Revised 4/11/2014
Lecture Notes for Chemistry 543, Part II
LECTURE 18: SYMMETRY ELEMENTS AND OPERATIONS
Reading: Chapters 2 and 3 of Cotton.
The symmetries of a pair wave function allow us to determine whether an optical transition between the states described by those functions is allowed. For an atom this is straight
forward; the states must have opposite parity for a dipole transition to be allowed. The
transition dipole matrix element is
∫∫∫
µ21 = e
ψ2∗ rψ1 r2 sin θdrdθdϕ,
(579)
where
r = r sin θ cos ϕî + r sin θ sin ϕk̂ + r cos θk̂.
(580)
We deduce that µ1s,2s = 0 because ψ1s and ψ2s are both even functions of r and µ = er is
an odd function. You can see this by evaluating the integrals over θ and ϕ. Since the wave
functions are independent of angle, the integrals along all three axes vanish. For example,
the y-component of this matrix element is
∫
2π
⟨ψ1s |y|ψ2s ⟩ ∼
sin ϕ dϕ = 0.
(581)
0
On the other hand, the 1s → 2pz transition is allowed because
∫ π
∫ 1
⟨ψ1s |z|ψ2pz ⟩ ∼
cos θ cos θ sin θdθ =
x2 dx = 2/3.
0
(582)
−1
We would like to generalize this concept to molecules and lattices. That is to say that
we would like to know whether the symmetry properties of the molecule impose mathematical constraints on the wave function so that certain matrix elements vanish. This is a
central question in spectroscopy, namely how to determine which transitions are forbidden
by symmetry and which are allowed. To do this we need a mathematical way of describing
the symmetry properties of an object. We start by defining a symmetry operation as a
geometric transformation that leaves the spatial orientation of the molecule unchanged, so
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that if you closed your eyes while the operation was executed you could not tell afterwards
whether anything occurred. For molecules, we consider only those operations that leave at
least one point in the molecule fixed in space.
Symmetry operations are generated by a set of symmetry elements of the molecule. In
analyzing the effect of a symmetry operation, we assume that the molecular coordinates
are transformed while leaving the symmetry elements fixed in space. For example, we may
rotate a water molecule about an axis that bisects the molecule, while the axis itself remains
fixed in space.
There are only four symmetry elements that a molecule may possess. Some molecules
have no symmetry; others may have only one symmetry element, while others may have as
many as all all four types of symmetry elements. They are as follows:
1. Axis of rotation. Example: C6 of benzene. The subscript 6 refers to a six-fold rotation.
In general, an n-fold rotation means a rotation of 360◦ /n.
2. Plane of reflection. Example: σv of H2 O. The subscript v denotes a “vertical” plane
which contains an axis of rotation (in this case C2 ).
3. Improper axis of rotation. Rotation about the axis followed by reflection in a plane
perpendicular to that axis. Example: S6 = σh C6 operating on staggered ethane. In this
case, neither C6 nor σh exists separately.
4. Center of inversion. x, y, z, → −x, −y, −z. Example: i operating on cyclobutane.
It is possible for a molecule to possess more than one example of a particular symmetry
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element. For example, H2 O has two different σv planes, each containing the C2 axis. Suppose
the molecule lies in the xz plane, with the C2 axis coincident with the z axis. A σv plane
coincident with the xz plane transforms y → −y, while a σv′ plane coincident with the yz
plane transforms x → −x.
In the case of benzene, there are three types of symmetry planes.
1. Three σv planes, each containing two carbon atoms and the C6 axis.
2. Three σd planes, each bisecting two C − C bonds and containing the C6 axis. The
subscript d denotes a “dihedral” plane (that is also vertical), which bisects some bonds.
3. One σh plane containing the molecular plane and perpendicular to the C6 axis. The
subscript h denotes a “horizontal” plane.
Benzene also has three sets of rotation axes:
1. One C6 axis perpendicular to the plane. Coincident with it are C2 , C3 and S6 axes.
2. Three C2′ axes in the molecular plane, each containing two carbon atoms.
3. Three C2′′ axes also in the molecular plane, each bisecting two carbon-carbon bonds.
Later we will study systematically how certain symmetry elements imply the existence
of others. We already saw how C6 implies a C2 axis. We also saw that σh C6 = S6 .
Another example is a tetrahedal molecule, such as methane. This molecule has three S4
axes conicident with the coordinate axes, three C2 axes, also conicident with the coordinate
axes, four C3 axes, each passing through an apex containing one of the atoms and perpendicular to the plane containing the other three atoms, and six σd planes, each perpendicular
to one of the cubic faces and containging two atoms in that plane.
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Another important property is commutativity. For example, C4 and i commute, whereas
σd and σv do not. We illustrate this for cyclobutane.
LECTURE 19: PROPERTIES OF GROUPS
Group theory provides the tools for describing the properties of symmetry operations. A
group is a set of objects and an operation (“multiplication”) that transforms one element
into into another. The set must satisfy the following four postulates:
1. Closure. The product of any pair of elements in the group is also an element of the
group.
2. Identity Operation. There exists one element that commutes with all the elements in
the group and leaves them unchanged. This element is called the “identity element.” The
usual symbol for this element is E.
3. Inverse. Every element in the group has an inverse, which means that multiplication
of an element by its inverse yields the identity. That is, AA−1 = E.
4. Associativity. Multiplication is commutative: A(BC) = (AB)C.
Let’s prove some useful theorems.
1. An element and its inverse commute. Suppose this statement is false. Choosing
AA−1 = E and assuming that [A, A−1 ] ̸= 0 leads to a contradiction:
AA−1 ̸= A−1 A
E ̸= A−1 A
−1
AE ̸= (AA )A = A
A ̸= A
(583)
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2. An element has only one inverse. Suppose B and C are both inverses of A.
AB = E
AC = E
AB = AC
(584)
A−1 AB = A−1 AC
B=C
3. The previous theorem proves that in general if AB = D and AC = D, then B = C.
AB = AC
A−1 AB = A−1 AC
(585)
B=C
4. The inverse of AB is B −1 A−1 .
(AB)(B −1 A−1 ) = A(BB −1 )A−1 = AA−1 = E
(586)
The number of elements in a group is called its order, h. There may exist (although there
need not be) more than one distinct group of a given order. Let’s construct a multiplication
table for h = 3. One possible solution is:
EAB
E E AB
AAB E
BB E A
We see that A and B are inverses of each other, and each element occurs once and only
once in each row and column. Theorem no. 3 tells us that the latter statement must be the
case. In other words, every element in the group may be generated by the product of two
elements in the group, and that product is unique. If say A occurs twice on a row, such
that BC = A and DC = A, then B = D. Knowing this, it is easy to show that there are
no other groups of order 3. The following is a failed attempt to find such a group.
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EAB
E E AB
AA E ?
BB ? A
Now let’s consider h = 4. One possible group is the cyclic group:
A2 = B
A3 = C
(587)
A4 = E
We readily construct its multiplication table. For example, BC = A2 A3 = A4 A = A.
E AB C
E E AB C
AAB C E
BB C E A
CC E AB
We see that all its elements commute. Such a group is called Abelian. We easily verify
that all of the postulates are satisfied:
1. Closure. Every binary product appears in the table.
2. E is one of the elements.
3. E occurs in every row and column, indicating that every element has an inverse.
4. We demonstrate associativity:
(AB)C = CC = B
(588)
A(BC) = AA = B
We can construct another group of order 4 by assuming that every element is its own
inverse. This does not work in general, but it does work for h = 4.
A2 = E
B2 = E
C2 = E
We readily construct its multiplication table:
(589)
89
E AB C
E E AB C
AA E C B
BB C E A
CC BA E
This group is also Abelian. It has a further property that some of its elements form
subgroups. There are in fact three subgroups of order 2: (A, E), (B, E) and (C, E). Another
useful theorem that we will state without proof is
Problem 48. Show that there does not exist a group of order 4 with exactly two subgroups
(not counting the trivial subgroup of just the identity operator).
5. The order of a subgroup is a factor of h. This theorem is proved easily in [1].
We see that the cyclic group of order 4 has only one subgroup, (B, E). Any other group
of order 4 having just one subgroup must be equivalent to this group, because interchanging
the names of elements does not constitute construction of a distinct group. Inspection shows
that there are no groups of order 4 having two subgroups. Hence, there are no other groups
of order 4. Extension of this reasoning leads to the following theorem:
6. There exits only one group for each prime order (i.e., if h is a prime number), and
that group is cyclic.
Let’s consider next groups of order 6. One example is the cyclic group:
A2 = B
A3 = C
A4 = D
A5 = F
A6 = E
We readily construct its multiplication table:
(590)
90
E A B CD F
E E A B CD F
AA B CD F E
BB CD F E A
CCD F E A B
DD F E A B C
F F E A B CD
We see that this group has subgroups (C, E) and (B, D, E).
Another group of order 6 has the following multiplication table:
E A B CD F
E E A B CD F
AA E D F B C
BB F E DCA
CCD F E A B
DDCA B F E
F F B CA E D
It has subgroups (A, E), (B, E), (C, E) and (D, F, E).
Problem 49. Find a way of constructing the multiplication table for the previous group.
(It was not by trial and error!) You may find this easy to do after Lecture 21.
Problem 50. Relable the elements of the last group so that its subgroups are (C, E),
(B, D, E) (A, E), and (F, E). Write out the multiplication table for the relabled group.
Problem 51. Construct a sixth order group having five subgroups of order 2.
Another relationship between group elements is the similarity transform. Let A and X
be two elements of a group. Then there exists an element B such that
B = X −1 AX.
We say that is B is a similarity transform of A and also that B is conjugate to A.
7. If A is conjugate to B, then B is conjugate to A.
(591)
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8. If A is conjugate to B, and B is conjugate to C, then A is conjugate to C.
Problem 52. Prove Theorems 7 and 8.
Some groups have the property that a subset of the elements are conjugate to each other
and to no other elements of the group. Such elements form a class. One can prove that:
9. The order of a class is a factor of the order of the group.
You may think of a similarity transform as a generalized coordinate transformation. X
transforms or “rotates” the object into a new coordinate system, operation A is performed,
and the object is then rotated back by X −1 . The same result could be obtained in the
original coordinate system by using operation B. In the noncyclic group of order 6 cited
above, (A, B, C) form one class and (D, F ) form another class. For example,
A−1 AA = A
B −1 AB = BD = C
C −1 AC = CF = B
(592)
D−1 AD = F B = B
F −1 AF = DC = C
A calculation of all the similarity transforms of B and C shows that A, B, and C belong to
the same class.
Problem 53. Show that elements D and F constitute a class of the non-cyclic group of
order 6.
LECTURE 20: RELATIONSHIPS BETWEEN SYMMETRY ELEMENTS
In this lecture we will show how the existence of one or more symmetry elements may imply the existence of other symmetry elements. We will approach this problem by examining
what symmetry operations they induce.
Consider first the consequences of individual symmetry elements.
1. A proper rotation axis of non-prime order implies the existence of lower order proper
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rotational axes along the same axis. That is,
Cmn =⇒ Cm + Cn
(593)
For example, a C2n axis implies the existence of a Cn axis. For example, the operations
(C62 , C64 ) are equivalent to (C3 , C32 ), implying the existence of a C3 axis. More generally,
an axis of order mn implies the existence of Cm and Cn axes. For example, the operations
(C6 , C62 , C63 , C64 , C65 , C66 ) are equivalent to (C6 , C3 , C2 , C32 , C65 , E), implying the existence of a
C2 and a C3 axis.
This is a special case of a general theorem in mechanics which states that a pair or
rotations of arbitrary angles about any two axes is equivalent to a single rotation about
some third axis. What is special here is that the three axes all belong to a symmetry group.
2. An improper rotation axis of even order implies the existence of a proper rotational
axis of half its order along the same axis. That is,
S2n =⇒ Cn .
(594)
To prove this, we must first prove that σh and Cn commute. This is true because the two
operations are orthogonal to each other. We may demonstrate this by seeing how they
operate on an arbitrary vector, r = xi + yj + zk. Suppose that r has a component r⊥ in the
xy plane, and that component makes and angle θ with respect to the x axis. It follows that
Cn (z)(xî + y ĵ + z k̂) = r⊥ cos(θ − 2π/n)î + r⊥ sin(θ − 2π/n)ĵ + z k̂
σh (xy)(xî + y ĵ + z k̂) = xî + y ĵ − z k̂.
(595)
We may apply this resullt to the operations generated by the S4 axis to demonstrate the
principle:
(S4 , S42 , S43 , S44 ) = (C4 σh , C42 σh2 , C43 σh3 , C44 σh4 ) = (S4 , C2 , S43 , E),
(596)
which shows that the C2 axis exists independently.
3. S2 = i.
Proof:
S2 (z)(xî+y ĵ+z k̂) = C2 (z)σh (xy)(xî+y ĵ+z k̂) = C2 (z)(xî+y ĵ−z k̂) = −xî−y ĵ−z k̂. (597)
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In general, if n is odd,
n
n
S2n
= C2n
σh = C2 σh = i.
(598)
We cannot draw this conclusion for an even value of n.
4. An improper rotational axis of odd order implies the existence of a proper rotational
axis of the same order and a horizontal plane. This follows from the identity
S2n+1 = C2n+1 σh .
(599)
(S3 , S32 , S33 , S34 , S35 , S36 ) = (C3 σh , C32 , σh , C3 , C32 σh , E),
(600)
For example,
showing that C3 and σh exist independently.
We consider next the consequences of two symmetry elements.
1. Two orthogonal C2 axes imply the existence of a third mutually orthogonal C2 axis.
C2 (y)C2 (x)(xî + y ĵ + z k̂) = C2 (y)(xî − y ĵ − z k̂) = −xî − y ĵ + z k̂
C2 (z)(xî + y ĵ + z k̂) = −xî − y ĵ + z k̂.
(601)
2. Two planes imply the existence of a C2 axis along their line of intersection.
σ(xy)σ(yz)(xî + y ĵ + z k̂) = σ(xy)(−xî + y ĵ + z k̂) = −xî + y ĵ − z k̂
C2 (y)(xî + y ĵ + z k̂) = −xî + y ĵ − z k̂.
(602)
3. A C4 axis and a perpendicular C2 axis imply the existence of a mutually perpendicular
C2 axis.
C4 (z) + C2 (x) =⇒ C2 (y).
(603)
This is true because a C4 rotation rotates the C2 axis of the molecule in the horizontal
plane. We may generalize it to a rotational axis of order 2n. That is, when we rotate a C2
axis in the xy plane by a 2π/2n angle n times, each rotation generates a new C2 axis, such
that
C2n (z) + C2 (xy plane) =⇒ nC2 (xy plane).
(604)
4. Similar reasoning leads to the existence of multiple vertical planes of symmetry:
C2n (z) + σv (xz) =⇒ nσv .
(605)
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5. We may use similar reasoning to demonstrate the existence of a pair of dihedral planes
generated by a C4 axis and a vertical plane σv First we prove a useful transformation:
C4 (z)(xî + y ĵ + z k̂) = y î − xĵ + z k̂.
(606)
We define a dihedral plane σd as a vertical plane lying at an angle of π/4 with respect to
the original σv plane and a second dihedral plane σd′ as a vertical plane lying at an angle of
−π/4 with respect to the original σv plane. It follows that
C4 (z)σv (yz)(xî + y ĵ + z k̂) = C4 (z)(−xî + y ĵ + z k̂) = y î + xĵ + z k̂.
(607)
But you can also show that
σd (xî + y ĵ + z k̂) = y î + xĵ + z k̂.
(608)
C4 (z)σv (yz) = σd .
(609)
It follows that
Identical reasoning may be used to generate the second dihedral plane. We start by demonstrating that
C43 (z)(xî + y ĵ + z k̂) = −y î + xĵ + z k̂.
(610)
It is then easy to show that
C43 (z)σv (yz) = σd′ .
(611)
Problem 54. Prove Eq. (611).
This result may be generalized to a C4n axis because a C4n axis automatically generates
a C4 axis.
6. Similarly, a C4 axis and an orthogonal C2 axis imply the existence of two additional
C2 axes at angles of ±π/4 with respect to the original C2 axis. Using vector notation to
designate the direction of the various axes, you can show that
(
)
C4 (k̂)C2 (ĵ) = C2 2−1/2 (î + ĵ) ,
with a similar equation for C4 (k̂)C2 (î).
Problem 55. Prove Eq. (612).
(612)
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7. We can also prove that a horizontal plane containing a C2 axis implies the existence
of a vertical plane containing that axis. That is,
σh (xy)C2 (x) = σv (xz),
(613)
Problem 56. Prove Eq. (613).
If we also have a Cn (z) axis, it follows that there are n vertical planes.
8. nσd planes and nC2 normal axes (i.e., in the xy plane) generate an S2n (z) axis. We
will illustrate this with the simplest case of n = 2. If σd and σd′ are defined as in result 5,
then the following are true:
C2 (x)σd = S4 (z)
C2 (y)σd = S43 (z)
C2 (x)σd′ = S43 (z)
(614)
C2 (x)σd′ = S4 (z)
Problem 57. Prove Eqs. (614).
9. Finally, the definition of an improper rotation implies that
Cn + σh =⇒ Sn .
(615)
We note that a necessary and sufficient condition for a molecule to be chiral (i.e., to have
a non-superposable mirror image; i.e., to be dissymmetric) is that it does not have an axis
of improper rotation. If it has an S2n+1 axis, then it automatically has a plane of symmetry,
and the molecule is therefore superimposable on it mirror image. If it has an S2n axis,
reflection through the plane perpendicular to that axis followed by a C2n rotation produces
a structure that is superposable on the original structure. In other words, even though the
mirror image of an S2n molecule cannot be superposed on the original molecule, a simple
Cn rotation of the mirrow image does allow it to be superposed.
The above statements establish that the absence of an Sn axis is a necessary condition for
a molecule to be dissymmetric. It is also a sufficient condition because if a molecule does not
have an imprpoper rotationl axis any symmetry opertation that can occur will not reflect
the molecule and therefore will not make it superimposable on its mirror image. Another
way of saying this is that a reflection (or a combination of a reflection and a proper rotation)
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does not produce a result that is superimposable on the original molecule. Keep in mind
that S1 is a simple reflection, so that water has an improper rotational axis and therefore
is not dissymmetric. Also, remember that S2 is equivalent to an inversion. The absence of
a plane of reflection or a centetr of symmetry are special cases of the more general criterion
of not having an improper axis of rotation.
LECTURE 21: SYMMETRY GROUPS
We now wish to show that the symmetry operations of an object satisfy the postulates
of a group, namely (1) closure, (2) identity, (3) inverse, and (4) associativity.
(1) We will invent any operation needed to guarantee closure. We may think as this as
the reason for including an improper rotation axis as a symmetry element. (2) The identity
postulate means that doing nothing leaves the object unchanged. (3) The inverse follows
from the property that any geometric transformation (e.g. xî+y ĵ+z k̂ → −xî−y ĵ−z k̂) may
be run backwards. For example, Snm = Cnm σh and its inverse is σh Cnn−m . (4) Associativity
follows from the fact that every symmetry operation may be described as a coordinate
transformation. Such transformations may be written as matrix operations on a vector,
and these matrices have the associative property. The same argument may be used for the
inverse.
It follows from this discussion that the number of symmetry operations equals the number of distinct orientations of the object, and this number in turn equals the order of the
group. This idea is best understood from a concrete example. We use an equilateral triangle for illustration. The triangle has 12 distinct orientations. Each orientation may
be generated from any starting point by one and only one of the following operations:
E, C3 , C32 , C2 , C2′ , C2′′ , σv , σv′ , σv′′ , σh , S3 , S35 . You can demonstrate that this set satisfies the
postulates of a group by examining how they transform the triangle.
Problem 58. Demonstrate the closure postulate for the D3h group using the equilateral
triangle as a model. Show geometrically that every pair of symmetry operations is equivalent
to a single operation. Report your results in the form of a 12 by 12 table in which each
operation appears once and only once in each row and column.
We now consider systematically all possible point groups.
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1. If a molecule has no symmetry elements, its only symmetry operation is E, and its
group is designated C1 .
Suppose the molecule has only one symmetry element (plus those that automatically
accompany that element). There are four possibilities:
2. If the only element is a plane, the group is of order 2, and it is designated as Cs .
3. If the only element is a center of inversion, the group is also of order 2, and it is
designated as i.
4. If the only element is a proper axis of rotation, the group is of order n, and it is
designated as Cn . It is cyclic and Abelian. Example: non-planar H2 O2 .
5a. If the only element is an improper axis of rotation of even order, the group is of order
n, and it is designated as Sn . For example, the symmetry operations for the S4 group are
E, S4 , S42 , S43 = E, S4 , C2 , S43 . We note that the S2 group is equivalent to i.
5b. If the only element is an improper axis of rotation of odd order, n, the group is of
order 2n, and it is designated as Cnh to emphasize the independent existence of a Cn axis and
a σh plane. For example, the symmetry elements of the C3h group are E, S3 , S32 , S33 , S34 , S35 =
E, S3 , C32 , σh , C3 , S35 . If n is an even number, the Cnh group does not have an odd order
rotational axis. The symmetry operations of the C2h group, for example, are C2 , σh , i, E.
Examples are trans-CHCl=CHCl and trans-planar H2 O2 .
We consider next molecules with two or more symmetry elements but with no perpendicular axes of order greater than 2.
6. We showed in Eq. (604) that the existence of a Cn axis and a normal C2 axis implies
the existence of a nC2 normal axes. The resulting group is of order 2n and is designated as
Dn . For example, the symmetry elements of the D4 group are E, C4 , C2 , C43 , 2C2′ , 2C2′′ .
7. We showed in Eq. (605) that Cn axis and a σv plane imply the existence of of n
vertical planes. The resulting group is of order 2n and is designated as Cnv . Examples:
cis-planar H2 O2 (C2v , symmetry elements E, C2 , σv , σv′ ) and N H3 (C3v , symmetry elements
E, C3 , C32 , σv , σv′ , σv′′ ) .
8. Now let’s add a plane of symmetry. Suppose we have a Cn axis, a σh plane, and nC2
axes. We showed in Eq. (613) that a σh plane and nC2 axes imply the existence of n vertical
planes. The Cn axis and σh plane also imply the existence of an Sn axis. Altogether we
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find that there are 4n symmetry operations. The group order is therefore 4n and the group
is designated as Dnh . Examples: benzene (D6h ) and cyclobutane (D4h ). The symmetry
elements of the latter are 4C4 , σh , 2σv , 2σd , 4C2 , S4 , i, S43 .
9. Suppose we have a Cn axis, nσd planes, and nC2 axes. The generalization of Eq. (614)
implies the existence of an S2n axis as well. The group order is 4n and is designated as Dnd .
3
7
9
Example: staggered ferrocene (symmetry elements 5C5 , 5C2 , 5σd , S10 , S10
, i, S10
, S10
).
10. A linear molecule belongs to either the C∞v group if does not have a σh plane
(example: CO) or to the D∞h group if it has a σh plane (example: CO2 ). Both have infinite
order rotation axes because the angle of rotation is 2π/n in the limit of n → ∞. Both groups
also have an infinite number of vertical planes.
We consider next molecules with rotational axes of order higher than 2 in additional to
the primary high order axis. These correspond to the five platonic solids: the tetrahedron,
cube, octahedron, dodecahedron, and the icosahedron. There are a total of 7 groups of this
type.
11. The tetrahedron has three S4 axes, four C3 axes, and six σd planes. The group order
is 24 and is designated as Td . The symmetry operations correspond to nine S4 rotations
including the C2 rotations, twelve C3 rotations including E, and six σd reflections. Example:
methane.
Problem 59. Methane is a tetrahedral molecule having four 3-fold axes and three 2-fold
axes of rotation. Assume that the four hydrogen atoms lie on the corners of a cube, with one
of located at (1, 1, 1). Write down the coordinates of the other three H atoms and identify
the directions of each axis.
12. The cube and the octahedron have 48 operations. The cube, for example, has three C4
axes passing through pairs of opposite faces, four C3 axes passing through pairs of opposite
apices, three planes bisecting opposite faces, six planes passing through pairs of opposite
apices, three S4 axes, and four S6 axes. The group designated as Oh .
Problem 60. Do Problem 15 on pages 64-65 of Cotton.
The following is a summary of which symmetry elements define each symmetry group.
(The entries for some of the high order groups are incomplete.) It might seem surprising at
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first that a group may be assigned by simply identifying the presence of a small number of
symmetry elements, considering that a group consists of a much larger niumber of elements,
typically 2n or 4n. The reason such a simple algorithm works is that the list of symmetry
operations designated in the decision tree imply the existence of all the others in the group,
using the theorems derived in the previous lecture.
group defining elements
derived elements
order
Cs
σ
2
i
i
2
Sn
Sn ( even n)
n
Cn
Cn
n
Cnh
Cn , σ h
Cnv
Cn , nσv
2n
Dn
Cn , nC2
2n
Dnh
Cn , nC2 , σh
Dnd
Cn , nC2 , nσd
S2n ; also i if n is odd
4n
C∞v
C∞
∞ σv
∞
D∞h
C∞ , σh
∞ σv , ∞ C2 , S∞
∞
Sn ; also i if n is even
2n
Sn , nσv ; also i if n is even 4n
T
4C3 , 3C2
12
Td
3S4 , 4C3 , 3C2
6σd
24
Th
4S6 , 3C2 , 3σh
4C3 , i
24
O
3C4 , 8C3
6C2
24
Oh
48
I
60
Ih
120
100
LECTURE 22: COORDINATE TRANSFORMATIONS
Reading: Cotton [1], chapter 4
Suppose we have a vector
A = xî + y ĵ.
(616)
If we rotate the coordinate axes counter-clockwise by an angle ϕ, the new unit vectors are
ˆ
i′ = î cos ϕ + ĵ sin ϕ
(617)
jˆ′ = −î sin ϕ + ĵ cos ϕ.
We invert these equations to get
î = î′ cos ϕ − ĵ′ sin ϕ
′
(618)
′
ĵ = î sin ϕ + ĵ cos ϕ.
We may then write A as
A = x(î′ cos ϕ − ĵ′ sin ϕ) + y(î′ sin ϕ + ĵ′ cos ϕ)
= î′ (x cos ϕ + y sin ϕ) + ĵ′ (−x sin ϕ + y cos ϕ)
(619)
= x′ î′ + y ′ ĵ′ ,
or


′
x
y′


=

cos ϕ sin ϕ
− sin ϕ cos ϕ


x
y

 = Û

x

(620)
y
Instead of rotating the coordinate axes counter-clockwise, we may rotate the vector clockwise. Suppose A makes an angle α with respect to the x-axis. That is,
x = r cos α
(621)
y = r sin α.
After a clockwise rotation by an angle ϕ,
x′ = r cos (α − ϕ) = r cos α cos ϕ + r sin α sin ϕ = x cos ϕ + y sin ϕ
y ′ = r sin (α − ϕ) = r sin α cos ϕ − r cos α sin ϕ = y cos ϕ − x sin ϕ,
(622)
which is ientical to Eq. (620). In other words, it makes no difference whether we rotate
the vector in one direction or the coordinate axes in the other direction. Rotating the vector is called an active transformation, whereas rotating the coordinate system is called a
101
passive transformation. Rotations, and any other symmetry operation, preserve the length
of the vector and the dot product of two vectors. A very important example of a coordinate transformation is the transformation from space-fixed to body-fixed coordinates. This
transformation is a sequence of three rotations using the Euler angles.
We next ask what happens to an operator Ŝ when we rotate the coordinate system. Let
Ŝ first operate on A in the old coordinate system,
B = ŜA.
(623)
Next we rotate the coordinate system, using a rotational operator X̂ = Û−1 . This is
equivalent to an active rotation of the vector by an operator Û. The vector B in the
transformed coordinates is
B′ = ÛB = Û(ŜA) = (ÛŜÛ−1 )ÛA = Ŝ′ (ÛA) = Ŝ′ A′ .
(624)
Comparing B and B′ . We see that in the new coordinate system the operator transforms
into
Ŝ′ = ÛŜÛ−1 = X̂−1 ŜX̂.
(625)
We see that we can arrive at B′ by two different routes. We could transform A into B
by operating on it with Ŝ in the original coordinate system, and then rotate B into B′ by
operating on it with Û. Alternatively, we could rotate A into A′ by operating on it with
Û, and then convert it into B′ by operating on it with Ŝ′ . (Of course, we are not restricted
to rotations and only used it as an illustration.)
Equation (625) is called a similarity transform. In group theory, two symmetry operations
belong to the same class if there exists a symmetry operation in the group that can generate a
coordinate transformation such that the first symmetry operation in the original coordinate
system is equivalent to the second operation in the new coordinate system. In other words,
Ŝ and Ŝ′ belong to the same class if there is a symmetry operation Û that transforms one
into the other. An example is C4 and C43 in the C4v group. They belong to the same class
because transforming the coordinate system with σd transforms (x, y) → (y, x), so that
a clockwise rotation of 90◦ (i.e., C4 ) in the original coordinate system is equivalent to a
counter-clockwise rotation of 90◦ (i.e., C43 ) in the transformed coordinate system. (In this
example, Ŝ = C4 , Ŝ′ = C43 , and Û = σd . A set of drawings will will be provided with the
102
following transformations:
A = xî + y ĵ
A′ = Ĉ4 A = y î − xĵ
(626)
A = xĵ′ + y î′
A′ = Ĉ4′ A = y ĵ′ − xî′ = y î − xĵ.
The first two expressions show the rotation of A in the original coordinate frame and the
second set in the new frame.
What are the properties of Û and Ŝ? Symmetry operations preserve the lengths of
vectors and the angles between them (i.e.,their dot products). The corresponding algebraic
property of the matrices that represent such operations is that they are unitary. The inverse
of a unitary matix equals its Hermitian adjoint; i.e.
ÛÛ† = 1
(627)
Û−1 = Û† .
(628)
or
The Hermitian adjoint of a matrix is the complex conjugate of its transpose; i.e.,
(Û† )ij = Uji∗ .
(629)
A Hermitian matrix is one that equals its Hermitian adjoint; i.e.,
Hij = Hji∗ .
(630)
A real unitary matrix is called an orthogonal matrix. An example of such a matrix is the
rotational matrix in Eq. (620).
A matrix Ŝ that induces a similarity transform does not have to be unitary, but it may
be. A very important example of a similarity transform is one that diagonalizes a matrix.
Here is a numerical example:


1 2 0




A=0 3 0


2 −4 2
(631)
103
Its eigenvalues are 3, 2 and 1, and its eigenvectors are


 


 −1 
0
 −1 


 


|1⟩ =  −1  ; |2⟩ =  0  ; |3⟩ =  0  .


 


2
1
2
The matrices that diagonalize A are


−1 0 −1




X =  −1 0 0  ;


2 1 2


X
−1
0 −1 0




=  2 0 1 ,


−1 1 0

such that
(632)
(633)

3 0 0




X−1 AX =  0 2 0  .


0 0 1
(634)
In this example, none of the matrices are unitary or Hermetian. If, however, A is Hermitian,
then X will be unitary, and as before its columns consist of the eigenvectors of A. Here is
a numerical example:


3 2 − i −3i




A =  2 + i 0 1 − i .


3i 1 + i 0
(635)
Its eigenvalues are −1, 6 and -2, and its normalized eigenvectors are


−1

1 


|1⟩ = √  1 + 2i  ;

7
1




X=

such that
−1
√
7
1+2i
√
7
√1
7
1−2i
√
728
6−9i
√
728
√13
728
1+3i
√
40
−2−i
√
40
√5
40


;




X−1 = 


1 + 3i




1 
 −2 − i 
|3⟩ = √ 


40 
5


1 − 21i

1 


|2⟩ = √
 6 − 9i  ;

728 
13
The matrices that diagonalize A are unitary:



(636)
.
−1
√
7
1+2i
√
728
1−3i
√
40

1−2i
√
7
6−9i
√
728
2+i
√
40
√1
7
√13
728
√5
40


,

(637)

1 0 0




−1
X AX =  0 6 0  .


0 0 −2
(638)
104
A Hermitian operator is not unitary. In fact, a unitary operator may be written as
Û = eiĤ , where Ĥ is Hermitian. Suppose we have an orthonormal basis set {ui }. The
members of that basis set could be, for example, the eigenfunctions of a Hermitian operator,
or they could be the unit vectors in three dimensional coordinate space. Now let’s transform
that basis to a new one, {vi }, with some operator, Û,
|vi ⟩ = Û|ui ⟩.
(639)
That is to say that we may expand the new basis vectors in terms of the original ones,
|vi ⟩ =
∑
Uji |uj ⟩.
(640)
j
One may prove with linear algebra that a necessary and sufficient condition for the new
basis set to also be orthonormal is that U is a unitary operator,
(U −1)ij = Uji∗ .
(641)
An example of a unitary operator in quantum mechanics is the time evolution operator,
Û(t2 , t1 ) = e−iĤt/~ ,
(642)
where H is the Hamiltonian. This operator transforms a wave function from its initial state
at time t1 to a state at a later time t2 ,
ψ(r, t2 ) = Û(t2 , t1 )ψ(r, t1 ).
(643)
Equation (642) is true for a conservative Hamiltonian (i.e., one that does not depend on time
explicitly), and may be derived by integrating the time-dependent Schrodinger equation.
The simplest application of the time evolution operator is the evolution of a stationary state
wave function, in which case Ĥ in the exponent is just the constant energy, E.
Problem 61. We showed that a symmetry operation S and a coordinate transformation U
may be performed in either order. That is, we may operate on A in the original coordinate
system,
B = SA,
(644)
and then transform the coordinates so that
B ′ = U B.
(645)
105
Alternatively, we may first transform coordinates
A ′ = U A,
(646)
B ′ = S ′A ′
(647)
S ′ = U SU −1 .
(648)
and then operate on A ′ ,
where
Suppose that S is the inversion operation in two-dimensions, i(xy), and U is a rotation
about the z-axis by an angle ϕ. Write down the 2 × 2 matrices for U, S, and S ′ and calculate
B ′ by both routes.
LECTURE 23: MATRIX REPRESENTATIONS OF A GROUP
Returning to group theory, we may show that orthogonal matrices corresponding to
symmetry operations satisfy the postulates of a group and are therefore said to represent
the group. Take for example the C2v group. We see by inspection that

E

x

=
y

C2 (z) 

=
σv (xz) 

x
=

σv (yz) 
x
y
0 −1

=

x
1 0
0 −1
−1 0
0 1

=
y



x

x

=
y


x
=


x
y
−x
−y
(649)

=


(650)


y


y


y


−1 0
y

1 0
0 1

x

x
−y
−x
y

(651)


(652)
106
We readily show that these 2 × 2 matrices satisfy the multiplication table of the group.
For example,

σv (xz)C2 (z) = 

1 0
0 −1

−1 0
0 −1


=
−1 0

 = σv (yz).
(653)
0 1
At this point it is useful to recognize that the C2v group is the same as (“isomorphic to”)
the non-Abelian group of order 4 that we discovered previously:
E C2 σ σ ′
E E C2 σ σ ′
C2 C2 E σ ′ σ
σ σ σ ′ E C2
σ ′ σ ′ σ C2 E
In choosing a representation, we do not have to restrict ourselves to coordinate transformations. For example, we may construct a four dimensional representation of the C2v group
from the four orientations of the hydrogen atoms in water, where Ā indicates that atom A
is “pointing” below the plane. The four operations are
E · AB → AB; C2 · AB → B̄ Ā; σ · AB → BA; σ ′ · AB → ĀB̄.
The corresponding basis vectors are:
 
 
1
0
 
 
0
1
 
 
AB =   ; BA =   ;
0
0
 
 
0
0
The symmetry operations are then



0
1 0 0 0



0
0 1 0 0



;
C
(z)
=
E=


2
0
0 0 1 0



1
0 0 0 1
 
0
 
0
 
ĀB̄ =   ;
1
 
0
represented by the matrices


0 1
0 0 1


1 0
0 1 0


;
σ(xz)
=


0 0

1 0 0

0 0
0 0 0
 
0
 
0
 
B̄ Ā =  
0
 
1
0 0


0 0

;
0 1

1 0
(654)

0 0 1 0



0 0 0 1


σ ′ (yz) = 
;
1 0 0 0


0 1 0 0
(655)
We readily show that these matrices also satisfy the multiplication table of the C2v group.
There are in fact an infinite number of representations of a group. This would seem to
make the concept of a representation of limited value, until we realize that there is a core
107
set of unique representations, out of which all others are built up. To determine how many
such unique representations exist for a particular group and what are their dimensions, we
use the following chain of reasoning:
1. First, we recognize that if a set of matrices form a representation of a group, then
any similarity transform of that set is also a representation of the group. We can prove this
statement by noting that if AB = C, then
(X −1 AX)(X −1 BX) = (X −1 (AB)X) = X −1 CX.
(656)
In other words, the similarity transforms of the matrices have the identical multiplication
table.
2. Next we assert without proof that if we have a set of matrices that represent a
group, we can find some matrix X such that the similarity transforms of all the matrices
in the representation are block-diagonal. Some of the matrices may happen turn out to be
completely diagonalized, but this will not always be the case. We know from linear algebra
that the diagonal elements are the eigenvalues of the matrix, and the columns of the matrix
used to perform the similarity transform are the eigenvectors of the original matrix. The
problem here is to diagonalize all the matrices in the group with the same transform. This
cannot generally be done. The best we can do is block diagonalize them. The dimensions
of the blocks for the lower order groups may be 1, 2, or 3. For the polyhedral groups they
may be as large as 5.
3. But now we recognize that block-diagonalized matrices have the unique property that
different blocks do not mix with each other when the matrices are multiplied. All the zeros
outside the blocks guarantee that this will be true. From this property it follows that each
set of blocks (one for each symmetry operation) constitutes a representation of the group.
If a block cannot be shrunk further, then it is said to be an irreducible representation of
the group. (If it can be shrunk, then it is a reducible representation.) These are the unique
representations that we seek.
4. We state without proof that the number of irreducible representations of a group
equals the number of classes in the group, and the sum of squares of the orders of all the
irreducible representations equals the order of the group. In other words, if there are J
108
classes with dimensions l1 , l2 , . . . lJ , then
J
∑
lj2 = h.
(657)
j
In the case of C2v , each element belongs to a class by itself. Since the order of the group is
4, there are 4 one-dimensional representations of the group. They are
E = 1, C2 = 1, σ = 1, σ ′ = 1
E = 1, C2 = 1, σ = −1, σ ′ = −1
(658)
E = 1, C2 = −1, σ = 1, σ ′ = −1
E = 1, C2 = −1, σ = −1, σ ′ = 1,
You can see, for example, that C2 σ = σ ′ for all four representations. For the C4v group there
are 5 irreducible representations, corresponding to the 5 classes of symmetry operations,
(E), (C4 , C43 ), (C2 ), (σv , σv′ ), and (σd , σd′ ) of orders 1, 1, 1, 1, and 2.
A sufficient, but not necessary condition that all irreducible representations be 1dimensional is that the group be cyclic. In that case each element belongs to a class by
itself. For a cyclic group of order h there must therefore be h classes, each of dimension 1.
Any higher dimensional representation of the group may be block-diagonalized into a
linear combination of the irreducible set. For example, the 4 dimensional representation of
C2v that we constructed may be transformed into

1 0 0 0


1 0 0
0


1 0 0 0


1 0
0 0









0 1 0 0
0 1 0 0 
 0 −1 0 0 
 0 −1 0 0 





 ′


E=
; C2 (z) = 
; σ(xz) = 
; σ (yz) = 
.
0 0 1 0
 0 0 −1 0 
0 0 1 0 
 0 0 −1 0 








0 0 0 1
0 0 0 −1
0 0 0 −1
0 0 0 1
(659)
The diagonal elements of these matrices are the 1-dimensional representations that we found
earlier.
5. The matrices in an irreducible representation have the additional wonderful property
that all we need to know about them is their traces, χ, (also called their “characters”) in
order to tell them apart. Let’s designate individual matrix elements for a representation i
of symmetry operation R by the symbol Γi (R)pq . (p and q label the row and column of that
109
matrix element.) Then the character of the matrix is defined by
χi (R) =
li
∑
Γi (R)pp .
(660)
p=1
6. Next we prove that matrices that are similarity transforms of each other have equal
characters. We can prove this as part of a more general property that the character of a
product of two matrices does not depend on which matrix comes first, even if the matrices
do not commute. The proof follows directly from the definition of matrix multiplication.
The pq element of the product of matrices A and B is given by
Γi (AB)pq =
li
∑
Γi (A)pm Γi (B)mq .
(661)
m=1
The character of AB is calculated by setting p = q and summing over p,
χi (AB) =
li ∑
li
∑
Γi (A)pm Γi (B)mp =
p=1 m=1
li ∑
li
∑
Γi (B)mp Γi (A)pm = χi (BA).
(662)
m=1 p=1
We may extend this result to a product of three or more matrices, using the associative
property of multiplication.
χ(ABC) = χ((AB)C) = χ(C(AB)) = χ(BCA).
(663)
Note that this proof works only for cyclic prermutations; i.e., χ(ABC) does not necessarily
equal χ(BAC). We may now apply this result to a similarity transform:
χ(X −1 AX) = χ(XX −1 A) = χ(A).
(664)
This result tells us that the characters of all the operations in a class are equal for any given
representation. If there are J classes in a group, all we need to know is the J characters to
describe a particular representation. A chracter table is a J × J square matrix. Each row
of the table contans the characters for a particular irreducible representation of the group,
while the columns contain the J classes of symmetry operations.
110
LECTURE 24: THE GROUP ORTHOGONALITY THEOREM
We now come to the key result that the set of characters for each representation are
coefficients of a set of orthonormal vectors of dimension h. The usefulness of this theorem
is that if we know the characters of some arbitrary representation, we may use the theorem
to project out the irreducible components of that representation, much the same as we can
express a vector in some orthonormal basis. In other words, we can write the reducible
representation as a linear combination of the irredicible representations for that group.
7. We start with the “Great Orthogonality Theorem,” may be written mathematically
as follows:
h
∑
h
Γi (R)pq Γj (R)∗p′ q′ = √ δij δpp′ δqq′ .
li lj
R=1
(665)
The sum is over all the symmetry operations in the group, and quatities that are being
multiplied in pairs are individual matrix elements of two different representations.
8. From this theorem we can prove that the sum of the squares of all the characters in a
representation equals the order of the group. Setting i = j, q = p, q ′ = p ′ and summing the
left hand side over p and p ′ gives the sum of the squares of the characters in representation
i. Summing the right hand side over p and p ′ gives
i ∑
i
i
h∑
h∑
δpp′ δpp′ =
δpp = h.
li p=1 p′ =1
li p=1
l
l
l
(666)
9. It also follows that the sum of the products of the characters from two different
representations is zero. Performing the same sums as before,
h
∑
χi (R)χj (R) = hδij .
R=1
We illustrate these principles for the tetrahedral group, Td . Its character table is
E 8C3 3C2 6S4 6σd
A1 1 1
1
1
1
A2 1 1
1
-1 -1
E 2 -1
2
0
0
T1 3 0
-1
1
-1
T2 3 0
-1
-1
1
(667)
111
The group has 24 elements separated into in 5 classes. The group accordingly has 5
irreducible representations, of which two are 1-dimensional, one is 2-dimensional, and two
are 3-dimensional. The dimension of the representation equals the character for the identity
operation. The sum of the squares of the dimensions of these representations equals the
order of the group:
12 + 12 + 22 + 32 + 32 = 24.
(668)
Note that there are J = 5 terms in this sum. The sum of the squares of the characters of
any representation also equals the order of the group, as proved by Eq. (666). In this sum
there are h terms. For example, for the T2 representation:
32 + (8 × (0)2 ) + (3 × (−1)2 ) + (6 × (−1)2 ) + (6 × 12 ) = 24.
(669)
The “dot product” between any two sets of characters is zero. For example, for the T1
and T2 representations,
(3 × 3) + (8 × 0 × 0) + (3 × (−1) × (−1)) + (6 × 1 × (−1)) + (6 × (−1) × 1) = 0.
(670)
Suppose we have some arbitrary representation with character χ(R) for each of the h
operations. We can express that representation as a linear combination of the J irreducible
ones,
χ(R) =
J
∑
aj χj (R).
(671)
j=1
We may project out the aj coefficients, as follows: Pick an irreducible representation, i.
Next, “project” the characters of this reperesentation onto those of the arbitrary reducible
representation,
h
∑
R
χ(R)χi (R) =
h ∑
J
∑
R=1 j=1
aj χj (R)χi (R) =
J
∑
j=1
aj
h
∑
χj (R)χi (R) = h
R=1
J
∑
aj δij = hai . (672)
j=1
In the next to last step we invoked Eq. (667). The ith irreducible component of the arbitrary
representation appears as ai blocks in the diagonalized representation, where
1∑
ai =
χ(R)χi (R).
h R=1
h
(673)
Suppose, for example, we have the following reducible representation of the Td group:
χ(E) = 20, χ(C3 ) = 2, χ(C2 ) = 8, χ(S4 ) = 4, χ(σd ) = 6.
(674)
112
The projection of the T2 component is
AT2 =
1
{(1 × 3 × 20) + (8 × 0 × 2) + (3 × (−1) × 8) + (6 × (−1) × 4) + (6 × 1 × 6)} = 2.
24
(675)
This means that the block-diagonalized representation contains two T2 blocks.
Another example is the 4-dimensional representaion of the C2v group shown in Eq. (655).
The characters of this representation are χ(E) = 4, χ(C2 ) = 0, χ(σ) = 0, χ(σ ′ ) = 0, and
we readily deduce this representation to A1 + A2 + B1 + B2 .
LECTURE 25: SYMMETRY PROPERTIES OF MATHEMATICAL FUNCTIONS
Reading: Cotton, Chapter 5.
In this lecture we will explore what representations are generated by various algebraic
functions. The core question is how different symmetry operations transform these functions.
For example, inversion changes the signs of all their coordinates. We can answer this question
without reference to any particular group. We then calculate the characters of each of the
matrices that describe the transformations. By comparing these characters with those of
the irreducible representations of a particular group, we can determine which representation
of that group is generated by the function of interest. We then say that this functions
“transform as” that representation.
This idea is best understood by working out an example. We have chosen the C4v group
for this purpose. We will consider the effects of its four classes of symmetry operations (in
addition to E) on nine algebraic functions. In doing so, it is sufficient to examine only one
operation from each class, since the characters of all the operations in a class are equal.
In the following table we have listed in each cell the transformed function and the character of the matrix that describes the transformation. In dealing with functions of x and
y, we recognize that some operations interchange them, thereby generating a 2-dimensional
representation.
113
C4 (z)
C2 (z)
σv (xz)
σd
rep
z
z, 1
z, 1
z, 1
z, 1
A1
x
y, 0
−x, −1
x, 1
y, 0
E
y
−x, 0
−y, −1
−y, −1
x, 0
E
z2
z2, 1
z2, 1
z2, 1
z2, 1
A1
x2 + y 2 x2 + y 2 , 1 x2 + y 2 , 1 x2 + y 2 , 1 x2 + y 2 , 1 A1
x2 − y 2 y 2 − x2 , −1 x2 − y 2 , 1 x2 − y 2 , 1 y 2 − x2 , −1 B1
xy
−xy, −1
xy, 1
−xy, −1
xy, 1
B2
xz
yz, 0
−xz, −1
xz, 1
yz, 0
E
yz
−xz, 0
xz, 0
E
−yz, −1 −yz, −1
We also want to see how rotations transform. We can do this geometrically by looking at
the effect of the symmetry operations on curly arrows wound around each axis. A better
way is to do this algebraically. The key is to notice that the direction of a rotation about a
particular unit vector is given by the cross product of the other two unit vectors. That is,
Rz ∼ î × ĵ = k̂
Rx ∼ ĵ × k̂ = î
(676)
Ry ∼ k̂ × î = ĵ
.
All we need to do is determine how each unit vector transforms and then take their cross
products. In doing so, keep in mind the following transformations:
C4 (z) : î → ĵ; ĵ → −î; k̂ → k̂
C2 (z) : î → −î; ĵ → −ĵ; k̂ → k̂
(677)
σv (xz) : î → î; ĵ → −ĵ; k̂ → k̂
σd (xy) : î → ĵ; ĵ → î; k̂ → k̂
C4 (z)
C2 (z)
Rz j × −i = k, 1 −i × −j = k, 1
σv (xz)
σd
rep
i × −j = k, −1 j × i = k, −1 A2
Rx −i × k = j, 0 −j × k = −i, −1 −j × k = −i, −1 i × k = −j, 0 E
Ry k × j = −i, 0 k × −i = −j, −1
k × i = j, 1
k × j = −i, 0 E
114
For example, C4 Ry = C4 (k × i) = (C4 (k) × (C4 (i)) = k × j = −i.
Problem 62. Two long-standing errors in the character tables were found just in 2007 by
R. Shirts, J. Chem. Ed. 84, 1882 (2007). In this article he claims that (Rx , Ry ) transform in
the S8 group according to E3 and not E1 . He also claims that the column headings for the
2S8 and 2S83 classes are reversed in the D8h character table. Verify that these statements are
correct. For the first statement it is sufficient to calculate the traces of just the S8 operator
on the three unit vectors and then take the appropriate cross products. For the second
statement it is sufficient to look at the transformation of (x, y) by the S8 operator. (Hint:
The character table contains two rows for the E representations. You should take the sum
of these rows, and all the numbers should be real. You should then compare the traces of
the operators on Rx , Ry with the characters of the E1 and E3 representations.)
Our next inquiry is how products of functions transform. If the individual functions generate one dimensional representations, their product will also generate a one dimensional
representation, the characters of which are just the products of the characters of the individual functions. A more interesting thing happens if the functions generate higher dimensional
representations. Our goal is to determine the character of the higher dimensional representation for each symmetry class. Suppose, for example, we have two functions, X1 and X2 , that
generate a 2D representation. The transformation produced by some particular operation,
S, is represented by a matrix, A, with coefficients aij given by
SX1 = X1′ = a11 X1 + a21 X2
SX2 = X2′ = a12 X1 + a22 X2
(678)
.
Now let’s suppose we have a second pair of functions, Y1 and Y2 . The transformation produced by the same symmetry operator is represented by another matrix, B, with coefficients
bij given by
SY1 = Y1′ = b11 Y1 + b21 Y2
SY2 = Y2′ = b12 Y1 + b22 Y2
(679)
.
We would like to know how the product of a pair functions, one taken from each set,
115
transform. For example,
SX1 Y1 = X1′ Y1′ = a11 b11 X1 Y1 + a11 b21 X1 Y2 + a21 b11 X2 Y1 + a21 b21 X1 Y1 .
(680)
The complete transformation matrix is given by

X1′ Y1′


a11 b11 a11 b21 a21 b11 a21 b21

X1 Y1


X1 Y1





 

X Y 
 X′ Y ′   a b a b a b a b   X Y 
1
2
1
2
11
12
11
22
21
12
21
22




 1 2 

 = Ĉ 

=

 X2 Y1 
 X ′ Y ′   a12 b11 a12 b21 a22 b11 a22 b21   X2 Y1 




 2 1 
′ ′
X2 Y2
X2 Y2
a12 b12 a12 b22 a22 b12 a22 b22
X2 Y2
(681)
We see that the trace of the transformation matrix, Ĉ, is
χ(C) = a11 (b11 + b22 ) + a22 (b11 + b22 ) = (a11 + a22 )(b11 + b22 ) = χ(A)χ(B).
(682)
We learn two things from this example. First, the dimension of the representation generated
by the products of two functions is the sum of the dimensions the individual representations,
and, second, the character of the new representation is the product of the original characters.
The new representation is called the direct product of the individual representations. Taking
the Td group as an example, the direct product of the E and T2 representations is written
as E ⊗ T2 . The characters of this product are 6, 0, −2, 0, 0. We see by inspection of the Td
character table that
E ⊗ T2 = T1 + T2 .
(683)
Problem 63. Use the previous result to determine the character of the representation of the
Td group generated by the product of a dx2 −y2 orbital, a dxy orbital, and the radial vector,
r. Write your answer as a linear combination of irreducible repreentations.
Problem 64. Determine the representations of x, y, z, x2 + y 2 , x2 − y 2 , z 2 , xy, xz, yz, Rx , Ry ,
and Rz for the C3v group. Do this by showing how each function is transformed by all
the symmetry operations in the group, and compare your results with the traces of the
irreducible representations of the group.
LECTURE 26: INFRARED AND RAMAN SELECTION RULES
We come now to one of the most important applications of group theory to spectroscopy,
namely a method for determining whether a transition is allowed. We know from earlier
116
lectures that the transition amplitude for a dipole-allowed transition is proportional to the
matrix element of the dipole operator µ in Eq. (579),
∫∫
µ21 =
ψ2∗ µ ψ1 dτnuc dτel ,
(684)
where the integral is over all electronic and nuclear coordinates.
The value of this integral is a number that does not change if we carry out a symmetry
operation on the molecule. This is true because the expectation value of an observable
should not be affected by a symmetry operation. To get a non-zero value, the integrand, or
some part of it, must be a basis for (“transform as”) the A1 representation. Our procedure
therefore is to take the direct product of the functions ψ1 , ψ2 , and the vector components
of µ and to determine whether it contains the A1 representation. More specifically, we take
the direct product of (ψ1 ψ2 ) with x, y, and z.
We can simplify the problem by taking advantage of the fact that all we need to know
is whether product contains an A1 component. Suppose in general we are interested in
the symmetry of the product of two functions, fA fB . Let χAB (R) be the character of the
representation generated by this product. As before, we express this character as a sum over
all irreducible representations,
χAB (R) =
J
∑
aj χj (R),
(685)
j=1
where the coefficients are given by
1∑
ai =
χAB (R)χi (R).
h R
h
(686)
In particular, the contribution of the A1 representation is
1∑
χAB (R),
h R
h
aA1 =
(687)
because χA1 (R) = 1 for all R. This last expression may be simplified further as
1∑
=
χA (R)χB (R),
h R
h
aA1
(688)
and this sum vanishes according to the Group Orthogonality Theorem unless A = B. For
a dipole transition, we choose fA = ψ1 ψ2 and f2 = x, y, or z. It follows that a transition
117
is dipole forbidden unless the representation generated by ψ1 ψ2 contains a representation
generated by x, y, or z. Moreover, if one of the wave functions, say ψ1 , corresponds to the
ground state, it is a function of x2 +y 2 +z 2 and is therefore totally symmetric (i.e., it belongs
to the A1 representation). In that case our criterion simplifies to requiring that ψ2 belong to
the same representation as one of the Cartesian coordinates. For atoms, the only symmetry
element is î, which leads to the selection rule that the only allowed trasitions are between a
gerade and an ungerade state. In other words, the only allowed dipole transitions for atoms
are between states of oposite parity. This result leads to the LaPorte rule that ∆l = ±1,
where l is the orbital angular momentum quantum number.
For molecules, we need to take into account vibrational and rotational degrees of freedom.
If we assume that the electronic and nuclear wave functions are separable (by the BornOppenheimer approximation), the transition dipole moment is given by
(∫
)
∫
∗
∗
∗
µ12 = ψv2 J2 m2
ψel,2 µ ψel,1 dτel ψv1 J1 m1 dτnuc ,
where
µ = µel + µnuc =
∑
eri +
∑
qα Rα ,
(689)
(690)
α
i
where the first vector sum is over electronic coordinates and the second is over nuclear
coordinates. The transion dipole matrix element may therefore be broken down into two
terms,
∫
µ12 =
ψv∗2 J2 m2
∫
+
(∫
)
∗
ψel,2
ψv∗2 J2 m2 µnuc
∗
µel ψel,1
dτel
(∫
∗
ψel,2
ψv1 J1 m1 dτnuc
)
∗
ψel,1 dτel ψv1 J1 m1 dτnuc .
(691)
The second integral vanishes unless ψel,1 = ψel,2 . For a transition within a single electronic
state, we may therefore write transition dipole matrix element as
∫
µ12 = ψv∗2 J2 m2 µ(R1 , R2 , . . . )ψv1 J1 m1 dτnuc ,
where
∫
µ(R1 , R2 , . . . ) =
∗
(µel + µnuc )ψel dτel .
ψel
(692)
(693)
A necessary condition for µ12 not to vanish is that ψtotal µ ψtotal contain a totally symmetric
component. We may go further and deduce selction rules for changes in the quantum
numbers by looking at the properties of the wave functions. Let’s consider first a purely
118
rotational transition (i.e., v2 = v1 ). The integral over electronic coordinates yields the
permanent dipole moment,
µ(R1 , R2 , . . . ) = µ0 (sin θ cos ϕî + sin θ sin ϕĵ + cos θk̂).
(694)
The selection rules on J and M are derived from the angular integrals over the rotational
wave functions, with dτnuc = sin θ dθ dϕ. For linear molecules,
∫ π ∫ 2π
1
YJ∗2 M2 (θ, ϕ)µ(θ, ϕ)YJ1 M1 (θ, ϕ) sin θ dθ dϕ
µ12 =
2π 0 0
∫ ∫
µ0 π 2π ∗
=
YJ2 M2 (θ, ϕ)YJ1 M1 (θ, ϕ)(sin θ cos ϕî + sin θ sin ϕĵ + cos θk̂) sin θ dθ dϕ.
2π 0 0
(695)
This integral leads to the selection rules ∆J = ±1, ∆M = 0, ±1. We will return to this
result when we discuss rotational spectroscopy.
For pure vibrational transitions (J2 , M2 = J1 , M1 ), it would appear that µ12 = 0 because
the vibrational wave functions are orthogonal, causing the nuclear integral (Eq. (692))
to vanish. (The rotational wave functions are also orthogonal, but there the molecular
orientation comes into play.) This is paradoxical because it seems to imply that molecules
do not have an infrared spectrum! We resolve this paradox by noting that the dipole
moment may change as the molecule vibrates. We express the vibrations as motion along
one or more of the 3N − 6 vibrational normal coordinates. Normal modes are harmonic
motions of a set of coupled oscillators. Each normal coordinate is a linear combination of
the Cartesian coordinates of the oscillators. The Schrodinger equation is separable in normal
coordinates, with each eigenvalue corresponding to one of the normal mode frequencies and
each eigenfunction being a function of the normal mode coordinate. Denoting the normal
coordinates by Qi , we may expand the dipole moment in a Taylor series,
µ = µ0 +
3N
−6 (
∑
i=1
∂µ
∂Qi
)
Qi + . . . ,
(696)
0
where the partial derivatives are evaluated at the equilibrium geometry of the molecule.
Inserting Eq. (696) into (689), and letting ψ1 and ψ2 be functions of one of the normal
coordinates, say Qj , only one term in the sum survives,
(
) ∫
∂µ
ψv2 (Qj )Qj ψv1 (Qj )dQj .
µ12 =
∂Qj 0
(697)
119
As before, the necessary condition for µ12 not to vanish is that the product of the wave
functions belong to the same representation of the Cartesian coordinates. Assuming again
a totally symmetric ground state wave function, the transition will be infrared active if the
excited normal mode belongs to the same representation as one of the Cartesian coordinates.
If the molecule is aligned in space, we may also determine the polarization of the light (x, y,
or z) needed for a photon to be absorbed. We note that, unlike the case of a pure rotational
transition, the molecule does not need to have a permanent dipole moment. The presence
of Qj in the integrand imposes a further restriction that v2 = v1 ± 1, neglecting the effects
of anharmonicity and vibrational-rotational coupling.
While the molecule does not need to have a permanent dipole moment to be IR active,
it is necessary that the derivavtive of the dipole moment not be zero; i.e.
(
)
∂µ
̸= 0.
∂Qj 0
(698)
This condition cannot be satisfied for a homonulcear molecule or for a totally symmetric
stretch of a molecule with a center of symmetry such as CO2 or SF6 .
We consider next a Raman transition. The first two terms of the Taylor series of the
dipole moment in powers of the electric field are
µ = µ0 + α · E + . . . ,
where α is the polarizability tensor,

α α α
 xx xy xz

α =  αyx αyy αyz

αzx αzy αzz
(699)



.

(700)
The induced dipole moment of a molecule in an electric field is therefore given by


 

E
α α α
µ
 x   xx xy xz   x 


 

(701)
 µy  =  αyx αyy αyz   Ey  ,


 

Ez
αzx αzy αzz
µz
where α is a tensor because in general the ploraization need not be parallel to the electric
field. The polarizability of a bulk medium is given by the average of the diagonal terms of
the polarizability,
1
ᾱ = (αxx + αyy + αzz ),
3
(702)
120
where the polarizability tensor is evaluated in a coordinate system coinciding with the principle axes of the molecule. This quantity is related to the bulk polarization by multiplying
it by the molecular density, ρ,
P = ρµ = χe E,
(703)
which in turn relates α to the susceptibility,
χe = ρᾱ.
(704)
The average energy of a molecule in an electric field is given by
W = W0 − µ · E −
1
E · α · E.
2
(705)
The factor of 1/2 may be derived classically and also comes out of perturbation theory.
How do we calculate the polarizability? The simplest approximation is to treat the
electron as a harmonic oscillator with frequency ω0 placed in a static electric field, E. The
restoring force on the atom just balances the Coulomb force of the field; i.e.,
−mω02 x = eE.
(706)
The dipole moment induced by the field is
µ = −ex =
e2
E,
mω02
(707)
giving for the polarizability
α=
e2
.
mω02
(708)
A collection of electrons with different frequencies ωj may be treated the same way, giving
α=
e2 ∑ fj
,
m j ωj2
(709)
where 0 ≤ fj ≤ 1 is the oscillator strength of frequency ωj . This is the DC polarizability.
To obtain the polarizability in an oscillating electric field, we use a damped forced oscillator model similar to the one in the Lorenz model. Solving the differential equation
yields
α(ω) =
e2 ∑
fj
.
2
m j ωj − ω 2 − iω/τ
(710)
121
The complex part of the polarizabiliy expresses the fact that the induced dipole may be out
of phase with respect to the driving field. The real and imaginary parts of α are related by
the Kramers-Kronig relations.
A quantum mechanical derivation of α uses time-dependent perturbation theory to calculate the time evolution of states ⟨2| and |1⟩ caused by a sinusoidal perturbation. The
perturbation for ⟨2| is epE0 (eiωt + e−iωt ), where p = x, y, or z. Similarly, the perturbation
for |1⟩ is eqE0 (eiωt + e−iωt ), where q = x, y, or z. It can then be shown that the matrix
element of the pq component of the α tensor is
[
]
1 ∑ ⟨1|µp |n⟩⟨n|µq |2⟩ ⟨1|µq |n⟩⟨n|µp |2⟩
.
⟨1|αpq |2⟩ =
−
~ n ω + ωn2 − 12 iΓn
ω − ωn1 − 12 iΓn
(711)
In the special case of Γ = 0 and |1⟩ = |2⟩,
[
]
2 ∑ ⟨1|µp |n⟩⟨n|µq |2⟩
⟨1|αpq |1⟩ =
.
~ n
ω + ωn2
(712)
The intermediate states in the sum arise from the perturbation expansion of the wave function in the unperturbed basis. This result reduces (nearly) to the classical result, Eq. (710),
when |2⟩ = |1⟩ (i.e. for the ground state polarizability). (We need to look more carefully at
the Γn term, which was introduced to account for the lifetimes of the intermediate states.)
We may interpret this expression as the sum over all pairs of transitions from the ground
state to an intermediate state and from the intermediate state to the final state. This is a
two-photon process and is just what we mean by a Raman transition.
Let us pause for a moment to consider the units of polarizability. Since α = µ/E, its units
are
Cm
V /m
′
= Cm2 /V . It is more common to refer to the polarizability volume, α = α/4πϵ0 ,
′
which has the units of avolume. In common usage, α and α are used interchangeably. It
can be shown, for example, that the ground state energy of the hydrogen atom in a static
electric field is
e2
(713)
− 9πϵ0 a30 E 2 ,
8πϵ0 a0
where a0 is the Bohr radius. The first term is just the 1s energy, and the second term
W0 = −
is caused by the quadratic Stark effect, calculated using second order time-independent
perturbation theory. The polarizability volume is seen to equal 92 a30 = 0.666Å3 (recall Eq.
(705)).
Returning now to the probability of a Raman transition, we see that the appearance
of products of dipole matrix elements in the perturbation expansion of the polarizability
122
shows that α transforms as a product of Cartesian coordinates. It follows that for a Raman
transition to be allowed, it is necessary that the excited state belong to the same symmetry
as a Cartesian product.
As before, we may obtain a selection rule for a vibrational Raman transition by expanding
the matrix element of αpq in a Taylor series of the normal coordinates,
αpq = (αpq )0 +
3N
−6 (
∑
i=1
∂αpq
∂Qi
)
Qi + . . .
(714)
0
Again, we project out just one term in dipole integral, Eq. (689). The remaining integral
vanishes unless v2 = v1 ± 1. A more detailed discussion is found in Herzberg, Chapter III
[4].
In general, it is possible for some normal modes to be both (or neither) ir and Raman
active. An example is the C3v group, where z, z 2 , and x2 +y 2 belong to the A1 representation,
and (x, y) and (xz, yz) belong to the E representation. An important exception is for
molecules that are centro-symmetric, i.e., ones that have a center of symmetry. In that case,
all representations are either gerade or ungerade. The Cartesian coordinates are all ungerade,
whereas their products are all gerade. It follows that for centro-symmetric molecules a
mode cannot be both ir and Raman active. Infrared transitions from the ground state may
terminate only on ungerade excited states, whereas Raman transitions state may terminate
only on gerade states.
The selection rule for pure vibrational (i.e. non-electronic) and rotational transitions
come from the orthonormal properties of the eigenfunctions. For the vibrational case, the
Hermite polynomials satisfy a recursion relation,
1
xHv (x) = vHv−1 (x) + Hv+1 (x),
2
so that the integral
(715)
∫
ψv2 (x) x ψv1 (x)dx = 0
(716)
unless v2 = v1 ± 1. This condition does not hold if ψv2 and ψv1 belong to different electronic
manifolds (with different equilbrium distances and force constants).
A similar argument works for the rotational wave functions, but without the restriction of being in the same electronic state. The rotational wave function is proportional to
123
PJM (cos θ)eiM ϕ . The legendre polynomials satisfy the recursion relations,
)
1 (
|M |
|M |
|M |
cos θPJ (cos θ) =
(J − |M | + 1)PJ+1 (cos θ) + (J + |M | + 1)PJ−1 (cos θ)
2J + 1
)
1 ( |M |+1
|M |
|M |+1
sin θPJ (cos θ) =
PJ+1 (cos θ) + PJ−1 (cos θ)
2J + 1
)
1 (
|M |−1
|M |−1
(J + |M |)(J + |M | − 1)PJ−1 (cos θ) + (J − |M | + 1)(J − |M | + 2)PJ+1 (cos θ)
=
2J + 1
.
(717)
Since the Cartesian coordinates are proportional to sin θ and cos θ, it follows that the nuclear
integral vanishes unless ∆J = ±1, ∆M = 0, ±1. The selection rule on M may also be derived
from the integral over ϕ.
LECTURE 27: SYMMETRY PROPERTIES OF THE NORMAL MODES
Reading: Cotton, Chapter 10, sections 1-4,6,7.
The central concept in vibrational spectroscopy of a polyatomic molecue having N atoms
is that the potential energy function, V (x1 , x2 , . . . , x3N ), may be transformed to a function
of 3N − 6 or 3N − 5 for a linear molecule) of mass-weighted coordinates, Qi , such that
near the potential energy minimum the kinetic and potential energies may be written as
quadratic forms,
3N −6
1 ∑
ki Q2i ,
V =
2 i=1
and
T =
3N −6
1 ∑ 2
Q̇ .
2 i=1 i
(718)
(719)
corresponding to 3N − 6 independent harmonic oscillators. The oscillations along each
coordinate is called a normal mode. The derivation of these coordinates and their characteristic frequencies may be found in many books (e.g., Goldstein). Our concern here is with
the symmetry properties of these modes. A key idea is that V and T are unchanged by
a symmetry operation. The coordinates themselves may be used to derive an irreducible
representation of the symmetry group of the molecule, and our goal is to determine what
those representations are.
One way to determine the possible representations of the different modes is to use the
coordinates of all the atoms in the molecule to generate a 3N -dimensional representation.
124
This is not as difficult as it may seem because all we need are the characters, and to calculate
those we may ignore all atoms that are not transformed into themselves by any symmetry
operation in the group. (The reason this is true is that if an atom moves as a result of
the operation, its coordinates will not lie on the trace of the matrix and therefore do not
contribute to the character of the representation.)
We will use NH3 as an illustration. The molecule belongs to the C3v group with symmetry
classes E, C3 , and σv . The character of E is the dimension of the representation, namely
12 (four atoms with three coordinates per atom). To calculate the character of C3 we need
consider only the transformation of the nitrogen atom. The rotation matrix for a 2π/3
rotation is



C3 (z) = 

− 12
√
3
2
√
3
2

0


− 12 0  ,

0 0 1
(720)
and its character is 0. Similarly, for σv we need consider only the N atom and one H atom.
Since each vertical plane contributes an equal amount to the character, we need to examine
only one of them:


1 0 0




σv (xz) =  0 −1 0  ,


0 0 1
(721)
and its character is 1. The total character of σv for both atoms is therefore 2. The set
of characters for all three classes is therefore (12, 0, 2). We can express the representation
described by these characters as the the sum of irreducible representations 3A1 +A2 +4E. Of
these, three degrees of freedom correspond to pure translation, which transform as (x, y, z)
with symmetry A1 + E, and three correspond to rotation, which transform as (Rx , Ry , Rz )
with symmetry A2 + E. By subtraction, the vibrations have symmetry 2A1 + 2E. We can
characterize these vibrations further as either stretches or bends. For the former, we use
the three N −H bonds as a basis for representing the group. For stretches, we need to see
which bonds transform into themselves. By inspection, χ(E) = 3, χ(C3 ) = 0, χ(σv ) = 1,
which reduce to A1 + E. Using the three HN H angles to generate a representation for the
bending degrees of freedom, we again get χ(E) = 3, χ(C3 ) = 0, χ(σv ) = 1, which also reduce
to A1 + E.
Next we would like to determine the geometric propereties of the normal modes. There
125
is a well-established procedure found in books on classical mechanics, and numerical results
may be obtained from electronic structure programs such as Gaussian. Frequently it turns
out that the normal coordinates are similar to the symmetry adapted coordinates, which
may be found by applying a projecion operator to some characteristic diatance or angle of
the molecule. We will show that the operator or “projector” that accomplishes this task is
the following:
P̂ j =
h
lj ∑ j
χ (R)R̂,
h R
(722)
where j is the label of the representation we wish to project out, lj is its dimension, and R̂
is a symmetry operator. To give a specific example, for the C4v group,
P̂ B1 = Ê − 2Ĉ4 + Ĉ2 + 2σ̂v − 2σ̂d .
(723)
The projection operator is derived as follows. (In first reading you may wish to skip this
derivation and go on to the next paragraph.) If we have some arbitary function that can be
expanded as a sum over basis funtions for all the representations,
J
∑
ϕi ,
(724)
P̂ j ψ = ϕj .
(725)
ψ=
i=1
then
The proof is as follows: Consider the transformation of a single basis function,
R̂ϕjt
=
h
∑
Γist (R)ϕis .
(726)
s=1
(The function ϕj is just one of these basis functions.) Left multiplying by a matrix element
and summing over symmetry operations,
∑
Γjs′ t′ (R)∗ R̂ϕjt
R
=
h
∑∑
R
Γjs′ t′ Γist (R)ϕis
s=1
=
h
∑
ϕis
∑
s=1
Γjs′ t′ Γist (R) =
R
h i
ϕ ′ δij δtt′ .
lj s
(727)
We define the partial projector
P̂sj′ t′ =
lj ∑ j
Γ ′ ′ (R)∗ R̂,
h R st
(728)
such that
P̂sj′ t′ ϕit = ϕjs′ δij δtt′ ,
(729)
126
and the full projector,
j
P̂ =
h
∑
t′ =1
P̂tj′ t′
h
lj ∑ ∑ j
lj ∑ i
=
Γt′ t′ (R)∗ R̂ =
χ (R)R̂.
h R t′ =1
h R
such that
P̂ j ϕit
=
h
∑
ϕjt δtt′ δij = ϕjt δij .
(730)
(731)
t′ =1
It follows that if ψ is a linear combination of functions taken from different represetations,
P̂ j projects out just the j th term.
We will ilustrate the methodology of projection opertators to find the symmetry coordinates of the square planar molecule XeF4 , which has D4h symmetry. The molecule has 5
atoms and therefore 9 normal coordinates. Using drawings provided in class, we show the
following:
1) There are 4 stretching coordinates of symmetries a1g + b1g + eu .
2) The in-plane bends have symmetries a2g + b2g + eu . Of these, the a2g mode corresponds
to a pure rotation.
3) The out-of-plane bends have symmetries a2u + b2u + eg . Of these, the eg modes
correspond to pure out-of-plane rotations.
Problem 65. Determine the symmetry types of all the normal modes of a tetrahedral
molecule with the formula AB4 . Characterize the modes as either ir or Raman active and
also as either stretches or a bends. Use the coordinates of all the atoms to generate a
representation of the group.
Problem 66. Repeat the previous problem for the pyrazine molecule. Assume that the
molecule lies in yz plane, with the nitrogen atoms lying along the z axis. You may check
your answer with the Figure 1 of J. Chem. Phys. 74, 4893 (1981).
Here is a subtle point. Until now we consoidered the symmetry of the normal coordinates,
which we obtained by analyzing the representation generated by the nuclear motion. But
what is the synnetry of the wave function itself? It depends on the vibrational quantum
number! The symmetry the vibtrational ground state (v = 0) wave function is A1 because
the function is a simple Gaussian. For v = 1 the symmetry is the same as that of the normal
coordinate because the wave function is the product of a Gaussian and the first Hermite
polynomial, which is linear in the coordinate. For pure ir transitions, that is all we need to
know. But for electronic transitions involving overtones, the situation is more complex. For
127
non-degenerate modes, even numbered states have A1 symmetry, and odd-numbered states
have the same symmetry as v = 1. The degenerate case is more complicated.
We note that the symmetries of normal modes may be used to calculate other types of
matrix elements. For example. the matrix element of the Hamiltonian, ⟨1|Ĥ|2⟩, vanishes
unless Γ1 Γ2 contains an A1 component. Another example is the coupling of two BornOppenheimer potential energy surfaces. The coupling strength is proportional to the square
of ⟨1|Qj |2⟩, where Qj is a normal mode known as a coupling mode. This matrix element
vanishes unless Γ1 Γ2 contains ΓQj .
Problem 67. The first two excited singlet states of pyrazine have B3u and B2u symmeties,
respectively. Which normal modes couple these states?
LECTURE 28: SYMMETRY ADAPTED COORDINATES AND LINEAR
COMBINATIONS OF FUNCTIONS
Reading: Cotton, Chapters 5, 6, and 7.3
A symmetry adapted coordinate is a coordinate that generates one of the irreducible
representations of the group. These coordinates are usually a good description of the normal
vibrational coordinate. We can find the symmetry adapted coordinates by operating on a
stretch or bend with the projection operator for the representation of interest. If the group
has a high order rotation axis, the problem may be simplified by using the characters of
the corresponding Cn cyclical subgroup. For XeF4 , that group is C4 group with operations
E, C4 , C2 , C43 and symmetry representations A, B, E. Using ∆s1 as a generating function,
its transformations by the symmetry operations are
Ê∆s1 = ∆s1
Ĉ4 ∆s1 = ∆s2
(732)
Ĉ2 ∆s1 = ∆s3
Ĉ43 ∆s1 = ∆s4 .
The projection operators may be constructed from the C4 character table after converting
128
the E pair to real values by taking sums and differences of the rows. The result is
P̂ A = Ê + Ĉ4 + Ĉ2 + Ĉ43
P̂ B = Ê − Ĉ4 + Ĉ2 − Ĉ43
P̂ E1 = Ê − Ĉ2
(733)
P̂ E2 = Ĉ4 − Ĉ43 .
The projections of ∆s1 are therefore
P̂ A ∆s1 = ∆s1 + ∆s2 + ∆s3 + ∆s4
P̂ B ∆s1 = ∆s1 − ∆s2 + ∆s3 − ∆s4
P̂
E1
∆s1 = ∆s1 − ∆s3
(734)
P̂ E2 ∆s1 = ∆s2 − ∆s4 .
Similar analyses are used to generate the in-plane and out-of-plane bends:
P̂ A ∆α1 = ∆α1 + ∆α2 + ∆α3 + ∆α4
P̂ B ∆α1 = ∆α1 − ∆α2 + ∆α3 − ∆α4
P̂ E1 ∆α1 = ∆α1 − ∆α3
(735)
P̂ E2 ∆α1 = ∆α2 − ∆α4 .
where α is the in-plane bending angle, and a similar set of results for the out of plane angle,
β.
The projection operators may be used to generate more complex symmetry-adapted functions, which are constructed from different generating functions. Consider the function x2
which has the symmetry of E ⊗ E = A1 + A2 + B1 + B2 in the C4v group. Let’s partition
it into basis functions for some or all of these irreducible representations. We first calculate
how x2 transforms under each operator.
Êx2 = x2
Ĉ4 x2 = y 2
Ĉ2 x2 = x2
σ̂v x2 = x2
σ̂d x2 = y 2
(736)
129
We now carry out all the projections:
1
1
1
P̂ A1 x2 = (x2 + 2y 2 + x2 + 2x2 + 2y 2 ) = x2 + y 2
8
2
2
1
P̂ A2 x2 = (x2 + 2y 2 + x2 − 2x2 − 2y 2 ) = 0
8
1
1
1
P̂ B1 x2 = (x2 − 2y 2 + x2 + 2x2 − 2y 2 ) = x2 − y 2
8
2
2
1
P̂ B2 x2 = (x2 − 2y 2 + x2 − 2x2 + 2y 2 ) = 0
8
2
P̂ E x2 = (2x2 − 2x2 ) = 0,
8
(737)
in agreement with the C4v character table.
Problem 68. Partition the function y 2 + z 2 into the different symmetry representations of
the C4v group.
Next we will use projection operators to find symmetry adapted molecular orbitals, using
the π orbitals of benzene as an example. The first thing we need to do is determine the
possible symmetries of the orbitals. Using the full D6h group, we assign the orbitals to
A2u +B2g +E1g +E2u . We can also obtain the coefficients of the LCAOs (“linear combination
of atomic orbitals”) by using the C6 cyclical sub-group. In this reduced symmetry, the
orbitals have A + B + E1 + E2 symmetry. We construct the LCAOs by applying the six
projection operators to any one of the π orbitals, say ϕ1 . The results of the individual
operations are:
Êϕ1 = ϕ1
Ĉ6 ϕ1 = ϕ2
Ĉ3 ϕ1 = ϕ3
Ĉ2 ϕ1 = ϕ4
Ĉ32 ϕ1 = ϕ5
Ĉ65 ϕ1 = ϕ6 .
(738)
130
We next construct the projection operators, using the C6 character table,
P̂ A = Ê + Ĉ6 + Ĉ3 + Ĉ2 + Ĉ22 + Ĉ65
P̂ B = Ê − Ĉ6 + Ĉ3 − Ĉ2 + Ĉ22 − Ĉ65
P̂ E1 = Ê + ϵĈ6 − ϵ∗ Ĉ3 − Ĉ2 − ϵĈ22 + ϵ∗ Ĉ65
∗
P̂ E1 = Ê + ϵ∗ Ĉ6 − ϵĈ3 − Ĉ2 − ϵ∗ Ĉ22 + ϵĈ65
(739)
P̂ E2 = Ê − ϵ∗ Ĉ6 − ϵĈ3 + Ĉ2 − ϵ∗ Ĉ22 − ϵĈ65
∗
P̂ E2 = Ê − ϵĈ6 − ϵ∗ Ĉ3 + Ĉ2 − ϵĈ22 − ϵ∗ Ĉ65 .
We then use these operators to construct the symmetry adapted orbitals.
P̂ A ϕ1 = ϕ1 + ϕ2 + ϕ3 + ϕ4 + ϕ5 + ϕ6 = ψA
P̂ B ϕ1 = ϕ1 − ϕ2 + ϕ3 − ϕ4 + ϕ5 − ϕ6 = ψB
P̂ E1 ϕ1 = ϕ1 + ϵϕ2 − ϵ∗ ϕ3 − ϕ4 − ϵϕ5 + ϵ∗ ϕ6 = ψ3
∗
P̂ E1 ϕ1 = ϕ1 + ϵ∗ ϕ2 − ϵϕ3 − ϕ4 − ϵ∗ ϕ5 + ϵϕ6 = ψ4
(740)
P̂ E2 ϕ1 = ϕ1 − ϵ∗ ϕ2 − ϵϕ3 + ϕ4 − ϵ∗ ϕ5 − ϵϕ6 = ψ5
∗
P̂ E2 ϕ1 = ϕ1 − ϵϕ2 − ϵ∗ ϕ3 + ϕ4 − ϵϕ5 + ϵ∗ ϕ6 = ψ6 ,
where
ϵ = e2πi/6 = cos (π/3) + i sin (π/3) =
1 i√
+
3.
2 2
(741)
Taking sums and differences of the complex LCAOs, we get
1
ψE1a = (ψ3 + ψ4 ) = 2ϕ1 + ϕ2 − ϕ3 − 2ϕ4 − ϕ5 + ϕ6
2
√
1
ψE1b = (ψ3 − ψ4 ) = 3(−ϕ2 − ϕ3 + ϕ5 + ϕ6 )
2i
1
ψE2a = (ψ5 + ψ6 ) = 2ϕ1 − ϕ2 − ϕ3 + 2ϕ4 − ϕ5 − ϕ6
2
√
1
ψE2b = (ψ5 − ψ6 ) = 3(ϕ2 − ϕ3 + ϕ5 − ϕ6 ).
2i
(742)
(In this example, we took sums and differences of the wave functions after projecting them
out, rather than sums and differences of the characters. The end result is the same). These
functions are orthogonal but not yet normalized.
[1] J. A. Cotton, Chemical Applications of Group Theory (Wiley, New York).
131
[2] D. C. Harris and M. D. Bertolucci, Symmetry and Spectroscopy Electrodynamics, (Oxford, New
York)
[3] H. Goldstein, Classical Mechanics, (Addison-Wesley, Reading)
[4] G. V, Molecular Spectra and Molecular Structure. II. Infrared and Raman Spectra of Polyatomic
Molecules, (Van Nostrand, New York)