prev

next

out of 17

View

230Download

1

Embed Size (px)

Vanishing integrals

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

[Eq. (2.18)]

(6.7) m* n dx

= mn

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

O

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

n

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

(6.10)

= qi

di

i

all charges

= 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

i

all charges

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

dipole-allowed.

Lets look at the infrared absorption by the molecular vibrations in water as an

example.

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-

coordinates.

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.

6.10).

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

0.5

1.0

V(x) in eV

x (m)

V(x) = 1/2 kx2

Harmonic Oscillator Probability Density

0

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 .

+ qi

i A1 y A1 d = qi

i A1 B2 A1 d = 0 .

+ qi

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