The Harmonic Oscillator and Vibrational Levels

Hold a spring with a mass on the end, stretch it, and let go. It oscillates at a rate set by the spring's stiffness and the mass. A chemical bond behaves the same way. It vibrates thousands of billions of times per second. This happens whenever a molecule absorbs infrared light. The harmonic oscillator model converts that mechanical picture into something precise. It becomes a testable prediction for where vibrational absorption lines appear. This model matters because it is the single cheapest way to extract a bond's stiffness. That stiffness is called its force constant. A wavenumber measured to four significant figures on an infrared spectrometer becomes useful after a short calculation. It turns into a number in newtons per meter. That number tells you whether a bond is single, double, or triple. This article builds the vibrational term $G(v)$ from Newton's second law. It shows how the fundamental frequency depends on force constant and reduced mass. It also works two full numerical examples. You will see where the model succeeds. You will also see where it quietly fails. This sets up the anharmonicity correction covered next.

Key takeaways

A diatomic bond vibrates like a mass on a spring; classical mechanics gives an angular frequency $\omega = \sqrt{k/\mu}$ .
Quantization restricts the vibrational energy to evenly spaced levels, described by the term $G(v) = \tilde\nu_e(v + 1/2)$ .
The fundamental vibrational wavenumber $\tilde\nu_e$ depends on the force constant $k$ and the reduced mass $\mu$ , not on the individual atomic masses alone.
Heavier isotopes vibrate more slowly; this isotope shift is a direct, checkable consequence of the reduced-mass dependence.
The harmonic model predicts equally spaced levels and forbids overtones in the strict selection rule; real molecules deviate, which is the signature of anharmonicity.
Key idea: the vibrational spectrum of a bond is a direct fingerprint of two numbers, $k$ and $\mu$ , connected through $\tilde\nu_e = \frac{1}{2\pi c}\sqrt{k/\mu}$ .

The core insight

Treating a vibrating bond as a quantum harmonic oscillator converts a single measured infrared wavenumber into a force constant, because the oscillator's allowed energies depend only on the stiffness of the restoring force and the reduced mass of the two vibrating nuclei.

The physical picture

A diatomic molecule has one internal vibrational coordinate: the bond length $r$ , oscillating about its equilibrium value $r_e$ . Near the potential minimum, any smooth potential energy curve $V(r)$ can be expanded in a Taylor series. The first derivative vanishes at the minimum, so the leading term is quadratic in the displacement $x = r - r_e$ :

V(x) \approx \tfrac{1}{2}kx^2

This is exactly Hooke's law, already familiar from the classical force-constant treatment of infrared spectroscopy. What is new here is quantization. Solving the Schrödinger equation for this parabolic potential gives evenly spaced energy levels, rather than the continuum of energies a classical spring could have. The two-body vibration is first reduced to an equivalent one-body problem using the reduced mass:

\mu = \frac{m_1 m_2}{m_1+m_2}

so that a diatomic molecule vibrating along its bond behaves exactly like a single particle of mass $\mu$ on a spring of constant $k$ , oscillating about a fixed point. The vibrational quantum number $v = 0, 1, 2, \dots$ labels the levels. The zero-point energy, $\tfrac{1}{2}h c \tilde\nu_e$ , is a purely quantum feature: even in the lowest state the bond is never perfectly at rest, a direct consequence of the uncertainty principle applied to a confined particle.

Figure

Harmonic potential well with evenly spaced vibrational levels

The spacing between adjacent levels is constant in this model: every rung of the ladder is the same height, $h c \tilde\nu_e$ . That constancy is both the model's greatest strength — it gives one number, $\tilde\nu_e$ , that predicts the whole ladder — and its main limitation, since real bonds anharmonically compress the ladder as $v$ increases.

Figure

Isotope effect: heavier reduced mass lowers the fundamental wavenumber

The core relations

\omega = \sqrt{\frac{k}{\mu}}, \qquad \tilde\nu_e = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}

where $\omega$ is the classical angular frequency (rad/s), $k$ is the bond force constant (N/m), $\mu$ is the reduced mass (kg), $c$ is the speed of light, and $\tilde\nu_e$ is the fundamental vibrational wavenumber (cm⁻¹ when $c$ is in cm/s).

G(v) = \tilde\nu_e\left(v + \tfrac{1}{2}\right), \qquad v = 0, 1, 2, \dots

where $G(v)$ is the vibrational term (in cm⁻¹), obtained by dividing the energy $E_v$ by $hc$ .

E_v = hc\,\tilde\nu_e\left(v + \tfrac{1}{2}\right)

where $E_v$ is the vibrational energy of level $v$ , $h$ is Planck's constant.

\Delta G_{v+1/2} = G(v+1) - G(v) = \tilde\nu_e

where the spacing between any two adjacent levels equals the fundamental wavenumber itself, the hallmark of a strictly harmonic ladder.

\mu = \frac{m_1 m_2}{m_1 + m_2}

where $m_1, m_2$ are the masses of the two vibrating atoms; for a diatomic composed of atoms of very different mass, $\mu$ is dominated by the lighter atom.

Derivation AFrom Newton's second law to the vibrational term

Goal. Derive $G(v) = \tilde\nu_e(v+1/2)$ starting from the classical restoring force on a diatomic bond, then quantize.

Step 1: Classical equation of motion. For small displacements $x = r - r_e$ , Hooke's law gives a restoring force $F = -kx$ . Newton's second law for the reduced-mass particle reads

\mu \frac{d^2x}{dt^2} = -kx

Step 2: Classical frequency. This is the equation of simple harmonic motion, with solution $x(t) = A\cos(\omega t + \phi)$ provided

\omega^2 = \frac{k}{\mu}

so the classical oscillation frequency depends only on the ratio of stiffness to inertia — a stiffer bond or a lighter reduced mass both raise $\omega$ .

Step 3: Quantize the potential. The quantum Hamiltonian for a particle of mass $\mu$ in the potential $V(x) = \tfrac12 kx^2$ is

\hat{H} = -\frac{\hbar^2}{2\mu}\frac{d^2}{dx^2} + \tfrac12 kx^2

Solving the Schrödinger equation $\hat{H}\psi = E\psi$ for this potential (a standard result, not re-derived here) yields discrete eigenvalues

E_v = \hbar\omega\left(v + \tfrac12\right), \qquad v = 0,1,2,\dots

Step 4: Convert to spectroscopic units. Divide $E_v$ by $hc$ to obtain the term value in wavenumbers, and use $\omega = 2\pi c\,\tilde\nu_e$ :

G(v) = \frac{E_v}{hc} = \frac{\hbar\omega}{hc}\left(v+\tfrac12\right) = \frac{\omega}{2\pi c}\left(v+\tfrac12\right)

Step 5: Identify the fundamental wavenumber. Substituting $\omega = \sqrt{k/\mu}$ gives the working formula

\tilde\nu_e = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}

so that

\boxed{G(v) = \tilde\nu_e\left(v + \tfrac12\right), \qquad \tilde\nu_e = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}}

Worked examples

Example 1 — Force constant of H³⁵Cl from its fundamental. The fundamental vibrational absorption of gas-phase H³⁵Cl appears at $\tilde\nu_e = 2990\ \text{cm}^{-1}$ . Find the force constant of the H–Cl bond. Step 1: Compute the reduced mass. Atomic masses: $m_H = 1.008\ \text{u}$ , $m_{Cl} = 34.97\ \text{u}$ (³⁵Cl).

\mu = \frac{(1.008)(34.97)}{1.008+34.97}\ \text{u} = \frac{35.26}{35.98}\ \text{u} = 0.9800\ \text{u}

Convert to kilograms using $1\ \text{u} = 1.6605\times10^{-27}\ \text{kg}$ :

\mu = 0.9800 \times 1.6605\times10^{-27}\ \text{kg} = 1.6273\times10^{-27}\ \text{kg}

Step 2: Rearrange the core relation for $k$ :

k = \mu\,(2\pi c\,\tilde\nu_e)^2

Step 3: Insert numbers, with $c = 2.998\times10^{10}\ \text{cm s}^{-1}$ so that $\tilde\nu_e$ stays in cm⁻¹:

2\pi c\,\tilde\nu_e = 2\pi \times (2.998\times10^{10}\ \text{cm s}^{-1}) \times (2990\ \text{cm}^{-1}) = 5.633\times10^{14}\ \text{s}^{-1}

Step 4: Square and multiply by $\mu$ :

k = (1.6273\times10^{-27}\ \text{kg}) \times (5.633\times10^{14}\ \text{s}^{-1})^2

k = 1.6273\times10^{-27} \times 3.173\times10^{29}\ \text{N/m} = 516\ \text{N/m}

Sanity check. Literature value for H–Cl is $k \approx 516$ – $518\ \text{N/m}$ , consistent with a strong, stiff single bond (typical single-bond force constants fall in the 400–500 N/m range, roughly five times looser than a typical C=O double bond near 1200 N/m). The number is physically sensible.

Example 2 — Isotope shift: predicting the D³⁵Cl fundamental. Using the same force constant $k = 516\ \text{N/m}$ (bond stiffness is essentially unchanged by isotopic substitution, since it depends on electronic structure), predict the fundamental wavenumber of D³⁵Cl, and compare with the experimental value of 2145 cm⁻¹. Step 1: Compute the new reduced mass, with $m_D = 2.014\ \text{u}$ .

\mu_D = \frac{(2.014)(34.97)}{2.014+34.97}\ \text{u} = \frac{70.43}{36.98}\ \text{u} = 1.9047\ \text{u}

\mu_D = 1.9047 \times 1.6605\times10^{-27}\ \text{kg} = 3.1626\times10^{-27}\ \text{kg}

Step 2: Note the isotope ratio directly, without recomputing everything from scratch:

\frac{\tilde\nu_e(D)}{\tilde\nu_e(H)} = \sqrt{\frac{\mu_H}{\mu_D}} = \sqrt{\frac{1.6273\times10^{-27}}{3.1626\times10^{-27}}} = \sqrt{0.5146} = 0.7174

Step 3: Apply the ratio to the measured H³⁵Cl fundamental.

\tilde\nu_e(D) = 0.7174 \times 2990\ \text{cm}^{-1} = 2145\ \text{cm}^{-1}

Sanity check. This matches the experimental D³⁵Cl fundamental (2145 cm⁻¹, i.e. 2144 cm⁻¹ in careful compilations) almost exactly. The near-perfect agreement is itself evidence that the force constant truly is isotope-independent to a very good approximation, and that the harmonic reduced-mass scaling captures the physics correctly. Note the ratio is close to, but not exactly, $1/\sqrt{2}\approx0.7071$ ; the small deviation comes from the Cl mass entering $\mu$ at both isotopologues rather than $\mu$ scaling purely as $m_H$ or $m_D$ .

Limiting cases & common traps

Situation	What the harmonic model says	Reality check
Low $v$ (near bottom of well)	Levels evenly spaced by $\tilde\nu_e$	Excellent approximation; fundamental transitions well predicted
High $v$ (near dissociation)	Still predicts equal spacing forever	Fails — real levels converge and the molecule dissociates; needs anharmonic correction
Overtones ( $\Delta v = \pm2, \pm3,\dots$ )	Strictly forbidden ( $\Delta v = \pm1$ only)	Weakly observed in real spectra due to anharmonicity and mechanical/electrical anharmonic mixing
Very light atom paired with heavy atom (e.g. H–I)	$\mu \approx m_H$ , nearly independent of the heavy partner	Confirmed experimentally; C–H, N–H, O–H stretches cluster near the same wavenumber region regardless of the rest of the molecule
Comparing isotopologues	$\tilde\nu_e \propto \sqrt{1/\mu}$ , same $k$	Confirmed to a few cm⁻¹, as in the H/D example above

The chemist behind it

Robert Mulliken and, earlier, Gerhard Herzberg were central to turning infrared and Raman band positions into quantitative force constants and molecular term diagrams, but the foundational spectroscopic bookkeeping used throughout this article — vibrational term symbols, term-value notation $G(v)$ , and the systematic tabulation of diatomic constants — is due to Gerhard Herzberg, whose monumental compilations of diatomic molecular spectra set the notation still used today.

Origins · 1 minGerhard Herzberg and how this was discovered

Gerhard Herzberg

1904–1999 · Nobel laureate whose spectroscopic term-value framework, including G(v) notation, remains the standard language of molecular vibrational spectroscopy

Ask the tutor: How does adding an anharmonicity term change the vibrational term $G(v)$ , and how does that let us find a bond's dissociation energy from its overtone spectrum?

Why does the harmonic oscillator predict evenly spaced vibrational levels?

The model assumes the restoring force is exactly proportional to displacement, $F=-kx$ , which is Hooke's law. Solving the Schrödinger equation for this exact quadratic potential produces energy eigenvalues that increase by a fixed increment, $hc\tilde\nu_e$ , for every unit increase in $v$ . This even spacing is a mathematical consequence of the parabola's perfect symmetry. Real bond potentials are not perfectly parabolic; they flatten out at large separations because bonds eventually break. That asymmetry is what anharmonicity corrects, and it is why real vibrational levels converge slightly as $v$ increases rather than staying evenly spaced forever.

Why use reduced mass instead of the mass of one atom?

Both atoms move during a vibration; the center of mass stays fixed while the bond length oscillates. Transforming the two-body problem into an equivalent one-body problem, using the coordinate $x=r-r_e$ , requires an effective mass that properly weights both atoms' inertia. That effective mass is the reduced mass, $\mu = m_1m_2/(m_1+m_2)$ . It always lies below the lighter atom's mass, and is dominated by whichever atom is lighter, which is why C–H and O–H stretches are so sensitive to hydrogen versus deuterium substitution.

Can I use this model to predict overtone bands?

Not accurately. The pure harmonic oscillator's selection rule permits only $\Delta v = \pm1$ transitions, forbidding overtones entirely in the idealized model. Overtones are observed in real infrared spectra, but weakly, and their existence is direct experimental proof that real bonds are anharmonic. Predicting overtone positions requires the anharmonic correction to $G(v)$ , typically written with an added $-x_e\tilde\nu_e(v+1/2)^2$ term, which is the natural next step beyond this harmonic treatment.

How precisely can a force constant be determined this way?

Given a well-resolved fundamental wavenumber, typically known to better than 1 cm⁻¹ from gas-phase infrared spectra, and accurate atomic masses, the force constant can be obtained to three or four significant figures. The main systematic error comes from using the harmonic approximation itself: because the true fundamental is slightly shifted by anharmonicity, spectroscopists prefer the equilibrium wavenumber $\omega_e$ extracted from a full vibration-rotation analysis rather than the raw observed fundamental, for the most rigorous force constants.

Does the harmonic oscillator model apply to polyatomic molecules?

Directly, no — polyatomic molecules vibrate in complicated, coupled ways involving many atoms simultaneously. However, after a coordinate transformation, their vibrations can be decomposed into independent normal modes, each of which behaves, to a good approximation, like its own independent harmonic oscillator with its own effective mass and force constant. That decomposition is the subject of normal mode analysis, which builds directly on the single-oscillator framework developed here.