = 1 2

0 e−iϕ/2



1 0



0 −1

e−iϕ/2 0

0 e+iϕ/2

1 0



X

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381

The off-diagonals are (Rz )12 = − (Rz )21 = =

1 Tr 2



i Tr 2



e+iϕ/2 0

e+iϕ 0



0 e−iϕ/2

0 −e−iϕ



0 1

e−iϕ/2 0

1 0



0

−i 0

0 i

e+iϕ/2



 = − sin ϕ

X

and finally R13 = R31 =

R23 = R32

1 Tr 2

1 = Tr 2



e+iϕ/2 0



e+iϕ/2 0



0

0 1

e−iϕ/2

0



0 i

e−iϕ/2



e−iϕ/2 0

1 0

−i 0



0 e+iϕ/2



e−iϕ/2 0



1 0

0 −1



0



1 0

e+iϕ/2

=0

0 −1

=0

(c) Using Eqn. (B.24): 1 Tr [(11 cos ϕ/2 + iˆ n·~ σ sin ϕ/2) σj (11 cos ϕ/2 − n ˆ ·~ σ sin ϕ/2) σk ] 2 X 1 1 = cos2 ϕ/2 Tr (σj σk ) + sin2 ϕ/2 n ˆmn ˆ ` Tr (σm σj σ` σk ) 2 2

Rjk =

`m

X i + sin ϕ/2 cos ϕ/2 n ˆ ` [Tr (σ` σj σk ) − Tr (σj σ` σk )] . 2 `

Now Tr (σj σk ) = 2δjk , so the first term is δij cos2 ϕ/2. Then using σj σ` = 11δj` +i twice, and noting that Trσj = 0, the second term becomes

P

 σ n j`n n

"

X X 1 1 sin2 ϕ/2 n ˆmn ˆ ` Tr (σm σj σ` σk ) = sin2 ϕ/2 n ˆmn ˆ ` Tr σm 2 2 `m

! δj` + i

X

`m

j`n σn

# σk

n

"

!

X X 1 n ˆmn ˆ ` Tr δj` σm σk + i j`n = sin2 ϕ/2 2 `m

= sin ϕ/2

X

#

n ˆmn ˆ ` δj` δmk −

`m

X

j`n mnp δpk

np

" = sin2 ϕ/2

X

#

n ˆmn ˆ ` δj` δmk −

`m

= sin2 ϕ/2

X

mnp σp

p

" 2

δmn + i

n

X

X

j`n kmn

n



n ˆmn ˆ ` δj` δmk − δjk δ`m − δjm δ`k




`m




= sin2 ϕ/2 2ˆ nj n ˆ k − δjk , where we’ve used

P `

n ˆ`n ˆ ` = 1. As similar calculation for the third term yields:

"

X X X i i sin ϕ/2 cos ϕ/2 n ˆ ` [Tr (σ` σj σk ) − Tr (σj σ` σk )] = sin ϕ/2 cos ϕ/2 n ˆ ` Tr 2i `jn σn σk 2 2 `

`

= −2 sin ϕ/2 cos ϕ/2

n

X

jk` n ˆ` .

`

382

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#

# σk

Putting it all together, using half-angle trig identities, finally gives




Rjk = cos2 ϕ/2 δij + sin2 ϕ/2 2ˆ nj n ˆ k − δjk − 2 sin ϕ/2 cos ϕ/2

X

jk` n ˆ`

`




= cos2 ϕ/2 − sin2 ϕ/2 δij + 2 sin2 ϕ/2 n ˆj n ˆ k − 2 sin ϕ/2 cos ϕ/2

X

jk` n ˆ`

`

= cos ϕ δij + (1 − cos ϕ) n ˆj n ˆ k − sin ϕ jk` n ˆ` , where we can drop the final sum over ` since j, k are fixed. This is the Rodrigues formula of Eqns. (B.4)-(B.5).

B.18 (a) From Eqn. (2.41), the unit sphere in R4 has spherical coordinates x = sin ψ sin θ cos φ y = sin ψ sin θ sin φ z = sin ψ cos θ w = cos ψ . where 0 ≤ φ < 2π, but both θ and ψ run from 0 to π. The north pole is most-easily identified with ψ = 0. Then the axis of rotation has the ordinary spherical coordinates n ˆ = (sin θ cos φ, sin ψ sin θ sin φ, cos θ) , and the sphere’s surface can be denoted by the four-dimensional vector (cos ψ, n ˆ sin ψ). The rotation angle then is ϕ = 2ψ, so that 0 ≤ ϕ < 2π. (b) If we take the familiar two-sphere to represent R3 rotations, then the direction a path takes out of the north pole (i.e, the longitude) gives the rotation axis, and the arc length of the path (i.e., the extent of a path’s colatitude) the rotation angle. But this requires that the south pole be at colatitude 2π in order for all 2π rotations to be equivalent. Thus a great circle from pole to pole and back is a 4π rotation.

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383

Appendix C The Bessel Family of Functions

C

C.1 From (39.18) and Bessel’s equation, we easily deduce that the Wronksian of the two solutions J±ν has the form W (Jν , J−ν ) = C/x for some constant C. To find it, compute W for small x. Using the expansion (C.2),

J±ν (x) =

∞ X

(−1)m m! Γ(m ± ν + 1)

 2m±ν x 2

x small

−−−−−→

1 Γ(1 ± ν)

 ±ν x 2

.

m=0

gives 1 2ν Γ(1 + ν)Γ(1 − ν) x 2ν 1 =− Γ(1 + ν)Γ(1 − ν) x 1 2 sin πν 1 2ν =− , =− νΓ(ν)Γ(1 − ν) x π x

0 W (Jν , J−ν ) = Jν J−ν − Jν0 J−ν = −

where we’ve used the identity Γ(a)Γ(1 − a) = sinππa . So C = −2 sin νπ/π, which vanishes for integer ν: J±ν are not independent for integer ν. A very similar calculation of the constant for the Wronskian of Jµ and Nν reveals C = 2/π. So Jµ and Nµ are independent for all ν.

C.2 The simplest thing to do is evaluate the limit — here for n = 0 to 10: In[]:= Out[]=

In[]:= Out[]=

Table[Limit[BesselJZero[n, k + 1] - BesselJZero[n, k], k  ∞], {n, 0, 10}] {π, π, π, π, π, π, π, π, π, π, π}

Table[Limit[BesselYZero[n, k + 1] - BesselYZero[n, k], k  ∞], {n, 0, 10}] {π, π, π, π, π, π, π, π, π, π, π}

But a visual presentation is more compelling — here for Jn , n = 0 to 2: In[]:=

ListPlot[{Table[BesselJZero[0, k + 1] - BesselJZero[0, k], {k, 1, 100}], Table[BesselJZero[1, k + 1] - BesselJZero[1, k], {k, 1, 100}], Table[BesselJZero[2, k + 1] - BesselJZero[2, k], {k, 1, 100}]}, AxesLabel  {k, None}, Ticks  {Automatic, {π}}]

Out[]=

π

20

40

60

80

100

k

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C.3 Adding and subtracting Eqns. (C.17) and letting ν → ν±1 yields (C.18a). Using these identities,

 d  ±ν x Jν (x) = ±νx±ν−1 Jν (x) + x±ν Jν0 (x) dx h i ±ν = x±ν Jν (x) + Jν0 (x) = ±x±ν Jν∓1 , x which is (C.18b).

x

1

C.4 With G(x, t) = e 2 (t− t ) , x

G(x, t) = e− 2t e =

xt 2

#" ∞ # "∞ X  x m tm X  −x n t−n n!

2

m!

2

J` (x) t` ,

`=m−n .

`=−∞

m=0

n=0

∞ X

≡

For ` ≥ 0, the coefficient function Jn can be found from (6.52) derived in Problem 6.20 — where the constraint n ≤ m now allows the sum over n to range from 0 to infinity:

J`≥0 (x) =

∞ X

an b`+n =

∞   X 1 −x n

n!

2

1 (` + n)!

 `+n x 2

n=0

n=0

=

∞  `+2n X (−1)n x

n!(` + n)!

2

,

n=0

which is (C.2) for integer ν.

C.5

X

Jn0 (x)tn =

∂G 1 = ∂x 2



t−

1 t



G(x, t)

n

=

1 2

∞ X

Jn (x) tn+1 − tn−1




n=−∞

=

1 2

∞ X

Jn (x) tn+1 −

n=−∞

=

∞ 1 X

2

∞ X

1 2

Jn (x) tn−1

n=−∞

[Jn−1 (x) − Jn+1 (x)] tn .

n=−∞

x

1

C.6 Verifying the Bessel identities in Table C.2: G(x, t) = e 2 (t− t ) – Observing that x

1

G(x, t) = e 2 (t− t ) = G(−x, 1/t) = G(x, −1/t) , we find ∞ X n=−∞

Jn (x)tn =

∞ X

n=−∞

Jn (−x)t−n =

∞ X

J−n (−x)tn

−→

Jn (x) = J−n (−x) ,

n=−∞

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385

and ∞ X

Jn (x)tn =

n=−∞

∞ X

Jn (x)(−t)−n =

n=−∞

∞ X

(−1)n J−n (x)tn

Jn (x) = (−1)n J−n (x) .

−→

n=−∞

Thus Jn (−x) = J−n (x) = (−1)n Jn (x). X – Setting t = 1 in the generating function: ∞ X

G(x, 1) = 1 =

Jn (x) .

X

n=−∞

– Using the Cauchy product (Problem 6.20), ∞ X

1

x

1

y

1

1

Jn (x + y) = e 2 (x+y)(t− t ) = e 2 (t− t ) e 2 (t− t )

n=−∞

" =

#"

∞ X

k

Jk (x)t

k=−∞

=

∞ X

#

∞ X

m

Jm (y)t

m=−∞

"

n=−∞

#

∞ X

Jk (x)Jn−k (y) tn .

X

k=−∞

– Setting t = eiϕ in the generating function: G(x, eiϕ ) = eix sin ϕ = cos(x sin ϕ) + i sin(x sin ϕ) =

∞ X

∞ X

Jn (x)einϕ =

n=−∞

Jn (x) [cos(nϕ) + i sin(nϕ)]

n=−∞

Equating real and imaginary parts,

cos(x sin ϕ) =

∞ X

Jn (x) cos(nϕ)

n=−∞

= J0 (x) +

∞ X

J2n (x) cos(2nϕ)

X

n=1

where we’ve and used the parity relation J−n (x) = Jn (−x) = (−1)n Jn (x). Similarly,

sin(x sin ϕ) =

∞ X

Jn (x) sin(nϕ)

n=−∞

=

∞ X

J2n−1 (x) cos((2n − 1)ϕ)

X

n=1

Incidentally, note that setting ϕ = π/2 gives Bessel expansions of cos x and sin x. – From the previous result, take ϕ = 0 in cos(x sin ϕ) and ϕ = π/2 in cos(x sin ϕ):

cos(x sin 0) = 1 =

∞ X

Jn (x)

X

n=−∞

386

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– Using the Cauchy product (Problem 6.20),

"

#"

∞ X

G(x, t)G(x, 1/t) = 1 =

n

Jn (x)t

n=−∞

Since 1 =

t0 ,

−m

Jn (m)t

m=−∞

!

∞ X

∞ X

n=−∞

k=−∞

=

#

∞ X

Jn+k Jk (x)

tn

only the n = 0 term contributes, ∞ X

1=

Jk2 (x) = J02 (x) + 2

k=−∞

∞ X

Jk2 (x)

X

k=1

and ∞ X

Jn+k (x)Jk (x) = 0 , n , 0.

k=−∞

– Using (C.18b),

 d  ±ν x Jν (x) = ±x±ν Jν∓1 (x) dx immediately gives

Z

xn+1 Jn (x)dx =

Z

 d  n+1 x Jn+1 (x) dx = xn+1 Jn+1 (x) dx

and

Z

x−(n−1) Jn (x)dx = −

Z

 d  −(n−1) x Jn−1 (x) dx = −x−n+1 Jn−1 (x) dx

X

C.7 (a) Perhaps the most elegant way to do this is to add and subtract the relations in (C.17) to obtain



n d − x dx



We can think of the operators





Jn = Jn+1 n x

∓

d dx

n d + x dx



Jn = Jn−1

as raising and lowering operators. In particular

d n+1 + x dx



n d − x dx



Jn = Jn .

Multiplying out the operator product — using the product run when appropriate! — reveals Bessel’s equation. (b) Adding/subtracting the two recursion relations gives n Jn ± Jn0 = Jn∓1 x

n→n±1

−−−−−→

n±1 0 Jn±1 ± Jn±1 = Jn x

X

Similarly,

 d  ±n x Jn (x) = ±nx±n−1 Jn (x) + x±n Jn0 (x) dx h i n = ±x±n Jn (x) ± Jn0 (x) = ±x±n Jn∓1 (x). x

X

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387

C.8 (a) We identify the generating function for the modified Bessel function In (x) as the function P GI (x, t) for which GI (x, t) = I (x)tn . To find it, start with in In (x) = Jn (ix) and Jn ’s n n generating function: GJ (ix, t) =

X

Jn (ix)tn = e

ix 2

1 (t− 1t ) = e x2 (it+ it )

n

≡

X

in In (x)tn =

X

n

In (x)(it)n .

n

So replacing it with τ , x

1

GI (x, τ ) = e 2 (τ + τ ) =

X

In (x)τ n .

n

(b) Observing that x

1

GI (x, t) = e 2 (t+ t ) = GI (x, 1/t) = GI (−x, −t) , we find ∞ X

In (x)tn =

n=−∞

∞ X

In (x)t−n =

n=−∞

∞ X

I−n (x)tn

−→

In (x) = I−n (x) ,

n=−∞

and ∞ X n=−∞

In (x)tn =

∞ X

In (−x)(−t)n =

n=−∞

∞ X

(−1)n In (−x)tn

−→

In (x) = (−1)n In (−x) .X

n=−∞

C.9 (a) Working backwards: eiπ(ν+1)/2 Iν (x) −

2 −iπν/2 2 π I−ν (x) − Iν (x) e Kν (x) = eiπ(ν+1)/2 Iν (x) − e−iπν/2 · π π 2 sin νπ
−iπν  eiπν/2  −iπν = Iν (x) i sin νπ + e −e I−ν (x) sin νπ  eiπν/2  = cos νπIν (x) − e−iπν I−ν (x) sin νπ  1  = cos νπ iν Iν (x) − i−ν I−ν (x) sin νπ 1 = [cos νπJν (ix) − J−ν (ix)] = Nν (ix) X sin νπ

(b) Using the result of part (a): (1)

Hν (ix) = Jν (ix) + iNν (ix)

h

= iν Iν (x) + i eiπ(ν+1)/2 Iν (x) − =

2 −iπν/2 e Kν (x) π

i

2 Kν (x) X πiν+1

Verifying the second identity is fairly easy to for integer ν, but a bit subtle (and tedious) for non-integer. Perhaps the most elegant approach is to via the complex conjugate of the Hankel functions. From the Taylor expansion of Jν , we discover Jν (ix)∗ = Jν (−ix). Then

h

(1)

i∗

Hν (ix)

= [Jν (ix) + iNν (ix)]∗ (2)

= Jν (−ix) − iNν (−ix) = Hν (−ix) Since Kν is a real function, the identity is verified.

388

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C.10 (a) Deriving the connection between Ai and K is challenging, but it helps to already know the √ answer. Start with the variable change t = 2 x sinh(u/3), 1 Ai(x) = 2π

∞

Z

1 3 +xt

ei( 3 t

−∞ ∞

Z

1 = 2π √ x = 3π

h 

exp i

Z−∞∞

) dt

i √ 2

8 3/2 x sinh3 (u/3) + 2x3/2 sinh(u/3) 3

h 

exp i −∞

3

x cosh(u/3)du

2 3/2 x sinh u cosh(u/3)du , 3



i

where we’ve used a (de Moivre-derived) identity for the sinh. Then letting s = ieu , √

Z

x Ai(x) = 3π √ x = 6π

∞

h 

exp i −∞ i∞

Z

h

exp − 0

2 3/2 x 3

1 2






1 u/3 −u/3
1 u/3 e + e−u/3 e +e du 2 2

2 3/2 x 3

i



s+

1 s

i 

(is)1/3 + (is)−1/3

 ds s

.

Motivated by the symmetry Kν = K−ν , note that since the exponential is invariant under s ↔ 1/s, the contributions from the s±1/3 terms are identical. Thus √

Z

i∞

1 x 1 2 3/2 ds x s+ (2 cos π/6) s−1/3 exp − 6π 0 2 3 s s √ Z ∞  i h  1 x 1 2 3/2 ds = √ x s+ s−1/3 , exp − 2 3 s s 2 3π 0

h

Ai(x) =





i

where in the last step we rotated the integration contour from the positive imaginary axis to the positive real axis. Then with one final substitution s = ew , Ai(x) =

√ x √ 2 3π

Z

∞

h 

exp − −∞

2 3/2 x cosh w e−w/3 dw , 3



i

which, as comparison with (C.28) reveals, is K 1 ( 23 x3/2 ) for real x > 0. 3

(b) The result emerges after adapting the recursion relations in Table C.3 for the modified Bessel function, 2ν Kν (x) = Kν+1 (x) − Kν−1 (x) x

2Kν0 (x) = − [Kν−1 (x) + Kν+1 (x)]

C.11 In terms of the family of functions φn (x) = sin(αmn nx/a), where αmn = mπ/n is the mth zero of φn , the familiar orthogonality relation

Z

a

sin(mπx/a) sin(`πx/a)dx = 0

a δm` 2

becomes

Z

a

φn (αmn nx/a)φn (α`n nx/a) dx = 0

a 02 φ (αmn )δm` , 2 n

φ0n (x)

where ≡ dφn (x)/d(nx) = cos(nx), which of course is ±1 for x a zero of the sine. This has the same Sturm-Liouville form as in (C.19) — though unlike Bessel functions, φ0n (α) is not equal to φn±1 (α). (See also Problems 40.22–40.23.)

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

389

C.12 For u(x) =

∞

Z

1 2

e−x cosh t−νt dt ,

x>0,

−∞

we find: d2 u 1 = dx2 2

Z

du 1 = dx 2

Z

∞

e−x cosh t−νt cosh2 t dt ,

−∞ ∞

e−x cosh t−νt cosh t dt .

−∞

The trick is the next term: integrating by parts — ν 2 u(x) =

1 2

1 = 2

∞

Z

e−x cosh t

Z−∞ ∞

x =− 2


d2 e−νt dt 2 dt


d2 e−x cosh t e−νt 2 dt

−∞ ∞

Z

e−x cosh t−νt cosh t dt +

−∞

x2 2

Z

∞

e−x cosh t−νt sinh2 t dt .

−∞

Plugging all of this into the modified Bessel equation, x2


du d2 u +x − x2 + ν 2 u(x) = 0 , 2 dx dx

and recalling that cosh2 − sinh2 = 1, confirms that u(x) is a solution. In the large x limit, 1 2

Z

∞

e−x cosh t e−νt dt =

−∞

Z

∞

e−x cosh t cosh νt dt

Z0 ∞ ≈

Z0 ∞ ≈

e−x cosh t dt e−x(1+

t2 2

)

dt = e−x

0

Z

∞

e−xt

2

/2

=

0

q

π −x e , 2x

which identifies this solution to the modified Bessel equation with Kν (x).

C.13 (a) In cylindrical coordinates, the Helmholtz equation is




∇2 + k 2 f =

1 ∂ (r ∂r f ) r r

+

1 2 ∂ f r2 φ

+ ∂z2 f + k2 f = 0 .

Inserting the variable-separated ansatz f (~ r ) = R(r)Φ(φ)Z(z), 1 1 Φ(φ)Z(z) ∂r (r ∂r R) + R(r)Z(z) 2 ∂φ2 Φ + R(r)Φ(φ)∂z2 Z + k2 R(r)Φ(φ)Z(z) = 0 . r r A bit of algebra gives 1 1 1 1 2 1 ∂r (r ∂r R) + ∂ Φ + k2 = − ∂z2 Z. Rr Φ r2 φ Z Since there’s no z dependence on the left, this equation can only be satisfied if both sides equal a constant — which we’ll denote γ 2 . Introducing β 2 ≡ k2 − γ 2 , a bit more algebra gives r 1 ∂r (r ∂r R) + β 2 r2 = − ∂φ2 Φ ≡ ν 2 , R Φ

390

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

where, by the same argument which gave us γ, we’ve introduced the constant ν. One last bit of algebra: substituting x ≡ βr leads to x∂x (x ∂x R) + (x2 − ν 2 )R = 0 , which is Bessel’s equation, (C.1). (b) In spherical coordinates, ∇2 f =

1 ∂ (r2 ∂r f ) r2 r

+

1 ∂ (sin θ r 2 sin θ θ

∂θ f ) +

1 ∂2 f r 2 sin2 θ φ

+ k2 f = 0 .

Inserting the variable-separated ansatz f (~ r ) = R(r)Θ(θ)Φ(φ), Θ(θ)Φ(φ)

1 1 ∂r (r2 ∂r R) + R(r)Φ(z) 2 ∂θ (sin θ ∂θ Θ) r2 r sin θ 1 +R(r)Θ(θ) 2 ∂ 2 Φ + k2 R(r)Θ(θ)Φ(φ) = 0 . r sin2 θ φ

A bit of algebra gives sin θ 1 1 sin2 θ∂r (r2 ∂r R) + ∂θ (sin θ ∂θ Θ) + k2 r2 sin2 θR = − ∂φ2 Φ R Θ Φ Since there’s no φ dependence on the left, this equation can only be satisfied if both sides equal a constant — which we’ll denote m2 . A bit more algebra gives 1 d R dr



r2

dR dr



+ k2 r2 = −

1 1 d Θ sin θ dθ



sin θ

dΘ dr



+

m2 . sin2 θ

Setting both sides equal to a constant λ gives the radial equation d dr



r2

dR dr






+ k2 r2 − λ R(r) = 0 ,

which is the spherical Bessel equation (C.35) with λ = `(` + 1).

C.14 Using j` (x) =

pπ 2x

J`+1/2 (x), (C.2) gives j0 (x) =

q

π 2x

∞  1/2 X

x 2

(−1)m m! Γ(m + 3/2)

 2m x 2

m=0

√

π 2

=

∞ X

(−1)m m! Γ(m + 3/2)

 2m x 2

.

m=0

Appealing to (8.81a), we have √

Γ(m + 3/2) = (m + 1/2)Γ(m + 1/2) = (m + 1/2)

π (2m)! . 22m m!

Thus √ j0 (x) =

π 2

 ∞ X (−1)m

22m m! √ π(m + 1/2)

m!

  2m x 2

m=0

=

∞ X (−1)m

(2m + 1)!

x2m =

1 x

m=0

n0 (x) =

π 2x

(2m + 1)!

x2m+1 =

sin x . x

X

m=0

With this in hand, we can find n` (x) ≡

q

∞ X (−1)m



pπ 2x

N`+ 1 (x): 2

cos(π/2)J1/2 (x) − J−1/2 (x) sin(π/2)

 =−

q

π J−1/2 (x) . 2x

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391

Employing (8.81a) once again,

r

2 x

J−1/2 (x) =

∞ X

(−1)m m!Γ(m + 1/2)

 2m x 2

r =

2 πx

m=0

∞ X

(−1)m

x2m . (2m)!

m=0

Thus n0 (x) = −

1 x

∞ X

(−1)m

x2m cos x =− . (2m)! x

X

m=0

From these, recursion relations yield the higher-order j’s and n0 s: sin x cos x d j0 (x) = − dx x2  x    3 sin x cos x sin x 3 cos x 3 3 sin x − − = − 1 − . j2 (x) = j1 (x) − j0 (x) = x x x2 x x x2 x x2

j1 (x) = −

Similarly, d sin x cos x n0 (x) = − − 2 dx x x    3 sin x cos x cos x 1 cos x 3 3 3 + + = − − − 2 sin x. n2 (x) = n1 (x) − n0 (x) = − x x x x2 x x2 x x x n1 (x) = −

C.15 With u(r) ≡ R(r)/ r

√

r:

d dR r dr dr

h

i


d d √ ru(r) dr dr h √ i 1 d d ru0 + √ u =r r dr dr 2 r h i √   d 1 √ 1 3/2 0 =r r u + ru = r r2 u00 + 2ru0 + u . dr 2 4 h

=r

i

Then Bessel’s equation (39.76) becomes r

d dr



r

dR dr






+ k2 r2 − ν 2 R(r) =

√



r r2 u00 + 2ru0 +


√ 1 u + k2 r2 − ν 2 ru = 0 4 

or d dr



r2

du dr





+ k2 r2 − ν 2 +

1 4



u=0.

With 2ν = 2` + 1 to replace half-integer ν with integer `, we get the spherical Bessel equation d dr



r2

du dr






+ k2 r2 − `(` + 1) u = 0 ,

which, due to the ansatz, must have as solutions the spherical Bessel and Neuman functions j` (kr) ≡

q

π J`+1/2 (kr) 2kr

n` (kr) ≡

q

π N`+1/2 (kr) . 2kr

C.16 Starting where the derivation of (C.8) left off, j` (x) =

q

π x large J`+1/2 (x) −−−−→ 2x =

392

q

π 2x

r

2 (` + 1/2)π π cos x − − πx 2 4



1 `π π cos x − − x 2 2





=

1 `π sin x − x 2



 

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.

The Neumann function is a bit more subtle: n` (x)

q

=

q

−

= x large

  π 1 cos(` + 1/2)π J`+1/2 (x) − J−`−1/2 (x) 2x sin(` + 1/2)π π 1 J−`−1/2 (x) 2x cos `π   1 (−` − 1/2)π π cos x − − cos `π 2 4     1 `π 1 `π cos x − + `π = − cos x − . cos `π 2 x 2

1 x 1 − x

−−−−→ − =

For small-enough x, only m = 0 contributes in (C.2). Using (8.81a) to get Γ(` + 3/2) = Γ ((` + 1) + 1/2)) = (` + 1/2)Γ(` + 1/2) √ √ (2` + 1)! π(2`)! = π 2`+1 = (` + 1/2) 2` 2 `! 2 `! yields x small

j` (x) −−−−−→ =

q

π 2x

 `+1/2

q

π 2x

 `+1/2

x 2

1 Γ(` + 3/2)

x 2

√

22`+1 `! 2` `! = x` . π(2` + 1)! (2` + 1)!

For the Neumann function, we use (8.81b) to get n` (x) = −

q

(2`)! π 1 x small J−`−1/2 (x) −−−−−→ − ` x−(`+1) . 2x cos `π 2 `!

C.17 Starting with (C.43) quickly reveals i0 (x) = j0 (ix) = i−`

sin(ix) sinh(x) = . ix x

The modified spherical k is more subtle: From (C.27)

r k0 (x) ≡

2 K1/2 (x) = πx

q

 π  I−1/2 (x) − I1/2 (x) 2x

Then, using the expansion (C.25),

I1/2 (x) =

∞ q X x

2

1 m! Γ(m + 3/2)

 2m x 2

m=0

=

∞ q X x 1

22m+1 m! √ m! π(2m + 1)!

2

 2m x 2

r =

∞ X

2x π

m=0

x2m . (2m + 1)!

m=0

and

r I−1/2 (x) =

r =

2 x 2 x

∞ X

1 m! Γ(m + 1/2)

 2m x 2

m=0 ∞

X 1 22m m!  x 2m m!

m=0

√

π(2m)!

2

r =

2 xπ

∞ X x2m

(2m)!

.

m=0

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393

Thus k0 (x) =

q

 π  I−1/2 (x) − I1/2 (x) 2x ∞  X 1 x2m

1 = √ x

√

x (2m)!

−

√

x

m=0 ∞

X

1 = x

x2m x2m+1 − (2m)! (2m + 1)!



x2m (2m + 1)!



e−x . x

=

m=0

C.18 h~k0 |~ki = (2π)3 δ(~k − ~k0 )

Z

~ ~0 ei(k−k )·~r d3 r = 4π)2

=

XX

0

ˆ0 )Y`m (k) ˆ i`−` Y`∗0 m0 (k

Z

j` (kr)j`0 (k0 r)r2 dr

Z

Y`∗0 m0 (ˆ r)Y`m (ˆ r) dΩr

`,m `0 ,m0 ∞

= (4π)2

X

ˆ0 )Y`m (k) ˆ Y`∗0 m0 (k

Z

j` (kr)j`0 (k0 r)r2 dr

0

`,m

This should be compared with the expansion of the delta function in spherical coordinates,

X 1 1 ∗ Y`m (θ, φ)Y`m (θ0 , φ0 ). δ(~k − ~k0 ) = 2 δ(k − k0 )δ(Ω − Ω0 ) = 2 δ(k − k0 ) k k `m

Thus we find (C.43), ∞

Z

j` (kr)j`0 (k0 r)r2 dr =

0

1 π 1 (2π)3 2 δ(k − k0 ) = δ(k − k0 ) (4π)2 k 2k2

C.19 With ` = n + 1/2, Eqn. (C.2) together with (8.81a) give j` (x) =

q

π J`+1/2 (x) = 2x

q

π 2x

∞  `+1/2 X

x 2

(−1)m m! Γ(m + ` + 3/2)

 2m x 2

m=0

√

∞  ` X

√

∞  ` X

π 2

=

x 2

(−1)m 1 m! Γ ((m + ` + 1) + 1/2)

 2m x 2

m=0

=

π 2

x 2

(−1)m 22(m+`+1) (m + ` + 1)! √ m! π [2(m + ` + 1)]!

 2m x 2

m=0

= 2`+1 x

∞ X (−1)m `

m!

(m + ` + 1)(m + `)! x2m [(2m + 2` + 2)(2m + 2` + 1)]!

m=0

= 2` x

  ∞ X (m + `)! (−1)m x2m ` m!

[2(m + `) + 1)]!

,

m=0

The large square brackets highlight the penumbra of an emerging series for sin x/x. To shine light, write x2m = (x2 )m =



d dx2

`

(x2 )m+` m! = (m + `)!/m! (m + `)!



1 d 2x dx

`

(x2 )m+` .

Note that since this gives zero for all m between −` and 0, we can shift the lower limit in the

394

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

sum to −` without penalty. Thus we get ∞ X (m + `)!

j` (x) = 2` x`

m!

m! (m + `)!



1 d 2x dx

` 

(−1)m (x2 )m+` [2(m + `) + 1)]!



` 



m=0

= x`



1 d x dx

`

"

∞ X (−1)m x2(m+`)

#

[2(m + `) + 1)]! m=−`

Shifting the summation index, k = m + `, finishes it off: j` (x) = x

`



1 d x dx

"∞ # ` X (−1)k−` x2k (2k + 1)!

= (−1)` x`



1 d x dx

sin x x

X

k=0

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

395