Physical Chemistry , 1st ed.

Example 13.11

Determine whether the following integrals are exactly zero or might be

nonzero strictly from symmetry considerations.

a.*A 1 A 2 din a Tdmolecule

b.A* 2 Edin a C3vmolecule

c.E*Edin a D2dmolecule

Solution

In each case, the characters of the irreducible representations must be multi-

plied together and the product evaluated for the presence of the totally sym-

metric representation of the respective point group.

a.In Td,A 1 A 2 A 2 , which is not the totally symmetric representation.

Therefore the integral must be exactly zero.

b.In C3v,A 2 EE, which is not the totally symmetric representation.

Therefore the integral must be exactly zero.

c.In D2d,EEyields a set of characters

E 2 S 4 C 2 2 C 

2 2 d

EE 4 0 4 0 0

This is not an irreducible representation and so must be broken down using

the GOT. One can show by applying equation 13.6 that

EEA 1 A 2 B 1 B 2

which does contain A 1. Therefore this integral may be nonzero. This is not a

guarantee that it is nonzero, but there is no symmetry reason requiring that

it be exactly zero.

As in the above example, simple symmetry considerations easily show many

integrals of wavefunctions to be exactly zero. This idea can be simplified further

for some conditions. For the product of only two irreducible representations,

the integral will be exactly zero unless both functions are of the same symmetry

species.In integrals that are combinations of more than two functions (that is,

two wavefunctions and an operator), the combination of two must yield a re-

sulting symmetry species that is the same as the third. Otherwise, the integral

is exactly zero. In showing this, it does not matter which two symmetry species

are combined. Thus, if symmetry considerations are applied first, many inte-

grals can be shown to be identically zero and will not even have to be evaluated.

Example 13.12

For a hydrogen atom whose symmetry is described by the Rh(3)point group,

the 1sorbital has the irreducible representation label Dg(0).The 2sorbital also

has the irreducible representation label Dg(0). The electromagnetic (EM) radi-

ation operator that causes an electron to go from the 1sstate to any other

state has the irreducible representation Du(1). Show that the integral

* 1 s(EM radiation operator)  2 sd

is exactly zero, implying that this electronic transition will not occur. You do

not need to consider the symmetry properties of the form ofd. You will have

to consult the character table for the Rh(3)point group.

442 CHAPTER 13 Introduction to Symmetry in Quantum Mechanics