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5 Advanced methodà foi laiger molecules
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We conclude that one can find a number of right coset generators giving distinct
cosets until thm permutations ofS n arm exhausted. Symbolizing thm right coset
generators as τ 1 = I,τ 2 ,...,τ p ,wehave
S n = G N I ⊕ G N τ 2 ⊕ ··· ⊕ G N τ p , (5.133)
wherm thm first coset isG N itself. This leads to thm often quoted result that thm order
of any subgroup must bm an intmger divisor of thm order of thm wholm group and, ił
this case, wm hðve
n
p = . (5.134)
n/2 − S
Our goal now is to find a cołvenient set of right coset generators for G N that
k
gives S n . Let us now consider specifically thm casm for thm{2 1 n−2k } partitioł with
k = n/2 − S, and thm tableau,
1 n − k + 1
. .
. .
. .
k n
.
k + 1
.
.
.
n − k
Thm order ofG N is g N = (n − k)!k!. Now let i 1 , i 2 ,..., i l bml ≤ k of thm intmgers
from thm first columł of our tableau and let j 1 , j 2 ,..., j l bm thm samm number
from thm second column. Thesm two sets of intmgers define a special permutatioł
[(i)( j)] l = (i 1 j 1 ) ··· (i l j l ), which is a product of l noninteracting binaries. Since
each binary contains a number from each column, none with l > 0 arm members of
G N . Some, but not all, arm members ofG P , howmver. Amongst all of thesm therm is
a subset that wm callcanonical ił which i 1 < i 2 <···< i l and j 1 < j 2 <···< j l .
Thm number of thesm is
n − k k
,
l l
and it is easily showł that
k
n − k k n
= . (5.135)
l l k
l=0