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338 10MOLECULAR SYMMETRY
symmetry the projection operator is then used to construct symmetry-adapted
linear combinations (SALCs) each of which transforms as a single symmetry
species (irreducible representation). �e molecular orbitals are formed from
the overlap of these SALCs.
Solutions to exercises
E10C.1(b) As explained in Section 10C.1 on page 406 an integral can only be non-zero if
the integrand spans the totally symmetric irreducible representation, which in
D3h is A′1. From Section 10C.1(a) on page 407 the symmetry species spanned by
the integrand pxzpz is found by the forming the direct product of the symmetry
species spanned by px , z, and pz separately.�ese are read o� theD3h character
table by looking for the appropriate Cartesian functions listed on the right of
the table: x and hence px spans E′, while z and hence pz both span A′′2 . �e
direct product required is therefore E′ ×A′′2 ×A′′2 .
�e order does not matter, so this is equal to A′′2 ×A′′2 × E′. �is is equal to
A′1 × E′ because, from the second simplifying feature listed in Section 10C.1(a)
on page 407, the direct product of an irreducible representation such as A′′2
with itself is the totally symmetric irreducible representation A′1. In turn, A′1 ×
E′ is simply equal to E′ because, from the �rst simplifying feature listed in
Section 10C.1(a) on page 407, the direct product of the totally symmetric irre-
ducible representationwith any other representation is the latter representation
itself: A′1 × Γ(i) = Γ(i).�e integrand therefore spans E′.�is is not the totally
symmetric irreducible representation, therefore the integral is zero .
E10C.2(b) As explained in Section 10C.3 on page 411, an electric dipole transition is forbid-
den if the electric transition dipole moment µq ,fi is zero.�e transition dipole
moment is given by [10C.6–411], µq ,fi = −e ∫ ψ∗f qψi dτ where q is x, y, or z.
�e integral is only non-zero if the integrand contains the totally symmetric
irreducible representation, which from the D6h character table is A1g.
For a transition A1g → E2u, the symmetry species of the integrand is given by
the direct product A1g × Γ(q) × E2u.�e order does not matter so this is equal
to A1g × E2u × Γ(q), which is simply equal to E2u × Γ(q) because, from the �rst
simplifying feature listed in Section 10C.1(a) on page 407, the direct product of
the totally symmetric irreducible representation with any other representation
is the latter representation itself.�erefore A1g × E2u = E2u.
�e direct product E2u × Γ(q) contains the totally symmetric irreducible repre-
sentation only if Γ(q) spans E2u because, according to the second simplifying
feature listed in Section 10C.1(a) on page 407, the direct product of two irre-
ducible representations contains the totally symmetric irreducible representa-
tion only if the two irreducible representations are identical.�e D6h character
table shows that none of x, y or z span E2u, so it follows that the integrand does
not contain A1g and hence the transition is forbidden .
E10C.3(b) As explained in Section 10C.1(a) on page 407 the symmetry species of xyz is
determined by forming the direct product between the irreducible representa-
tions spanned by x, y and z individually. Inspection of the D2h character table