e S 11 ¼ ξ 1 jξ 1
h
i¼
ð
ϕ 1 þ ϕ 2
ð
Þ
à ϕ 1 þ ϕ 2
ð
Þ dτ ¼
ð
ϕ 1 þ ϕ 2
ð
Þ
2 dτ
¼
ð
ϕ
2
1 dτ þ
ð
ϕ
2
2 dτ þ 2
ð
ϕ 1 ϕ 2 dτ ¼ 2 þ 2
ð
ϕ 1 ϕ 2 dτ,
ð19:97Þ
where we used the fact that ϕ 1 and ϕ 2 have been normalized. Also following the
convention, we denote
S
ð
ϕ 1 ϕ 2 dτ ¼ S 12 ,
ð19:98Þ
where S is called overlap integral. Thus, we have
e S 11 ¼ 2 1 þ S
ð
Þ:
ð19:99Þ
Similarly, we get
e S 22 ¼ 2 1 À S
ð
Þ:
ð19:100Þ
Substituting (19.95) and (19.96) along with (19.99) and (19.100) for (19.90), we
get as the energy eigenvalue
λ 1 ¼
α þ β
1 þ S
and λ 2 ¼
α À β
1 À S
:
ð19:101Þ
From (19.97), we get
j ξ 1
j j j¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
ξ 1 jξ 1
h
i
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 1 þ S
ð
Þ
p
:
ð19:102Þ
Thus, for one of MOs corresponding to an energy eigenvalue λ 1 , we get
Ψ 1 ¼
j ξ 1 i
j ξ 1
j j j
¼
ϕ 1 þ ϕ 2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 1 þ S
ð
Þ
p
:
ð19:103Þ
Also, we have
j ξ 2
j j j¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
ξ 2 jξ 2
h
i
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 1 À S
ð
Þ
p
:
ð19:104Þ
For another MO corresponding to an energy eigenvalue λ 2 , we get
19.4 MO Calculations Based on π-Electron Approximation
753