Group theory, representations and their applications in solid state

TMS 2018

1

Outline of the course

1. Introduction: Symmetries, degeneracies and representations.

2. Irreducible representations as building blocks. Application to molecular vibrations.

3. Operations with representations: Physical properties and spectra.

4. Spin and double valued representations. Splitting of atomic orbitals in crystals.

5. Representation theory and electronic bands.

2

Reducible and irreducible representations

V (E) =

0

@1 0 00 1 00 0 1

1

A

V (C+3 ) =

0

B@� 1

2 �p32 0p

32 � 1

2 00 0 1

1

CA

V (C�3 ) =

0

B@� 1

2

p32 0

�p32 � 1

2 00 0 1

1

CA

V (�d1) =

0

@1 0 00 �1 00 0 1

1

A

V (�d2) =

0

B@� 1

2 �p32 0

�p32

12 0

0 0 1

1

CA

V (�d3) =

0

B@� 1

2

p32 0p

32

12 0

0 0 1

1

CA

V (x, y, z) = A1(z) + E(x, y)

13

In group theory language, we say that ~a belongs to thevector representation V and write

V (ax

, ay

, az

)

Invariants can be constructed by taking inner products of objectsthat belong to the same representation, so that their variationscancel out.

System invariant under the crystal point group C3v.The hamiltonian H(~a) depends on one vector variable.

~a = (ax

, ay

, az

)

vskip1cm The pyramid is invariant under 2⇡3 rotations

and 3 vertical reflection planes.The hamiltonian H(~a) depends on one vector variable.

The pyramid is invariant under 2⇡3 rotations

and 3 vertical reflection planes.

V (C+3 ) =

0

@� 1

2 �p32 0p

32 � 1

2 00 0 1

1

A , V (�a

) =

0

@�1 0 00 1 00 0 1

1

A

The group C3v is generated by a 2⇡3 rotation around

the OZ axis and a vertical plane:

The vector representation V decomposes into the twoirreducible representations (IRs) A1 and E

V (ax

, ay

.az

) = A1(az) + E(ax

, ay

)

3

Reducible and irreducible representations

W (E) =

0

@1 0 00 1 00 0 1

1

A

W (C+3 ) =

0

B@� 1

2 �p34 �

p34p

32

14 � 3

4p32 � 3

414

1

CA

W (C�3 ) =

0

B@� 1

2

p34

p34

�p32

14 � 3

4

�p32 � 3

414

1

CA

W (�d1) =

0

@1 0 00 0 �10 �1 0

1

A

W (�d2) =

0

B@� 1

2 �p34 �

p34

�p32

34 � 1

4

�p32 � 1

434

1

CA

The representation W decomposes into the two irreducible

representations A1 and E

W (x, y, z) = A1(y � z) + E(2x, y + z)

W (�d3) =

0

B@� 1

2

p34

p34p

32

34 � 1

4p32 � 1

434

1

CA

4

Equivalent representations

T2(g) = AT2(g)A�1 8g 2 G

In our case, with

A =

0

@1 0 00 1 10 1 �1

1

A

W (g) = AV (g)A�1 8g 2 C3v

Thus W(g) is just V(g) in the coordinate system (x0, y

0, z

0)

Two representations and of a group G are said to be equivalent if there is a non-singular matrix A such that

T1 T2

This can be interpreted as a change of coordinates, with , i.e. u0 = Au x

0 = x, y

0 = y + z, z

0 = y � z

5

CharactersTraces are invariant under changes of basis

A representation is thus fully characterized by the set of traces of the elements of the group G.

tr(AT (g)A�1) = tr(T (g))

As a consequence, equivalent representations have identical traces. The converse is also true

T1 ⌘ T2 , tr(T1(g)) = tr(T2(g)) 8g 2 G

The set of traces of T(g) for all the elements of the group G is kown as the character of the representation T.

6

Character tablesC3v E C+

3 C�3 �d1 �d2 �d3

A1 1 1 1 1 1 1A2 1 1 1 -1 -1 -1E 2 -1 -1 0 0 0

V 3 0 0 1 1 1

7

Reducible and irreducible representations

V (E) =

0

@1 0 00 1 00 0 1

1

A

V (C+3 ) =

0

B@� 1

2 �p32 0p

32 � 1

2 00 0 1

1

CA

V (C�3 ) =

0

B@� 1

2

p32 0

�p32 � 1

2 00 0 1

1

CA

V (�d1) =

0

@1 0 00 �1 00 0 1

1

A

V (�d2) =

0

B@� 1

2 �p32 0

�p32

12 0

0 0 1

1

CA

V (�d3) =

0

B@� 1

2

p32 0p

32

12 0

0 0 1

1

CA

8

Character tablesC3v E C+

3 C�3 �d1 �d2 �d3

A1 1 1 1 1 1 1A2 1 1 1 -1 -1 -1E 2 -1 -1 0 0 0

V 3 0 0 1 1 1

There are three classes in C3v

{E}, {C+3 , C�

3 }, {�d1,�d2,�d3}

V = A1 + E , �V = �A1 + �E

Two elements and of G belong to the same class if there is an element h in G such that g2 = h � g1 � h�1

g1 g2

9

Character tablesC3v E C±

3 �di

A1 1 1 1A2 1 1 -1E 2 -1 0

V 3 0 1

V = A1 + E , �V = �A1 + �E

There are three classes in C3v

{E}, {C+3 , C�

3 }, {�d1,�d2,�d3}

Two elements and of G belong to the same class if there is an element h in G such that g2 = h � g1 � h�1

g1 g2

10

How many non-equivalent IRs?

THEOREM:The number of non-equivalent irreducible representations for a finite group G is equal to the number of classes in G.

THEOREM:The sum of the squares of the dimensions of the non-equivalent irreducible representations of a finite group G is equal to the order of the group (i.e., the number of elements of G).

d21 + d22 + . . .+ d2C = N

COROLLARY: The IRs of abelian groups are always one-dimensional.

11

Decomposing a reducible representation

(�1,�2) =1

N

X

g2G

�⇤1(g)�2(g)

THEOREM: If and are irreducible representations of a group G, then

⌧i ⌧j

(�i,�j) = �ij

DEFINITION: Given two representations and of a group G, we define the following scalar product:

T1 T2

12

Character tablesC3v E C+

3 C�3 �d1 �d2 �d3

A1 1 1 1 1 1 1A2 1 1 1 -1 -1 -1E 2 -1 -1 0 0 0

V 3 0 0 1 1 1

13

Decomposing a reducible representationTo find the decomposition of a reducible representation

T = m1⌧1 +m2⌧2 + . . .mc⌧cjust take the scalar product of

�T = m1�1 +m2�2 + . . .mc�c

This is the famous “magic formula” that gives the multiplicities of the different IRs

mj =1

N

X

g2G

�⇤(g)�j(g)

with to get (�T ,�j) =X

i

mi(�i,�j) = mj�j

14

Decomposing the vector representationC3v E C±

3 �di

A1 1 1 1A2 1 1 -1E 2 -1 0

V 3 0 1

mA1 =1

6(3⇥ 1 + 0⇥ 1⇥ 2 + 1⇥ 1⇥ 3) = 1

mA2 =1

6(3⇥ 1 + 0⇥ 1⇥ 2 + 1⇥ (�1)⇥ 3) = 0

mE =1

6(3⇥ 2 + 0⇥ (�1)⇥ 2 + 1⇥ 0⇥ 3) = 1

V = A1 + E

mj =1

N

X

g2G

�⇤(g)�j(g)

15

Computing characters

If the matrices of the representation are known, we just take the traces.

But, in practice, we don’t need all the actual matrices in order to compute the character of a representation. Instead, we use the following tricks:

a) Characters are class functions. Therefore, choose the “easiest” element from each class.

b) Traces are independent of the coordinates. Therefore, choose the most convenient coordinates for each element.

16

V (E) =

0

@1 0 00 1 00 0 1

1

A

V (C+3 ) =

0

B@� 1

2 �p32 0p

32 � 1

2 00 0 1

1

CA

V (C�3 ) =

0

B@� 1

2

p32 0

�p32 � 1

2 00 0 1

1

CA

V (�d1) =

0

@1 0 00 �1 00 0 1

1

A

V (�d2) =

0

B@� 1

2 �p32 0

�p32

12 0

0 0 1

1

CA

V (�d3) =

0

B@� 1

2

p32 0p

32

12 0

0 0 1

1

CA

Computing the character of V

17

Computing the character of VGeneral formulas Besides the identity, a point group may contain only four types of elements

RotationsReflection planesInversionRoto-reflections

I�Cn

Sn

By choosing the appropriate coordinates, one easily finds�V (Cn) = 1 + 2 cos ✓n

�V (Sn) = �1 + 2 cos ✓n

�V (�) = 1

�V (I) = �3

18

C3v E C±3 �di

A1 1 1 1A2 1 1 -1E 2 -1 0

V 3 0 1

Computing the character of V

�V (Cn) = 1 + 2 cos ✓n

�V (Sn) = �1 + 2 cos ✓n

�V (�) = 1

�V (I) = �3

19

Computing the character of RGeneral formulas

The rotational representation tells how pseudovectors (a.k.a. axial vectors) transform.

Axial vectors transform like vectors under rotations, but with a relative minus sign under “improper”operations.

Thus the formulas for the vector representation imply

�R(Cn) = 1 + 2 cos ✓n

�R(Sn) = 1� 2 cos ✓n

�R(�) = �1

�R(I) = 3

20

Computing the character of M

The mechanical representation tells how general mechanical deformations of a molecule transform.The matrices for the mechanical represention have a block structure where the blocks are just the matrices of the vector representation.

General formula

21

�d1

�d2

�d3

C

Computing the character of M

M(C+3 ) =

0

BBBB@

0 0 V (C+3 ) 0 0

V (C+3 ) 0 0 0 0

0 V (C+3 ) 0 0 0

0 0 V (C+3 ) 0

0 0 0 0 V (C+3 )

1

CCCCA

u =

0

BBBB@

~u1

~u2

~u3

~u4

~uC

1

CCCCA

ug = M(g)u

H2

H3

H1D4

22

�d1

�d2

�d3

C

Computing the character of M

u =

0

BBBB@

~u1

~u2

~u3

~u4

~uC

1

CCCCA

ug = M(g)u

H2

H3

H1D4

M(�d1) =

0

BBBB@

V (�d1) 0 0 0 00 0 V (�d1) 0 00 V (�d1) 0 0 00 0 0 V (�d1) 00 0 0 0 V (�d1)

1

CCCCA

23

Computing the character of MGeneral formula The mechanical representation tells how general mechanical deformations of a molecule transform.The matrices for the mechanical represention have a block structure where the blocks are just the matrices of the vector representation.

The blocks are on the diagonal when the corresponding atoms are invariant under the symmetry operation. Thus

�M (g) = ninv(g)�V (g)

where is the number of atoms invariant under g. ninv(g)

24

C3v E C±3 �di

ninv 5 2 3

Computing the character of M

�d1

�d2

�d3

C

H2

H3

H1D4

25

Computing the character of MC3v E C±

3 �di

A1 1 1 1A2 1 1 -1E 2 -1 0

V 3 0 1ninv 5 2 3

M 15 0 3

mj =1

N

X

g2G

�⇤(g)�j(g)

mA1 =1

6(15⇥ 1 + 0⇥ 1⇥ 2 + 3⇥ 1⇥ 3) = 4

mE =1

6(15⇥ 2 + 0⇥ (�1)⇥ 2 + 3⇥ 0⇥ 3) = 5

mA2 =1

6(15⇥ 1 + 0⇥ 1⇥ 2 + 3⇥ (�1)⇥ 3) = 1

M = 4A1 +A2 + 5E

26

Vibrational spectrum of CH3DGROUP THEORY COMPUTATIONS:

M = 4A1 +A2 + 5E

V = A1 + E

R = A2 + E

V ib = M � V �R = 3A1 + 3E

IR A1 A2 Edim 1 1 2

are irreducible representations of the point group C3v(3m){A1, A2, E}

V ib = 3A1 + 3E

9 = 3⇥ 1+ 3⇥ 2

27

SummaryCOMPUTING CHARACTERS

�V (Cn) = 1 + 2 cos ✓n

�V (Sn) = �1 + 2 cos ✓n

�V (�) = 1

�V (I) = �3

�R(Cn) = 1 + 2 cos ✓n

�R(Sn) = 1� 2 cos ✓n

�R(�) = �1

�R(I) = 3

�M (g) = ninv(g)�V (g)

DECOMPOSING REPRESENTATIONS

mj =1

N

X

g2G

�⇤(g)�j(g) V ib = M � V �R

28