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Analysis of Molecular Vibrations

Marc Alomar

April 25, 2013

Abstract

The goal of this exercise is to show how finite group theory can simplify the

analysis of molecular vibrations. This problem was first considered by E. Wigner1,

and most of this analysis is based on his work. In particular, I analyze the vi-

brations of a triangular molecule, first with the classical Lagrangian formalism,

and then with the formalism of group theory. Both approaches lead to the same

results with regard to the normal modes.

1 Lagrangian Formalism

In this section we will study the planar vibrations of a triangular molecule. We candescribe this system by a set of three masses m, sitting on the vertices of an equilateraltriangle and connected by springs of constant k. Following a classical dynamics ap-proach, we will use the Lagrangian formalism to find the frequencies and normal modes

of this system.

2

13

y

x

1

1

The equilibrium positions in terms of the center of mass reference frame are8>:

r1 =1

2ex 1

2p3

ey

r2 =1p3

ey

r3 = 12 ex 12p3 ey1Nachr. Ges. Wiss. Gttingen 133 (1930). An English translation can be found in the book

Applied Group Theory, by Arthur P. Cracknell.

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It is convenient to describe the departure from equilibrium in terms of the displace-ment vector ~ri = (xi, yi). These coordinates set the position of the masses withrespect to its equilibrium position, as described in the figure. Therefore, we can write

the position vector of each mass as~r

= ri +~r

i.

A general expression for the potential energy is

V (xi, yi) =1

2k

(|~r1 ~r2| 1)2 + (|~r2 ~r3| 1)2 + (|~r1 ~r3| 1)2

If we limit our analysis to small amplitude vibrations, we can write V in terms ofa Taylor expansion about the equilibrium point. The zero and first order will be zero,and we are left with the second order term,

V =1

2

~r TH~r

where ~r T = (x1 y1 x2 y2 x3 y3) is the displacement vector and H is the Hessian matrixevaluated at the equilibrium point ~r = 0. The Hessian matrix is given by

H =1

4

0BBBBBBB@

5 p

3 1p

3 1 0p3 3

p3 3 0 0

1p

3 2 0 1 p

3p3 3 0 3 p3 3

1 0 1 p

3 3p

3

0 0 p

3 3p

3 3

1CCCCCCCA

We can find the normal modes and frequencies if we solve the eigenvalue problemH ~u = !2 ~u. The solutions are8>>>>>>>>>>>>>>>:

!1 = 3 ~u1 =p3, 1, 0, 2, p3, 1

!2 = 1.5 ~u2 = 0.5 (p

3, 1, p

3, 1, 0, 2)!3 = 1.5 ~u3 = 0.5 (1,

p3, 1,

p3, 2, 0)

!4 = 0 ~u4 = (0, 1,p

3, 0, 0, 1)

!5 = 0 ~u5 = (1, 0, 1, 0, 1, 0)

!6 = 0 ~u6 = (0, 2, p

3, 1, 0, 0)

The zero frequency modes represent the translational and rotational modes. Thus,we have found that this system has one non-degenerate and two degenerate normalmodes.

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2 Group theory analysis

We now proceed to the analysis of the same problem using the tools of group the-ory. For a given molecule, we first identify its symmetry group G, i.e. the group oftransformations that leave the equilibrium positions invariant. Then, we study howthese transformations modify the displacement vector ~r. It turns out that when wetransform a normal mode under G, its frequency doesnt change, because the relativepositions of the molecule are the same. In fact, the normal modes form an irreduciblerepresentation of the group. We only have to identify the completely reducible represen-tation D that transforms the displacement vectors, and decompose it. Each irreduciblerepresentation corresponds to a normal mode with a characteristic frequency. For eachnormal mode, the dimension of the irreducible representation is equal to the number oflinearly independent normal modes.

Lets return to our example. The first step is to identify the point group thatdescribes the symmetry of the triangle. There are six symmetry operations that leavethe equilibrium positions invariant: rotations through 120 and 240 ( C3 and C

03),

three reflections i in a plane that contains the atom i and bisects the opposite edge,and the identity operation E. In fact, each of these three different kinds of operationsform a conjugacy class. An important result of group theory states that the numberof conjugacy classes is equal to the number of irreducible representations. This grouphas three irreducible representations: two one-dimensional, A1 and A2, and one two-dimensional, E1. The character table is shown below.

E 2 C3 3

A1 1 1 1 z x2 + y2, z2A2 1 1 -1 RzE1 2 -1 0 x, y x

2 y2, xyRx, Ry xz, yz

D 6 0 0

Table 1: Character table.

The normal modes transform according to the irreducible representations of thegroup. Non-degenerate modes correspond to one-dimensional representations, whereasdegenerate modes correspond to representations of dimension greater than 1. Our job isto find a representation that describes the transformations of the displacement vectors.This representation will be completely reducible, and our task is to decompose it. Oncewe know the number of times that each irreducible representation appears, we will knowthe number of normal modes and its degeneracy.

It turns out that its not necessary to find the exact expression of the six 6x6matrices; we only need their characters to decompose D in terms of the irreducible

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terms. It is also well known that symmetry elements that belong to the same class havethe same character, and then our problem is reduced to obtain the character of threeoperations: rotations, reflections, and the identity. The character of the identity matrix

is 6, and the rotation matrices have character 0, because none of the atoms remains atthe same position (i.e., the diagonal only contains zeros). The reflection matrices alsohave character 0. We can verify this result with the matrix 2,

2 =

0BBBBBB@

0 0 0 0 1 00 0 0 0 0 10 0 1 0 0 00 0 0 1 0 01 0 0 0 0 00 1 0 0 0 0

1CCCCCCA

The second atom remains at the same position, but the reflection along the plane

yz changes x0 ! x and y0 ! y.

The number of times mDa that each irreducible representation Da appears in D isobtained using the orthogonality relation,

mDa =1

N

Xg2G

Da(g)D(g)

where N is the number of elements of the group and D(g) is the character of the elementg in the representation D. In this case we find mA1 = 1, mA2 = 1 and mE1 = 2.Thus,the decomposition of D is

D = A1 + A2 + 2 E1

However, if we only want the vibrational modes we have to substract the transla-tional and rotational modes. Their irreducible representation can be identified usingtable 1. Translations along the x and y axis are described by E1. This representationhas dimension 2, so we only have to substract one E1. Rotations correspond to A2, andwe are left with

Dvib = A1 + E1

We conclude that the system has three normal vibrational modes: one non-degenerate,u1, and two degenerate, u2 and u3. However, we also know their symmetry properties.This information will allow us to obtain the eigenvectors.

The eigenvector of the non-degenerate mode is very easy to obtain. In one-dimensionalrepresentations, the characters are equal to the representations. In this case, A1 is thetrivial representation (D(g) = 1, 8 g). Thus, this normal mode must be symmetric

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under any symmetry transformation, i.e. DA1(g) = , 8 g. The only vector thatsatisfies this relation is

~u1 =

1

2

1, p3, 0, 2, 1, p3This mode is usually known as the breathing mode.The degenerate eigenvectors are more difficult to obtain. However, the matrix rep-

resentation of E1 tells us how these modes transform. Most of the matrices are non-diagonal, but the matrix for 2 is particularly easy. We then know that, under thereflection 2 , the normal modes u2 and u3 transform as

DE1 (2)

u2u3

=

u2u3

The mode u2 is antisymmetric under 2, whereas u3 is symmetric. Therefore, the

motion of particle 2 must be along the x axis in u2 , and along the y axis in u3. Themotion of particles 1 and 3 must follow precise symmetry relations: (x3, y3) = (x1, y1)in u2, and (x3, y3) = (x1, y1) in u3. We can make a reasonable guess using theserelations and the fact that both vectors are linearly independent,(

~u2 =1

2

1, p3, 2, 0, 1, p3~u3 =

1

2

p3, 1, 0, 2, p3, 1It turns out that these vectors also transform correctly under rotations, and we

conclude that they are the normal modes of E1.

u uu1 2 3

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