Group Theory with Applications in Chemical Physics

by Patrick Jacobs

Overview

Group Theory is an indispensable mathematical tool in many branches of physics and chemistry. Providing a detailed account of Group Theory and its applications to chemical physics, the first half of this book focuses on elementary topics, such as molecular and crystal symmetry. The latter, more advanced portion includes proven examples. The book's comprehensive coverage makes it suitable for a two-semester course at the advanced undergraduate and graduate levels.

Product Details

ISBN-13:	9780521642507
Publisher:	Cambridge University Press
Publication date:	10/18/2005
Edition description:	New Edition
Pages:	506
Product dimensions:	7.09(w) x 10.00(h) x 1.18(d)

About the Author

Professor Jacobs is Emeritus Professor of Physical Chemistry at the University of Western Ontario. He has taught extensively in the area of physical chemistry, in particular group theory and has lectured on the subject across North America, Europe and the former USSR. He has authored over 315 publications, mainly in the fields of solid state chemistry and physics and his work has been awarded with the Solid State Medal of the Royal Society of Chemistry.

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Cambridge University Press
0521642507 - Group Theory With Applications in Chemical Physics - by P. W. M. Jacobs
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1 The elementary properties of groups

1.1 Definitions

All crystals and most molecules possess symmetry, which can be exploited to simplify the discussion of their physical properties. Changes from one configuration to an indistinguishable configuration are brought about by sets of symmetry operators, which form particular mathematical structures called groups. We thus commence our study of group theory with some definitions and properties of groups of abstract elements. All such definitions and properties then automatically apply to all sets that possess the properties of a group, including symmetry groups.

Binary composition in a set of abstract elements {g_i}, whatever its nature, is always written as a multiplication and is usually referred to as "multiplication" whatever it actually may be. For example, if g_i and g_j are operators then the product g_i g_j means "carry out the operation implied by g_j and then that implied by g_i." If g_i and g_j are both n-dimensional square matrices then g_i g_j is the matrix product of the two matrices g_i and g_j evaluated using the usual row × column law of matrix multiplication. (The properties of matrices that are made use of in this book are reviewed in Appendix A1.) Binary composition is unique but is not necessarily commutative: g_i g_j may or may not be equal to g_j g_i. In order for a set of abstract elements {g_i} to be a G, the law of binary composition must be defined and the set must possess the following four properties.

(i) Closure. For all g_i, with g_j ∈ {g_j},

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Because g_k is a unique element of {g_i}, if each element of {g_i} is multiplied from the left, or from the right, by a particular element g_j of {g_i} then the set {g_i} is regenerated with the elements (in general) re-ordered. This result is called the rearrangement theorem

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Note that {g_i} means a set of elements of which g_i is a typical member, but in no particular order. The easiest way of keeping a record of the binary products of the elements of a group is to set up a multiplication table in which the entry at the intersection of the g_ith row and g_jth column is the binary product g_i g_i = g_k, as in Table 1.1. It follows from the rearrangement theorem that each row and each column of the multiplication table contains each element of G once and once only.

Table 1.1. Multiplication table for the group G = {g_i} in which the product g_i g_j happens to be g_k.

G	gi	gj	gk	...
gi	gi2	gk	gi gk
gj	gj gi	gj2	gj gk
gk	gk gi	gk gj	gk2
⋮

(ii) Multiplication is associative. For all g_i, g_j, g_k ∈ {g_i},

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(iii) The set {g_i} contains the identity element E, with the property

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(iv) Each element g_i of {g_i} has an inverse g_i^-1 ∈ {g_i} such that

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The number of elements g in G is called the order of the group. Thus

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When this is necessary, the order of G will be displayed in parentheses G(g), as in G(4) to indicate a group of order 4.

Exercise 1.1-1 With binary composition defined to be addition: (a) Does the set of positive integers {p} form a group? (b) Do the positive integers p, including zero (0) form a group? (c) Do the positive (p) and negative (-p) integers, including zero, form a group? [Hint: Consider the properties (i)-(iv) above that must be satisfied for {g_i} to form a group.]

The multiplication of group elements is not necessarily commutative, but if

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then the group G is said to be Abelian. Two groups that have the same multiplication table are said to be isomorphous. As we shall see, a number of other important properties of a group follow from its multiplication table. Consequently these properties are the same for isomorphous groups; generally it will be necessary to identify corresponding elements in the two groups that are isomorphous, in order to make use of the isomorphous property. A group G is finite if the number g of its elements is a finite number. Otherwise the group G is infinite, if the number of elements is denumerable, or it is continuous. The group of Exercise 1.1-1(c) is infinite. For finite groups, property (iv) is automatically fulfilled as a consequence of the other three.

If the sequence g_i, g_i², g_i³, ... starts to repeat itself at g_i^c+1 = g_i, because g_i^c = E, then the set {g_i, g_i², g_i³ ... g_i^c = E}, which is the period of g_i, is a group called a cyclic group, C. The order of the cyclic group C is c.

Exercise 1.1-2 (a) Show that cyclic groups are Abelian. (b) Show that for a finite cyclic group the existence of the inverse of each element is guaranteed. (c) Show that ω = exp(-2πi/n) generates a cyclic group of order n, when binary composition is defined to be the multiplication of complex numbers.

If every element of G can be expressed as a finite product of powers of the elements in a particular subset of G, then the elements of this subset are called the group generators. The choice of generators is not unique: generally, a minimal set is employed and the defining relations like g_i = (g_j)^p (g_k)^q, etc., where {g_j g_k} are group generators, are stated. For example, cyclic groups are generated from just one element g_i.

Example 1.1-1 A permutation group is a group in which the elements are permutation operators. A permutation operator P rearranges a set of indistinguishable objects. For example, if

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then P is a particular permutation operator which interchanges the objects a and b. Since {a b ...} is a set of indistinguishable objects (for example, electrons), the final configuration {b a c ...} is indistinguishable from the initial configuration {a b c ...} and P is a particular kind of symmetry operator. The best way to evaluate products of permutation operators is to write down the original configuration, thinking of the n indistinguishable objects as allocated to n boxes, each of which contains a single object only. Then write down in successive rows the results of the successive permutations, bearing in mind that a permutation other than the identity involves the replacement of the contents of two or more boxes. Thus, if P applied to the initial configuration means "interchange the contents of boxes i and j" (which initially contain the objects i and j, respectively) then P applied to some subsequent configuration means "interchange the contents of boxes i and j, whatever they currently happen to be." A number of examples are given in Table 1.2, and these should suffice to show how the multiplication table in Table 1.3 is derived. The reader should check some of the entries in the multiplication table (see Exercise 1.1-3).

The elements of the set {P₀ P₁ ... P₅} are the permutation operators, and binary composition of two members of the set, say P₃ P₅, means "carry out the permutation specified by P₅ and then that specified by P₃." For example, P₁ states "replace the contents of box 1 by that of box 3, the contents of box 2 by that of box 1, and the contents of box 3 by that of box 2." So when applying P₁ to the configuration {3 1 2}, which resulted from P₁ (in order to find the result of applying P₁² = P₁ P₁ to the initial configuration) the contents of box 1 (currently 3) are replaced by those of box 3 (which happens currently to be 2 - see the line labeled P₁); the contents of box 2 are replaced by those of box 1 (that is, 3); and finally the contents of box 3 (currently 2) are replaced by those of box 2 (that is, 1). The resulting configuration {2 3 1} is the same as that derived from the original configuration {1 2 3} by P₂, and so

Table 1.2. Definition of the six permutation operators of the permutation group S(3) and some examples of the evaluation of products of permutation operators.

In each example, the initial configuration appears on the first line and the permutation operator and the result of the operation are on successive lines. In the last example, the equivalent single operator is given on the right.

The identity P0 = E
	1	2	3	original configuration (which therefore labels the "boxes")
P0	1	2	3	final configuration (in this case identical with the initial configuration)
The two cyclic permutations
	1	2	3					1	2	3
P1	3	1	2				P2	2	3	1
The three binary interchanges
	1	2	3			1	2	3			1	2	3
P3	1	3	2		P4	3	2	1		P5	2	1	3
Binary products with P1
					1	2	3
				P1	3	1	2	P1
			P1	P1	2	3	1	P2
			P2	P1	1	2	3	P0
			P3	P1	3	2	1	P4
			P4	P1	2	1	3	P5
			P5	P1	1	3	2	P3

Table 1.3. Multiplication table for the permutation group S(3).

The box indicates the subgroup C(3).

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so that P₁ P₁ = P₂. Similarly, P₂ P₁ = P₀, P₃ P₁ = P₄, and so on. The equivalent single operators (products) are shown in the right-hand column in the example in the last part of Table 1.2. In this way, we build up the multiplication table of the group S(3), which is shown in Table 1.3. Notice that the rearrangement theorem (closure) is satisfied and that each element has an inverse. The set contains the identity P₀, and examples to demonstrate associativity are readily constructed (e.g. Exercise 1.1-4). Therefore this set of permutations is a group. The group of all permutations of N objects is called the symmetric group S(N). Since the number of permutations of N objects is N!, the order of the symmetric group is N!, and so that of S(3) is 3! = 6.

Exercise 1.1-3 Evaluate the products in the column headed P₃ in Table 1.3.

Exercise 1.1-4 (a) Using the multiplication table for S(3) in Table 1.3 show that (P₃ P₁)P₂ = P₃(P₁ P₂). This is an example of the group property of associativity. (b) Find the inverse of P₂ and also the inverse of P₅.

Answers to Exercises 1.1

Exercise 1.1-1 (a) The set {p} does not form a group because it does not contain the identity E. (b) The set {p 0} contains the identity 0, p + 0 = p, but the inverses {-p} of the elements {p}, p + (-p) = 0, are not members of the set {p 0}. (c) The set of positive and negative integers, including zero, {p □ 0}, does form a group since it has the four group properties: it satisfies closure, and associativity, it contains the identity (0), and each element p has an inverse □ or -p.

Exercise 1.1-2 (a) g_i^pg_i^q = g_i^p+q = g_i^q+p = g_i^qg_i^p. (b) If p < c, g_i^p g_i^c-p = g_i^c = E. Therefore, the inverse of g_i^p is g_i^c-p. (c) ωⁿ = exp(-2πi) = 1 = E; therefore {ω ω² ... ωⁿ = E} is a cyclic group of order n.

Exercise 1.1-3

	P0	1	2	3
	P3	1	3	2	P3
P1	P3	2	1	3	P5
P2	P3	3	2	1	P4
P3	P3	1	2	3	P0
P4	P3	2	3	1	P2
P5	P3	3	1	2	P1

Exercise 1.1-4 (a) From the multiplication table, (P₃ P₁) P₂ = P₄ P₂ = P₃ and P₃ (P₁ P₂) = P₃ P₀ = P₃. (b) Again from the multiplication table, P₂ P₁ = P₀ = E and so P₂^-1 = P₁; P₅ P₅ = P₀, P₅^-1 = P₅.

1.2 Conjugate elements and classes

If g_i, g_j, g_k ∈ G and

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then g_k is the transform of g_j, and g_j and g_k are conjugate elements. A complete set of the elements conjugate to g_i form a class, □_i. The number of elements in a class is called the order of the class; the order of □_i will be denoted by c_i.

Exercise 1.2-1 Show that E is always in a class by itself.

Example 1.2-1 Determine the classes of S(3). Note that P₀ = E is in a class by itself; the class of E is always named □₁. Using the multiplication table for S(3), we find

P0 P1 P0-1 = P1 P0 = P1,

P1 P1 P1-1 = P2 P2 = P1,

P2 P1 P2-1 = P0 P1 = P1,

P3 P1 P3-1 = P4 P3 = P2,

P4 P1 P4-1 = P5 P4 = P2,

P5 P1 P5-1 = P3 P5 = P2.

Hence {P₁ P₂} form a class □₂. The determination of □₃ is left as an exercise.

Exercise 1.2-2 Show that there is a third class of S(3), □₃ = {P₃ P₄ P₅}.

Answers to Exercises 1.2

Exercise 1.2-1 For any group G with g_i ∈ G,

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Since E is transformed into itself by every element of G, E is in a class by itself.

Exercise 1.2-2 The transforms of P₃ are

P0P3P0-1 = P3P0 = P3,

P1P3P1-1 = P5P2 = P4,

P2P3P2-1 = P4P1 = P5,

P3P3P3-1 = P0P3 = P3,

P4P3P4-1 = P2P4 = P5,

P5P3P5-1 = P1P5 = P4.

Therefore {P₃ P₄ P₅} form a class, □₃, of S(3).

1.3 Subgroups and cosets

A subset H of G, H ⊂ G, that is itself a group with the same law of binary composition, is a subgroup of G. Any subset of G that satisfies closure will be a subgroup of G, since the other group properties are then automatically fulfilled. The region of the multiplication table of S(3) in Table 1.3 in a box shows that the subset {P₀ P₁ P₂} is closed, so that this set is a subgroup of S(3). Moreover, since P₁² = P₂, P₁³ = P₁ P₂ = P₀ = E, it is a cyclic subgroup of order 3, C(3).

Given a group G with subgroup H ⊂ G, then g_r H, where g_r ∈ G but g_r ∉ H unless g_r is g₁ = E, is called a left coset of H. Similarly, H g_r is a right coset of H. The {g_r}, g_r ∈ G but g_r ∉ H, except for g₁ = E, are called coset representatives. It follows from the uniqueness of the product of two group elements (eq. (1.1.2)) that the elements of g_r H are distinct from those of g_s H when s ≠ r, and therefore that

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where t is the index of H in G. Similarly, G may be written as the sum of t distinct right cosets,

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If H g_r = g_r H, so that right and left cosets are equal for all r, then

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and H is transformed into itself by any element g_r ∈ G that is not in H. But for any h_j ∈ H

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Therefore, H is transformed into itself by all the elements of G; H is then said to be an invariant (or normal) subgroup of G.

Exercise 1.3-1 Prove that any subgroup of index 2 is an invariant subgroup.

Example 1.3-1 Find all the subgroups of S(3); what are their indices? Show explicitly which, if any, of the subgroups of S(3) are invariant.

The subgroups of S(3) are

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Inspection of the multiplication table (Table 1.3) shows that all these subsets of S(3) are closed. Since g = 6, their indices t are 2, 3, 3, and 3, respectively. C(3) is a subgroup of S(3) of index 2, and so we know it to be invariant. Explicitly, a right coset expansion for S(3) is

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The corresponding left coset expansion with the same coset representative is

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Note that the elements of G do not have to appear in exactly the same order in the left and right coset expansions. This will only be so if the coset representatives commute with every element of H. All that is necessary is that the two lists of elements evaluated from the coset expansions both contain each element of G once only. It should be clear from eqs. (5) and (6) that H g_r = g_r H, where H = {P₀ P₁ P₂} and g_r is P₄. An alternative way of testing for invariance is to evaluate the transforms of H. For example,

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Similarly for P₃ and P₅, showing therefore that C(3) is an invariant subgroup of S(3).

Exercise 1.3-2 Show that C(3) is transformed into itself by P₃ and by P₅.

H₁ = {P₀ P₃} is not an invariant subgroup of S(3). Although

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showing that H₁ is a subgroup of S(3) of index 3,

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so that right and left cosets of the representative P₁ are not equal. Similarly,

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Consequently, H₁ is not an invariant subgroup. For H to be an invariant subgroup of G, right and left cosets must be equal for each coset representative in the expansion of G.

Exercise 1.3-3 Show that H₂ is not an invariant subgroup of S(3).

Answers to Exercises 1.3

Exercise 1.3-1 If t = 2, G = H + g₂ H = H + H g₂. Therefore, H g₂ = g₂ H and the right and left cosets are equal. Consequently, H is an invariant subgroup.

Exercise 1.3-2 P₃{P₀ P₁ P₂}P₃^-1 = {P₃ P₄ P₅}P₃ = {P₀ P₂ P₁} and P₅{P₀ P₁ P₂}P₅^-1 = {P₅ P₃ P₄}P₅ = {P₀ P₂ P₁}, confirming that C(3) is an invariant subgroup of S(3).

Exercise 1.3-3 A coset expansion for H₂ is

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The right coset for P₁ is {P₀ P₄}P₁ = {P₁ P₅}, while the left coset for P₁ is P₁{P₀ P₄} = {P₁ P₃}, which is not equal to the right coset for the same coset representative, P₁. So H₂ is not an invariant subgroup of S(3).

1.4 The factor group

Suppose that H is an invariant subgroup of G of index t. Then the t cosets g_r H of H (including g₁ H = H) each considered as one element, form a group of order t called the factor group,

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Each term in parentheses, g_r H, is one element of F. Because each element of F is a set of elements of G, binary composition of these sets needs to be defined. Binary composition of the elements of F is defined by

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where the complete set {g_r} contains g₁ = E as well as the t -1 coset representatives that ∉ H. It follows from closure in G that g_p g_q ∈ G. Because H is an invariant subgroup

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This means that in F

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which is the necessary and sufficient condition for H to be the identity in F.

Exercise 1.4-1 Show that g₁ g₁ = g₁ is both a necessary and sufficient condition for g₁ to be E, the identity element in G. [Hint: Recall that the identity element E is defined by

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Thus, F contains the identity: that {F} is indeed a group requires the demon-stration of the validity of the other group properties. These follow from the definition of binary composition in F, eq. (2), and the invariance of H in G.

Closure: To demonstrate closure we need to show that g_p g_q H ∈ F for g_p, g_q, g_r ∈ {g_r}. Now g_p g_q ∈ G and so

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Associativity:

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and so multiplication of the elements of {F} is associative.

Table 1.4. Multiplication table of the factor group
F = {E^′P^′}.

F	E′	P′
E′	E′	P′
P′	P′	E′

Inverse:

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so that the inverse of g_r H in F is g_r^-1 H.

Example 1.4-1 The permutation group S(3) has the invariant subgroup H = {P₀ P₁ P₂}. Here g = 6, h = 3, t = 2, and

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where the elements of F have primes to distinguish E^′ = H ∈ F from E ∈ G.

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E^′ is the identity element in F, and so the multiplication table for the factor group of S(3), F = {E^′ P^′}, is as given in Table 1.4.

Exercise 1.4-2 Using the definitions of E^′ and P^′ in eq. (13), verify explicitly that E^′ P^′ = P^′, P^′ E^′ = P^′. [Hint: Use eq. (2).]

Exercise 1.4-3 Show that, with binary composition as multiplication, the set {1 -1 i -i}, where i² = -1, form a group G. Find the factor group F = G/H and write down its multiplication table. Is F isomorphous with a permutation group?

Answers to Exercises 1.4

Exercise 1.4-1

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and so E E = E is a necessary consequence of the definition of E in eq. (1.1.5). If g₁ g₁ = g₁, then multiplying each side from the left or from the right by g₁^-1 gives g₁ = E, which demonstrates that g₁ g₁ = g₁ is a sufficient condition for g₁ to be E, the identity element in G.

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© Cambridge University Press

Table of Contents

1. The elementary properties of groups; 2. Symmetry operators and point groups; 3. Matrix representations; 4. Group representations; 5. Bases of representations; 6. Molecular orbitals; 7. Crystal field theory; 8. Double groups; 9. Molecular vibrations; 10. Transitions between electronic states; 11. The rotation group; 12. Projective representations; 13. Time-reversal symmetry; 14. The crystallographic magnetic point groups; 15. Macroscopic properties of crystals; 16. Space groups; 17. Electronic energy states in crystals; 18. The vibration of atoms in crystals; Appendices.

Editorial Reviews

“The mathematical development is clear and concise, and examples interspersed throughout clearly illustrate the different principles considered. For practicing chemists or physicists with previous exposure to group theory, this book is indispensable. As a primary resource for special-topics graduate courses, this book would be outstanding.”
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