116
5 The Vibrations of Polyatomic Molecules
for each atom
Atom X 0
Atom Y 1
Atom Y 2
(γ x ) 0 = 0
(γ x ) 1 = 2b 1 ω 3
(γ x ) 2 = −2b 1 ω 3
(γ z ) 0 = 2a 0 ω 3
(γ z ) 1 = −2a 1 ω 3
(γ z ) 2 = −2a 1 ω 3
It is interesting to compare this Coriolis interaction with the bending vibration ω 2
of symmetry A 1 . If we draw the vectors representing these Coriolis accelerations, we
find that when the molecule vibrates with the frequency ω 3 , the Coriolis acceleration
induces the bending vibration Q 2 but with the frequency ω 3 .
To determine whether a Coriolis interaction is possible, it is enough to take into
account the fact that the term T rv must be totally symmetric. A simple calculation
shows that
T rv =
r,s
α
α Q r ζ
α
rs
˙
Q s
(5.35)
In this equation, ζ
α
rs is a Coriolis coupling constant. α has the symmetry of the
rotation R α which can be found in the table of characters of the group, see Table 5.2.
If the product of the symmetry species of two vibrational modes contains the species
of rotation, Coriolis interaction takes place between these two modes (Jahn 1939).
For instance, for H 2 O, Q 3 is of symmetry B 1 and Q 2 of symmetry A 1 . The product
of the symmetry species of these two modes is B 1 ×A 1 = B 1 ; this is the symmetry
of a rotation around axis c and the Coriolis coupling constant ζ
c
23 , which can be
calculated from the harmonic force field, is different from zero as shown above.
Another example of Coriolis interaction is found in formaldehyde, H 2 CO, see
Fig. 5.3. The υ 4 = 1 and υ 6 = 1 states are of symmetry B 1 and B 2 , respectively. As
B 1 × B 2 = A 2 , this product has the symmetry of a rotation around the symmetry axis
a, see Table 5.2. Actually, it is not necessary to use the character table to identify
Coriolis interactions. As an example, we will consider the vibration υ 6 = 1. For a
rotation about the symmetry axis z (principal axis a), (5.33) shows that the Coriolis
acceleration of the mode υ 4 = 1 describes the mode υ 6 = 1, see Fig. 5.3.
When the frequencies of the interacting states are very different, the Coriolis
interaction is taken into account by a perturbation calculation. However, when these
two frequencies are close, the two interacting vibrational states have to be treated
Fig. 5.3 Vibrations υ 4 = 1
and υ 6 = 1 for formaldehyde
5 The Vibrations of Polyatomic Molecules
for each atom
Atom X 0
Atom Y 1
Atom Y 2
(γ x ) 0 = 0
(γ x ) 1 = 2b 1 ω 3
(γ x ) 2 = −2b 1 ω 3
(γ z ) 0 = 2a 0 ω 3
(γ z ) 1 = −2a 1 ω 3
(γ z ) 2 = −2a 1 ω 3
It is interesting to compare this Coriolis interaction with the bending vibration ω 2
of symmetry A 1 . If we draw the vectors representing these Coriolis accelerations, we
find that when the molecule vibrates with the frequency ω 3 , the Coriolis acceleration
induces the bending vibration Q 2 but with the frequency ω 3 .
To determine whether a Coriolis interaction is possible, it is enough to take into
account the fact that the term T rv must be totally symmetric. A simple calculation
shows that
T rv =
r,s
α
α Q r ζ
α
rs
˙
Q s
(5.35)
In this equation, ζ
α
rs is a Coriolis coupling constant. α has the symmetry of the
rotation R α which can be found in the table of characters of the group, see Table 5.2.
If the product of the symmetry species of two vibrational modes contains the species
of rotation, Coriolis interaction takes place between these two modes (Jahn 1939).
For instance, for H 2 O, Q 3 is of symmetry B 1 and Q 2 of symmetry A 1 . The product
of the symmetry species of these two modes is B 1 ×A 1 = B 1 ; this is the symmetry
of a rotation around axis c and the Coriolis coupling constant ζ
c
23 , which can be
calculated from the harmonic force field, is different from zero as shown above.
Another example of Coriolis interaction is found in formaldehyde, H 2 CO, see
Fig. 5.3. The υ 4 = 1 and υ 6 = 1 states are of symmetry B 1 and B 2 , respectively. As
B 1 × B 2 = A 2 , this product has the symmetry of a rotation around the symmetry axis
a, see Table 5.2. Actually, it is not necessary to use the character table to identify
Coriolis interactions. As an example, we will consider the vibration υ 6 = 1. For a
rotation about the symmetry axis z (principal axis a), (5.33) shows that the Coriolis
acceleration of the mode υ 4 = 1 describes the mode υ 6 = 1, see Fig. 5.3.
When the frequencies of the interacting states are very different, the Coriolis
interaction is taken into account by a perturbation calculation. However, when these
two frequencies are close, the two interacting vibrational states have to be treated
Fig. 5.3 Vibrations υ 4 = 1
and υ 6 = 1 for formaldehyde