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Let $r$ be the instantaneous bond length, $r_e>0$ its equilibrium value, and $q=r-r_e$ the stretching displacement. Assume one fixed electronic quantum state, a smooth potential energy $V(r)$ with a nondegenerate local minimum at $r_e$, and omit molecular rotation initially. The Taylor series has no linear term because $V^{\prime }(r_e)=0$:Keeping only the quadratic term is the harmonic approximation near a potential minimum. The bond force constant $k$ has units $\mathrm{N}{\mathrm{m}}^{-1}$. It measures local stiffness; it is not the bond-dissociation energy, which concerns the depth and distant shape of the entire well.

The two-particle relative coordinate explains the reduced mass. For stretching along a fixed axis in a stationary centre of mass frame, let $u_1,u_2$ be the nuclear displacements from equilibrium. Then $u_2-u_1=q$ and $m_1{u}_1+m_2{u}_2=0$, soHere $T$ is the internal kinetic energy, and $m_1,m_2>0$ are the two nuclear masses. The lighter partner moves farther; the reduced mass is smaller than either individual mass. The restoring force $-dV/dq\simeq -kq$ gives simple harmonic motion:The angular frequency is $\omega$, the ordinary frequency is $\nu$, and ${\tilde{\nu }}_{\mathrm{harm}}$ is the corresponding spectroscopic wavenumber; $c$ is the speed of light. Use $c$ in $\mathrm{cm}{\mathrm{s}}^{-1}$ to obtain ${\tilde{\nu }}_{\mathrm{harm}}$ in $\mathrm{c}{\mathrm{m}}^{-1}$. The symbol $k$ here is a bond force constant, not an angular wavenumber.

For an original dimensionless example, choose $m_1=m_2=2$ and $k=4$ in consistent model units. Then $\mu =1$ and $\omega =2$. Releasing at $q(0)=0.1$ with $\dot{q}(0)=0$ gives $q(t)=0.1\cos (2t)$ and $u_1=-q/2$, $u_2=q/2$. Each atomic nucleus moves only half the stretching amplitude, while the conserved internal mechanical energy above $V_0$ is $\frac{1}{2}k(0.1{)}^2=0.02$. Fixing one partner instead would give $\omega =\sqrt{2}$, demonstrating why a one-body calculation changes the model.

Quantizing the stretching coordinate gives a quantum harmonic oscillator with reduced mass $\mu$:Here $v$ is the vibrational quantum number and $\hslash =h/(2\pi )$ the reduced Planck constant. Consecutive energy eigenvalues have gap $\hslash \omega$; the ground state has zero-point energy $\hslash \omega /2$ above the well minimum. The offset $V_0$ cancels from every transition energy.

A short proof uses creation and annihilation operators: with $\ell =\sqrt{\hslash /(\mu \omega )}$, write $q=\ell (a+a^{†})/\sqrt{2}$. The Quantum Hamiltonian becomes $\hat{H}-V_0=\hslash \omega (a^{†}a+1/2)$. Nonnegativity of $a^{†}a$ gives the lowest level; repeated application of $a^{†}$ raises its energy by $\hslash \omega$ each time. This also gives $⟨q{⟩}_v=0$ andThe ground state therefore has a nonzero spread of bond lengths, even though its mean stretching displacement vanishes. A stationary energy eigenstate has a time-independent position probability density; the classical oscillation curve is not a trajectory of an atomic nucleus in that state.

The parabolic model formally extends $q$ to the whole real line, whereas an actual bond length satisfies $r>0$. It is useful for low levels concentrated near $r_e$, with typical stretching displacements much smaller than $r_e$. It fails near breakup of the chemical bond and when neglected anharmonicity becomes substantial.

An actual infrared spectroscopy route is to compare gaseous hydrogen chloride containing ordinary hydrogen with its deuterium counterpart. A documented teaching experiment uses a stopcock-equipped gas cell with potassium bromide windows and a Nicolet Avatar instrument for Fourier transform spectroscopy. A nitrogen-purged flask and drying tube supply the mixture; the sample also contains sulfur dioxide. An empty sample-chamber background precedes the filled-cell scan. The absorbance trace and peak-picked spectroscopic wavenumbers are the readout. The analysis assigns and indexes ${}^1{\mathrm{H}}^{35}\mathrm{Cl}$ and ${}^2{\mathrm{H}}^{35}\mathrm{Cl}$ transitions separately and propagates transition uncertainties. The instructions distinguish resolvable chlorine isotope peaks from adequate precision for the intended parameter analysis. ^[ref]

A separate published teaching dataset reports a Nicolet Nexus 670 instrument, a $10\mathrm{cm}$ gas cell at $20\mathrm{Torr}$, and chlorine isotope peaks resolved to baseline. These apparatus details describe that dataset, not the preceding experiment. ^[ref]

The inference connects several models. Fit correctly assigned transitions to separate molecular rotation from molecular vibration, then compare the inferred vibrational scales using isotope scaling of diatomic vibration. Agreement supports the approximate separation between a nearly unchanged chemical bond stiffness and a changed nuclear mass. It does not make a real molecule an exact classical spring: anharmonicity shifts the fundamental transition away from the local harmonic oscillator scale, and coupling to molecular rotation shifts individual lines.

For an independent test, calculate the reduced masses from the specified isotopes and compare the inferred values $k=\mu (2\pi c{\tilde{\nu }}_{\mathrm{harm}}{)}^2$. Agreement within uncertainty tests the common-curvature assumption; choosing separate $k$ values just to fit each frequency would not test it. The documented background addresses instrument and ambient contributions, but does not by itself subtract window absorption, identify contaminating species, or calibrate the spectroscopic wavenumber scale. Those require separate controls. Finite spectral resolution, blended peaks, incorrect isotope assignments, and departure from the harmonic approximation near a potential minimum limit the comparison. No numerical measured line positions are assumed here.

The minima occur at $x=(2j+1)\pi$, where $V=-2V_0$. Writing $x=(2j+1)\pi +\xi$ givesThe harmonic approximation near a potential minimum therefore has frequencybecause $\frac{1}{2}m{\omega }_0^2{\xi }^2=\frac{5}{2}{V}_0{\xi }^2$. The two lowest localized energies are

In the tight-binding model, tunnelling between neighbouring wells broadens each localized level into a band,For large $V_0$, the hopping amplitudes $J_n$ are exponentially small compared with the oscillator spacing. Hence the gap between the first two bands isup to exponentially small tight-binding bandwidth corrections.

Take nearest-neighbour separation $d$, a common spring constant $K$, and one scalar displacement per atom. For a monatomic chain, the harmonic approximation near a potential minimum, Newton's second law and a normal mode giveThe monatomic-chain dispersion relation has one acoustic phonon branch, with $\omega \simeq d\sqrt{K/m}|k|$ near the origin. The first Brillouin zone ends at $\boxed{k=\pm \pi /d}$, where $\omega =2\sqrt{K/m}$ and the group velocity vanishes.

For alternating masses $m,M$, choose a two-atom cell of length $2d$. The coupled normal modes give the diatomic-chain dispersion relation with equal springs,The first Brillouin zone now ends at $\boxed{k=\pm \pi /(2d)}$. The minus branch is the acoustic phonon branch; its two atoms move together at long wavelength. The plus branch is the optical phonon branch; at $k=0$ the atoms move oppositely with a stationary centre of mass and ${\omega }_{+}^2=2K(1/m+1/M)$. For $M>m$ the edge frequencies are $\sqrt{2K/M}$ and $\sqrt{2K/m}$, with a gap between them.

If $M=m$, the two branches reduce to $2\sqrt{K/m}|\sin (kd/2)|$ and $2\sqrt{K/m}|\cos (kd/2)|$, touching at $\sqrt{2K/m}$. This is zone folding in an equal-mass diatomic chain: identical sites actually have a primitive unit cell of length $d$. A doubled cell folds the monatomic branch but opens no gap. Equal masses alone do not specify all possible spring models; the sketches assume equal nearest-neighbour spring constants.