permutation group s(n) and young diagrams c v...
TRANSCRIPT
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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
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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
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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.
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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
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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.
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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
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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
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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
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.
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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
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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:
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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
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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-|
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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
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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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