1

Permutation Group S(N) and Young diagrams

C3v: All operators = reflections or products ofreflections

S(3): All operators = permutations =transpositions (=exchanges) or products of transpositions

They are isomorphous

S(3)= C3v

2

3 3

2

3 3

2

3 3

2

3 3

2

3 3

2

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

C E C

E C C

C E C

C C E

a

b

cx

S(N) : order= N! huge representations but allowsgeneral analysis, with many applications. Example

3

2

3

, ,

, ,

( , , )

( , , )

( , , )

( , , )

a

b

c

E a b c

C b c a

C c a b

a c b

c b a

b a c

Permutations of Group elements are the basis of the regular representation of any Group.

In C3v reflections transpositions.

Permutation Group S(N) and Young diagrams

S(N) : order= N! huge representations but allows general analysis, with many applications.

Young diagrams are in one-to one correspondence with the irreps of S(N)

Alfred Young (1873-1940)

Rule: partition N in not increasing integers: e.g.

8=3+2+2+1

Draw a diagram with 3 boxes, and below two boxes twice and finally one box, all lined up to the left

This corresponds to an irrep of S(8)

3

Young Diagrams for S(3)= C3v and partitions of 3 in not increasing integers

(lower rows cannot be longer)

3

3

2+1

1+1+1

Diagrams that are obtained from each other by interchanging rows and columns are conjugate diagrams . The representations are said conjugate, like these.

Each Young Diagram for S(N) corresponds to an irrep

4

3v 3 v

1

2

2C 3 6

1 1 1 symmetric

1 1 1 antisymmetric

2 1 0 mixed

C I g

A

A

E

4

Young Diagrams for S(3)= C3v and correspondence to irreps

symmetrization

an

tisym

me

trizatio

n

A1

E

A2

1

Theorem: if P S(N) (is a permutation) and and conjugate irreps of S(N),

( ) ( ) ( )P

jk kjD P D P

55

Young Tableaux (Tables)

The Young tables or Young tableaux for S(N) are obtained from the Young diagrams

by inserting numbers from 1 to N so that they grow along every line and every column.

1 2 3

1 2

3

1 3

2

1

2

3

6

Young Projectors

The Young tables or Young tableaux are associated to symmetrization along lines and antisymmatrization along columns. In this way one projects onto irreps

123

symmetrizerS

6

1 2 3

12 13 13 12A S S A 1 3

2

13 12A S1 2

3

123 12 13 23 12 13 13 12

(1,2,3) [1 ] (1,2,3)S f P P P P P P P f

12 13 12

(1,2,3) [ (1,2,3) (3,2,1)] (1,2,3) (3,2,1) (2,1,3) (3,1,2)A S f A f f f f f f

13 12 13

(1,2,3) [ (1,2,3) (2,1,3)] (1,2,3) (2,1,3) (3, 2,1) (2,3,1)A S f A f f f f f f

7

Rule: first, symmetrize.

There are two tables with mixed permutation symmetry

(i.e. not fully symmetric or antisymmetric) due to degeneracy 2 of the irrep E. One can show that this is general. In the Young tables for

S(N), the m-dimensional irreps occur in m differentt tableaux.

7

123A

1

2

3

123 12 13 23

12 13 23 12 13 13 12

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

[1 ] (1,2,3)

A f P P P f

P P P P P P P f

Antisummetrizer

Counting the number of diagrams can be long, but there is a shortcut

Dimension of a representation and hook-length formula: An example for SN with N=13

Hook length of a box= 1+ number of boxes on the same line on the right of it + number of boxes in the same column below it

9 7 5 4 2 1

1246

3 1

1

n=number of boxes =13

product of hook lengths .3 362880

(red from first line, blue

6

9.7.5.4.2

from seco

.. 4.2

nd line)

!dimension=number of tableaux of this shape 17160

n

10

Projection operators: verification for S(3)

3v 3 v

1

2

2C 3 6

1 1 1

1 1 1

2 1 0 ,

z

C I g

A z

A R

E x y

All operators = reflections or products of reflectionsS(3)= C3v

2

3 3

2

3 3

2

3 3

2

3 3

2

3 3

2

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

C E C

E C C

C E C

C C E

a

b

cx

x

y

3

2

3 3

irrep :

1 0( ) ( ) ( )

0 1

1 0( ) ( ) ( )

0 1

2 1 2 3cos( ) , sin( )

3 2 3 2

v

a c b

E of 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

10

11

* 2

3 3

2 1 1( ) 1

6 2 2 E E

xx xx a b c

R

P D R R C C

a

b

cx

x

y

2

3 3

exchanges b and c; reflections are exchanges.a

a b a cC C

11

Projection on component x of E

Now write rotations in terms of reflections

11

2 2

(1 )(1 )2

E axx b c a b c

b c

a

P

In C3v reflections transpositions, i.e., exchanges.

12

23 23

Now we can introduce symmetrizer and antisymmetrizer

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

2 2 2 2

a aP PS b c S A b c A

In S(N), transpositions are the basis of the regular representation. So weare projecting from the regular representation to irreps of S(3).

Similar rules apply for S(N)

Recall: the permutations of N objects are the basis of the Regular

Representation of S(N)

12

23 12 13( )E

xxP A S S 1 3

2

1 2

3

11

2 2

(1 )(1 )2

E axx b c a b c

b c

a

P

13

1

3

4

2 1

2

4

3 1

2

3

4

3d irrep1d irrep

1 2 3

4

1 2 4

3

1 3 4

2

3d irrep

1

2

3

4

1 2 3 4

1d irrep

Young Tableaux for S(4)

1 2

3 4

1 3

2 4

2d irrep

conjugate representations (conjugate diagrams) are obtained from each other by exchanging rows with columns.

The number of tableaux for each diagram is the degeneracy of the irrep.

14

24 13 34 12

1 2Example: projection onto

3 4

1 2 P( )=

3 4A A S S

24 13 34 12 24 131 2 3 4 1 2 1 2 3 4 3 4 A A S S A A

24

Antisymmetrize on 1 and 3 and get

( 1 2 1 2 3 4 3 4

3 2 3 2 1 4 1 4 )

A

Antisymmetrize on 2 and 4 and get the final result:

1 2 1 2 3 4 3 4

3 2 3 2 1 4 1 4

1 4 1 4 3 2 3 2

3 4 3 4 1 2 1 2

15

Young tableaux and spin eigenfunctions

Consider the eigenstates | S,MS > obtained by solving the eigenvalue

problems for S2

and Sz. Several eigenstates of S

2and S

zwith the

same quantum numbers can occur.

Any permutation of the spins sends an |S,MS

>

eigenfunction into a linear combination of the

eigenfunctions with the same eigenvalues S,MS; in other

terms, the S,MS

quantum numbers label subspaces of

functions that do not mix under permutations.

15

Example: N=3 electron spins

23=8 states, maximum spin = 3/2 4 states.

Hilbert space: 1 quartet and 2 doublets.

Within each permutation symmetry subspace, by a technique based on

shift operators we shall learn to produce S

and MS

eigenfunctions that besides bearing the spin labels also form a

basis of irreps of S(N).

invariantThe reason is that is under permutations

of .

i

i

i

S S

S

We can use the example of N=3 electron spins

23=8 states, maximum spin = 3/2 4 states.

Hilbert space: 1 quartet and 2 doublets.

17

With MS

= 3/2

the only state is quartet | ↑↑↑ >, which is invariant for any permutation.

Acting on | with S

1

get | 3/2 , ½> = (| > +| > + | > ).3

This is invariant for spin permutations, too, and belongs to the A1 irrep of S(3).

The (total-symmetric) shift operators preserve the permutation symmetry, and all the

2MS

+1 states belong to the same irrep.

17

Acting again with we get

1 | 3/2 , -½> = (| > +| > + | > )

3

| 3/2 , -3/2> = | >

S

The ortogonal subspace involving one down

spin yields two different doublets with MS=1/2

1 1 1 1 1 1, , ,

2 2 2 22 2

We can orthonormalize the doublets :

1 1 1 1 1 1, , , 2

2 2 2 22 6

Why two? What good quantum number distinguishes these two states ?It is the permutation symmetry, which admits a degenerate representation.

18

1

1 1The squartet | 3/2 , > = (| > +| > + | > )

2 3

|involves | > ,| > , | > according to A of S(3).

Out of these we can also build a 3d subspace with  Ms 1/ 2.

two-d subspace is orA1

thogonal to | 3/2 , >.2

19

3we recognize irrep of ( odd for 2 3)

1 1, , , 2

2 6

within subspace with  Ms 1/ 2

vE C x

E x E y

Moreover, by the spin shift operators each yields

its |1/2 ,−1/2 > companion.

A quarted and 2 doublets exhaust all the 23

states

available for N=3, and there is no space for the A2

irrep.

This is general:

since spin 1/2 has two states available, any spin wave function

belongs to a Young diagram with 1 or 2 rows.

19

Looking at the doublets :

1 1 1 1 1 1, , , 2

2 2 2 22 6

1

32

x

y

Conclusion: a system consisting of N spins 1/2.

The set of spin configurations, like

.... can be used to build a

representation of the permutation Group S(N)

One can build spin eigenstates by selecting the

number of up and down spins accoding to Ms

and then projecting with the Young tableaux.

21

Problems

4 2 4

1

2

2 2

1

2

2 2 2 8

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

2 1 0 0 0 ( , )

v v d

z

C I C C g

A z

A R

B x y

B xy

E x y

For the CuO4 model cluster

(1) Find the irreps of the one-electron orbitals.

(2) Consider this cluster with 4 fermions, in the Sz = 0 sector. Classify the

4-body states with the irreps of the Group.

(1)

Consider the Group operators acting on the basis of atomic orbitals (1,2,3,4,5). Atoms that do not move contribute 1 to the character. The characters of the

representation Γ(1) with one electron are

χ(E) = 5, χ(C2) = 1, χ(2C4) = 1, χ(2σv) = 3, χ(2σd) = 1. Applying the

LOT one finds Γ(1) = 2A1 + E + B1.

1

2

5

3

4

21

CuO4

22

1

2

5

3

4

25 5 5

choices of i,j choices of k,l 100 configurations2 2 2

C2: (4, 2, 4, 2), (4, 2, 5, 3), (5, 3, 4, 2), (5, 3, 5, 3) are

invariant, +1 to character each

C4: none is invariant

4 Fermion case

4 2 4

1

2

2 21

2

2 2 2 8

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

2 1 0 0 0 ( , )

v v d

z

C I C C g

A z

A R

B x y

B xy

E x y

invariant configurations:

22

Basis: (i,j,k,l) i j k l

4The Model ClusterCuO

23

1

2

5

3

4

4 2 4

1

2

2 21

2

2 2 2 8

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

2 1 0 0 0 ( , )

v v d

z

C I C C g

A z

A R

B x y

B xy

E x y

23

Basis: (i,j,k,l) i j k l

  2, 1, 5,3 , 2, 1, 3,5 2, 1, 5,3

change sign since order requires exchange of creation operators; also,

4, 1, 5, 3 , 4, 2, 5, 3 , 5, 3, 2, 1 , 5, 3, 4, 1 , 5,3, 4, 2 -1 each

x :   2, 1, 2, 1 , 2, 1, 4, 1 , 2, 1, 4, 2 , 4, 1, 2, 1 , 4, 1,4, 1 , 4, 1, 4, 2 ,

4, 2, 2, 1 , 4, 2, 4, 1 , 4, 2, 4, 2 , 5, 3, 5, 3 invariant, 1 each

24

ne

: (5, 2, 5, 2), (5, 2, 4, 3), (4, 3, 5, 2), (4, 3, 4, 3) invariant

1

2

5

3

4

4 2 4

1

2

2 21

2

2 2 2 8

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

2 1 0 0 0 ( , )

v v d

z

C I C C g

A z

A R

B x y

B xy

E x y

( ) *1( ) ( )i

i

R GG

n R RN

1 2 1 24 15 11 24 13 13A A E B B

(4) 100 4 0 4 4

4 2 4

1

2

2 21

2

2 2 2 8

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

2 1 0 0 0 ( , )

v v d

z

C I C C g

A z

A R

B x y

B xy

E x y

24

25

5 4The Model ClusterCu O

Classify the 4-holes states in the Sz=0 sector by the C4v irreps

(4) 1296 16 0 64 16

4 2 4

1

2

2 21

2

2 2 2 8

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

2 1 0 0 0 ( , )

v v d

z

C I C C g

A z

A R

B x y

B xy

E x y

1 2 1 2(4) 184 144 320 176 152A A E B B

Basis: (i,j,k.l)= Ii+j+k-m-|

29

12962

configurations

Symmetric Group and many-electron states

1

1

Let , , =N electron amplitude depending on space coordinates only,

We take , , component i of irrep of S ,m times degenerate.

Then if permutation S

i N

i N N

N

x x

x x

P

1 1 , , , , ( ).

m

i N j N ji

j

P x x x x D P

1

1

Now let , , =N electron amplitude depending on spin coordinates only.

We take , , component q of irrep of S ,m times degenerate.

Then if permutation S

q N

q N N

NP

1 1 , , , , ( ).

m

q N n N nq

n

P D P

1 1

The full electron wave function must be of the form:

, , , , , with α and β same degeneracy, such that

with P ( 1) .

m

k N k N

k

P

x x

26

1 1

1 1

1 1

, , , , ( )

, , , , ( )

, , , , ,α and β same

Putting all together:

degeneracy

m

i N j N ji

j

m

q N n N nq

n

m

k N k N

k

P x x x x D P

P D P

x x

1 1

1 1

1 1

with P [ , , ][ , , ]

, , ( ) , , ( )

, , , , ( ) ( )

m

k N k N

k

m m m

j N jk n N nk

k j n

m m m

j N n N jk nk

j n k

P x x P

x x D P D P

x x D P D P

1

1

one finds that P ( 1) needs: ( ) ( ) ( ) .

This is true if ( ) ( ) ( ) because then

( ) ( ) ( ) ( ) ( ) ( ) ( )

mP P

jk nk nj

k

P

jk kj

m mP P

jk nk kj nk nj

k k

D P D P

D P D P

D P D P D P D P D E

27

1Recall the Theorem: ( ) ( ) ( ) if conjugate irreps.P

jk kjD P D P

Spin eigenfunctions can have Young tableaux of 1 or 2 lines since the spin states are only 2. Here is a possible [N-M,M] tableau:

a1 a2

b1 bM

… … … … aN-M

b2 …

aN-M

b2a2

a1 b1

…

…

…

bM

… …Then this is the [2M,1N-M] conjugate tableau:

28

2929

Irreducible Tensor Operators

We can consider (x1, x2, x3) as a set of functions or as the

components of an operator, actually the two rules differ by a

matter of notation.

( ) if we think of functionsi i ik kx Rx D R x

†

†

if we think of operators, then

just halts the action of R on the right.

i ix Rx R

R

But the linear combinations are the same!

†

†

†

ˆ ˆLet , wavefunctions, suppose , some operator.

Now let R be a symmetry, 1.

ˆActing with : .

ˆSo the rules are: , ,

ˆ

ˆ .

A A

R R

R R RA R RRAR

RARR R A

Symmetry operators act on wave functions and on any other operator

3030

D matrices make a representation of G. If the representation is the

irrep α, we can speak of the irreducible tensor operator T (α)

irreducible implies that all its components are mixed by the Group

operations, and one cannot find linear combinations than are not mixed.

More generally, a tensor is a set of components that are

sent into linear combinations by every symmetry S in

Group G

The Group multiplication table is OK since

† † †: ( ) ( ) ( )i i j ji k kj ji

j k

RS T RSTS R R T D S R T D R D S

( )k kik

T D RS

† ( )i i j ji

j

T STS T D S

( x1, x2, x3 ) is a vector operator ( ) i i ik kx Rx D R x

Tensors of GL(n) in Cartesian Form

In GL(n) one considers linear transformations in n-dimensional space Rn.

A vector x transforms according to a law of the form :

' i (1,... ) x,x' Rn

n

i ij j

j

x a x n

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

' ...

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

r r r rr

T a a a T

k n k n

In the case of multi-index tensors the linear transformation law

is the same as if it were a product of coordinates

3 2 2 2

1 2 3In R , from one obtains the scalar TrT= x x x and other

combinations that transform like quadrupole tensor components.

ij i iT x x

linear

3232

Tensors of rank 2

Symmetry and antisymmetry are conserved under GL(n)

transformations!

mnjnimij TaaT

mninjmji TaaT

( ) symmetric

( ) antisymmetric

ij ij ji jm in jn im mn

ij ij ji jm in jn im mn

S T T a a a a T

A T T a a a a T

ij ij ji

ij

ij ij ji

S T TT

A T T

In general, the tensors of GL(n) are reduced into irreducible parts by

taking linear combinations according to the irreps of the permutation

Group S(r)where r=rank of T. Further reduction is possible when

considering subgroups of GL(n).

One can build symmetric and

antisymmetric tensors of rank 2 (2 indices)