Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

Solutions Manual to accompany

AN INTRODUCTION TO GROUPS AND THEIR MATRICES FOR SCIENCE STUDENTS KOLENKOW

Copyright c⃝ 2021 by Robert Kolenkow

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

2

Acknowledgment The style code used for typesetting this manual was written by the experienced and skillful Ali Woollatt for use by Cambridge University Press.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

CONTENTS

1

FUNDAMENTAL CONCEPTS

1

2

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

17

3

MOLECULAR VIBRATIONS

40

4

CRYSTALLINE SOLIDS

53

5

BOHR’S QUANTUM THEORY AND MATRIX MECHANICS

65

6

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

74

7

ROTATION

88

8

QUANTUM ANGULAR MOMENTUM

97

9

THE STRUCTURE OF ATOMS

112

10

PARTICLE PHYSICS

127

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

1.1 Apply Tx = –x.

Tex = e(–x) = e–x !

Here is another approach. Apply the operator to ex expressed in series form. x2 x3 T ex T 1 x = + + + + ...

2! 3! !

(– x)2 (– x)3 + = 1 – x + 2! + ...

3! ! x2 x3 = 1–x+ – +...

2! 3! = e–x The operator has changed the sign of the odd terms.

2

FUNDAMENTAL CONCEPTS

1.2 Apply Tx = –x.

T cos (x) = cos (–x) = cos (x) This result can also be understood by2 considering the series expansion of cos (x). x x4 x6 cos (x) 1 = – + – + ...

2! 4! 6!

T cos (x) = cos (–x) (–x)2 (–x)4 (–x)6 = 1– + – + ...

2!

4!

6! x2 x4 x6 =1– + – + ...

2! 4! 6! = cos (x) The series expansion of cos (x) contains only even terms, so Tx = –x changes nothing.

1.3 According to the group axioms (AB)(AB)–1 = E Multiply by A–1 from the left.

A–1AB(AB)–1 = A–1E B(AB)–1 = A–1 Multiply by B–1 from the left.

B–1B(AB)–1 = B–1A–1 (AB)–1 = B–1A–1

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

FUNDAMENTAL CONCEPTS

3

1.4 Check whether the axioms that define a group (Sec. (1.3) of the text) are satisfied for positive and negative real integers under addition.

The sum of two integers is also an integer, a member of the set, satisfying axiom (ii) that the group operation does not result in an element outside the set.

The sum of an integer N and 0 is equal to N, so 0 acts as the identity element under addition, satisfying axiom (iii) that the group must have an identity element.

The sum of an integer N and its negative –N is N – N, equal to the identity element 0. Every integer N has an inverse element –N under addition, so axiom (iv) is satisfied that the group must contain an inverse under the group operation for every element.

Trivially, the addition of integers is associative: N1 + (N2 + N3) = (N1 + N2) + N3, satisfying axiom(v).

1.5 Check whether the axioms that define a group (Sec. (1.3) of the text) are satisfied by the set of real integers under multiplication.

The product of two integers is also an integer, a member of the set, satisfying axiom (ii) that the group operation does not result in an element outside the set.

The product of an integer N and 1 is equal to N, so 1 acts as the identity element under multiplication, satisfying axiom (iii) that the group must have an identity element.

Axiom (iv) requires that N × N–1 = 1. It follows that N–1 = 1N. The inverse of a real integer under multiplication is not an integer except for 1, so the real integers are not a group because the set of real integers does not include the inverses under multiplication.

4

FUNDAMENTAL CONCEPTS

1.6 The product table for the group {E, A, B} is E E A B

E A B

A A B E

B B E A

The product table shows that B2 = BB = A. Hence the cyclic set {E, B, B2} has the same elements as the set {E, B, A}.

1.7 The group must have an identity element E. Let the group’s two elements be E and A, where A is distinct from E. A2 can only be A or E. If A2 = A, then A2 = A AA = A Multiplying by A–1 from the left gives

A=E so for this case A = E and A is not distinct. Hence the only possibility is AA = E so the product table must be E A

E E A

A A E

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

FUNDAMENTAL CONCEPTS

5

1.8 Products of the operations can be found by adding their rotations. For example, CC = 270◦ + 270◦ = 540◦ = 180◦ + 360◦ = BE = B. EE = 0◦ + 0◦ = E

EA = 0◦ + 90◦ = A

EB = 0◦ + 180◦ = B

EC = 0◦ + 270◦ = C

AE = 90◦ + 0◦ = A

AA = 90◦ + 90◦ = B

AB = 90◦ + 180◦ = C

AC = 90◦ + 270◦ = E

BE = 180◦ + 0◦ = B

BA = 180◦ + 90◦ = C BB = 180◦ + 180◦ = E

BC = 180◦ + 270◦ = A

CE = 270◦ + 0◦ = C CA = 270◦ + 90◦ = E

CB = 270◦ + 180◦ = A CC = 270◦ + 270◦ = B

The product table is therefore E E A B C

E A B C

A A B C E

B B C E A

C C E A B

The group is cyclic – for example AE = A AA = B AB = AAA = C AC = AAAA = E and similarly for the other group elements.

1.9 Consider the hypothetical product table for the group of distinct operations {E, P, Q, R} where the Q column contains the operation P twice.

E P Q R

E E P Q R

P P Q R E

Q Q R P P

R R E P Q

According to the product table QQ = P RQ = P QQ = RQ Multiplying by Q–1 from the right gives Q = R. This is a contradiction because the operations are assumed to be distinct.

6

FUNDAMENTAL CONCEPTS

1.10 The product table for S3 is E P2 P3 P4 P5 P6

E E P2 P3 P4 P5 P6

P2 P2 E P5 P6 P3 P4

P3 P3 P6 E P5 P4 P2

P4 P4 P5 P6 E P2 P3

P5 P5 P4 P2 P3 P6 E

P6 P6 P3 P4 P3 E P5

According to the product table for S3 the set {E, P5, P6} has the product table E E P5 P6

E P5 P6

P5 P5 P6 E

P6 P6 E P5

The set {E, P5, P6} satisfies the group axioms. Cayley’s theorem says that S 3 has every subgroup of order 3. {E, P5, P6} is the only subgroup of order 3 found in S3 and there are no others.

1.11

E K L M

E E K L M

K K E M L

L L M E K

M M L K E

Check that the group axioms are satisfied: (ii) The product of any two operations is a member of the set. (iii) There is an identity element E such that KE = K, etc. (iv) Every element of the set has an inverse. KK = E so that K–1 = K etc.

The group is Abelian; all the elements commute. KL = LK = M etc.

The group is not cyclic. K2 = E etc. The group elements cannot be obtained by taking powers of one element.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

FUNDAMENTAL CONCEPTS

7

1.12 The product table for a group {E, A} of order 2 is E A

E E A

A A E

For distinct elements it follows that AA = E. Inspection of the product table for S3 (see Problem 1.10) shows that PnPn = E for n = 2, 3, 4. The three subgroups of order 2 are therefore {E, P2}, {E, P3} and {E, P4}.

1.13 (a) 1 2 3 23 1 For the interchange 3 ↔ 1 the bottom row becomes (213).

A second interchange 2 ↔ 1 converts the bottom row to standard form (123).

Because two interchanges are required, this permutation is even. (b) 1 2 3 4 2 3 1 4 For the interchange 3 ↔ 1 the bottom row becomes (2134).

A second interchange 2 ↔ 1 converts the bottom row to standard form (1234).

Because two interchanges are required, this permutation is even. (c) 1 2 3 4 4 3 2 1 Converting to standard form requires 2 interchanges: 4 ↔ 1. The bottom row becomes (1324).

3 ↔ 2. The bottom row becomes (1234).

This permutation is even.

8

FUNDAMENTAL CONCEPTS

1.14 In cycle notation for permutations an entry maps the entry on its right. Hence for the cycle (536142), 6 ›→ 1.

1.15 In cycle notation the entry at the end of the cycle maps around to the entry at the beginning. Hence for the cycle (326415), 5 ›→ 3.

1.16

Let c be the length of the side of the equilateral triangle. The “flip” in this problem is a rotation about aa by u.

Before the flip the coordinates are ,

xt = 0, yt = 32 c (top apex) xr = c2, yr = 0 (bottom right apex) x4 = – c 2, y4 = 0 (bottom left apex) After the flip the coordinates are ,

xtJ = 0, ytJ = 3 2 c J c J xr = – 2, y r = 0 xJ = + c , yJ = 0 4

2

4

continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

FUNDAMENTAL CONCEPTS

9

The representation matrix has the general form ! ! ! xiJ A11 A12 xi yJ = A A y 21

i

22

i

Written out, the matrix form is xiJ = A11 xi + A12yi yiJ = A21 xi + A22yi To start evaluating the terms Ai j, note that yiJ is independent of xi. Therefore A21 = 0. Because yiJ = yi, A22 = 1.

Similarly A12 = 0 because xiJ is independent of yi. A11 = –1 because xr and x4 change sign from the flip. The result for the representation matrix is –1 0 0

1

1.17 The product table for the group Γ = {E, A, B} is

E A B

E E A B

A A B E

B B E A

A correct homomorphism must give product results that agree with the product table. For example, according to the product table AB = E. The corresponding homomorphism product (-1) (-1) = (1) agrees.

However, AA = B. The homomorphism (-1) (-1) = (1) does not agree. The given matrices are not a homomorphism for Γ.

10

FUNDAMENTAL CONCEPTS

1.18 Consider C = AB where A and B are n × n diagonal matrices. The elements of A and B satisfy Ai j = 0

i≠ j

Bk4 = 0

k≠4

The elements of C are n X

Ci4 =

Ai j Bj4 j

Ai j = 0 if i ≠ j.

Ci4 = Aii Bi4 Bi4 = 0 if i ≠ 4 so C has only diagonal elements Cii = AiiBii.

1.19 The group operations are E: rotate by 0◦ A: rotate by 90◦ B: rotate by 180◦ C: rotate by 270◦ The required matrix was derived in Sec. (1.5.3).

1 0

cos θ

sin θ

– sin θ

cos θ –1 0 0 –1

0 –1

01

0 1 –1 0

D(E)

D(A)

D(B)

D(C)

1 0

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

FUNDAMENTAL CONCEPTS

11

1.20 A matrix has an inverse only if its determinant is not 0. Otherwise, it would be possible to use Cramer’s rule to calculate an inverse. (a) det

1 0 = (1)(–1) = –1 0 –1

This matrix has an inverse. (b) det

1 –1 1 –1

= (1)(–1) – (1)(– 1) = 0

This matrix does not have an inverse. (c) 1 3 2 det 3 0 –1 = (1)(–2) – (3)(3 + 2) + (2)(–6) = –29 2 –2 1 This matrix has an inverse.

12

FUNDAMENTAL CONCEPTS

1.21 Let C be the resultant matrix in each case.

C = A+B Ci j = Ai j + Bi j 3 2 –3 5 2 –3 C = –2 3 2 + 1 3 1 1 0 4 –2 0 5 8 4 –6 =

–1 6 –1 0

3 9

C = A–B Ci j = Ai j – Bi j –2 0 C=

0

–3 0 1 3 0 –1

C = AB X Ci j = Aik Bk j k

(3)(5) + (2)(1) + (–3)(–2)

(3)(2) + (2)(3) + 0

(3)(–3) + (2)(1) + (–3)(5)

C = (–2)(5) + (3)(1) + (2)(–2) (–2)(2) + (3)(3) + 0 (–2)(–3) + (3)(1) + (2)(5) (1)(5) + 0 + (4)(–2) (1)(2) + 0 + 0 (1)(–3) + 0 + (4)(5) 23 12 –22 = –11 5 19 –3 2 17 C = BA X Ci j = Bik Ak j k

(5)(3) + (2)(–2) + (–3)(1) (5)(2) + (2)(3) + 0 (5)(–3) + (2)(2) + (–3)(4) C=

(1)(3) + (3)(–2) + (1)(1) (–2)(3) + 0 + (5)(1)

(1)(2) + (3)(3) + 0 (–2)(2) + 0 + 0

(1)(–3) + (3)(2) + (1)(4) (–2)(–3) + 0 + (5)(4)

8

=

16 –23 –2 11 7 –1 –4 26

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

FUNDAMENTAL CONCEPTS

13

1.22

Let C be the resultant matrix in each case.

C = A+B Ci j = Ai j + Bi j 2 –2 –1 –1 5 4 C = –3 3 4 + 5 1 4 2 0 –3 –3 0 2 1 =

3

3

2 4 8 –1 0 –1

C = A–B Ci j = Ai j – Bi j –3 7 C=

5

–8 2 0 5 0 –5

C = AB X Ci j = Aik Bk j k

(–1)(2) + (5)(5) + (4)(–3) (–1)(–2) + (5)(1) + 0 (–1)(–1) + (5)(4) + (4)(2) C = ( –3)(2) + (3)(5) + (4)(–3) ( –3)(–2) + (3)(1) + 0 ( –3)( –1) + (3)(4) + (4)(2) (2)(2) + 0 + (–3)(–3)

11 =

7

(2)(–2) + 0 + 0

(2)(–1) + 0 + (–3)(2)

29

–3 9 23 13 –4 –8

C = BA X Ci j = Bik Ak j k

C= 2 = 0

(2)(–1) + (–2)(–3) + (–1)(2) (2)(5) + (–2)(3) + 0 (2)(4) + (–2)(4) + (–1)(–3) (5)(–1) + (1)(–3) + (4)(2) (5)(5) + (1)(3) + 0 (5)(4) + (1)(4) + (4)(–3) (–3)(–1) + 0 + (2)(2) (–3)(5) + 0 + 0 (–3)(4) + 0 + (2)(–3) 4 28

3 12

7 –15 –18

14

FUNDAMENTAL CONCEPTS

1.23

2 –1 2 A= 2 0 3 3 –1 –1 1 2 –2 B = 0 3 –1 2 1 –2 AB =

6 3 –7 8 7 –10 1 2 –3

det(A) = (2)(3) – (–1)(–11) + (2)(–2) = –9 det(B) = (1)(–5) – (2)(2) + (–2)(–6) =3 det(A) det(B) = (–9)(3) = –27 det(AB) = (6)(–1) – (3)(–14) + (–7)(9) = –27

1.24 A˜ is the transpose of matrix A so its elements are (A˜)i j = Aji. ˜ i j = (AB) ji (AB) X = Ajk Bki k X

=

A˜k jB̃ ik

k X

=

B̃ik A˜k j

k

= B˜ A˜

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

FUNDAMENTAL CONCEPTS

1.25

1 5 2 A = 3 0 –1 4 –2 1 (A∗)i j = Ai∗j 1 5 2 = 3 0 –1 4 –2 1 (A˜)i j = Aji 1 3 4 = 5 0 –2 2 –1 1 (A†)i j = (A∗) ji 1 3 = 5 0 2 –1

4 –2 1

15

16

FUNDAMENTAL CONCEPTS

1.23

1 3

3+i 0

2i 1 – + 2i

4 – 3i

–2

1–i

A=

(A∗)i j = Ai∗j =

1 3–i –2i 3 0 –1 – 2i 4 + 3i –2 1+i

(A˜)i j = Aji 1 = 3+i

3 0

4 – 3i –2

2i

–1 + 2i

1–i

1 = 3–i

3 0

4 + 3i –2

–2i

–1 – 2i

1+i

(A†)i j = (A∗) ji

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

2.1 In the problem the reference to Sec. (1.4) should instead be to Sec. (1.5.3).

Let T be rotation by θ and let ø1 = x, ø2 = y.If x and y are truly basis functions for this operation they must satisfy Tøj = Tx = D11 x + D12y Tø2 = Ty = D21 x + D22y In Sec.(1.5.3) Eqs. (1.1) and (1.2), the expression for rotation of the x – y axes is Tx = x cos θ + y sin θ Ty = –x sin θ + y cos θ The operation of T on x and y give terms involving only x and y. Hence x and y qualify as basis functions. continued next page =→

18

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

D11 = cos θ D12 = sin θ D21 = – sin θ D22 = cos θ cos θ D(T) = – sin θ

sin θ cos θ

This is the same as the matrix in Sec. (1.5.3) for rotated axes.

The matrix D(T) for θ = 120◦ is therefore

D(T) =

–,2

1

– 23

, 3 2 – 21

2.2 In the problem the reference to Sec. (1.4) should instead be to Sec. (1.5.3).

In Sec.(1.5.3) Eqs. (1.1) and (1.2), the expression for rotation of the x – y axes is Tx = x cos θ + y sin θ Ty = –x sin θ + y cos θ Tx2 = (x cos θ + y sin θ)2 = x2 cos2 θ + y2 sin2 θ + 2xy cos θ sin θ Ty2 = (–x sin θ + y cos θ)2 = x2 sin2 θ + y2 cos2 θ – 2xy cos θ sin θ T(x2 – y2) = (x2 – y2)(cos2 θ – sin2 θ) + 4xy cos θ sin θ T(xy) = (x cos θ + y sin θ)(–x sin θ + y cos θ) = –(x2 – y2) cos θ sin θ + xy(cos2 θ – sin2 θ) T(x2 –y2) and T(xy) contain only the functions (x2 –y2) and (xy) so these are indeed basis functions referring to the same subspace. continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

19

Let ø1 = (x2 – y2) and let ø2 = (xy). Thus 2 X

Tøj =

Di jøi

i=1

Tø1 = D11ø1 + D21ø2 Tø2 = D12ø1 + D22ø2 It follows that

D11 = cos2 θ – sin2 θ D12 = – cos θ sin θ D21 = 4 cos θ sin θ D22 = cos2 θ – sin2 θ cos2 θ – sin2 θ D(T) = 4 cos θ sin θ

– cos θ sin θ

(1)

cos2 θ – sin2 θ

From the product table, AB = E where A is rotation by 120◦ and B is rotation by 240◦. The matrices in a valid matrix representation must obey the same relation as the product table. As a check using Eq. (1) , 3

D(A)D(B) =? D(E) 1

,

1 – 3 1 0 –2 4 2– , , 14 = 1 3 –2 0 1 – 3 –2

= D(E) as required.

This representation is equivalent to the representation in Sec. (1.5.3) because {x, y} and {x2 – y2, xy} are in the same function subspace. The representations are related by a similarity transformation.

20

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

2.3

A flip about the y axis by operation A causes x → –x, y → y, z → –z. Let ø1 = (x2 + y2) and ø2 = z. Under the transformation (–x)2 + (y)2 → ø1 and (–z) → –ø2. øJ1

ø2 J

=

1

0 0 –1

ø1

ø2

The representation matrix D(A) is therefore D(A) =

1 0 0 –1

2.4

A flip about the y axis by operation A causes x → –x, y → y, z → –z. Let ø1 = (x2 – y2) and ø2 = yz. Under the transformation (–x)2 – (y)2 → ø1 and (–yz) → –ø2. øJ1

øJ2

=

1

0 0 –1

ø1

ø2

The representation matrix D(A) is therefore D(A) =

1 0 0 –1

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

21

2.5 Let the elements of S be S i j.

D(β) = SD(α)S–1

(1)

Multiply the similarity transformation Eq. (1) by S from the right.

–1 0 S1 0 1 S21 –S11 S 21

S12 S11 = S2 S21

S 12 S 22

–S12 = S 22

–S 12 –S 22

S11 S 21

1 0 0 –1

-S 11 = S 11 so S 11 = 0. Similarly S 22 = –S 22 so S 22 = 0. S therefore has the form S=

0

S 12

S 21

0

If S is indeed unitary it must satisfy 0

S 21

S 12

0

S–1 = SS–1 = E 0

S 12

S 21

0

0 S12

S 21 = 1 0 0 1

0

Hence S 12S 12 = 1 and S 21S 21 = 1. There are four possible forms for S: 0 1 10

0 –1

0 1 –1 0

1 0

0

–1 –1 0

They are all unitary and they all satisfy the similarity transformation.

2.6

SABS–1 = SAS–1SBS–1 = (SAS–1)(SBS–1) = RT

22

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

2.7 The product table for the group {E, T} is E E T

E T

T T E

The regular representation for T is therefore 0 1 D (T) = 1 0

(reg)

Consider the reduced form D(red) (T) =

1 0 0 a

If a = +1 the block diagonal irreducible matrix D(red)(T) is the same as for E, so the elements are not distinct. Hence a = -1.

As another proof, consider the similarity transformation D(red)(T) = SD(reg)(T)S–1 and let the elements of S be S i j.

Multiply the similarity transformation from the right by S. 1 0 S1 0 a S 21 S1 aS 21

S12 S 11 S12 = S 22 S21 S 22 S12 = aS 22

S12 S 22

0 1 1 0

S 11 S 21 continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

23

The equality of the two matrices requires the relations S 11 = S 12 aS 21 = S 22 aS 22 = S 21 a2S 22 = S 22 so a = ±1. S therefore has the form S=

S 11 S 11 S ±S 21

21

If S is indeed unitary it must satisfy S–1 =

S 11

S 21

S 11 ±S 21

so that

SS–1 = E 1 0 01

S11 S11 S11 S21 = S 21 ±S 21 S 11 ±S 21 This matrix relation requires 2 2S 11 =1

1 S 11 = ± , 2 2 2S 21 = 1 1 S 21 = ± , 2 S 11(S 21 ± S 21) = 0 The solution to Eq. (1) must take the negative sign S 21 – S 21 = 0. S is therefore S=

1

, 2

1 1 –1

1

(1)

24

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

2.7

2,

S=

1

– ,2 1

1 , 2

1 , 2

1

1 –1

=, 2 1

1

a) As an example, consider the product of the first row with itself.

– ,12 1 , 2

1 – ,2 1

1 , 2

1 , 2

+

= 21+ 21= 1 Next consider the product of the first and second columns.

– ,2 + ,12 ,12 = –2 1+ 2 =1 0 These and similar examples demonstrate the

orthogonality of S. b) Because S is orthogonal S–1 = S̃ .

DJ(G1) = SD(G1))S̃ 1 1 –1 1 0 1 = 2 1 1 0 1 1 2 0 = 2 0 2

=

1 –1 1

1 0 0 1

DJ(G2) = SD(G2)S̃ 1 1 –1 1 1 = , 2 1 1 2 3

, 3

1

–1

–1 1

1

, , 1 –1 ,1 – 3 1 + 3, = 4 ,1 1 3+1 3–1 1 –3 1 = , 2 1 3 1

DJ(G3) = SD(G3)S̃ 1 1 –1 1 –1 = , 2 1 1 2 –3

, 3 –1

1

1 –1 1

, , 1 –1 –1, – 3 –1 + 3, = 4 1 1 , – 3+1 – 3–1 1 –1 3 = , 2 –3 –1 1

continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

25

c) The character χ of a matrix is the sum of its diagonal elements. χ (D(G1)) = 1 + 1 = 2 . χ DJ(G1) = 1 + 1 = 2 1 1 χ (D(G2)) = – = 0 2 2 , , 3 3 . J χ D (G2) = – + =0 12 1 2 χ (D(G3)) = – – = –1 2 .

1 2 χ DJ(G3) = 1 = –1 – – 2 2 J The invariance χ (D(G)) = χ (D (G)) is not surprising because as proved in Sec. (2.6) a similarity transformation does not change character.

2.9

T=

, 3 2 1 2

, 1 3

–1 ,2 3 2

=

2

–1 , 3

1

a) As an example, consider the product of the first row with itself.

,

3

2

,

3 2

+

– 12 – 1 2 = 3 +4 1 =41 Next consider the product of the first and second columns. ,

3 2

1

1 2

,

3

, 3

, 3

–2 + 2 = – 4 + 4 = 0 These and similar examples demonstrate the orthogonality of T. b) Because T is orthogonal T–1 = T̃.

DJ(H1) = TD(H1))T̃ , 1 3 –1 –1 0 = , 4 1 3 0 , –3 1 –1 , = 2 –3 1

, 3 1 –1

1 , 3

continued next page =→

26

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

DJ(H2) = TD(H2)T̃ , 1 3 –1 1 = , , , 1 3 , –3 18 3 –1, 3 = 4 1 3 –1 1 4 0 = 4 0 –4

=

, –3

, 3

–1 –1 , –3

1 –1

, 3

1 0 0 –1

DJ(H3) = TD(H3)T̃ , , , 1 3 –1 –1 –3 3 = , , –1 –1 8 ,1 3 3 1 3 ,–1 0 –4 = 8 1 3 4 0 , 1 –1 –3 , = 2 3 –1

1 , 3

c) The character χ of a matrix is the sum of its diagonal elements. χ (D(H1)) = –1 + 1 = 0 .

1 1 χ DJ(H1) = – + = 0 12 12 χ (D(H2)) = – = 0 2 2 . J χ D (H2) = 1 – 1 = 0 1 1 χ (D(H3)) = – – = –1 2 .

1 2 χ DJ(H3) = 1 = –1 – – 2 2 J The invariance χ (D(H)) = χ (D (H)) is not surprising because as proved in Sec. (2.6) a similarity transformation does not change character.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

27

2.10 The character χ of a matrix is the sum of the diagonal elements. a) χ = (–2)11 = -2 b) χ = 2 - 1 = 1 c) χ = a + d d) χ = -1 + i + 2 + 1 = 2 + i e) χ = 1 + 1 + 1 + 1 = 4

2.11 E E A B

E A B

B B E A

A A B E

Columns A and B have been interchanged to place E along the main diagonal so that D(reg)(E) is the identity matrix.

D

(reg)

1 0 0 (E) =

0 1 0 0 0 1

0 0 1 D(reg)(A) =

1 0 0 0 1 0

0 1 0 D(reg)(B) = 0 0 1 1 0 0

2.12

E K L M

E E K L M

K K E M L

L L M E K

M M L K E

The product table already has E along the main diagonal to make D(reg)(E) the identity matrix.

0 1 0 0 1 0 0 0 D(reg)(K) = 0 0 0 1 0 0 1 0

28

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

2.13

E A B C

E E A B C

C C E A B

B B C E A

A A B C E

Columns A and C have been interchanged to place E along the main diagonal so that D(reg)(E)is the identity matrix.

0 0 0 1 1 0 0 0 D(reg) (A) = 0 1 0 0 0 0 1 0

2.14 a) According to Theorem 4 (stated in Sec. (2.7.3) and near the end of Chapter 2) the number of nonequivalent irreducible representations of a group is equal to the number of classes. The 422 group has five classes, hence it has five nonequivalent irreducible representations. b) According to Theorem 7 (stated in Sec. (2.7.4) and near the end of Chapter 2) the sum of the squares of the dimensions of the irreducible representations is equal to the order of the group. There are five irreducible representations so there must be five terms in the sum equal to 8. The trial (2)2 + (2)2 = 8 has only two terms and is not acceptable. The only valid possibility is (1)2 + (1)2 + (1)2 + (1)2 + (2)2 = 8.

The 422 group therefore has four 1-dimensional irreducible representations and one 2-dimensional.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

29

2.15

E K L M

E E K L M

K K E M L

L L M E K

M M L K E

a) A group’s classes are the sets of conjugates XSX–1 where X and S are all the group’s elements. To find the inverse of a group element use the product table: E–1 = E, K–1 = K. L–1 = L. M–1 = M. The conjugates are thus EEE–1 = E

EKE–1 = K

KEK–1 = E

KKK–1 = K

LEL–1 = E

LKL–1 = K

MEM–1 = E

MKM–1 = K

ELE–1 = L

EME–1 = M

KLK–1 = L

KMK–1 = M

LLL–1 = L

LML–1 = M

MLM–1 = L

MMM–1 = M

There are four distinct classes: {E}, {K}, {L}, {M}. Note that E is a class of one for any group, because XEX–1 = E. b) According to Theorem 4 (stated in Sec. (2.7.3) and near the end of Chapter 2) the number of nonequivalent irreducible representations of a group is equal to the number of classes. Hence this group has four irreducible representations. c) According to Theorem 7 (stated in Sec. (2.7.4) and near the end of Chapter 2) the sum of the squares of the dimensions of the irreducible representations is equal to the dimension of the group. There are four irreducible representations so there must be four terms adding to 4. The only possibility is (1)2 + (1)2 + (1)2 + (1)2 = 4.

This group therefore has four 1-dimensional irreducible representations.

30

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

2.16

E A B C

E E A B C

A A B C E

B B C E A

C C E A B

a) A group’s classes are the sets of conjugates XSX–1 where X and S are all the group’s elements. To find the inverse of a group element use the product table: E–1 = E, A–1 = C. B–1 = B. C–1 = A. The conjugates are thus EEE–1 = E

EAE–1 = A

AEA–1 = E

AAA–1 = A

BEB–1 = E

BAB–1 = A

CEC–1 = E

CAC–1 = A

EBE–1 = B

ECE–1 = C

ABA–1 = B

ACA–1 = C

BBB–1 = B

BCB–1 = C

CBC–1 = B

CCC–1 = C

There are four distinct classes: {E}, {A}, {B}, {C}. Note that E is a class of one for any group, because XEX–1 = E. b) According to Theorem 4 (stated in Sec. (2.7.3) and near the end of Chapter 2) the number of nonequivalent irreducible representations of a group is equal to the number of classes. Hence this group has four irreducible representations. c) According to Theorem 7 (stated in Sec. (2.7.4) and near the end of Chapter 2) the sum of the squares of the dimensions of the irreducible representations is equal to the dimension of the group. There are four irreducible representations so there must be four terms adding to 4. The only possibility is (1)2 + (1)2 + (1)2 + (1)2 = 4.

This group therefore has four 1-dimensional irreducible representations.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

31

2.17 Start from the definition of a representation matrix in Eq. (2.1), Sec. (2.2). hα X

Tøj =

D(α) (T)øk kj

(1)

k=1

Now use the orthogonality relation in Theorem 5 (stated is Sec. (2.7.4) and near the X end of Chapter 2). n J J D(α)∗(T)D(β) (T) (2) = δαβδi i δ j j ij iJ jJ h T The value = 0 unless α = β. Hence h → hα. the dimension of the α representation (α) because both representations of the group.

X are D . n is the order n α α ( )∗ ( ) D (T)D (T) = δ δ J (3) i iJ j j ij iJ jJ hα T The task is to use Eqs.(1) and (3) to prove hα∗ X (α) øi = D (T)Tøj n T

(4)

There is no sum over j in Eq. (4).

Multiply Eq. (1) by D(αi )∗j (T)øj and sum over group elements T. X XX D(α)∗(T)Tøj = D(α)∗(T)D(α)(T)øk ij T

ij k

kj

T

Now use the orthogonality X relation Eq. (2). n X D(α)∗(T)Tøj = δi k δ j j øk ij h α T k n X δi køk = hα k n = øi hα Hence hα X (α)∗ øi = D (T)Tøj n T ij

32

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

2.16 The 32 group has two 1-dimensional irreducible representations (A1, A2) and one 2-dimensional irreducible peresntation Γ. Representation A1 1

1

1

1

1

1

E

A

B

C

D

F

1

–1

–1

–1

1

1

E

A

B

C

D

F

1 0

–1 0 0

, 3 2 – 12

–1

0 1

1 2 ,3 2

Representation A2

Representation Γ

E

1 A

, 3 2 – 12

B

1 2 ,3

–

– 2

, 3 2 – 12

2 ,3 – 2

C

D

,

– 12 – 23

,3 2

– 12

F

The orthogonality relation Theorem 5 states X n D(α)∗(R)D(β) (R) J J J J = δ δ δ ij i j h αβ ii j j R

(1)

h is the dimension of the representation and n=6 is the order of the 32 group. hA1 = 1

hA2 = 1

hΓ = 2

Let (lhs) be the left-hand side of Eq. (1) and let (rhs) be the right-hand side. All quantities in the given matrices are real. continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

33

Example 1 Let α = A1 and let β = A2. Both representations are 1-dimensional so i = iJ = j = jJ =1 X (lhs) = D(A1)(R)D(A2)(R) 11 11 R = D(A1)(E)D(A2)(E) + D(A1)(A)D(A2)(A) + D(A1)(B)D(A2)(B) 11

11

11

11

11

11

11

11

11

11

11

11

+ D(A1)(C)D(A2)(C) + D(A1)(D)D(A2)(D) + D(A1)(F)D(A2)(F) = (1)(1) + (1)(–1) + (1)(–1) + (1)(–1) + (1)(1) + (1)(1) =0 n (rhs) = δA1 A2 δ1 1δ1 1 h =0 Example 2 Let α = Γ and let β =X Γ. Let i = iJ = 1, j = jJ = 2. (lhs) = D(Γ)(R)D(Γ)(R) 12 12 R = D(Γ)(E)D(Γ)(E) + D(Γ)(A)D(Γ)(A) + D(Γ)(B)D(Γ)(B) 12 12 12 12 12 12 + D(1Γ)2(C)D(1Γ)2(C) + D(1Γ)2(D)D(1Γ)2(D) + D(1Γ)2(F)D(1Γ)2(F) , , , , = (0)(0) + (0)(0) + 3 3 + – 3 – 3 2 2 2 2 , , , , 3 3 3 3 + 2 + – 2 – 2 2 3 =0+0+ + 3 + 3 + 3 4

4

4

4

=3 !

6 (rhs) = δ δ δ 2 ΓΓ 11 22 =3 continued next page =→

34

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

Example 3 Let α = A2 andX let β = Γ. Let i = iJ = j = jJ = 1. (lhs) = D(A2)(R)D(Γ)(R) 11 R (A2)

=D

11

11

(E)D(Γ)(E) + D(A2)(A)D(Γ)(A) + D(A2)(B)D(Γ)(B) 11

11

11

11

11

+ D(1A12)(C)D(1Γ1)(C) + D(1A12)(D)D(1Γ1)(D) + D(1A12)(F)D(1Γ)1(F) = (1)(1) + (–1)(–1) + (–1) 1 + (–1) 1 + (1) – 1 + (1) – 1 2

2

2

2

=0 n (rhs) = δA2 Γ δ1 1δ1 1 h =0

2.19 Here is the character table of the 32 group. {E} 1 1 2

1 A Γ

{A, B, C} 1 -1 0

{D, F} 1 1 -1

Theorem 6 (stated in Sec. (2.7.4) and near the end of Chapter 2) for the orthogonality χ of the characters of irreducible representations is X

χ(α)∗(R)χ(β)(R) = n δα β

(1)

R

where R is an element of the group and n is the order of the group. n = 6 for the 32 group. In this problem all quantities are real. continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

Let (lhs) be the left-hand side of Eq. (1) and let (rhs) be the right-hand side.

Example 1 Let α = 1 and let β = A. X (lhs) = χ(1)(R)χ(A)(R) R

= χ(1)(E)χ(A)(E) + χ(1)(A)χ(A)(A) + χ(1)(B)χ(A)(B) + χ(1)(C)χ(A)(C) + χ(1)(D)χ(A)(D) + χ(1)(F)χ(A)(F) = (1)(1) + 3(1)(–1) + 2(1)(1) =0 (rhs) = n δ1 A =0 Example 2 Let α = A and let β = A. X (lhs) = χ(A)(R)χ(A)(R) R

= χ(A)(E)χ(A)(E) + χ(A)(A)χ(A)(A) + χ(A)(B)χ(A)(B) + χ(A)(C)χ(A)(C) + χ(A)(D)χ(A)(D) + χ(A)(F)χ(A)(F) = (1)(1) + 3(–1)(–1) + 2(1)(1) =6 (rhs) = n δA A =6 Example 3 Let α = A and let β = Γ.

X (lhs) = χ(A)(R)χ(Γ)(R) R

= χ(A)(E)χ(Γ)(E) + χ(A)(A)χ(Γ)(A) + χ(A)(B)χ(Γ)(B) + χ(A)(C)χ(Γ)(C) + χ(A)(D)χ(Γ)(D) + χ(A)(F)χ(Γ)(F) = (1)(2) + 3(–1)(0) + 2(1)(–1) =0 (rhs) = n δA Γ =0

35

36

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

2.20 Here is the character table of the 32 group. {E} 1 1 2

1 A Γ

{A, B, C} 1 -1 0

{D, F} 1 1 -1

Theorem 7 (stated in Sec. (2.7.4) and near the end of Chapter 2) for the squared sum of characters χ of irreducible representations is X

2

χ(α)(R) = n

(1)

R

where R is an element of the group and n is the order of the group. n = 6 for the 32 group. In this problem all quantities are real.

Let (lhs) be the left-hand side of Eq. (1) and let (rhs) be the right-hand side.

Example 1 Let α = 1. (lhs) =

X

χ(1)(R)

R (1)

2

2 (1)

2

(1)

2

(1)

2

(1)

2

(1)

= χ E + χ A + χ B + χ C + χ D + χ F

2

= (1)2 + 3(1)2 + 2(1)2 =6 (rhs) = n =6 Example 2 Let α = Γ. (lhs) =

X R

χ(Γ)(R) (Γ)

2

2 (Γ)

2

(Γ)

2

(Γ)

2

(Γ)

2

(Γ)

= χ E + χ A + χ B + χ C + χ D + χ F

2

= (2)2 + 3(0)2 + 2(–1)2 =6 (rhs) = n =6 Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

37

2.21 This problem can be solved with educated guesswork using results from Chapters 1 and 2.

From Sec. (1.4.3) the permutation group of order n (also called the symmetric group S n of order n) has n! elements. Thus the group S 4 is of order 4! = 4 · 3 · 2 · 1 = 24.

According to Theorem 8 (stated in Sec. (2.7.4) and toward the end of Chapter 2) the sum of the squares of the dimensions of all the irreducible representations is equal to the order of the group. Like every group, S n has a 1-dimensional homomorphic identity representation and a 1-dimensional homomorphic antisymmetric representation. The task is therefore to find a sum of squares with dimensions > 1 that give 24 – (1)2 – (1)2 = 22.

The trial (5)2 = 25 does not work. For the trial (4)2 = 16 it would be necessary to find a sum of squares equal to 22 – 16 = 6 but this condition cannot be satisfied for dimensions > 1.

The trial (3)3 = 9 can give 22 – 9 – 9 = 4 = (2)2, and the same result is found for the trial (2)2 = 4. Thus there are two 3-dimensional irreducible representations and one 2-dimensional irreducible representation in addition to the two original 1-dimensional representations.

38

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

2.22 The given representation matrices are , 3 2 – 21

1 –,2 – 23

1 0 0 1 E

, 3

1 –,2 3 2

A

– 2 – 21 B

Consider the matrix products –1 , 4 –3 1 –1 D(B)D(A) = , 4 3

D(A)D(B) = 1

, 3 –1 , –3 –1

, –3

–1 , 3

–1 , 3

–1 , –3

1 0

=

–1

=

0 1 10 0 1

Because matrix representations agree with the product table it follows that AB = BA. The group is Abelian, as expected from 120◦ + 240◦ = 240◦ + 120◦. From Theorem 9 (stated in Sec. (2.7.4) and near the end of Chapter 2) all irreducible representations of Abelian groups are 1-dimensional. Hence the given 2-dimensional representation is reducible.

2.23 The Kronecker (Direct) product A ⊗ B of two matrices A and B is defined in Sec. (2.8). Consider as an example the Kronecker product of two 2 × 2 matrices A (elements ai j) and B (elements bi j).

A⊗B=

a11 a12 ⊗ a21 a22

b11 b12 b21 b22

a11b11 a11b12 a b a b = 11 21 11 22 a21 b11 a21 b12 a21b21 a21b22 =

a12b11 a12b21 a22 b11 a22b21

a12b12 a12b22 a22 b12 a22b22

a11 a B 12 B a21B a22B continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MATRIX REPRESENTATIONS OF DISCRETE GROUPS

39

According to this result a Kronecker product can be written as a single matrix whose elements are the individual elements of the first matrix times the entire second matrix.

In the following calculation let C ≡ AAJ so that the matrix elements of C are . ci j = k aikaJk j. a11B . . . a1nB aJ11BJ . . . aJ1nBJ . ... .. . . . .

(A ⊗ B)(AJ ⊗ BJ) = J J J an1B . . . annB a n1B . . . a nnBJ .n .................... n J J BBJ J k=1 a1ka k1 k=1 a1ka kn BB ... . = . .n J .................... n J BBJ k=1 ankaJk1 BB k=1 anka kn =

c11BBJ . . . c1nBBJ ... . . cn1BBJ . . . cnnBBJ

= (C) ⊗ (BBJ) = (AAJ) ⊗ (BBJ)

3.1 The bi j coefficients are obtained from the equations of motion. Omitting the common factor m1 throughout we have x¨1 + (kJ + k) x1 – (k) x2 = 0 x¨2 – (k) x1 + (kJ + k) x2 = 0 b11 = kJ + k b12 = –k b21 = –k b22 = kJ + k From Sec. (3.2.2) the vibration frequencies are ω12 = kJ ω22 = kJ + 2k From Eq. (3.5) with n = 1 (b11 – ω12)c11 + b21c12 = 0 kc11 – kc12 = 0 c12 = c11

as in Sec. (3.2.2). continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MOLECULAR VIBRATIONS

41

From Eq. (3.5) with n = 2 (b11 – ω22)c21 + b12c22 = 0 –kc21 – kc22 = 0 c21 = –c22

as in Sec. (3.2.2).

Hence 11

c

C=

c12 = c21 c22

11

c c11 –c22 c22

C B =? Ω C c11 –c22

c11 c22

kJ + k –k

–k k +k J

kJc11

kJc11

–(kJ + 2k)c22

(kJ + 2k)c22

=?

0

=

0 k + 2k

c11 c11 –c 22 c22

kJc11

kJc11

–(kJ + 2k)c22

(kJ + 2k)c22

kJ

J

3.2 Set kJ = k.The bi j coefficients are obtained from the equations of motion. Omitting the common factor m1 throughout we have x¨1 + (2k) x1 – (k) x2 = 0 x¨2 – (k) x1 + (2k) x2 = 0 b11 = 2k b12 = –k b21 = –k b22 = 2k From Sec. (3.2.2) the vibration frequencies with kJ = k are ω12 = k ω22 = 3k From Eq. (3.5) with n = 1 (b11 – ω12)c11 + b21c12 = 0 kc11 – kc12 = 0 c12 = c11

as in Sec. (3.2.2). continued next page =→

42

MOLECULAR VIBRATIONS

From Eq. (3.5) with n = 2 (b11 – ω22)c21 + b12c22 = 0 –kc21 – kc22 = 0 c21 = –c22

as in Sec. (3.2.2).

Hence C=

11

c

c12 = c21 c22

11

c c11 –c22 c22

C B =? Ω C c11 –c22

c11 c22

2k –k ?

–k 2k =

k 0 0 3k

kc11

kc11

kc11

kc11

–3k c22

3k c22

–3k c22

3k c22

=

c1 –c22

c11 c22

3.3 (a) Let BF 3 be in the x – y plane.

1) rotate about z by 0◦ 2) rotate about z by 120◦ 3) rotate about z by 240◦ 4) flip by 180◦ about any B-F axis 5) reflection by a mirror in the x – y plane 6) reflection by a mirror in a plane containing the z axis and a B-F bond b) in 1) all atoms stay at home in 2) and 3), only B stays at home in 4) B and one F stay at home in 5) all atoms stay at home in 6) B and one F stay at home

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MOLECULAR VIBRATIONS

43

3.4 (a)

Let the H atoms be in the x – y plane.

1) rotate about z by 0◦ 2) rotate about z by 120◦ 3) rotate about z by 240◦ 4) reflection by a mirror in a plane containing the z axis and a N-H bond b) in 1) all atoms stay at home in 2) and 3) only N stays at home in 4) N and one H stay at home

3.5 The relation between basis functions and the matrix representation is given by Eq. (2.1) in Sec. (2.2). hα X Tøj = D(α) (T)øi (1) ij i=1

A B

1 1 1

2z 1 -1

For the identity operation 1 (x) ›→ x. The operation 2z is a rotation about the z ◦ axis by 360 = 180◦ so that 2z (x) ›→ –x.

2 Using Eq. (1), 1(x) = (1)x 2z(x) = (–1)(x) The results are in accord with Eq. (1) and show that x is a basis function for irreducible representation B.

44

MOLECULAR VIBRATIONS

3.6 The relation between basis functions and the matrix representation is given by Eq. (2.1) in Sec. (2.2). hα X Tøj = D(α) (T)øi (1) ij i=1

A B1 B2 B3

1 1 1 1 1

2x 1 -1 -1 1

2y 1 -1 1 -1

2z 1 1 -1 -1 ◦

The operation 2n is a rotation about the n axis by 360 = 180◦.

2 1(x) ›→ x = (1)x 2x(x) ›→ x = (1)x 2y(x) ›→ –x = (–1)x 2z(x) ›→ –x = (–1)x The results are in accord with Eq. (1) and show that x is a basis function for irreducible representation B3 of group 222.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MOLECULAR VIBRATIONS

45

3.7 The relation between basis functions and the matrix representation is given by Eq. (2.1) in Sec. (2.2). hα X Tøj = D(α) (T)øi (1) ij i=1

A B1 B2 B3

1 1 1 1 1

2x 1 -1 -1 1

2y 1 -1 1 -1

2z 1 1 -1 -1

Angular momentum in determinant form is jˆ k̂ ˆi Lx Ly = x y z Lz mx˙ my˙ mz˙ so that Lz = m xẏ – mx˙y. Hence Lz transforms like xy. ◦ The operation 2n is a rotation about the n axis by 360 = 180◦.

2 1(xy) ›→ xy = (1)xy 2x(xy) ›→ x(–y) = (–1)xy 2y(xy) ›→ (–x)y = (–1)xy 2z(xy) ›→ (–x)(–y) = (1)xy The results are in accord with Eq. (1) and show that xy is a basis function for irreducible representation B1 of group 222.

46

MOLECULAR VIBRATIONS

3.8

a) Each mass can have three displacements δx, δy, δz in 3-dimensional space so there are 3 × 3 = 9 total displacements. b) Three displacements are needed to describe zero frequency translation and three displacements are needed to describe zero frequency rotation, leaving 9 - 3 - 3 = 3 nontrivial normal modes.

c) The identity irreducible representation has the property that any group operation leaves the displacement pattern unchanged. Symmetry group operations include rotation about the z axis by 0◦, 120◦, 240◦ and mirror reflections in planes that contain the z axis and the a – a, b – b and c – c axes.

Consider the mirror reflection in the plane determined by the z axis and the a – a axis. To remain unchanged by the reflection, the displacement of the mass at the apex a must be along the a – a line. Similarly for the other reflections, as shown in the sketch for the normal mode that is a basis function for the identity representation.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

MOLECULAR VIBRATIONS

47

3.7 basis functions x2 + y2, z2 Lz z x, y, x2 – y2, xy xz, yz, Lx, Ly

A1 A2 B1 B2 E1 E2

1 1 1 1 1 2 2

mz 1 1 -1 -1 2 -2

3z 1 1 1 1 -1 -1

2y 1 -1 1 -1 0 0

mx 1 -1 -1 1 0 0

With three masses there are 3 × 3 displacements, which will give rise to a 9 × 9 matrix for each operation in the reducible representation. The rows and the columns in each matrix are labeled δx1, δy1, . . . δy3, δz3.

As an example, the identity operation 1 leaves all displacements unchanged so δxi ›→ δxi, δyi ›→ δyi, δzi ›→ δzi for i = 1,2,3. Hence the matrix has entries (δxi, δxi) = 1, (δyi, δyi) = 1, (δzi, δzi) = 1 for i = 1,2,3. All other entries are 0. mz is reflection in a plane perpendicular to the z axis. Its effect on the displacements is δxi ›→ δxi, δyi ›→ δyi, δzi ›→ –δzi, i = 1,2,3.

D(1) =

100000000 010000000 001000000 000100000 000010000 000001000 000000100 000000010 000000001

D(mz) =

10 0 00 0 00 0 01 0 00 0 00 0 0 0 –1 0 0 0 0 0 0 00 0 10 0 00 0 00 0 01 0 00 0 00 0 000 0000 –1 0 01 00 00 00 0 00 0 01 0 0 0 0 0 0 0 0 0 –1

3z is a single rotation about the z axis by 120◦. Its effect on the displacements is ,

δx1 ›→ – 12 δx2 – 3 δy2 2 ,

δx2 ›→ – 12 δx3 – ,3 δy3 2 3 1 δx3 ›→ – δx1 – δy1 2

2

,

δy1 ›→ 23 δx2 – 12 δy2 , δy2 ›→ ,23 δx3 – 12 δy3 δy3 ›→ 3 δx1 – 1 δy1 2 2

δz1 ›→ δz2 δz2 ›→ δz3 δz3 ›→ δz1

continued next page =→

48

MOLECULAR VIBRATIONS

0 0 0 0

D(3z) =

, 1 0 – – 30 0 0 0 2 , 2 1 – 0 0 0 0 3 0 0 2 2 1 0 0 0 0 0 0 0 , 1 3 0 0 0 0 0–2 –2 0 , 1 3 0 0 0 0 0 0 0 0 0 0 2 –2 , 0 0 0 1 3 0

0 0 1

–2 –2 0 0 ,3 1 0 0 2 –2 0 0 1 0

0

0 0

0

0

0 0

0 0 0 0

0 0

0 0

To understand the structure of these matrices, it may be helpful to write a portion of D(3z) as the matrix transformation of δx1J , δyJ1, δzJ1 from δx2, δy2, δz2. , 3 0 0 0 –12 – 2 0 0 0 0 , 3 1 000 – 2 2 0 000 000 0 0 1000

δx1J δyJ1 δzJ1 δxJ 2J δy 2

δx1 δy1 δz1 δx2 δy2 δz2

.. . . . .2 . . . . . . . . . . ◦ 2y is a 180 flip about the y axis. Its effect on the displacements is δzJ

=

δx1 ›→ –δx1

δy1 ›→ δy1

δz1 ›→ –δz1

δx2 ›→ –δx3

δy2 ›→ δy3

δz2 ›→ –δz3

δx3 ›→ –δx2

δy3 ›→ δy2

δz3 ›→ –δz2

mx is reflection in a plane perpendicular to the x axis. Its effect on the displacements is

D(2y) =

δx1 ›→ –δx1

δy1 ›→ δy1

δz1 ›→ δz1

δx2 ›→ –δx3

δy2 ›→ δy3

δz2 ›→ δz3

δx3 ›→ –δx2

δy3 ›→ δy2

δz3 ›→ δz2

–1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 –1 0 0 0 0 0 0 0 0 0 –1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 –1 0 0 0 0 0 0 0 0 0 –1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 –1

D(mx) =

–1 0 0 0 0 0 0 0 0 0 10 0 00 0 00 01 0000 0 00001 00 00 00 – 0 00 0 00 0 10 0 00 0 00 0 01 0 0 0 –1 0 0 0 0 0 0 00 0 10 0 00 0 00 0 01 0 00

continued next page =→

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MOLECULAR VIBRATIONS

49

Character χ is the sum of the diagonal elements of the matrices. χ(1) = 9 χ(mz) = 3 χ(3z) = 0 χ(2y) = -3 χ(mx) = 1

3.10 basis functions x2 + y2, z2 Lz z x, y, x2 – y2, xy xz, yz, Lx, Ly

A1 A2 B1 B2 E1 E2

1 1 1 1 1 2 2

mz 1 1 -1 -1 2 -2

3z 1 1 1 1 -1 -1

2y 1 -1 1 -1 0 0

mx 1 -1 -1 1 0 0

a) Physical translations transform like x, y, z. From the character table, these basis functions transform like irreducible representations B2 + E1. b) Physical rotations transform like Lx, Ly, Lz. From the character table, these basis functions transform like irreducible representations A2 + E2.

3.11 a) From Problem 10, the zero-frequency translation and rotation transform like irrreducible representations A2 + B2 + E1 + E2. Subtracting these from A1 + A2 + B2 + 2E1 + E2 leaves A1 + E1 for the nontrivial normal modes. b) There are two distinct nontrivial mode frequencies. One normal mode transforms like A1. The normal mode that transforms like the 2-dimensional representation E1 is degenerate; the two modes have the same frequency.

50

MOLECULAR VIBRATIONS

3.12 Rotation about the z axis is not a symmetry operation for this structure.

Let the mass 2m be located at the apex on the y axis. Symmetry operations are then 2y flip by 180◦ about the y axis mz reflection by a mirror perpendicular to the z axis mx reflection by a mirror perpendicular to the x axis (in the y – z plane)

3.13 a) There are 5 atoms in the molecule and each atom has 3 possible displacements δxi, δyi, δzi. Hence there are 3 × 5 = 15 displacement coordinates. b) Three displacements are needed to describe zero frequency translation and three displacements are needed to describe zero frequency rotation, leaving 15 - 3 - 3 = 9 nontrivial normal modes.

c) The identity irreducible representation has the property that any group operation leaves the displacement pattern unchanged. Symmetry group operations of Xe F 4 include rotation about the z axis by 0◦, 90◦, 180◦, 270◦ and mirror reflections in planes that contain the z axis and a Xe-F bond, mirror reflection in the x – y plane, and 180◦ flips about F-Xe-F axes, and other operations. None of the symmetry operations change the pattern of the “breathing” normal mode in the sketch.

3.14 Each of the 5 atoms has two displacements for motion in the x – y plane, for a total of 2 × 5 = 10. In the x – y plane two displacements are needed to describe physical translation and two displacements are needed for physical rotation. Hence the number of in-plane normal modes is 10 - 2 - 2 = 6.

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51

3.15 a) Each atom in NH 3 has three possible displacements, for a total of 3 × 4 = 12.

Three displacements are needed for physical translations and three for physical rotations, to give 12 - 3 - 3 = 6 nontrivial normal modes.

In another approach, the normal modes have irreducible representations 2A1 + 2E. Each A1 contributes one normal mode and each E contributes two normal modes, for a total of 2(1) = 2(2) = 6 normal modes as before. b) The character table of NH 3 is basis functions z, x2 + y2, z2 Lz x, y, x2 – y2, xy, xz, yz, Lx, Ly

A1 A2 E

1 1 1 2

3z 1 1 -1

mx 1 -1 0

As the character table shows, z is a basis function for A1 and x,y are basis functions for E. All normal modes of ammonia are IR active. c) Each A1 mode contributes an IR frequency. Each twofold E mode is degenerate – the two frequencies of a E mode are the same, giving a total of four distinct frequencies.

3.16 basis functions x2 + y2, z2 Lz z x, y, x2 – y2, xy xz, yz, Lx, Ly

A1 A2 B1 B2 E1 E2

1 1 1 1 1 2 2

mz 1 1 -1 -1 2 -2

3z 1 1 1 1 -1 -1

2y 1 -1 1 -1 0 0

mx 1 -1 -1 1 0 0

The BF 3 molecule is planar and has the same character table as the triangle structure in Problem 9. The normal modes have irreducible representation A1 + B2 +2E1. a) As the character table shows, z transforms like B2 hence this mode has a timevarying electric dipole moment and is IR active. Similarly x,y transform like E1 so these modes are also IR active. continued next page =→

52

MOLECULAR VIBRATIONS

b) Modes are Raman active if they transform like products such as xy, xz, z2 etc.

From the character table, z2 transforms like A1 and xy transforms like E1, so all these modes are Raman active. B2 is not Raman active. c) Only products transform like A1, so this mode is Raman active but not IR active.

3.17 The character table for H 2O is basis functions z, x2, y2, z2 xy, Lz x, xz, Ly y, yz, Lx

A1 A2 B1 B2

1 1 1 1 1

2z 1 1 -1 -1

my 1 -1 1 -1

mx 1 -1 -1 1

The nontrivial normal modes transform like 2A1 + B2. According to the character table, z2 transforms like A1, giving a Raman active mode from each A1. Products yz transform like B2, so this normal mode is also Raman active. All three normal modes of water are Raman active.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

4.1 From Bragg’s Law Eq. (4.1) in Sec. (4.3) λ = 2d d≥ dmin =

θ n

≤1

λ 2 0.154 2

= 0.077 nm

4.2 a) From the scalar product of two vectors, b·c cos θ = bc b · c = (0.322)(0.344) + (0.692)(0.241) = 0.278 , b = (0.322)2 + (0.692)2 = 0.763 , c = (0.344)2 + (0.241)2 + (0.896)2 = 0.990 0.278 cos θ = = 0.368 (0.763)(0.990) θ = 68.4◦ continued next page =→

54

CRYSTALLINE SOLIDS

b) V = a · (b × c) Because a has only a component along ˆi only the ˆi component of b × c is needed.

b × c = (0.692)(0.896) ˆi + . . .

V = (0.741)(0.692)(0.896) = 460 nm3

4.3 All the lattice angles are 90◦ so the crystal axes lie along the Cartesian axes. Use the notation of Eq. (4.4) in Sec. (4.5). a1 = a = 0.50 ˆi a2 = b = 0.80 ˆj a3 = c = 0.57 k̂ Note that the denominators in Eq. (4.4) are all equal. denominator = a1 · (a2 × a3) = 0.50 ˆi · (0.80)(0.57) ˆi = 0.23 (0.80)(0.57) ˆ i = 1.98 ˆi 0.23 (0.57)(0.50) ˆ b2 = j = 1.24 ˆj 0.23 (0.50)(0.80) b3 = k̂ = 1.74 k̂ 0.23 The reciprocal lattice is also orthorhombic, with all angles 90◦. b1 =

4.4 a) The primitive call of the simple cubic Bravais lattice is the Bravais simple cubic lattice itself, a special case. Each corner site of a simple cubic lattice is shared among 8 lattices, so all 8 corner sites of a simple cubic comprise only 1 net site, as required for a primitive cell. continued next page =→

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CRYSTALLINE SOLIDS

55

The primitive cell for a bcc lattice or for a fcc lattice is not the same as the cubic structure of the lattice, because such a primitive cell would include more than one lattice site. b) In general, a primitive cell with axes a, b, c has volume V = a · (b × c). The axes of a simple cubic primitive cell all have the same length a. The axes form an orthogonal triad, so b × c = a ˆj × a k̂ = a2 ˆi, hence V = (a2) ˆi · a ˆi = a3.

4.5 a) The volume of the cell is V = a · 3(b ! × c) = c · (a × b) 3 a a ˆ ˆ ˆ ˆ = (i + j + k) · k = 2 2 b) V = a · (b × c) ! a3 ˆ · .(iˆ – jk) ˆˆ × (iˆ + ˆj – k) ˆ (–iˆ + jˆ + k) = 8 !

3 a3 ˆ · (2 jˆ + 2 k) ˆ =a (–iˆ + jˆ + k) = 8 2 The primitive cells in a) and b) are different but they have the same volume, a general property. Any primitive cell can be stacked to form the lattice and all have the same minimum volume.

4.6 Use the notation of Eq. (4.4) in Sec. (4.5). a1 = a = 0.741 ˆi a2 = b = 0.322 ˆi + 0.692 ˆj a3 = c = 0.344 ˆi + 0.241 ˆj + 0.896 k̂ continued next page =→

56

CRYSTALLINE SOLIDS

Note that the denominators in Eq. (4.4) are all equal. denominator = a1 · (a2 × a3) = 0.741 ˆi · (0.692)(0.896) ˆi + . . . = 0.459 (0.692)(0.896) ˆi – (0.322)(0.896) ˆj + ((0.322)(0.241) – (0.344)(0.692)) k̂ 0.459 = 1.351 ˆi – 0.629 ˆj – 0.350 k̂ (0.896)(0.741) ˆj + (0.241)(0.741) k̂ b2 = – 0.459 = –1.446 ˆj + 0.389 k̂ (0.741)(0.692) k̂ b3 = 0.459 = 1.117 k̂ b1 =

4.7

T1 ø(r) = ø(r + τ1) T1 ø(r + τ1) = T21ø(r) = ø(r + τ1 + τ1) = ø(r + 2τ1) T2 ø(r) = ø(r + 2τ1) T2 = T21

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CRYSTALLINE SOLIDS

57

4.8 For a simple cubic lattice, the physical lattice and the reciprocal lattice are essentially the same except for the overall factors a and 1a . Hence Cartesian unit vectors can be used.

The Miller index (001) specifies a plane normal to the vector (0) ˆi + (0) ˆj + (1) k̂ = k̂ The specified plane is shown shaded in the sketch.

4.9

For a simple cubic lattice, the physical lattice and the reciprocal lattice are essentially the same except for the overall factors a anda 1 .

Hence Cartesian unit vectors can be used. The Miller index (110) specifies a plane normal to the vector (1) ˆi + (1) ˆj + (0) k̂ = ˆi + ˆj.

The specified plane is shown shaded in the sketch.

4.10 For n=1, Bragg’s Law is d=

λ 2 sin θ

=

0.0714 2 sin (8.90◦)

= 0.231

From Table (4.1) in Sec. (4.5.1) for Miller index (111) , a = 3 d = 0.400 nm close to the modern value 0.405 nm.

58

CRYSTALLINE SOLIDS

4.11 For n=1, Bragg’s Law is d=

λ 2 sin θ

=

0.0714 2 sin (17.25◦)

= 0.120

From Table (4.1) in Sec. (4.5.1) for Miller index (311), , a = 11 d = 0.399 nm close to the modern value 0.405 nm.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

CRYSTALLINE SOLIDS

59

4.12 For n=1, Bragg’s Law is d=

λ 2 sin θ

=

0.0714 2 sin (14.63◦)

= 0.141

From Table (4.1) in Sec. (4.5.1) for Miller index (110) , a = 2 2 d = 0.400 nm The volume V occupied by one mole of aluminum is 26.98 g · mole–1 2.70 g · cm–3 = 9.99 cm3 · mole–1

V=

= 9.99 × 1021nm3 · mole–1 Each lattice occupies volume v

v = a3 = 0.064 nm3 V 9.99× 1021 = v 0.064 = 1.56 × 1023 lattices per mole As discussed in Sec. (4.2.1) each fcc lattice contributes four atoms because of sharing between neighboring lattices. Hence

NA = (4)(1.56 × 1023) = 6.24 × 1023 atoms per mole The modern value is NA ≈ 6.02 × 1023.

60

CRYSTALLINE SOLIDS

4.13 basis functions r2 x, y, z, Lx, Ly, Lz 3z2 – r2, x2 – y2, xy, yz, zx

A F1 F2 G H

1 1 3 3 4 5

12C5 1 η 1-η -1 0

12C52 1 1-η η -1 0

20C3 1 0 0 1 -1

15C2 1 -1 -1 0 1

a) The operations listed in Table (4.6) for the icosahedral group include five-fold rotations C5 and C52, which are not proper rotations for the point group of a lattice. b) As discussed in Sec. (4.9.2) if inversion does not change the sign of the basis functions of an irreducible representation, the representation is even (gerade) (+). If inversion changes the sign of the basis functions of an irreducible representation, the representation is odd (ungerade) (-). From the character table A is (+); r2 does not change sign under inversion F1 has undefined (±); x, y, . . . change sign under inversion, Lx . . . do not change sign H is (+); xy, . . . do not change sign under inversion

4.14

The stereogram on the left is for a group with only the identity element. The stereogram on the right is for a group with two elements, the identity and inversion. Under inversion, a point in the northern hemisphere (filled circle) is projected onto the southern hemisphere (open circle).

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

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61

4.12

The four-fold rotations and flips of a square (422 group) has the symmetry operations: identity four-fold rotation about the z axis 180◦ flips about the a – a, b – b, c – c, and d – d axes

4.16

Several different styles have been devised for drawing stereograms. Both stereograms in the sketch show the symmetry group of a square (422 group) with the symmetry elements listed in Problem 4.15.

The stereogram on the left uses the symbols introduced in Sec. (4.7.1). The square in the center symbolizes the four-fold rotation about the z axis. The oval shapes at the ends of each axis symbolize 180◦ flips about the axis.

The stereogram on the right uses filled circles for locations in the northern hemisphere and open circles for the southern. This stereogram shows the four-fold rotation about the z axis implicitly. 180◦ flips about the x and y axis are clearly shown, but flips about the diagonal axes are implicit.

4.17 The symmetry operations for an orthorhombic lattice (222 group) shown by the stereogram are identity two-fold rotation about the z axis 180◦ flip about the x axis and also about the y axis

62

CRYSTALLINE SOLIDS

4.13 basis functions z2, x2 + y2 z, Lz x, y, x2 – y2, xy, xz, yz, Lx, Ly

A1 A2 E1

An irreducible representation is gerade (+) if its basis functions do not change sign under inversion and it is ungerade (-) if its basis functions change sign under inversion.

A1 is (+) because z2 etc. do not change sign under inversion.

A2 has undefined (±) because z changes sign under inversion but Lx does not.

E1 has undefined (±) because some basis functions change sign under inversion and some do not.

4.19 basis functions z2, x2 + y2 z, Lz x, y, xz, yz, Lx, Ly x2 – y2, xy

A1 A2 B1 B2 E1 E2

An irreducible representation is gerade (+) if its basis functions do not change sign under inversion and it is ungerade (-) if its basis functions change sign under inversion.

A1 is (+) because z2 etc. do not change sign under inversion A2 has undefined (±) because z changes sign under inversion but Lx does not E1 has undefined (±) because x changes sign under inversion but xz etc. do not E2 is (+) because xy etc. do not change sign under inversion

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

CRYSTALLINE SOLIDS

63

4.12 Equations (4.15) and (4.16) for the linear chain model are optical branch: ω2 = +

C

, m + M + m2 + M2 + 2mM cos (qa)

(4.15)

mM

acoustic branch: , C m + M – m2 + M2 + 2mM cos (qa) mM Setting m = M, the equations become ω2+ =

optical branch: ω2 = +

C

(4.16)

, 2M + 2M2(1 + cos (qa))

M2

acoustic branch: ω–2 =

C

, 2M – 2M2(1 + cos (qa))

M2 The maximum value of q occurs for q = au so cos(qa) = cos(u) = –1. The right hand side of the equations for the optical and acoustic branches are then equal. , C ω2 = 2M – 2M2(1 + cos (u)) M2 C = (2M) M2!

2C = M , 2C ω= M

64

CRYSTALLINE SOLIDS

4.13 A sound wave is propagating along a linear diatomic chain. The equation for the acoustic branch is , C ω2– = m + M – m2 + M2 + 2mM cos (qa) (4.16) mM dω– a) The maximum possible frequency of propagation is found from dq = 0. Assume M > m.

C , dω– = – –amM sin (qa) 2ω 2 – dq mM m + M2 + 2mM cos (qa) = (aC) , sin (qa)

m2 + M2 + 2mM cos (qa) The derivative is 0 for qa = ±u. Hence from Eq. (4.16) with qa = u , C m + M – m2 + M2 + 2mM cos (u) mM , C = m + M – m2 + M2 – 2mM mM , C = mM m + M – (M – m)2 !

2C = ,M 2C ω| = – max M 2

ω– max =

ω– has 0 slope at the zone boundaries, as seen in the figure Sec. (4.10.1). The slope must be 0 at the zone boundaries to maintain smooth continuity with the neighboring zones, as shown in the sketch for M = 3m.

b) As discussed in Sec. (4.10.1) the phase velocity of an acoustic wave propagating along a linear diatomic chain is v = dωdq– . The maximum possible frequency of

ω– = 0 as shown in a), so the speed of the maximum frequency propagation is for ddq sound wave is 0.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

5.1 The Rydberg formula Eq. (5.5) Sec. (5.2) for the hydrogen spectrum is 1 !

1 1 = R∞ 2 – λ n m2 where m > n.

(5.5)

For the Paschen series n = 3, Eq. (5.5) becomes !

1 1 1 = R∞ – λ 9 m2 λ is longest when the right-hand side is smallest, in this case for m = 4. !

1 1 1 = R∞ – λ 9 16 ≈ 1.097 × 10–2 (0.0486) nm–1 λ |max ≈ 1875 nm The visible spectrum, where the human eye is sensitive, is considered to be from 400 nm (deep violet) to 700 nm (deep red). The Paschen line at 1875 nm is in the infrared and is not visible. The series limit of the Paschen series is !

1 1 = R∞ 9 λ ≈ 1.097 × 10–2 (0.1111) nm–1 λ |lim ≈ 820 nm None of the Paschen series is visible.

66

BOHR’S QUANTUM THEORY AND MATRIX MECHANICS

5.2 The Rydberg formula Eq. (5.5) Sec. (5.2) for the hydrogen spectrum is 1 !

1 1 = R∞ 2 – λ n m2 where m > n.

(5.5)

For the Lyman series n = 1, Eq. (5.5) becomes

!

1 1 1 = R∞ – λ 1 m2 λ is longest when the right-hand side is smallest, in this case for m = 2. Lβ, the second longest wavelength, is for m = 3. !

1 1 1 = R∞ – λβ 1 9 ≈ 1.097 × 10–2 (0.8889) nm–1 λβ ≈ 102.6 nm The visible spectrum, where the human eye is sensitive, is considered to be from 400 nm (deep violet) to 700 nm (deep red). Lβ at 102.6 nm is in the ultraviolet and is not visible.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

BOHR’S QUANTUM THEORY AND MATRIX MECHANICS

67

5.3 Bohr’s theory postulates that a particle of mass m moving with constant speed v in a circular orbit of radius r has quantized angular momentum L = nk: mvr = L = nk From de Broglie’s relation

mv =

h λ

Hence ! h r = nk λ nλ r= 2u 2ur = nλ

5.4 According to Bohr’s theory, the energy level En of a hydrogen atom is ! ! µ02c4mee4 1 En = – 8h2 n2 ∆E = En+1 – En ! !

1 µ2c4mee4 1 – =– 0 2 8h (n + 1)2 n2 ! ! µ02c4mee4 n2 – (n + 1)2 =– 8h2 n2(n + 1)2 !

2 4 4!

For n » 1 ∆E ≈ µ0 c mee 2n 8h2 n4 ! ! µ02c4mee4 1 = 4h2 n3

(5.2)

(1)

In these equations, Eq. (5.2) has been multiplied by e to convert electron volts to joules. continued next page =→

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BOHR’S QUANTUM THEORY AND MATRIX MECHANICS

Classically, mv2 r

=

e2

4us0 r2 1 e2 E = mv2 – 2 4us0 r

(2) (3)

From Eqs. (2) and (3)

e2 mv2 = 4us0 r 2 ! =– 1 e 2 4us0 r ! e2 ∆E ≈ ∆r 8us0 r2 ∆r = rn+1 – rn

E

(4)

Now use Bohr’s theory to convert this classical result, for comparison with Eq. (1). ∆r = a0 (n + 1)2 – n2 For n » 1

∆r ≈ 2a0 n

Equation (4), with r = a0 n2, becomes ! e2 ∆E ≈ a0 n 4us 0 a20 n4 !

2 e = 4us0 a0n3 ! ! ! e2 umee2 1 = 4us0 s0h2 n3 !

4 ! = me2e 2 13 4s h n 0 ! !

2 µ0 c4mee4 1 = 4h2 n3 in agreement with Eq. (1).

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

BOHR’S QUANTUM THEORY AND MATRIX MECHANICS

69

5.5 The Hamiltonian 7 of a system is the sum of the kinetic and potential energies.

Taking into account only Coulomb forces the Hamiltonian for a single electron in a hydrogenic atom with fixed nucleus is p2 e2 k2 2 e2 – 7(r) = – =– ∇ 4us0r 2me 4us0r 2me

5.6 Taking into account only Coulomb forces the Hamiltonian 7 for electrons 1 and 2 in a helium atom (Z = 2) with fixed nucleus is ! !

2e2 2e2 p2 e2 p2 + + – – 7(r1, r2) = 2me (2) 4us0 r1 4us0r 2 4us0|r1 – r 2| 2me (1) The first two terms on terms on the right-hand side are the kinetic energies of the electrons. The third and fourth terms are the Coulomb interactions of each electron with the Z=2 nucleus. The fifth term is the Coulomb interaction of the two electrons with each other.

5.7

∞ X

y(t) =

øneinωt

n=–∞

= . . . + ø–1e–iωt + ø0 + ø1eiωt + . . .

(1)

Take the complex conjugate of Eq. (1), noting that y(t) is real.

y(t) = . . . + ø∗ 1 eiωt + ø0 + ø∗1e–iωt + . . .

–

(2)

Under the stated conditions Eqs. (1) and (2) are equal, so the coefficients of the corresponding exponentials must be equal. In general ø–n = ø∗n.

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BOHR’S QUANTUM THEORY AND MATRIX MECHANICS

5.8 The quantization condition in matrix mechanics is pq – qp = –ik1

(5.13)

pq = qp – ik1 Multiply from the right by q and use Eq. (5.13).

pq2 – qpq = –ikq pq2 – q2p + ikq = –ikq pq2 – q2p = –2ikq

5.9 2

The Hamiltonian 7 for a free particle is 7 = p2m. From Eq. (5.18), the commutator of any operator a with the Hamiltonian is 7a – a7 = –ikȧ

(5.18)

For a = p

!

1 2m

7p – p7 = –ikṗ p2p – pp2 = –ikṗ

!

1 p3 – p3 = –ikṗ 2m 0 = –ikṗ

The result ṗ = 0 shows that p does not change with time. The linear momentum of a free particle is conserved.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

BOHR’S QUANTUM THEORY AND MATRIX MECHANICS

71

5.5 The Hamiltonian 7 for any central force motion is !

1 7= p2 + U(r) 2m Let the orbit lie in the x – y plane so that the electron has only the z component Mz of angular momentum. From Sec. (5.5.3) Mz = (qxpy – pxqy). To begin, consider the commutator of Mz with the kinetic energy term.

2

?

2

2

2

(px + py)(qxpy – pxqy) = (qxpy – pxqy)(px + py) ? qxpyp2 + qxp3 – pxqyp2 – pxqyp2 p2xqxpy – p3xqy + p2yqxpy – p2ypxqy = x y x y

(1)

According to the commutation rules in Sec. (5.4.2), qxpy = pyqx and qypx = pxqy.

From Eqs. (5.14) and (5.15) qxqy = qyqx and pxpy = pypx. Using these commutation rules px3qy – pxqyp2x = p3xqy – p3xqy = 0 p2qxpy – qxp3 = qxp3qx – qxp3 = 0 y

y

y

y

Equation 1 reduces to py(p2x qx – qxp2x) – px(p2yqy – qyp2y) ?= 0

(2)

Evaluate the first term in parentheses (px2qx –qxp2)x using the quantization condition Eq. (5.13). Multiply the quantization condition by px from the left, then multiply it by px from the right and add. p2x qx – pxqxpx = –ikpx pxqxpx – qxp2x = –ikpx p2x qx – qxp2x = –2ikpx Using the same method to evaluate the second term in parentheses gives p2qy – qyp2 = –2ikpy y

y

Using these results in Eq. (2) gives –2ikpypx + 2ikpxpy = 0 Hence Mz commutes with the kinetic energy term of 7. continued next page =→

72

BOHR’S QUANTUM THEORY AND MATRIX MECHANICS

The remaining task is to show that Mz commutes with the potential energy function U(r). The potential energy function U(r) is a function only of coordinates so qiU(r) = U(r)qi. According to the commutation rules in Sec. (5.4.2), qxpy = pyqx.

U(r)Mz =? MzU(r) ?

U(r) qxpy – pxqy = qxpy – pxqy U(r) ?

qx U(r)py – pyU(r) = (U(r)px – pxU(r)) qy Using Eq. (5.17) qx

bU ? bU bqy = bq q

y

(2)

x

, For central force motion in the x – y plane U = U(r) where r = qx2 + qy2. bU bU br bU qx = = bqx br bqx br r3 bU bU qy = bqy br r3 Equation (2) becomes bU bU = qyqx br br Hence Mz commutes with the potential energy term of 7.

The results together show that Mz commutes with 7. As shown in Sec. (5.4.3) the commutator of an operator with 7 is the time derivative of the operator. Here Mz commutes with 7, so its time derivative is 0. Mz is independent of time – it is conserved. An operator that commutes with the Hamiltonian is conserved. qxqy

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

BOHR’S QUANTUM THEORY AND MATRIX MECHANICS

73

5.5 Consider the 1-dimensional harmonic oscillator according to matrix mechanics.

The key to proving d is a diagonal matrix is to show ḋ = 0. The reason is that in matrix mechanics, off-diagonal elements have complex exponential time dependence and only diagonal elements are real. (See Problem 5.7.) d = pq – qp ḋ = ṗ q + p q̇ – q̇ p – q ṗ Using the Hamilton equations of motion Eq. (5.8),

ḋ = –

b7

q+p

b7

–

b7

p+q

b7

bq bp! bp bq ! b7 b7 b7 b7 = p – p + q – q bp bp bq bq

The Hamiltonian for the 1-dimensional harmonic oscillator is ! !

1 k 2 2 7= p + q 2m 2 ! b7 1 bp = m p b7 = (k) q bq Using these results in Eq. (1) gives

1 ḋ = m =0

! p2 – p2 + (k) q2 – q2

(1)

6.1 In one dimension Schrödinger’s wave equation Eq. (6.4) with V = 0 is ! k2 d2ψ – = Eψ 2m dx2 The solution is ψ = Aeikx

! k2 2 k ψ = Eψ 2m 2mE k2 = k2

6.2 a) The normalization condition is , ψ∗ψdr = 1 , is over the applicable , ∞ n-dimensional where the integration space. x2 1 1 , ∗ – ψ ψdr = e dx = uβ β , , uβ –∞ uβ =1 continued next page =→ Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

b) The expectation value (x) of x is

75

,

(x) =

ψ∗ xψdx , ∞ x2 1 (x) = xe– dx β , uβ –∞ =0

The result (x) = 0 is expected because of the antisymmetric integrand. As suggested by the sketch, the negative portion exactly compensates the positive portion.

6.3 The one dimensional Schrödinger wave equation Eq. (6.4) with V = k2x2 is ! k 2 k2 ! d2ψ x ψ = E1ψ – + 2m dx2 2

(1)

The given wave function for the first excited state is x2

ψ = Axe– 2β x2 ! x2 = A 1 – β e– 2β dx ! d2ψ 3x + x3 – x2β2 = A – dx2 β β2 e dψ

Use these results in Eq. (1). A and the exponential are common factors and cancel. ! ! ! k2 3x x3 k 3 – – + 2 + x = E1 x 2m β β 2 ! ! ! k2 3 x2 k 2 – – + 2 + x = E1 2m β β 2 2 !

2!

2 !

2 2! k x k m ω x2 – 2m β2 = – 2m k2 ! ω2m 2 = – k x2 =– x 2 2 continued next page =→

76

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

This cancels the potential energy term, leaving ! ! k2 3 E1 = – – 2m β !

2 !

3mω = k 2m k 3 = kω 2 In general !

1 En = n + kω 2 Even at absolute zero, the ground state n = 0 has vibrational energy.

6.4

The wave function for the ground state of hydrogen is a simple decreasing exponential ! r 1 with wave function ψ = , 3 ea0 . a0 u

It is spherically symmetric, with no angular dependence. The spherically symmetric volume element is dr = 4ur2dr. a) Normalization requires

,

ψ∗ψdr = 1 !, ∞ , ∞ 4u 2 2r ∗ 2 ψ ψ4ur dr = r ea0 dr a30u 0 0 ! !

4u a30 = 3 a0u 4 =1 continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

b) (r) = 4u =

, ∞

4u

ψ∗rψr2dr 0! , ∞ 3

a30u 0 !

3 = 2 a0

2r

r ea0 dr

6.5

r

The wave function is ψ = N 2 – ar0 e–2 a0 where N = ,1 . The wave function is spherically 4

2a30u

symmetric so the volume element is dr = 4ur2dr. a) Normalization requires

, ψ∗ψdr = 1

,

! !, ∞ r 2 2 r 4u – a0 ψ∗ψdr = 23– 4 ! r er dr (16)(2)a30u a0r 0 4r e– dr !, ∞ a0 2– = 13 + a0 a02 8a!0 0 4r !

1 4 = (4)(2)a3 – (4)(6)a0 (24)a50 + 0 a0 a20 8a30 =1 continued next page =→

77

78

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

b) (r) = 4u

, ∞

ψ∗rψr2dr

0

= = =

1 8a!30 1

!, ∞ 0

! r

e – dr 4r4 a0 3 4r – a0 + 2 a 0

(4)(6)a4 –

8a3 !

10

8a03 = 6a0

r5

0

(4)(24)a5 (120)a6 !

0 0 + 2 a0 a 0

(48a4) 0

6.6

r

The wave function is ψ = N 2 – ar0 e–2 a0 where N = ,1 3 . The wave function is spherically 4

2a0u

symmetric. The common factor N cancels throughout.

From Sec. (5.2) a0 =

s0h2

(1a)

umee2 h2 = umee2µ0c2

(1b)

2

e The Schrödinger equation with V = –(4us is 0)r ! k2 e2 – ∇2ψ – ψ = E2ψ 2me (4us0)r

continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

79

Taking Laplacian from Sec. (6.4.1) the equation becomes ! the radial part ! of the k2 1 b 2 bψ e2 – r – ψ = E2ψ (2) 2me r2 br br (4us )r 0 ! ! bψ 1 r – 2ra 0 br = a0 –2 + 2a0 e ! ! 2r2 3 ! r r 1 = – + bψ e– 2a0 2 br a 2a 0 0 r ! ! ! ! !

12 b 1 1 b –2r 2 bψ r3 – 2ar0 + 2a r br r2 br = r2 a0 br 0 e ! r 4 5 r = – + 2 – 3 e– 2a0 a0r 2a0 4a0 The exponential is a common factor in the calculation and cancels throughout. ! ! ! ! !

Equation (2) becomes r k2 1 4 5 r e2 r E2 2 – 2 – – + – – =– a0 r 2a0 4a20 (4us0)r – a0 2me a0 Using a0 from Eq. (1a) k2 2me a0

=

k2

umee2

!

2me s0h2 2 = e ! !

8us0e2 ! 4 5umee2 u2m2e4 ! umee2 e2 e4ume E2 2 – r r – – + =– – + 2s0h2 s0h2 2us0r 4us02h2 4s02eh4 8us0 r e2 5mee4 um2e6 e2 mee4 e + r– = –– + 2us0r 2us0r 4s02h2 16s20h2 32s03h4 ume2e6 mee4 =– + 16s2h2 32s3h4 r 0 !

0 ! ! mee4 um ee2 –2 + = r 32s20h2 s0h2 mee4 E =– 2 32s20h2 1

Use c = ,s0µ0 .

E2 = –

me µ02 c4 e4

32h2 Dividing Eq. (3) by e converts E2 from joules to eV to compare with Eq. (5.2).

E 2= –

meµ2c4e3 0

32h2 in agreement with Eq. (5.2) for n = 2.

= –3.4eV

(3)

80

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

6.7 a) As shown in Sec. (6.3.2) the inner product (A|B) is a scalar, as in the example ø1 = ξ∗ø1 + ξ∗ø2 ξ ξ 1

2

ø2

1

2

Hence (A|B) = 3i 4 –2i

3 –2i 0

= –9i – 8i = –17i b) As shown in Sec. (6.3.2), the ø1 outer product |B) ø1ξ1∗ (A| ø1ξis∗ a matrix, as in the example ξ 1 ξ2 = ø2 ø2ξ1∗ ø2ξ2∗2 Hence 3 |B) (A| = –2i

–9i 12 6i –8i 4

3i 4 –2i = –6 0

0

0

0

6.8 a) As shown in Sec. (6.3.2) the inner product (A|B) is a scalar, as in the example ø1 = ξ∗ø1 + ξ∗ø2 ξ ξ 1

2

ø2

1

2

Hence (A|B) = 2 2i –3i

5 3 –2i

= 10 – 6i + 6 = 16 – 6i continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

81

b) As shown in Sec. (6.3.2), the ∗ (A| is a matrix, as in the example ø1 outer product ø|B) 1ξ1 ø1ξ∗ ξ 1 ξ = ø2 ø2ξ1∗ ø2ξ2∗2 2

Hence 5 |B) (A| =

3 –2i

10 = 6

2 2i –3i

–10i 15i –6i 9i –4i –4

6

6.9 The dumbbell consists of two equal masses M linked by a rigid massless rod of total length ρ. The dumbbell rotates in a circle of radius 2ρ by angle ø in the x-y plane about a fixed axis along z through the center of mass. The potential function is V = 0.

As developed in Sec. (6.4.1) the wave function is ψ = Aeinø where n is an integer.

To normalize the wave function, evaluate , 2 A ψ∗ψdr = 1

(1)

ψ has a value only along the orbit, so dr = dø. Equation (1) becomes , 2u 1 = A2

0

2uA2 = 1 1 A=, 2u

e–in øein ødø

82

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

6.10 The wave function for one of the excited states of a freely rotating rigid diatomic molecule Sec. (6.4.2) is , 3 iø ψ= e sin θ 8u The angle dependent part! of the Schrödinger!equation is ! k2 1 b bψ 1 b2ψ sin θ + – = Eψ 2µρ2 sin θ bθ sin2 θ bø2 bθ M2 where µ is the reduced mass = MM1+M and ρ is the bond length. The normalization 1

factor cancels throughout.

1

b2eiø sin θ

sin2 θ

bø2

2

eiø sin θ

= –

The complex exponential in ø cancels throughout, leaving !

b sin θ !

1 !

1 b θ sin θ bθ sin bθ – sin θ ! !

2 1 cos 2 θ – sin 2 θ – 1 =– k2 sin θ 2µρ sin θ ! !

2 k 1 1 2 1 – 2 sin θ – =– sin θ 2µρ2 sin θ ! k2 = (2 sin θ) 2µρ2

k2 E sin θ = – 2µρ2

Note that µρ2 = I, the moment of inertia about the center of mass. k2 I There are three states with the same energy, corresponding to magnetic quantum numbers m = +1,0,-1. The factor eiø in this wave function shows that it belongs to m = +1.

E=

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

83

6.11 From Eqs. (6.23) and (6.24) in Sec. (6.5.1) using the eigenvalues E1 = 8, E2 = –5, S 12 –4 = 6i S 2i 11 S 12 = S 11 3 S 22 9 = 6i S 21 2i S 21 = S 22 3 The S matrix becomes

S 11 2iS 22 3

Its determinant is

2i S 3 11 S 22

4 S 11S 22 + S 11S 22 = 1 9 9 S 11S 22 = 13

9 . For example, take Take any values for S 11 and S 22 as long as their product is 13 S 11 = 1. To normalize the determinant to 1, multiply the remaining elements by 913

1 S = 18i

39

As a check, take the determinant.

9 13

+

36 (3)(39)

=1

2i 3 9 13

84

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

6.10 To diagonalize the matrix use a similarity transformation with matrix S .

S

0 1 –1 d1 S = 10 0

0 d2

S 11 S12 S 21 S 22

01 d1 1 0 = 0

d2

S1 S 21

d1S 11

d1S 12

d2S 21

d2S 22

S 12

S 11

S 22

S 21

=

0

S 12 S 22

S 12 = d1S 11 S 11 = d1S 12 S 22 = d2S 21 S 21 = d2S 22 The solutions d2 = 1, d2 = 1 follow directly. The diagonalized matrix with distinct 1

2

d1, d2 is 1 0 0 –1 The distinct irreducible representations of T are therefore (1) and (-1).

6.13 a) To diagonalize the matrix use a similarity transformation with matrix S .

S S 11 S12 S21 S 22

2 –1 0 1

S –1 =

d1

2 –1 0 1=

d1

0

0 d2

0

0 d2

S1 S 21

S 12 S 22

continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

85

Equating corresponding matrix elements gives the equations (2 – d)S 11 = 0 –S 11 + (1 – d)S 12 = 0 (2 – d)S 21 = 0 –S 21 + (1 – d)S 22 = 0 The determinant of coefficients is 2–d –1

0 1–d

The secular equation is (2 - d)(1 - d) = 0. The roots are obviously 2, 1. The diagonalized matrix is therefore

2 0 0 1

(1)

b) The trace of a matrix is the sum of its diagonal elements. The original matrix has trace 2 + 1 = 3, and the diagonalized matrix in Eq. (1) has trace 2 + 1 = 3.

Section (2.6) gives a general proof that the trace is invariant under a similarity transformation.

6.14 Consider the wave function 1 |ΨAB) = , (|↓)A ⊗ |†B) – |†)A ⊗ |↓)B) 2 It is convenient to switch from ket notation to matrices. |†)A =

a 0

|↓)A =

0 a

|†)B =

b 0

|↓)B =

(1)

0 b

continued next page =→

86

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

|ΨAB) =

1

0

, ⊗ 2

b

a

– a

Ψ1 1 Ψ2 = , Ψ3 2 Ψ4

⊗ 0

0 0

!

0

0 ab0

1

ab 0

–,

b

2 0 0

0 1 –ab = , 2 ab 0 Using the test for entanglement from Sec. (6.7.1) Ψ1Ψ4 – Ψ2Ψ3 = (ab)2 ≠ 0 The nonzero result shows that Eq. (1) is entangled. In the entangled state, a measurement of state A affects the state B.

6.15 Consider the Kronecker product of the 2 × 2 matrices S and T . The result is a 4 × 4 matrix. Using the definitions in Sec. (2.8), S ⊗T =

a b

⊗

A

B

c d

C D

=

aA aB aC aD cA cB cC cD

bA bC dA dC

=

aT bT cT dT

bB bD dB dD

The last step follows because as discussed in Sec. (1.6.1) if matrix B is a multiple of matrix A every element of A is multiplied by the factor.

B = cA bi j = cai j Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

WAVE MECHANICS, MEASUREMENT, AND ENTANGLEMENT

87

6.16 Consider the Kronecker product of the 2 × 2 matrices A and B. The result is a 4 × 4 matrix. Using the definitions in Sec. (2.8), 12 6 8 4 A⊗B=

3 2 4 2 ⊗ = 2 1 31

9 8

3 6 2 4 4 2

6

2 3 1

12 8 6 4 B⊗A=

A ⊗ B ≠ B ⊗ A.

4 2 3 2 ⊗ = 3 1 21

8 9

4 4 2 6 3 2

6

3 2 1

7.1 From Eq. (7.1) with θ = 60◦ AJx AJ y AJz

0.500 –0.866 0 = 0.866 0.500 0 0

2.000 0 –4.000 1 3.000

4.464 0.268 – 3.000

= AJ = 4.464 ˆi – 0.268 ˆj + 3.000 k̂

7.2 From Eq. (7.1) with θ = –30◦ AJx AJy J

Az

0.866 0.500 0 2.000 = –0.500 0.866 0 –4.000 0 0 1 3.000 =

–0.268 –4.464 3.000

AJ = –0.268 ˆi – 4.464 ˆj + 3.000 k̂

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

ROTATION

89

7.3

The sketch shows axes rotated clockwise about x. Because this is a righthanded coordinate system, the x axis points out of the paper. Using vector analysis ˆjJ = cos θ ˆj – sin θ k̂ k̂ J = sin θ ˆj + cos θ k̂ The matrix for clockwise rotation of the axes about x is therefore ˆi ˆiJ 1 0 0 J ˆj = 0 cos θ – sin θ j ˆ 0 sin θ

kĴ

cos θ

(1)

k̂

The matrix for counterclockwise rotation of a vector A (equivalently, counterclockwise rotation of points x, y, z ) about x is identical to the matrix Eq. (1).

AJx

1

0

0

A y = 0 cos θ – sin θ AJz 0 sin θ cos θ J

Ax Ay Az

7.4

The sketch shows axes rotated clockwise about y. Because this is a righthanded coordinate system, the y axis points into the paper. Using vector analysis ˆiJ = cos θ ˆi + sin θ k̂ k̂ J = – sin θ ˆi + cos θ k̂ continued next page =→

90

ROTATION

The matrix for clockwise rotation of the axes about y is therefore ˆiJ ˆjJ =

cos θ 0

0 sin θ 1 0

k̂J

– sin θ

0 cos θ

ˆi

(1)

ˆj k̂

The matrix for counterclockwise rotation of a vector A (equivalently, counterclockwise rotation of points x, y, z ) about y is identical to the matrix Eq. (1).

AJx AJy =

cos θ 0

0 sin θ 1 0

AJz

– sin θ

0 cos θ

Ax Ay Az

7.5 The matrix for rotation about z is cos θ – sin θ 0 sin θ cos θ 0

(1)

0 0 1 A unit vector along the line y = x has its endpoint at x = about z by 45◦ changes the end point coordinates to xJ, yJ:

,1 , y = 2

,1 . Rotation 2

cos θ – sin θ 0 x yJ = sin θ cos θ 0 y zJ 0 0 1 z xJ

=

1 , 2 1 , 2

0

1

– ,2 0 1 , 2

0

0 1

1 2 1 , 2

,

z

which gives xJ = 0, yJ = 1, zJ = z. Thus the endpoint of the unit vector is now along the y axis.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

ROTATION

91

7.6

Because we want to express x, y in terms of xJ, yJ it is convenient to use the inverse of the rotation matrix. As proved in Sec. (7.2.3) the inverse of a rotation matrix is its transpose. cos θ R–1(θ, z) = – sin θ 0

sin θ 0 cos θ 0 0 1

Taking θ = 90◦ x 0 1 0 xJ y = –1 0 0 yJ z 0 0 1 zJ which gives x = yJ y = –xJ z = zJ The equation for the rotated hyperbola is thus yJ2 2 – xJ = 1 4

7.7 From Sec. (7.6), the Pauli matrices are σ1 =

0 1 10

σ2 =

0 –i i 0

σ3 =

1

0

0 –1

There are two reasons why the three Pauli matrices do not form a group. First, there is no identity element. Second, the product of a Pauli matrix with itself is the identity, which is not a member of the Pauli matrix set – the products of members of a group must always give a member of the group.

92

ROTATION

7.8 From Sec. (7.6), the Pauli matrices are σ1 =

01 1 0

0 –i

σ2 =

0

i

1

σ3 =

0

0 –1

a) From Sec. (1.7.4) a Hermitian matrix S is a matrix that is equal to its adjoint (its transpose complex conjugate) S = S0 †.–i 0 1 1 0 † σ= = σ1 σ† = = σ2 σ† = = σ3 1

2

10

b) σ2 = 1 2

σ=

01

01

σ23 =

i

0 1 0 0 –1

01 1 0

= 0

01

1 0 = 0 –1

1 0 01

i

0 –1

10

=

1 0 1 0 0 –i 0 –i

2

3

0

i

7.9 From Sec. (7.6), the Pauli matrices are σ1 =

01 1 0

σ2 =

0 –i i

0

σ3 =

1

0

0 –1

Prove σiσj = iPσk where the subscripts i, j, k are all different. P is +1 if the subscripts are an even permutation and -1 if the permutation is odd, as defined in Sec. (1.4.3).

There are six cases to consider. continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

ROTATION

σ1σ2 = σ1σ3 = σ2σ1 =

0 1

0 –i 1 0

0

i

0 1

1 1 0

0 1 0

i σ2σ3 =

1 0

0 –i i

σ3σ1 = σ3σ2 =

0

1 0 1 0

0 –1 0 1

0 –1 1 0

1 0 0 –i

0 –1

1 0

(1 3 2) is odd

0

= –iσ3

0 –1 0 1 10

0 –1 0 0 1

(2 1 3) is odd (2 3 1) is even

= +iσ1

1

= –i

(1 2 3) is even

= +iσ3

= –iσ2

1

= +i

=

0

i

0 –1

= –i 0

0 0 –1

=

0 –1

0 –i

1

= +i

93

(3 1 2) is even

= +iσ2

= –iσ1 10

(3 2 1) is odd

7.10 From Sec. (7.6) the Pauli matrices are σx =

01 1 0

k2 2 σS = 4

σy =

0 –i 0

i

1 0 0 –1

σz =

σx ˆi + σy jˆ + σz kˆ · σx ˆi + σy jˆ + σz kˆ

Keep in mind that a constant times a matrix multiplies every element of the matrix.

Note also that the square of any Pauli matrix is the identity matrix.

k2

σ2S =

4

0

0 2

k 4

k2

+

4

0

0 2

k 4

k2

+

4

0

0 2

k 4

3k2 4

=

0 2

0 3k 2 4

The matrix is already diagonal so the eigenvalues are 3k . This result has the form 4

m(m + 1)k2 → 21

3 2

k2.

94

ROTATION

7.11 This problem is another illustration of how a 2-dimensional SU(2) matrix can give rise to a 3-dimensional physical rotation matrix. From Eq. (7.12) cos 2( β ) – sin ( β )2 Uy(β) = sin ( β2)

cos ( β2)

Uy is a real rotation matrix so its inverse is just the transpose (as can be verified by calculation of the product UyUy–1).

Uy–1(β) =

cos ( β2 )

sin ( β2 )

– sin ( β2 ) cos ( 2β ) V =

z x – iy x + iy –z

Perform a similarity transformation of V by Uy. VJ = UyVUy–1 For convenience let s ≡ sin ( β ) and c ≡ cos ( β ).

2 2 zJ xJ – iyJ c –s = z x – iy c xJ + iyJ

–zJ

s c =

–z

x + iy

s

(1)

–s c

(c2 – s2)z – 2(sc)x 2

2

(2sc)z + (c2 – s2)x – i(c2 + s2)y

2

2

2

(2sc)z + (c – s )x + i(c + s )y

2

–(c – s )z + 2(sc)x (2)

Use the trigonometric identities c2 + s2 = cos2 ( β ) + sin2 ( β ) = 1

2 2 β β c2 – s2 = cos2 ( 2) – sin2 ( 2 ) = cos β (2sc) = 2 sin ( β ) cos ( β ) = sin β 2

2

With these identities the right-hand side of Eq. (2) becomes (cos β) z – (sin β) x (sin β) z + (cos β) x + iy

(sin β) z + (cos β) x – iy –(cos β) z + (sin β) x

(3)

Note that Eq. (3) does not refer to half-angles. continued next page =→

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

ROTATION

95

The object is to express xJ, yJ, zJ in terms of x, y, z. Set corresponding elements equal between matrix VJ on the left-hand side of (1) and matrix (3) to give zJ = (cos β) z – (sin β) x

(4)

xJ – iyJ = (sin β) z + (cos β) x – iy

(5)

xJ + iyJ = (sin β) z + (cos β) x + iy

(6)

–zJ = –(cos β) z + (sin β) x

(7)

Equations (4) and (7) are identical and give zJ = –(sin β) x + (cos β) z Adding Eqs. (5) and (6) gives xJ = (cos β) x + (sin β) z Subtracting Eqs. (5) and (6) gives yJ = y In matrix form the final result is xJ yJ = zJ The 3 × 3 matrix is R(β, y).

cos β 0

0 sin β 1 0

– sin β 0 cos β

x y z

96

ROTATION

7.11 The matrix in this problem is k2 times the Pauli matrix σ2.

0

–i k

ik

0

2

2

Let d be an eigenvalue. The secular equation is then det

–d –ik2 i k2

–d

=0 d2 =

k2

4 k d= ± 2

The distinctive diagonalized matrix is k 2

0

0 – k2

As we shall see in Chapter 8 the eigenvalues ±k2express directional quantization of a spin- 12 particle. The spin angular momentum can only point either “up” or “down”, as shown by the matrix symbolism up –→

k 2

0

down –→

0 k

2

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

8.1 From Sec. (8.3) ! !

A2 B2 e e = 1 – iA – + . . . 1 – iB – +...

2 2 B2 A2 – iA – AB – = 1 – iB – +...

2 B2 2 2 A = 1 – i(A + B) – AB – – 2 2 2 (A + B) e–i(A+B) = 1 – i(A + B) – + ...

2 A2 B2 (AB + BA) + . . .

– = 1 – i(A + B) – – 2 2 2 Equations (1) and (2) do not agree unless AB = BA. –iA –iB

(1)

(2)

8.2 Here are two ways of showing that quantum mechanical angular momentum operators are Hermitian. Both methods use the definition that an operator S is Hermitian if it is equal to its transpose complex conjugate S † = S . continued next page =→

98

QUANTUM ANGULAR MOMENTUM

The first method uses the matrix representations in Sec. (8.4).

0 0 0 Ix = k

0 I˜x = k

0 0 –i 0 i 0

0 0 i Iy = k 0 0 0

0

0 0 i 0 –i 0

0 0 –i I˜y = k 0 0 0

–i 0 0

0

i

0 –i 0 Iz = k i 0 0 0 0

0

I˜z = k

0

0

0 i 0 –i 0 0 0

0 0

0 0 0 ∗ † ˜ I =I =k 0 0 i =I x

x

x

0 –i 0 0

0 i

I˜y∗ = Iy† = k

0

0 0 –i 0 0

Iz∗ = Iz† = k

0 –i 0 i 0 0 = Iz

˜

0 0

= Iy

0

The second method uses matrix mechanics. Recall that in matrix mechanics coordinate operators and linear momentum operators are Hermitian and that a coordinate commutes with linear momentum except its conjugate linear momentum. Also, the transpose of a product is the product of the transposes in reverse order.

Ix = qypz – qzpy

Ix† = pz†qy† – py†qz† = pzqy – pyqz = qypz – qzpy = Ix

Iy = qzpx – qxpz

Iy† = p†xqz† – pz†q†x = pxqz – pzqx = qzpx – qxpz = Iy

Iz = qxpy – qypx

Iz† = py†q†x – p†xqy† = pyqx – pxqy = qxpy – qypx = Iz

8.3 Consider the first-order approximation to U = eiθG. U ≈ 1 + iθG

(1)

Multiply Eq. (1) by H from the left.

HU ≈ H + iθHG Multiply Eq. (1) by H from the right.

UH ≈ H + iθGH

(2)

Subtract Eqs. (1) and (2).

HU – UH = iθ (HG – GH) continued next page =→

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QUANTUM ANGULAR MOMENTUM

99

Because HU = UH the left-hand side is 0. Therefore HG = GH As shown in Sec. (5.4.3) an operator that commutes with the Hamiltonian is constant in time – it is a conserved quantity. An important consequence for quantum mechanics is that such a commuting operator is measurable and can be labeled with a quantum number.

8.4 An operator S is Hermitian if it is equal to its transpose complex conjugate S † = S . Consider the ladder operators I±. By definition I± = Ix ± iIy The angular momentum operators Ix and Iy are Hermitian. A Hermitian operator changes the sign of i.

I+† = Ix† + (–i)Iy† = Ix – iIy ≠ I+ I–† = I†x – (–i)I†y = Ix + iIy ≠ I– The ladder operators are not Hermitian.

100

QUANTUM ANGULAR MOMENTUM

8.5 Factors of k have been omitted. I2 = I2 + I2 + I2. Using the matrix representations x

y

z

from Sec. (8.4) I2 =

0 0 i 0 0 i 0 0 0 0 0 0 0 –i 0 0 0 –i 0 0 –i + 0 0 0 0 0 0 + i 0 0 0 i 0 0 i 0 –i 0 0 –i 0 0 0 0 0

0 –i 0 i 0 0 0 0 0

0 0 0 1 0 0 1 0 0 = 0 1 0 + 0 0 0 + 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 = 2 0 1 0 = 2E 0 0 1

where E is the identity matrix. Because I±E = I± [I2, I±] = I± – I± = 0

8.6 Let d be an eigenvalue. The secular equation is –d det 0 0

0 –d i

0 –i = –d(d2 – 1) = 0 –d

(1)

The roots of the secular equation (1) are d = 1, 0, –1, which are the magnetic quantum numbers for angular momentum j = 1.

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QUANTUM ANGULAR MOMENTUM

101

8.7 a) The dimension of a representation matrix D( j) is 2 j + 1.

2j +1 = 7 Hence j = 3. b) The magnetic quantum number for j = 3 has possible values 3, 2, 1, 0, –1, –2, –3.

Note that there are seven values, equal to the dimension of D( j).

8.8 a) The dimension of a representation matrix D( j) is 2 j + 1.

2j +1 = 4 Hence j = 32 . b) The magnetic quantum number for j = 3 has possible values 3 , 1 , –1 , –3 . Note 2

that there are four values, equal to the dimension of D( j).

2 2

2

2

8.9 From Eqs. (8.27) and (8.28) , I+ | j m) = j( j + 1) – m(m + 1) | j m + 1) , I– | j m) = j( j + 1) – m(m – 1) | j m – 1) 1 Ix | j m) = (I + I ) | j m) – 2 + , 1 , = j( j + 1) – m(m + 1) | j m + 1) + j( j + 1) – m(m – 1) | j m – 1) 2

102

QUANTUM ANGULAR MOMENTUM

8.10 From Eqs. (8.27) and (8.28) , I+ | j m) = j( j + 1) – m(m + 1) | j m + 1) , I– | j m) = j( j + 1) – m(m – 1) | j m – 1) i Iy | j m) = – (I – I ) | j m) – 2 + , , i =– j( j + 1) – m(m + 1) | j m + 1) – j( j + 1) – m(m – 1) | j m – 1) 2

8.11 Let N be a normalization factor omitting factors of k. Using Eq. (8.14) ! b 1 b I = eiø + + bθ tan θ bø I+u10 = NI+ cos θ iø

a –e sin θ a Y11

8.12 To calculate uJ1 1, the rotated version of u1 1, we need only the first row of D(1).

Using material from Sec. (8.7.1) uJ 11 ··· =

e–iα 1+cos β e–iγ 2

···

–e–iα sin β, e–iα 1–cos β eiγ ···

2

2

···

u1 1 u1 0

···

u1–1 ··· ··· ··· ! ! ! sin β –iα 1 – cos β J –iγ –iα –iα 1 + cos β e u1 1 – e u1 0 + e eiγu1 –1 u11 = e , 2 2 2 Set γ = 0, β = u = 45◦, α = u = 90◦. Equation (1) becomes , 4 2 , 2–1 –i u 2+1 u uJ = –i i + u 1 –1 1 0 11 , , 11 2 2 2 2 2

(1)

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QUANTUM ANGULAR MOMENTUM

103

8.13 From Eq. (8.25) I2 = Iz2 + Iz + I–I+ I 2 u j m = Iz2 u j m + Izuj m + I–I+uj m From Eqs. (8.29) and (8.27)

I 2 u j m = m2uj m + muj m +

,

j( j + 1) – m(m + 1) I–uj m+1

From Eq. (8.28) , , I 2 u j m = m(m + 1)uj m + j( j + 1) – m(m + 1) j( j + 1) – m(m + 1) u j m = m(m + 1) + j( j + 1) – m(m + 1) u j m = j( j + 1) u j m

8.14 Proceeding from Y2–2(θ, ø) to Y2–1(θ, ø) requires one I+ step in Eq. (8.35). With 4 = 2 and using Eq. (8.13) ! i b 2 b + sin2 θ e–2iø I+ sin θ e–iø = eiø ! ! bθ tan θ bø i sin2 θ iø iø iø 2 2 e 2 sin cos e ( 2i) e – – = θ θ + – tan θ = 4e–iø sin θ cos θ a Y2–1(θ, ø) in agreement with listings of spherical harmonics.

104

QUANTUM ANGULAR MOMENTUM

8.10 Proceeding from Y3–3(θ, ø) to Y3–2(θ, ø) requires one I+ step in Eq. (8.35). With 4 = 3 and using Eq. (8.13) ! i b 3 b i i + sin3 θ e–3iø I+ sin θ e– ø = e ø ! ! tan θ bø 3 bθ i sin θ iø i i e 3 sin2 cos e 3 ø ( 3i) e 3 ø – = θ θ – + – tan θ = e–2iø 3 sin2 θ cos θ + 3 sin2 θ cos θ = 6e–2iø sin2 θ cos θ a Y3–2(θ, ø) in agreement with listings of spherical harmonics.

8.16 ! a) Expand e–iøIz in Taylor’s series.

2 (–iøI ) z i I e– ø jm 1 ( i I) jm z | )= + –ø z + +... | ) 2!

From Eq. (8.29) Iz | jm) = m | jm) so Iz is replaced by m everywhere it appears in the expansion. e–iøI z

jm 1 ( i m) | )= + –ø +

(–iøm)2 2!

! jm +... | )

= e–imø | jm)

b) Using the result from a), ( jmJ|e–iøIz | jm) = ( jmJ|e–imø| jm) e–imø is not an operator and can be removed from the matrix element.

( jmJ|e–iøIz | jm) = e–imø ( jmJ| jm) = δmJme–imø by orthogonality of the | jm).

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

QUANTUM ANGULAR MOMENTUM

105

8.13 Magnetic quantum numbers add directly. For a spin 12 state there are 2 possible values of the magnetic quantum number2 1 , –21 with maximum value m = 12 . For a spin 1 state there are 3 values of the magnetic quantum number 1, 0, –1, with maximum mJ = 1. Together there are 6 possible sums of magnetic quantum numbers. The maximum possible sum is m + mJ = 21 + 1 = 3 2, which can arise from an angular momentum state j = 3 . For j = 3 there are 4 magnetic quantum num-

2 2 1 3 3 1 bers 2 , 2 , –2 , – 2 . The remaining states arise from j = 12, with 2 magnetic quantum numbers 21 , – 12 .

8.18 1

Table (F.2) in Appendix F lists the 3-j1 coefficients for the combinations D (1) ⊗D( 2). ( )

Reading down the column labeled W 2

–1/2

1 ( )

2 = W –1/2

,

gives

, 1 1 1 u(1)v( 2 ) 2 u(1)v( 2 ) 3 0 –1/2 – 3 –1 1/2

As a check, note that the magnetic quantum number subscripts all sum to –1/2.

8.19 Table (F.3) in Appendix F lists the 3-j coefficients for the combinations D(1) ⊗D(1).

Reading across the row labeled u1 (1)0v(1) gives , , (1) (1) (2) 1 u1 v0 = 2 W1 + 12 W(1) 1

8.20 Table (F.3) in Appendix F lists the 3-j coefficients for the combinations D(1) ⊗D(1).

Reading across the row labeled u0 (1)1v(1) gives , , u0(1)v1(1) = 12 W1(2) – 12 W(1) 1

106

QUANTUM ANGULAR MOMENTUM

8.10

Possible values of j are 3, 2, 1 as shown in the vector model sketches.

Further discussion may be instructive. Consider a system of two bound particles 1 and 2. Particle 1 has spin function u( j1) and particle 2 has spin function v( j2).

The spin function of the system as a whole is W( j). As discussed in Sec. (8.10) and following the W( j) spin functions can be expressed in terms of the uv pairs multiplied by appropriate 3-j coefficients. For j1 = 2 there are (2 j1 + 1) = 5 spin states and for j2 = 1 there are (2 j2 + 1) = 3, making a total of (5)(3) = 15 pairs.

The possible spin states of W( j) must total the same.

X (2 j1 + 1)(2 j2 + 1) = (2 j + 1) (1) j

The sum is over all possible values of j. For j = 3, 2, 1 the sum is 7 + 5 + 3 = 15 in agreement with the number of pairs and also in agreement with the vector model.

The left-hand side of Eq. (1) is a fixed number, so if Eq. (1) is not satisfied there is an error in the assumed values of j.

8.22

Possible values of j are 2, 1 as shown in the vector model sketches.

Further discussion may be instructive. Consider a system of two bound particles 1 and 2. Particle 1 has spin function u( j1) and particle 2 has spin function v( j2). The spin function of the system as a whole is W( j). continued next page =→

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QUANTUM ANGULAR MOMENTUM

107

As discussed in Sec. (8.10) and following the W( j) spin functions can be expressed in terms of the uv pairs multiplied by appropriate 3-j coefficients. For j1 = 32 there are (2 j1 + 1) = 4 spin states and for j2 = 21 there are (2 j2 + 1) = 2, making a total of (4)(2) = 8 pairs. The possible spin states of W( j) must total the same.

X (2 j1 + 1)(2 j2 + 1) = (2 j + 1) (1) j

The sum is over all possible values of j. For j = 2, 1 the sum is 5 + 3 = 8 in agreement with the number of pairs and also in agreement with the vector model.

The left-hand side of Eq. (1) is a fixed number, so if Eq. (1) is not satisfied there is an error in the assumed values of j.

8.23 J

Consider a two-particle system with spin functions u( j) and v( j ) for the individm mJ J) ( ual particles and spin function WM for the system as a whole. They satisfy the

orthogonality relation Eq. (8.43): X ( j jJ m M – m|J M)( j jJ m MJ – m|JJ MJ) = δJJJ δMMJ

(8.43)

m 1

Table (F.2) is for D (1) ⊗ D( 2 ) so take j = 1 and jJ = 12. m can have three values 1, 0 ,-1. Note that J, M, JJ, MJ are constants. J and JJ can have possible values 3 or 1 .

2

2

Case 1 : Let J = JJ = 3 and M = MJ = 1 . There are three terms m = 1, 0, –1 in the 2 2 sum. Note that the third term does not contribute to the sum because jJ = 12 cannot have mJ = M – m = 1 – (–1) = 3 .

2

2

X

1

= 1 12 1 – 12 32 2 1 12 1 – 12 32 12

m

+ 1 21 0 12 32 21 1 21 0 12 32 21 + 1 21 –1 32 32 21 1 21 –1 32 32 12 , , , , 1 1 2 2 1 2 = + = + =1 3 3 3 3 3 3 in agreement with δJJJ δMMJ = δ JJ δ MM = 1. continued next page =→

108

QUANTUM ANGULAR MOMENTUM

Case 2 : Let J = 3 ,2 JJ = 1 2and M = MJ = 1 . There are three terms m = 1, 0, –1 2 in the sum. Note that the third term in the sum does not contribute because because jJ = 1 cannot have mJ = M – m = 1 – (–1) = 3 .

2

2

X

2

1

= 1 12 1 – 12 32 2 1 12 1 – 12 12 12

m

+ 1 21 0 12 32 21 1 21 0 12 12 21 + 1 21 –1 32 32 21 1 21 –1 32 12 12 , , , , 1 2 2 1 = 3 3– 3 3=0 in agreement with δJJJ = 0 for J ≠ JJ.

8.24 J

Consider a two-particle system with spin functions u( j) and v( j ) for the individm1 m2 J) ( ual particles and spin function WM for the system as a whole. They satisfy the

orthogonality relation Eq. (8.44): X ( j jJm1m2 |JM)( j jJmJ1mJ2|JM) = δm1mJ1 δm2mJ2

(8.44)

J,M

Table (F.3) is for D(1) ⊗ D(1) so j = jJ = 1. J can have possible values 2, 1, 0.

Case 1 : Let J = 2. m1 = mJ1 = 1 and m2 = mJ2 = –1. There is only one value M = m1 + m2 = 0.

X

= (1 1 1 – 1| 2 0) (1 1 1 – 1| 2 0)

J,M

+ (1 1 1 – 1| 1 0) (1 1 1 – 1| 1 0) + (1 1 1 – 1| 0 0) (1 1 1 – 1| 0 0) , , , , , , 1 1 1 1 1 1 = + + 6 6 2 2 3 3 1 1 1 = + + =1 6 2 3 in agreement with δm1m1 δm2m2 = 1. continued next page =→

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QUANTUM ANGULAR MOMENTUM

109

Case 2 : Let m1 = 1, mJ1 = 0 and m2 = –1, mJ2 = 0. Note that m1, mJ1, m2, mJ2, j, jJ, are constants in the sum. J has three values 2, 1. 0 and M has two values 0, 1, giving 6 possible terms in the sum. However, terms with M = 1 do not contribute because M = m1 + m2 = 0, M = mJ1 + mJ2 = 0, leaving only the three terms with M = 0.

X = (1 1 1 – 1| 2 0) (1 1 0 0| 2 0) J,M

+ (1 1 1 – 1| 1 0) (1 1 0 0| 1 0) + (1 , 1 1, – 1| 0 0) ,(1 1 0,0| 0,0) 1 2 1 1 1 = + (0) – =0 6 3 2 3 3 J in agreement with δm1mJ1 = 0 for m1 ≠ m 1 and δm2mJ2 = 0 for m2 ≠ mJ2.

8.25 One approach is to apply selection rules to the proposed electric dipole transition Y1 –→ Y0.

2

1

rule ∆4 = ±1

2 –→ 1 ∆4 = –1

rule ∆m = 0, ±1

rule is satisfied

1 –→ 0 ∆m = –1

rule is satisfied

The transition obeys the selection rules and is allowed.

Another approach is the spherical harmonics method illustrated in Sec. (8.12.1).

The matrix element „ . m∗4 JJ Y (θ, ø) (x, y, z)Ym4 (θ, ø) dr expresses an electric dipole transition between states 4, m and 4J, mJ. The spherical harmonics are orthogonal with respect to 4 and to m: , 2u , u J Y4mJ ∗(θ, ø)Y4m(θ, ø) sin θ dθ dø = δmmJ δ44J 0

0

Omitting the factor r that plays no role in rotation, x a cos ø sin θ

y a sin ø sin θ

z a cos θ

x ± iy a e±iø sin θ

Consider the matrix element of x + iy = eiø sin θ and use eiø sin θ Y0 a Y1.

1 2 „ „ Y 1∗ eıø sin θ Y10 dr a Y 1∗ Y12 dr ≠ 0 2 2 The electric dipole transition with the matrix element of x + iy is allowed.

110

QUANTUM ANGULAR MOMENTUM

8.26 Let the initial wave function be ψi and let the final wave function be ψf . ψ is the product of a space function u and spin wave function v. For l = 1, m = 0 and ms = 12, ψi = u(1) v(s) 0 1/2 where s must be half odd integer = 1 , 3 , . . . .

2 2

Note that the space operator Lz operates only on u and the spin operator S z operates only on v.

Lzψi = Lzu(1) v(s) 0 1/2 (s) = (0)u0(1)v1/2 (s) 2S zψi = 2u0(1) S zv1/2 (s) = 2u0(1) 21 v1/2 (Lz + 2S z) ψi = u(1)v(s) 0

1/2

The matrix element for a z-component magnetic dipole transition is , , ∗ ψ∗f (Lz + 2S z ) ψi dr = u(lmJ)v(smJs) u(1) v(s) 0 1/2 dr J

J

(1)

The angular part of u is a spherical harmonic. Spherical harmonics satisfy the orthogonality relation , Ylm∗J Yml dr = δllJ δmmJ J

It follows that in this problem the matrix element for the z-component magnetic dipole transition in Eq. (1) is ≠ 0 only if lJ = l = 1, mJ = m = 0. Similarly, the summation over spin functions in Eq. (1) is ≠ 0 only if mJs = mJ = 12. Thus for the z-component magnetic dipole transition ∆l = 0, ∆m = 0, ∆ms = 0.

As can be inferred from Eqs. (8.27) and (8.28), matrix elements for x-component and y-component transitions lead to the additional selection rules ∆m = ±1 and ∆ms = ±1 for allowed transitions. The selection rule ∆l = 0 holds for all components.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

QUANTUM ANGULAR MOMENTUM

111

8.27 a) The terms x2, y2, z2 do not change sign under inversion x → –x, y → –y, z → –z so Q has even parity. b) The matrix element (ψJ|T|ψ) for a transition with operator T is 0 if it changes sign under inversion. The electric quadrupole moment Q does not change sign under inversion, so its matrix element will not change sign if the wave functions ψ, ψJ have the same parity – both even (+) or both odd (-).

8.28 The electric quadrupole operator Q is a second-rank tensor with k = 2 and q = 2, 1, 0, –1, –2. According to the Wigner-Eckart theorem Sec. (8.12.2) the matrix elements of Q are proportional to the 3-j coefficient (|Q|) a ( jJ2 mJq| jm)

(1)

The 3-j coefficient in Eq. (1) is 0 unless mJ + q = m. It follows that ∆mj = q = 0, ±1, ±2 A second requirement is that the 3-j coefficient in Eq. (1) is 0 unless jJ, j, 2 satisfy the triangle rule. It follows from the triangle rule that ∆ j = 0, ±1, ±2 There are exceptions to the ∆ j selection rule because if j or jJ are too small the triangle rule with k = 2 cannot be satisfied. j = 0 → jJ = 0 not allowed 1 1 j = → jJ = not allowed 2 2 j = 1 → jJ = 0 or jJ = 1 → j = 0 not allowed

9.1 a) According to Eqs. (9.4) and (9.5) the solutions for x and y are x = Ax (sin (ωt) + ø) y = Ay (cos (ωt) + ø) x2 + y2 = A2 + A2 = A2 x

y

The radius r = A of the motion is constant. Consider the case ø = 0. It follows that r = x ˆi + y ˆj dr dx ˆ dy ˆ dt = dt i + dt j = rω cos (ωt) ˆi – sin (ωt) ˆj r·

dr = rω sin (ωt) cos (ωt) – cos (ωt) sin (ωt) dt =0

This result shows that the tangential velocity ddtr is perpendicular to the radius vector r as expected for circular motion. continued next page =→

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THE STRUCTURE OF ATOMS

113

b) The particle is in a circular orbit in the x – y plane so its magnetic moment is in the +z (ccw) or –z (cw) direction. For ccw motion the particle’s magnetic moment is parallel to B k̂ so the magnetic energy µ · B is maximum – thus the ccw motion corresponds to the higher energy (higher frequency) ω+ as developed in Lorentz’s classical theory Sec. (9.2.1). The electron cw motion has magnetic moment antiparallel to B k̂ and the magnetic energy is minimum – thus the cw motion corresponds to the lower frequency ω–. c) As observed perpendicular to the magnetic field B k̂ the electron oscillates along k̂ with the unshifted frequency ω0 and oscillates in the x – y plane with the higher and lower frequencies ω±.

9.2 a) From Eq. (9.15) the Lande´ g-factor is J(J + 1) – L(L + 1) + S (S + 1) 2J(J + 1) 2J(J + 1) J(J + 1) – L(L + 1) + S (S + 1) = + 2J(J + 1) 2J(J + 1) 3 –L(L + 1) + S (S + 1) = + 2 2J(J + 1) 2 3 –L – L + S 2 + S = + 2 2J(J + 1) 3 S – L + S 2 – L2 = + 2 2J(J + 1) 3 S – L + (S – L)(S + L) = + 2 2J(J + 1) 3 (S – L)(S + L + 1) = + 2 2J(J + 1)

gJ = 1 +

b) If L = S then gJ = 3 2for all allowed values of J. The sum over J ranges from J = L – L = 0 to J = L + L = 2L. There are 2L + 1 terms each equal to 32.

X 2L 0

gJ =

3

(2L + 1) 2 continued next page =→

114

THE STRUCTURE OF ATOMS

c) For L = 3 and S = 1, J has three possible values 2, 3, 4.

J=2 3 (1 – 3)(1 + 3 + 1) 2 g2 = + = 2 4(3) 3

J=3 g3 =

3 (1 – 3)(1 + 3 + 1) 13 + = 2 6(4) 12

J=4 3 (1 – 3)(1 + 3 + 1) 5 + = 2 8(5) 4 The sum is 8 + 13 + 15 = 3. Pauli showed that in general the sum for L > S is g4 =

12

12

12

2S + 1, in agreement with this example where S = 1.

9.3 The Lande´ g-factor Eq. (9.15) is gJ = 1 +

J(J + 1) – L(L + 1) + S (S + 1) 2J(J + 1)

If S = 0, then J = L + S = L and the g-factor becomes gJ = 1 +

L(L + 1) – L(L + 1) =1 2L(L + 1)

9.4 The notation for a term symbol has the form 2S +1LJ. a) For 1S 0, 2S + 1 = 1 so S = 0. The symbol S means L = 0. Hence J = L + S = 0 as shown in the lower right subscript. b) For 2S 1/2, 2S +1 = 2 so S = 1 . The symbol S means L = 0. Hence J = L +S = 1 2

2

as shown in the lower right subscript. c) For 5D4, 2S + 1 = 5 so S = 2. The symbol D means L = 2. According to the vector model J could be 4, 3, 2, 1, 0. J = 4 in the lower right subscript agrees with the possible value 4.

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THE STRUCTURE OF ATOMS

115

9.5 The notation for a term symbol has the form 2S +1LJ. a) For 1S 0, 2S + 1 = 1 so S = 0. The symbol S means L = 0. Hence J = L + S = 0 as shown in the lower right subscript. b) For 2P1/2, 2S + 1 = 2 so S = 1 2. The symbol P means L = 1. According to the vector model J = L + S could be J = 3 or 1 . J = 1 in the lower right subscript 2 2 2 1 agrees with the possible value 2. c) For 3D3, 2S + 1 = 3 so S = 1. The symbol D means L = 2. According to the

vector model J could be J = 3, 2, 1. J = 3 in the lower right subscript agrees with the possible value 3.

9.6 The notation for a term symbol has the form 2S +1LJ. a) For 5D4, 2S + 1 = 5 so S = 2. The symbol D means L = 2. J = 4 according to the lower right subscript. The Lande´ g-factor Eq. (9.15) is J(J + 1) – L(L + 1) + S (S + 1) 2J(J + 1) (4)(5) – (2)(3) + (2)(3) = 1+ 2(4)(5) 1 3 =1+ = 2 2 b) For 3F2, 2S + 1 = 3 so S = 1. The symbol F means L = 3. J = 2 according to the lower right subscript. The Lande´ g-factor Eq. (9.15) is gJ = 1 +

J(J + 1) – L(L + 1) + S (S + 1) 2J(J + 1) (2)(3) – (3)(4) + (1)(2) = 1+ 2(2)(3) 1 2 =1– = 3 3

gJ = 1 +

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THE STRUCTURE OF ATOMS

9.7 The notation for a term symbol has the form 2S +1LJ. a) For 3P2, 2S + 1 = 3 so S = 1. The symbol P means L = 1. J = 2 according to the lower right subscript. The Lande´ g-factor Eq. (9.15) is J(J + 1) – L(L + 1) + S (S + 1) 2J(J + 1) (2)(3) – (1)(2) + (1)(2) = 1+ 2(2)(3) 1 3 =1+ = 2 2 b) For 3P1, 2S + 1 = 3 so S = 1. The symbol P means L = 1. J = 1 according to the lower right subscript. The Lande´ g-factor Eq. (9.15) is gJ = 1 +

J(J + 1) – L(L + 1) + S (S + 1) 2J(J + 1) (1)(2) – (1)(2) + (1)(2) = 1+ 2(1)(2) 1 3 =1+ = 2 2 3 c) For P0, 2S + 1 = 3 so S = 1. The symbol P means L = 1. J = 0 according to the lower right subscript. The Lande´ g-factor Eq. (9.15) is gJ = 1 +

gJ = 1 +

J(J + 1) – L(L + 1) + S (S + 1) 2J(J + 1)

L = S so J(J + 1) 2J(J + 1) 1 3 =1+ = 2 2

gJ =

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THE STRUCTURE OF ATOMS

117

9.8 For 3S 1, 2S + 1 = 3 so the spin quantum number is S = 1. The central symbol S means L = 0. According to the lower right subscript J = 1. The g-factor for this state is J(J + 1) – L(L + 1) + S (S + 1) 2J(J + 1) (1)(2) – 0 + (1)(2) =1+ 2(1)(2) =1+1 =2

gJ = 1 +

For 3P1, S = 1, L = 1, J = 1. The g-factor is J(J + 1) – L(L + 1) + S (S + 1) 2J(J + 1) (1)(2) – (1)(2) + (1)(2) = 1+ 2(1)(2) 1 3 =1+ = 2 2

gJ = 1 +

In a relatively weak magnetic field B the Zeeman effect ∆EB = µBgJ BMJ splits the 3S 1 state into three levels and splits the 3P1 into three levels as shown in the sketch.

Because of the selection rule ∆MJ = 0, ±1 only the seven transitions shown are allowed.

The chosen field 3.2 T was the strength of Zeeman’s strongest magnet. continued next page =→

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THE STRUCTURE OF ATOMS

The longest wavelength (smallest ∆E) is for the transition -1 → 0. The shortest wavelength (largest ∆E) is for the transition 1 → 0. As inferred from the energy level diagram the energy difference between the two transitions is the energy difference between the MJ = 1 and MJ = –1 levels of 3S 1. ∆E = h∆ν 2gJµBB ∆ν = h 9.27× 10–24 J · T–1 ! =4 (3.2 T) = 1.79 × 1011 s–1 6.63 × 10–34 J · s–1 ∆ν ∆ν 2 ∆λ = λ= λ ν c !

2 1.79 × 1011 s–1 –9 = 480 × 10 m = 0.138 nm 3 × 108 m · s–1

9.9 In a relatively weak magnetic field the 2P3/2 state is split into four levels M3/2 = 3 1 , , –12 , –32 and the lower 2S 1/2 is split into two levels M1/2 = 12, –12. With no fur2 2 ther consideration there would be eight transitions, but the actual number is only six because ∆MJ = 0, ±1 is not satisfied for two of the potential transitions.

The transitions listed are from the upper state → the lower state.

3 3

2

→

1 2 1

: ∆MJ = –1

allowed

→ – : ∆MJ = –2 not allowed 2 1 1 → : ∆MJ = 0 allowed 2 2 1 1 → – : ∆MJ = –1 allowed 2 2 1 1 – → : ∆MJ = +1 allowed 2 2 1 1 – → – : ∆MJ = 0 allowed 2 2 3 1 – → : ∆MJ = +2 not allowed 2 2 3 1 – → – : ∆MJ = +1 allowed 2 2 2

The figure for the D1 spectrum in Sec. (9.3) is traced from the photo of an experiment and shows six lines.

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9.8 In the strong field regime orbital angular momentum L and spin angular momentum S are decoupled to a good approximation and they remain good quantum numbers. The energy of a level in field B is E = (gL ML + gS MS )µBB where gL = 1 is the orbital g-factor and gS ≈ 2 is the spin g-factor.

The 2D → 2 P transitions are from a D state L = 2, S = 12 to a P state L = 1, S = 1 2. The Paschen-Back selection rule for ML is ∆ML = 0, ±1. The selection rule for MS is ∆MS = 0 so allowed transitions occur only between levels with the same MS . The list shows the allowed transitions using ket notation |ML, MS ) and the net magnetic energy of the transition.

1

1

1

1

1

1

1

1

1

1

|2, 2 ) – |1, 2 ) → 3µBB – 2µBB

= µBB

|1, 2 ) – |1, 2 ) → 2µBB – 2µBB

= 0

|1, 2 ) – |0, 2 ) → 2µBB – µBB

= µBB

|0, 2 ) – |1, 2 ) → µBB – 2µBB

= –µBB

|0, 2 ) – |0, 2 ) → µBB – µBB

= 0

|0, 2 ) – |–1, 2 ) → µBB – 0

= µBB

|–1, 2 ) – |0, 2 ) → 0 – µBB

= –µBB

|–1, 2 ) – |–1, 2 ) → 0 – 0

=0

|–2, 2 ) – |–1, 2 ) → –µBB – 0

= –µBB

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

|2, – 2 ) – |1, – 2 ) → µBB – 0

= µBB

|1, – 2 ) – |1, – 2 ) → 0 – 0

= 0

|1, – 2 ) – |0, – 2 ) → 0 + µBB

= µBB

|0, – 2 ) – |1, – 2 ) → –µBB – 0

= –µBB

|0, – 2 ) – |0, – 2 ) → –µBB + µBB

=0

1

1

|0, – 2 ) – |–1, – 2 ) → –µBB + 2µBB = µBB 1

1

|–1, – 2 ) – |0, – 2 ) → –2µBB + µBB = –µBB 1

1

1

1

|–1, – 2 ) – |–1, – 2 ) → –2µBB + 2µBB = 0 |–2, – 2 ) – |–1, – 2 ) → –3µBB + 2µBB = –µBB There are only three spectral lines – a “normal” Zeeman triplet – with ∆E = µBB, 0, –µBB relative to the unshifted line. There are six transitions for each line.

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THE STRUCTURE OF ATOMS

9.11 In the Paschen-Back strong field regime the spectrum consists of a “normal” Zeeman triplet with energies µBB, 0, –µBB relative to the unshifted line. The energy separation between adjacent lines is therefore µBB. ∆E = h∆ν = µBB µBB ∆ν = h ! h∆ν hc B= = ∆λ µB µBλ2 (6.63 × 10–34 J · s)(3 × 108 m · s–1) !

(0.42 × 10–9 m) = (9.27 × 10–24 J · T– 1)(404.6 × 10–9 m)2 = 55 T = 550 kG

9.12 Equation (9.16) gives the fine structure energy for a level J, L, S .

1 [J(J + 1) – L(L + 1) – S (S + 1)] (9.16) 2 Treating L and S as constant, the energy difference J → J – 1 is 1 ∆ ( | L · S | ) = [J(J + 1) – (J – 1)J] 2 1 = (2J) = J 2 The Lande´ interval rule for J treats L and S as constant so it is not exact but experimentally the error is only 10% or less. (|L · S|) =

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121

9.

121 Section (9.6.2) gives the hyperfine energy for a level F, J, I. 1 [F(F + 1) – J(J + 1) + I(I + 1)] 2 Treating I and J as constant, the energy difference F → F – 1 is 1 ∆ ( | I · F | ) = [F(F + 1) – (F – 1)F] 2 1 = (2F) = F 2 (|I·F|) =

9.14 a) From Eq. (9.20) and other results in Sec. (9.6), α ∆7 = [F(F + 1) – I(I + 1) – J(J + 1)] 2 where I is the nuclear spin and F = I + J.

(1)

The ground state of sodium is 2S 1/2 so spin S = 1 and L = 0. Thus J = L +S = 1 .

2

Hence F = I ± 1 .2The difference " ∆ν between ! !the two F!levels is!#therefore α 1 3 1 1 ∆ν = I+ I+ – I– I+ 2" 2 2 2 # 2 α 2 3 2 1 = I + 2I + – I + 2 4 4 α = (2I + 1) 2 From the given data

2I + 1 =

(2)(1.771 × 109) =4 885.8 × 106

so that I = 32in agreement with atomic beam experiments (1933). b) F=I±J 3 1 = ± = 2, 1 2 2

2

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THE STRUCTURE OF ATOMS

9.15 a) From Sec. (9.6.1), the hyperfine splitting energy ∆E in 1H from F = 1 to F = 0 is

∆E = h∆ν = 6.626 × 10–34 J · s 1420 × 106 s–1 = 9.409 × 10–25 J The magnetic energy for µB » µN is magnetic energy ≈ gJ MJµBB In the ground state of 1H J = 1 , L = 0, S = 1 so that gJ = 2 and gJ MJ = (2) 1 = 2

2

1. I and J are substantially decoupled if

2

hyperfine splitting energy ≈ 1.0 magnetic energy !

9.409 × 10–25 J = 1.0 9.274 × 10–24 J · T–1 B B ≈ 0.1T = 1000 G b) If I and J are decoupled F = I + J so that MF = MI + MJ For F = 1, MF = 1, 0, –1. For F = 0, MF = 0. The level MF = 1 must have MI = 12 , MJ = 1 . The level MF = –1 must have MI = – 1 , MJ = – 1 . The magnetic energy 2

2

for µB » µN is

2

magnetic energy ≈ gJ MJµBB = (2) ± 12 µBB This case is analogous to the Zeeman effect including spin-orbit coupling as discussed in Sec. (9.5), here with J · I taking the place of L · S. The hyperfine matrix element is (|I · J|) As shown in Sec. (9.5) the analogous matrix element has only diagonal elements a MI MJ, independent of B. The levels 1 , 1 and – 1 , – 1 therefore vary linearly with B.

2 2

2

2

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123

9.

123 The matrix of a permutation is square and is typically expressed as a permutation of the rows of the identity matrix E or alternatively as a permutation of the columns.

Like E, the elements of a permutation matrix will be either 0 or 1, with one 1 in each row and each column and 0 otherwise. Consider the permutation P2.

P2 =

1 2 3 2 1 3

To form the permutation matrix using the row method: 1 → 2 row 1 of P2 is row 2 of E 2 → 1 row 2 of P2 is row 1 of E 3→3

row 3 of P2 is row 3 of E

The resulting matrix for P2 is the 3 × 3 matrix: 0 1 0 1 0 0 0 0 1 The P2 matrix applied to the standard column vector |A) produces the permutation of the elements of |A).

01 0 P2 |A) = 1 0 0 0 0 1

2 1 2 = 1 3 3

The trace of a matrix is the sum of its diagonal elements, so the trace of the P2 matrix is 1.

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THE STRUCTURE OF ATOMS

9.15 The matrix of a permutation is square and is typically expressed as a permutation of the rows of the identity matrix E or alternatively as a permutation of the columns.

Like E, the elements of a permutation matrix will be either 0 or 1, with one 1 in each row and each column and 0 otherwise. Consider the permutation P5.

P5 =

1 2 3 3 1 2

To form the permutation matrix using the row method: 1 → 3 row 1 of P5 is row 3 of E 2 → 1 row 2 of P5 is row 1 of E 3→2

row 3 of P5 is row 2 of E

The resulting matrix for P5 is the 3 × 3 matrix: 0 0 1 1 0 0 0 1 0 The P5 matrix applied to the standard column vector |A) produces the permutation of the elements of |A).

00 1 P5 |A) = 1 0 0 0 1 0

3 1 2 = 1 3 2

The trace of a matrix is the sum of its diagonal elements, so the trace of the P5 matrix is 0.

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125

9.

125 Consider two electrons labeled 1,2 and having wave functions ψα, ψβ. Their Slater determinant wave function is 1 ψα(1) ψα(2) Ψ(1, 2) = , ψ (1) ψβ (2) 12 β = ψ (1)ψ (2) – ψ (2)ψ (1) , α β α β 2 Interchange 1 and 2.

1 ψα(2) ψα(1) Ψ(2, 1) = , 2 ψβ (2) ψβ (1) 1 = , ψ α(2)ψβ(1) – ψ α(1)ψβ (2) 2 = –Ψ(1, 2)

9.19 a) The two 1s electrons contribute 1s2 to the configuration. The two 2s electrons contribute 2s2 and the four 2p electrons contribute 2p4. The electron configuration is therefore 1s2 2s2 2p4. b) The closed inner shells 1s2 and 2s2 have paired electrons with L = 0 and contribute nothing to J. To develop the term symbol for the ground state of 8O first consider 7N which has three p electrons outside of closed shells. A p electron has three possible values of m4 = 1, 0, -1. Thus each of the three p electrons can have the same ms and still obey the Pauli exclusion principle if each p electron has a different m4. continued next page =→

126

THE STRUCTURE OF ATOMS

Returning to 8O , an additional p electron is added to 7N . Now all four p electrons cannot have the same ms without violating Pauli exclusion. The fourth electron must pair with another, leaving two unpaired p electrons. For maximum multiplicity according to Hund’s first rule the unpaired electrons must have the same ms, so S = 1 → 2S + 1 = 3. The parallel spins are a symmetric spin state, requiring an antisymmetric orbital state. As the sketches in Sec. (9.9.1) show the only possible antisymmetric orbital state for two p electrons is L = 1. For L = 1 and S = 1 the possible values of J are 2, 1, 0. c) The p shell would be completely filled with six p electrons. In 8O the four p electrons make the shell more than half-filled and according to Hund’s third rule the state with largest J lies lowest, which points to J = 2. Hence the ground state of 8O is predicted to be 3P2, which agrees with observation.

9.20 a) The closed shells in 13Al contain two 1s electrons, two 2s electrons, and six 2p electrons. The complete electron configuration is therefore 1s2 2s2 2p6 3s2 3p. b) The closed shells make no contribution to J, L, S . The single outer p electron can only contribute S = 12 , so that the multiplicity is 2S + 1 = 2. The single p electron can only contribute L = 1, so the ground state must be 2P. The possible values of J are 3 , 1 . According to Hund’s third rule if the outer shell is no more than half-filled 2 2

the ground state has the lowest possible J. For Al this would be J = 12 so the term symbol is predicted to be 2P 1 , which agrees with observation.

2

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

10.1 222 The atomic masses are 24He → 4.002603 u, 226 88 Ra → 226.025408 u, 86Rn → 222.017576 u The mass difference between the radium atom and the decay products is 0.00525 u. Using Table (10.1) the mass-energy released is (0.00525 u)(931.494 MeV · u–1) = 4.89 MeV. The classical conservation laws of momentum and energy apply to good accuracy. If the radium atom is at rest the initial momentum is 0. The alpha particle (mass Mα and velocity vα) and the radon atom (mass MRn , velocity vRn ) fly off back-to-back to conserve momentum = 0.

Mαvα = MRn vRn !

Mα vRn = v MRn α 1 1 M v2 + M v2 = 4.89 MeV 2 α α 2 Rn Rn Using Eq. (1)

1+

Mα MRn

!

!

1 v2 = 4.89 MeV M 2 α α 1 4.89 MeV 4.89 MeV M v2 = = = 4.80 MeV 2 α α 1 + MMRnα 1 + 4.00 222.0

(1)

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PARTICLE PHYSICS

10.2 Use data from Problem 10.1 as needed. a) .

BE = 2 mp + 2 mA – mα c2 = (2)(938.272) + (2)(939.565) – (931.494)(4.0026) = 27.28 MeV 27.28 MeV BE/nucleon = = 6.82 MeV 4 b) Let Mi be the mass of the initial nucleus before the decay, let M f be the mass of the residual nucleus, and let Md be the mass of the decay particle. The decay releases energy ∆E. ∆E = Mi – M f – md c2 MeV If ∆E > 0 the decay particle is emitted with positive energy. Consider hypothetical proton decay from MRa. MRa → MFr + mp ∆E = (226.025 – 225.026))(931.494) – (938.272) = –7.7 MeV The hypothesized proton decay requires energy and cannot occur spontaneously.

An α particle is strongly bound and alpha decay therefore provides 27 MeV more energy than decay into individual nucleons 2mp + 2mA. For many heavy nuclei the lower mass of mα makes ∆E > 0, allowing spontaneous decay.

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10.3 The base units of the natural system and their dimensions are [k] : L2 MT –1 [c] : LT –1 [me] : M [s0] : L–3 M–1T 4I2

[4] = [k]α[c]β[me]γ L1 = (L2αMαT –α)(LβT –β)(Mγ) = L(2α+β) M(α+γ)T –(α+β) 2α + β = 1

α+γ =0

– (α + β) = 0

Solutions are α = 1, β = –1, γ = –1 so that k 1.05 × 10–34 [4] = ≈ m ec (9.11 × 10–31)(3.00 × 108) = 3.84 × 10–13 m This length falls between the Bohr radius ≈ 5×10–11 m and the diameter of a proton ≈ 2 × 10–15 m.

10.4 The base units of the natural system and their dimensions are [k] : L2 MT –1 [c] : LT –1 [me] : M [s0] : L–3 M–1T 4I2 [E] = [k]α[c]β[me]γ. Because energy E a mv2 the dimensions of E are [E] = M1L2T –2 = (L2αMαT –α)(LβT –β)(Mγ) = L(2α+β) M(α+γ)T –(α+β) 2α + β = 2

α+γ =1

– (α + β) = –2 continued next page =→

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PARTICLE PHYSICS

Solutions are α = 0, β = 2, γ = 1 so that [E] = mec2 ≈ (9.11 × 10–31)(3.00 × 108)2 = 8.20 × 10–14 J 8.20 × 10–14 = 1.60 × 10–19 ≈ 0.51 MeV This is the rest mass of the electron expressed in MeV.

10.5 The fine structure constant α is α=

1 e2 ≈ 4u s0 k c 137

α is dimensionless so it has the same value in any units system. In natural units s0 = 1, k = 1, c = 1. In natural units e2 4u , , 4u e = 4uα ≈ = 0.303 (dimensionless) 137

α=

In SI units e = 1.60 × 10–19 C so in natural units 1.60 × 10–19 –19 e= = 5.28 × 10 C 0.303

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131

10.6 In the relation E = hν = kω ω has units s–1.

1 ω

=

k E

In natural units k = 1. In GeV units, E = GeV. s=

1 GeV

In SI units s=

1.05 × 10–34 J · s

k GeV

=

1.60 × 10–10 J

–25

= 6.56 × 10

s

10.7 From the energy-time uncertainty principle ∆E∆t ≈ k A particle with speed c crosses the interaction range = 1.5 fm in time ∆t.

1.5 × 10–15 m ∆t ≈ 3.0 × 108 m · s–1 = 5 × 10–24 s In time ∆t the energy can change by ∆E ≈

1.05 × 10–34 J · s 5 × 10–24 s ∆t 2 × 10–11 eV ≈ 2 × 10–11 J = 1.6 × 10–19 = 125 MeV k

=

This result is comparable to the masses of pions.

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PARTICLE PHYSICS

10.8 Consider the reactions (a) ë + p → ë + p

cross section σ

(aJ) ë + p → ů + A cross section σJ Assume that incident energies are less than a few hundred MeV and assume that 1

isospin in conserved. Let u(1) be pion wave functions and let v(2 ) be nucleon wave functions. From Table (F.2) in Appendix F the total wave functions are , , 1 3 1 ( ) ( ) (1) ( ) 1 2 2 2 2 ë + p → u–1 v 1/2 = 3 W–1/2 – 3 W–1/2 , 3 ( 2) ≈ 13W –1/2 , , 3 1 1 ( ) ( ) (1) ( ) 2 1 W2 + 2 W2 ů + A → u,0 v –1 3 = 3 –1/2 3 –1/2 /2 ( ) ≈ 23W–1/2 2 1 ( ) Terms in W 2 can be neglected at moderate energies.

For reaction (a)

*,

3 ( ) W 2 3 –1/2

1

, 7

3 + ( ) 1 W 2 = ×/

1

3

–1/2

3

where / is the part of the matrix element independent of isospin. Hence

For reaction (aJ)

*,

1

3 ( ) W 2 3 –1/2

, 7

2

σ=

1 ×/2 9

3 + ( ) W 2 =

, 2 ×/ 3

σJ =

2 ×/2 9

3

–1/2

The ratio of cross sections is 2

σJ 9 = 1 =2 σ 9

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PARTICLE PHYSICS

133

10.6 Consider the reactions (a) p + p → ã + d+

cross section σ

(aJ) A + p → ů + d+

cross section σJ

Assume that incident energies are less than a few hundred MeV and assume that 1

isospin in conserved. Let u (1) be pion wave functions and let v(2 ) be nucleon wave functions. The deuteron wave function is w0(0) = 1. From the tables in Appendix F the total wave functions are 1 1 ( ) ( ) (1) 2 v 2 = W p + p → v 1/2 1/2 1 ã + d+ → u(1) w(0) = W,(1) 1 1 0 1 1 1 (1) ( ) ( ) A + p → v 2/ v 2 = W0 – –1 2 1/2 2

,

1 (0) W0 2

ů + d+ → u(1) w(0) = W(1) 0 0 0 For reaction (a) D

E W1(1) 7 W(1) =/ 1 σ = /2

For reaction (aJ) D *,

E W0(1) W0(0) + = 0,

(1) 1 (1) W0 7 W0 = 2

σJ = The ratio of cross sections is 1 σJ 1 2 = 1 = 2 σ 1

1 2

×/

1 ×/2 2

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PARTICLE PHYSICS

10.10 Consider the reactions (a) ã + d+ → p + p

cross section σ

(aJ) ë + d+ → A + A

cross section σJ

As discussed in Sec. (10.3.2), the deuteron is an isospin singlet with isospin 0.

Assume that incident energies are less than a few hundred MeV and assume that 1

isospin in conserved. Let u(1) be pion wave functions and let v(2 ) be nucleon wave functions. The deuteron wave function is w0(0) = 1. From the tables in Appendix F the total wave functions are ã + d+ → u(1) w(0) = W(1) 1 0 1

1 1 ( ) ( ) (1) 2 v 2 = W p + p → v 1/2 1/2 1 (1) (0) (1) + ë + d → u –1w 0 = W –1 1 1 ( ) ( ) (1) 2 2 A + A → v /2v /2 = W–1 –1 –1

For reaction (a)

D

E W1 (1) 7 W(1) =1×/ 1

where / is the part of the matrix element independent of isospin.

For reaction (aJ) D E W–1(1) 7 W(1) =1×/ –1 Hence σ = (1)2 × / 2 = σJ. This result is consistent with experiments measuring total cross sections for pions passing through cells of “heavy water” D 2O.

10.11 a) In standard SI units the equation is E2 = p2c2 + m20c4. b) For a particle with zero rest mass m0 = 0 so E2 = p2c2 and p = Ec . Despite having no mass the photon nevertheless carries momentum according to the relativistic formulation. Some satellites intended for interplanetary exploration have been launched with “solar sails” – sheets of thin low mass plastic with large area when unfurled. Fuel is saved by gaining momentum from solar photons.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

PARTICLE PHYSICS

10.

135

a) As discussed in Sec. (10.6) the Lagrangian L is kinetic energy - potential energy. In this problem KE = 1 mv2 = 1 m(x˙2 + y˙2) 2

2

PE = mgy L = KE – PE = 12m(x˙2 + y˙2) – mgy The Lagrangian equations of motion Eq. (10.3) are ! d bL bL 0= – dt bq˙i bq1 Applied to coordinate x ! d bL bL 0= – dt bx˙ bx = mx¨ – 0 x¨ = 0 Applied to coordinate y ! d bL dL 0= – dt by˙ dy = my¨ + mg y¨ = –g Newton’s laws applied to this example are ma = F mx¨ = Fx = 0 my¨ = Fy = –mg identical to the Lagrange equations of motion.

135

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10.10 The neutron is a baryon so according to the quark model it consists of three quarks. From Table (10.6), the u quark carries electric charge +2/3 and the d quark carries -1/3 charge. The neutron has zero charge so in terms of u and d quarks its quark content must be udd to give +2/3 -1/3 -1/3 = 0.

10.14 According to the quark model a meson consists of a quark-antiquark pair. From Table (10.6) u and c quarks have charge +2/3 and their corresponding antiquarks therefore have charge -2/3. The d and s quarks have charge -1/3 and their antiparticles have charge +1/3. A neutral meson with quark-antiquark pair content can only be a combination u and c and their antiparticles or a combination of d and s and their antiparticles.

A neutral meson made from u and c can only have quark content uū, uc¯, ūc, or cc¯.

A neutral meson made from d and s can only have quark content dd¯, ds¯, d¯s, or ss¯.

Thus there are only eight distinct neutral mesons made from u, d, c, s.

10.15 ë is the antiparticle of ã so its quark content is ū d, which gives the correct charge -2/3 - 1/3 = -1.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow

PARTICLE PHYSICS

137

10.

137 Consider the reaction 0

ë+p→Λ +K electric charge : On the left-hand side ë has charge Q = -1 and p has charge Q = +1 for a net charge of 0. On the right-hand side both particles are neutral, so charge is conserved 0 = 0. baryon number : On the left-hand side ë is a meson and has B = 0. p is a baryon and has B = +1 for a net baryon number B = +1. On the right-hand side Λ0 is a baryon B = +1 according to Table (10.10) and K is a meson with B = 0. Hence baryon number is conserved +1 = +1. lepton number : From Table (10.7) none of the particles in the reaction is a lepton so lepton number is conserved 0 = 0. strangeness : On the left-hand side ë has quark content ū d so its strangeness number is S = 0. The quark content of p is uud, so it also has S = 0. According to Table (10.10) Λ0 on the right-hand side has quark content uds hence its strangeness number is S = -1. K has quark content us¯ so its strangeness number is S = +1 for net strangeness -1 + 1 = 0. Hence strangeness is conserved 0 = 0.

10.17 Consider j decay j → ē + vµ + v̄ e electric charge : On the left-hand side j has charge Q = -1. On the right-hand side ē has charge Q = -1 and the neutrinos are neutral. Hence charge is conserved -1 = -1. baryon number : None of the particles is a baryon so baryon number is conserved 0 = 0. lepton number : From Table (10.7) and Sec. (10.10) all particles in the reaction are leptons. On the left-hand side j has lepton number +1. On the right-hand side ē has lepton number +1, vµ has lepton number +1, and v̄ e has lepton number -1.

Hence lepton number is conserved +1 = +1 +1 -1.

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10.10 Consider the decay Ξ– → Σ– + ů electric charge :On the left-hand side Ξ– has charge Q = -1. On the right-hand side Σ– has charge Q = -1 and ů has charge Q = 0. Charge is conserved -1 = -1 + 0. baryon number : From Table (10.10) Ξ– on the left-hand side is a baryon B = +1.

On the right-hand side Σ– is a baryon B = +1. ů is not a baryon so it has B = 0.

Baryon number is conserved +1 = +1 + 0. strangeness : From Table (10.10) Ξ– on the left-hand side has quark content dss so S = -2. On the right-hand side Σ– has quark content dds S = -1. ů has quark content a combination of uu¯ and dd¯ so it has S = 0. Strangeness is not conserved: -2 ≠ -1. Failure of strangeness conservation prevents decay by the strong interaction.

10.19 Consider the decay Λ0 → p + ë electric charge : On the left-hand side Λ0 has charge Q = 0. On the right-hand side p has charge Q = +1 and ë has charge Q = -1, so charge is conserved 0 = +1 -1. baryon number : From Table (10.10) Λ0 on the left-hand side is a baryon B = +1.

On the right-hand side p is a baryon B = +1. ë is a meson so it has B = 0. Baryon number is conserved +1 = +1 + 0. strangeness : From Table (10.10) Λ0 on the left-hand side has quark content uds so S = -1. On the right-hand side p has quark content uud according to Sec. (10.9) so contributes S = 0. ë is a meson with quark content ū d so contributes S = 0.

Strangeness is not conserved in the decay: S = -1 ≠ 0 + 0. Failure of strangeness conservation prevents decay by the strong interaction.

Solution Manual for An Introduction to Groups and their Matrices for Science Students 1st Edition By Robert Kolenkow