11

2

3 3

irrep :

1 0( ) ( ) ( )

0 1

1 0( ) ( ) ( )

0 1

2 1 2 3cos( ) , sin( )

3 2 3 2

a c b

E

c s c sD E D C D C

s c s c

c s c sD D D

s c s c

c s

How can the Dirac characters commute with every D?

2

3 3

1 0( ) ( )

0 1R

c s c sD C D C

s c s c

1 0 0 0( ) ( ) ( )

0 1 0 0a c b

c s c sD D D

s c s c

2

2

3 3

2

3 3

2

3

3 3

2

3 3

2

3 3

3

a b c

c a b

b c a

a b c

b c a

c a b

E C C

C C E

E C C

C E C

C

C

C

C

E

E

Matrices of Dirac Characters within an irrep

22

1 0 0 0

0 1 0 0R

How do constant matrices arise?

Diagonalizing H within each subspace one finds eigenstates;

Those eigenstates carry the Irrep label.

Consider a pair of eigenfunctions (ψx, ψy) transforming like (x,y); they belongto the Irrep E.

x x x y y yH = , H = .

x x x y

- x x x

x x

x

S G : S =a +ß ,

but since [S,H] = 0, HS =SH = S

S must belong to the same energy eigenvalue as .

0 = = , a constant matrix.

0

Why are the characters also const

x

yH

ant?

constant matrices!

Indeed, quite the same reasoning can be repeated for the Ω matrices.

x x x y - x x x

x x

S G : S =a +ß , but since [S, ] = 0, S =S = S

S must belong to the same character eigenvalue as .

0.

0

To sum up,

the matrices of H and those of Dirac characters in the subspace spanned by an irrep have

the remarkable property of being constants, that is, identity matrices I m×m

multiplied by numbers:

H = Im×m ΩC = ωC Im×m

3

44

D Matrices are not unique, e.g. one can rotate frame

By any unitary transformation (not necessarily a symmetry) we can change basis and matrices within the E irrep. Let U be any rotation: then for all R,D(R) → UD(R)U−1 is consistent with the multiplication table and does

not modify the Dirac characters, which are eigenvalues.

Up to this unitary transformation, the Irreps with the

same characters are to be identified.

a

c b

What makes different representations really distinct?

5

Example: we are familiar with the representing matrices of E

2

3 3

irrep :

1 0( ) ( ) ( )

0 1

1 0( ) ( ) ( )

0 1

2 1 2 3cos( ) , sin( )

3 2 3 2

a c b

E

c s c sD E D C D C

s c s c

c s c sD D D

s c s c

c s

3Find a new representation D in which D (C ) is diagonal.

†

1 i

2 2U= is :

1 -i

2 2

1 1 1 i

1 02 2 2 2

-i i 1 -i 0 1

2 2

unitar

2

y

2

U U

a

c b

Det U = -i U is no rotation

atomic orbitals:

1 i

12 2U =

1 -i 2

2 2

x x x y

y y x y

z

p p p ip

p p p ip

L eigenstates

Diagonal

66

2

3†

3 3 2

3

1 i 1 1

02 2 2 2D (C ) =UD (C ) U

1 -i -i i0

2 2 2 2

i

i

c s e

s ce

†0

=UD ( ) UBut then1

( )1 0

a a is off diagonaD l

The rotations are diagonalized:

Reduction of a representation:

one should use the smallest possible D matrices.

† †

Unitary transformations U leave traces invariant since:

Tr(U DU)=Tr(UU D)=Tr(D)

77

Joning bases of irreps large block-diagonal D matrices can be obtained

D( R )=

A block-diagonal matrix can be transformed

to a non- block-diagonal form by a generic unitary U

U+

U =

Reducible representation

But one should try to do the reverse, until the representation is irreducible!

88

Traces and Characters of an Irrep

Consider a basis of a m-dimensional irrep i and the representative matrices

1

†

For class C, define the character ( ) ( ) ( ), ;

since S X is unitary ( ) depends on

is independent

Dirac's character also depends on ; Tr .

Equivaently, o

( ) ( )C

C

m

n

n

C mXm C

C

R

R

C TrD R D R R C

SX C

R

R I C

C C C

m

n

1 1

ne can work out TrR= (C) and multiply by the n

( ) ( ) ( )

( )( ) character mean eigenvalue

umber n

Cn m m

C C

C

C

C

R

C

C C

C n C R m

m CC R

n n m

Identity: (E)=m

2

We shall prove that irrep dim are obtained from irrep characters by:

G

C

C C

Nm

n

99

Character Tables of common point groups

Below we shall prove: Character tables are square

1010

1111

It is important to realize that Dirac’s characters =constant matrices by a mere algebraic fact.

Therefore I shall present an algebraic proof, which is very enlightening.

1212

Schur Lemma . Any m×m matrix M commuting with all the representative matrices

D(Ri), i = 1, · · ·NG of an m-dimensional Irrep must be proportional

to the Identity matrix Im×m.

What characterizes the irrep is that the matrices cannot be taken simultaneously to the same block form, because otherwise the

representation would be reduced. We first prove the lemma for a 2 × 2 diagonal matrix, then generalize to any matrix of any size.

1313

Proof for diagonal matrices2x2 matrix

1 11 12 12 2 1

2 21 22 21 2 1

0 0 ( )[A, ]

0 ( ) 0

d a a a d dM A M

d a a a d d

12 2 1

21 2 1

0 ( ) 0 0Condition for [A, ]

( ) 0 0 0

a d dM A irrep

a d d

If d2 were different from d1, this would require a diagonal A. Since one cannot diagonalize simultaneously all the D(Ri) i = 1, · · ·NG), we conclude

that d2 = d1.

1 0

0 1M d

1414

3x3 matrix

1 11 12 13

2 21 22 23

3 31 32 33

12 2 1 13 3 1

21 1 2 23 3 2

31 1 3 32 3 2

0 0

0 0

0 0

0 ( ) ( )

[ , ] ( ) 0 ( )

( ) ( ) 0

d a a a

M d A a a a

d a a a

a d d a d d

A M a d d a d d

a d d a d d

11

22 23

32 33

0 0

0

0

a

A a a

a a

which cannot be true for all the representative matrices in an irrep, because they cannot have a common block form. This is generalized formally to matrices of any size with

the conclusion that M = λIm×m for some λ.

If all the di are different, this vanishes only for diagonal A. If d1 is different

from the other diagonal elements,

1515

nxn diagonal matrix: similar proof

Extension to Hermitean matrices

Assuming M hermitean we know that by a unitary transformation U it can be diagonalized. U sends M to U · M · U† = diagonal and the set A to the set U · A· U†

which still represent the same irrep and commute with U · M · U† . Then U · M · U† = diagonal= λIm×m and actually this implies that

M is such in any basis since U † · Im×m · U =Im×m .

1616

Extension to any matrix

†

† †

: basis of irrep, [ ,A] 0

: :

First we show that if 0 then also commute

Assumption

Thesis Proof

s,

.

mxm

A M

M I

MA AM M

AM M A

† † † †

† 1 † 1

1 1 †

, 0 .

Since , 0 implies [ ,A ] 0, .

But since A ( ) ( ) basis of irrep, [ ,A] 0, .

Clearly MA AM A M M A

A A MA AM A M A

R A R M A

Out of these we can make two further commuting matrices:

†

1

†

2 ( )

H M M

H i M M

These are Hermitean, hence they are constants, and

1 2

1( )

2mxmM H iH I

1717

Many Groups of fundamental importance in Physics are Lie Groups.

They are continuous (elements can be labeled by parameters) and continuously connected ( for every pair of elements a continuous path in parameter space

can be found that joins them). Moreover, the parameters of the products are

C1 functions of those of the factors.

A compact Lie Group has all parameters that vary over a closed interval;

the Lorentz Group and the Group of all translations are noncompact Lie

Groups, while the rotations are a compact Lie Group.

Marius Sophus Lie1842-1899

Example:

' ,

'

x ax by c

y dx ey f

In order to be invertible : det 0a b

c d

Typical element of Group: g(a,b,c,d,e,f)

Identity element of Group: g(a=1,b=0,c=0,d=1,e=0,f=0)

Group of 2d linear coordinate transformations : ( , ) ( ', ')a b x e

x y x yc d y f

Transformation of functions: ( , ) ( ', ') under thex y x y

effect of Group element g: ( , ) ( ( , ))x y g x y

for example:

0

( , ) ( , )Hence, ( , ) lim ( , )a

x x y x yX x y x x y

x

0

(1 ,0,0,1,0,0) (1,0,0,1,0,0)generator lim

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

0 1

( ', ') ( , ) ((1 ) , ) ( , ) ( , )

a

g gX

xx y x y x y x y

y

x y ax by c dx ey f x y x y x x yx

One cannot study element by element, but one can study the local structure of the Group around the identity (then we can go to any other spot with the application of a Group element). Introduce generators that do infinitesimal transformations:

Typical element of Group g: ( , ) ( ( , )) with ( , , ), , ,g a b c dx y g x y e f

: ( , ) ( ', ') ( , )

Expand around identity : a=d=1, b=c=e=f=0

a b x eg x y x y ax by e cx dy f

c d y f

0

( , ) ( , )Hence, ( , ) lim ( , ),

In similar way:

a

c d f

x x y x yX x y x x y

x

X X y Xx y y

It is well known that translations c and f yield momentum components.

0

(1, ,0,1,0,0) (1,0,0,1,0,0)generator lim

1( , ) ( , )

0 1

( , ) ( , ) ( , )

b

b

g gX

xx y x y y

y

x y x y y x y yx

X yx

k

ij[ c =structure constant] s, k

i j ij k

k

X X c X

General theorem for Lie Groups:

Recall: where generates O(3)iL

L L i L L R e

The SO(3) Group of

rotations in 3d (SO=Special Orthogonal) is a continuous Group. An element may be represented as a vector φ directed along the axis and

with length equal to the angle of (say, counterclockwise) rotation φ; this corresponds to a sphere of radius π where,

however, each point of the surface is equivalent to the opposite one.

All the

rotations with the same |φ| belong to the same class..

The generators of SO(4) :

[I ,I ]=i I

[K ,K ]=i with [I , ] 0.

i j ijk k

i j ijk k i j

obey

K K

Recall: where generates O(3)

( ) and cyclic permutations (3 components).

iL

x z y

L L i L L

R e

L i yp zp

many operators generate O(4)?

46 generators are needed.

2

How

This is the algebra of SO(4). Fore more details see L.Schiff Quantum Mechanics page 235

2323

Double valued spin representation and Covering Group

(2) covering Group of (3) : identity for 4SU O

fill a sphere of radius 2 in SU(2), in SO(3)

2

20,0

.

( 2 )1 0( 2 ) 20 12

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

2

,

2

1For , rotation matrix around z is

2

; however,

0is 2 valu

0

0

More gener

ed

a

:

0

lly,

z

i S

ii

i

i

i

i

J

R e

eR R e R

e

eR e

e

The angular momentum operator L is the generator of infinitesimal rotations. For integer L one finds

2L+1 spherical harmonics YLM(θ, φ) that are simultaneous eigenvectors of

L2 and Lz. Only the harmonics of a given L mix under rotations; they are

the basis of an irrep labelled by L.

' '

'

' , 'j

j j

m m m m

m j

R jm jm D D jm R jm

Symmetry and degeneracy

Let m= degeneracy (=number of basis functions) of irrep i:

by acting with off-diagonal R and orthogonalization.

From one can obtain all the m)(

1

irrep irrep

j

For instance in the triangle Group the degenerate irrep E has basis functions that can be taken to transform like (x,y),The x function is taken by rotations and reflections to a linear combination of x and y.

This implies that the energy eigenvalues whose eigenfunctions belong to an irrep must be degenerate m

times: Can they be even more degenerate?

2525

Accidental degeneracy

It is possible that by accident two states unrelated by symmetry come so

close in energy to appear degenerate in low resolution experiment; however

a mathematically exact degeneracy with no symmetry reason is miraculous.

Simply, one was unaware of using a Subgroup of the actual Group, because

some symmetry had still to be discovered

2626

Famous example: hydrogen atom2

2

p kH

m r

Extra degeneracy on L explained by conservation of Laplace- Runge-Lenz vector. In classical physics, the conserved vector is:

p L rR k

m r

33

The angular momentum L is conser

( ) central forc

ved since:

e

0

1.

( ) .

r k rF k

r r

dL d dr dpr p p r r F

dt dt dt

r

dt

r

Why are the Hydrogen eigenvalues independent of L?

One cannot produce 2p orbitals by rotating 2s !

Carl David Tolmé Runge (German: [ˈʀʊŋə]; 1856–1927)

Wilhelm Lenz (February 8, 1888 in Frankfurt am Main – April 30, 1957 in Hamburg)

During the motion, p ∧L is in the xy plane. At perihelion, p ⊥ r ⇒ p ∧ L is parallel to r.

Since it is conserved, R is pinned at the aphelion-perihelion direction, and this is why the orbits are

closed .

Orbits are closed in the 3d harmonic oscillator,

in which case, separate energy conservation

along the 3 axes is the extra symmetry.

Pierre-Simon Laplace, marchese di Laplace (Beaumont-en-Auge, 23 marzo 1749 – Parigi, 5 marzo 1827),

perihelion

p L rR k

m r

pL

constant of motion: proofp L r

R km r

3

1

r

r r

2

3 3 3

1 1The derivative of the second term: start by v.

1 1 v. v v. r v- (v. )v. .

This can be simplified:

d

dt r r

d r d r r r rr

dt r r r dt r r r r

2

x y

x y3 3 3

2

x

2

x x

x y

2

3 3 3

y

( )v -x( +yv )(yv

xv-xv )

( )v -y(xv yv+ )(-yv +xv )

yd x y Ly

dt r r r mr

xd y x Lx

dt r r

y

r mr

x

3

( , ,0)

d r L y x

dt r mr

3Force (0,0, )

rF k L L

r

( , ,0) (0,0, )x yF F Ld p L F L

dt m m m

0dR

dt

28

3

( , ,0) ( , ,0).

y xF F Ld p L y x L

dt m m mr

3 3 3 30

0 0

i j k

x y Ly Lxi j

r r r r

L

2929

Wolfgang Pauli in 1926 first solved the SE for the H atom using the SO(4) symmetry. SO(4) is a Lie Group (see below).

W. Pauli, “On the hydrogen spectrum from the standpoint of the new quantum mechanics,”Z. Physik 36, 336-363 (1926).

Therefore we must say more about Lie Groups. Some are already familiar from the

theory of angular momentum in Quantum Mechanics.

Short Outline of the Pauli treatment of the H spectrum of bound states

The classical is replaced by .2

p L r p L L p rR k M k

m r m r

2 [ , ] ( )

[L , ]

i j ijk k

i j ijk k

HM M i L

m

M i M

2 2 2 2

. 0 M.L=0

2=( ) (L + ) +k

L M

HM

m

[L , ]i j ijk kL i L

2 2 2 22=( ) (L + ) +k , where H=Hamisu ltch that onian

HM

m

2To build the generators since [ , ] ( )

for E<0 introduce ' . 2

[ ' , ' ]

[L , ' ] '

i j ijk k

i i

i j ijk k

i j ijk k

HM M i L

m

mM M Then

E

M M i L

M i M

' 'The generators of SO(4) are: I ,K .

2 2

[I ,I ]=i I

[K ,K ]=i with [I , ] 0.

i j ijk k

i j ijk k i j

L M L M

K K

3131

1 2 3 1 2 3

_

_

_

46 generators in 3d , ,

2

, extends those in 3 :

,

, 0

x y z

i j ijk k

i j ijk k

i j

A A A B B B L L L

A A i A d L L i L

B B i B

A B

Extension to SO 4 rotations in 4d, relevant to H atom :

6 planes ij 

i 1,4 , j 1,4 .

One finds:

The extra symmetry is the reason why there is an extra conserved quantity

3232

Quantum Runge-Lenz vector :

and one finds after long algebra that [ , ] 02

p L L p rR k R H

m r

2 2

do

Therefore,

[ , ] 0 does not bel

es not depend on

ong to

LM LM LM LM

LM

H E HR ER

R L R L

E L

22

2 2

52

1 1

2 2

[ , ]_ 0

Biederharn-Johnson-Lippman peudoscalar operator

.( ) , [ . ]

E does not depend on sign of K (2 ) (2 )

eH c p mc Z H B

r

i e e rB K c p Z Z K L

mc r c r

E s E p

Dirac’s equation relativistic Runge-Lenz vector

33

My own simple example0 1 1 0

1 0 0 1

1 0 0 1

0 1 1 0

h

2

0,1

0,2

2

1 1 1 12 ( , , , )

2 2 2 2

1 1( ,0,0, )

2 20

1 1(0, , ,0)

2 2

1 1 1 12 ( , , , )

2 2 2 2

matrice delle adiacenze

Autovalori e autofunzioni di h

34

Now we deform the square, e.g. by doubling the hopping integrals connecting site 3 to 1 and 4. Now, the Hamiltonian reads

0 1 2 0

1 0 0 1

2 0 0 2

0 1 2 0

h

10

0,1

0,2

10

1 1 2 110 ( , , , )

2 5 210

1 1( ,0,0, )

2 20

2 1(0, , ,0)

5 5

1 1 2 110 ( , , , )

2 10 5 2

Accidental degeneracy!

3535

2

0 0 1 0 0 1 3 0

1 0 0 0 3 0 0 11,

0 0 0 1 1 0 0 310

0 1 0 0 0 3 1 0

S

T

rotationS S S hS h

h hS

is a symmetry. To explain this degeneracy we want a non-Abelian Group, however.

Generalized rotation with ST=S-1

0,1 0,1 0,2 0,2

0 0 0 1

0 1 0 0

0 0 1 0

1 0 0 0

10

0,1

0,2

10

1 1 2 110 ( , , , )

2 5 210

1 1( ,0,0, )

2 20

2 1(0, , ,0)

5 5

1 1 2 110 ( , , , )

2 10 5 2

3636

The deformed problem has a lower

geometrical symmetry, but actually is still C4v symmetric because of a hidden dynamical symmetry.

S4 = 1

S does not produce a mere

permutation of sites; however:

S mixes ψ0,1and ψ0,2 does not commute

with σ and explains the degeneracy.

Generalized rotation with ST=S-1

10

0,1

0,2

10

1 1 2 110 ( , , , )

2 5 210

1 1( ,0,0, )

2 20

2 1(0, , ,0)

5 5

1 1 2 110 ( , , , )

2 10 5 2

37

( )( ) *( ) ( )GN

ji Gij

R G i

ND R D R

m

Great Orthogonality Theorem (GOT)

We know we must diagonalize H on a symmetry-adapted basis, labelled by Dirac’scharacters. The practical method to do this results from this abstract theorem!

Each element of each D(R) in each irrep is a symmetry type and this leads to orthogonality when integrating

over the Group.

3838

Remark number 1

( ) * ( ) 1( ) ( )i iD R D R

3

2

3 3

Recall irrep of :

1 0( ) ( ) ( )

0 1

1 0( ) ( ) ( )

0 1

2 1 2 3cos( ) , sin( )

3 2 3 2

v

a c b

E C

c s c sD E D C D C

s c s c

c s c sD D D

s c s c

c s

Note that C23 is the inverse of C3 and its matrix is D(C3)T .

More generally, the operators are unitary and

3939

Orthogonality on irreps i,j : different good quantum numbers (Dirac characters).

Remark number 2 is following theorem:

Let total-symmetric operator : , 0,O O R R G

and component of irrepj j

then ( )i j

ijO O i

Orthogonality on components:

For i = j the matrix {Oμν} represents an operator that commutes with everything, so it must commute with all the D matrices; so Schur’s lemma applies and diagonal elements are equal.

Dependence on i: the diagonal element depends on the irrep.

For example, a spherically symmetric potential V(r) has vanishing matrix elements between states of different L and within a given L it has vanishing matrix elements between states of different ML; the

diagonal matrix elements are independent of ML and depend on L.

Another example: matrix of the square of the angular momentum on the basis of spherical harmonics.

4040

Theorem:

Remark number 3 on symmetrization

So indeed O commutes with every T

11

G

1

R

OT TT OT TT R R T

1 1

1from rearrangement the

( )

(

( )

orem)

R G RT G

R G

T RT RT T RT RT

T R R TO

1

R G

arbitrary operator O = invariant, , :R R T G OT TO

Proof:

41

3

2

3 3

Recall irrep of :

1 3 1 31 0 2 2 2 2( ) ( ) ( )0 1 3 1 3 1

2 2 2 2

1 3 1 31 0 2 2 2 2( ) ( ) ( )

0 1 3 1 3 1

2 2 2 2

v

a b c

E C

D E D C D C

D D D

Let us form 6-component vectors with the elements of the D matrices.

Remark number 4: example on the working of GOT

42

3

2

3 3

Recall irrep of :

3 3

0 2 2( ) ( ) ( )

0 1 3 1 3 1

2 2 2 2

3 3

0 2 2( ) ( ) ( )

0 1 3 1 3 1

1 1

1 2 2

1 1

1 2 2

2 2 2 2

v

a b c

E C

D E D C D C

D D D

Remark number 4: example of GOT

11 11 11

1 1 1 1 6(1, , , 1, , ), . 3

2 2 2 2 2E E E G

Nv v v

m

43

Remark number 4: example of GOT

3

2

3 3

irrep of :

1 11 2 2( ) ( ) ( )0 1 3 1 1

2 2 2

1 11 2 2( ) ( ) ( )

0 1 3 1 3 1

2 2

3 30 2 2

3 30 2

3

2

2 2

2

v

a b c

E C

D E D C D C

D D D

12 12 12 12 11

3 3 3 3(0, , ,0, , ), . 3, . 0

2 2 2 2E E E E Ev v v v v

11 11 11

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

2 2 2 2E E E GNv v v

m

44

Remark number 4: example of GOT

12 12 12 12 11

3 3 3 3(0, , ,0, , ), . 3, . 0

2 2 2 2E E E E Ev v v v v

3

2

3 3

irrep of :

1 3 1 31 0 2 2 2 2

( ) ( ) ( )1 1 1

2 2

1 3 1 31 0 2 2 2 2

( ) ( ) ( )

0 3 3

2 2

0 3 3

22

1

2 2

1 1

v

a b c

E C

D E D C D C

D D D

21 21 21 21 11 21 12

3 3 3 3(0, , ,0, , ), . 3, . 0, . 0

2 2 2 2E E E E E E Ev v v v v v v

11 11 11

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

2 2 2 2E E E GNv v v

m

45

Remark number 4: example of GOT

12 12 12 12 11

3 3 3 3(0, , ,0, , ), . 3, . 0

2 2 2 2E E E E Ev v v v v

3

2

3 3

irrep of :

1 3 1 31 0 2 2 2 2

( ) ( ) ( )0 3 3

2 2

1 3 1 31 0 2 2 2 2

1 1 1

2 2

1 1 1

2

( ) ( ) ( )0 3 3

2 22

v

a b c

E C

D E D C D C

D D D

22 22 22 22 11 22 12 22 21

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

2 2 2 2E E E E E E E E Ev v v v v v v v v

21 21 21 21 11 21 12

3 3 3 3(0, , ,0, , ), . 3, . 0, . 0

2 2 2 2E E E E E E Ev v v v v v v

11 11 11

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

2 2 2 2E E E GNv v v

m

4646

Example of GOT: including the other irreps

1

2

1 1 1 111 (1, , . 1, , ) 3

2 2 2 2

3 3 3 312 (0, , ,0, , ) 3

2 2 2 2

3 3 3 321 (0, , ,0, , ) 3

2 2 2 2

1 1 1 122 (1, , .1, , ) 3

2 2 2 2

11 (1,1,1,1,1,1) 6

11 (1,1,1, 1, 1, 1) 6

element irrep list squares

E

E

E

E

A

A

Each element of each D(R) in each irrep is a symmetry type and this leads to orthogonality when integrating

over the Group.

4747

*

Proof of GOT :i j G

ij

R G i

ND R D R

m

1

R G

Then let basis function of irrep j component :

Work ou

)

t

(

i j

ij

j

i j

O

R R

O i

( )inserting ( )jm

j j jR D R

† ( ) *and ( )

jm

i i iR D R

( ) * ( )

R G ,

( ) * ( )

( ) ( ) ( )

We want ( ) ( ), must get rid of

jm

i j i j

ij

i j i j

D R D R O i

D R D R

1

R G

Let some operator, to be specified later; build the invariant O = .R R

4848

We are still free to choose as we please, so pick

the irrep&component switch operator i j

i j i i j j

( ) * ( )

R G ,

We get ( ) ( ) ( )jm

i j

ijD R D R O i

( ) * ( )

R G

that is, ( ) ( ) ( ) .

( ) is independent of , and we got the orthogonality on , .

We need to show yet:

( ) is independent of ,

e got the orthogonality on , as well, we

i j

ijD R D R O i

O i

O i

W

may insert .

( ) * ( )

R G ,

( ) * ( )

( ) ( ) ( )

We want ( ) ( ), must get rid of

jm

i j i j

ij

i j i j

D R D R O i

D R D R

49

*

( )i j

ij

R G

D R D R O i

Orthogonality on first indices , proved by recalling

( ) * ( ) 1( ) ( )i iD R D R

Clearly, no dependence of on is allowed.

(they are second indices, now)

O i

1

*1 1 , but

This exchanges first and second indices,

i j

ij

R G R G R G

D R D R O i

Still, we must find O(i).

( ) * ( )

R G

( ) ( ) ( ) and also i j

ijD R D R O i

50

To determine O(i) we set i = j, μ = ν and α = β. We get

* 1( )

i i i i

R G R G

D R D R O i D R D R

Summing over α from 1 to mi ,

1 1

1 1 1

( ) ( )

X identity 1

order of G

i im mi i i

i

R G R G

i i i

i i G

R G

O i m O i D R D R D R R

D R R m m D R R D R R N

( ) 1 ( ) Gi G

R G i

Nm O i N O i

m

*i j G

ij

R G i

ND R D R

m

CVD

*

( )i j

ij

R G

D R D R O i

51

( )( ) *( ) ( )GN

ji Gij

R G i

ND R D R

m

Great Orthogonality Theorem (GOT)

I presented the proof for discrete Groups, however the GOT extends to continuousones.

2( )( ) *

' ' ' ' '

( )

'

( )

'

2 2

0 0 0

8( ) ( )

2 1

Euler angles Wigner matrices

( ) '

sin( )

jJ

ab a b jj aa bb

J

mm

J

mm

d D Dj

D

D jm R jm

d d d d

Each element of each D(R) in each irrep is a symmetry type and this leads to orthogonality when integrating

over the Group.

The line of nodes (N) is defined as the intersection of the xy and

the XYcoordinate planes.