Taking the derivatives of Eq ( 20.6 ) with respect to c1 and c2, we get the following two equations.
(
1 -E )c1 + ( - ES ) c2 = 0
(20.8a)
( - ES)c1 + ( 2 - E) c2 = 0
(20.8b)
To solve (20.8) the determinant called the secular determinant of equation 20.9 has to be zero.
(20.9)
For identical atoms
= and the two roots of the quadratic equation (20.9) for E ( which give Eb and Ea ) are
(20.10a)
and Ea =
(20.10b)
Substituting Eb in eq (20.8) we get c1 = c2 for the bonding orbital and substituting Ea in eq (20.8) we get c1 = -c2 for the antibonding orbital. Thus the coefficients in 20.1 can be determined.
We can extend the argument to
orbitals of ethylene by taking
S O in eq (20.9) and
The secular determinant becomes
(20.11)
and the energies of the bonding and antibonding orbitals are
Eb = and Ea =
(20.12)
Here
is negative and hence the bonding orbital lies lower in energy compared to the antibonding orbital.
1 -E )c1 + ( 
- ES ) c2 = 0 
- ES)c1 + ( 
2 - E) c2 = 0 
= 
and the two roots of the quadratic equation (20.9) for E ( which give Eb and Ea ) are 
orbitals of ethylene by taking
S 
O in eq (20.9) and
The secular determinant becomes 
and Ea = 
is negative and hence the bonding orbital lies lower in energy compared to the antibonding orbital. 
