Part II Chemistry 2009–10 Course B7: Symmetry and ... · • P. R. Bunker and P. Jensen,...

Part II Chemistry 2009–10

Course B7: Symmetry and Perturbation Theory

Symmetry

Outline

The purpose of this course is

• To explore the foundations of the theory in a bit more detail;

• In particular, to introduce the Great Orthogonality Theorem and to show how itleads to the important theorems of representation theory;

• To extend the methods introduced in previous courses;

• To investigate a wider range of applications.

Assumed:

• Basic practical use of group theoretical methods as outlined in IB Symmetryand Bonding;

• Matrix methods from the Part IA NST Maths course (there is a reminder ofbasic definitions in Appendix A);

• Some Perturbation Theory results from the first half of this course.

Not assumed:

• Anything from the Part IB Maths course.

1

Recommended books

• A. Vincent , ‘Molecular Symmetry and Group Theory ’ (Wiley)An excellent elementary introduction. Relatively cheap. (If you don’tknow most of the material in this book you should review it beforecontinuing this course.)

• A. Cotton , ‘Chemical Applications of Group Theory ’ (Wiley)Expensive but useful for chemical applications. Don’t use the firstedition, which contains many errors.

• P. R. Bunker and P. Jensen , ‘Fundamentals of Molecular Symmetry ’,(IoP). Discusses the Molecular Symmetry Group.

• P. W. Atkins and R. S. Friedman , ‘Molecular Quantum Mechanics’(O.U.P.). A very good account of theoretical methods and applications,with useful coverage of Group Theory. Recommended for theoreticalspecialists.

• J. P. Elliott and P. G. Dawber , ‘Symmetry in Physics’, vol. 1 (Macmillan)Good on basic principles but intended for physicists.

2

1. What is a symmetry operator?

We define a symmetry operator generally as any operator R that has an inverse andthat commutes with the Hamiltonian:

RH = HR (1.1)

or, equivalently, that leaves the Hamiltonian unchanged:

RHR−1 = H. (1.2)

Then, if we have a wavefunction ψ that satisfies the Schrödinger equation, withenergy W:

Hψ = Wψ,

the wavefunction Rψ also satisfies it, with the same energy:

H(Rψ) = RHψ = RWψ = W(Rψ). (1.3)

3

Symmetry and degeneracy

If Hψ = Wψ then H(Rψ) = W(Rψ).

Rψ may be the same function as ψ (possibly multiplied by −1 or some other number).For example, we might find that for some functions Rψ = ψ and for others Rψ = −ψ.This allows us to classify functions according to their symmetry behaviour.

But if ψ and Rψ are different, symmetry leads to degeneracy: the functions ψ and Rψhave the same energy.

We postulate that if a set of wavefunctions is degenerate, the degeneracy must be aconsequence of some kind of symmetry. Another way of expressing this postulate is

Accidental degeneracy doesn’t happen.

By ‘accidental degeneracy’ we mean degeneracy that isn’t due to symmetry.

4

What kinds of symmetry are there?

We need to look for operators that don’t change the Hamiltonian. We consider isolatedmolecules.

1. Any translation of all particles by the same distance.

If we move all particles by the same amount, then all distances between particles staythe same, so the potential energy doesn’t change.

If x′ = x−d then ∂/∂x′ = ∂/∂x (by the chain rule), so the ∇2 kinetic energy terms inH don’t change either.

So the Hamiltonian is unchanged if we move the whole molecule to a new position.

This means that moving a molecule to a new position doesn’t change its energy.Consequently the force on the molecule (derivative of energy with respect to position)is zero, and so there is no change in momentum.

Thus this symmetry leads to conservation of linear momentum.

5

Symmetry operations

2. Any rotation of the whole system.

This also leaves all distances unchanged and doesn’t affect the potential energy.

Rotation doesn’t affect ∇2 either (this too can be shown using the chain rule), so thekinetic energy operators stay the same.

We conclude that rotating the whole molecule doesn’t change its energy.Consequently the torque (derivative of energy with respect to orientation) is zero, andthere is no change in angular momentum.

Thus rotational symmetry of isolated molecules leads to conservation of angularmomentum.

If Rz(δϕ) is a rotation about the z axis, we can see that

limδϕ→0

Rz(δϕ)−Eδϕ

ψ =∂

∂ϕψ ∝ Jzψ,

where Jz is the usual angular momentum operator. Consequently the theory ofrotational symmetry and the theory of angular momentum are one and the same.

6

Symmetry operations: electron permutations

3. Any permutation of the electrons.

All electrons are equivalent — the Hamiltonian has a kinetic energy term of the sameform for each one, and the potential energy terms also have the same form for everyelectron. Therefore permuting the electrons has no effect on the Hamiltonian, and anysuch permutation is a symmetry operation.

Experimental observations can only be explained by quantum mechanics if we requirethat all wavefunctions are antisymmetric with respect to exchange of electrons. This isthe Pauli principle.

More generally, it is found experimentally that all wavefunctions have to beantisymmetric with respect to exchange of identical fermions of any kind, andsymmetric with respect to exchange of identical bosons of any kind. For chemistry, weare concerned only with electrons and nuclei. Nuclei with even mass number arebosons, and those with odd mass number are fermions.

7

Symmetry operations: nuclear permutations

4. Any permutation of a set of identical nuclei.

Just as for electrons, the Hamiltonian is unchanged in form if we permute the labels ofa set of identical nuclei.

5. Parity: inversion of every particle in the centre of mass

Again, this leaves particle–particle distances and ∇2 terms unchanged and leaves theHamiltonian invariant. It is called the parity operation, and denoted E∗, to distinguish itfrom the inversion operation of finite symmetry groups, usually denoted ı.

These are the operations that make up the elements of the ordinary molecularsymmetry group.

8

Example: the water molecule

We need to understand the relationship between these operations of inversion andnuclear permutation and the conventional geometrical symmetry operations. This willgive us a deeper and more secure understanding of what symmetry operations do.

We are mainly interested in internal degrees of freedom — electronic and vibrational— but also sometimes in overall rotation of the molecule. We need a local molecularcoordinate frame (x,y,z), fixed in the molecule. Internal degrees of freedom arereferred to this frame, while the rotational wavefunction depends on the orientation ofthis local frame relative to the global frame (X,Y,Z).

This local frame can usually be defined only in terms of the nuclear labels. If the labelschange as the result of a nuclear permutation, the axes may change.

9

Symmetry operations of the water molecule

The local axes have to take account of the fact that the molecule is constantlyvibrating, so that at any particular instant its particles are in an unsymmetricalconfiguration.

For the water molecule, we define theorigin of the local axis system as thecentre of mass. The z axis is parallel tothe bisector of the HOH angle. The yaxis is perpendicular to the z axis, inthe molecular plane, and in the generaldirection from proton a to proton b. Thex axis completes a right-handed set.(Other definitions are possible.)

b

bb

z

y

x

a b

The symmetry operations are (ab) (i.e. exchange of proton labels a and b), theinversion E∗, and their product (ab)E∗, usually abbreviated to (ab)∗. Theidentity operation E completes the group.

10

Symmetry operations of the water molecule: (ab)

b

bb

z

y

x

a b

2px

1sa

(ab)b

bb

z

y x

b a

−2px

1sb

Cz2

b

bb

z

y

x

a b

−2px

1sb

Because the y axis changes directionwhen a and b are exchanged, the x axisalso has to change direction to maintain aright-handed coordinate frame.

The effect of the (ab) operation on internalcoordinates is the same as that of Cz

2. Wesee that Cz

2 rotates functions andcoordinates but not the nuclear labels.

11

Symmetry operations of the water molecule: E∗

b

bb

z

y

x

a b

2px

1sa

E∗b

bb

z

y x

ab

−2px

1sa

σyzv

b

bb

z

y

x

a b

−2px

1sa

The E∗ operation corresponds to reflectionof the internal coordinates in the plane ofthe nuclei.

12

Symmetry operations of the water molecule: (ab)∗

b

bb

z

y

x

a b

2px

1sa

(ab)∗b

bb

z

y

x

ba

2px

1sb

σxzv

b

bb

z

y

x

a b

2px

1sb

The product (ab)E∗ is written (ab)∗ forbrevity.

(ab) corresponds to Cz2 and E∗ to σyz

v , so(ab)∗ corresponds to their product, whichis σxz

v .

13

Ortho and para hydrogen

The Pauli principle applies in a more general form to nuclear permutations as well asto electrons. The complete molecular wavefunction has to be antisymmetric withrespect to exchange of identical nuclei if they are fermions (i.e. if they havehalf-odd-integer spin) and symmetric if they are bosons (i.e. if they have integer spin).

The wavefunction of an H2 molecule is a product of electronic, vibrational, rotationaland nuclear-spin factors:

Ψ = ψelecψvibψrotψns.

The symmetry operation (ab) which exchanges the proton labels can in principle

affect these factors. The ground electronic state of H2 is 1Σ+g , i.e. totally symmetric, so

it is unchanged. ψvib depends only on the bond length |ra−rb|, which is unchangedby exchange of a and b.

However ψrot is a spherical harmonic YJM(θ,ϕ),where θ and ϕ specify the direction of themolecular axis. This axis is reversed when a andb are interchanged, and the rotationalwavefunction changes sign if J is odd, and isunchanged if J is even.

b

b

zb

ab

b

z

a

b

(ab)

14

Ortho and para hydrogen contd.Finally the nuclear spin function is either a singlet function

√

12(αaβb−βaαb)

(antisymmetric under exchange of a and b) or one of the triplet functions αaαb,√

12(αaβb +βaαb) and βaβb (symmetric).

Because the nuclear spins interact only very weakly with the environment, the spinstate doesn’t change easily, and we speak of ortho hydrogen (triplet nuclear spin) andpara hydrogen (singlet nuclear spin).

For the overall wavefunction to be antisymmetric under exchange of a and b, therotational wavefunction can only have odd J for ortho hydrogen, and even J for parahydrogen. Because of the nuclear-spin degeneracy, there is three times as muchortho hydrogen as para at equilibrium at high temperatures, and the line intensities inthe rotational Raman spectrum alternate accordingly.

Carbon dioxide

In CO2, the O atoms have zero spin, so their nuclear spin function is just ψns = 1.As they are bosons (integer spin) the overall wavefunction must be unaffected byexchange of the O atom labels. Consequently only even-J rotational functions areallowed, and rotational Raman or vibration–rotation transitions involving the odd Jlevels are absent.

15

Oxygen

The oxygen molecule, O2, has a 3Σ−g ground

state. The (ab) operation which exchanges the Onuclei switches the direction of the local z axis,which we take from nucleus a to b. This has theeffect of changing the sign of internal ϕcoordinates, which are measured corkscrew-wisearound the z axis.

The dependence of the 3Σ−g wavefunction on the

ϕ coordinates of the unpaired electrons issin(ϕ1−ϕ2). This is the reason for the minus signon the term symbol, denoting that thewavefunction changes sign under reflection inplanes containing the molecular axis.

b

b

z

ϕ

b

ab

b

z

ϕ

a

b

(ab)

Consequently the electronic wavefunction changes sign under (ab), while thenuclear wavefunction is unchanged, as for CO2. In order for the overallwavefunction to be unchanged, the rotational wavefunction must have odd J,and the even-J levels are absent.

16

2. GroupsA group is a set of elements, together with a rule for combining them, which togethersatisfy the group postulates:

Postulate 1: (Closure) The result R∗Sof combining any two elements is defined andis an element of the set.

Postulate 2: (Associativity) Combination is associative: (R∗S)∗T = R∗ (S∗T).

Postulate 3: (Identity) The set contains an identity E such that E ∗R= R∗E = R forany element R of the set.

Postulate 4: (Inverses) Every element R has an inverse R−1 such thatR∗R−1 = R−1∗R= E.

For symmetry operators, the product R∗S (usually written simply as RS) is defined asthe operator whose effect is that of operating first with Sand then with R:

(RS)ψ ≡ R(Sψ). (2.1)

Associativity follows at once:

((RS)T)ψ = (RS)(Tψ) = R(S(Tψ)) = R((ST)ψ) = (R(ST))ψ,

while the Identity and Inverse postulates are satisfied because of the nature ofsymmetry operations as coordinate transformations.

17

Multiplication tablesA group is summarized in its multiplication table, shown below for D 3. Theproduct RSappears in the row labelled Rand the column labelled S. If RS= SRfor all pairs of elements R and S, the group is said to be Abelian. This isevidently not the case for D 3.

a

b

c

C3

first

D 3 E C3 C3−1 C2

a C2b C2

c

second

E E C3 C3−1 C2

a C2b C2

c

C3 C3 C3−1 E C2

c C2a C2

b

C3−1 C3

−1 E C3 C2b C2

c C2a

C2a C2

a C2b C2

c E C3 C3−1

C2b C2

b C2c C2

a C3−1 E C3

C2c C2

c C2a C2

b C3 C3−1 E

18


a

b

c

C3

first

D 3 E C3 C3−1 C2

a C2b C2

c

second

E E C3 C3−1 C2

a C2b C2

c

C3 C3 C3−1 E C2

c C2a C2

b

C3−1 C3

−1 E C3 C2b C2

c C2a

C2a C2

a C2b C2

c E C3 C3−1

C2b C2

b C2c C2

a C3−1 E C3

C2c C2

c C2a C2

b C3 C3−1 E

C3C2a = C2

c

18


a

b

c

C3

C3C2a = C2

c

first

D 3 E C3 C3−1 C2

a C2b C2

c

second

E E C3 C3−1 C2

a C2b C2

c

C3 C3 C3−1 E C2

c C2a C2

b

C3−1 C3

−1 E C3 C2b C2

c C2a

C2a C2

a C2b C2

c E C3 C3−1

C2b C2

b C2c C2

a C3−1 E C3

C2c C2

c C2a C2

b C3 C3−1 E

C2aC3 = C2

b

18

Generators

The number of possible groups is severely limited by the group postulates.

For example, if a group contains Cn, then it must contain CnCn = Cn2,

Cn2Cn = Cn

3, . . . , and Cnn = E. The set of n elements obtained from Cn in this way is

a group (called Cn); we say that Cn generates the group Cn.

Many groups need two or more generators. Not every possible set of generatorsdefines a finite group; in all but a few cases more and more elements are generatedad infinitum. The possible finite symmetry groups, for rigid molecules, are given inAppendix B. There are also two infinite groups, C∞v and D∞h, which are appropriatefor linear molecules. Note that the choice of generators is not unique; also that anytwo of the four C3 axes in the cubic groups generate the others, as do any two of thesix C5 axes for the icosahedral groups.

19

Direct product groups

If two groups G (with nG elements gi ) and H (with nH elements h j ) have no elementin common except the identity, and have the property that every gi commutes withevery h j , then the products gih j form a group of order nGnH , called the direct product

group and written G ⊗H .

For example, D 3h = D 3⊗Cs:

• Every element of D 3 commutes with every element of Cs.

• Each element of D 3h is the product of an element of D 3 with an element of Cs.

D 3 E C3 C3−1 C2

a C2b C2

c

Cs E E C3 C3−1 C2

a C2b C2

c

σh σh S3 S3−1 σa

v σbv σc

v

If a group can be expressed as a direct product of two smaller groups, the symmetriesdescribed by the two groups can be treated independently.

20

Classes: equivalent operations

Two symmetry operators R and Sare said to be equivalent (notation R∼ S) if there isan element Q in the group such that S= QRQ−1.

The operator QRQ−1 can be thought of as the operator obtained by applying Q to theoperator R itself; if R is a rotation through α about an axis in the direction of the vector

ξ, then QRQ−1 is a rotation through the same angle α about the rotated axis Qξ.

In D 3,

Ca2C3(C

a2)

−1 = (C3)−1

C3Ca2(C3)

−1 = Cb2 = (C3)

−1Cc2C3

so (C3)−1 is equivalent to C3 and Cb

2 is equivalent to Ca2 and Cc

2.

However, in C2v, the two reflections, σxzv and σyz

v , are not equivalent, even though theyare both reflections, because there is no symmetry operator in the group thattransforms one into the other.

21

Classes• If S∼ R, so that S= QRQ−1, then R∼ S, because R= Q−1SQ.

• If R∼ Sand S∼ T, then R∼ T.

• It follows that the elements of the group can be divided into equivalenceclasses (or just classes), so that all the elements in each class are equivalent toeach other and to no other element of the group.

• Notice that the identity E always forms a class by itself, because for anyelement Q in the group, QEQ−1 = E.

• Similarly, every element of an Abelian group forms a class by itself, because in

that case QRQ−1 = RQQ−1 = R, for any Q.

22

3. Representations

A vector r describing some fixed physical quantity can be specified by setting up acoordinate system in terms of vectors i, j and k. In practice the basis is oftensuppressed, but when we write

r =

x

y

z

we really mean

r = ix+jy+kz=(

i j k

)

x

y

z

. (3.1)

This is a representation of vectors r in the basis (i,j,k).

It is usually convenient to choose the basis vectors to be orthogonal and normalized,but this is not essential.

23

Wavefunctions as basis functions

Wave functions are dealt with in the same way. If we are studying BF3, we may beinterested in boron p orbitals lying in the plane of the molecule (the xy plane). We maywrite any such orbital in the form

ϕ =(

px py

)

(

cx

cy

)

,

where cx and cy are numerical coefficients which may be complex.

The set of all functions which can be written in this way (all p orbitals, normalized ornot, that lie in the xy plane) is called a vector space. In this case it is the vector spacespanned by the basis (px, py).

24

Representation of symmetry operators

Now consider the effect of a symmetryoperator on such a function. Because

C3px = −12 px +

√3

2 py,

C3py = −√

32 px− 1

2 py,(3.2)

we see that

b b

b

b

px

pyC3px

C3py

C3ϕ = C3

(

px py

)

(

cx

cy

)

=(

px py

)

(

−12 −

√3

2√3

2 −12

)(

cx

cy

)

=(

px py

)

D(C3)

(

cx

cy

)

, (3.3)

so for functions of this nature the symmetry operator C3 can be represented by thematrix D(C3). In a similar way we can set up representations of the other symmetryoperators in the group.

25

Products of symmetry operators

If we treat a product RSas a single element, we obtain

(RS)(

px py

)

(

cx

cy

)

=(

px py

)

D(RS)

(

cx

cy

)

.

However we can also treat R and Sseparately:

(RS)(

px py

)

(

cx

cy

)

= R(

px py

)

D(S)

(

cx

cy

)

=(

px py

)

D(R)D(S)

(

cx

cy

)

,

and we see that D(RS) = D(R)D(S) — the matrices multiply in the same wayas the symmetry operations.

26

Some more jargon

This procedure is one way to generate a matrix representation of a group, which isdefined as a set of matrices corresponding to the elements R of the group, andsatisfying

D(R)D(S) = D(RS) (3.4)

for all pairs of elements R and S in the group.

The matrices of a representation need not be all different; if they are, therepresentation is said to be faithful .

The dimension of a representation is the dimension of each of its matrices.

The set of matrices D(R) forms a group, and the relationship expressed byeq. (3.4) between the elements of the original group and the representation matricesis a homomorphism.

27

Representations contd.

The homomorphism condition is

D(R)D(S) = D(RS). (3.4)

From (3.4) we see that

D(R) = D(ER) = D(E)D(R),

D(E) = D(RR−1) = D(R)D(R−1).

That is, D(E) is always the unit matrix, and D(R−1) =(

D(R))−1

.

There is always a trivial or symmetric representation, for which D(R) = (1) (the1×1 unit matrix) for all R. ‘Trivial’ is the mathematicians’ term; it is particularlyimportant as far as applications are concerned. It may be called A1, A1g, A′

1, Σ+g , etc,

in different symmetry groups. Here we shall use the notation Γ1 as a general symbolfor the symmetric representation.

28

Matrix representations

If we choose to start with a different set of basis functions we shall get differentmatrices.

In general the matrix D(R) representing a symmetry operator R in a basis(ϕ1,ϕ2, . . . ,ϕn) is defined by the equation

Rϕi = ∑j

ϕ jD ji (R). (3.5)

If the representation matrices D(R) are all unitary then we have a unitaryrepresentation.

(See Appendix A for a summary of matrix terminology.)

29

Using generators

In working out representation matrices, it isn’t necessary to use eq. (3.5) forevery R. Instead, we can use the group generators (see slide 19 andAppendix B) together with the homomorphism, eq. (3.4).

For D 3 the generators can be taken as C3 and Ca2. (The choice is not unique.)

Then, for example,

C−13 = C2

3,

Cb2 = Ca

2C3.

Since the matrices multiply the same way as the operations, according toeq. (3.4), this means that

D(

C−13

)

= D(

C3)2

,

D(

Cb2

)

= D(

Ca2

)

D(

C3)

.

So once we have the matrices for C3 and Ca2, it is easy to work out all the

others.

30

Representation matrices for D 3

D 3 E C3 C3−1 Ca

2 Cb2 Cc

2

A1 (s) (1) (1) (1) (1) (1) (1)

A2 (pz) (1) (1) (1) (−1) (−1) (−1)

E (px, py)

(

1 0

0 1

)

− 12 −

√3

2√

32 − 1

2

− 12

√3

2

−√

32 − 1

2

(

1 0

0 −1

)

− 12 −

√3

2

−√

32

12

− 12

√3

2√

32

12

(p1, p−1)

(

1 0

0 1

) (

ω∗ 0

0 ω

) (

ω 0

0 ω∗

) (

0 −1

−1 0

) (

0 −ω−ω∗ 0

) (

0 −ω∗−ω 0

)

Γs (sa,sb,sc)

1 0 0

0 1 0

0 0 1

0 0 1

1 0 0

0 1 0

0 1 0

0 0 1

1 0 0

1 0 0

0 0 1

0 1 0

0 0 1

0 1 0

1 0 0

0 1 0

1 0 0

0 0 1

The basis functions are all atomic orbitals for BF3. sa, sb and sc are fluorine 2s orbitals and the

rest are boron orbitals. The functions p1 and p−1 are defined by

p1 = − 1√2(px + ipy), p−1 = + 1√

2(px− ipy),

and ω = exp(2πi/3) = − 12 +

√3

2 i.

31

Equivalent representations

When we set up a representation of ordinary three-dimensional vectors, the basisvectors (i,j,k) are arbitrary, so long as they are linearly independent. If T is any3×3 matrix possessing an inverse, we can write

(

i′ j ′ k′)

=(

i j k

)

T−1 and

x′

y′

z′

= T

x

y

z

,

and then

(

i′ j ′ k′)

x′

y′

z′

=(

i j k

)

T−1T

x

y

z

=(

i j k

)

x

y

z

= r.

The new representation is equivalent to the old one; we can still describe any vector r

in terms of it, though the numbers will be different. This procedure is just a simplecoordinate transformation.

32

We can apply a similar procedure in the case of group representations. Write therelationship between (p1, p−1) and (px, py) in the form

(

p1 p−1

)

=(

px py

)

T−1 =(

px py

)

−1/√

2 1/√

2

−i/√

2 −i/√

2

.

Then if

ϕ =(

px py

)

(

cx

cy

)

we can equivalently write

ϕ =(

px py

)

T−1T

(

cx

cy

)

=(

p1 p−1

)

(

c1

c−1

)

.

That is, any cxpx +cypy can be expressed just as well in terms of p1 and p−1. Thetwo basis sets span the same vector space.

33

Similarly, from the definition of the representation matrices for the basis (px, py), i.e.

Rϕ =(

px py

)

D(R)

(

cx

cy

)

,

we can obtain the representation matrices for the basis (p1, p−1) = (px, py)T−1:

Rϕ =(

px py

)

T−1TD(R)T−1T

(

cx

cy

)

=(

p1 p−1

)

D′(R)

(

c1

c−1

)

,

so that D′(R) = TD(R)T−1 is the representation matrix for the basis (p1, p−1).Thus the representation matrices D′(R) for (p1, p−1) are related to the matrices D(R)for (px, py) by

D′(R) = TD(R)T−1.

This relationship is called a similarity or equivalence transformation, and the tworepresentations D(R) and D′(R) are said to be equivalent . In this example T isunitary, but any non-singular matrix can be used to generate such a transformation.

34

In this case T is unitary, so T−1 = T†, and, for example,

D(Ca2) =

(

1 0

0 −1

)

,

D′(Ca2) = TD(Ca

2)T−1

=

−√

12 i

√

12

√

12 i

√

12

(

1 0

0 −1

)

−√

12

√

12

−i√

12 −i

√

12

=

−√

12 i

√

12

√

12 i

√

12

−√

12

√

12

i√

12 i

√

12

=

(

0 −1

−1 0

)

.

Characters

Much of the information in the representation matrices therefore depends on thechoice of basis, and is irrelevant to the physics of the problem. However the trace of amatrix is invariant under equivalence transformations:

trace(A′) = ∑i

A′ii = ∑

i jk

Ti j A jk(T−1)ki = ∑

i jk

(T−1)kiTi j A jk = ∑k

Akk = trace(A). (3.6)

The set of traces χ(R) of the matrices D(R) is called the character of therepresentation, and it contains all the information that is normally needed.

35

Characters for D 3

D 3 E C3 C3−1 Ca

2 Cb2 Cc

2

A1 (s) 1 1 1 1 1 1

A2 (pz) 1 1 1 −1 −1 −1

E (px, py) 2 −1 −1 0 0 0

(p1, p−1) 2 −1 −1 0 0 0

Γs (sa,sb,sc) 3 0 0 1 1 1

Notice that the characters of the equivalent representations based on (px, py) and(p1, p−1) are the same.

Notice also that C3 and C−13 always have the same character, as do Ca

2, Cb2 and Cc

2.

36

Characters for equivalent operations

If two elements R and Sare in the same class, so that S= QRQ−1, thenbecause of the homomorphism property of group representations theirrepresentation matrices are related by

D(S) = D(QRQ−1) = D(Q)D(R)D(Q−1)

= D(Q)D(R)D(Q)−1,

where the last step is again aconsequence of the homomorphism, andthen

χ(S) = traceD(S) = traceD(R) = χ(R).

Because of this it is usual to collect all theelements of each class together in thecharacter table.

D 3 E 2C3 3C2

A1 (s) 1 1 1

A2 (pz) 1 1 −1

E (px, py) 2 −1 0

Γs (sa,sb,sc) 3 0 1

37

Reducible and irreducible representations

We might choose to set up a representation for D 3 using the basis (i,j,k) of unit vectors in three

dimensions.

However the operations of the group all leave the vector k unchanged or merely change its sign, while

vectors in the ij plane are changed into other vectors in the ij plane. The one-dimensional space of

vectors kz forms an invariant subspace of the original three-dimensional space, as does the

two-dimensional space of vectors of the form ix+jy. This is apparent from the form of the representation

matrices, which are

E C3 C3−1 Ca

2 Cb2 Cc

2

1 0 0

0 1 0

0 0 1

− 12 −

√3

2 0√

32 − 1

2 0

0 0 1

− 12

√3

2 0

−√

32 − 1

2 0

0 0 1

1 0 0

0 −1 0

0 0 −1

− 12 −

√3

2 0

−√

32

12 0

0 0 −1

− 12

√3

2 0√

32

12 0

0 0 −1

Comparison with Table 3 shows that all these matrices are of the form

DE 0

0

0 0 DA2

.

38

The matrices generated from the basis (i,j,k) are all of the form

DE 0

0

0 0 DA2

.

They are said to be in block diagonal or direct sum form (symbol ⊕): DE ⊕DA2.

The 3-dimensional representation generated from the basis (i,j,k) is said to bereducible.

39

Reducible representations contd.

The relationship between the representation Γs based on the functions (sa,sb,sc) andthe representations A1, A2 and E goes one stage further. None of these functionsforms a basis for an invariant subspace; all of them are transformed into some otherfunction by several of the symmetry operations. However the function sa +sb +sc isleft unchanged by all the symmetry operations, so it corresponds to an invariantsubspace, and the representation Γs must be reducible.

We can define some new basisfunctions:

ϕ1 =√

13(sa +sb +sc),

ϕ2 =√

16(2sa−sb−sc),

ϕ3 =√

12(sb−sc).

This is a similarity transformationusing the matrix

T =

√

13

√

13

√

13

√

23 −

√

16 −

√

16

0√

12 −

√

12

,

and T here is orthogonal so that

T−1 = TT .

40

Using this transformation we can construct a new representation Dϕ = TDsT−1

equivalent to Ds.

In this case the matrices are

E C3 C3−1 Ca

2 Cb2 Cc

2

1 0 0

0 1 0

0 0 1

1 0 0

0 − 12 −

√3

2

0√

32 − 1

2

1 0 0

0 − 12

√3

2

0 −√

32 − 1

2

1 0 0

0 1 0

0 0 −1

1 0 0

0 − 12 −

√3

2

0 −√

32

12

1 0 0

0 − 12

√3

2

0√

32

12

These matrices are again in direct sum form, this time with Dϕ = DA1 ⊕DE.

41

Direct sums

The notation Dϕ = DA1 ⊕DE expresses the direct sum form of Dϕ.

Similarly we can write the basis as a direct sum:

(ϕ1,ϕ2,ϕ3) = (ϕ1)⊕ (ϕ2,ϕ3).

ϕ1 is left unchanged by all symmetry operations, while ϕ2 and ϕ3 together transformin exactly the same way as px and py.

We can write Γϕ = A1⊕E too.

We also write Γs∼ A1⊕E, indicating that Γs is equivalent to a direct sum of A1 and E.

In practice we often just write this as Γs = A1 +E.

If a representation cannot be reduced to direct sum form (that is, if the space spannedby its basis contains no invariant subspace) then it is irreducible.

42

The character of a reducible representation

The character of the direct sum is evidently just the arithmetic sum of the charactersof the component representations:

χϕ(R) = χA1(R)+χE(R).

Since the characters of equivalent representations are identical, it follows thatχs(R) = χA1(R)+χE(R) also.

How do we find the matrix T that transforms the representation matrices to blockdiagonal form? That is just the standard procedure that we use, for example, to findsymmetry orbitals.

43

4. The Great Orthogonality Theorem

This is the fundamental theorem from which most of the useful results of group theoryare derived. It states that if Γa and Γb are two irreducible unitary representations, thentheir representation matrices satisfy

1h ∑

RDa

ip(R)∗Dbjq(R) =

0 if Γa 6∼ Γb,

1na

δi j δpq if Γa = Γb,(4.1)

• h is the number of elements in the group;

• na is the dimension of representation Γa.

We do not consider the case where Γa and Γb are equivalent but not identical; a morecomplicated formula applies in that case, but we shall not need it.

An outline of the proof is given in Appendix C. (It is not examinable.)

44

Example: Irreducible representation matrices for D 3

D 3 E C3 C3−1 Ca

2 Cb2 Cc

2

A1 (1) (1) (1) (1) (1) (1)

A2 (1) (1) (1) (−1) (−1) (−1)

E

1 0

0 1

− 12 −

√3

2√

32 − 1

2

− 12

√3

2

−√

32 − 1

2

1 0

0 −1

− 12 −

√3

2

−√

32

12

− 12

√3

2√

32

12

• 1h ∑

RDE

11(R)∗DE21(R)

=16

(

1.0+(−12).

√3

2 +(−12).(−

√3

2 )+1.0+(−12).(−

√3

2 )+(−12).

√3

2

)

= 0;

• 1h ∑

RDE

11(R)∗DE11(R)

=16

(

1.1+(−12).(−1

2)+(−12).(−1

2)+1.1+(−12).(−1

2)+(−12).(−1

2))

=12

=1

nE.

The first orthogonality theorem for characters

From the Great Orthogonality Theorem we immediately deduce the first orthogonalitytheorem for the characters of irreducible representations:

1h ∑

Rχa(R)∗χb(R) =

1h ∑

R∑ik

Daii (R)∗Db

kk(R)

=

0 if Γa 6∼ Γb,

1na

∑ki

δikδik =1na

∑k

δkk = 1 if Γa = Γb.(4.2)

The second case in fact applies when Γa and Γb are equivalent, not necessarilyidentical, since the characters of equivalent representations are the same. Thereforewe can write this result in the form

1h ∑

Rχa(R)∗χb(R) = δΓaΓb, (4.3)

where δΓaΓb = 1 if Γa ∼ Γb, 0 otherwise.

45

The first orthogonality theorem for characters contd.

The theorem states that

1h ∑

Rχa(R)∗χb(R) = δΓaΓb, (4.3)

where δΓaΓb = 1 if Γa ∼ Γb, 0 otherwise.

If we use the fact that all elements in a class have the same character, we canexpress eq. (4.3) in the form

1h ∑

chcχa(c)∗χb(c) = δΓaΓb. (4.4)

In eq. (4.4) the sum is taken over classes c, and hc is the number of elements in classc.

46

Example for D 3

D 3 E 2C3 3C2

A1 1 1 1

A2 1 1 −1

E 2 −1 0

1h ∑

chcχA1(c)∗χA2(c) =

16

(

1.1.1+2.1.1+3.1.(−1))

= 0;

1h ∑

chcχE(c)∗χE(c) =

16

(

1.2.2+2.(−1).(−1)+3.0.0)

= 1

Reducible representations

If we have a reducible representation Γ = m1Γ(1) ⊕m2Γ(2) ⊕·· · , its character is

χ(c) = m1χ(1)(c)+m2χ(2)(c)+ · · · .

To find how many times a particular representation Γa occurs, multiply by

(hc/h)χ(a)(c)∗ and sum over c:

1h ∑

chcχ(a)(c)∗χ(c) =

1h ∑

chcχ(a)(c)∗∑

imiχ(i)(c)

= ∑i

miδΓaΓi

= ma.

We have obtained the familiar reduction formula:

ma =1h ∑

chcχ(a)(c)∗χ(c) (4.5)

47

The second orthogonality theorem for characters

Eq. (4.4) shows that the rows of the character table are orthogonal, and therefore thatthere cannot be more rows (i.e. irreducible representations) than columns (i.e.classes). The second orthogonality theorem for characters (not proved here) statesthat the columns are orthogonal:

hc

h ∑Γ

χΓ(c)∗χΓ(c′) = δcc′ , (4.6)

where Γ runs over the irreducible representations.

From eq. (4.6) we deduce that the number of columns cannot exceed the number ofrows, and thus (4.4) and (4.6) together tell us that

the number of irreducible representations is equal to the number of classes.

We also observe, by putting c = c′ = E in eq. (4.6), and noting that hE = 1 and

χΓ(E) = nΓ, that

∑Γ

(nΓ)2 = h. (4.7)

48

D 3 E 2C3 3C2

A1 1 1 1

A2 1 1 −1

E 2 −1 0

hC3

h ∑Γ

χΓ(C3)∗χΓ(C2) =26

(

1.1+1.(−1)+(−1).0)

= 0

hC3

h ∑Γ

χΓ(C3)∗χΓ(C3) =26

(

1.1+1.1+(−1).(−1))

= 1

hE

h ∑Γ

χΓ(E)∗χΓ(E) =16

(

1.1+1.1+2.2)

= 1

The last of these can be written in the form of eq. (4.7):

∑Γ

(nΓ)2 = 12 +12 +22 = 6.

Character tables for Abelian groups

For an Abelian group,

• Multiplication is commutative: RS= SR, for all pairs of elements in the group.

• Therefore RSR−1 = SRR−1 = S for any R, so every element is in a class byitself, and there are h classes;

• Therefore there are h representations,

• and because ∑hΓ=1(nΓ)2 = h, every representation is one-dimensional.

49

The projection formulaConsider the following operator, in which the coefficient of symmetry operator R is acomponent of the representation matrix for the irreducible unitary representation Γa:

Pai =

na

h ∑R

Daii (R)∗R, (4.8)

We apply this operator to a function ϕbj which transforms according to component j of

irreducible representation Γb:

Pai ϕb

j =na

h ∑R

Daii (R)∗Rϕb

j (no summation convention)

=na

h ∑R

Daii (R)∗∑

k

ϕbkDb

k j(R)

= ∑k

ϕbkδΓaΓbδikδi j using the GOT, eq. (4.1)

=

ϕai if Γa = Γb and j = i,

0 otherwise.(4.9)

50

Projection formula examples: C2v

C2v E C2 σxzv σyz

v

A1 1 1 1 1

A2 1 1 −1 −1

B1 1 −1 1 −1

B2 1 −1 −1 1

b

b

b

O

HH

sasb

z

y

PA1sa =14(1.sa +1.sb +1.sb +1.sa) = 1

2(sa +sb)

PB2sa =14(1.sa +(−1).sb +(−1).sb +1.sa) = 1

2(sa−sb)

A1 B2

sa = 12(sa +sb) + 1

2(sa−sb)

PA1sa = 12(sa +sb) + 0

PB2sa = 0 + 12(sa−sb)

Projection formula examples: D 3

D 3 E 2C3 3C2

A1 1 1 1

A2 1 1 −1

E 2 −1 0

sa

sb

sc

y

x

PA1sa =16

(

1.sa +1.sb +1.sc +1.sa +1.sc +1.sb)

=13(sa +sb +sc)

PExsa =26

(

1.sa +(−12).sb +(−1

2).sc +1.sa +(−12).sc +(−1

2).sb)

=13(2sa−sb−sc)

PEysa =26

(

1.sa +(−12).sb +(−1

2).sc +(−1).sa +(12).sc +(1

2).sb)

= 0

A1 Ex Ey

sa = 13(sa +sb +sc) + 1

3(2sa−sb−sc) + 0

PA1sa = 13(sa +sb +sc) + 0 + 0

PExsa = 0 + 13(2sa−sb−sc) + 0

PEysa = 0 + 0 + 0

The usual projection formula

The operator Pai leaves unchanged a function that transforms according to component

i of representation Γa, but annihilates everything else.

This is a powerful result, but is only useful when the representation matrices aretabulated. If however we sum over i, we get a formula that can be used when only thecharacter is available:

Pa =na

h ∑R

∑i

Daii (R)∗R=

na

h ∑R

χa(R)∗R. (4.10)

This is the formula most commonly used. It has the disadvantage, when Γa is notone-dimensional, that it yields only a single function, which in general is a mixture ofthe components of Γa.

However the other components may be obtained by Schmidt orthogonalisation.

51

Schmidt orthogonalisation

Suppose that the projection formula gave a function ϕ1. If Γa is not one-dimensionalthere must be a symmetry operator R which gives a different function Rϕ1 = ϕ′. It iseasy to normalise both functions so that

〈ϕ1|ϕ1〉 = 〈ϕ′|ϕ′〉 = 1,

but ϕ1 and ϕ′ may not be orthogonal. In this case a function ϕ2 which is orthogonal toϕ1 (though not normalized) is given by

ϕ2 = ϕ′−〈ϕ1|ϕ′〉ϕ1.

We can normalise this to obtain a function ϕ2 that is normalised and orthogonal to ϕ1.

This Schmidt orthogonalisation process can be continued if nΓ > 2 until a completeset of orthogonal basis functions has been obtained.

52

Schmidt orthogonalisation: example

Using the projection operator for E in D 3 we obtain from the function sa the

projected function ϕ1 =√

16(2sa−sb−sc) (after normalisation). Applying the

C3 operator produces the function ϕ′ =√

16(2sb−sc−sa), also normalised,

but 〈ϕ1|ϕ′〉 = 16(−2−2+1) = −1

2 .

Construct

ϕ2 = ϕ′−〈ϕ1|ϕ′〉ϕ1

=√

16(2sb−sc−sa)+ 1

2

√

16(2sa−sb−sc)

=√

16(3

2sb− 32sc)

so ϕ2 =√

12(sb−sc).

sa

sb

sc

y

x

5. Spherical harmonics and the full rotation group

For an isolated atom or molecule, any rotation about any axis through the centre ofmass is a symmetry operation, and the symmetry group comprising all these rotationsis the Full Rotation Group, also called SO(3), the Special Orthogonal Group in 3dimensions. (If spin is involved, it is necessary to use a different group called SU(2),but that is beyond the scope of the present course.)

The theory of the Full Rotation Group is the theory of angular momentum:

• Any rotation is a symmetry operator, i.e. it doesn’t change the energy.

• Therefore there are no torques on the system.

• Therefore angular momentum is conserved.

53

Rotating functionsSuppose that we want to rotate a functionψ through α about the z axis, giving a newfunction ψ′:

R(α,z)ψ = ψ′.

From the figure, we see thatψ′(r′) = ψ(r), or (in polar coordinates)

ψ′(r,θ,ϕ+α) = ψ(r,θ,ϕ).

That is,

R(α,z)ψ(r,θ,ϕ) = ψ′(r,θ,ϕ) = ψ(r,θ,ϕ−α).

In general,

Rψ(r) = ψ(R−1r). (5.1)

To apply the operation R to a function, we apply the inverse operation R−1 to itsarguments.

x

y

ϕ ϕ+αr

r′

ψ

ψ′ 54

Angular momentum operators

Construct the operator

Jz = i~ limα→0

R(α,z)−Eα

. (5.2)

where R(α,z) is the operator for a rotation through α about the z axis.

This operator certainly commutes with the Hamiltonian, since both R(α,z) and Ecommute with the Hamiltonian.

Its effect on some function ψ(ϕ) is

Jzψ(ϕ) = i~ limα→0

R(α,z)−Eα

ψ(ϕ) = i~ limα→0

ψ(ϕ−α)−ψ(ϕ)

α= −i~

∂ψ∂ϕ

, (5.3)

so Jz is the angular momentum operator, as the notation suggests.

Consequently the eigenfunctions of the Hamiltonian can be chosen to be alsoeigenfunctions of angular momentum.

55

Implications

The fact that angular momentum can be treated by the methods of group theory hasimplications that go far beyond the scope of the present course, but some of them areuseful here and are quoted without proof:

• The rotation R(α,ξ) through an angle α about an axis ξ belongs to the sameclass as a rotation R(α′,ξ′) if and only if α = α′. Consequently the class C(α)of SO(3) contains all rotations through the angle α, about any axis.

• The representations of SO(3) are labelled by the quantum number J of angularmomentum theory. Representation J has dimension 2J+1.

• The spherical harmonics YJM are basis functions for representation J.

56

Character of representation JThe angular dependence of the spherical harmonics takes the form

YJM(θ,ϕ) = ΘJM(θ)

√

12π

eiMϕ. (5.4)

When such a function is rotated through α about the z axis, we find the new functionby replacing ϕ by ϕ−α, so that

R(α,z)YJM(θ,ϕ) = ΘJM(θ)

√

12π

eiM(ϕ−α) = e−iMαYJM(θ,ϕ). (5.5)

The character χJ(α) for the set of functions with a given value of J and withM = J,J−1, . . . ,−J is then

χJ(α) =J

∑M=−J

e−iMα.

57

Character of representation J contdThe character χJ(α) for the set of functions with a given value of J and withM = J,J−1, . . . ,−J is

χJ(α) =J

∑M=−J

e−iMα

=exp[(J+1)iα]−exp[−Jiα]

exp[iα]−1,

using the geometric series formula. Multiplying top and bottom by exp(−12 iα),

χJ(α) =sin(J+ 1

2)αsin1

2α. (5.6)

This result is true for rotations through α about anyaxis, because all such rotationsare in the same class. It allows us to determine the behaviour of a set of atomicorbitals under the rotations of a finite group.

58

Example: character of f functions in Td

E 8C3 3C2 6S4 6σd

= iC4 = iC2

7 f 7

There are 7 functions, so the character of E is 7.


E 8C3 3C2 6S4 6σd

= iC4 = iC2

7 f 7 1


χ(C3) =sin7

22π3

sin12

2π3

=sin7π

3

sin π3

=sin(

2π+ π3

)

sin π3

= 1


E 8C3 3C2 6S4 6σd

= iC4 = iC2

7 f 7 1 −1


χ(C3) =sin7

22π3

sin12

2π3

=sin7π

3

sin π3

=sin(

2π+ π3

)

sin π3

= 1

χ(C2) =sin7

2πsin1

2π=

sin7π2

sin π2

=sin(

4π− π2

)

sin π2

= −1


E 8C3 3C2 6S4 6σd

= iC4 = iC2

7 f 7 1 −1 +1


χ(C3) =sin7

22π3

sin12

2π3

=sin7π

3

sin π3

=sin(

2π+ π3

)

sin π3

= 1

χ(C2) =sin7

2πsin1

2π=

sin7π2

sin π2

=sin(

4π− π2

)

sin π2

= −1

χ(S4) = χ(iC4) = −χ(C4) because all the f functions change sign on inversion. Thus

χ(S4) = −sin72

π2

sin12

π2

= −sin7π4

sin π4

= −sin(

2π− π4

)

sin π4

= +1


E 8C3 3C2 6S4 6σd

= iC4 = iC2

7 f 7 1 −1 +1 +1


χ(C3) =sin7

22π3

sin12

2π3

=sin7π

3

sin π3

=sin(

2π+ π3

)

sin π3

= 1

χ(C2) =sin7

2πsin1

2π=

sin7π2

sin π2

=sin(

4π− π2

)

sin π2

= −1

χ(S4) = χ(iC4) = −χ(C4) because all the f functions change sign on inversion. Thus

χ(S4) = −sin72

π2

sin12

π2

= −sin7π4

sin π4

= −sin(

2π− π4

)

sin π4

= +1

χ(σd) = χ(iC2) = −χ(C2) = +1

6. The symmetric representation

Every group possesses a one-dimensional symmetric representation. In thisrepresentation the matrix of every operator is (1) and the character is 1. We shall useΓ1 as a general symbol for this representation; in different groups the conventionalsymbol may be A1 (as in D 3), or A1g, A′

1, Σ+g , S, etc.

If in the G.O.T., eq. (4.1), we put Γa = Γ1 (so that i = p = 1) we have DΓ111(R)∗ = 1,

and we get another useful formula:

1h ∑

RDΓ

jq(R) =

0 if Γ 6= Γ1,

1 if Γ = Γ1.(6.1)

59

Physical properties—zero or non-zero?We consider next the behaviour of physical properties such as charge or dipolemoment. There may be just one such quantity for a particular molecule, as in the caseof the charge, or there may be several, as for instance the three components of thedipole moment. A symmetry operator will change such a quantity into a linearcombination of the several components. However a symmetry operation, by its nature,cannot change the value of any physical quantity. Thus

µi = Rµi = ∑j

µjDΓµji (R),

and if we now sum over the operations of the group we get

hµi = ∑R

Rµi = ∑j

µj ∑R

DΓµji (R). (6.2)

If Γµ is irreducible we can use eq. (6.1) to deduce that

µi = 0 unless Γµ is the symmetric representation.

60

Physical properties contd.

If Γµ is indeed the symmetric representation, then we just get µi = µi , so thesymmetry tells us only whether a quantity is zero or non-zero; it tells us nothing aboutthe magnitude of non-zero quantities.

If Γµ is reducible, it is necessary first to reduce the representation, and then to applythe argument to each irreducible component in turn.

This leads to the conclusion that the number of independent nonzero components isthe number of times that the symmetric representation occurs in Γµ.

61

Example: dipole moment of H2O

b

b

b

O

H

H

Eµy = σyzµy = µy

Cz2µy = σxzµy = −µy

Consider the y component of the dipolemoment of water. Eq. (6.2) tells us that4µy = ∑RRµy = (E +Cz

2 +σxz+σyz)µy.But the figure shows that this sum is zero.

The components of the dipole momenttransform as A1⊕B1⊕B2, according tothe character table, so there is onenon-zero component, i.e. the A1component µz.

We hardly need group theory to arrive atthis result. However the same approach isuseful in less obvious cases.

Example: polarizability of BF3

The polarizability describes the dipole moment induced by an electric field:∆µi = ∑ j αi j E j , where i and j can be x, y or z. The component αi j transforms

like i j , so we can look up the symmetries in the D 3h character table:

αzz transforms like z2 A′1 so αzz 6= 0

αxx+αyy x2 +y2 A′1 αxx+αyy 6= 0

αxx−αyy x2−y2 αxx−αyy = 0

αxy xy

E′

αxy = 0

αxz xz αxz = 0

αyz yz

E′′

αyz = 0

Thus αxx+αyy 6= 0 but αxx−αyy = 0.

We see that there are two independentnon-zero components, correspondingto the two occurrences of A′

1: αzz= α‖and αxx = αyy = α⊥.

b b

b

b

B F

F

F

x

y

62

When is an integral non-zero?

In quantum mechanics we often encounter integrals, typically of the form∫

ψ1∗Qψ2dτ. Sometimes, as in the determination of selection rules for spectroscopy,we don’t need to know the exact value of the integral — we just need to know whetherit is zero or not. Even when we do need the value, we can avoid the trouble of workingout integrals that the group theory tells us are zero.

Consider first a simpler integral,∫

FΓi dτ, whose integrand FΓ

i transforms according tocomponent i of representation Γ. The integral is over all space, so rotating orreflecting the function makes no difference. Therefore

h∫

FΓi dτ = ∑

R

∫

RFΓi dτ = ∑

R

∫

FΓj DΓ

ji (R)dτ.

But ∑RDΓji (R) = 0 unless Γ is the symmetric representation.

We see that the integral can only be non-zero if the integrand is symmetric.

To deal with the more general kind of integral that we started with, we need to be able

to determine whether an integrand like ψ1∗Qψ2 is symmetric. To do this we needdirect products.

63

7. Direct product representations

Consider first the case of non-degenerate representations.Suppose that we want to know whether the electronic transitionfrom the HOMO of H2O (b1 symmetry) to the LUMO (a1) isallowed in z polarization. The integral is

∫

ψa1∗µzψb1dτ. A

symmetry operator R affects each factor in the integrandindividually, so

R(

ψa1∗µzψb1

)

=(

Rψa1

)∗(Rµz)(

Rψb1

)

= ψa1∗χA1(R)µzχA1(R)ψb1χB1(R)

= ψa1∗µzψb1

(

χA1(R))2χB1(R).

We get the character for the whole integrand just by multiplyingtogether the characters of the individual factors. In this case theresult is χB1(R), so the integrand transforms according to B1,which is not symmetric. Consequently the integral is zero andthe transition is forbidden.

b2∗

a1∗

b1

a1

b2

a1

z

64

Direct products of degenerate representations

For non-degenerate representations, we find the character of a product of functions bysimply multiplying together the characters of the factors. If the functions belong todegenerate representations, it turns out that this is still true, but the proof is a littlemore complicated.

Suppose that we have two sets of functions: ξ1, ξ2, . . . ,ξm, a basis forrepresentation Γa, and η1, η2, . . . ,ηn, a basis for Γb, and consider the set ofproducts ξiη j . Operating with R gives

R(ξiη j) = ∑k

ξkDΓaki (R)∑

l

ηl DΓbl j (R) = ∑

kl

(ξkηl )DΓaki (R)DΓb

l j (R) (7.1)

We can also view the mnproduct functions, labelled by the double index i j , as a basisfor a new representation of dimension mn:

R(ξiη j) = ∑kl

(ξkηl )DΓa⊗Γbkl,i j (R). (7.2)

This is the direct product representation Γa⊗Γb.

65

The character of the direct product

For the direct product, we have

R(ξiη j) = ∑kl

(ξkηl )DΓa⊗Γbkl,i j (R) = ∑

kl

(ξkηl )DΓaki (R)DΓb

l j (R),

so the representation matrices for the direct product are

DΓa⊗Γbkl,i j = DΓa

ki (R)DΓbl j (R). (7.3)

To find the character χΓa⊗Γb we set kl equal to i j and sum over i and j :

χΓa⊗Γb(R) = ∑i j

DΓaii (R)DΓb

j j (R) = χΓa(R)χΓb(R). (7.4)

So the character is very easily determined, and the standard reduction proceduregives the irreducible components.

66

ExamplesD 3 E 2C3 3C2

A1 1 1 1

A2 1 1 −1

E 2 −1 0 (x,y)

E⊗E 4 1 0 A1⊕A2⊕E

Td 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

T1⊗T2 9 0 1 −1 −1 A2⊕E⊕T1⊕T2

How many symmetric components in a direct product?

It is often important to know how many times the symmetric representation occurs in adirect product, for example to determine whether an integral is non-zero.

If Γa and Γb are both irreducible, then the direct product Γa∗⊗Γb (for which the basisfunctions are ξi∗η j ) contains the symmetric representation onceif Γa ∼ Γb, andotherwise not at all.

To prove this, we use eq. (4.5) to find the number of occurrences of Γ1:

m1 =1h ∑

chcχΓ1(c)∗χΓ∗

a⊗Γb =1h ∑

chcχΓ1(c)∗

(

χΓa(c)∗χΓb(c))

=1h ∑

chcχΓa(c)∗χΓb(c)

= δΓaΓb. (7.5)

67

The symmetric component of a direct productIn the case where the representations forming the direct product are identical (not justequivalent) so that Γa = Γb = Γ, the totally symmetric component is given by theprojection formula, eq. (4.10):

(ξi∗η j)Γ1 =

1h ∑

RDΓ1

11(R)R(ξi∗η j)

=1h ∑

R∑kl

ξk∗ηl DΓki(R)∗DΓ

l j (R)

= ∑kl

(ξk∗ηl )1nΓ

δklδi j

so finally

(ξi∗η j)Γ1 = δi j

1nΓ

∑k

ξk∗ηk. (7.6)

The δi j in this equation means that ξi∗η j contains no totally symmetric componentunless i = j .

68

Calculation of integrals

Now we can find general rules for determining whether integrals are zero or not. One common

type of integral is the overlap integral∫

(ξai )∗ηb

j dτ, where ξai transforms according to

component i of irreducible representation Γa and ηbj according to component j of

representation Γb. The integrand here is a component of the direct product Γa∗⊗Γb, which

contains Γ1 only if Γa ∼ Γb. Therefore

∫

(ξai )∗ηb

j dτ = 0 unless Γa ∼ Γb. (7.7)

If Γa = Γb = Γ (identical representations, not just equivalent) then from eq. (7.6) we see that

∫

(ξΓi )∗ηΓ

j dτ = 0 unless i = j. (7.8)

Eq. (7.6) also shows that if i = j then∫

(ξΓi )∗ηΓ

i dτ =1nΓ

∫

∑k

(ξΓk )∗ηΓ

k dτ (7.9)

and is the same for all i.

69

Calculation of integrals contd.

Next we consider integrals such as∫

(ξai )∗Qc

kηbj dτ, which involve an operator. Here Qc

k is one

of a set of operators transforming according to the irreducible representation Γc. The functions

Qckηb

j form a basis for the direct product representation Γc⊗Γb, so

∫

(ξai )∗Qc

kηbj dτ = 0 unless Γc⊗Γb ∋ Γa. (7.10)

A special case of eq. (7.10) arises when Γc = Γ1—that is, the operator is totally symmetric.

Then Γc⊗Γb = Γ1⊗Γb = Γb, and (7.10) reduces to a formula like (7.7):

∫

(ξai )∗QΓ1ηb

j dτ = 0 unless Γa ∼ Γb, (7.11)

and if Γa = Γb = Γ (identical representations) then

∫

(ξΓi )∗QΓ1ηΓ

j dτ = 0 unless i = j. (7.12)

Here again we see that∫

(ξΓi )∗QΓ1ηΓ

i dτ is independent of i.

70

Integrals for M.O. theoryThe application of these results to M.O. theory is now clear. The Hamiltonian, bydefinition, is unaffected by any symmetry operator and so is totally symmetric.Therefore:

• Overlap integrals and Hamiltonian integrals between symmetry orbitals ofdifferent symmetries are zero. That is, orbitals of different symmetry don’t mix.

• If symmetry orbitals of the same symmetry are constructed so that theircomponents match (the representations are identical, not just equivalent) thenintegrals involving different components of the same representation are alsozero.

• Moreover for different components i and j , we have

∫

(ξai )∗Hηa

i dτ =∫

(ξaj )∗Hηa

j dτ,

and similarly for the overlap integrals, so that the secular equations forcomponents i and j are identical.

71

The symmetrized and antisymmetrized square

When the two bases ξ1, ξ2, . . ., ξn and η1, η2, . . ., ηn in a direct product span thesame representation Γ, it is often convenient to consider as basis functions for thedirect product the symmetrized products ξiη j +ηiξ j (symmetric with respect to

exchange of ξ and η) and the antisymmetrized products ξiη j −ηiξ j , rather than the

simple products ξiη j .

The symmetrized products transform only among themselves, and they therefore form

a basis for a representation, of dimension 12n(n+1), which is called the symmetrized

square of Γ and denoted (Γ⊗Γ)+. Similarly the antisymmetrized products form a

basis for the antisymmetrized square (Γ⊗Γ)−, with dimension 12n(n−1). The

characters of these two representations are given by the formula

χΓ⊗Γ±(R) = 12

[(

χΓ(R))2±χΓ(R2)

]

. (7.13)

If the two bases are the same, so that ξi = ηi , then evidently the antisymmetrizedproducts vanish.

72

The direct product E⊗E

If we have two sets of basis functions, ξx,ξy and ηx,ηy, both transforming

according to E, the direct product basis is ξxηx,ξxηy,ξyηx,ξyηy.

ξxηx and ξyηy are already symmetric. From the other two products we can construct

ξxηy +ηxξy (symmetric) and ξxηy−ηxξy (antisymmetric).

The direct product transforms as A1⊕A2⊕E, and the symmetry-adapted functionsare

A1 : ξxηx +ξyηy symmetric

A2 : ξxηy−ηxξy antisymmetric

E : ξxηy +ηxξy symmetric

ξxηx−ξyηy symmetric

The symmetrized square is (E⊗E)+ = A1⊕E, and the antisymmetrized square is(E⊗E)− = A2.

73

Electronic terms

If we have two electrons in the same degenerate set of orbitals, then we can takeξi = ψi(1) and ηi = ψi(2). We can construct orbitally-symmetric functions

ξiη j +ηiξ j = ψi(1)ψ j(2)+ψi(2)ψ j(1), (7.14)

and orbitally-antisymmetric ones

ξiη j −ηiξ j = ψi(1)ψ j(2)−ψi(2)ψ j(1). (7.15)

The orbitally-symmetric functions, i.e. the components of the symmetrized square,have to be combined with antisymmetric (singlet) spin functions, and the orbitallyantisymmetric ones with symmetric (triplet) spin functions, in order that the overallwavefunction shall be antisymmetric.

Thus the configuration e2 for a molecule, with two electrons in an E set of orbitals,

leads to 1A1, 3A2 and 1E molecular terms.

In octahedral symmetry, the symmetrized square (T2g⊗T2g)+ = A1g⊕Eg⊕T2g, while

the antisymmetrized square (T2g⊗T2g)− = T1g. Therefore the configuration t22g leads

to the molecular terms 1A1g, 1Eg, 1T2g and 3T1g.

74

Direct products in the Full Rotation Group

In the Full Rotation Group, the direct product Γ(J1) ⊗Γ(J2) reduces as follows(see ex. 6.2.):

Γ(J1) ⊗Γ(J2) = Γ(J1+J2) ⊕Γ(J1+J2−1) ⊕·· ·⊕Γ|J1−J2|.

In the language of angular momentum, this means that a system having twocomponents, one with angular momentum J1 and one with angular momentum J2, hasstates with total angular momentum J1 +J2, J1 +J2−1, . . . |J1−J2|. (This is calledthe Clebsch–Gordan series.)

With J1 = J2 this becomes

Γ(J) ⊗Γ(J) = Γ(2J) ⊕Γ(2J−1) ⊕·· ·⊕Γ(0).

It can be shown that the terms in this direct sum are alternately symmetric andantisymmetric, so that

(

Γ(J) ⊗Γ(J))

+= Γ(2J) ⊕Γ(2J−2) ⊕·· · ,

(

Γ(J) ⊗Γ(J))

− = Γ(2J−1) ⊕Γ(2J−3) ⊕·· · .

75

Atomic terms

To find the allowed terms for the p2 configuration, we need

(P⊗P)+ = Γ(2) ⊕Γ(0) = D⊕S, (P⊗P)− = Γ(1) = P.

To construct antisymmetric states to satisfy the Pauli principle, the symmetric spacefunctions have to be combined with the antisymmetric, i.e. singlet, spin functions, and

vice versa, so the allowed terms are 1D, 3P and 1S.

Similarly, the allowed terms for the d2 configuration are 1G, 3F , 1D, 3P and 1S.

76

8. The Jahn–Teller theoremWe consider a molecule in an orbitally degenerate state of symmetry Γ, so that thereare n degenerate many-electron wavefunctions ψ1, ψ2, . . . ,ψn.

The Jahn–Teller theorem states that a non-symmetric distortion of symmetry ΓQ willlower the energy of at least one of the states, provided that (Γ⊗Γ)+ ∋ ΓQ.

If the molecule is distorted away from the symmetrical geometry, the electronicHamiltonian will change, introducing a perturbation VQ of the same symmetry ΓQ asthe distortion itself. Perturbation theory shows that the energy changes are, to firstorder in the distortion, the eigenvalues εi of the matrix V with elementsVi j =

∫

ψi∗VQψ j dτ. We assume that the functions ψi are all real (which can alwaysbe arranged in the case of an orbital degeneracy) and note that the perturbation VQ isjust a change in potential—the kinetic energy terms in the electronic Hamiltonian donot change—and that it too is real. Consequently we can write

Vi j =∫

(ψiψ j)VQdτ,

and we see that all the Vi j vanish unless the representation spanned by the (ψiψ j)

(that is, the symmetrized square (Γ⊗Γ)+) contains ΓQ.

77

The Jahn–Teller theorem contd.

We can now prove the following results:

1. The sum of the energy changes is zero unless the distortion is totally symmetric.

To see that this is so, remember that the sum of the eigenvalues of the matrix Vis equal to its trace. That is,

∑i

εi = ∑i

Vii =∫

(

∑i

ψ2i

)

VQdτ (8.1)

which vanishes unless ΓQ = Γ1, because ∑i ψ2i is symmetric (eq. (7.6)).

However we are interested only in unsymmetrical distortions, because asymmetric distortion does not change the symmetry of the electronicHamiltonian and so cannot lift the degeneracy.

78

The Jahn–Teller theorem contd.1. The sum of the energy changes is zero unless the distortion is totally

symmetric.

2. The sum of the squares of the energy changes is non-zero provided that(Γ⊗Γ)+ ∋ ΓQ.

If V has eigenvalues εi , then the matrix V2 has eigenvalues ε2i . So the sum of

squares of the energy changes is

∑i

ε2i = trace(V2) = ∑

i jVi jVji = ∑

i j|Vi j |2, (8.2)

and this will be non-zero if any of the elements of V is non-zero. Therequirement for this is that (Γ⊗Γ)+ ∋ ΓQ, as noted above.

79

The Jahn–Teller theorem contd.1. The sum of the energy changes is zero unless the distortion is totally

symmetric.

2. The sum of the squares of the energy changes is non-zero provided that(Γ⊗Γ)+ ∋ ΓQ.

If the sum of the energy changes is zero, but the sum of their squares is non-zero,then at least one must be negative, and the Jahn–Teller theorem follows.

It can be shown that for non-linear molecules there is always a distortion satisfying theJahn–Teller condition. Proving it is much more difficult, and was not achieved until 20years after Jahn & Teller proved their theorem. For a discussion, see Landau &Lifshitz, Quantum Mechanics (Non-relativistic Theory), 3rd edn., pp. 405–410.

It is much easier to show that for linear molecules there is never a distortion satisfyingthe Jahn–Teller condition.

80

Jahn–Teller distortion in octahedral d9 complexes

The classical example of Jahn–Teller distortion is the octahedral d9 complex. Here theelectronic state is Eg, with symmetrized square A1g⊕Eg, so the distortion has to beEg.

The mode in which the z ligands move out and the x and y ones move in is onecomponent of an Eg vibration; the other component has the x ligands moving out andthe y ligands moving in, with the z ligands not moving at all. (Compare dz2 and dx2−y2

orbitals, also of Eg symmetry.)

In fact any linear combination of these modes leads to a splitting of the eg orbitals infirst order, with one orbital lowered in energy and the other raised. For largedistortions of this type the repulsion between the ligands and the central ion causesthe energy to rise.

The potential energy surface isapproximately a parabola rotated aroundan axis parallel to its own axis, and has aring-shaped well. There are normally threeminima on this ring, corresponding to thegeometries in which two ligands move outand the others in.

81


b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

82


In d8 complexes, the electron configuration is e2g and the allowed terms are

1Eg, 1A1g and 3A2g. The lowest-energy state in the octahedral geometry is 3A2g,

according to Hund’s first rule,with the 1Eg above it and the 1A1g above that, but

the 1Eg state is subject to Jahn–Teller distortion, which may reduce the energybelow that of the triplet.

Consequently in d8 we expect either a symmetrical triplet state or a distortedsinglet.

1Eg

3A2g

83

The dynamical Jahn–Teller effect

In some Jahn–Teller molecules the energy associated with the Jahn–Teller distortionis quite small compared with kT, and then the distortion is not static but dynamic.

This does not occur in octahedral d9 complexes, where there is a strong interactionbetween the eg orbitals and the ligands, but it is common in complexes where only thet2g orbitals are occupied, since these interact much more weakly with the ligands.

84

9. Vibrational coordinatesVibrational coordinates can be symmetry-adapted in much the same way as atomicorbitals. Each atom has 3 cartesian displacement coordinates, and together theyprovide 3N symmetry coordinates. 3 of these correspond to translational motions, and3 to rotations (or 2 in the case of linear molecules). Subtracting these off, we are leftwith 3N−6 (or 3N−5) true vibrations.

In the case of BF3, the 3N cartesian displacement coordinates transform asA′

1⊕A′2⊕2A′′

2 ⊕3E′⊕E′′. The translations transform as A′′2 ⊕E′ and the rotations as

A′2⊕E′′, leaving the true vibrations A′

1⊕A′′2 ⊕2E′.

Of these, the A′1 mode is the symmetric combination

√

13(r1 + r2 + r3) of the fluorine

radial displacement coordinates.

The A′′2 mode is a linear combination of zB and the symmetric combination of the

fluorine z coordinates√

13(z1 +z2 +z3):

QA′′2= c1zB +c2

√

13(z1 +z2 +z3). (9.1)

The coefficients have to be chosen so that the motion leaves the centre of massstationary.

85

Vibrational coordinates for BF3

D 3h E 2C3 3C2 σh 2S3 3σv

A′1 1 1 1 1 1 1

A′2 1 1 −1 1 1 −1 Rz

E′ 2 −1 0 2 −1 0 (x,y)

A′′1 1 1 1 −1 −1 −1

A′′2 1 1 −1 −1 −1 1 z

E′′ 2 −1 0 −2 1 0 (Rx,Ry)

B (x,y) 2 −1 0 2 −1 0 E′

B z 1 1 −1 −1 −1 1 A′′2

F (z1,z2,z3) 3 0 −1 −3 0 1 A′′2 ⊕E′′

F (r1, r2, r3) 3 0 1 3 0 1 A′1⊕E′

F (t1, t2, t3) 3 0 −1 3 0 −1 A′2⊕E′

Cartesian: A′1 A′

2 2A′′2 3E′ E′′

Translations: A′′2 E′

Rotations: A′2 E′′

Vibrations: A′1 A′′

2 2E′

r1t1

t2

r2

t3r3

b b

b

b

B F

F

F

xyb b

b

b

B F

F

F

86

E′ modes for BF3The E′ symmetry coordinate constructed from the r i is essentially a stretching mode.We need to add an x or y displacement of the B atom so that the centre of mass doesnot move.

b b

b

b

B F

F

F

b b

b

b

B F

F

F q3x

b b

b

b

B F

F

F

b b

b

b

B F

F

F q3y

Similarly the E′ symmetry coordinate constructed from the ti is a bending mode.

b b

b

b

B F

F

F

b b

b

b

B F

F

F q4x

b b

b

b

B F

F

F

b b

b

b

B F

F

F q4y

These diagrams are schematic only. See Herzberg, Infrared and Raman Spectra, p.179, for a diagram of the calculated modes, allowing also for mixing between q3 and q4.

87

Allowed harmonic terms in the vibrational potential

Normally the potential energy for nuclear motion is expanded in a Taylor series aboutthe equilibrium geometry. The linear terms are zero, so the first terms to consider arethe quadratic (harmonic) terms. However the Hamiltonian is totally symmetric, so onlysymmetric terms can appear.

If we have a set of vibrational coordinates (qa1,qa2, . . .) belonging to irreduciblerepresentation Γa, and another set (qb1,qb2, . . .) belonging to Γb, possible quadraticterms are the elements qaiqb j of the direct product. This set of elements includes asingle symmetric component if and only if Γa ∼ Γb. If Γa = Γb, the symmetriccomponent is proportional to ∑i qaiqbi. (Eq. (7.6).)

Thus the only quadratic terms that may appear in the vibrational potential of BF3 are

q21, q2

2, q23x +q2

3y, q24x +q2

4y and q3xq4x +q3yq4y. (9.2)

Here q1 is the A′1 mode and q2 the A′′

2 one. Note that the components (q3x,q3y) and(q4x,q4y) have been defined so that they match — the representations are identical,not just equivalent, so that we can use eq. (7.6) to find the symmetric component ofthe direct product (q3x,q3y)⊗ (q4x,q4y).

88

Allowed anharmonic terms

There are more possibilities for anharmonic terms.

(i) Any of the harmonic terms can be multiplied by q1, which is symmetric, toobtain an allowed cubic term.

(ii) If we could construct a quadratic term of A′′2 symmetry, we could multiply it by

q2 to obtain a symmetric term. However the only such combination is q1q2,

which would give q1q22, and this has been included under (i) above.

(iii) We can construct a quadratic E′ term from (q3x,q3y). It has the components

(q23x−q2

3y,−2q3xq3y). (They have to be defined like this so that they transform

in the standard way, i.e. like (x,y).) The A′1 component of the direct product of

this with (q3x,q3y) is q33x−3q3xq2

3y, which may therefore appear as a cubic

anharmonic term. Other terms of similar form may be constructed frommixtures of q3 and q4 components.

No other cubic terms may appear in the Hamiltonian.

Anharmonic terms can lead to coupling between vibrational states, and hence toperturbations of their energies. Because the anharmonic terms must be symmetric,they can only couple states of the same symmetry.

89

Scaling the harmonic oscillatorThe harmonic oscillator Hamiltonian is

H = − ~2

2m∂2

∂q2 + 12kq2. (9.3)

Define x = q/s, where s is some length, to be determined. We get

H = − ~2

2ms2∂2

∂x2 + 12ks2x2, (9.4)

orH

ks2 = − ~2

2mks4∂2

∂x2 + 12x2. (9.5)

Choose s=(

~2/mk

)1/4, and we get ks2 = ~

√

k/m= ~ω and

H

~ω= −1

2∂2

∂x2 + 12x2. (9.6)

90

Vibrational ladder operators

To deal with the symmetries of excited vibrational states it is helpful to introduce somenew operators.

If we use the scaled units, with the energy unit as ~ω, the harmonic oscillatorHamiltonian is

H = −12

∂2

∂x2 + 12x2.

Define the operators

Q =√

12

(

x+∂∂x

)

, Q† =√

12

(

x− ∂∂x

)

. (9.7)

For any ψ,

QQ†ψ = 12

(

x+∂∂x

)(

x− ∂∂x

)

ψ = 12

(

x2− ∂2

∂x2 +1

)

ψ = (H+ 12)ψ. (9.8)

Therefore QQ† = H+ 12 . Similarly Q†Q = H− 1

2 .

91


Now suppose that there is a wavefunction ψn that is aneigenfunction of H: Hψn = Enψn. Consider the new

function Q†ψn. We find

H(Q†ψn) = (Q†Q+ 12)Q†ψn = Q†(QQ† + 1

2)ψn

= Q†(H+1)ψn = (En +1)(Q†ψn). (9.9)

So Q†ψn is an eigenfunction of H with eigenvalueEn +1. (Remember that the energy unit is ~ω.)

In the same way, we can show that Qψn is aneigenfunction with energy En−1.

Q and Q† are called ladder operators or raising andlowering operators, because they move up and down aladder of vibrational states.

En +1

En

En−1

Q† Q

Q† Q

92


The energy of a harmonic oscillator can’t be negative, so there must be some state

ψ0 for which Qψ0 = 0. The energy of that state is given by

Hψ0 = (Q†Q+ 12)ψ0 = 1

2 (9.10)

(in units of ~ω). To find the wavefunction ψ0 we solve Qψ0 = 0:√

12

(

x+∂∂x

)

ψ0 = 0, (9.11)

with the solution

ψ0 = exp(−12x2) = exp

(

−√

km~2 q2

)

. (9.12)

Now we can get all the other vibrational wavefunctions using Q†:

ψn = (Q†)nψ0, (9.13)

and ψn has energy (n+ 12)~ω.

93

State symmetries for non-degenerate vibrations

The ground state vibrational wavefunction is exp(−12x2), and if x is a

non-degenerate vibrational coordinate then x2 is symmetric, so ψ0 issymmetric.

ψ1 = Q†ψ0, and Q† has the same symmetry as x, so ψ1 also has the samesymmetry as x.

ψ2 = (Q†)2ψ0, which is symmetric because (Q†)2 is symmetric.

Thus for non-degenerate vibrations, the vibrational states ψn alternatebetween symmetric (even n) and the same symmetry as the vibrationalcoordinate (odd n).

94

State symmetries for degenerate vibrations

For a degenerate vibration, with coordinates x1, x2, . . . , there is a pair of ladder

operators Qi and Q†i for each xi . The ground state has zero quanta in each mode; it is

ψ00... = exp(

−12(x2

1 +x22 + . . .)

)

. The exponent is symmetric (see eq. (7.6)) so ψ00...

is symmetric.

There is a set of singly-excited states Q†i ψ00.... These transform like the Q†

i andtherefore like the xi .

The doubly-excited states are Q†i Q†

j ψ00.... These transform like the products Q†i Q†

j ,

but Q†i Q†

j = Q†j Q

†i , so we only get the components of the symmetrized square.

Thus for an E vibration in D 3, the ground state has symmetry A1, the first excitedstate is doubly degenerate, with symmetry E, and the second excited state hassymmetry (E⊗E)+ = A1 +E. The A1 and E components will have different energies(no accidental degeneracy!) due to anharmonic interactions involving other vibrationalstates.

95

Fermi resonance in CO2

The bending vibration, ν2, in CO2 has

symmetry Πu. The first excited state 0110has the same symmetry.(The notation here indicates that there is one

quantum of excitation in ν2 and that the resulting

angular momentum about the molecular axis is ±1.)

The doubly-excited state of this vibrationhas symmetry (Πu⊗Πu)+ = Σ+

g ⊕∆g.

These states are denoted by 0200 and0220 respectively: 2 quanta in ν2, leadingto angular momentum 0 or ±2.

The Σ+g 0200 state has the same symmetry, and very nearly the same energy,

as the state 100with a single excitation in the ν1 symmetric stretch. There is

an anharmonic term in the Hamiltonian proportional to Q1(Q22x +Q2

2y), and this

is responsible for the mixing between the two Σ+g states and the resulting

splitting.

Σ+g 000

Πu 0110

∆g 0220

Σ+g 0200

Σ+g 100

96

10. More on symmetry operations

In ammonia, there are six possible permutations of the three protons, namely E,(abc), (acb), (ab), (ac) and (bc), each of which can be combined with the inversionE∗ to give six permutation–inversions. Thus there are twelve operations altogether.However the usual symmetry group, C3v, has only six elements. Which of the twelvedo these correspond to, and what do the other six do?

Permutation–inversions correspond to improper operations, and true permutations toproper ones, and it is not difficult to establish the mapping of (abc) and (acb) to C3

and C−13 , and of (ab)∗, (bc)∗ and (ac)∗ to the three reflections. What does E∗ do?

We define axes with z perpendicular to the plane of the H atoms, x so that the xzplane contains atom a, and y in the direction from c to b. The direction of the z axis ischosen to complete a right-handed frame.

Then the effect of E∗ is to reflect everything, including the nuclear configuration, in theplane of the H atoms. We can regard it as a σh operation.

97

Symmetry operations for NH3: E∗

b

b

b

b

z

y

x

ab

c

2pz

1sa

E∗

b

b

b

b

z

y

x

ab

c

−2pz1sa

σh

b

b

b

bz

y

x

ab

c

−2pz

1sa

98

Inversion of ammonia

The new configuration of ammonia generated by the E∗ operation is notsuperimposable on the original one — it is a mirror image. It is a secondversion of the equilibrium geometry, distinguishable from the first if we takenuclear labels into account. Moreover it is accessible from the original

configuration via a planar configuration, with a low barrier of about 25kJmol−1.

Vibrational wavefunctions of the two versions can mix by tunnelling through thebarrier, and this leads to a small but observable splitting of the energy levels.The wavefunctions are either symmetric or antisymmetric with respect to theE∗ operation, so we need it to get a complete symmetry classification.

99

Feasibility

In ammonia, the two versions of the equilibrium geometry are easily accessible fromeach other via a low barrier. The operations that convert one version into the other aresaid to be feasible, and we need them for a complete classification of the energylevels.

In the methane molecule, there are 4! = 24 permutations of the four H atoms, and 24permutation–inversions.

However some of these (in fact exactly half of them) lead to an inverted configurationof methane that can only be reached by crossing a very high energy barrier. There isno mixing between wavefunctions of these two versions of methane, and thesymmetry operations that convert one into the other do not add any usefulinformation. Such non-feasible operations can be ignored.

The Molecular Symmetry Group is the set of feasible permutations andpermutation–inversions.

100

Non-rigid molecules

Ammonia is an example of a non-rigid molecule. Such molecules have morethan one version of the minimum-energy geometry, distinguished from oneanother by nuclear labels, and connected to each other by low-energypathways on the potential energy surface.

An important class of such non-rigid molecules comprises weakly-bound ‘Vander Waals molecules’.

In the water dimer, for example, one molecule acts as a proton donor to theother to form a hydrogen bond. The resulting structure has Cs symmetry. TheH atom in the hydrogen bond can be any of the four; moreover the two Hatoms in the acceptor molecule can change places by a feasible motion.

Consequently there are eight versions of the equilibrium structure, each havingCs symmetry, so there are sixteen symmetry operations in the molecularsymmetry group. All of these must be included in order to classify thevibrational states.

Often the symmetry groups for molecular clusters like this do not correspond toany of the standard tabulated symmetry groups.

101

Water dimer

Eb

b

b

34

bb

b

1

2(34)

b

b

b

43

bb

b

1

2(12)

b

b

b

34

bb

b

2

1(12)(34)

b

b

b

43

bb

b

2

1

(13)(24)b

b

b

12

bb

b

3

4(1324)

b

b

b

21

bb

b

3

4(1423)

b

b

b

12

bb

b

4

3(14)(23)

b

b

b

21

bb

b

4

3

102

Approximate symmetry and descent in symmetry

In the complex [Ni(en)3]2+ (en = ethylene diamine) the six N atoms act as ligands to

the metal ion. Formally, the complex has D 3 symmetry only, and the metal d orbitalsare classified in this group as A1⊕E⊕E.

This information does not in itself tell us much about the relative energies of the dorbitals. However the immediate environment of the metal ion is approximatelyoctahedral. In Oh symmetry we know that the d orbitals split into Eg⊕T2g, with the Eg

orbitals higher in energy by the crystal-field splitting ∆, which we can estimatereasonably reliably from our knowledge of true octahedral complexes.

To determine what happens in [Ni(en)3]2+ we use descent in symmetry.

We start by classifying the orbitals according to the approximate symmetry Oh. Thenwe investigate the behaviour of the Oh symmetry orbitals under the symmetryoperations that remain in D 3.

103

Descent in symmetryOf the 48 symmetry elements in Oh, onlysix survive in D 3: the identity, one of the C3operations and its inverse, and three of thesix dihedral C2 operations. In the Oh

character table, we delete all the columnsreferring to the other symmetry operations,leaving three columns only. For eachrepresentation Γ of Oh, these threecolumns give the character for thesurviving elements in D 3. We have toreduce this character to determine thecorresponding symmetry species in D 3.

Oh E C3 C2 D 3

A1g 1 1 1 A1

A2g 1 1 −1 A2

Eg 2 −1 0 E

T1g 3 0 −1 A2⊕E

T2g 3 0 1 A1⊕E

For [Ni(en)3]2+ we see that the Eg orbitals

remain degenerate (E in D 3) while the T2g

orbitals split into A1⊕E. We expect thissplitting to be small. Group theory alonecannot predict the magnitude of thesplitting or even its sign, but often theapplication of perturbation theory cananswer such questions.

eg

t2g

e

e

a1

104

11. Selection rules

Time-dependent perturbation theory tells us that the transition probability betweenstates |Ψ′〉 and |Ψ′′〉 when a perturbation V cosωt is applied is proportional to∣

∣〈Ψ′|V|Ψ′′〉∣

∣

2, provided that ~ω = |E′−E′′|.

Usually V is proportional to the electric dipole moment operator (electric dipoletransitions) but it can be the magnetic dipole operator (in magnetic resonance), thepolarizability (in Raman scattering) or various other quantities. We consider only theelectric dipole case, where the perturbation is the scalar product −µ ·E cosωt of theelectric dipole operator and the electric vector of the radiation field.

The fundamental selection rule is then simply that if |Ψ′〉, |Ψ′′〉 and µ transformaccording to Γ′, Γ′′ and Γµ respectively, then the transition probability is zero (the

transition is forbidden) unless Γµ⊗Γ′′ ∋ Γ′.

105

Selection rules for molecular spectroscopy

The general derivation of selection rules for molecular spectroscopy is summarizedvery briefly as follows. Write the wavefunction for the molecule as Ψ = ψrotψint, whereψrot is the rotational function and ψint is the internal, i.e. vibrational-electronic orvibronic, function. Suppose that the light is polarized in the Z direction (fixed in space)so that V = −µZEZ is proportional to the component µZ of the dipole operator. Wehave µZ = ∑α lαZµα, where the lαZ are direction cosines depending on the orientationΩ of the molecule, and the µα are components of µ referred to axes fixed in themolecule.

Then we can factorise the golden rule integral into a part that depends only on themolecular orientation, and a part that depends only on the internal coordinates:

〈Ψ′|µZ|Ψ′′〉 =∫

ψ′rot∗ψ′

int∗µZ ψ′′

rotψ′′int dτ

= ∑α

∫

ψ′rot∗ lαZ ψ′′

rot dΩ∫

ψ′int∗µα ψ′′

int dQdq, (11.1)

where Q refers to vibrational coordinates and q to electronic.

106

Pure rotational spectroscopy

Here the transition is between rotational levels of the same vibronic state, usually theground state, so ψ′

int = ψ′′int = ψ0

int and the second integral in (11.1) is the expectation

value of the dipole moment in the ground state ψ0int. At least one component must be

non-zero for pure rotational transitions to occur. The first factor of (11.1) then gives thedetailed selection rules ∆J = 0,±1 etc.

For example, consider a linear molecule. Here µx = µy = 0, so the only non-zero termin (11.1) has α = z. If the molecular axis z lies in the direction (θ,ϕ), then lzZ = cosθ,which is proportional to the spherical harmonic Y10, while the rotational wavefunctionsare also spherical harmonics.

Consequently the rotational factor has the form

∫

YJ′M′∗Y10YJ′′M′′dΩ, (11.2)

which is zero unless Γ(J′′) ⊗Γ(1) ∋ Γ(J′), i.e. unless J′′ = J′ or J′±1. However wemust also take parity into account: YJM changes sign under the parity operation if J isodd but not if J is even, so the integral vanishes if J′ +J′′ +1 is odd, which it is ifJ′ = J′′. Consequently the selection rule is J′ = J′′±1.

107

Vibration-rotation spectroscopy

Write ψint = ψelec(q,Q)ψvib(Q) and integrate over the electronic coordinates q. Thetransition is between vibration-rotation levels of the same electronic state, soψ′

elec = ψ′′elec = ψn

elec and the second factor in (11.1) becomes

∫

ψ′vib∗µn

α(Q)ψ′′vib dQ, (11.3)

where µn(Q) is the dipole moment in state ψnelec:

µnα(Q) =

∫

ψnelec(q,Q)∗µα(q,Q)ψn

elec(q,Q)dq. (11.4)

µnα remains a function of Q; expand it in a Taylor series about the equilibrium

geometry:

µnα(Q) = (µn

α)eq+∑k

Qk

(

∂µnα

∂Qk

)

eq+ · · · . (11.5)

108

Vibration-rotation spectroscopy

µnα(Q) = (µn

α)eq+∑k

Qk

(

∂µnα

∂Qk

)

eq+ · · · . (11.5)

Now the terms in this equation must all transform in the same way. µnα is a vector

component and transforms as such; the normal mode Qk transforms according tosome irreducible representation Γk; while the coefficients (µα)eq and (∂µα/∂Qk)eq arejust numbers and so are totally symmetric.

Thus (µnα)eq must be zero unless µα is symmetric, and similarly (∂µn

α/∂Qk)eq must bezero unless Qk has the same symmetry as µα.

The first factor of (11.1) gives the rotational selection rules in the same way as before,but we may now have parallel vibrations in which µz varies, and perpendicularvibrations in which µz or µy vary. For the former, the rotational selection rules are asfor pure rotational spectroscopy, so that the spectrum shows P and R branches but noQ; for the latter, it turns out that the Q branch is allowed as well. For details, seeHollas ‘High Resolution Spectroscopy ’ or Bernath ‘Spectra of Atoms and Molecules’.

The second term in (11.5) is responsible for vibrational transitions. . .

109

Vibrational selection rule

The vibrational integral is now

∫

ψ′vib∗µn

α(Q)ψ′′vib dQ =

∫

ψ′vib∗(

(µnα)eq+∑

k

Qk

(

∂µnα

∂Qk

)

eq+ · · ·

)

ψ′′vib dQ

= (µnα)eq

∫

ψ′vib∗ψ′′

vib dQ

+∑k

(

∂µnα

∂Qk

)

eq

∫

ψ′vib∗Qkψ′′

vib dQ+ · · · (11.6)

The first term is an overlap integral between vibrational eigenfunctions, and is zerounless ψ′

vib = ψ′′vib, in which case we’re back to pure rotational spectroscopy.

In the second term, we can write Qk in terms of the ladder operators:

Qk =√

12(Qk + Q†

k). (11.7)

Operating on ψ′′vib this increases or reduces the vibrational quantum number in mode

k by 1 (for harmonic oscillator functions), so we obtain the vibrational selection rule∆vk = ±1.

110

Electronic spectroscopyHere ψ′

elec 6= ψ′′elec, but we can still follow the same procedure to get an expansion of

the transition dipole between the electronic states (m and n, say)

µmnα (Q) =

∫

ψmelec(q,Q)∗µα(q,Q)ψn

elec(q,Q)dq

= (µmnα )eq+∑

k

Qk

(

∂µmnα

∂Qk

)

eq+ · · · . (11.8)

If (µmnα )eq 6= 0, the transition is allowed; if (µmn

α )eq = 0, so that the transition iselectronically forbidden, the second term may still be non-zero (so that the transitiondipole becomes non-zero when the molecule is distorted) and we get a simultaneousvibrational–electronic transition. Such transitions are substantially weaker than normalallowed transitions.

111

Vibrationally allowed electronic transitions

In octahedral transition-metal complexes, thelow-lying excited states arise from d–d transitionsbetween the t2g and eg levels. Being g↔ gtransitions they are forbidden by the Laporte(parity) selection rule.

Consider however a transition in a d1 complexfrom the 2T2g electronic ground state in its ground

vibrational state to the 2Eg electronic excited statewith (say) a T1u vibration excited. The upper statehas symmetry Eg⊗T1u = T1u⊕T2u.

The dipole operator transforms according to T1u, so transitions areallowed to excited states that transform like some component ofT2g⊗T1u = A2u⊕Eu⊕T1u⊕T2u. Consequently transitions to both theT1u and T2u components of the excited state are allowed.

T2g

Eg

T1u,T2u

v = 0

v = 1

112

Part II Chemistry 2009–10 Course B7: Symmetry and ... · • P. R. Bunker and P. Jensen,...

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