Physical Chemistry , 1st ed.

The minimization condition represented by equation 12.26 must now be
fulfilled according to variation theory. Because of the substitutions made to ar-
rive at equation 12.29, the derivatives with respect to ca,1and ca,2contain a lot
of terms but are relatively straightforward to determine. We would find (after
substituting for the expression for Eitself in the expression for the derivative):







c

E

a,1

 0 (H 11 ES 11 )ca,1(H 12 ES 12 )ca,2 0

(12.30)







c

E

a,2

 0 (H 21 ES 21 )ca,1(H 22 ES 22 )ca,2 0

We get two equations in terms of the coefficients ca,1and ca,2, which also have
as part of them the energy integrals H 11 ,H 12 (which equals H 21 ), and H 22 , and
also the overlap integrals S 11 ,S 12 (which equals S 21 ), and S 22. In order for both
derivatives to be zero, each of the equations 12.30 must be satisfied at the same
time. We must solve these two simultaneous equations.You should recognize
that we would have more equations to solve simultaneously if we were using
an example having more terms, and therefore more constants ci,jin it.
For a set of simultaneous equations that are all equal to zero (like equations
12.30), there are mathematical ways of finding solutions. Linear algebra allows
for two possibilities. The first is that all of the constants, in this case ca,1and
ca,2, are exactly zero. Although this possibility would satisfy the equations
12.30, it provides a useless, or trivial, solution (0 exactly: a wavefunction
that has been rejected previously for its uselessness). The other possibility can
be defined in terms of the coefficients on the c’s in equations 12.30, the ex-
pressions involving the H’s and the S’s. Linear algebra allows for a nontrivial
solution of the simultaneous equations 12.30 if the determinant formed from
the coefficient expressions of equations 12.30 is equal to zero:



H 11 ES 11 H 12 ES 12

H 21 ES 21 H 22 ES 22 ^0 (12.31)

The above determinant is called a secular determinant. Linear variation theory
rests on equation 12.31: if the secular determinant formed from the energy and
overlap integrals and the energy eigenvalues (which are the unknowns!) is
equal to zero, then the equations 12.30 will be satisfied and the energy will be
minimized.
There seems to have been a change of focus. First we were concentrating on
the coefficients ca,1and ca,2, and now we are concerned with a determinant in
terms of integrals and energies. This is simply a consequence of linear algebra.
Do not be confused by this change in focus. The minimum energy of the sys-
tem is still the ultimate goal. Recognize, however, that the 2 2 determinant
in equation 12.31 can easily be evaluated but will yield an equation that has a
term in E^2. Therefore, in solving for E, we will get two answers (using the qua-
dratic formula). The ground-state energy is the lower of the two. Generally, if
one has a trial function ithat has nexpansion coefficients, one will get an
n nsecular determinant of the form



H 11 ES 11 H 12 ES 12 ... H 1 nES 1 n



H 21 ES 21 H 22 ES 22 ... H 2 nES 2 n

... ... ... ... (12.32)

Hn 1 ESn 1 Hn 2 ESn 2 ... HnnESnn

 0

where Eis the unknown. (Remember,Hijand Sijare the energy and overlap
integrals in terms of basis functions. The values of these integrals should be

12.8 Linear Variation Theory 399