The Harmonic Oscillator Model of Vibration

Every diatomic molecule is held together by a potential energy curve that the Born-Oppenheimer approximation hands to us ready-made: solve the electrons at each fixed nuclear separation $R$ and the electronic energy $V(R)$ becomes the potential the nuclei move in. That curve has a minimum at the equilibrium bond length $R_e$ , where attraction and repulsion cancel and the bond is most stable. The harmonic oscillator is what happens when you keep only its behaviour right at the bottom of the well. The move is a Taylor expansion. The value $V(R_e)$ is an energy offset, the slope $V'(R_e)$ vanishes at a minimum, and the first term that matters is the curvature $\tfrac12 V''(R_e)(R-R_e)^2$ . That is a parabola, and a parabola is a spring: the restoring force is proportional to displacement with force constant $k=V''(R_e)$ . Separating the two-body problem into centre-of-mass drift and internal stretching, the stretch obeys a one-dimensional oscillator equation with the reduced mass $\mu$ in place of a single particle's mass. Quantise that oscillator and two features fall out that no classical spring possesses. The energies form a perfectly uniform ladder, $E_v=(v+\tfrac12)\hbar\omega$ with $\omega=\sqrt{k/\mu}$ , so every jump between neighbouring rungs costs the same $\hbar\omega$ and produces one sharp spectral line. And the lowest rung sits not at the bottom of the well but a finite $\tfrac12\hbar\omega$ above it — the zero-point energy, the molecule's irreducible quantum tremble that survives even at absolute zero. This single parabola is the reference frame for all of vibrational spectroscopy: infrared bands, vibration-rotation branches, isotope shifts, and Franck-Condon overlaps all begin here and are corrected afterwards by anharmonicity.

Key takeaways

Expand any molecular potential $V(R)$ about its minimum $R_e$ and the first surviving term is quadratic, $V\approx\tfrac12 k(R-R_e)^2$ with force constant $k=V''(R_e)$ — every bound bond looks like a spring up close.
The two-body problem separates so the stretch obeys a one-dimensional oscillator with reduced mass $\mu=m_1m_2/(m_1+m_2)$ ; its natural frequency is $\omega=\sqrt{k/\mu}$ .
Quantising gives an evenly spaced ladder $E_v=(v+\tfrac12)\hbar\omega$ , so a single fundamental frequency governs absorption and emission under the selection rule $\Delta v=\pm1$ .
The ground state carries zero-point energy $\tfrac12\hbar\omega$ , a purely quantum consequence of the uncertainty principle — the nuclei never sit still, even at $T=0$ .
For H–Cl the model gives $k\approx516\ \mathrm{N/m}$ and a zero-point energy of $1495\ \mathrm{cm^{-1}}$ (0.185 eV) from the harmonic wavenumber $\tilde\nu\approx2990\ \mathrm{cm^{-1}}$ ; the model is the reference the Morse potential later corrects.
Key idea: any smooth potential looks parabolic near its minimum, so small-amplitude nuclear vibration is universally a harmonic oscillator with $\omega=\sqrt{k/\mu}$ , equally spaced levels $(v+\tfrac12)\hbar\omega$ , and an irreducible zero-point energy $\tfrac12\hbar\omega$ .

Where this lives today

The harmonic oscillator is the interpretive backbone of every infrared and Raman spectrometer in use today. A Fourier-transform infrared instrument records a sample's fundamental bands in seconds, and the software that identifies functional groups — the C=O stretch near 1700 cm⁻¹, the O–H near 3300 cm⁻¹, the C≡N near 2250 cm⁻¹ — is reading off harmonic frequencies set by $\sqrt{k/\mu}$ . Environmental monitoring uses the same physics in reverse: because CO₂, CH₄, and N₂O have infrared-active vibrations while N₂ and O₂ do not, their harmonic fundamentals are the absorption fingerprints that satellite spectrometers track to map greenhouse gases. At the frontier, the harmonic picture underpins vibrational strong coupling in optical microcavities, where a molecular vibration is tuned into resonance with a cavity photon to form vibrational polaritons that can alter chemical reactivity. Cold-molecule metrology pushes it hardest: by cooling diatomics to microkelvin temperatures and probing individual vibrational levels with frequency-comb lasers, experiments use the molecule's own ladder as a clock sensitive to a possible drift in the proton-to-electron mass ratio.

The physical picture

Zoom in on the bottom of any smooth potential well and the curve becomes indistinguishable from a parabola. This is a theorem about minima, not a special property of molecules: the first non-vanishing term of a Taylor series at a minimum is the quadratic one, because the constant is an offset and the linear slope is zero. So the harmonic oscillator is not a model you choose but the universal small-amplitude limit every bound vibration is forced into near equilibrium. The full molecular curve rises steeply as the atoms are pushed together and flattens to a dissociation plateau as they are pulled apart, but for the low-lying levels thermal molecules occupy, only the curvature at the bottom is felt.

Figure

Any bound potential V of R becomes a parabola near its minimum: the true curve for H–Cl in green, R_e = 1.27 Å, and its harmonic fit one-half k times the squared displacement in blue coincide at the bottom, where levels are equally spaced by ħω ≈ 2990 cm⁻¹ and the v = 0 state floats one-half ħω ≈ 1495 cm⁻¹ above the minimum

Because the parabolic force is strictly linear in displacement, a classical particle in it swings with a period that does not depend on amplitude. Quantum mechanically this becomes the most striking feature of the spectrum: the levels are a perfectly uniform ladder, every rung separated from its neighbour by the same $\hbar\omega$ . That equal spacing is why a heteronuclear diatomic absorbs at one sharp fundamental frequency, and departures from it at high excitation are the signature of the real curve peeling away from its parabola.

Figure

The equally spaced harmonic ladder on the left versus the crowding real rungs of H–Cl on the right: the Δv = +1 selection rule allows only neighbour jumps, so absorption piles into a single fundamental band at ω̃ = 2990 cm⁻¹ while anharmonicity squeezes the true levels together toward dissociation

Zero-point energy is forced on the ground state by the uncertainty principle. Pinning the nuclei exactly at $R_e$ — zero spread in position — would demand infinite momentum spread and therefore infinite kinetic energy, which the well cannot supply. The molecule settles instead on a Gaussian cloud of finite width whose kinetic and potential energies balance to leave it $\tfrac12\hbar\omega$ above the bottom. For H–Cl that residual amounts to a root-mean-square stretch of only about 0.076 Å, roughly 6 percent of the 1.27 Å bond, yet it never switches off. A stiffer bond (larger $k$ ) or lighter atoms (smaller $\mu$ ) raise $\omega$ , which is why C–H and O–H stretches ring at high infrared frequencies while heavy-atom stretches lie low.

The core equations

The starting point is the Taylor expansion of the Born-Oppenheimer potential about its minimum. The constant term is a reference offset, the linear term vanishes at a minimum, and the surviving quadratic term defines the spring; cubic and higher terms are the anharmonic corrections we discard here and restore later through the Morse potential.

V(R)=V(R_e)+\tfrac12\,V''(R_e)\,(R-R_e)^2+\tfrac16\,V'''(R_e)\,(R-R_e)^3+\dots \qquad (1)

where: $V(R)$ — molecular potential energy curve, the Born-Oppenheimer electronic energy; $R_e$ — equilibrium internuclear separation, the location of the minimum; $V''(R_e)\equiv k$ — curvature at the minimum, which is the force constant; $V'''(R_e)$ — leading cubic term, neglected in the harmonic model. Near $R_e$ every smooth bond curve is a parabola of stiffness $k$ ; the whole harmonic edifice rests on truncating after the quadratic term. With the parabola in hand and the two-body problem reduced to the relative coordinate $x=R-R_e$ carrying the reduced mass $\mu$ , the internal motion obeys a one-dimensional Schrödinger equation — the exactly solvable problem whose spectrum defines the model.

-\frac{\hbar^2}{2\mu}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}+\tfrac12 k x^2\,\psi=E\,\psi,\qquad x=R-R_e \qquad (2)

where: $\mu=m_1 m_2/(m_1+m_2)$ — reduced mass of the two nuclei; $x$ — displacement of the bond from equilibrium; $k$ — force constant $V''(R_e)$ ; $\psi$ — vibrational wavefunction. The reduced mass is the payoff of separating centre-of-mass motion: the diatomic stretch is identical to one particle of mass $\mu$ on a spring of stiffness $k$ . Solving equation (2) — whether by the Hermite-function series or by the ladder operators of the first derivation below — yields the celebrated evenly spaced spectrum, the reason vibrational spectroscopy is so clean.

E_v=\left(v+\tfrac12\right)\hbar\omega,\qquad v=0,1,2,\dots,\qquad \omega=\sqrt{\frac{k}{\mu}} \qquad (3)

where: $v$ — vibrational quantum number, a non-negative integer; $\omega=\sqrt{k/\mu}$ — angular vibration frequency in rad s⁻¹; $\hbar\omega$ — the constant spacing between any two adjacent levels. Every transition $v\to v+1$ costs the same $\hbar\omega$ , so absorption and emission collapse onto a single fundamental frequency. Setting $v=0$ in equation (3) isolates the feature that most sharply separates the quantum oscillator from a classical spring: the ground state does not rest at the bottom of the well, but carries a zero-point energy present at any temperature down to absolute zero.

E_0=\tfrac12\hbar\omega \qquad (4)

where: $E_0$ — energy of the $v=0$ ground state above the potential minimum. It is no small correction: it shifts dissociation energies, drives kinetic isotope effects, and is directly observed in every molecular spectrum. The eigenfunctions accompanying these energies are Gaussian-weighted Hermite polynomials, $\psi_v(x)\propto H_v(\sqrt\alpha\,x)\,e^{-\alpha x^2/2}$ with $\alpha=\mu\omega/\hbar$ : the $v=0$ state is a pure Gaussian centred on $R_e$ , whose width sets the zero-point amplitude, and the higher states have $v$ nodes. Spectroscopists rarely quote $\omega$ in rad s⁻¹; they use the wavenumber, the reciprocal vacuum wavelength of the absorbed photon, which is the number that actually appears on an infrared spectrum.

\tilde\nu=\frac{\omega}{2\pi c}=\frac{1}{2\pi c}\sqrt{\frac{k}{\mu}} \qquad (5)

where: $\tilde\nu$ — vibrational wavenumber in cm⁻¹, the standard spectroscopic measure; $c$ — speed of light. Read backwards, a measured band position delivers the bond force constant directly — exactly what the second derivation exploits. Absorption requires more than an allowed energy gap: the molecule must present an oscillating dipole to the field. Expanding the electric dipole moment about $R_e$ and keeping the linear term gives both the requirement for infrared activity and the selection rule.

\mu_{\mathrm{el}}(R)\approx\mu_{\mathrm{el}}(R_e)+\left(\frac{\mathrm{d}\mu_{\mathrm{el}}}{\mathrm{d}R}\right)_{R_e}\!x\ \Rightarrow\ \Delta v=\pm1 \qquad (6)

where: $\mu_{\mathrm{el}}(R)$ — electric dipole moment versus bond length; $(\mathrm{d}\mu_{\mathrm{el}}/\mathrm{d}R)_{R_e}$ — slope of the dipole at equilibrium, which must be non-zero for infrared activity. Because $x$ connects only neighbouring Hermite functions, the harmonic selection rule is strictly $\Delta v=\pm1$ , and homonuclear diatomics, whose dipole is zero for all $R$ , have no infrared fundamental.

Derivation AThe energy ladder from raising and lowering operators

The goal is to obtain the entire spectrum $E_v=(v+\tfrac12)\hbar\omega$ and the zero-point energy algebraically, without solving a differential equation, by factorising the Hamiltonian into operators that step between levels. This exposes the physics: the spectrum is uniform because a single operation adds exactly one quantum $\hbar\omega$ , and the ground state is fixed by a first-order condition rather than a boundary condition.

Step 1. Define $a=\sqrt{\mu\omega/2\hbar}\,(x+ip/\mu\omega)$ and $a^\dagger=\sqrt{\mu\omega/2\hbar}\,(x-ip/\mu\omega)$ . Using $[x,p]=i\hbar$ , a short computation gives $[a,a^\dagger]=1$ . Why: this single commutation relation is the engine of everything that follows — it is what makes each rung differ from the next by a fixed amount rather than a variable one.

Step 2. Rewrite $\hat H=p^2/2\mu+\tfrac12 k x^2$ in terms of these operators; the cross terms and commutator combine to $\hat H=\hbar\omega\,(a^\dagger a+\tfrac12)$ , with number operator $\hat N=a^\dagger a$ . Why: the stray $\tfrac12$ is no bookkeeping choice — it falls out of the commutator when $x^2$ and $p^2$ are symmetrised, and it is the algebraic origin of the zero-point energy.

Step 3. The ground state obeys $a\,\psi_0=0$ , since no state can have negative occupation; acting with $\hat H$ gives $\hat H\psi_0=\tfrac12\hbar\omega\,\psi_0$ , so $E_0=\tfrac12\hbar\omega$ . Why: the lowest energy is set by an annihilation condition, not the well bottom — exactly why the molecule cannot sit at $R_e$ with zero energy.

Step 4. Because $[\hat H,a^\dagger]=\hbar\omega\,a^\dagger$ , applying $a^\dagger$ raises any eigenstate's energy by exactly $\hbar\omega$ : $\hat H(a^\dagger\psi_v)=(E_v+\hbar\omega)(a^\dagger\psi_v)$ . Climbing from $\psi_0$ gives $E_v=(v+\tfrac12)\hbar\omega$ . Why: each step adds the same quantum, the algebraic statement of the equal spacing and the reason the fundamental band is a single line. *Verified: matches the Hermite-function solution of equation (2); standard in Loudon, The Quantum Theory of Light, and Bransden and Joachain Ch. 10.*

Go deeper · Derivation B · 1 minForce constant and isotope shift from a measured band

Derivation BForce constant and isotope shift from a measured band

The goal is to run the model in reverse: given only the fundamental wavenumber $\tilde\nu$ and the atomic masses, extract the bond force constant $k$ and predict how the frequency and zero-point energy shift under isotopic substitution. This is the everyday working use of the harmonic model, turning the abstract $k$ into a number in newtons per metre that characterises the chemical bond.

Step 1. Invert equation (5). Squaring $\tilde\nu=(1/2\pi c)\sqrt{k/\mu}$ and solving for $k$ gives $k=\mu(2\pi c\,\tilde\nu)^2$ , with $\mu$ computed from the two atomic masses converted to kilograms. Why: this turns a spectral line position into a mechanical bond property and lets one compare bond stiffness across molecules — CO at about 1900 N/m is far stiffer than H–Cl at about 516 N/m.

Step 2. Substitute a different isotope. The force constant $k=V''(R_e)$ is purely electronic and identical for two isotopologues, so only $\mu$ changes, and equation (5) gives $\tilde\nu_2/\tilde\nu_1=\sqrt{\mu_1/\mu_2}$ . Why: this clean mass scaling is a stringent test of the harmonic picture and the basis of isotopic labelling in spectroscopy — replacing H by D nearly doubles $\mu$ for an X–H bond and drops the frequency by close to $\sqrt2$ .

Step 3. Track the zero-point energy. Since $E_0=\tfrac12\hbar\omega\propto\mu^{-1/2}$ , a heavier isotopologue sits lower in its well and is therefore slightly more strongly bound. Why: this differential zero-point energy is the physical origin of the kinetic isotope effect, where a C–D bond reacts more slowly than a C–H bond simply because it starts from a lower rung. *Verified: dimensional check $\mu(2\pi c\tilde\nu)^2$ has units $\mathrm{kg\,s^{-2}=N\,m^{-1}}$ ; H–Cl at $\tilde\nu=2990\ \mathrm{cm^{-1}}$ gives $k=516\ \mathrm{N/m}$ and the H–Cl to D–Cl ratio 0.717, both reproduced in the worked examples below.*

Limiting cases

The harmonic result is exact only for a true parabola; real molecules depart from it, and how they depart is itself informative.

Case	Result	Meaning
Small $v$ (bottom of well)	$E_v\approx(v+\tfrac12)\hbar\omega$	the parabolic approximation is excellent; the low-lying levels of most diatomics are nearly equally spaced and thermally populated at room temperature
Large $v$ (near dissociation)	$E_v=(v+\tfrac12)\hbar\omega-\hbar\omega x_e(v+\tfrac12)^2+\dots$	levels crowd together and converge to the dissociation limit; the harmonic model fails and the Morse potential is required
Heavy-atom limit	$\omega=\sqrt{k/\mu}\to0$	as $\mu\to\infty$ the spacing collapses toward a classical continuum; isotopic substitution shifts $\omega$ by $\sqrt{\mu_1/\mu_2}$
Classical correspondence	$\hbar\omega\ll E_v$	at large $v$ neighbouring levels are close relative to the total energy and the quantum ladder merges with a classical oscillator of frequency $\omega/2\pi$

Common trapThe vibrating mass is the reduced mass, and equal spacing is only the harmonic idealisation

Two errors travel together here. First, the mass in $\omega=\sqrt{k/\mu}$ is the reduced mass $\mu=m_1m_2/(m_1+m_2)$ , never a single atomic mass — for H–Cl, $\mu=0.980$ u sits just below the hydrogen mass because chlorine is so much heavier, and using 1.008 u or 35 u instead throws the force constant off by a large factor. Second, the perfectly even ladder $E_v=(v+\tfrac12)\hbar\omega$ holds only for the parabola. Real molecules are anharmonic: the levels crowd together as $v$ rises, the overtone $0\to2$ falls noticeably below twice the fundamental, and the ladder terminates at dissociation rather than climbing forever. Treating high- $v$ levels as equally spaced misplaces overtones and hides the very dissociation the Morse potential is built to capture. The harmonic model is the correct leading-order description near the bottom, not a description of the whole well.

Worked examples

1 · Force constant and zero-point energy of hydrogen chloride

The harmonic (equilibrium) vibrational wavenumber of H–35Cl is $\tilde\nu\approx2990\ \mathrm{cm^{-1}}$ . Using atomic masses $m_{\mathrm H}=1.008$ u and $m_{\mathrm{Cl}}=34.97$ u, find the reduced mass, the bond force constant, and the zero-point energy in cm⁻¹, eV, and kJ/mol.

Step 1. Reduced mass. $\mu=\dfrac{m_{\mathrm H}m_{\mathrm{Cl}}}{m_{\mathrm H}+m_{\mathrm{Cl}}}=\dfrac{1.008\times34.97}{35.98}=0.980\ \mathrm{u}$ . Converting, $\mu=0.980\times1.6605\times10^{-27}=\mathbf{1.627\times10^{-27}\ kg}$ . Note it lies just below the hydrogen mass, as it must when one partner is far heavier.

Step 2. Angular frequency. With $c=2.998\times10^{10}\ \mathrm{cm\,s^{-1}}$ , $\omega=2\pi c\,\tilde\nu=2\pi(2.998\times10^{10})(2990)=\mathbf{5.63\times10^{14}\ rad\,s^{-1}}$ .

Step 3. Force constant. $k=\mu\omega^2=(1.627\times10^{-27})(5.63\times10^{14})^2=\mathbf{516\ N/m}$ . This stiffness is the electronic property that stays fixed under isotope substitution.

Step 4. Zero-point energy. $E_0=\tfrac12\hbar\omega=\tfrac12(1.0546\times10^{-34})(5.63\times10^{14})=2.97\times10^{-20}\ \mathrm J$ . In spectroscopic units $\tfrac12\tilde\nu=\mathbf{1495\ cm^{-1}}$ ; dividing by $1.602\times10^{-19}$ gives $\mathbf{0.185\ eV}$ , and multiplying by Avogadro's number gives $\mathbf{17.9\ kJ/mol}$ — locked into the bond even at absolute zero, several times the room-temperature thermal energy $RT\approx2.5\ \mathrm{kJ/mol}$ .

2 · The deuteration isotope shift, H–Cl to D–Cl

Deuterium substitution replaces H by D ( $m_{\mathrm D}=2.014$ u) while leaving the electronic force constant $k$ unchanged. Predict the D–Cl fundamental wavenumber and the ratio of zero-point energies.

Step 1. New reduced mass. $\mu_{\mathrm{DCl}}=\dfrac{2.014\times34.97}{36.98}=\mathbf{1.904\ u}$ — not quite twice $\mu_{\mathrm{HCl}}=0.980$ u (the ratio is $1.94$ ), because chlorine's finite mass keeps the doubling from being exact.

Step 2. Frequency ratio. Since $k$ is unchanged, $\dfrac{\tilde\nu_{\mathrm{DCl}}}{\tilde\nu_{\mathrm{HCl}}}=\sqrt{\dfrac{\mu_{\mathrm{HCl}}}{\mu_{\mathrm{DCl}}}}=\sqrt{\dfrac{0.980}{1.904}}=\sqrt{0.514}=\mathbf{0.717}$ . So D–Cl vibrates at $2990\times0.717=\mathbf{2144\ cm^{-1}}$ , matching the measured band near $2145\ \mathrm{cm^{-1}}$ .

Step 3. Interpret the near- $\sqrt2$ drop. The naive $1/\sqrt2=0.707$ (from exactly doubling the mass) slightly overshoots the observed 0.717, because the shift is governed by the reduced mass, not the hydrogen mass. The zero-point energies scale identically: $E_0^{\mathrm{DCl}}/E_0^{\mathrm{HCl}}=0.717$ , so D–Cl sits about $1495\times(1-0.717)=\mathbf{423\ cm^{-1}}$ deeper in its well — the extra binding that slows deuterated reactions and separates isotopes spectroscopically.

Physics of the portrait

The harmonic oscillator of a molecule is where the abstract quantum ladder meets the laboratory: someone had to measure thousands of diatomic bands, extract their force constants, and tabulate the constants that made the model quantitative. That labour, more than any single equation, is why $\omega=\sqrt{k/\mu}$ is a working tool rather than a curiosity.

Origins · 1 minGerhard Herzberg and how this was discovered

Gerhard Herzberg

1904–1999 · molecular spectroscopist who tabulated the vibrational and rotational constants of diatomic molecules

Gerhard Herzberg turned molecular spectroscopy from an art into a reference science. A German-born physicist who fled to Canada in 1935 and built the spectroscopy laboratory of the National Research Council in Ottawa into the world's leading centre for the field, he spent decades measuring the band spectra of diatomic and small polyatomic molecules. His treatise Molecular Spectra and Molecular Structure — the first volume, on diatomic molecules, is the one every spectroscopist still reaches for — laid out the theory of the vibrating-rotating molecule and tabulated the harmonic constants $\omega_e$ , the anharmonicity $\omega_e x_e$ , and the derived force constants for a vast catalogue of bonds. Those tables are what let a chemist read a force constant off a band position using exactly the harmonic relation $k=\mu(2\pi c\tilde\nu)^2$ . Herzberg also identified molecular species in stellar and cometary spectra and made the first laboratory detection of several free radicals, work for which he received the 1971 Nobel Prize in Chemistry. The harmonic oscillator is the first line of his diatomic volume; everything anharmonic is a correction to it.

History

1900

— Planck introduces quantised oscillator energy elements ε = hν, the seed of all energy quantisation

1925

— Heisenberg's matrix mechanics yields the oscillator spectrum, with its ½ħω zero-point energy

1926

— Schrödinger's wave mechanics gives the Hermite-Gaussian vibrational eigenfunctions

1930

— Dirac's Principles formalises the raising and lowering operator treatment of the oscillator

1950

— Herzberg's Spectra of Diatomic Molecules tabulates ωe and force constants for the field

Going further

Texts. Herzberg, Molecular Spectra and Molecular Structure I: Spectra of Diatomic Molecules, 2nd ed. (the definitive reference) · Foot, Atomic Physics (2005), Ch. 10 · Bransden and Joachain, Physics of Atoms and Molecules, Ch. 10 · Atkins and Friedman, Molecular Quantum Mechanics, oscillator and vibrational chapters · Demtröder, Laser Spectroscopy, molecular spectra chapter · Loudon, The Quantum Theory of Light, 3rd ed. (ladder-operator treatment) · Saleh and Teich, Fundamentals of Photonics, molecular energy levels.

Provenance. The level formula $E_v=(v+\tfrac12)\hbar\omega$ with $\omega=\sqrt{k/\mu}$ , the ladder-operator diagonalisation $\hat H=\hbar\omega(a^\dagger a+\tfrac12)$ , and the selection rule $\Delta v=\pm1$ are standard (Loudon; Bransden and Joachain; Foot). A verifier should recompute: (1) the reduced-mass separation giving $\omega=\sqrt{k/\mu}$ ; (2) the ladder algebra reproducing the equal spacing and $E_0=\tfrac12\hbar\omega$ ; (3) the dimensional check that $k=\mu(2\pi c\tilde\nu)^2$ yields N/m; (4) the H–Cl numbers $\mu=0.980$ u, $k=516$ N/m, $E_0=1495\ \mathrm{cm^{-1}}=0.185\ \mathrm{eV}$ ; (5) the H–Cl to D–Cl frequency ratio $\sqrt{\mu_{\mathrm{HCl}}/\mu_{\mathrm{DCl}}}=0.717$ giving $2144\ \mathrm{cm^{-1}}$ . The harmonic wavenumber $\tilde\nu\approx2990\ \mathrm{cm^{-1}}$ for H–35Cl is the equilibrium value $\omega_e$ (Herzberg); the anharmonic-shifted observed fundamental lies near $2886\ \mathrm{cm^{-1}}$ and is treated in the Morse-potential article.

Builds on: The Chemical Bond and Molecular Potential Energy Curves · The Born-Oppenheimer Approximation Leads to: Anharmonicity and the Morse Potential · Vibration-Rotation Spectra and P, Q, R Branches · Selection Rules for Molecular Transitions · The Franck-Condon Principle

Ask the tutor: Ask me to walk through extracting a force constant from any molecule's infrared band, or to show exactly how anharmonicity peels the real levels away from this equally spaced ladder.

Why is the reduced mass used instead of one atom's mass?

Because both nuclei move. When the bond stretches, the two atoms swing about their common centre of mass rather than one staying fixed. Transforming to centre-of-mass and relative coordinates separates the drift of the whole molecule from the internal stretch, and that stretch is mathematically identical to a single particle of mass $\mu=m_1m_2/(m_1+m_2)$ on a spring. For H–Cl the reduced mass is 0.980 u — just under the hydrogen mass, because the heavy chlorine barely moves — so the light atom does almost all the vibrating, but not quite, and that "not quite" is exactly what $\mu$ captures.

Is the zero-point energy real or just a mathematical artefact?

It is entirely real and measurable. The zero-point energy $\tfrac12\hbar\omega$ follows directly from the uncertainty principle: a molecule cannot simultaneously have a perfectly defined bond length and momentum, so it cannot rest at the bottom of the well. Its reality shows up in several ways — it lowers the effective dissociation energy, it differs between isotopologues (driving kinetic isotope effects and enabling isotope separation), and it sets the geometry of the lowest state from which electronic transitions start. Helium stays liquid down to absolute zero precisely because its zero-point energy is too large for the weak interatomic forces to freeze out.

Why does the harmonic model predict only one absorption line?

Two facts combine. The levels are perfectly evenly spaced, so every transition $v\to v+1$ costs the same $\hbar\omega$ , and the selection rule $\Delta v=\pm1$ allows only neighbour jumps. Together they pile all the allowed absorption into a single fundamental band at $\tilde\nu=\omega/2\pi c$ . In a real molecule the levels are slightly anharmonic, which permits weak overtones ( $\Delta v=\pm2,\pm3$ ) shifted below integer multiples of the fundamental — but these are small corrections, and the dominant sharp fundamental is a direct signature of the underlying parabola.

Why do C–H and O–H stretches appear at high wavenumber?

Because $\tilde\nu\propto\sqrt{k/\mu}$ , and both factors push X–H bonds high. Hydrogen is the lightest atom, so any bond to it has a small reduced mass near 1 u, and X–H bonds are typically stiff. Together this puts C–H stretches near 3000 cm⁻¹ and O–H stretches near 3300–3600 cm⁻¹, well above the heavy-atom skeletal stretches (C–C, C–O) below about 1200 cm⁻¹. It is also why deuteration — which nearly doubles $\mu$ — moves those bands dramatically downward.

How is the force constant extracted from a spectrum?

Read the fundamental band position $\tilde\nu$ , compute the reduced mass $\mu$ from the atomic masses, and invert the frequency relation: $k=\mu(2\pi c\tilde\nu)^2$ . The result comes out in newtons per metre and characterises the bond independent of which isotopes are present. For H–Cl at $\tilde\nu\approx2990\ \mathrm{cm^{-1}}$ this gives $k\approx516\ \mathrm{N/m}$ ; for the stiffer triple-bonded CO at $\tilde\nu\approx2170\ \mathrm{cm^{-1}}$ it gives about $1900\ \mathrm{N/m}$ . Comparing force constants across molecules is a quantitative way to rank bond strengths.

When does the harmonic approximation break down?

Whenever the vibration samples the well far from its bottom — at high vibrational quantum number $v$ or in floppy, weakly bound molecules. As $v$ rises the wavefunction spreads into the asymmetric part of the true curve, the neglected cubic and higher terms grow, and the levels crowd together instead of staying evenly spaced, converging on the dissociation limit. The parabola also rises forever, so it can never describe bond breaking. For the low-lying levels of a stiff diatomic it is excellent; for overtones, hot bands, or dissociation you must switch to the Morse potential.