# 5.5: The Harmonic Oscillator and Infrared Spectra

Skills to Develop

• The Harmonic Oscillator Accounts for the Infrared Spectrum of a Diatomic Molecule

Infrared (IR) spectroscopy is one of the most common and widely used spectroscopic techniques employed mainly by inorganic and organic chemists due to its usefulness in determining structures of compounds and identifying them. Chemical compounds have different chemical properties due to the presence of different functional groups.

### Introduction

Infrared (IR) spectroscopy is one of the most common and widely used spectroscopic techniques. Absorbing groups in the infrared region absorb within a certain wavelength region. The absorption peaks within this region are usually sharper when compared with absorption peaks from the ultraviolet and visible regions. In this way, IR spectroscopy can be very sensitive to determination of functional groups within a sample since different functional group absorbs different particular frequency of IR radiation. Also, each molecule has a characteristic spectrum often referred to as the fingerprint. A molecule can be identified by comparing its absorption peak to a data bank of spectra. IR spectroscopy is very useful in the identification and structure analysis of a variety of substances, including both organic and inorganic compounds. It can also be used for both qualitative and quantitative analysis of complex mixtures of similar compounds.

### Diatomic Molecular Vibration

The absorption of IR radiation by a molecule can be likened to two atoms attached to each other by a massless spring. Considering simple diatomic molecules, only one vibration is possible. The Hook's law potential on the other hand is based on an ideal spring

$F = -kx \label{5.5.1}$

$F(x) = -\dfrac{dV(x)}{dx} \label{5.5.2}$

this results in one dimensional space

$V(r) = \dfrac{1}{2} k(r-r_{eq}) \label{5.5.3}$

One thing that the Morse and Harmonic oscillator have in common is the small displacements ($$x=r-r_{eq}$$) from the equilibrium. Solving the Schrödinger equation for the harmonic oscillator potential results in the energy levels results in

$E_v = \left(v+\dfrac{1}{2}\right)h \nu \label{5.5.4}$

with $$v=0,1,2,3 \cdots \infty$$

$v_e = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{5.5.5}$

When calculating the energy of a diatomic molecule, factors such as anharmonicity (has a similar curve with the harmonic oscillator at low potential energies but deviates at higher energies) are considered. The energy spacing in the harmonic oscillator is equal but not so with the anharmonic oscillator. The anharmonic oscillator is a deviation from the harmonic oscillator. Other considered terms include; centrifugal stretching, vibrational and rotational interactions have to be taken into account. The energy can be expressed mathematically as

$E_v = \underset{\text{Harmonic Oscillator}}{\left(v+\dfrac{1}{2}\right)hv_e} - \underset{\text{anharmonicity}}{\left(v+\dfrac{1}{2}\right)^2 X_e hv_e} + \underset{\text{Rigid Rotor}}{B_e J (J+1)} - \underset{\text{centrifugal stretching}}{D_e J^2 (J+1)^2} -\alpha_e \underset{\text{rovibrational coupling}}{\left(v+\dfrac{1}{2}\right) J(J+1)} \label{5.5.6}$

The first and third terms represent the harmonicity and rigid rotor behavior of a diatomic molecule such as HCl. The second term represents anharmonicity discussed previously. The fourth and fifth terms represents centrifugal stretching and the interaction between the vibration and rotational interaction of the molecule.

#### The Deduction of Frequency

The second law of Newton states that

$F = ma\label{5.5.7}$

where m is the mass and a is the acceleration, acceleration is a 2nd order differential equation of distance with respect to time. Thus "a" can be written as

$a = \dfrac{d^2 y}{d t} \label{5.5.8}$

Substituting this into Equation \ref{5.5.1} gives

$\dfrac{m\, d^2 y}{d t^2}= - k y \label{5.5.9}$

the 2nd order differential equation of this equation is equal to $$\dfrac{-k}{m}$$ displacement of mass and time can be stated as

$y = A \cos 2 \pi \nu_m t \label{5.5.10}$

where $$\nu_m$$ is the natural vibrational frequency and $$A$$ is the maximum amplitude of the motion. On differentiating a second time the equation becomes

$$\dfrac{d^2 y}{d t^2} = - 4 \pi^2 \nu_m^2 A \cos 2 \pi \nu_m t \label{5.5.11}$$

substituting the two equations above into Newton's second law for a harmonic oscillator,

$m \left (-4\pi^{2}\nu_{m}^{2} A \cos 2\pi\nu_{m}t \right ) = -k \left ( A \cos 2\pi\nu_{m}t \right ) \label{5.5.12}$

If we cancel out the two functions $$y$$,

$4m\pi^{2}\nu_{m}^{2} = k$

from above, we obtain the natural frequency of the oscillation.

$\nu_m = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}} \label{5.5.13}$

$$\nu_m$$ is the natural frequency of the mechanical oscillator and depends on the force constant of the spring $$k$$ and the reduced mass of the attached body $$\mu$$ and is independent of energy imparted on the system. When there are two masses involved in the system then the reduced mass used in the above equation becomes

$\mu = \dfrac{m_1 m_2}{m_1+m_2} \label{5.5.14}$

The vibrational frequency can be rewritten as

$\nu_m = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{5.5.15}$

#### IR Spectroscopy

Transitions between vibrational energy levels can be induced about by absorption or emission of radiation. To understand this, knowledge of both the initial and final eigenstates is needed. The energy of a harmonic oscillator can be written as

$E = \left(v+\dfrac{1}{2}\right) \dfrac{h}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{5.5.16}$

where

• $$h$$ is Planck's constant and
• $$v$$ is the vibrational quantum number and ranges from 0,1,2,3.... infinity.

$E = \left(v+\dfrac{1}{2}\right)h \nu_m \label{5.5.17}$

where $$\nu_m$$ is the vibrational frequency. Transitions in vibrational energy levels can be brought about by absorption of radiation, provided the energy of the radiation exactly matches the difference in energy levels between the vibrational quantum states and provided the vibration causes a change in dipole moment. This can be expressed as

${\Delta E} = h \nu_m = \dfrac{h}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{5.5.18}$

The majority of molecules are in the ground state $$v = 0$$ so let's only consider this as the initial eigenstate, then from Equation $$\ref{5.5.17}$$

$E_o = \dfrac{1}{2}hv_m \label{5.5.19}$

and when a molecule absorbs energy to be promoted to the first excited state ($$v=1$$) in Equation $$\ref{5.5.17}$$.

$E_1 = \dfrac{3}{2} hv_m \label{5.5.20}$

$\Delta E = \left(\dfrac{3}{2} hv_m - \dfrac{1}{2} hv_m \right) = hv_m \label{5.5.21}$

The frequency of radiation $$\nu$$ that will bring about this change is identical to the classical vibrational frequency of the bond $$v_m$$ and can be expressed as

$E_{radiation} = hv = {\triangle E} = hv_m = \dfrac{h}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{5.5.22}$

The cm-1 is the wavenumber scale and it can also be defined as 1/wavelength in cm. A linear wavenumber is often used due to its direct relationship with both frequency and energy. The frequency of the absorbed radiation causes the molecular vibrational frequency for the absorption process:

$\bar{v}(cm^{-1}) = \dfrac{1}{\lambda(\mu m)} \times 10^4 (\dfrac{\mu m}{cm}) = \dfrac{v(Hz)}{c(cm/s)} \label{5.5.28}$

Equation \ref{5.5.22} can be modified so that the radiation can be expressed in wavenumbers

$\widetilde{\nu} = \dfrac{1}{2\pi c} \sqrt{\dfrac{k}{\mu}} \label{5.5.23}$

where

• $$c$$ is the velocity of light (cm s-1) and
• $$\widetilde{\nu}$$ is the wave number of an absorption maximum (cm-1)

IR deals with the interaction between a molecule and radiation from the electromagnetic region ranging (4000- 40 cm-1). The IR region of the electromagnetic spectrum ranges in wavelength from 2 -15 µm. Conventionally the IR region is subdivided into three regions (Table $$\PageIndex{1}$$): near IR, mid IR and far IR. Most of the IR used originates from the mid IR region.

Table $$\PageIndex{1}$$: Region of the IR Spectrum
Region Wavelength Wavenumbers ($$\widetilde{\nu}$$), cm-1 Frequencies (v), HZ
Near 0.78 -2.5 12800 - 4000 3.8 x 1014 - 1.2 x 1014
Middle 2.5 - 50 4000 - 200 3.8 x 1014 - 1.2 x 1014
Far 50 -100 200 -10 3.8 x 1014 - 1.2 x 1014
Most Used 2.5 -15 4000 -670 3.8 x 1014 - 1.2 x 1014

Example $$\PageIndex{1}$$

The force constants for typical diatomic molecules are in the range between 400 to 2000 $$N \cdot m^{-1}$$.

Molecule HF HCl HBr HI CO NO
Force constant, $$k$$ (N.m-1) 970 480 410 320 1860 1530

For the diatomic molecules listed above, calculate the following:

1. angular frequency ($$\text{rad} \cdot s^{-1}$$)
2. natural frequency (Hz)
3. period (s)
4. separation between energy levels
5. wavelength $$\lambda$$ of the electromagnetic radiation absorbed in the transition $$v=0 \rightarrow v=1$$.

Solution

For $$HCl$$:

1. angular frequency $\omega = 5.45 \times 10^{14} \text{rad} \cdot s^{-1}$
2. natural frequency: $\nu = \dfrac{\omega}{2pi} = 8.68 \times ^{13}$
3. period (s): $T =\dfrac{1}{\nu} = 1.15 \times 10^{-14}$
4. separation between energy levels: $\Delta E = E_{v=1} - E_{v=0} = \hbar \omega = 5.75 \times 10^{-21} \; J$
5. wavelength $$\lambda$$ of the electromagnetic radiation absorbed in the transition $$v=0 \rightarrow v=1$$: $3.46 \times 10^{-5} \, m$

The electromagnetic radiation released (and absorbed) for vibrations is primarily in the infrared (IR) part of the spectrum.

Calculating the above properties for the other molecules remains as an exercise.

### Selection Rules for Transitions

Photons can be absorbed or emitted, and the harmonic oscillator can go from one vibrational energy state to another. Which transitions between vibrational states are allowed? For IR absorption to occur two conditions must be met:

1. There must be a change in the dipole moment of the molecule as a result of a molecular vibration (or rotation). The change (or oscillation) in the dipole moment allows interaction with the alternating electrical component of the IR radiation wave. Symmetric molecules (or bonds) do not absorb IR radiation since there is no dipole moment.
2. If the frequency of the radiation $$\nu$$ matches the natural frequency of the vibration (Equation $$\ref{5.5.22}$$), the IR photon is absorbed and the amplitude of the vibration increases.

Furthermore, there are selection rules that describes whether particular transitions is allow. These results from evaluating the following transition moment integral that expresses the probability of a transition form the $$v$$ to the $$v'$$ eigenstates:

$\mu_T = \int \limits _{-\infty}^{\infty} \Psi_{v'}^* (Q) \hat {\mu} (Q) \Psi _v (Q) dQ \label {5.5.29}$

To evaluate this integral we need to express the dipole moment operator, $$\hat {\mu}$$, in terms of the magnitude of the vibration $$Q$$. The dipole moment operator is defined as

$\hat {\mu} = \sum _{electrons} er + \sum _{nuclei} qR \label {5.5.30}$

where the two sums are over all the electrons and nuclei and involve the particle charge (-e or q) multiplying the position vector ($$\vec{r}$$ or $$\vec{R}$$). We can obtain this dipole moment operator in terms of the magnitude of the normal coordinate, Q, in a simple way by using a Taylor series expansion for the dipole moment.

$\mu (Q) = \mu _{Q=0} + \left ( \dfrac {d \mu (Q)}{dQ} \right ) _{Q=0} Q + \left ( \dfrac {d^2 \mu (Q)}{dQ^2} \right ) _{Q=0} Q^2 + \cdots \label {5.5.31}$

Retaining only the first two terms and substituting into Equation $$\ref{5.5.29}$$ produces

$\mu _T = \mu _{Q=0} \int \limits _{-\infty}^{\infty} \psi _{v'} (Q) \psi _v (Q) dQ + \left ( \dfrac {d \mu (Q)}{dQ} \right ) _{Q=0} \int \limits _{-\infty}^{\infty} Q\psi _{v'}^* (Q) \psi _v (Q) dQ \label {5.5.32}$

In the above expressions,

$\mu _{Q=0} = 0$

is the dipole moment of the molecule when the nuclei are at their equilibrium positions, and

$\left (\dfrac {d\mu (Q)}{dQ} \right )_{Q=0}$

is the linear change in the dipole moment due to the displacement of the nuclei in the normal mode. The derivative is the linear change because it multiplies $$Q$$ and not a higher power of $$Q$$ in Equation $$\ref{5.5.31}$$. Both $$\mu$$­ and

$\left (\dfrac {d\mu (Q)}{dQ}\right )_{Q=0}$

are moved outside of the integral because they are constants that no longer depend on $$Q$$ because they are evaluated at $$Q = 0$$.

The integral in the first term in Equation $$\ref{5.5.32}$$ is 0 because any two harmonic oscillator wavefunctions are orthogonal. The integral in the second term of Equation is zero except when $$v' = v \pm 1$$ as demonstrated in Exercise $$\PageIndex{1}$$. Also note that the second term is zero if

$\left (\dfrac {d\mu (Q)}{dQ}\right )_{Q=0} = 0 \label {5.5.33}$

Exercise $$\PageIndex{1}$$

Use one of the Hermite polynomial recursion relations to verify that the second integral in Equation \ref{5.5.32} is 0 unless $$v' = v \pm 1$$.

If we are to observe absorption of infrared radiation due to a vibrational transition in a molecule, the transition moment cannot be zero. This condition requires that the dipole moment derivative Equation $$\ref{5.5.33}$$ cannot be zero and that the vibrational quantum number change by one unit. The normal coordinate motion must cause the dipole moment of the molecule to change in order for a molecule to absorb infrared radiation. If the normal coordinate oscillation does not cause the dipole moment to change then $$\mu _T = 0$$ and no infrared absorption is observed.

$\text { For allowed transitions} \Delta v = \pm 1 \label {5.5.34}$

Consider oxygen and nitrogen molecules. Because they are symmetrical, their dipole moments are zero, $$\mu = 0$$. Since the vibrational motion (only bond stretching for a diatomic molecule) preserves this symmetry, the change in the dipole moment due to the vibrational motion also is zero, $$\dfrac {d\mu (Q)}{dQ} = 0$$. Consequently, oxygen and nitrogen do not absorb infrared radiation as a result of vibrational motion.

Exercise $$\PageIndex{2}$$

Explain why the absorption coefficient in Beer's Law is larger for some normal modes than for others.

The case v' = v + 1 corresponds to going from one vibrational state to a higher energy one by absorbing a photon with energy hν. The case v' = v − 1 corresponds to a transition that emits a photon with energy hν. In the harmonic oscillator model infrared spectra are very simple; only the fundamental transitions, $$\Delta = \pm 1$$, are allowed. The associated transition energy is $$\hbar \omega$$, according to Equation6-30). The transition energy is the change in energy of the oscillator as it moves from one vibrational state to another, and it equals the photon energy.

$\Delta E = E_{final} - E_{initial} = hv_{photon} = \hbar \omega _{oscillator} \label {5.5.35}$

In a perfect harmonic oscillator, the only possibilities are $$\Delta = \pm 1$$; all others are forbidden. This conclusion predicts that the vibrational absorption spectrum of a diatomic molecule consists of only one strong line since the energy levels are equally spaced in the harmonic oscillator model. If the levels were not equally spaced, then transitions from v = 0 to 1 and from v = 1 to 2, etc. would occur at different frequencies.

Only the fundamental transitions, $$\Delta = \pm 1$$, are observed in infrared spectra within harmonic oscillator model.

The actual spectrum is more complex, especially at high resolution. There is a fine structure due to the rotational states of the molecule. These states will be discussed in the next chapter. The spectrum is enriched further by the appearance of lines due to transitions corresponding to $$\Delta = \pm n$$ where $$n > 1$$. These transitions are called overtone transitions and their appearance in spectra despite being forbidden in the harmonic oscillator model is due to the anharmonicity of molecular vibrations. Anharmonicity means the potential energy function is not strictly the harmonic potential. The first overtone, Δv = 2, generally appears at a frequency slightly less than twice that of the fundamental, i.e. the frequency due to the Δv = 1 transition.

Exercise $$\PageIndex{3}$$

Compute the approximate transition frequencies in wavenumber units for the first and second overtone transitions in HCl given that the fundamental is at 2,886 cm-1.

Generally each intense peak that is seen in an infrared spectrum of a polyatomic molecule corresponds to the fundamental transition of a different normal mode because the overtones are forbidden in the harmonic approximation and hot bands are weak at room temperature. However in a polyatomic molecule, combination bands that involve the excitation of two normal modes also can be intense. They arise from a higher order term in the expansion of μ­(Q); namely,

$\left ( \dfrac {\partial ^2 \mu}{\partial Q_A \partial Q_B} \right)_{Q=0} Q_A Q_B \label{5.5.36}$

This derivative gives the change in the dipole moment due to the motion of two normal modes simultaneously.