1 1

The Oh Group is the point Group of many interesting solids, including complexes

like CuSO4 · 5H2O and FeCl3 where a transition metal ion at the center of an Octahedron

A LCAO model of their properties is often called ligand field theory.

Ligand Group Orbitals

Il solfato di rame anidro (bianco) ridiventa pentaidrato (blu) aggiungendo acqua.

2 2

if A...F represent s orbitals they produce a representation G with following characters:

E

A

B

C

D

F

Exact or approximate symmetry

of many complexes

Binding occurs between central atom orbitals and ligand orbitals of like symmetry. For instance,

3 3

C4,C2 2 unmoved atoms: character=2

4 2 2 3 4 6

2 2 2

1

1

2

2

2 2 2 2 2

1

1

6 3 6 ' 8 6 3 6 8 48

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

2 0 2 0 1 2 0 2 0 1 ( ,2 )

2 0 2 0 1 2 0 2 0 1

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0 ( ,

h h d

g

u

g

u

g

u

g x y z

u

O E C C C C i S S g

A x y z

A

A

A

E x y z x y

E

T R R R

T x

2

2

, )

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0

g

u

y z

T xy xz yz

T

4 2 2 3 4 6

2 2 2

1

1

2

2

2 2 2 2 2

1

1

6 3 6 ' 8 6 3 6 8 48

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

2 0 2 0 1 2 0 2 0 1 ( ,2 )

2 0 2 0 1 2 0 2 0 1

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0 ( ,

h h d

g

u

g

u

g

u

g x y z

u

O E C C C C i S S g

A x y z

A

A

A

E x y z x y

E

T R R R

T x

2

2

, )

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0

6 2 2

g

u

y z

T xy xz yz

T

A F basis

G

4 4

n. unmoved atoms

4 2 2 3 4 6

2 2 2

1

1

2

2

2 2 2 2 2

1

1

6 3 6 ' 8 6 3 6 8 48

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

2 0 2 0 1 2 0 2 0 1 ( ,2 )

2 0 2 0 1 2 0 2 0 1

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0 ( ,

h h d

g

u

g

u

g

u

g x y z

u

O E C C C C i S S g

A x y z

A

A

A

E x y z x y

E

T R R R

T x

2

2

, )

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0

6 2 2 0

g

u

y z

T xy xz yz

T

A F basis

G

A

B E

D

F

C

S4 0 unmoved atoms

4 2 2 3 4 6

2 2 2

1

1

2

2

2 2 2 2 2

1

1

6 3 6 ' 8 6 3 6 8 48

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

2 0 2 0 1 2 0 2 0 1 ( ,2 )

2 0 2 0 1 2 0 2 0 1

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0 ( ,

h h d

g

u

g

u

g

u

g x y z

u

O E C C C C i S S g

A x y z

A

A

A

E x y z x y

E

T R R R

T x

2

2

, )

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0

6 2

g

u

y z

T xy xz yz

T

A F basis

G

5 5

n. unmoved atoms

C’2 0 unmoved atoms

4 2 2 3 4 6

2 2 2

1

1

2

2

2 2 2 2 2

1

1

6 3 6 ' 8 6 3 6 8 48

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

2 0 2 0 1 2 0 2 0 1 ( ,2 )

2 0 2 0 1 2 0 2 0 1

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0 ( ,

h h d

g

u

g

u

g

u

g x y z

u

O E C C C C i S S g

A x y z

A

A

A

E x y z x y

E

T R R R

T x

2

2

, )

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0

6 2 2 0 0

g

u

y z

T xy xz yz

T

A F basis

G

4 2 2 3 4 6

2 2 2

1

1

2

2

2 2 2 2 2

1

1

6 3 6 ' 8 6 3 6 8 48

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

2 0 2 0 1 2 0 2 0 1 ( ,2 )

2 0 2 0 1 2 0 2 0 1

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0 ( ,

h h d

g

u

g

u

g

u

g x y z

u

O E C C C C i S S g

A x y z

A

A

A

E x y z x y

E

T R R R

T x

2

2

, )

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0

6 2 0

g

u

y z

T xy xz yz

T

A F basis

G

6

n. unmoved atoms

4 2 2 3 4 6

2 2 2

1

1

2

2

2 2 2 2 2

1

1

6 3 6 ' 8 6 3 6 8 48

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

2 0 2 0 1 2 0 2 0 1 ( ,2 )

2 0 2 0 1 2 0 2 0 1

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0 ( ,

h h d

g

u

g

u

g

u

g x y z

u

O E C C C C i S S g

A x y z

A

A

A

E x y z x y

E

T R R R

T x

2

2

, )

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0

6 2 0 0

g

u

y z

T xy xz yz

T

A F basis

G

4 2 2 3 4 6

2 2 2

1

1

2

2

2 2 2 2 2

1

1

6 3 6 ' 8 6 3 6 8 48

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

2 0 2 0 1 2 0 2 0 1 ( ,2 )

2 0 2 0 1 2 0 2 0 1

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0 ( ,

h h d

g

u

g

u

g

u

g x y z

u

O E C C C C i S S g

A x y z

A

A

A

E x y z x y

E

T R R R

T x

2

2

, )

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0

6 2 0 0 0 0 0

g

u

y z

T xy xz yz

T

A F basis

G

C3,S6 0 unmoved atoms

7 7

n. unmoved atoms

h 4 unmoved atoms

4 2 2 3 4 6

2 2 2

1

1

2

2

2 2 2 2 2

1

1

6 3 6 ' 8 6 3 6 8 48

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

2 0 2 0 1 2 0 2 0 1 ( ,2 )

2 0 2 0 1 2 0 2 0 1

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0 ( ,

h h d

g

u

g

u

g

u

g x y z

u

O E C C C C i S S g

A x y z

A

A

A

E x y z x y

E

T R R R

T x

2

2

, )

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0

6 2 2 0 0 0 0 4 0

g

u

y z

T xy xz yz

T

A F basis

G

A

B E

D

F

C

8 8

n. unmoved atoms

d 2 unmoved atoms

1 1g g uA E TG

*1 i

i

RG

n R RN

Number of times irrep i is present in basis:

Let us see the Projection of orbital A into T1u

4 2 2 3 4 6

2 2 2

1

1

2

2

2 2 2 2 2

1

1

6 3 6 ' 8 6 3 6 8 48

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

2 0 2 0 1 2 0 2 0 1 ( ,2 )

2 0 2 0 1 2 0 2 0 1

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0 ( ,

h h d

g

u

g

u

g

u

g x y z

u

O E C C C C i S S g

A x y z

A

A

A

E x y z x y

E

T R R R

T x

2

2

, )

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0

6 2 2 0 0 0 0 4 2 0

g

u

y z

T xy xz yz

T

A F basis

G

9 9

( ) *Projector : ( )i i

R

P R R

4 2 2 3 4 6

1

6 3 6 ' 8 6 3 6 8

3 1 1 1 0 3 1 1 1 0

h h d

u

O E C C C C i S S

T

E

A

B

C

D

F

1( )

4

2

2

3

4

6

contribution

3 3

6 2 1 2

3 2 1 2

6 ' 2 1 (2 )

8 2( ) 0 0

3 3

6 2 1 (2 )

3 2 1 2

6 2 1 2

8 2( )

uT

R C

h

d

Class RA

E A A

C A B C E F A B C E F

C A D A D

C D B C E F D B C E F

C B C E F

I D D

S D B C E F D B C E F

A D A D

A B C E F A B C E F

S B C E F

0 0

Projection of orbital A into T1u

4

2

2

:

2 operations with axis AD :

2 operations with axis EB : ,

2 operations with axis CF : ,

:

1 operations with axis AD :

1 operations with axis EB :

1 operations with axis CF :

' :

with axis : , with axis :

C

A A

A C F

A B E

C

A A

A D

A D

C

AB A B AC

with axis : , with axis :

with axis : , with axis :

1 operation with axis CF :

A C

AE A E AF A F

CB A D EC A D

A D

10

E

A

B

C

D

F

The normalized T1u projection is

ψ1 =( A−D)/ √2.

Operating in the same way

on D we again get ψ1.

Operating on the other functions, we obtain

ψ2 =(B−E )/√2

and ψ3 = (C−F )/√2.

In this way one easily builds the ligand group orbitals.

Ψ1 can make bonds with px orbitals of central atom Ψ2 can make bonds with py orbitals of central atom Ψ3 can make bonds with pz orbitals of central atom

11 11

Crystal field theory

Number of d electrons:

2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8

Ti V Cr Mn Fe Co Ni Cu

M

M

M O

Hund’s Rule 1: ground state has maximum S, highest L compatible with S

Hund’s Rule 2: ground state has maximum J for > half filling, lowest J for < half filling

Friedrich Hund, 1896- 1997 (aged 101)

transition metal ion M in an Oxygen cage with Oh

symmetry. In many cases the main effect is the

splitting of the ion levels by the crystal field. In some

cases the on-site interactions are important, in other

cases they can be neglected in a first approximation.

2

2 62 6

4

6

( ) green,paramagnetic(3 )

( ) yellow,diamagnetic

Fe H OFe d

Fe CN

This is explained by crystal field theory

For instance, with 3 electrons ground state is

4| 2 1 0 |, 3g LM F

1 2 3 4 5 6 7 8 9

2 3 4 5 6 5 4 3 2

electron number

ground state

d d d d d d d d d

D F F D S D F F D

The same ion can behave in very different ways in different compounds: Ferrous Fe

:

2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8

Number of d electrons

Ti V Cr Mn Fe Co Ni Cu

M

M

Hund’s rule prompts the ground state quantum numbers:

M O

13 13

Crystal field theory-independent electron approximation

For the 5 d orbitals (or even many-body D) states, in octahedal symmetry , using

one finds

1 5sin[ ] sin[ ]2 2( ) , 2 ( ) ,

sin sin2 2

j

j

jim

m j

j

e j

G

4 2 2 3 4 66 3 6 ' 8 6 3 6 8

5 1 1 1 1h h d

d

O E C C C C i S S

4

5 1 1Sin[ ] Sin[ ] 1

2 4 42 2C

2

5Sin[ ] 1 Sin[ ] 1 1

2 2C

2

2 5 3 3Sin[ ] Sin[ ] 1

3 3 2 3 2C

M O

Inversion is like E for d states. For the 5 d orbitals (or even D) states, i= parity=E.

14 14

1 5sin[ ] sin[ ]2 2Since ( ) , and 2, ( ) 1

sin sin2 2

j

j

G

4 2 2 3 4 66 3 6 ' 8 6 3 6 8

5 1 1 1 1 5 1 1h h d

d

O E C C C C i S S

Hence we can deduce the character of reflections,as follows. For the 5 d orbitals (or even D) states, parity=+ , so a reflection is like a rotation.

6 6 4 4

5 5sin[ ] sin[ ]

2 3 2 2( ) ( ) 1, ( ) ( ) 1.

sin sin6 4

S R S R

Inversion i is the result of a rotation and a reflection

Since a reflection is a factor 1,improper rotations are like properones:

1sin[ ]

2( )

sin2

j

j

jim

m j

j

e

15 15

one finds

G

4 2 2 3 4 66 3 6 ' 8 6 3 6 8

5 1 1 1 1 5 1 1 1 1h h d

d

O E C C C C i S S

4 2 2 3 4 6

2 2 2

1

1

2

2

2 2 2 2 2

1

1

6 3 6 ' 8 6 3 6 8 48

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

2 0 2 0 1 2 0 2 0 1 ( ,2 )

2 0 2 0 1 2 0 2 0 1

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0 ( ,

h h d

g

u

g

u

g

u

g x y z

u

O E C C C C i S S g

A x y z

A

A

A

E x y z x y

E

T R R R

T x

2

2

, )

3 1 1 1 0 3 1 1 1 0 ( , , )

3 1 1 1 0 3 1 1 1 0

5 1 1 1 1 5 1 1 1 1

g

u

d

y z

T xy xz yz

T

d basis

G

2d g gE TG

d2 2 2 2 2( ,2 )x y z x y

( , , )xy xz yz

*1 i

i

RG

n R RN

Number of times irrep i is present in basis:

16 16

T2g

Eg

D

In complexes usually the energy splitting is Δ = E(Eg)−E(T2g) > 0, since the T2g orbitals stay far from the negative ligands.

Thus in the absence of Coulomb interactions in the ion one would fill the available levels according to the aufbau principle, starting with T2g.

M O

17 17

T2g

Eg

D

In crystal field theory one tries to predict the magnetic properties by diagonalizing a many-electron Hamiltonian which is the sum of the isolated ion Hamiltonian and the one of the crystal field.

Many papers have been published on the electron spectroscopies of transition metal compoounds using Group theory methods.

If Δ << U, one treats Δ as a perturbation of the isolated ion multiplet: Hund rule and paramagnetism

If Δ >> U, Hund’s rule holds (high spin is preferred) within the degenerate T2g and Eg levels, but Eg starts being filled only after T2g is full, and 6 electrons yield a

diamagnetic complex.

M O

Crystal field theory-interactions

18 18

1 1 1 N N N 1 3v= x , , ,... x , , ( ,... )Ny z y z v v

(Born-Oppenheimer)

i

i

vv

vi

Um

v nuclear displacement vector

Groups help with any symmetric secular problem

2

i

i

vv

vi

Um

2

i

i

1 1 1Equation of motion: v v v v

v 2pq p q ij j

pq ji i

U Um m

Normal modes of molecular vibration: classical motion of i-th nucleus

Harmonic approx to potential:

Uij force matrix

1v v

2ij i j

ij

U U

Applications of Group Theory to vibrations

19 19

2

i

1Equation of motion: v vij j

ji

Um

People prefer to put masses into force matrix

2

i

1v vi ij j

ji

m Um

2

iv vij

i j j

j i j

Um m

m m

i

2 2

Introducing v one has the secular problem

( ) 0

ij

i i ij

i j

i ij j

j

UQ m W

m m

Q W Q Det W I

αQ eigenvectors of W =normal modes

αω eigenfrequencies of W

0 : 3 translations and 3 rotations

(2 rotations for linear molecules)

0 :

20 20

Each nucleus has 3 displacements Cartesian reference

which may be rotated/reflected

R: unitary matrix that rotates / reflects the whole molecule leading to an identical geometry. R sends each atomic displacement to a linear combination

of atomic displacement. This associates to R a matrix D(R).

R is a symmetry if [R,W]=0.

The set of matrices is a representation of the symmetry Group.

Vibrations belonging to different irreps are orthogonal.

W commutes with all the R and cannot mix vibrations belonging to different irreps

Reducing D to irreps, secular determinant is put in block form.

Group Theory and classification of vibrations

αQ eigenvectors of Wαω eigenfrequencies of W

21 21

Program: To diagonalize W simultaneously with as many D as possible, + Dirac’s characters W

Practical use: reduce W to block diagonal form by linear combinations of the Q components:

U +

U =

x

z

y

Example:Water Molecule on xz plane

22 22

0 0

( ( )

1 0 0

with ) has a block str 0 1 0

0 0 1

ucture 0 0 ,

0 0

b

D xz b b

b

0 0 1 0 0

( ( )) also has a block structure 0 0 , with 0 1 0

0 0 0 0 1

b

D yz b b

b

2

0 0

(

1 0 0

with ) has a block structu 0 1 0

0 0

re 0

0 1

0 ,

0

b

b b

b

D C

2

1 0 0

0 1 0

0 0 1

1 0 0

0 1 0

0 0 1

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

( ) 0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0

1

0 0

0 0

0 1 0

0 0 1

D C

2 : 1 7, 2 8, 3 9,C

23 23

Using the 9X9 matrices :

However, D matrices are needed to find eigenvectors (see below)

We could arrive to this result without writing the D matrices, taking into

account that for each operation:

– the atoms that change position contribute 0 to the character;

– each arrow (cartesian movement) that it remains invariant contributes

+1, and every arrow that changes sign contributes -1,

– more generally, the cartesian shifts of an atom that does not change position

behave like (x, y, z), so if the arrow is rotated by θ the contribution

is cos(θ).

2 2 ( ) ( )

9 1 1 3vC E C yz xz

G

24 24

2

9 unmoved 9

1 unmoved 2 reversed 1

Oxygen arrows moved

E

C

6 unmoved 3 reversed 3

2 unmoved 1 reversed 1

xz

yz

2 2 ( ) ( )

9 1 1 3vC E C yz xz

G

One can find the characters

without writing the D matrices:

25 25

2 1 2

:

:

i j j k

k

rot

Note R x x xx

A B B

G

2 2 ( ) ( )

9 1 1 3vC E C yz xz

G

2 2

1

2

1

2

4

1 1 1 1

1 1 1 1 ,

1 1 1 1 ,

1 1 1 1 ,

v xz yz

z

y

x

C I C g

A z

A xy R

B x R

B y R

2 2 ( ) ( )

3 1 1 1v

trasl

C E C yz xz

G

2 2 ( ) ( )

3 1 1 1v

rot

C E C yz xz

G

1 1 2

read from Table:

:trasl A B BG

26 26

Gvibr = 2 A1 + B1

*1 i

i

RG

n R RN

Analysis in irreps:

Breathing mode A1 in all molecules

2 2

1

2

1

2

4

1 1 1 1

1 1 1 1 ,

1 1 1 1 ,

1 1 1 1 ,

v xz yz

z

y

x

C I C g

A z

A xy R

B x R

B y R

G

G

G

G

2 2

By difference:

4

9 1 3 1

3 1 1 1

3 1 1 1

3 1 3 1

v xz yz

trasl

rot

vibr

C I C g

What kind of vibration is B1 ?

Remark. A similar analysis applies in the application to molecular orbitals in the LCAO method when p prbitals are involved.

27

What kind of vibration is B1 ?

2

1 2 2

2

1 0 ( )

( ) ( ) ( ) 0 1 ( ) 0

( ) 0 1

b xz b C b yz

P B E C xy yz b xz b C b yz

b C b yz b xz

0 0 1 0 0

( ( )) 0 0 , ( ( )) 0 1 0 . The block matrix for the projector is:

0 0 0 0 1

b

D xz b b xz

b

0 0 1 0 0

( ( )) 0 0 , ( ( )) 0 1 0

0 0 0 0 1

b

D yz b b yz

b

2 2

0 0 1 0 0

( ) 0 0 , ( ) 0 1 0

0 0 0 0 1

b

D C b b C

b

1 2

Projection operator on arrow basis:

( ) ( ) ( )

Recall the block matrices:

P B E C xy yz

2 2

1

2

1

2

4

1 1 1 1

1 1 1 1 ,

1 1 1 1 ,

1 1 1 1 ,

v xz yz

z

y

x

C I C g

A z

A xy R

B x R

B y R

28 28

1

0

0

0

0

0

1

0

0

2

0

0

0

0

0

0

1

0

0

)( 1BP

1

0

1

0

0

( ) 0.0

0

0

0

0

P B

The only arrows are 3 and 9, and are opposite. One H shifts up along the molecular axis and the other goes down; such a vibration indeed changes sign

under C2 and σ(yz).

2

1 2

2

1 0 ( )

( ) 0 1 ( ) 0

( ) 0 1

b xz b C b yz

P B b xz b C b yz

b C b yz b xz

2 2

1

2

1

2

4

1 1 1 1

1 1 1 1 ,

1 1 1 1 ,

1 1 1 1 ,

v xz yz

z

y

x

C I C g

A z

A xy R

B x R

B y R

29 29

3v 3 v

1

2

2C 3 6

1 1 1

1 1 1

2 1 0 , , ,

z

x y

C I g

A z

A R

E x y R R

3Vibrations of NH

Movements of N like (x,y,z) : A1+E

G(N) has characters 3 0 1

3

3

v

( ) :

9 unmoved 9

0 unmoved 0

2 unmoved 1 reversed 1

H

E

C

G

3 3

3

3

2 3

( ) 3 0 1

( ) 9 0 1

( ) 12 0 2

3 0

6 0 2

1

3 0 1

G

G

G

G

G

G

v

tras

v

l

ot

ibr

r

C E C

N

H

NH

a

b

c

12 2vibr A EG

30

3 2 4

21

2

2 2 2 2

1

2

8 3 6 6 24

1 1 1 1 1

1 1 1 1 1

2 1 2 0 0 (3 , )

3 0 1 1 1 ( , , )

3 0 1 1 1 ( , , )

d d G

x y z

T E C C S N

A r

A

E z r x y

T R R R

T x y z

Vibrations of Methane (CH4)

3

2

4

1

With 5 atoms χ(E) = 15.

– 8C3 :

all the atoms move except one H and the C: for each, take an arrow along the rotation axis (χ= +1),while the other two, on the perpendicular plane, are transformed as the coordinates (x,y) of

this plane and contribute

TrD(R) = 2cos(2π/3 ) = −1. Therefore χ(8C3) = 0.

– 3C2

all H moved. For the atom of C:

2 arrows of the C change sign and the third does not move: χ = −-1

3

2

4

1

– S4

all H moved. For the atom of C: /2 rotation around z, (x, y, z) → (y,−x, z); then reflection →

(y,−x,−z). So, χ = −1.

3

2

4

1

– 6σd

CH2 remains in place; each atom has 2 arrows in plane and one reflected and χ =

3.

3 2 4

21

2

2 2 2 2

1

2

8 3 6 6 24

1 1 1 1 1

1 1 1 1 1

2 1 2 0 0 (3 , )

3 0 1 1 1 ( , , )

3 0 1 1 1 ( , , )

d d G

x y z

T E C C S N

A r

A

E z r x y

T R R R

T x y z

3 2 4

2

1

1 2

8 3 6 6 24

15 0 1 3 1

3 0 1 1 1

3 0 1 1 1

9 0 1 3 1 2

d d G

tot

trasl

rot

vibr

T E C C S N

T

T

A E T

G

G

G

G

Thus the characters of the representation of vibrations are:

6 6 3 2 2 2 3 6

1

2

1

2

1

2 22

1

2

2 2 3 ' 3 '' 2 2 3 3 24

1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1

2 1 1 2 0 0 2 1 1 2 0 0 ( , )

2 1 1 2 0 0 2 1 1 2 0 0 ( , )

1 1 1 1 1 1 1 1 1 1 1 1

1 1 1

h h d v

g

g z

g

g

g x y

g

u

u

D E C C C C C i S S g

A

A R

B

B

E R R

E x y xy

A

A

1

2

1

2

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1

2 1 1 2 0 0 2 1 1 2 0 0 ( , )

2 1 1 2 0 0 2 1 1 2 0 0

u

u

u

u

z

B

B

E x y

E

C’2

C’’2

Benzene D6h Group

33

Benzene C6H6 There are 12 atoms and 36 coordinates, therefore χ(E) = 36.

The rotations

C2, C3 and C6 around the vertical axis move all the atoms and have character 0.

Rotation C’2 around to a diagonal of the hexagon leaves 4 atoms in place:

for each one arrow is invariant and the others two change sign. Therefore,

χ(C2) = −4.

Rotation C”2 around an axis ⊥ to opposite sides has character 0.

S3 and S6

move all the atoms and have character 0.

C’2

C’’2

34

The reflections

σh in the plane of the hexagon leaves two arrows invariant for every atom and changes sign to

the third, therefore χ(σh) = 12.

The reflection for a plane containing the C”2 axis has character 0 .

The reflection χ(σv) for a plane containing C’2 leaves 4

atoms in place, with two arrows invariant and one changed of sign for every

atom. Therefore χ(σv) = 4.

The characters of Γtrasl are the sums of those of A2u and E1u;

those of Γrot are the sums of those of A2g and E1g.

C’2

C’’2

35

C’2

C’’2

6 6 3 2 2 2 3 6

1

2

1

2

1

2 22

1

2

2 2 3 ' 3 '' 2 2 3 3 24

1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1

2 1 1 2 0 0 2 1 1 2 0 0 ( , )

2 1 1 2 0 0 2 1 1 2 0 0 ( , )

1 1 1 1 1 1 1 1 1 1 1 1

1 1 1

h h d v

g

g z

g

g

g x y

g

u

u

D E C C C C C i S S g

A

A R

B

B

E R R

E x y xy

A

A

1

2

1

2

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1

2 1 1 2 0 0 2 1 1 2 0 0 ( , )

2 1 1 2 0 0 2 1 1 2 0 0

36 0 0 0 4 0 0 0 0 12 4 0

3 2 0 1 1 1 3 2 0 1 1 1

3 2 0 1 1 1 3 2 0 1 1 1

30 4 0 2 2 2 0 0 0 12 4 0

u

u

u

u

tot

trasl

rot

vibr

z

B

B

E x y

E

G

G

G

G

G 1 2 2 1 2 2 1 1 2 2

2 2 2 3 4 2vibr g g u u g u g u g u

A A A B B B E E E E

The characters of Γtransl are the sums of those of A2u and E1u;

those of Γrot are the sums of those of A2g and E1g.

36

t1, t2, t3 primitive translation vectors

One-electron approximation: the electron moves in

a periodic crystal potential V (x )

{Ti} set of lattice translation operators = combinations of primitive translations

with integer coefficients

Periodic boundary conditions: for some N>> 1, TiN = 1.

Space-Time Symmetries of Bloch States in solids

Halite (sodium chloride)

- a single, large crystal .

ii.k ·t

i

N

with T unitary, eigenvalue C=e

: Born-Von Karman b.c. C 1

i k k kH,T _=0 with H (x)= (x) and

ii.k ·t

i k k i kBloch wave function T (x) = (x+t ) = e (x),

37

i

i

1 2 3

j

i j

i.N.k

ij

·t 2 requires k · t = * integer; therefore,

Npg + qg + rg

k= N

with p, q, r Z, and the reciprocal lattice basis vectors g

defined by

t · g = 2

C 1=e

.

N

Elementary aspects:

iti.

i

i k

k ·

k

k ·

x

k

k

Bloch's functions are:

H and

(x) = e u (x ),

with u (x ) lattice

T =

p

,

e d

e

rio ic.

k k k

k

ik

2

k k k

·x ik ·xˆ ˆFor each k, since (with =1) p e =e (p+k)

the Schrödinger eq

(p + k )[ + V (x)]u (x) =

ua

u

tion has one solution:

within the unit c(x)2m

ell.

38

k is the label of an irrep, neither H nor

translations can mix different k

Group theory aspects: No degeneracy is predicted, since Abelian Groups have only one-

dimensional representations.

Abelian group Each translation Ti by a lattice vector is a class.

The degeneracy of 1 d chain spectrum

is  not explained in this way: the inversion symmetry is

( ) 2 c

invo

os

lved

(see

( )

l

(

b ow)

)

e

k t k k

( ) . ( ) .( ) 1-dim representation ( ) character

One solution for each ,

k ik t k ik t

k

D t e t e

k

39

( ') ( , ')i t t k

C

k BZ

e N t t

* ( ')1: ( ) ( ) ( ')i j i k k t

G ij

R G t Bravaiscells

LOT R R N e k kN

Second character orthogonality theorem

( ) ( ) *'( ) ( ')i i G

CCi C

NC C

n

( ) ( ) .Since ( ) ( ) is the character,k k ik tD t t e

BZ integration cell localization!

2

2 4

3 2 2 2

Dirac's equation

yields relativistic corrections via a transformation:

2 8 4

A A

B B

eAi eV c p

t cmc

eAc p i eV

c t

Foldy Wouthuysen

p p eH eV

m m c m c

2

2 2.( )

8

eE p divE

m c

Relativistic corrections for low speeds

The correction which lifts degeneracy: SO interaction

2 2 2 2 2 2In atomic physics .( ) .( ) .

4 4 2SO

e e dV e dVE p r p S L H

m c m c r dr m c r dr

2 2 2 2Also, .( ) .

4

1

4V pE

m

ep

m c c

40 How does it change the symmetry in crystals?

41

Eigen-spinors can still be taken of the Bloch form,

i

ik ·x

, k,

i.

k

t

k k

,

k ·

i

H and

(x) = e u (x ), u (x ) lattice p

e

e

T =

ri

, with

odic spinor.

k k k

k

2

2 2

( ) 1( ) .( ) ( ) ( ),

2 4

and the solution ( ) is cell-periodic.

k k k

k

p kV x V p k u x u x

m m c

u x

Adding the Spin-orbit interaction lifts trivial spin degeneracy:

2 2

1' .

4SOH V p

m c

2

( ) '2

SO

pH V x H

m

2

The equation for the periodic function

( )( ) ( ) ( ) with the spin-orbit interaction becomes

2 k k k

p kV x u x u x

m

2

2 2

2 2

2 2

2 2

( ) 1( ) .( ) can be written

2 4

1( ) . . , with

2 2 41

.4

p kH V x V p k

m m c

p kH V x V p k

m m m cp

Vm m c

The . Methodk p

If the problem is solved at k=0 one can expand for small k. (This is also used in the nonrelativistic limit). The perturbation mixes different bands.

2

2

, 0 . , 0 , 0 . , 010

0 0 2

k k k k k k kk

m m

* *

1 10 inverse effective mass tensork k k

m m

The diagonal elements of at k=0 vanish by symmetry since <V>=0. Then one treats in second order and k2/2m in first-order obtaining for the band

k.p theory- NonDegenerate bands

Sometimes,

small gap strong interaction large inverse mass small effective mass.

* 4

1Example: 0.136 0.006 4*10x xCd Hg Te x gap eV m m

k.p theory-Degenerate bands-No Spin-orbit case

Let us recall the results of degenerate perturbation theory in second order (Landau-Lifschitz paragraph 39). If V is the perturbation and n,n’ belong to the degenerate level, while m denotes the other states, one must solve the secular equation

'' '(0) (0)

| | 0nm mnnn nn

m n m

V VDet V E

E E

Consider a cubic crystal (Group Oh) with a band of symmetry Eu interacting with a higher band of symmetry A1 separated by a gap EG.

(0) (0)

n m GE E E

Inversion is a symmetry. Since V is proportional to p which is odd, the first-order energy vanishes by parity.

'

'| | 0nm mn

mnn

G

V V

Det EE

'

'| | 0 Eigenvaues ofnm mn

mnn

G

V V

Det EE

2

2

| |

., where .

| |

xs syxs

G G

ys sx ys

G G

V VV

E E k pV

mV V V

E E

22

2 2

2 2 2

E (x,y). By symmetry, 0, while . Then,

| || |set , , ,

u x y x y

y y y y y yx x

x y x y

G G G

s p y s p x s p x s p y

s k p y y k p s s k p xs k p xAk Ak Ak k

m E m E m E

2

2

2 has eigenvalues 0,

x x y

x y y

Ak Ak kk

Ak k Ak

One finds two isotropic bands with different effective masses.

46

Space Inversion operation (0)P x x

Lactic acid enantiomers are mirror

images of each other

Stereoisomers that are mirror images are callend enantiomers derived from

'ἐνάντιος', opposite, and 'μέρος', part or portion.

2

, , ,Let ( ) ( ), spinor index, ( ) '

2k k k SO

pH x x H V x H

m

2(0) (0) (0) (0) ( ) ( )

(0) (0) (0) (0)

, , , ,is the same for enantiome

Indeed formally: 1, ( ) '2

( ) ( ) s, r

i i

SO

k k k k

pP P P HP H V x H

m

P HP P x P x

2( ) ( )

( ) (0)

(0

(0)

( )

,

)

2 2

Consider inverted crystal: same problem with ( ) ' ,2

( ) ( ) ( ) where parity.

is the same

Intuitiv

1

ely: inverted proble

( , , ) ' .:

m (

4S

i i

O

SO

i

i

k

p p H V pm

pH V x H

m

V x V x P V x

Pc

P

x

, ,

) ( ) with same .k k

x

Two enantiomers of a

generic aminoacid

Site at origin

47

If the crystal is enantiomer of itself,that is, [P(0) ,H]− = 0, adding this element to the translations produces a non-Abelian Group which implies degeneracy.

(0)P

Effect of Parity: nothing happens

(0)P t

Effect of Parity and

then up translation

Space Inversion symmetry

t

Effect of up translation

(0)tP

Effect of up translation then Parity

48

,

(0)

, , ,

, ,

, ,

Yes, ( ) ( ) ( )

since ( ) ( ).

They belong to different eigenvalues of the unitary translation operato

belongs to irrep

The

r.

n ( ) ( )

ikx

k k k

ikt

k

k k

k

k

x P x k

H x

e u x

t x x

x

e

2(0

(0) (0) (0) (0

) (0) (0)

, ,

, , ,

)

, , ,

Yes : ( ) ' ( ) ( )2

( ( ) ( ) .) ( )

SO k k

k k kk k kP HP

pP HP H V x H

P x

P x xm

HP x xx

, ,Are ( ) and ( ) orthogonal?

k kx x

, ,Are ( ) and ( ) degenerate?

k kx x

Crystals with Space Inversion symmetry

Two orthogonal eigenfun

degenerac

Deg

y (in simp

enerac

le chai

ct

n,

ion

y?

s

too).

49

, ,

, ,

, , , From (( ) ( ) and

we conclude

) ( )k k k k k

k k

H x xx x

,

,

, ,

(

,

.

,

)

( ) ( ) is the o

( ) ( ) is the o

( ) ( ) a

Yes. Since ( ) is a 1 dim representation

nly solution for , and

nly solution f

nd

or

moreover

( ) ( )

,

ikx

k k

ikx

k

k ik t

k

kk

k k

k k

x e u x

x e u

D t e

k

kx

x x

x x

, , , ,( ) ( ). ( ) ( )ikt

k k k kx e u x u x u x

, ,Question: ( ) ( ) ?k kx x

Remark: Different quantum numbers label the degenerate eigenfunctions.

The simple chain analogy helps here.

( ) ( ) ( ), with Real ( )

Can we use the knowledge of ( ) to solve time-rev

Yes, we can.

Set ' . Can we solve by sett

ersed dynamics:

'( )( ) '( ) ?

ing '( ') ( ) ?

ti H t t H t

t

t

ti H t t

t t t

t

t

No, but we shall find the time-reversal operator T such that '( ') ( ).t T t

Time Reversal operator Suppose we can solve Schrödinger equation with no

magnetic field

*Introduce Kramers operator :'

K K Ki it t

K K 50

1ˆ ˆand Time-reversed operators: 'A TAT

**( ) ( )

( ) ( ) .) ( ( )Ki KH i H

Write for t

**( )

Now set and get ( ) ( )t

t i H t tt

'( )

compare to time-reversed dynamics ( ) '( )t

i H t tt

*'( ) ( )t t

yields '( ') ( ) with t'=-t

(Not only t -t, not only K, but also reverse spin)

T K t T t

51

1Time-reversed operators: ' .p TpT KpK p

Time reversal operator for Pauli equation with B

H0 = spin independent part of the Hamiltonian,

0

ei [ ( ) . ( )] ( ), =

t 2m,

cH t B t t

Primes are needed: indeed the currents change sign under time reversal, hence the vector potential and the magnetic field also

change sign. Thus, B’=-B, A’=-A: Therefore the imaginary part of H0 (term in Ap) changes sign.

*

0

'( )time-reversed dynamics [ ( ) . ( )] '( )

ti H t B t t

t

0

'( )Time-reversed dynamics [ '( ) . '( )] '( )

ti H t B t t

t

22 2

2

0

2( ) ( )

( ) ( )2

e ep A t A t p

ccH t V tm

52

to solve the time-reversed dynamics? Yes.

Is complex conjugation still su

an we

ffici

use

ent?

.

( )

No

C t

One should expect that the spin must be reversed by time reversal .

2 2 2

0 1 1 0Note: , ( )( )

1 0 0 1i i i

*

2Next, I show that  '( ) ( )t i t

0

eOriginal problem: i [ ( ) . ( )] ( ), =

t 2,

mcH t B t t

K must obviously occur. Write for t

0

** *

* * *

0

*

0

,

,

( )c.c of Pauli equ

( )i [ ( ) . ( )] ( ), ap

ation: i ( ) * ( ) . ( ) (

ply

( )i [ ( ) * . ( )]

)

( ) ,

.

H B K

H B s

tH t t B t t

t

et t

0

* *

*'( )[ ( ) . ( )] '( ) :

( ) does not work bec

Compare to time-reversed dynamics

insaus teae of d o f -

ti H t B t t

t

t

53

** * *

0

( )c.c. Pauli equation: i ( ) * ( ) . ( ) ( )

tH t t B t t

tMultiply on the left by –i2

* * * *

2 0 2 2i ( ) ( ) ( ) * ( ) ( ) ( ) . ( ) ( ) .i t H t i t i B t tt

*

2 2Next, note that ( ) ( ) in fact,i i

* *

2 1 2 2 1 2 2 1 2 2 2 1 1

* *

2 2 2 2 2 2 2 2 2 2

* *

2 3 2 2 3 2 2 3 2 2 2 3 3

( ) ( )

( ) ( )

( ) ( ) .

i i

i i

i i

* * *

2 0 2 2i ( ) ( ) ( ) * ( ) ( ) . ( )( ) ( )i t H t i t B t i tt

*

0

'( )compare time-reversed dynamics [ ( ) . ( )] '( )

ti H t B t t

t

*

2'( ) ( ) (Not only t -t, not only K, but also reverse spin).t i t 54

* * * *

2 22 0 2 2i ( ( )) ( ) ( ) * ( ) ( ) ( ) . (( ( .)) )i t H t i t i B it tt

i

and using (–i2 )(–i2 ) =-1

Thus, the time reversal operator is

yT i K

K K

0 1Inverse of :

1 0yT i K K

2 10 1 0 1 0 1 0 1

11 0 1 0 1 0 1 0

T K K T T

55

Time-reversal of matrix elements: also for spinors

2

0 1

1 0i

56

10 1 0 1

' ( )1 0 1 0

p TpT Kp K KpK p

1 0 1 0 1' ( )

1 0 1 0L TLT KL K KLK L

Time-reversal of dynamical variables:

* 1

2 2

1

( ) ( ) ,

. . . invariant

i i TST S

TLST LS LS

Spin-orbit interaction is time-reversal invariant:

Dirac’s Theory is.

In the case of P, we found that if it is a symmetry, it bears

degeneracy.

What about T?, when is it a symmetry? If it is, does it imply

degeneracy?

T is a symmetry when H is time independent and there is no B.

Kramers theorem: in Pauli theory,time-reversal symmetry (i.e. H with no

magnetic field and no time dependence) implies degeneracy.

2

( ) '2

SO

pH V x H

m

58

1

_[ , ] 0

yT i K T H H THT

H E TH ET HT ET

Now I show that has twofold degeneracy (even with the spin-orbit interaction) since an H eigenspinor and its time-inverted spinor have the same energy and

are orthogonal. The time-reversed spinor has the same energy because T is a symmetry:

* *

* *

Moreover, .

Proof that :

0 10

1 0

(note: T flips spin, indeed). This implies degeneracy.

T

T

T T

59

*( ) ikx

k y kT x ie u belongs to –k and has the same energy

reverses spin Indeed, I show that z reverses under T.

2and that

. Using this property,

y y yT T i K i K K K

K K

k z k z k k z k k k z k

anticommutation Hermiticity of z

T T T T T T

, ,

, ,

In conclusion : ( ) ( )

[ , ] 0 also with spin-orbit.

k k

k k

T x x

T H

Note that implies since [ , ] 0y z z y z

T i K T T

60

Summary:

0 0

, ,( ) ( ) , 0k k k kParity P x x P H

Time reversal ( ) ( ) , 0k k k kT x x T H

k -k

k -k

61

0

0[ , ] 0, [ , ] 0 [ , ] 0

C P T

P H T H C H

(0) (0)( ) ( ) ( ) ( )k k k k

k k k

C x P T x P x x

spin degeneracy at every k point even with SO

k -k k -k

Coniugation