Vibrations of Molecules

Molecular bonds act like springs

Every stable diatomic molecule — such as HCl, CO, or H$_2$ — has an equilibrium bond length $r_0$ at which the potential energy is a minimum. If the bond is stretched or compressed by a small displacement $x = r - r_0$ from that equilibrium, the restoring force is approximately linear:

where $k$ is the bond force constant (units N/m). This is Hooke's law, and it means the potential near equilibrium is parabolic:

This parabolic form is precisely the quantum harmonic oscillator potential $V = \frac{1}{2}m\omega^2 x^2$, with the angular frequency given by $\omega = \sqrt{k/\mu}$, where $\mu$ is the reduced mass of the two atoms.

Reduced mass

For a diatomic molecule with atom masses $m_1$ and $m_2$, the two-body vibration problem reduces to an equivalent one-body problem with a single particle of reduced mass

This substitution converts the relative-coordinate Schrödinger equation into the standard harmonic oscillator equation. For example, for HCl ($m_\text{H} \approx 1.008\,\text{u}$, $m_\text{Cl} \approx 35.45\,\text{u}$):

Because hydrogen is so much lighter than chlorine, the reduced mass is dominated by the hydrogen mass — the chlorine barely moves while the hydrogen oscillates back and forth.

Vibrational energy levels

Substituting the reduced mass into the oscillator energy formula immediately gives the vibrational energy levels of the molecule:

where $\omega = \sqrt{k/\mu}$. Molecular spectroscopists usually quote $\hbar\omega$ in wavenumbers (cm$^{-1}$) or in eV. Converting via $E = hc\tilde{\nu}$ with $hc = 1.240 \times 10^{-4}\,\text{eV·cm}$:

for the HCl stretching mode, where $\tilde{\nu} = 2886\,\text{cm}^{-1}$ is the measured fundamental absorption wavenumber. Every upward transition $n \to n+1$ absorbs a photon of this energy.

The zero-point energy is observable

Even at absolute zero a molecule does not sit still at the bottom of its potential well. The ground-state energy

is the zero-point energy — a direct consequence of the Heisenberg uncertainty principle. For HCl, $E_0 \approx 0.179\,\text{eV}$, which is large enough to affect the equilibrium bond length and isotope-dependent properties. Replacing the hydrogen with deuterium (D) raises $\mu$ and therefore changes $\omega$ and $E_0$, shifting every absorption line. This isotope shift was one of the earliest precise tests of the quantum oscillator model.

Infrared spectroscopy

A diatomic molecule can absorb or emit infrared radiation when its electric dipole moment changes during vibration. The selection rule for a harmonic oscillator is $\Delta n = \pm 1$, so the spectrum consists of a single line at energy $\hbar\omega$ (the fundamental). Real molecules are not perfectly harmonic — anharmonicity makes the potential asymmetric and slightly softens the ladder, producing weak overtone lines near $2\hbar\omega$, $3\hbar\omega$, and so on, each slightly below the harmonic prediction. The lowest-order correction comes from the Morse potential, but for small $n$ the harmonic approximation is excellent.

Try it

This is a numerical exercise — your code should return a number. A diatomic molecule has a vibrational quantum $\hbar\omega = 0.36\,\text{eV}$ (close to the HCl stretching mode). Compute the energy (in eV) of the first excited vibrational level $n = 1$ using $E_n = \hbar\omega(n + \frac{1}{2})$.

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