Physical Chemistry , 1st ed.

Solution
According to equation 12.31, the nontrivial solution of the simultaneous
equations found by minimizing the energy will be given by



 15 E 1  1 E 0

 1 E 0  4 E 1 ^0

where we have explicitly included the facts that, for the orthonormal wave-
functions,S 11 S 22 1 and S 12 S 21 0. This secular determinant becomes



 15 E  1 E

 1  4 E^0

which can be expanded to get the quadratic equation

E^2  19 E 59  0

Using the quadratic formula, the two possible values ofEthat will satisfy this
equation are
E 1 15.09 and E 2 3.91

What has happened is that one energy level is now slightly lower than the ideal
lowest energy level (given by H 11 ) and the other is now slightly higher than
the higher of the two ideal energy levels (given by H 22 ). The coefficients ca,1
and ca,2can also be evaluated for the lower-energy state (where E15.09):



c

c

a

a

,

,

1

2

 

0.

1

09



0.09ca,1ca,2

So to normalize the wavefunction:



j

c^2 i,j 1 c^2 a,1c^2 a,2

c^2 a,1(0.09c^2 a,1) 1
ca,10.996
and accordingly,
ca,20.0896

The complete wavefunction for the ground state, where E15.09 (which
is the more negative, or lower,energy), is

a0.996a,10.0896a,2

The coefficients for the first excited state are given by using E3.91
(which is the higherof the two energy values) in either of the two ratios in
equation 12.33:



c

c

a

a

,

,

1

2







4

1





(

(





3

3

.

.

9

9

1

1

)

)

(

(

1

0

)

)



0.

1

09



ca,10.09ca,2

Normalizing provides the following values for the two coefficients:

ca,10.0896 and ca,20.996

The wavefunction whose energy is 3.91 is

a0.0896a,10.996a,2

 1 (15.09)(0)



 15 (15.09)(1)

12.8 Linear Variation Theory 401