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Solutions to the Problems in
Group Theory with Applications in Chemical Physics
by
Patrick Jacobs
(Cambridge University Press, 2005)
[ P.W.M. Jacobs 2005]
Equation numbers, figure numbers and table numbers are of the form
(n1.n2.n3) but start with SP, so that eq. (SP1.6.3), for example, means eq. (3) in
Problem 1.6 . Equation, figure or table numbers without SP refer to the above
text which contains these problems. Within the solution to a particular problem,
equation numbers are abbreviated and are referred to simply as (n3) or eq. (n3).
Many of the smaller tables are not numbered but are incorporated in the solutions.
Any general references in these solutions are included in the list of references in
the text, but some specific references are given in situ. As far as possible the
notation follows that in the text. In my MS I used Coronet font C as the nearest
approximation in MSWord to a script capital C. By the time I typed out these
solutions, Word had dropped the Coronet font, so readers will just have to make
do with italic capital C instead. I am much indebted to Professor B. Zapol for
reading these solutions and for his helpful comments. If readers detect any
typographical or other errors, would they kindly write to me at: [email protected].
These solutions are provided by the Publisher for the private use of
readers of this book and are Copyright.
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Problems 1
1.1 Show that the inverse of gi gj is gj 1 gi 1 .
Solution: The inverse of gi , written g, is the element which, when multiplied by
gi , gives the identity E. Now
(gi gj)( g j 1 g i1 ) = gi g i1 = E
Therefore the inverse of gi gj is g j 1 g i1 .
1.2 Prove that if each element of a group G commutes with every other element of
G (so that G is an Abelian group) then each element of G is in a class by itself .
Solution: If gi gj = gj gi ,  gi , gj  G, then for any gj  G
gi gj g i1 = gj gi g i1 = gj ,  gi  G.
Therefore, in an Abelian group, any element gj of G is in a class by itself.
1.3 Find a generator for the group of Exercise 1.4-3.
Solution: For G = { 1 1 i i}, i2 = 1, i3 = i, i4 = 1. Therefore, i is a
generator of G.
1.4 Show that {P1 P3} is a generator for S(3).
Solution: In S(3), P12 = P2 , P13 = P0 , P1 P3 = P5 , P12 P3 = P2 P3 = P4 .
Therefore, the set {P1 , P3} is a generator of S(3).
1.5 Show that conjugation is transitive, that is if gk is the transform of gj and gj is
the transform of gi , then gk is the transform of gi .
Solution: gk = gp gj g p1 = gp gq gi g q1 g p1 = (gp gq) gi ( g p g q )-1 = gr gi g r1 .
1.6 Show that conjugation is reciprocal, that is if gk is the transform of gj then gj
is the transform of gk .
Solution: If gk is the transform of gj , then for some gi  G,
gk = gi gj g i1
(1)
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gj = g i1 gk gi
1
(2)
Let gp be the inverse of gi . Then
gj = gp gk g p1
2
(3)
Therefore, it makes no difference whether a class is defined by (1), as in this book,
or by (2). [cf. Altmann, S.L. (1986) p. 42].
1.7 Prove that binary composition is conserved by conjugation.
Solution Suppose that gi gj = gk and let gk(p) denote the conjugate of gk by gp.
Then, gk(p) = gp gk g p1 = gp gi gj g p1 = gp gi g p1 gp gj g p1 = gi(p) gj(p) .
1.8 There are only two groups of order 4 that are not isomorphous and so have
different multiplication tables.
groups, G 14 and G 24 .
Derive the multiplication tables of these two
[Hints: First derive the multiplication table of the cyclic
group of order 4. Call this group G 14 . How many elements of G 14 are equal to
their inverse? Now try to construct further groups in which a different number of
elements are equal to their own inverse. Observe the Rearrangement Theorem.]
Solution Let A  G 14 ; then the cyclic group of order 4 is {A4 = E, A , A2 = B, A3 =
C}. The multiplication table of this (Abelian) group is
G 14
E
A
B
C
E
E
A
B
C
A
A
B
C
E
B
B
C
E
A
C
C
E
A
B
Only E and B are equal to their inverses. Now try to construct a group of order 4
in which a different number of elements are equal to their inverses. You will soon
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realize that, because of the Rearrangement Theorem, the only other possibility is
the group G 24 , with muliplication table:
G 24
E
A
B
C
E
E
A
B
C
A
A
E
C
B
B
B
C
E
A
C
C
B
A
E
In this group, each of the four elements is equal to its inverse.
1.9 Arrange the elements of the two groups of order 4 into classes.
Solution:
The multiplication tables of both these groups G 14 and G 24 are
symmetrical about their principal diagonals and therefore they are both Abelian
groups. Consequently, in both groups, each element is in a class by itself.
1.10 Identify the subgroups of the two groups of order 4.
Solution In G 14 the set {E B} is closed, but there are no other subgroups. In G 24 ,
the sets {E A}, {E B} and {E C} are closed and therefore they are subgroups of
G 24 .
1.11 Write down a coset expansion of S(3) on its subgroup H3 = {P0 P5} .
Show that H3 is not an invariant subgroup of S(3).
Solution: In S(3), using Table 1.1-3,
{P0 P5} + {P0 P5}P1 + {P0 P5}P2 = {P0 P5 P1 P3 P2 P4} = S(3)
The R coset {P0 P5}P1 = P1 P3 . But the corresponding L coset P1{P0 P5} =
P1 P4 , which is not equal to the R coset and therefore H3 is not an invariant
subgroup of S(3).
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1.12 The inverse class of a class C j = {gj} is C j = { g j 1 }. Find the inverse class
of the class {P1 P2} in S(3).
Solution: In S(3) the inverse class of {P1 P2} is { P11 P21 } = {P2 P1} = {P1 P2}.
Therefore, in S(3) The class {P1 P2} is ambivalent (equal to its inverse class).
1.13 The classes of S(3) are C1 = {P0}, C2 = {P1 P2}, C3 = {P3 P4 P5}.
Prove that 3 2 = 23 .
Solution: In S(3), 3 2 = (P3  P4  P5)(P1  P2) = P4  P5  P3  P5  P3  P4 = 23
1.14 Prove that for S(3), c3 g1
Solution: c3 g1
=
=
3
6
1
2

g i S( 3 )

g i S( 3 )
g i P3 g i1 = 3 .
g i P3 g i1
[P0 P3 P01  P1 P3 P11  P2 P3 P21  P3 P3 P31  P4 P3 P41  P5 P3 P51 ]
[P3  P5 P2  P4 P1  P3  P2 P4  P1 P5] =
1
2
[P3  P4  P5  P3  P5  P4] = 3

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Problems 2
2.1 Prove the following results by using projection diagrams.
(a) Show that R( m) and R( n) commute when m is normal to n.
(b) Show that y x = C2z .
(c) Two planes 1, 2 intersect along n and make an angle /2 with one
another. Show that 2 1 = R( n ). Do 1 and 2 commute?
(d) Show that R( x) R( z) = R( z) R( x).
Solution: (a) With no loss in generality, choose the y axis coincident with m. In
Figure SP2.1-1(a) [on the following page], the points 1, 2, 3, 4 show the effect of
the following operators on the representative point E:1 R( m) ; 2 R( n) R( m) ; 3 R( n) ; 4 = 2 R( m) 2 R( n) .
(b) See Figure SP2.1-1(b).
(c) See Figure SP2.1-1(c), which shows the unit vectors 1 and 2 and the reflecting
planes 1 and 2 . 1 E = A , 2 1 E = B, R( n) E = B . 1 and 2 do not
commute.
(d) See Figure SP2.1-1(d). R( z) E = A, R( x) R( z) E = B; R( x) E = C,
R( z) C = R( z) R( x) E = B .
2.2 Identify the set of symmetry operators associated with the molecule transdichloroethylene (Figure 2.4-2).
Set up the multiplication table for these
operators and hence show that they form a group.
Name this symmetry group.
[Hint: Set up a right-handed system of axes with y along the C=C bond and z
normal to the plane of the molecule.]
Solution: The molecule trans-dichloroethylene is shown in Figure 2.4-2. Set up
a right-handed set of axes with y along the carbon-carbon double bond and z
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Figure SP2.1-1
x y E
1 mE
n
n
2, 4
x
y x = C2z
3
(a)
(b)
1
A
B
2
E
/2 /2
y E

A
x

1
2
(c)
B
C
(d)
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normal to the plane of the paper. The multiplication table for the symmetry
operators for this molecule, is :G
E
C2z
I
z
E
E
C2z
I
z
C2z
C2z
E
z
I
I
I
z
E
C2z
z
z
I
C2z
E
The set G = {E C2z I z} is closed, contains E, each element has an inverse, and
associativity holds. For example, for this randomly chosen set of three operators,
C2z(zI) = C2z C2z = E and (C2zz)I = I I = E. The same result will be found for
any other threesome, showing that the set is associative. Therefore, G is a group,
in fact an Abelian group of order 4, isomorphous with G 24 .
2.3 Determine the symmetry elements of the following molecules and hence
identify the point group to which each one belongs. [Hints: Adhere to the
convention stated in §2.1. Many of these structures are illustrated in Figure 2.20.
Sketching the view presented on looking down the molecular axis will be found
helpful for (k) and (l).]
(a) NH3 (non-planar)
(b) H3CCCl3 (partly rotated)
(c) CHFClBr
(d) C5H5 (planar)
(e) C6H6 (planar)
(f) [Ti F6 ]3 (octahedral)
(g) allene
(h) [NbF7]2
(i) Pd2Cl6
(j) hydrogen peroxide
(k) bis(cyclopentadienyl)iron or ferrocene (staggered configuration)
(l) dibenzenechromium (like ferrocene, a 'sandwich compound', but the
two benzene rings are in the eclipsed configuration in the crystal)
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Solution: (a) NH3 has one C3 axis, no C2  C3 , three vertical planes v and
therefore belongs to the point group C3v or 3m.
(b) H3CCCl3 , in the partly rotated configuration (which is not the ground state of
this molecule) has one C3 axis, no C2  C3 , and does not have a h plane.
Therefore, it belongs to C3 or 3.
(c) CHFClBr [Figure 2.20(c)] has no symmetry elements and therefore belongs to
C1 or 1.
(d) C5H5 is planar and therefore has one C5 axis, five C2 axes, and a h .
Therefore, it belongs to D5h or 1 0 m2 .
(e) Benzene C6H6 is planar and has a C6 axis, six C2 axes in the plane of the
molecule, and a h . The point group is therefore D6h or 6/mmm.
(f) Octahedral [TiF6]3 has four C3 axes, three C4 axes and an inversion centre and
therefore belongs to the point group Oh or m3m .
(g) Allene [Figure 2.20(g)] has a C2 axis which is also the S4 principal axis, two
C2 axes, no h , no v , but two d and it therefore belongs to D2d or 42m .
(h) [NbF7]2 [Figure 2.20(h)] has a C2 principal axis, no C2 axes, and two vertical
planes of symmetry and therefore belongs to C2v or 2mm.
(i)
Pd2Cl6 [Figure 2.20(i)] has a C2 principal axis, two C2 axes, and a h , and
therefore has the point group D2h or mmm.
(j) H2O2 [Figure 2.20(j)] has a C2 principal axis, no C2 axes, and no planes of
symmetry and therefore has the point group C2 .
(k) Ferrocene [Figure 2.20(k)] has a C5 axis which is also the S10 principal axis,
five C2 axes, no h , and five d planes that bisect the angles between the C 2 axes.
Ferrocene therefore belongs to the point group D5d or 5m .
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(l) The eclipsed configuration of dibenzenechromium has a C6 principal axis, six
C2 axes, and a h . In this configuration, this molecule therefore belongs to the
point group D6h .
2.4 List a sufficient number of symmetry elements in the molecules sketched in
Figure 2.21 to enable you to identify the point group to which each belongs. Give
the point group symbol in both Schönflies and International notation.
Solution: (a) Octahedral ML6 has four C3 axes, three C4 axes, and a centre of
symmetry I : therefore its point group is Oh or m3m .
(b) ML5L has one C4 axis, no C3 axes, no C2 axes and four v planes: therefore
its point group is C4v or 4mm .
(c) ML4L2 has one C4 axis, no C3 axes, four C2 axes  the C4 axis, and a h plane.
Its point-group symmetry is therefore D4h or 4/mmm .
(d) fac-ML3L3 has one C3 axis, no C2 axes and three v planes of symmetry. Its
point group is therefore C3v or 3m .
(e) mer-ML3L3 has one C2 axis, no C2 axes, and two vertical v planes, so that its
point group is C2v or mm2 .
2.5 Show that each of the following sets of symmetry operators is a generator for
a point group. State the point group symbol in both Schönflies and International
notation. [Hints: The use of projection diagrams is generally an excellent method
for calculating products of symmetry operators.
See Figure 2.10(a) for the
location of the C2a axis in (d) and (g).]
(a) {C2y C2z}
(e) {C4z x}
(b) {C4z I}
(f) { 6 }
(c) {S4z C2x}
(d) {C3z C2a I}
(g) {S3z C2a}.
Solution: [It is recommended that the reader sketch projection diagrams that
illustrate the point group operations.]
(a) {C2y C2z} C2y C2y = E , C2y C2z = C2x , {E C2z C2x C2y} = D2 or 222.
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(b) {C4z I} I 2 = E , C 4z C 4z = C2z , I C2z = z , C 4z C2z = C 4z , I C 4z = S 4z ,
I C 4z = S 4z , {E C 4z C2z I S 4z z} = C4h or 4/m .
(c) {S4z C2x} S 4z S 4z = C2z , S 4z C2z = S 4z , C2x C2z = C2y , C2x S 4z = a ,
C2x S 4z = b , {E S 4z C2z C2x C2y a b} = D2d or 4 2m .
(d) With the aid of Figure 2.10(a) and a projection diagram, I C 3z = S 6z , I C2a = a ,
I C2b = b , I C2c = c .
The reflecting plane a bisects the angle between the C2
axes C2b and C2c and similarly (Figure 2.10(a)).
Thus there are three dihedral
planes and the point group is {E 2C3 3C2 I 2S6 3d} = D3d or 3m .
(e) {C4z x } C 4z C 4z = C2z , C 4z C2z = C 4z , C2z C2z = E, C 4z x = a , where a =
[1 1 0] and the reflecting plane a contains z and [ 1 1 0]. C 4z  x = b , where b =
[ 1 1 0] and the reflecting plane b contains z and [1 1 0]. x C2z = y , y C2z = x .
x and y are vertical planes of symmetry and a and b are dihedral planes.
Therefore, the point group is {E 2C4 C2 2v 2d} = C4v or 4mm.
(f) { 6 } I C 6z = S 3z , (I C 6z )2 = C 3z , (I C 6z )3 = h , (I C 6z )4 = C 3z ,
(I C 6z )5 = S 3z , (I C 6z )6 = E. {E 2C3 h 2S3} = C3h or 3/m
(g) { S 3z C2a} [Figure 2.10(a) should help.] S 3z S 3z = C 3z , S 3z C 3z = h ,
S 3z h = C 3z , C 3z C 3z = E, C2a S 3z = e , C2a C 3z = C2b , C2a C 3z = C2c .
The symmetry planes d, e, f contain the three 3C2 axes C2a, C2b, C2c and are
therefore vertical planes. The point group is {E 2C3 3C2 h 2S3 3v} = D3h or
6 m2 .
2.6 List a sufficient number of symmetry elements (and also significant absences)
in the following closo B n H n 2 ions that will enable you to determine the point
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group to which each belongs. The shapes of these molecules are shown in Figure
2.22.
2
2
(a) B5 H 52 (b) B6 H 62 (c) B9 H 92 (d) B10 H10
(e) B12 H12
.
Solution:
(a) B5 H 52 One C3 axis, three C2 axes, h : D3h or 6 m2.
(b) B6 H 62 Three C4 axes, four C3 axes, inversion centre I : Oh or m3m .
(c) B9 H 92 One C2 axis through 8 and the mid-point of 2-3, no C2 axes, two
vertical planes through 1, 6, and 8, and through 2, 3 and 8: C2v or 2mm . [The
triangular faces are not equilateral triangles.]
2
(d) B10 H10
One C4 axis (1-10), four C2 axes, four dihedral planes: D4d or 8 2m .
[Hint: If you do not see the C2 axes sketch the view presented on looking down
the C4 axis from 1 to 10. You will see two squares 2, 3, 5, 6 and 6, 7, 8, 9 rotated
by /4 with respect one another. One C2 axis bisects 5-9 and 3-7.
Locate the
other three C2 axes. The dihedral planes are vertical planes that bisect the angles
between the C2 axes, that is they pass through 1, 10 and 7, 9 or 6, 8, or 3, 5 or
2, 4.]
2
(e) B12 H12
There are six C5 axes (joining opposite apices like 1, 12) and ten C 3
axes (through the centres of opposite pairs of triangular faces). There is a centre
of symmetry, I , so the point group is Yh .
2.7 Evaluate the following direct products showing the symmetry operators in
each group. [Hint: For (a)-(e), evaluate products using projection diagrams. This
technique is not useful for products that involve operators associated with the C 3
axes of a cube or tetrahedron so in these cases study the transformations induced
in a cube.] Explain why the direct products in (d)-(f) are semi-direct products.
(a) D2  Ci (b) D3  Ci (c) D3  Cs (d) S4  C2 (C2 = {E C2x})
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(e) D2  C2 (C2 = {E C2a}) (f) D2  C3 (C3 = {E C31 }).
Solution:
(a) D2  Ci = {E C2z C2x C2y}  {E I} = {E C2z C2x C2y z x y}
= D2h or mmm
(b) D3  Ci = {E C 3z C2a C2b C2c}  {E I}
= {E C 3z C2a C2b C2c I S 6z a b c}
= {E 2C3 3C2 I 2S6 3d} = D3d or 3m
The reflecting planes a , b , c bisect angles between C2 axes normal to the C3
principal axis and so are dihedral planes.
(c) D3  Cs = {E C 3z C2a C2b C2c}  {E z}
= {E C 3z C2a C2b C2c z S 3z d e f}
= {E 2C3 3C2 h 2S3 3v} = D3h or 6 m2.
The reflecting planes d , e , f contain the horizontal axes and so are vertical
planes of symmetry.
(d) S4  C2 = {E S 4z C2z S 4z }  {E C2x} = {E S 4z C2z S 4z C2x a C2y b }
= {E 2S4 C2 2C2 2d } = D2d or 42m
D2d = {E C2x} + S 4z {E C2x} + C2z {E C2x} + S 4z {E C2x}
= {E C2x S 4z b C2z C2y S 4z a}
D2d = {E C2x} + {E C2x} S 4z + {E C2x}C2z + {E C2x} S 4z
= {E C2x S 4z a C2z C2y S 4z b}
The second and fourth terms in the left and right coset expansions are not the
same, and therefore {E C2x} is not an invariant subgroup of D2d .
Consequently,
the direct product of S4 and C2 is a semi-direct product.
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(e) D2  C2 = {E C2z C2x C2y}  {E C2a }
= {E C2z C2x C2y C2a C2b C 4z C 4z }
= {E 2C4 C2 4C2} = D4 or 422
This is a semi-direct product because D2 but not C2 is an invariant subgroup of D4.
The left coset expansion of D4 on C2 is
D4 = {E C2a } + C2z{E C2a } + C2x{E C2a } + C2y{E C2a }
= { E C2a C2z C2b C2x C 4z C2y C 4z }
The right coset expansion on C2 is
D4 = { E C2a C2z C2b C2x C 4z C2y C 4z }
The third and fourth terms of the left and right coset expansions are not equal and
therefore C2 is not an invariant subgroup of D4 .
(f) D2  C3 = {E C2z C2x C2y}  {E C31 C31 }
This problem involves the C3 axes of the cube and Figure 2.12 will be found to be
extremely useful. Remember that symmetry elements are defined with respect to
fixed axes and so remain fixed, when the unit sphere in configuration space is
rotated by symmetry operators. The drawing of small cubes is a useful device for
keeping track of the configurations of the cube during symmetry operations. The
first drawing is the original configuration: label alternate apices 1, 2, 3, 4 as in
Figure 2.12.
Now mentally carry out the operation C31 (a positive rotation
through 2/3 about the axis through O1 and label your next drawing with the
resulting configuration.
(1423).
Calling the original configuration (1234), C31 (1234) =
Your labelled cube after C31 should correspond to (1423).
Similarly,
C31 (1234) = (1342). Next, C2z C31 (1234) = C2z(1423) = (4132). You must now
decide which C3 rotation of the original configuration would give this
configuration (4132). [Hint: Look for the apex that has remained in its original
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position, since this will be the one that lies on the rotation axis.] Only 3 is in its
original position, so the product C2z C31 is a C33 rotation.
negative rotation?
But is it a positive or
Imagine the view presented on looking down the axis O3.
One sees an inverted equilateral triangle with 4 on the right, 1 on the left and 2
below. With the same sequence of labelling, the original configuration is 124 and
it is a positive (anti-clockwise) rotation of the original configuration 124 that
yields the final configuration 412. Therefore, C2z C31 (1234) = (4132) = C 33 (1234)
and so C2z C31 = C 33 .
Continuing in the same manner: C2x C31 = C 34 ,
C2y C31 = C 32 , C2z C31 = C 34 , C2x C31 = C 32 , C2y C31 = C 33 . All the other necessary
products can be evaluated in a similar fashion, so that
{E C2z C2x C2y}  {E C31 C31 } = { E 4 C 3 3C2} = T or 23
The left coset expansion of T on {E C31 C31 } is
T = {E C31 C31 } + C2z{E C31 C31 } + C2x{E C31 C31 } + C2y{E C31 C31 }
= {E C31 C31 C2z C 33 C 34 C2x C 34 C 32 C2y C 32 C 33 }
The right coset expansion is
T = {E C31 C31 } + {E C31 C31 }C2z + {E C31 C31 }C2x + {E C31 C31 }C2y
= {E C31 C31 C2z C 34 C 33 C2x C 32 C 34 C2y C 33 C 32 }
The left and right cosets are not all equal and therefore {E C31 C31 } is not an
invariant subgroup of T. Be sure to check all these products in the left and right
coset expansions in order to develop confidence in your ability to calculate
products involving C3 operations about cube axes.
Problems 3
3.1 Show by evaluating [(R)]T (R), where R is the proper rotation R( z ), that
(R) is an orthogonal matrix and hence write down [(R)]1.
Also write down
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(R( z)). Is this the same matrix as [(R( z)]1 and if so, is this the result
you would expect ? Evaluate det (R( z)) and det (S( z)).
Solution: An orthogonal matrix A is a matrix with the property A1 = AT.
 c s 0
[R( z)] =  s c 0


0 0 1
(1)
 c s 0   c s 0  1 0 0 
(R)T(R) =  s c 0  s c 0 = 0 1 0


 
 
0 0 1 0 0 1 0 0 1
(2)
Therefore, (R)T = (R)1, and (R) is a real orthogonal matrix.
1, 2
 c s 0
[R( z)] =  s c 0 = ([R( z)] )1


0 0 1
(3)
A negative rotation through  is the operation that is the inverse of R( z).
Since the MRs of symmetry operators obey the same multiplication table as the
operators themselves, (3) is the expected result.
1
det [(R( z)] = 1(c2  s2) = 1
c s
0
det [S( z)] = s c 0 = 1(c2 + s2) = 1
0 0 1
Similarly, (S) is a real, orthogonal matrix.
3.2 Find the matrix representative (R) for R = R(2/3 n) with n a unit vector
from O along an axis that makes equal angles with OX, OY, OZ . What is the
trace of (R)? Find x y z = (R) x y z and write down the Jones symbol for
this operation. [Hints: Consider the effect of R(2/3 n) by noting the action of R
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on e1 e2 e3  as you imagine yourself looking down n towards the origin. The
trace of a matrix is the sum of its diagonal elements.]
Solution:
0 0 1 
R(2/3 n) e1 e2 e3 = e2 e3 e1 = e1 e2 e3 1 0 0 = e1 e2 e3 (R)


0 1 0
(1)
Tr (R) = 0
1
0 0 1   x 
 x y z = (R) x y z = 1 0 0  y  =  z x y
0 1 0  z 
The Jones symbol for (R) is zxy.
3.3 (a) Find the matrix representative (R) of R for R(/2 z) and hence find the
matrix (I) (R) .
(b) Using PDs find the single operator Q that is equivalent to I R ; show also
that I and R commute. Give the Schönflies symbol for Q.
(c) Find the matrix representative (Q) from Q e1 e2 e3 = e1 e2 e3 =
e1 e2 e3 (Q).
(d) What can you deduce from comparing (Q) from part (c) with (I) (R)
from part (a)?
Solution:
 0 1 0
(a) R(/2 z) e1 e2 e3 =  e 2 e1 e3 = e1 e2 e3  1 0 0 = e1 e2 e3(R)
 0 0 1
(1)
 1 0 0   0 1 0  0 1 0 




(I)(R) =  0 1 0   1 0 0 = 1 0 0 
 0 0 1   0 0 1 0 0 1 
(2)
1
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From a projection diagram (draw it!) Q = I R = R I, so that I commutes with R.
Also, from this PD, Q = I R = I C 4 = S 4 .
0 1 0 


Q e1 e2 e3 =  e 2 e1 e3  = e1 e2 e3 1 0 0 
0 0 1 
(3)
Therefore, Q = I R and
(Q) = (I)(R)
2, 3
so that again, in this example, the MRs obey the same multiplication rule as the
operators. [These have been illustrative examples. The proof for the general case
is in §3.4 .]
3.4 Find the MRs of the operators a , b for the basis  e1 e2 e3  where
a = 2½[1 1 0], b = 2½[ 1 1 0]. Evaluate (a)( b). Using a PD find
Q = a b . Find the MR of Q and compare this with (a)( b). What can you
conclude from this comparison?
Solution:
 0 1 0


a e1 e2 e3 =  e2 e1 e 3  = e1 e2 e3  1 0 0
 0 0 1
(1)
0 1 0 
b e1 e2 e3 = e2 e1 e3 = e1 e2 e3 1 0 0 .
0 0 1
(2)
From a PD, Q = a b = C2z and
 1 0 0


Qe1 e2 e3 =  e1 e2 e 3  = e1 e2 e3  0 1 0 = e1 e2 e3 (Q)
 0 0 1
(3)
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 0 1 0 0 1 0   1 0 0




(a)(b) =  1 0 0 1 0 0 =  0 1 0 = (Q)


 0 0 1 0 0 1  0 0 1
1, 2, 3
We conclude that Q = a b = C2z and (Q) = (a)(b) = ( C2z) . This is
another example of the MRs obeying the same multiplication table as the
symmetry operators.
3.5 Find the MRs of the operators E, C 4z , C 4z , x , y for the basis  e1 e2 e3 .
Solution:
0 1 0 


C 4z e1 e2 e3 =  e 2 e1 e 3  = e1 e2 e3 1 0 0
0 0 1
 0 1 0
C 4z e1 e2 e3 =  e2 e1 e 3  = e1 e2 e3  1 0 0


 0 0 1
 1 0 0


σ x e1 e2 e3 =  e1 e 2 e 3  = e1 e2 e3  0 1 0
 0 0 1
1 0 0
σ y e1 e2 e3 =  e1 e2 e 3  = e1 e2 e3 0 1 0
0 0 1
3.6 Write down the Jones symbols for R  C4v and then the Jones symbols for
{R1}.
[Hints: You have enough information from problems 3.4 and 3.5 to do
this very easily. Remember that the MRs of {R} are orthogonal matrices.] Write
down the angular factor in the transforms of the five d orbitals under the
operations of the point group C4v. [Hint: This may be done immediately by using
the substitutions provided by the Jones symbols for R1.]
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Solution: See Table SP3.6-1.
3.7 Find the MR of R(2/3 [1 1 1]) for the basis  e1 e2 e3 . Hence write down
the Jones representations of R and of R1.
Find the transformed d orbitals R̂ d,
when d is dxy , dyz , or dzx . [Hint: Remember that the unit vectors e1 e2 e3 are
oriented initially along OX, OY, OZ but are transformed under symmetry
operations.
Observe the comparative simplicity with which the transformed
functions are obtained from the Jones symbol for R1, instead of trying to visualize
the transformation of the contours of these functions under the configuration space
operator R. ]
Solution: Draw a cube and label alternate apices as in Figure 2.12. Then label
the remaining four corners such that the numbering n, n of diagonally opposite
corners satisfies the relation n = n + 4.
Check by noting that the unit vector
which is the axis of rotation in this problem, lies along the diagonal 8-4.
Then
looking down the axis of rotation from 8 you will see the axes oriented such the e1
lies on your right, e3 points vertically upwards, and e 2 is on your left.
A
clockwise rotation gives
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Table SP3.6-1
C4v
E
C 4z
C 4z
C2z
σx
σy
 1 0 0


 0 1 0
 0 0 1


 1 0 0


 0 1 0
 0 0 1


1 0 0


0 1 0 
0 0 1
a
0

1
0

1 0

0 0
0 1 
b
1 0 0
(R) 0 1 0
0 0 1
0 1 0


1 0 0
0 0 1


 0 1 0


 1 0 0
 0 0 1
1 0 0


0 1 0
0 0 1
 0 1 0


 1 0 0
 0 0 1
0 1 0 


1 0 0 
0 0 1 


 1 0 0


 0 1 0
 0 0 1


 1 0 0


 0 1 0
 0 0 1


1 0 0


0 1 0 
0 0 1
J(R) xyz
yxz
yxz
xyz
x yz
xyz
yxz
yxz
J(R-1) xyz
yxz
yxz
xyz
x yz
xyz
yxz
yxz
xy
xy
-xy
-xy
xy
-xy
-xy
xy
xy
yz
yz
-zx
zx
-yz
yz
-yz
-zx
zx
zx
zx
yz
-yz
-zx
-zx
zx
-yz
yz
-(x2-y2)
-(x2-y2)
x2-y2
x2-y2
x2-y2
-(x2-y2)
-(x2-y2)
z2
z2
z2
z2
z2
z2
z2
(R-1)
a
x2-y2 x2-y2
b z2
a
z2
0

1
0

1 0

0 0
0 1 
0 1 0


1 0 0
0 0 1
0 1 0


1 0 0
0 0 1
In these problem solutions I am using J(R) to denote the Jones symbol for R.
b The
angle-dependent factor in this d orbital is 3z2 - r2.
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 0 0 1
R(2/3 [1 1 1])e1 e2 e3 =  e2 e3 e1  = e1 e2 e3  1 0 0 = e1 e2 e3(R)


 0 1 0
Therefore the Jones symbol for this rotation is given by
 0 0 1  x 
(R) x y z =  1 0 0  y  = zxy . On transposing (R),

 
 0 1 0  z 
0 1 0   x 


(R1) x y z = 0 0 1   y  = yzx .
1 0 0   z 
Since R̂ f (x y z) = f (R1[x y z]), the transformed d functions dxy , dyz , dzx are
dyz ,  dzx ,  dxy respectively. Observe the comparative simplicity with which the
transformed orbitals are obtained, without the strain of trying to visualize the
rotation of the contours of the functions when the rotation axis is not normal to the
plane determined by the orbital subscript.
3.8 (a) List the symmetry operators of the point group D 2 . Show in a PD their
action on a representative point E.
find the classes of D2 .
Complete the multiplication table of D 2 and
[Hint: This can be done without evaluating transforms
QRQ1, Q  D2 .]
(b) Evaluate the direct product D2  Ci = G and name the point group G. Study
the transformation of the basis  e1 e2 e3  under the symmetry operators R  G =
{R}.
Use the MRs of R1 to find the Jones symbols for {R1} and hence write
down the transformed d orbitals when the symmetry operators of G act on
configuration space.
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Solution: (a) First, write down the multiplication table of D2 . [Hint: Draw a PD
in the xy plane normal to z , and use it to evaluate the necessary products.]
D2
E
C2z
C2x
C2y
E
E
C2z
C2x
C2y
C2z
C2z
E
C2y
C2x
C2x
C2x
C2y
E
C2z
C2y
C2y
C2x
C2z
E
The multiplication table is symmetrical about the principal diagonal (from top left
to lower right) and therefore the group is Abelian. Consequently, each element of
D2 is in a class by itself.
(b) D2  Ci = { E C2z C2x C2y}  {E I}
= { E C2z C2x C2y I z x y} = D2h
The necessary products are readily evaluated by using a PD. [You may check
your diagram from Figure 2.15(a).] The derivation of the Jones symbols J(R-1) is
explained in Table SP3.8-1. This is an unusual case, in that all the symmetry
operators (apart from E) are proper or improper binary rotations and therefore the
MRs are all diagonal matrices. This means that (RT) = (R) and consequently
J(R-1) = J(R).
3.9 For R( x) draw a figure like Figure 3.3 but with OX normal to the plane of
the paper and OZ pointing vertically towards the top of the page. Rotate the basis
e1 e2 e3  about x through an angle . The transformed vectors {e1 e2 e3 } are
each the sum of their projections on {e1 e2 e3 } :
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Table SP3.8-1
R
E
e1e2e3
C2z C2x
z
I
x
y
 e1e2 e3 
 e1e 2 e3 
 e1e2 e3 
 e1e 2 e3   e1e 2e 3 
 1 0 0


 1 1 0
 0 0 1


1 0 0 


0 1 0 
0 0 1 
1 0 0 


0 1 0
0 0 1 


1 0 0 


0 1 0
0 0 1 


1 0 0 
0 1 0 


0 0 1 
 1 0 0


 0 1 0
 0 0 1


1 0 0
0 1 0 


0 0 1
J(R-1) xyz
xyz
xyz
xyz
xyz
xyz
x yz
xyz
xy
xy
xy
-xy
-xy
yz
yz
-yz
yz
-yz
yz
-yz
yz
-yz
zx
zx
-zx
-zx
zx
zx
-zx
-zx
zx
x2-y2
x2-y2
x2-y2
x2-y2
x2-y2
x2-y2
z2
z2
z2
z2
z2
1 0 0
(R) 0 1 0
0 0 1
x2-y2 x2-y2
a z2
a The
z2
 e1e2e3 
C2y
xy
xy
-xy
 e1e2e 3 
-xy
z2
x2-y2
z2
angle-dependent factor in this d orbital is 3z2 - r2.
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Full file at http://TestBankSolutionManual.eu/Solution-Manual-for-Group-Theory-with-Applications-in-Chemical-Physics-byJacobs
e1 = e1(1)  e2(0)  e3(0)
e2 = e1(0)  e2(cos )  e3(sin )
e3 = e1(0)  e2( sin )  e3(cos )
0
0 
1

e1 e2 e3  = e1 e2 e3  0 cos   sin   = e1 e2 e3  (R( x))


0 sin  cos  
For R( y) draw a similar figure, but with OY normal to the plane of the paper and
OX (and therefore x) pointing to your left. Operate on the basis e1 e2 e3  with
R( y). Then
e1 = e1(cos )  e2(0)  e3( sin )
e2 = e1(0)  e2(1)  e3(0)
e3 = e1(sin )  e2(0)  e3(cos )
 cos  0 sin  
 = e e e  (R( x))
e1 e2 e3  = e1 e2 e3   0
1
0
1 2 3


 sin  0 cos 

25
Full file at http://TestBankSolutionManual.eu/Solution-Manual-for-Group-Theory-with-Applications-in-Chemical-Physics-byJacobs
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