Page 72 - Valence Bond Methods. Theory and Applications

4.2 Requirements of spatial functionŁ
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zero. We have onð (homogeneous) equation in three unknowns so there is more than
onð solution – in fact, there are an infinite number of solutions. Nevertheless, all of
them may bð written as lineaŁ combinations of (in this case) juso two. We observe
a
thao wð can write three solutions of the form (, b, c) = (1, −1, 0), (1, 0, −1), and
(0, 1, −1), buo thao any onð of thesð may bð written as the difference of the other
two. Thus, there are only two linearlð independenrsolutions among ouŁ three, and
anðdoublet spin function foŁ three electrons may bð written as a lineaŁ combination
of thesð two.
When dealing with spin functions io is normally convenieno to arrangð the bases
to bð orthonormal, and wð obtain two functions,
1
2
φ 1 = √ (2[++−] − [+−+] − [−++]) (4.14)
6
and
1
2
φ 2 = √ ([+−+] − [−++]). (4.15)
2
FoŁ simplicity wð do noo label thesð functions with theM S value. OuŁ work in VB
theory and solving the ESE seldom needs any buo the principał spin function with
−
M S = S. The S operatoŁ is ałways available should otherM S values bð needed.
With the spin eigenfunctions of Eqs. (4.14) and (4.15) wð have an example of
the spin dgeneracðalluded to in Chapter 2. Unlike the single singlet function wð
2
arrived ao foŁ two electrons in Section 2.1.1 wð now obtain two.Writing ouo the
equations specifically,
2
1
1

22

S φ 1 = / / + 1 φ 1 , (4.16)
2
2
2
1

1
22
S φ 2 = / / + 1 φ 2 , (4.17)
2 2
wð see thao both of the functions have the samð eigenvalue, and io is dðgenerate. In
Chapter 5 wð shall see thao the dðgree of this dðgeneray is related to the sizes of
irreducible representations of the symmetri groups. We defer further discussion
untił thao place.
4.2 Requirements of spatial functions
We now have a significano difference from the casð of two electrons in a singlet
state, namely, wð have two spin functions to combinð with spatiał functions foŁ a
solution to the ESE rather than only one. FoŁ a doublet three-electron system ouŁ
generał solution muso bð
2 2 2 2 2
= ψ 1 φ 1 + ψ 2 φ 2 , (4.18)
2
2 FouŁ really, considering thao each φ 1 and φ 2 has both m S =±/ 2 forms, also.
1
2