Stone M., Goldbart P. Mathematics for Physics: A Guided Tour for Graduate Students

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14.2 Representations 515

From g

αβ

= e

γβ

γα

) and the matrix-element orthogonality, it follows that

αβ

γδ

dim J

|G|



g∈G

αβ

(g)

∗

γδ

dim J

|G|



g∈G

αβ

(g)

∗

γ

(g)e

δ

= δ

α

βγ

δ

= δ

βγ

αδ

. (14.34)

For each J , this multiplication rule of the e

αβ

is identical to that of matrices having zero

entries everywhere except for the (α, β)-th, which is a “1”. There are (dim J )

of these

αβ

for each n-dimensional representation J , and they are linearly independent. Because

(dim J )

=|G|, they form a basis for the algebra. In particular every element of G

can be reconstructed as

g =



(g)e

. (14.35)

We can also define the useful objects



dim J

|G|



g∈G

(g)

∗

g. (14.36)

They have the property

= δ



= I, (14.37)

where I is the identity element of C(G). The P

are therefore projection operators

composing a resolution of the identity. Their utility resides in the fact that when D(g) is

a reducible representation acting on a linear space

V =

, (14.38)

then setting g → D(g) in the formula for P

results in a projection matrix from V onto

the irreducible component V

. To see how this comes about, let v ∈ V and, for any fixed

p, set

= e

v, (14.39)

where e

v should be understood as shorthand for D(e

)v. Then

D(g)v

= ge

v = e

(g) = v

(g). (14.40)

516 14 Groups and group representations

We see that the v

, if not all zero, are basis vectors for V

. Since P

is a sum of the

, the vector P

v is a sum of such vectors, and therefore lies in V

. The advantage of

using P

over any individual e

is that P

can be computed from character tables, i.e.

its construction does not require knowledge of the irreducible representation matrices.

The algebra of classes

If a conjugacy class C

consists of the elements {g

, g

, ...g

}, we can define C

to be

the corresponding element of the group algebra:

+ g

+···g

). (14.41)

(The factor of 1/d

is a conventional normalization.) Because conjugation merely per-

mutes the elements of a conjugacy class, we have g

−1

g = C

for all g ∈ C(G).

The C

therefore commute with every element of C(G ). Conversely any element

of C(G) that commutes with every element in C(G) must be a linear combination:

C = c

+ c

+ ... The subspace of C(G) consisting of sums of the classes is

therefore the centre Z[C(G)] of the group algebra. Because the product C

commutes

with every element, it lies in Z[C(G)], and so there are constants c

such that



. (14.42)

We can regard the C

as being linear maps from Z[C(G)] to itself, whose associated

matrices have entries (C

)

= c

. These matrices commute, and can be simultaneously

diagonalized. We will leave it as an exercise for the reader to demonstrate that





. (14.43)

Here χ

≡ χ

{e}

= dim J . The common eigenvectors of the C

are therefore the pro-

jection operators P

, and the eigenvalues λ

= χ

/χ

are, up to normalization, the

characters. Equation (14.43) provides a convenient method for computing the characters

from knowledge only of the coefficients c

appearing in the class multiplication table.

Once we have found the eigenvalues λ

, we recover the χ

by noting that χ

is real and

positive, and that

|χ

=|G|.

Exercise 14.17: Use Schur’s lemma to show that for an irrep D

(g) we have



g∈C

(g) =

dim J

and hence establish (14.43).

14.3 Physics applications 517

14.3 Physics applications

14.3.1 Quantum mechanics

When a group G ={g

} acts on a mechanical system, then G will act as a set of linear

operators D(g) on the Hilbert space H of the corresponding quantum system. Thus H

will be a representation

space for G. If the group is a symmetry of the system then the

D(g) will commute with the Hamiltonian

H . If this is so, and if we can decompose

H =

irreps J

(14.44)

into

H -invariant irreps of G then Schur’s lemma tells us that in each H

the Hamiltonian

H will act as a multiple of the identity operator. In other words every state in H

will

be an eigenstate of

H with a common energy E

This fact can greatly simplify the task of finding the energy levels. If an irrep J

occurs only once in the decomposition of H then we can find the eigenstates directly by

applying the projection operator P

to vectors in H. If the irrep occurs n

times in the

decomposition, then P

will project to the reducible subspace

⊕ H

⊕···H

A BC D

copies

= M ⊗ H

Here M is an n

-dimensional multiplicity space. The Hamiltonian

H will act in M as

an n

-by-n

matrix. In other words, if the vectors

|n, i≡|n⊗|i∈M ⊗ H

(14.45)

form a basis for M ⊗H

, with n labelling which copy of H

the vector |n, ilies in, then

H |n, i=|m, iH

D(g)|n, i=|n, jD

(g). (14.46)

Diagonalizing H

provides us with n

H -invariant copies of H

and gives us the energy

eigenstates.

Consider, for example, the molecule C

(buckminsterfullerene) consisting of 60

carbon atoms in the form of a soccer ball. The chemically active electrons can be treated

in a tight-binding approximation in which the Hilbert space has dimension 60 – one π -

orbital basis state for each carbon atom. The geometric symmetry group of the molecule

The rules of quantum mechanics only require that D(g

)D(g

) = e

iφ(g

)

D(g

). A set of matrices that

obeys the group multiplication rule “up to a phase” is called a projective (or ray) representation. In many

cases, however, we can choose the D(g) so that φ is not needed. This is the case in all the examples we

discuss.

518 14 Groups and group representations

Table 14.3 Character table for the group Y .

Typical element and class size

YeC

Irrep 1 12 12 15 20

A 11111

3 τ

−1

−τ −10

3 −ττ

−1

−10

G 4 −1 −101

H 5001−1

L=4

L=3

L=2

L=1

L=0

–3

–2

–1

Figure 14.3 A sketch of the tight-binding electronic energy levels of C

is Y

= Y ×Z

, where Y is the rotational symmetry group of the icosahedron (a subgroup

of SO(3)) and Z

is the parity inversion σ : r (→−r. The characters of Y are displayed

in Table 14.3. In this table τ =

(

√

5 − 1) denotes the golden mean. The class C

is the set of 2π/5 rotations about an axis through the centres of a pair of antipodal

pentagonal faces, the class C

is the set of of 2π/3 rotations about an axis through the

centres of a pair of antipodal hexagonal faces and C

is the set of π rotations through the

midpoints of a pair of antipodal edges, each lying between two adjacent hexagonal faces.

The geometric symmetry group acts on the 60-dimensional Hilbert space by permuting

the basis states concurrently with their associated atoms. Figure 14.3 shows how the

60 states are disposed into energy levels.

Each level is labelled by a lower-case letter

specifying the irrep of Y , and by a subscript g or u standing for gerade (German for

After R. C. Haddon, L. E. Brus, K. Raghavachari, Chem. Phys. Lett., 125 (1986) 459.

14.3 Physics applications 519

even)orungerade (German for odd) that indicates whether the wavefunction is even or

odd under the inversion σ : r (→−r.

The buckyball is roughly spherical, and the lowest 25 states can be thought of as being

derived from the L = 0, 1, 2, 3, 4 eigenstates, where L is the angular momentum quantum

number that classifies the energy levels for an electron moving on a perfect sphere. In

the many-electron ground-state, the 30 single-particle states with energy below E < 0

are each occupied by pairs of spin up/down electrons. The 30 states with E > 0 are

empty.

To explain, for example, why three copies of T

appear, and why two of these are T

and one T

, we must investigate the manner in which the 60-dimensional Hilbert space

decomposes into irreducible representations of the 120-element group Y

. Problem 14.23

leads us through this computation, and shows that no irrep of Y

occurs more than three

times. In finding the energy levels, we therefore never have to diagonalize a matrix

bigger than 3-by-3.

The equality of the energies of the h

and g

levels at E =−1isanaccidental

degeneracy . It is not required by the symmetry, and will presumably disappear in a

more sophisticated calculation. The appearance of many “accidental” degeneracies in an

energy spectrum hints that there may be a hidden symmetry that arises from something

beyond geometry. For example, in the Schrödinger spectrum of the hydrogen atom all

states with the same principal quantum number n have the same energy although they

correspond to different irreps L = 1, ..., n −1ofO(3). This degeneracy occurs because

the classical Kepler-orbit problem has symmetry group O(4), rather than the naïvely

expected O(3) rotational symmetry.

14.3.2 Vibrational spectrum of H

The small vibrations of a mechanical system with n degrees of freedom are governed by

a Lagrangian of the form

L =

˙x

M ˙x −

V x (14.47)

where M and V are symmetric n-by-n matrices, and with M being positive definite. This

Lagrangian leads to the equations of motion

M ¨x = V x. (14.48)

We look for normal mode solutions x(t) ∝ e

iω

, where the vectors x

obey

−ω

M x

= V x

. (14.49)

The normal-mode frequencies are solutions of the secular equation

det (V − ω

M ) = 0, (14.50)

520 14 Groups and group representations

and modes with distinct frequencies are orthogonal with respect to the inner product

defined by M,

x, y=x

M y. (14.51)

We are interested in solving this problem for vibrations about the equilibrium config-

uration of a molecule. Suppose this equilibrium configuration has a symmetry group G.

This gives rise to an n-dimensional representation on the space of x’s in which

g : x (→ D(g)x (14.52)

leaves both the intertia matrix M and the potential matrix V unchanged:

[D(g)]

MD(g) = M , [D(g)]

VD(g) = V . (14.53)

Consequently, if we have an eigenvector x

with frequency ω

−ω

M x

= V x

(14.54)

we see that D(g)x

also satisfies this equation. The frequency eigenspaces are therefore

left invariant by the action of D(g) and, barring accidental degeneracy, there will be

a one-to-one correspondence between the frequency eigenspaces and the irreducible

representations occurring in D(g).

Consider, for example, the vibrational modes of the water molecule H

O (Figure 14.4).

This familiar molecule has symmetry group C

which is generated by two elements: a

rotation a through π about an axis through the oxygen atom, and a reflection b in the

plane through the oxygen atom and bisecting the angle between the two hydrogens. The

product ab is a reflection in the plane defined by the equilibrium position of the three

atoms. The relations are a

= b

= (ab)

= e, and the characters are displayed in

Table 14.4.

The group C

is abelian, so all the representations are one dimensional.

Figure 14.4 Water molecule.

14.3 Physics applications 521

Table 14.4 Character table of C

Class and size

eabab

Irrep1111

1111

11−1 −1

1 −11−1

1 −1 −11

To find out what representations occur when C

acts, we need to find the character

of its action D(g) on the nine-dimensional vector

x = (x

, y

, z

, x

, y

, z

, x

, y

, z

). (14.55)

Here the coordinates x

, y

, z

etc. denote the displacements of the labelled atom

from its equilibrium position.

We take the molecule as lying in the xy-plane, with the z pointing towards us. The

effect of the symmetry operations on the atomic displacements is

D(a)x = (−x

, +y

, −z

, −x

, +y

, −z

, −x

, +y

, −z

D(b)x = (−x

, +y

, +z

, −x

, +y

, +z

, −x

, +y

, +z

D(ab)x = (+x

, +y

, −z

, +x

, +y

, −z

, +x

, +y

, −z

Notice how the transformations D(a), D(b) have interchanged the displacement coor-

dinates of the two hydrogen atoms. In calculating the character of a transformation we

need look only at the effect on atoms that are left fixed – those that are moved have

matrix elements only in non-diagonal positions. Thus, when computing the compound

characters for ab, we can focus on the oxygen atom. For ab we need to look at all three

atoms. We find

(e) = 9,

(a) =−1 + 1 − 1 =−1,

(b) =−1 + 1 + 1 = 1,

(ab) = 1 + 1 − 1 + 1 + 1 − 1 + 1 + 1 − 1 = 3.

By using the orthogonality relations, we find the decomposition

⎛

⎜

⎝

−1

⎞

⎟

⎠

= 3

⎛

⎜

⎝

⎞

⎟

⎠

⎛

⎜

⎝

−1

⎞

⎟

⎠

+ 2

⎛

⎜

⎝

−1

⎞

⎟

⎠

+ 3

⎛

⎜

⎝

−1

⎞

⎟

⎠

(14.56)

522 14 Groups and group representations

= 3χ

+ χ

+ 2χ

+ 3χ

. (14.57)

Thus, the nine-dimensional representation decomposes as

D = 3A

⊕ A

⊕ 2B

⊕ 3B

. (14.58)

How do we exploit this? First we cut out the junk. Out of the nine modes, six cor-

respond to easily identified zero-frequency motions – three of translation and three

rotations. A translation in the x-direction would have x

= x

= ξ , all other

entries being zero. This displacement vector changes sign under both a and b, but is

left fixed by ab. This behaviour is characteristic of the representation B

. Similarly

we can identify A

as translation in y, and B

as translation in z. A rotation about the

y -axis makes z

=−z

= φ. This is left fixed by a, but changes sign under b

and ab, so the y rotation mode is A

. Similarly, rotations about the x- and z-axes cor-

respond to B

and B

, respectively. All that is left for genuine vibrational modes is

⊕ B

We now apply the projection operator

[(χ

(e))

∗

D(e) + (χ

(a))

∗

D(b) + (χ

(b))

∗

D(b) + (χ

(ab))

∗

D(ab)]

(14.59)

to v

, a small displacement of H

in the x-direction. We find

− v

+ v

)

− v

). (14.60)

This mode is an eigenvector for the vibration problem.

If we apply P

to v

and v

O,y

we find

+ v

O,y

= v

O,y

, (14.61)

but we are not quite done. These modes are contaminated by the y translation direction

zero mode, which is also in an A

representation. After we make our modes orthogonal

to this, there is only one left, and this has y

= y

=−y

/(2m

) = a

, all other

components vanishing.

14.3 Physics applications 523

We can similarly find vectors corresponding to B

+ v

)

− v

)

O,x

= v

O,x

and these need to be cleared of both translations in the x-direction and rotations about

the z-axis, both of which transform under B

. Again there is only one mode left and it is

=−y

= αx

= βx

= a

(14.62)

where α is chosen to ensure that there is no angular momentum about O, and β to

make the total x linear momentum vanish. We have therefore found three true vibration

eigenmodes, two transforming under A

and one under B

as advertised earlier. The

eigenfrequencies, of course, depend on the details of the spring constants, but now that

we have the eigenvectors we can just plug them in to find these.

14.3.3 Crystal field splittings

A quantum mechanical system has a symmetry G if the Hamiltonian

H obeys

−1

(g)

HD(g) =

H , (14.63)

for some group action D(g) : H → H on the Hilbert space. If follows that the

eigenspaces, H

, of states with a common eigenvalue, λ, are invariant subspaces for the

representation D(g).

We often need to understand how a degeneracy is lifted by perturbations that break

G down to a smaller subgroup H .Ann-dimensional irreducible representation of G

is automatically a representation of any subgroup of G, but in general it is no longer

irreducible. Thus the n-fold degenerate level is split into multiplets, one for each of

the irreducible representations of H contained in the original representation. The man-

ner in which an originally irreducible representation decomposes under restriction to a

subgroup is known as the branching rule for the representation.

Aphysically important case is given by the breaking of the full SO(3) rotation symme-

try of an isolated atomic Hamiltonian by a crystal field. Suppose the crystal has octohedral

symmetry. The characters of the octohedral group are displayed in Table 14.5.

The classes are labelled by the rotation angles, C

being a twofold rotation axis

(θ = π ), C

a threefold axis (θ = 2π/3), etc.

The character of the J = l representation of SO(3) is

(θ) =

sin(2l + 1)θ/2

sin θ/2

, (14.64)

524 14 Groups and group representations

Table 14.5 Character table of the octohedral group O.

Class (size)

OeC

(8) C

(3) C

(6) C

(6)

11111

11 1−1 −1

E 2 −1200

30−11−1

30−1 −11

Table 14.6 Characters evaluated on rotation classes.

Class (size)

leC

(8) C

(3) C

(6) C

(6)

011111

13 0 −1 −1 −1

25 −111−1

37 1 −1 −1 −1

490111

and the first few χ

’s evaluated on the rotation angles of the classes of O are dsiplayed

in Table 14.6.

The ninefold degenerate l = 4 multiplet therefore decomposes as

⎛

⎜

⎝

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

⎛

⎜

⎝

−1

⎞

⎟

⎠

⎛

⎜

⎝

−1

⎞

⎟

⎠

⎛

⎜

⎝

−1

⎞

⎟

⎠

, (14.65)

SO(3)

= χ

+ χ

. (14.66)

The octohedral crystal field splits the nine states into four multiplets with symmetries

, E, F

, F

and degeneracies 1, 2, 3 and 3, respectively.

We have considered only the simplest case here, ignoring the complications introduced

by reflection symmetries, and by two-valued spinor representations of the rotation group.