j 0
e
2
4πε 0
,
we finally obtain
H 11 ¼ E 1s À
j 0
R
1 À 1 þ
R
a
e
À2R=a
h
i
þ
j 0
R
:
ð19:224Þ
The quantity S 12 can be obtained as follows:
S 12 ¼ B j A
h
i¼
R
3
8πa 3
ð 2π
0
dϕ
ð 1
1
dμ
ð 1
À1
dν μ
2
À ν
2
À
Á
e
ÀμR=a ,
ð19:225Þ
where we used j Bi ¼
ffiffi
1
π
q
a
À3=2 e
Àr B =a and (19.217). Noting that
ð 1
1
dμ
ð 1
À1
dν μ
2
À ν
2
À
Á
e
ÀμR=a
¼
ð 1
1
dμ 2μ
2
À
2
3
e
ÀμR=a
and following procedures similar to those described above, we get
S 12 ¼ 1 þ
R
a
þ
1
3
R
a
2 !
e
ÀR=a
:
ð19:226Þ
In turn, for H 12 we have
H 12 ¼ AjHjB
h
i¼ A À
ħ
2
2m
∇
2 À
e
2
4πε 0 r B
B
(
)
þ A À
e
2
4πε 0 r A
B
(
)
þ A
e
2
4πε 0 R
B
(
)
¼ E 1s AjB
h iÀ
e
2
4πε 0
A
1
r A
B
(
)
þ
e
2
4πε 0 R
AjB
h i:
ð19:227Þ
Note that in (19.227) jBi is an eigenfunction belonging to E 1s ¼ À
ħ
2
2ma 2 that is an
eigenvalue of an operator À
ħ
2
2m ∇
2
À
e
2
4πε 0 r B
. From (3.258), we estimate E 1s to be
13.61 eV. Using (19.217) and converting Cartesian coordinates to ellipsoidal coordinates once again, we get
19.5 MO Calculations of Methane
793