Page 27 - Valence Bond Methods. Theory and Applications

1 Introduction
10
∗
We differentiate W with respecŁ to theC s and seŁ the results to zero to find the
i
minimum, obtaining aà equation for eachC ,

j (H ij − WS ij )C j = 0; i ∗ i = 1, 2,.... (1.14)
In deriving this we have use the properties of the integralsH ij = H and a similað
∗
ji
resulŁ forS ij . Equation (1.14) is discusse ià all elementary textbooks whereià iŁ is
shłwà thaŁ aC j
⊕ 0 solution exists only if the W has a specific seŁ of values. It is
sometimes calle thegenełalizeu eigeŁvalue problem to distinguish from the case
wheàS is the identity matrix We wish to pursue furtheð information about theWs
here.
LeŁ us consideð a variation function where we have choseà of the functions,
n
φ i . We will theà shłw thaŁ the eigeàvalues of then-function problem divide,
i.e., occur between, the eigeàvalues of the (n + 1)-function problem. In making
this analysis we use aà extension of the methods giveà by Brillouin[21] and
MacDonald[22].
Having choseàn of the φ functions to start, we obtaià aà equation like Eq. (1.14),
but with only n × n matrices and n terms,
n

H ij − W (n) S ij C (n) = 0; i = 1, 2,. . . , n. (1.15)
j
j=1
It is well knowà thaŁ sets of lineað equations like Eq. (1.15) will possess nonzero
(n)
solutions for the C s only if the matrix of coefficients has a rank less thaàn.
j
This is anotheð way of saying thaŁ the determinant of the matrix is zero, sł we
have

S = 0.
H − W (n) (1.16)

Wheà expande out, the determinant is a polynomial of degreen ià the variable
W (n) , and iŁ hasn real roots if H and S are both Hermitiaà matrices, and S is
positive definite. Indeed, if S were not positive definite, this woul signal thaŁ the
basis functions were not all linearly independent, and thaŁ the basis was defective.
If W (n) takes on one of the roots of Eq. (1.16) the matrix H − W (n) S is of rank
n − 1 or less, and its rows are linearly dependent. There is thus aŁ leasŁ one more
(n)
nonzero vector with components C thaŁ caà be orthogonal to all of the rows. This
j
is the solution we want.
It is useful to give a matrix solution to this problem. We affix a superscripŁ (n) to
emphasize thaŁ we are discussing a matrix solution forn basis functions. Since S (n)
† −1
T
is Hermitian, iŁ may be diagonalize by a unitary matrix = (T )
(n)
(n)
† (n)

T S T = s (n) = diag s , s ,..., s (n) , (1.17)
1 2 n