1 Permutation Group S(N) and Young diagrams C 3v : All operators = reflections or products of reflections S(3): All operators = permutations =transpositions (=exchanges) or products of transpositions They are isomorphous S(3)= C 3v 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 c x S(N) : order= N! huge representations but allows general analysis, with many applications. Example 3 2 3 ,, ,, (,,) (,,) (,, ) (,,) a b c E abc C bca C cab acb cba bac Permutations of Group elements are the basis of the regular representation of any Group. In C3v reflections transpositions.
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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.
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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.
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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.
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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