= Diatomic bond stretch
{title2=$q=r-r_e,\quad\omega=\sqrt{k/\mu}$}
= Diatomic stretching vibration
{synonym}
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 $V'(r_e)=0$:
$$
V(r_e+q)=V_0+\frac12kq^2+O(q^3),\qquad
V_0=V(r_e),\qquad k=V''(r_e)>0.
$$
Keeping only the quadratic term is the <harmonic approximation near a potential minimum>. The <bond force constant> $k$ has units $\mathrm{N\,m^{-1}}$. 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 $m_1u_1+m_2u_2=0$, so
$$
u_1=-\frac{m_2}{m_1+m_2}q,\qquad
u_2=\frac{m_1}{m_1+m_2}q,\qquad
T=\frac12m_1\dot u_1^2+\frac12m_2\dot u_2^2=\frac12\mu\dot q^2,\qquad
\mu=\frac{m_1m_2}{m_1+m_2}.
$$
Here $T$ 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>:
$$
\mu\ddot q=-kq,\qquad
q(t)=A\cos(\omega t+\phi),\qquad
\omega=\sqrt{\frac{k}{\mu}},\qquad
\nu=\frac{\omega}{2\pi},\qquad
\widetilde\nu_{\rm harm}=\frac{\omega}{2\pi c}.
$$
The <angular frequency> is $\omega$, the ordinary <frequency> is $\nu$, and $\widetilde\nu_{\rm harm}$ is the corresponding <spectroscopic wavenumber>; $c$ is the <speed of light>. Use $c$ in $\mathrm{cm\,s^{-1}}$ to obtain $\widetilde\nu_{\rm harm}$ in $\mathrm{cm^{-1}}$. 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 $\frac12k(0.1)^2=0.02$. Fixing one partner instead would give $\omega=\sqrt2$, demonstrating why a one-body calculation changes the model.
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