P
À1 HP ¼ Q
À1 CQ or PQ
À1
À
Á À1 H PQ
À1
À
Á ¼ C:
Namely, H and C are similar, i.e., the representation is equivalent. Thus, recalling
Schur’s First Lemma, Eq. (19.184) results.
Our next task is to construct SALCs, i.e., proper basis vectors using projection
operators. From the above, we anticipate that the MOs comprise a linear combination of the hydrogen-associated SALC and carbon-associated SALC that belongs to
the same irreducible representation (i.e., A 1 or T 2 ). For this purpose, we first find
proper SALCs using projection operators described as
P
α
ð Þ
l l
ð Þ ¼
d α
n
X
g
D
α
ð Þ
ll g
ð Þ
à g:
ð18:156Þ
We apply this operator to Span{H 1 H 2 H 3 H 4 }. Taking, e.g., H 1 and operating
both sides of (18.156) on H 1 , as a basis vector corresponding to a one-dimensional
representation of A 1 we have
P
A 1
ð Þ
1 1
ð Þ H 1 ¼
d A 1
n
X
g
D
A 1
ð Þ
11 g
ð Þ
à gH 1 ¼
1
24
1 Á H 1
ð
Þþ 1 Á H 1 þ 1 Á H 1 þ 1 Á H 3
ð
½
þ1 Á H 4 þ 1 Á H 3 þ 1 Á H 2 þ 1 Á H 4 þ 1 Á H 2 Þ þ 1 Á H 4 þ 1 Á H 3 þ 1 Á H 2
ð
Þ
þ 1 Á H 3 þ 1 Á H 2 þ 1 Á H 2 þ 1 Á H 4 þ 1 Á H 4 þ 1 Á H 3
ð
Þ
þ 1 Á H 1 þ 1 Á H 4 þ 1 Á H 1 þ 1 Á H 2 þ 1 Á H 1 þ 1 Á H 3
ð
Þ Š
¼
1
24
6 Á H 1 þ 6 Á H 2 þ 6 Á H 3 þ 6 Á H 4
ð
Þ
¼
1
4
H 1 þ H 2 þ H 3 þ H 4
ð
Þ :
ð19:185Þ
The case of C2s is simple, because all the symmetry operations convert C2s to
itself. That is, we have
P
A 1
ð Þ
1 1
ð Þ C2s ¼
1
24
24 Á C2s
ð
Þ¼C2s:
ð19:186Þ
Regarding C2p, taking C2p x for instance, we have
P
A 1
ð Þ
1 1
ð Þ C2p x ¼
1
24
1 Á C2p x
ð
Þ
f
þ 1 Á C2p y þ 1 Á C2p z þ 1 Á ÀC2p y
À
Á
Â
þ1 Á ÀC2p z
À
Á þ 1 Á ÀC2p y
À
Á þ 1 Á C2p z þ 1 Á C2p y þ 1 Á ÀC2p z
À
Á Š
þ 1 Á C2p x þ 1 Á ÀC2p x
ð
Þþ1 Á ÀC2p x
ð
Þ
ð
Š þ
½
½ 1 Á ÀC2p x
ð
Þ
þ1 Á ÀC2p x
ð
Þþ1 Á ÀC2p z
À
Á þ 1 Á C2p z þ 1 Á C2p y þ 1 Á ÀC2p y
À
Á Š
19.5 MO Calculations of Methane
783