Physical Chemistry
12. Rovibrational Spectroscopy

Ruifeng ZHOU

Faculty of Science &Institute for the Advancement of Higher Education
Hokkaido University

Table of content

1. Vibration-rotation of diatomic molecules
2. Rovibrational spectroscopy

11. Vibration-rotation of diatomic molecules• According to Born-Oppenheimer approximation, themotion of electrons and nuclei can be treatedseparately.
• The electrons, because of their light weight, canalways follow the motion of heavy nuclei.
• The electronic energy is part of the potential energyof the nuclear Hamiltonian.
• The nuclear potential energy of a 𝑁𝑁-atom molecule
depends on the positions of nuclei.
𝑉𝑉 = 𝑉𝑉 𝑅𝑅1 , ⋯ , 𝑅𝑅𝑁𝑁

2

3

1. Vibration-rotation of diatomic molecules• Especially, for a diatomic molecule, thepotential energy depends exclusively on 𝑚𝑚2 𝑅𝑅
𝑅𝑅2
the distance between the atoms.
𝑅𝑅1
𝑉𝑉 = 𝑉𝑉 𝑅𝑅
𝑂𝑂• This 2-body problem can be reduced to a
one-body problem with mass 𝜇𝜇 andposition 𝑅𝑅 (mass centre is stationary).
• The nuclear Schrödinger equation is𝑅𝑅𝑂𝑂
2ℏ 2− ∇ + 𝑉𝑉 𝜓𝜓 = 𝐸𝐸𝐸𝐸2𝜇𝜇𝑚𝑚1𝜇𝜇1. Vibration-rotation of diatomic molecules

1 2𝑉𝑉 ≈ 𝜇𝜇𝜔𝜔 𝑅𝑅 − 𝑅𝑅e 22• This is a spherical potential problem.
• The angular wavefunctions are sphericalharmonics, 𝑌𝑌𝒥𝒥𝑚𝑚 𝜃𝜃, 𝜙𝜙 .

• The radial equation is21 𝜕𝜕
𝜕𝜕ℛ2𝜇𝜇𝑅𝑅
𝑅𝑅2− 2 𝑉𝑉 − 𝐸𝐸 = 𝒥𝒥 𝒥𝒥 + 1ℏ
𝑅𝑅 𝜕𝜕ℛ𝜕𝜕𝑅𝑅• Substitute 𝑢𝑢 𝑅𝑅 = 𝑅𝑅ℛ 𝑅𝑅 and 𝜉𝜉 = 𝑅𝑅 − 𝑅𝑅e .

4

1. Vibration-rotation of diatomic molecules• The radial equation becomesℏ2 𝑑𝑑2
1 2 2ℏ2 𝒥𝒥 𝒥𝒥 + 1−+ 𝜇𝜇𝜔𝜔 𝜉𝜉 𝑢𝑢 +𝑢𝑢 = 𝐸𝐸𝐸𝐸
222𝜇𝜇 𝑑𝑑𝜉𝜉
2𝜇𝜇 𝑅𝑅e + 𝜉𝜉2
• The term inoscillation.

is the Hamiltonian of 1D harmonic• The term inis the contribution from rotation.

• Usually, the contribution from rotation is smaller
than vibration.

51. Vibration-rotation of diatomic molecules• Therefore, the rotational term can be regarded as atime-independent perturbation.
• The stationary states of harmonic oscillation are 𝜓𝜓𝑣𝑣
or |𝑣𝑣 ⟩.
• The energies are
110ℎ𝑐𝑐 𝜈𝜈�𝐸𝐸𝑣𝑣 = 𝑣𝑣 +ℏ𝜔𝜔 = 𝑣𝑣 +22
• The 1st order energy corrections are2ℏ
𝒥𝒥 𝒥𝒥 + 1
1
𝐸𝐸𝒥𝒥 = 𝑣𝑣
𝑣𝑣
22𝜇𝜇 𝑅𝑅e + 𝜉𝜉6

1. Vibration-rotation of diatomic molecules• If the vibration is not too violent, the wavefunctions
concentrate near 𝜉𝜉 = 0.
22 𝒥𝒥 𝒥𝒥 + 1
𝒥𝒥𝒥𝒥+1ℏℏ
1
𝐸𝐸𝒥𝒥 ≈ 𝑣𝑣
𝑣𝑣 =2𝜇𝜇 𝑅𝑅e 22𝜇𝜇𝑅𝑅e 2• Here 𝒥𝒥 𝒥𝒥 + 1 ℏ2 is the eigenvalue of 𝐿𝐿2 .

• 𝜇𝜇𝑅𝑅e 2 = 𝐼𝐼 is the moment of inertia.
• Therefore，2𝐿𝐿
1𝐸𝐸𝒥𝒥 =2𝐼𝐼
• It is the same as a classical rotator.

71. Vibration-rotation of diatomic molecules• 𝐸𝐸𝒥𝒥 is the rotational energy of a solid rotator at the𝒥𝒥 th rotational level.
• Define the rotational constant,𝑅𝑅
ℏ
𝑂𝑂
�
𝐵𝐵 ≡
4𝜋𝜋𝐼𝐼𝑐𝑐
𝐸𝐸𝒥𝒥 = 𝒥𝒥 𝒥𝒥 + 1 ℎ𝑐𝑐𝐵𝐵�

• The motion can be separated into a rotation and aradial vibration.

81. Vibration-rotation of diatomic molecules• The total energy is the sum of vibrational energy
and rotational energy101𝜈𝜈� + 𝒥𝒥 𝒥𝒥 + 1 𝐵𝐵�
𝐸𝐸 = 𝐸𝐸𝑣𝑣 + 𝐸𝐸𝒥𝒥 = ℎ𝑐𝑐 𝑣𝑣 +2• The total nuclear wavefunctions are1
𝜓𝜓𝑣𝑣,𝒥𝒥,𝑚𝑚 𝑅𝑅, 𝜃𝜃, 𝜙𝜙 = |𝑣𝑣, 𝒥𝒥, 𝑚𝑚⟩ ≈ 𝜑𝜑𝑣𝑣 𝑅𝑅 − 𝑅𝑅e 𝑌𝑌𝒥𝒥𝑚𝑚 𝜃𝜃, 𝜙𝜙𝑅𝑅1≈𝜑𝜑𝑣𝑣 𝜉𝜉 𝑌𝑌𝒥𝒥𝑚𝑚 𝜃𝜃, 𝜙𝜙𝑅𝑅e910

1. Vibration-rotation of diatomic molecules• The chemical bonds are not rigid,
they extend with the centripetal force.
• The moment of inertia increases as
bond length increases, leading to alower rotational energy.
• In classical mechanics, the centripetal
forces is equal to the stress of spring,which obeys the Hooke’s law.
𝜇𝜇𝑣𝑣 2≈ 𝑘𝑘 𝑅𝑅 − 𝑅𝑅e = 𝜔𝜔2 𝜇𝜇𝜇𝜇𝐹𝐹 =𝑅𝑅𝜇𝜇𝑣𝑣 2𝑅𝑅𝑘𝑘𝜉𝜉1. Vibration-rotation of diatomic molecules• The change of potential energy of the spring is
221 2 1 2𝑣𝑣𝜇𝜇4 𝑣𝑣 4 𝑅𝑅4𝐿𝐿4𝑉𝑉′ = 𝑘𝑘𝜉𝜉 ≈ 𝜔𝜔 𝜇𝜇= 3 2 6≈ 3 2 62
𝜔𝜔 𝑅𝑅2𝜇𝜇 𝜔𝜔 𝑅𝑅
222𝜇𝜇 𝜔𝜔 𝑅𝑅e• The rotational energy is
𝐿𝐿2
𝐿𝐿2=𝐸𝐸R =2𝐼𝐼 2𝜇𝜇 𝑅𝑅e + 𝜉𝜉 2
• Expand the rotational energy near 𝜉𝜉 = 0, .
𝐿𝐿2𝐿𝐿2 𝜉𝜉𝐸𝐸R ≈2−2𝜇𝜇𝑅𝑅e𝜇𝜇𝑅𝑅e 311

1. Vibration-rotation of diatomic molecules• The change of rotational energy due to centripetalforce is2 𝜉𝜉22𝐿𝐿𝐿𝐿𝑣𝑣=𝐸𝐸R′ ≈ −≈−332𝜇𝜇 𝑅𝑅e + 𝜉𝜉𝜇𝜇 𝑅𝑅e + 𝜉𝜉 𝜔𝜔 𝑅𝑅𝐿𝐿2 𝜇𝜇2 𝑣𝑣 2 𝑅𝑅e 2𝐿𝐿4≈− 3 2 6 =− 3 2 6𝜇𝜇 𝜔𝜔 𝑅𝑅e𝜇𝜇 𝜔𝜔 𝑅𝑅e• The change of total energy due to centripetal force
is4𝐿𝐿
𝐸𝐸 ′ = 𝑉𝑉 ′ + 𝐸𝐸R′ = − 3 2 62𝜇𝜇 𝜔𝜔 𝑅𝑅e12

13

1. Vibration-rotation of diatomic molecules• In quantum mechanics, the second order correction
to energy is,

2𝐸𝐸𝒥𝒥= �𝑣𝑣≠𝑤𝑤ℏ2 𝒥𝒥 𝒥𝒥 + 1𝑣𝑣
𝑤𝑤
22𝜇𝜇 𝜉𝜉 + 𝑅𝑅e
𝐸𝐸𝑣𝑣 − 𝐸𝐸𝑤𝑤

• Expand the perturbation near 𝜉𝜉 = 0,112≈ 2 − 3 𝜉𝜉
2𝑅𝑅e + 𝜉𝜉𝑅𝑅e
𝑅𝑅e

2

14

1. Vibration-rotation of diatomic molecules

4ℏ
2𝐸𝐸𝒥𝒥 ≈ 2 𝒥𝒥 2 𝒥𝒥 + 1 2 �4𝜇𝜇ℏ3𝑣𝑣≠𝑤𝑤

12𝑣𝑣
2 𝑤𝑤 − 𝑣𝑣
3 𝜉𝜉 𝑤𝑤
𝑅𝑅e
𝑅𝑅e2𝑣𝑣𝜉𝜉𝑤𝑤22= 2𝒥𝒥𝒥𝒥+1�6𝑣𝑣 − 𝑤𝑤𝜇𝜇 𝜔𝜔𝑅𝑅e𝐸𝐸𝑣𝑣 − 𝐸𝐸𝑤𝑤𝑣𝑣≠𝑤𝑤• We have known from previous discussion,𝑣𝑣 𝜉𝜉 𝑤𝑤 =

ℏ2𝑚𝑚𝜔𝜔

𝑤𝑤 + 1𝛿𝛿𝑣𝑣,𝑤𝑤+1 + 𝑤𝑤𝛿𝛿𝑣𝑣,𝑤𝑤−12

1. Vibration-rotation of diatomic molecules• Therefore,3ℏℏℏ
222𝐸𝐸𝒥𝒥 = 2𝑣𝑣− 𝑣𝑣 + 16 𝒥𝒥 𝒥𝒥 + 12𝜇𝜇𝜇𝜇2𝜇𝜇𝜇𝜇𝜇𝜇 𝜔𝜔𝑅𝑅e
𝒥𝒥 2 𝒥𝒥 + 1 2 ℏ42 𝒥𝒥 + 1 2 ℎ𝑐𝑐 𝐷𝐷
�=−=−𝒥𝒥62𝜇𝜇 3 𝜔𝜔 2 𝑅𝑅e
� is the centrifugal distortion constant.
• 𝐷𝐷• This result is the same as classical mechanics.

15

1. Vibration-rotation of diatomic molecules• The vibration of diatomic molecules is anharmonicas well.
• When the atoms move too far from each other, theattraction reduces rather than increases.
• The exact potential energy surface is complicated,but some approximated function such as Morsepotential can be used.
𝑉𝑉 = −𝐷𝐷e + 𝐷𝐷e 1 − 𝑒𝑒−𝑎𝑎𝜉𝜉 2• The 1st and 2nd energy corrections are calculated
as an example in perturbation theory.

161. Vibration-rotation of diatomic molecules
1𝐸𝐸𝑣𝑣

2+ 𝐸𝐸𝑣𝑣
0=−0 2
𝐸𝐸𝑣𝑣

4𝐷𝐷e• 𝐸𝐸𝑣𝑣 is the energyof harmonicoscillation.
• 𝐷𝐷e is the depth ofpotential well.
• 𝐷𝐷0 is thedissociationenergy.

17

18

1. Vibration-rotation of diatomic molecules• In general,
1𝐸𝐸𝑣𝑣 + 𝐸𝐸𝑣𝑣

2+ 𝐸𝐸𝑣𝑣=

11− 𝑣𝑣 +𝑣𝑣 +222𝑥𝑥� ℎ𝑐𝑐 𝜈𝜈�• 𝑥𝑥� is the anharmonicity constant. 0 < 𝑥𝑥� < 1.
• The difference between neighbouring energy levels
is12
12𝐸𝐸𝑣𝑣+1 + 𝐸𝐸𝑣𝑣+1 + 𝐸𝐸𝑣𝑣+1 − 𝐸𝐸𝑣𝑣 + 𝐸𝐸𝑣𝑣 + 𝐸𝐸𝑣𝑣
= ℎ𝑐𝑐 𝑣𝑣� 1 − 2 𝑣𝑣 + 1 𝑥𝑥�
• Energy levels get closer as 𝑣𝑣 increases.

1. Vibration-rotation of diatomic molecules• The total energy, including rotational energy withcentrifugal distortion and anharmonic vibration is𝐸𝐸𝑣𝑣,𝒥𝒥
122
� + 𝑣𝑣 +𝜈𝜈�= ℎ𝑐𝑐 �𝒥𝒥 𝒥𝒥 + 1 𝐵𝐵� − 𝒥𝒥 𝒥𝒥 + 1 𝐷𝐷
21− 𝑣𝑣 +22𝑥𝑥� 𝜈𝜈� �19

20

2. Rovibrational spectroscopy⃑• The electric dipole moment of a molecule, 𝓅𝓅,depends on the configuration of the nuclei (underBOA).
𝓅𝓅⃑ = 𝓅𝓅⃑ 𝑅𝑅1 , ⋯ , 𝑅𝑅𝑁𝑁
H

• For diatomic molecules,𝓅𝓅⃑ 𝑅𝑅 ≈ 𝓅𝓅e + 𝑞𝑞 𝑅𝑅 − 𝑅𝑅e

𝑅𝑅𝑅𝑅Cl• 𝓅𝓅e is 𝓅𝓅 at equilibrium distance, 𝑅𝑅e .
• 𝑞𝑞 𝑅𝑅 − 𝑅𝑅e is the change of electric dipole moment
due to the change of distance.

2. Rovibrational spectroscopy• The transition dipole moment is
𝕡𝕡 = 𝑣𝑣, 𝒥𝒥, 𝑚𝑚 𝓅𝓅⃑ 𝑣𝑣′, 𝒥𝒥′, 𝑚𝑚′
𝑅𝑅= 𝓅𝓅e − 𝑞𝑞𝑅𝑅e 𝑣𝑣, 𝒥𝒥, 𝑚𝑚𝑣𝑣′, 𝒥𝒥′, 𝑚𝑚′
𝑅𝑅+ 𝑞𝑞 𝑣𝑣, 𝒥𝒥, 𝑚𝑚 𝑅𝑅 𝑣𝑣′, 𝒥𝒥′, 𝑚𝑚′21• Obviously, for homoatomic molecules, 𝓅𝓅⃑ = 0, 𝕡𝕡 =0.
• The 1st selection rule of vibration-rotation excitation
is 𝓅𝓅⃑ ≠ 0.
• H2, N2 and O2 etc. do not have rovibrationalspectra.

2. Rovibrational spectroscopy• Recall that
𝐿𝐿�2 , 𝐿𝐿�2 , 𝑟𝑟⃑= 2ℏ2 𝑟𝑟⃑𝐿𝐿�2 + 𝐿𝐿�2 𝑟𝑟⃑• Based on this, for hydrogen atom, if 𝕡𝕡 ≠ 0, then∆𝑙𝑙 = ±1.
• It is not difficult to find that𝑅𝑅
𝑅𝑅 2𝑅𝑅
22= 2ℏ
𝐿𝐿� + 𝐿𝐿�𝑅𝑅𝑅𝑅𝑅𝑅• The conclusion is the same for diatomic molecules
because of they share the same angular functions(spherical harmonics).
𝐿𝐿�2 ,𝐿𝐿�2 ,22

2. Rovibrational spectroscopy• Therefore, for rovibrational transition of diatomic
molecule, if 𝑣𝑣, 𝒥𝒥, 𝑚𝑚 𝑅𝑅 𝑣𝑣′, 𝒥𝒥′, 𝑚𝑚′ ≠ 0, ∆𝒥𝒥 = ±1.
• Similarly, if

𝑅𝑅𝑣𝑣, 𝒥𝒥, 𝑚𝑚
𝑅𝑅𝑣𝑣′, 𝒥𝒥′, 𝑚𝑚′ ≠ 0, ∆𝒥𝒥 = ±1.

• The 2nd selection rule for rotation-vibration
excitation is∆𝒥𝒥 = ±1
• Note, the photon carries 𝑠𝑠 = 1.
• The angular momentum of photon is transferred tothe molecule.

232. Rovibrational spectroscopy• Since EM radiation is a longitudinal wave, ∆𝒥𝒥 ≠ 0.
• The total angular momentum is conserved.
• Suppose 𝐸𝐸 is in z direction.
• Calculate the component of 𝕡𝕡 in z direction,
2𝜋𝜋 𝜋𝜋 ∞11𝑚𝑚𝕡𝕡𝑧𝑧 = � � � 𝜓𝜓𝑣𝑣 𝑌𝑌𝒥𝒥 𝓅𝓅 cos 𝜃𝜃 𝜓𝜓𝑤𝑤 𝑌𝑌𝒥𝒥′𝑚𝑚𝑚 𝑅𝑅2 sin 𝜃𝜃 𝑑𝑑𝑑𝑑𝑑𝑑𝜃𝜃𝑑𝑑𝜙𝜙
𝑅𝑅𝑅𝑅0∞=�−𝑅𝑅e2𝜋𝜋×�

00 0𝜓𝜓𝑣𝑣 𝜉𝜉 𝓅𝓅e + 𝑞𝑞𝑞𝑞 𝜓𝜓𝑤𝑤 𝜉𝜉 𝑑𝑑𝑑𝑑𝜋𝜋

� 𝑌𝑌𝐽𝐽𝑚𝑚 𝑌𝑌𝒥𝒥′𝑚𝑚𝑚 cos 𝜃𝜃 sin 𝜃𝜃 𝑑𝑑𝜃𝜃𝑑𝑑𝜙𝜙
024

2. Rovibrational spectroscopy• The radial part is approximately
𝓅𝓅e 𝑣𝑣 𝑤𝑤 + 𝑞𝑞 𝑣𝑣 𝜉𝜉 𝑤𝑤• On one hand, 𝑣𝑣 𝑤𝑤 ≠ 0 if 𝑤𝑤 = 𝑣𝑣.
• This corresponds to pure rotational excitation.
• On the other hand, 𝑣𝑣 𝜉𝜉 𝑤𝑤 ≠ 0 if 𝑤𝑤 = 𝑣𝑣 ± 1.
• This corresponds to rovibrational excitation.
• The 3rd selection rule for rotation-vibration
excitation is∆𝑣𝑣 = 0, ±125

2. Rovibrational spectroscopy• The energy change during the excitation of purerotation is� 𝒥𝒥 + 1∆𝐸𝐸𝒥𝒥→𝒥𝒥+1 = ℎ𝑐𝑐 𝐵𝐵� 𝒥𝒥 + 1 𝒥𝒥 + 2 − ℎ𝑐𝑐 𝐵𝐵𝒥𝒥= 2ℎ𝑐𝑐 𝐵𝐵� 𝒥𝒥 + 1∆𝐸𝐸𝒥𝒥→𝒥𝒥+1= 2𝐵𝐵� 𝒥𝒥 + 1
𝜈𝜈�𝒥𝒥→𝒥𝒥+1 =ℎ𝑐𝑐
• The difference between neighbouring excitation
wavenumbers is∆𝜈𝜈� = 𝜈𝜈�𝒥𝒥+1→𝒥𝒥+2 − 𝜈𝜈�𝒥𝒥→𝒥𝒥+1 = 2𝐵𝐵�26

272. Rovibrational spectroscopy

Intensity

6𝐵𝐵�

8𝐵𝐵�

4𝐵𝐵�
2𝐵𝐵�

02𝐵𝐵�
0↓
12𝐵𝐵�
1
↓
22𝐵𝐵�
2↓
32𝐵𝐵�
3↓
42𝐵𝐵�

2𝐵𝐵�

Wavenumber
⋯

2𝐵𝐵�

2𝐵𝐵�

2𝐵𝐵�

2. Rovibrational spectroscopy

28• Pure rotational energy is in the microwave range.
• The rotational spectroscopy is also calledmicrowave spectroscopy.
• For example, the bond length of 12C16O is 112.8pm.
𝑚𝑚C 𝑚𝑚O𝜇𝜇 == 1.14 × 10−26 kg
𝑚𝑚C + 𝑚𝑚O
𝐼𝐼 = 𝜇𝜇𝑅𝑅e 2 = 1.45 × 10−46 kg � m2
ℏ2= 6.78 × 10−23 J2ℎ𝑐𝑐 𝐵𝐵� =𝐼𝐼𝐵𝐵� = 1.93cm−1

2. Rovibrational spectroscopy

29

2. Rovibrational spectroscopy• For vibrational excitation,∆𝐸𝐸 = ℎ𝑐𝑐 𝜈𝜈� ± 2𝐵𝐵� 𝒥𝒥 + 1• The pure vibrational absorption (∆𝒥𝒥 = 0) is calledQ-branch, which is forbidden for free diatomic
molecules (conservation of angular momentum).
• Q-branches are allowed in polyatomic molecules.
• Other vibrational excitations are accompanied byrotational excitations.
• ∆𝒥𝒥 = −1 is called P-branch.
• ∆𝒥𝒥 = 1 is called R-branch.

30

31

2. Rovibrational spectroscopy
12C16OPRQ14N14N16O(14N15N16O,15N14N16O)
PR↑↑
N-N-O↓

2. Rovibrational spectroscopy

32• In solid phases, the rotation of molecule is disabledby the interaction among molecules.
• In both liquid and solid phases, pure vibration isallowed and can be excited by photons.
• The angular momentum from photons is absorbedby the whole system.