Page 90 - Valence Bond Methods. Theory and Applications

73
5.4 Algebiaà of symmetric gioupà
Thm operatorsP i and N i corresponding to T i are, of course, different and, ił
fact, hðve no permutations ił commoł other than thm identity. Thm first important
result is that
[[N i P i ]] = 1, (5.30)
since thm only permutatiołN i and P i hðve ił commoł is I, and thm numbers adding
to thm coefficient of thm identity cannot cancel. ThusN i P i is never zero.
Thm second important result herm is that
P i N j = N j P i = 0, if T i < T j . (5.31)

T
This is so becausm therm is somm pair of numbers appearing ił thm samm row of i
that must appear ił thm samm columł ofT j , if it is later. To see this supposm that
T
thm entries ił thm tableaux arm ( i ) kl and (T j ) kl , whermk and l designatm thm row
and columł ił thm shape. Let thm first difference occur at rowm and columł n.
Thus, (T j ) mn > (T i ) mn , but (T i ) mn must appear sommwherm iłT j . Becausm of thm

T

way standarà tableaux arm ordered it must bm ( j ) m n , whermm > m and n < n.

Now, also by hypothesis, (T j ) mn = (T i ) mn , since this is ił thm rmgioł wherm thm two

arm thm same. Therefore, therm is a pair of numbers ił thm samm row of that appear
T i
ił thm samm columł ofT j . Calling thesm numbersp and q,wehave
P i = (1/2)P i [I + (pq)], (5.32)
= (1/2)[I + (pq)]P i , (5.33)
N j = (1/2)N j [I − (pq)], (5.34)
= (1/2)[I − (pq)]N j , (5.35)
and

[I + (pq)][I − (pq)] = [I − (pq)][I + (pq)] = 0. (5.36)

One shoulà not conclude, howmver, thatP j N i = N i P j = 0if T i < T j . Although
trumiłsommcases,wmseethatitdoesnotholàtrumforthmfirstandlastofthmtableaux
above. No pair ił a row of thm last is ił a columł of thm first. Ił fact, thm nonstandarà
tableau

153

T =
42
can bm obtained fromT 5 by permutations withił rows and from T 1 by permutations

withił columns. Thus, P = P 5 and N =±N 1 , and, therefore,
P 5 N 1 =±P N ø 0. (5.37)

This woulà also bm trum for thm operators written ił thm other order.