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

S(3) and C3v 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

bc

x

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 (do you recall Burnside?).

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

Do this in all possible ways

To represent this particular partition 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 irreducible representations are said conjugate, like these. They have the same m.

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

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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

Rule: first, symmetrize.

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Rule:

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 different tableaux;

In other terms, there are m ways to put integers in the diagram following the rules.

7

8

In S3 there is an antisymmetric irrep

8

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

Antisymmetrizer

Counting the number of tableaux (=m of the irrep) can be very long!

9 7 5 4 2 1

1246

3 1

1

To count the number of tableaux there is a shortcut

dimension=number of tableaux of this shape 17160

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

11

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.

12

24 13 34 12

1 2Example: projection a product onto .

3 4

1 2 The projection operator is P( )=

1

3 4

2 3 4 of

A 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

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Young tableaux and spin eigenfunctions

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

eigenvalue problems for S2 and Sz. Several eigenstates of S2 and Sz with 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. 13

Example: N=3 electron spins

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

Therefore, Hilbert space consists of 1

quartet and 2 doublets. Which quantum

number distinguishes the 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.

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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.

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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.

16

1

1 1The quartet | 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 ortA1

hogonal to | 3/2 , >.2

17

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

17

1 Looking at the doublets with M = :

2

1 1 1 1 1 1, , , 2

2 2 2 22 6

S 1

32

x

y

Starting with two up spins and a down spin, we could have obtained

these states direcly by projecting on the permutation symmetry

eigenstates by the use of the Young theory.

18

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.

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1 Looking at the doublets with M = :

2

1 1 1 1 1 1, , , 2

2 2 2 22 6

S 1

32

x

y

Conclusion:

Consider 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 the z

component of spin and then projecting with the

Young tableaux.

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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

20

CuO4

21

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:

21

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

4The Model ClusterCuO

22

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

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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

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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

23

24

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

25

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

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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:

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