The utility of the direct product of irreducible representations is extensive. Lets
look, for example, at applications of the direct product in applications of quantum
mechanics. Revisiting the property of orthonormal basis sets of eigenfunctions we found
(6.7) m* n dx
A symmetry operation leaves the molecule in an indistinguishable orientation with the
same electron density e m* m( ) and energy as the original. For the case of non-
degenerate eigenvalues the effect of the symmetry operator on the wavefunction is to
leave it unchanged or at most change its sign
(6.8) O m = 1( ) m .
Moving through each of the symmetry operations in a group we arrive at an irreducible
representation for m . Thus, the wavefunctions in the integral in Eq. (6.7) can be
replaced by one of the irreducible representations for the group. In Chapter 2 we found
that, at least for one-dimensional problems, the integrand m* n had to be an even
function of x otherwise the integral would vanish. The test for an even function was that
upon the inversion of the x-coordinate the function remained the same, i.e. ,
f (x) = f (x) , and f (x) is symmetric with respect to inversion of x . The same holds
true for the molecular wavefunctions in that m* n must be symmetric with respect to
every operation in the group, i.e., the totally symmetric representation A1, otherwise the
integral in Eq. (6.7) vanishes. Since m* n is a direct product of irreducible
representations the direct product must be A1 symmetry or vanish. The only direct
product of irreducible representations that is A1 symmetry is a product of two identical
representations, e.g. B1 x B1 = A1. If instead m* = B1 and n = B2 ,
then the direct
product is B1 x B2 = A2, is antisymmetric with respect to xz and yz, and the integral is
odd and vanishes. Thus, replacing the wavefunctions in the integrand with their
irreducible representations results in mn as usual. That is, the irreducible representations
in a character table make up an orthonormal basis set for the molecule.
Other integrals that we frequently encounter in quantum mechanics are those
where an operator splits the two wavefunctions as in the expectation value for the energy
E = m* H md . For the result to be non-vanishing (or even) we again require the
integrand to be totally symmetric with respect to all operations of the symmetry group.
We need the reducible representation for the Hamiltonian operator H . Because the
energy of a molecule will not change with a symmetry operation that leaves the molecule
in an indistinguishable orientation H must transform as the totally symmetric
representation. One way to think about this is to realize that it is the Hamiltonian that
decides the atomic arrangement in the molecule that minimizes the energy and that
arrangement is the one we looked at to decide what symmetry elements the molecule had.
So, the Hamiltonian has each of those symmetry elements as well and belongs to the
totally symmetric representation. For the direct product m* H m to be totally symmetric
requires again that m* and belong to the same irreducible representation.
Symmetry and transition moments
An important integral used in spectroscopy to determine the intensity of light
absorption or emission is the transition moment
(6.9) Mmn = m* n d .
On absorption or emission of light the energy of a molecular system changes by
E ' =
E when the dipole moment of a molecule
projects onto the electric field of
the light and work is done [Eq. (5.1)]. The wavefunction of the molecule jumps from the
initial state, n , to a final state m . The intensity of the transition is proportional to the
transition moment. For a transition to be dipole-allowed the direct product m* n in the
integrand of Eq. (6.9) must again transform as the totally symmetric species in the
molecular point group. Therefore we need to know the irreducible representation for the
dipole moment operator . The dipole moment for a molecule will be charge q times the
vector distance from a fixed point d with all the charges in the molecule added up
= qixii + qiyi
i + qizi = x
i + y + z .
For m* n to transform as the totally symmetric irreducible representation in the point
group at least one of the dipole components x ,z or z will have to belong to the same
irreducible representation as m* n .
(6.11) Mmn = m* n d = qi m* xi n + m* yi n + m* zi n
Since the charges qi are constants we require x, y, or z to transform as the same
irreducible representation as the direct product m* n to assure that at least one of the
integrals in Eq. (6.11) is totally symmetric (even) and the transition from n to m
Lets look at the infrared absorption by the molecular vibrations in water as an
Water has three atoms and with three degrees of freedom on each atom (an x, y, and z
coordinate system at each atom) we have a total of nine degrees of freedom. Of these,
three are translation of the entire molecule along the x-, y-, and z-coordinates and three
are rotations about
the x-, y-, and z-
Fig. 6.8 Double-click figure to observe rotations of water.
The remaining three degrees of freedom are the vibrations of water.
Fig. 6.9 Double-click figure to observe vibrations of water.
To get a better sense of the symmetry and infrared activity of the three vibrations of water
we draw the vibrations with arrows to indicate the displacement from equilibrium (Fig.
symmetric stretch bend asymmetric stretch
Fig. 6.10 Schematic representations of displacement from equilibrium for each of the
three vibrations in water.
A look at the wavefunctions for the harmonic oscillator (Table 3.2) reveals that the
wavefunction for the ground state harmonic oscillator 0 has a maximum probability
density 0* 0 at the undisplaced equilibrium position (x = 0) while the first excited state
1 is displaced away from equilibrium.
- 3. x 10- 11- 2. x 10- 11- 1. x 10- 11 1. x 10- 11 2. x 10- 11 3. x 10- 11
V(x) in eV
V(x) = 1/2 kx2
Harmonic Oscillator Probability Density
Fig. 6.11 Undisplaced (x = 0) ground state harmonic oscillator probability density
0* 0 and displaced first excited state harmonic oscillator probability density
1* 1 .
Thus, if we are interested in whether the transition from the ground vibrational state
n = 0 to the first excited vibrational state m = 1 is dipole allowed then we must
determine the irreducible representations for the undisplaced molecule and for the
molecule after it is displaced along the arrows in Fig. 6.10. The ground state
wavefunction n = 0 is water with each of its atoms in the equilibrium position, i.e., the
same arrangement we looked at to consider the symmetry elements of water. The ground
state vibrational wavefunction n = 0 belongs therefore to the totally symmetric
representation A1 of the C2v symmetry group. This will be true for the ground state
wavefunctions for all vibrations in all molecules, not just water. The ground state
vibrational wavefunction always belongs to the totally symmetric representation. The
symmetry of the excited state wavefunction, however, depends on the particular
displacement. We can learn the symmetry of the displaced water molecule by
considering the symmetry of the arrows in Fig. 6.10. For the symmetric stretch vibration,
the arrows are transformed into indistinguishable arrangement by each of the symmetry
elements in C2v. That is, the first excited state of the symmetric stretch vibration in water
is transforms as the totally symmetric irreducible representation A1. The transition
moment for the vibrational excitation of the water symmetric stretch becomes
(6.12) Mmn = m
* n d = A1 A1 d = qii A1 x A1 d = qi
i A1 B1 A1 d = 0 .
i A1 y A1 d = qi
i A1 B2 A1 d = 0 .
i A1 z A1 d = qi
i A1 A1 A1 d 0 .
and will be non-vanishing only if a component of , x, y, or z, is also A1. We see from
the character table for C2v (Table 6.8) that z transforms as A1. Therefore, the symmetric
stretch for water is said to be infrared active and z-polarized.
Looking now at the bend motion we realize that the arrows indicating
displacement in the excited vibrational state again transform as A1 because an
indistinguishable arrangement of arrows results from each operation in the group. The
bend is therefore similarly IR-active and z-polarized.
Considering finally the asymmetric stretch of water the behavior of the
displacement arrows with the C2z, yz, and xy operations is shown in Fig. 6.12.
Fig. 6.12 Operation of C2z, yz, and xy on the asymmetric stretch of water.
The asymmetric stretch is antisymmetric with respect to the C2z and xy operations but is
symmetric with respect to the yz operation, matching the B1 irreducible representation of
the C2v poin