17
Vanishing integrals The utility of the direct product of irreducible representations is extensive. Let’s 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) ψ m * ψ n dx −∞ = δ mn A symmetry operation leaves the molecule in an indistinguishable orientation with the same electron density eψ m * ψ 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) ˆ Oψ m = ± 1 ( ) ψ m . Moving through each of the symmetry operations in a group we arrive at an irreducible representation for ψ 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 ψ m * ψ 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 ψ m * ψ n must be symmetric with respect to every operation in the group, i.e., the totally symmetric representation A 1 , otherwise the integral in Eq. (6.7) vanishes. Since ψ m * ψ n is a direct product of irreducible ˆ O

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Page 1: Vanishing Integrals

Vanishing integrals

The utility of the direct product of irreducible representations is extensive. Let’s

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) ψ m* ψ n dx

−∞

∫ = δmn

A symmetry operation leaves the molecule in an indistinguishable orientation with the

same electron density eψ m* ψ 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) Oψ m = ±1( )ψ m .

Moving through each of the symmetry operations in a group we arrive at an irreducible

representation for ψ 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 ψ m* ψ 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 ψ m* ψ 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 ψ m* ψ n is a direct product of irreducible

O

Page 2: Vanishing Integrals

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 ψ m* = B1 and ψ n = B2 ,

then the direct

product is B1 x B2 = A2, is antisymmetric with respect to σxz and σyz, and the integral is

odd and vanishes. Thus, replacing the wavefunctions in the integrand with their

irreducible representations results in δmn 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 = ψ m* H∫ ψ mdτ . 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 ψ m* Hψ m to be totally symmetric

requires again that ψ m* and belong to the same irreducible representation.

Symmetry and transition moments

ψ n

Page 3: Vanishing Integrals

An important integral used in spectroscopy to determine the intensity of light

absorption or emission is the transition moment

(6.9) Mmn = ψ m*∫ µψ n dτ .

On absorption or emission of light the energy of a molecular system changes by

E ' = −µ ⋅E when the dipole moment of a molecule

µ 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, ψ n , to a final state ψ m . The intensity of the transition is proportional to the

transition moment. For a transition to be dipole-allowed the direct product ψ m* µψ 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 µ . 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)

µ = qi

di

i

all charges

∑ = qixii∑ + qiyi

i∑ + qizi = µx

i∑ + µy + µz .

For ψ m* µψ n to transform as the totally symmetric irreducible representation in the point

group at least one of the dipole components µx ,µz or µz will have to belong to the same

irreducible representation as ψ m*ψ n .

(6.11) Mmn = ψ m*∫ µψ n dτ = qi ψ m

* xiψ n + ψ m* yiψ n + ψ m

* ziψ 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 ψ m*ψ n to assure that at least one of the

Page 4: Vanishing Integrals

integrals in Eq. (6.11) is totally symmetric (even) and the transition from ψ n to ψ m

dipole-allowed.

Let’s 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.

Page 5: Vanishing Integrals

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 ψ 0 has a maximum probability

density ψ 0*ψ 0 at the undisplaced equilibrium position (x = 0) while the first excited state

ψ 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)

s��s�

s��s�

V(x) = 1/2 kx2

Harmonic Oscillator Probability Density

0

Page 6: Vanishing Integrals

Fig. 6.11 Undisplaced (x = 0) ground state harmonic oscillator probability density

ψ 0*ψ 0 and displaced first excited state harmonic oscillator probability density

ψ 1*ψ 1 .

Thus, if we are interested in whether the transition from the ground vibrational state

ψ n = ψ 0 to the first excited vibrational state ψ m = ψ 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 ψ n = ψ 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 ψ n = ψ 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 = ψ m

*∫ µψ n dτ = A1∫ µ A1 dτ = qii∑ A1∫ x A1 dτ = qi

i∑ A1∫ B1 A1 dτ = 0 .

Page 7: Vanishing Integrals

+ qi

i∑ A1∫ y A1 dτ = qi

i∑ A1∫ B2 A1 dτ = 0 .

+ qi

i∑ A1∫ z A1 dτ = qi

i∑ A1∫ A1 A1 dτ ≠ 0 .

and will be non-vanishing only if a component of µ , 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, σyz, and σxy operations is shown in Fig. 6.12.

Page 8: Vanishing Integrals

Fig. 6.12 Operation of C2z, σyz, and σxy on the asymmetric stretch of water.

The asymmetric stretch is antisymmetric with respect to the C2z and σxy operations but is

symmetric with respect to the σyz operation, matching the B1 irreducible representation of

the C2v point group. Because the x-component of the dipole moment operator also

transforms as the B1 irreducible representation of the C2v the transition moment for the

asymmetric stretch is non-vanishing, IR-active and x-polarized.

(6.13)

Mmn = ψ m

*∫ µψ n dτ = qii∑ B1∫ µ A1 dτ = qi

i∑ B1∫ x A1 dτ = qi

i∑ B1∫ B1 A1 dτ ≠ 0

Page 9: Vanishing Integrals

+ qi

i∑ B1∫ y A1 dτ = qi

i∑ B1∫ B2 A1 dτ = 0

.

+ qi

i∑ B1∫ z A1 dτ = qi

i∑ B1∫ A1 A1 dτ = 0 .

We have found that all three vibrations of water are infrared active.

Now let’s consider the transition moments for the intensity of Raman scattered

light for the same vibrations. Raman scattering occurs when radiation incident upon a

molecule induces a dipole in the molecule. The intensity depends on the electronic

polarizability, α , of the molecule. Raman scattering is induced with light of wavelengths

shorter than the infrared, typically ultraviolet, visible and near infrared wavelengths. The

energy of light in this wavelength range is sufficient to perturb electron density in the

molecule, move electrons around, such that a new dipole charge arrangement can result

even if there was no dipole in the molecule before irradiation. Thus, a dipole is induced,

µind , by the electronic polarization effects of the electric field of light

E .

(6.14) µind = α

E

The induced dipole has components along the Cartesian coordinates as does the electric

field

E . Connecting these two vectors is the polarizability tensor α .

(6.15)

µx

µ y

µz

⎜⎜⎜

⎟⎟⎟=

α xx α xy α xz

α yx α yy α yz

α zx α zy α zz

⎜⎜⎜⎜

⎟⎟⎟⎟

E x

E y

Ez

⎜⎜⎜

⎟⎟⎟

A dipole can be induced along y, µy, with light polarized along z (Ez ≠ 0) if the αyz is

non-zero. The components of the polarizability tensor transform as the direct product of

the coordinates. For example, αyz transforms as the direct product of the symmetry

Page 10: Vanishing Integrals

representations for y and z. In the C2v point group (Table 6.8) y transforms as B2 and z as

A1 such that yz transforms as B2 x A1 = B2 as evidenced by the appearance of the yz direct

product in the rightmost column of Table 6.8.

The Raman intensity is proportional to the transition moment

(6.16) αmn = ψ m*∫ α ψ n dτ .

The Raman activity of each of the vibrational modes of water is found by again replacing

the wavefunctions with their irreducible representations. For the symmetric stretch and

the bend, both with A1 initial and final state wavefunctions,

(6.17) αmn = ψ m*∫ α ψ n dτ = A1∫ α A1 dτ ≠ 0 , only if α is also A1.

If a component of α can be found that transforms as A1 then the integral in Eq. (6.17) is

totally symmetric and non-vanishing. We see from Table 6.8 that the xx, yy, and zz

components of α all transform as A1 leaving these symmetric stretch and bend vibrations

Raman-active. They are also said to be polarized because an electric vector polarized

along y, for example, will induce a dipole along y.

By contrast, the asymmetric stretch in water belongs to the B1 irreducible

representation in C2v. Any Raman intensity of this mode depends upon finding a

component of α that is also B1. The xz component of α transforms as B1 (Table 6.8)

making the asymmetric stretch in water Raman-active and depolarized.

(6.18) αmn = ψ m*∫ α ψ n dτ = B1∫ α A1 dτ ≠ 0 , because the xz component of α is B1.

Thus, all of the vibrations in water are both IR-active and Raman-active.

We can extend the use of molecular symmetry to decide whether electronic

transitions are dipole-allowed. Consider, for example, the π* ←π transition in cis-

Page 11: Vanishing Integrals

dichloroethylene. This molecule also enjoys C2v symmetry. The highest occupied

molecular orbital (HOMO) is a π -state while the lowest unoccupied molecular orbital is

a π* -state. The transition between the two states is depicted in Fig. 6.13.

Fig. 6.13 Cis-dichloroethylene electronic transition from the HOMO to the LUMO.

Like colors indicate like sign on the wavefunction.

Both the initial and final state in the transition belong to the B2 irreducible representation

of the C2v point group. Notice that, unlike the ground state of molecular vibrations, the

ground electronic state does not necessarily belong to the totally symmetric

representation. We again examine the transition moment to decide whether the transition

is dipole-allowed.

(6.19) Mmn = ψ m

*∫ µψ n dτ = B2∫ µ B2 dτ = qii∑ B2∫ xB2 dτ = qi

i∑ B2∫ B1 B2 dτ = 0

+ qi

i∑ B2∫ yB2 dτ = qi

i∑ B2∫ B2 B2 dτ = 0

/ /��(B2)�(B2)

Page 12: Vanishing Integrals

+ qi

i∑ B2∫ zB2 dτ = qi

i∑ B2∫ A1 B2 dτ ≠ 0 .

We find the transition is dipole-allowed and z-polarized.

Reducible representations

We have shown above how to decide the irreducible representation for each of the

nine degrees of freedom in water: the three translations along x, y, and z, the three

rotations about x, y, and z, and the three vibrations. Well, the form of the vibrations

(arrows on each atom) was given and we just figured out the symmetry of each. If we do

not know the form of the vibrations up front we can nonetheless determine the

symmetries of all the vibrations in a molecule. We do this by finding a reducible

representation for all nine degrees of freedom in the molecule. We then project this

reducible representation onto the character table of irreducible representations in a

manner very analogous to projecting an unknown vector onto Cartesian space to

determine its components along x, y, and z.

A reducible representation for the 3N degrees of freedom in a N-atom molecule

can be constructed by placing a Cartesian coordinate system on each atom as we have

done for water in Fig. 6.1. One representation could be the transformation matrices for

each operation in the group. We have already constructed matrix representations for the

symmetry operations C2z, σyz , and σxz, in the point group C2v, Eqs. (6.2)-(6.4). That for

the identity operation, E, is a simple extension of that found in Eq. (6.1) to include three

atoms. The matrices are

Page 13: Vanishing Integrals

E =

1 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 00 0 0 1 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 1 0 0 00 0 0 0 0 0 1 0 00 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 1

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

C2z =

0 0 0 0 0 0 −1 0 00 0 0 0 0 0 0 −1 00 0 0 0 0 0 0 0 10 0 0 −1 0 0 0 0 00 0 0 0 −1 0 0 0 00 0 0 0 0 1 0 0 0−1 0 0 0 0 0 0 0 00 −1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

σyz =

−1 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 00 0 0 −1 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 1 0 0 00 0 0 0 0 0 −1 0 00 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 1

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

σxz =

0 0 0 0 0 0 1 0 00 0 0 0 0 0 0 −1 00 0 0 0 0 0 0 0 10 0 0 1 0 0 0 0 00 0 0 0 −1 0 0 0 00 0 0 0 0 1 0 0 01 0 0 0 0 0 0 0 00 −1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

These matrices are too unwieldy to fit inside the character table. Let’s instead gather the

characters of the matrix. Remember that the character of a matrix χ is the sum of

diagonal elements top-left to bottom-right with only atoms that remain unshifted by the

operation contributing to the diagonal. Thus, χ(E) = 9, χ(C2z) = -1, χ(σyz) = 3, χ(σxz) = 1.

We use these characters to now define a convenient reducible representation of the 3N

degrees of freedom in water, Γ3N. We project the reducible representation onto each of

the irreducible representations by multiplying the character of the reducible

representation for a particular operation by the character of the irreducible representation

for the same operation, Γ3N .A1, for example, in Table 6.9. The resultant characters are

Page 14: Vanishing Integrals

then summed. Summing is equivalent to the integration of ψ *ψ over space introduced in

Chapter 2. The total is divided by the order of the group h, i.e., the number of symmetry

elements in the group that are summed (h = 4 for C2v). Division by h is equivalent to

normalization of a wavefunction. If the total/h is other than an integer then an error has

occurred. Additionally, the number of irreducible representations found to make up the

irreducible representation for the 3N degrees of freedom should equall 3N as they do in

Table 6.9 for water.

Table 6.9 Reducing Γ3N about the irreducible representations of C2v for water.

C2v E C2z σyz σxz

Γ3N 9 -1 3 1 total total/h

A1 1 1 1 1 z x2, y2, z2

A2 1 1 −1 −1 Rz xy

B1 1 −1 1 -1 x, Ry xz

B2 1 −1 -1 1 y, Rx yz

Γ3N .A1 9 −1 3 1 12 3

Γ3N .A2 9 −1 -3 -1 4 1

Γ3N .B1 9 1 3 -1 12 3

Γ3N .B2 9 1 -3 1 8 2

total degrees of freedom 9 = 3N

Our reducible representation for the nine degrees of freeom in water reduces down to a

combination of the irreducible representations.

Page 15: Vanishing Integrals

(6.20) Γ3N = 3 A1 + 1 A2 + 3 B1 + 2 B2

If we remove the irreducible representations for the three translational degrees of freedom

(x = B1 , y = B2 , z = A1) and remove the three rotational degrees of freedom (Rx = B2 , Ry

= B1, Rz = A2 ) what remain are the irreducible representations for the 3N – 6 = 3

vibrations.

(6.21) Γ3N - 6 = 2 A1 + B1

matching those representations found given the arrow-form of the vibrations (Fig. 6.10.)

Thus, we have found a way to determine systematically the reducible representations for

the vibrational modes of a molecule. The dependence of the reducible representation on

the summing of diagonal elements of a transformation matrix is simplified by only

considering contributions to the diagonal from unshifted atoms. As we have seen, atoms

that shift during an operation are transformed by off-diagonal elements in the

transformation matrix. Finding the character of the matrix is streamlined by multiplying

the number of unshifted atoms in an operation by the character per unshifted atom.

Earlier we determined the contribution per unshifted atom for some types of symmetry

element an in Table 6.10 we tabulate those and others

Table 6.10 Contribution per unshifted atom for each type of symmetry element

Symmetry  element   Contribution  to  the  character  per  unshifted  atom  

E   3

C2   -1

C3   0

C4   1

C6   2

σ 1

Page 16: Vanishing Integrals

i   -3

S3   -2

S4   -1

S6   0

Applying this simplified method to constructing the reducible representation for the

degrees of freedom in water we find

Table 6.11 Streamlined determination of the reducible representation for the degrees

of freedom for water.

C2v E C2z σyz σxz

# unshifted

atoms 3 1 3 1

contribution

per

unshifted atom

3 -1 1 1

Γ3N 3 x 3 = 9 1 x (-1) = -1 3 x 1 = 3 1 x 1 = 1

Raman and Infrared Activity of combinations and overtones of molecular vibrations

We can use symmetry considerations to predict the Raman and infrared activities

of overtones and combinations of molecular vibrations. In Problem 6.3 you will discover

that the selection rule for infrared transitions between harmonic oscillator energy levels

requires that the vibrational quantum number, v, change by ± 1, i.e., Δv = ± 1.

Anharmonicity of vibrations in real molecules leads to observations of weak

spectroscopic transitions wherein Δv = ± 2, ± 3, ± 4, etc. These lines are called the first

Page 17: Vanishing Integrals

overtone (Δv = ± 2), second overtone (Δv = ± 3), and so forth. Because the energy levels

in an anharmonic oscillator get closer together with increasing v the frequency of the

overtone is invariably close to, but lower than, the expected multiple of the fundamental

vibrational frequency.

Similarly, combinations of fundamentals may also appear in the infrared and

Raman spectra. For example, in the infrared spectrum of gaseous CS2, shown in Fig.

6.14, IR-active fundamentals are shown at 397 cm-1 and 1535 cm-1. Using simple

arithmetic the weak line appearing at 2332 cm-1 is likely the transition to a combination

(2ν2 + ν3) = (2 x 397 cm-1 + 1535 cm-1) = 2329 cm-1.

Fig. 6.14 Infrared spectrum of gaseous carbon disulfide.