Calculating e.g., terms of
∂
∂y 0 2 and
∂
∂z 02 , we have similar results. Then, collecting all
those 27 terms, we get
s 11
2
þ s 21
2
þ s 31
2
À
Á ∂
2
∂x
2
¼
∂
2
∂x
2
,
ð19:35Þ
where we used an orthogonal relationship of S. Cross terms of
∂
2
∂x∂y
,
∂
2
∂y∂x
, etc. all
vanish. Consequently, we get
∂
∂x 0 2 þ
∂
∂y 0 2 þ
∂
∂z 0 2 ¼
∂
2
∂x
2
þ
∂
2
∂y
2
þ
∂
2
∂z
2
:
ð19:36Þ
Defining ∇
02 as
∇
0 2
∂
∂x 0 2 þ
∂
∂y 0 2 þ
∂
∂z 0 2 ,
ð19:37Þ
we have
∇
0 2 ¼ ∇
2
:
ð19:38Þ
Notice that (19.38) holds with not only the symmetry operation of the molecule,
but also any rotation operation in ℝ
3 (see Sect. 17.4).
As in the case of (19.28), we have
R∇
2
Â
à ψ r
0
ð Þ ¼ R ∇
2 ψ r
0
ð Þ
Â
à ¼ ∇
0 2 ψ
0 r
0
ð Þ ¼ ∇
2 ψ
0 r
0
ð Þ
¼ ∇
2 Rψ r
0
ð Þ ¼ ∇
2 R
Â
Ã
ψ r
0
ð Þ:
ð19:39Þ
Consequently, we get
R∇
2 ¼ ∇
2 R:
ð19:40Þ
Adding both sides of (19.29) and (19.40), we get
R ∇
2 þ V
À
Á ¼ ∇
2 þ V
À
Á
R:
ð19:41Þ
From (19.24), we have
RH ¼ HR:
ð19:42Þ
19.2 Method of Molecular Orbitals (MOs)
737