Lecture 5

The harmonic oscillatorClassical and quantum mechanical

NC State University

Chemistry 431

Classical Vibration of a DiatomicAs was the case for rotation, we can consider a simplemodel of a mass on a spring attached to a wall of infinitemass and a diatomic molecule as two simple examples.

Mass on a spring Diatomic

μ = m1m2m1 + m2

mm2m1

Reduced mass

kk

k is the force constant

Harmonic approximation

V(Q) = V(Q0) + ∂V∂Q (Q – Q0) + 1

2∂2V∂Q2 (Q – Q0)

2 + ...

At equilibrium∂V∂Q = 0

Assume terms higher than quadratic are zeroBy definition

∂2V∂Q2 = k, the force constant

Classical approach to vibration

V(Q)k force

constant

Q

Solution is oscillatoryAny energy is possible

μ2

∂Q∂t

2

+ k2Q2 = 0

Q

Classical vibrational motion

• A particle undergoes harmonic motion if it experiences a restoring force that is proportional to its displacement, x.

• F = -kQ (k is a force constant) • F = -dV/dQ and V = 1/2kQ2.• The classical harmonic oscillator can also be written as:

• Solutions have the form of sin(ωt) or cos(ωt) depending on the initial conditions.

• These solutions imply that

ma = kQ , – μ∂2Q∂ t2 + kQ = 0

ω = kμ

Classical potential function• The potential is V = 1/2kQ2, which is a parabolic function.• This potential is called a harmonic potential.

• The force constant k has units of Newtons/meter (N/m) or Joules/meter2 (J/m2).

• The angular frequency ω = 2πν, ν is the frequency in Hz.

Quantum approach to vibration

Solution is GaussianEnergy is quantized

– h2

2μ∂2Ψ∂Q2 + k

2Q2Ψ = EΨ

v is the quantum numberAllowed transitionsv’ → v + 1, v’ → v - 1

E = v + 12 hν

Q

Vibrational wavefunctions

• Energy levels are given by Ev = (v +1/2)hω

• Wavefunctions are χv = NvHve-y2/2 where Hvis the Hermite polynomial

• Typical energies are of the order of 0 - 3000 cm-1

Solutions to harmonic oscillator

Hv(y) = (–1)vey 2 ddy

v

e–y 2

The Hermite polynomials are derivatives of a Gaussian, y = α1/2Q

The normalization constant is

α = μkh

v Hv(y) Nv

0 1 α/π 1/4

1 2y 1/ 2 α/π 1/4

2 4y 2 – 2 1/ 8 α/π 1/4

3 8y 3 – 12y 1/ 48 α/π 1/4

Nv = 12vv!

απ

1/4

The bonding electronic state gives rise to a potential energy surface for the nuclear motion

Harmonic approximation

V Q = 12kQ2

χ0 Q = απ

1/4e– αQ2

χ1 Q = απ

1/42αQe– αQ2

...

Ev = v + 12 hω

χ2

There is a potential energy surface that corresponds to each electronic state of the molecule

The shift in the nucleardisplacement arises fromthe fact that the bondlength increases in theσ* state compared to theσ state. We will show thatthe overlap of the vibra--tional wave functions iskey to understanding theshape of absorption bands.

The zero point energy

• The lowest level is E0 = 1/2hω• The lowest vibrational level is not zero in energy.• This is consistent with the uncertainty principle. If

atoms were completely still at absolute zero then we would know both their position and moment to arbitrary accuracy.

• The width of the wavefunction is related to positional uncertainty of an atom.

• We call E0 the zero point energy.

Polyatomic Molecules• There are 3N total degrees of freedom in a molecule

that contains N atoms.• There are three translational degrees of freedom. These

correspond to motion of the center of mass of the molecule.

• In a linear molecule there are two rotational degrees of freedom. In a non-linear molecule there are 3 rotational degrees of freedom.

• The remaining degrees of freedom are vibrational.

There are 3N-6 vibrational degrees of freedom in a molecule with N atoms

Three degrees of freedom are required for translation.Three degrees of freedom are required for rotation.

For example, in H2O there are 9 total degrees of freedomand 3 vibrational degrees of freedom.In C6H6 there are 36 degrees of freedom and 30 vibrationaldegrees of freedom.

Exception: In linear molecules there are only 2 rotationaldegrees of freedom and therefore the number of vibrationsis 3N - 5.

The vibrational degrees of freedom can be expressed as normal

modes.All normal modes have the same form for the harmonicoscillator wavefunction and differ only in the forceconstant k and mass m. The total wavefunction is a product of normal modes. The total nuclear wavefunction for water is χ1χ2χ3.The normal mode wavefunctions of water correspondto the symmetric stretch, bend, and asymmetric stretch.These are linear combinations of the stretching and bending internal coordinates of H2O.

Normal modes - water

Symmetric stretch ν1 3650 cm-1

Asymmetric stretch ν3 3750 cm-1

Bend ν2 1600 cm-1

There are 3 normal modes (3N - 6). All of them are infrared active since all show a dipole moment change in their motion. The harmonic approximationcan be applied to each normal mode.

Symmetric Stretchν1 3825 cm-1

Asymmetric Stretchν3 3935 cm-1

Bendν2 1654 cm-1

Normal modes - CO2

Symmetric stretch(infrared inactive)

Asymmetric stretch ν3 2349 cm-1 Bends

ν2 667 cm-1

.+.

(infrared active)

There are 4 normal modes (3N - 5). Three of them are infrared active since they show a dipole moment change in their motion.

Symmetric stretchν1 2289 cm-1

(Raman active)

Asymmetric stretchν3 2349 cm-1

(IR active)Bends

ν2 667 cm-1

Asymmetric stretchν3 2349 cm-1

(IR active)(IR active)

VibrationalTransitions

Vibrational transitionsVibrational transitions arise because of the oscillation of the molecule about its equilibrium bond configuration. As the molecule oscillates infrared radiation can interact to alter the quantum state.

Mvib = ∂μ∂Q χv + 1

* QχvdQ– ∞

∞

μ Q = μ0 + ∂μ∂Q Q + ...

Transition dipolesIn order for infrared light to be absorbed the polarization must be aligned with the directionof the transition moment. For a vibrational modethis is determined by the directional change inthe dipole moment. This is shown below forthe bending mode of H2O.

H

O

H H

O

H

Transition dipolesThe change in ground state dipole momentduring vibration interacts with light.

The first term is static and does not contributeto the transition. Calling the vibrational wave-functions χi the transition moment is:

μ = μg +∂μg

∂Q Q +

μ10 =∂μg

∂Q χ1Qχ0dQ,

Dipole derivativesThe vibrational wavefunctions χi are Gaussians,thus the transition moment for transition fromvibrational state 0 to vibrational state 1 is:

The transition dipole moment is proportional tothe dipole derivative. This is true for any normal mode of vibration (i.e. harmonic).

μ10 =∂μg

∂Q e– αQ2/2Q2e– αQ2/2dQ– ∞

∞

= 12α

∂μg

∂Q

Vibrational transitionsAs an example we can calculate the transitionmoment between the state v = 0 and v = 1.

χ0 = απ

1/4e– αQ2/2 , χ1 = α

π1/4

2αQe– αQ2/2

Mvib = ∂μ∂Q

απ

1/22α e– αQ2/2Q2e– αQ2/2dQ

– ∞

∞

= ∂μ∂Q

απ

1/2 2α π2α3/2 = ∂μ

∂Q12α

Vibrational transitionsNote that this result is a statement of the vibrational selection rule. Within the harmonicapproximation transitions can only occur between states separated by one quantum number (Δv = 1 or Δv = -1).

This general rule can be seen by consideringintegrals of the type shown in the previous slide.

Vibrational Transition

Vibrational Transition

v’ = v + 1

v’ = v - 1

v = 0

v = 1

v = 2

Q

Vibrational Selection Rule

Comparison of harmonic and anharmonic potentials

Overtones of water

ν1 symmetric stretch 3825 cm-1

ν2 bend 1654 cm-1

ν3 asymmetric stretch 3935 cm-1

Even in water vaporν1 ≈ ν3, but symmetriesare different, Γ1 ≠ Γ3.However, the third overtoneof mode 1 has the same symmetry as the combination bandΓ1 Γ1 Γ1 = Γ1 Γ3 Γ3 .Strong anharmonic couplingleads to strong overtonesat 11,032 and 10,613 cm-1.These intense bands givewater and ice their blue color.

Frequency shift due to molecular interactions

Hydrogen bonding lowers O-H force constantand H-O-H bending force constant.

vapor → liquidν1 3825 → 3657ν2 1654 → 1595ν3 3935 → 3756

Question

Which expression is correct?

A. k = μω

B. k = 1μω

C. k = μωD. k = μω2

Question

Which expression is correct?

A. k = μω

B. k = 1μω

C. k = μωD. k = μω2

ω = kμ

ω2 = kμ

k = μω2

Question

How many normal modes of vibrationdoes methane have?

A. 15B. 12C. 9D. 6

Question

How many normal modes of vibrationdoes methane have?

A. 15B. 12C. 9D. 6

Analysis of isotope effectsVibrational spectra are analyzed within theharmonic approximation.

m1 m2k

μ∂2x∂t2 + kx = 0

Classical harmonic oscillator equation

μ = m1m2

m1 + m2

ω = kμx = Acos ωt

Reduced mass

The vibrational partition functionThe vibrational energy levels are evenly spaced

with a separation of hν or hω.  If we take the zero‐‐point level as our “zero of energy” then εv = vhω.

qv = e– βvhcνΣv = 1

∞

= e– βhcν v= 1

1 – e– βhcνΣv = 1

∞

This is the partition function that we found for aninfinite ladder of energy levels.For high frequency modes ε >> kT and qv ≈ 1For very low frequency modes ε << kT and qv ≈ 1

βhcν

v