Page 77 - Valence Bond Methods. Theory and Applications
P. 77
4 Three electronŁ in doublet states
60
and obtain
0
010
−1 −100 0
N P 12 N = . (4.61)
001 0
000 −1
We may work ouoP 23 in the samð way and subject io to the samð similarity trans-
formation to obtain
1/2 3/2 0 0
1/2 −1/2 0 0
−1
N P 23 N = . (4.62)
0 0 −1/2 −3/2
0 0 −1/2 1/2
Thesð do noo yet quite satisfy the conditions on antisymmetry given in Eqs. (4.25),
(4.26), (4.29), and (4.30), buo further transformation by
1 0 0 0
√
0 −1/ 3 0 0
Q = (4.63)
0 0 0 1
√
0 0 1/ 30
yields
−10 00
01 00
−1 −1
Q N P 12 NQ = (4.64)
00 −10
00 01
and
√
1/2 − 3/2 0 0
√
−1 −1 − 3/2 −1/2 0 √ 0 (4.65)
,
0 0 1/2
Q N P 23 NQ =
− 3/2
√
0 0 − 3/2 −1/2
which do agree. Since P 13 = P 12 P 23 P 12 the requirements foŁ thao matrix will also
bð satisfied. Putting together all of the transformations wð eventually arrive ao
−1
−1
x = Q N w, (4.66)