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For one normal mode, the fundamental is $v=0\to 1$, with other modes initially in their ground levels. In the linear-dipole harmonic approximation, the transition moment is nonzero for $\Delta v=\pm 1$, so a vibrational fundamental is normally much stronger than its overtones, provided the mode has vibrational infrared activity.

Let $u=r-r_e$ measure displacement from the equilibrium bond length and put the minimum of potential energy at zero. The harmonic model and the Morse potential areThe harmonic approximation is a symmetric parabola and has no dissociation limit. The Morse potential has a steep compressed side and a shallower extended side, approaching $D_e$ as $r\to \infty$. Matching the curvatures at the minimum gives $k=2D_e{\alpha }^2$. With $\tilde{\omega }$ denoting a spectroscopic wavenumber rather than an angular frequency, the energy levels areHere $x_e$ is the dimensionless spectroscopic anharmonicity constant. The harmonic spacing is constant, while the Morse oscillator spacing is $E_{v+1}-E_v=hc\tilde{\omega }[1-2x_e(v+1)]$ and decreases as the levels approach dissociation. Only finitely many Morse oscillator levels are bound. The exactly solvable potential was introduced to model this anharmonic behaviour.^[ref]

For the harmonic model put $q=\sqrt{\mu \omega /\hslash }u$, with $\omega =2\pi c\tilde{\omega }$, and $N={\pi }^{-1/4}$. The first three normalized reduced-coordinate wavefunctions areTheir numbers of nodes are zero, one and two, and their parities are even, odd and even. The $v=2$ nodes are at $q=\pm 1/\sqrt{2}$. The diagrams below use analytic curves and arbitrary illustrative scales, with both potentials given the same local curvature.

Set $l=m=g=1$. At $({\theta }_1,{\theta }_2)=(0,0)$ the spring has $d=r$, so its potential energy has zero gradient: ${\partial }_j{V}_s=k(d-r){\partial }_{j}d=0$. The gravitational potential energy has ${\partial }_j{V}_g=\sin {\theta }_j=0$ there. Zero velocities therefore satisfy the Euler-Lagrange equations, proving this is an equilibrium.

For fixed $r>0$, let $\epsilon =\max (|{\theta }_1|,|{\theta }_2|)$ and $\delta ={\theta }_2-{\theta }_1$. The Taylor series giveThus, near the downward equilibrium, the horizontal separation is positive andIn particular, $V_s=k{\delta }^2/2+O({\epsilon }^4)$ and $V_g=-2+({\theta }_1^2+{\theta }_2^2)/2+O({\epsilon }^4)$. Dropping the additive constant and terms beyond quadratic order gives the harmonic approximationFor physical spring constant $k\ge 0$, the quadratic potential energy is a positive-definite quadratic form. The downward position is therefore a stable equilibrium in Lagrangian mechanics, and its nearby motion admits small oscillations.

Write $w=\sqrt{1+2k}$. The initial conditions on the normal mode coordinates are $u(0)=v(0)=0$ and $\dot{u}(0)=\dot{v}(0)=k$. Solving the two harmonic oscillator equations and recovering the original angles gives the exact solution within the harmonic approximation:Their angular velocities are ${\dot{\theta }}_1=k(\cos t+\cos wt)/2$ and ${\dot{\theta }}_2=k(\cos t-\cos wt)/2$, which verify all four initial conditions.

To expose the beat, define $\Delta =w-1$ and $\Omega =(w+1)/2$. Replacing $1/w$ by $1$ changes each angle by $O(k^2)$ uniformly in time. The sum-to-product identities then giveThe displayed angular velocities are exact in the linear system. The absolute and signed beat envelopes have leading amplitudes $k|\cos (\Delta t/2)|$ and $k|\sin (\Delta t/2)|$. Because $\Delta =k+O(k^2)$, at $\tau =\pi /k$ we have $\Delta \tau /2=\pi /2+O(k)$ and $\Omega \tau =\tau +\pi /2+O(k)$. HenceThe first simple pendulum has negligible oscillation amplitude. Its kinetic energy is $T_1(\tau )=O(k^4)$, while the second has $T_2(\tau )=k^2{\cos }^2\tau /2+O(k^3)$. Thus, to the retained order, all instantaneous kinetic energy belongs to the second simple pendulum. The fast phase can make that kinetic energy small: the whole initial energy $k^2/2$ is transferred as the second simple pendulum's combined kinetic energy and gravitational potential energy, sinceThis is energy transfer after an impulse in two weakly coupled oscillators. It does not require $\cos \tau =1$.

At $2\tau$, $\Delta (2\tau )/2=\pi +O(k)$, so the beat has returned the oscillation to the first simple pendulum:The energy localization returns, although the fast oscillation phase need not equal its initial phase. The exact first minimum of the leading beat envelope is at $\pi /\Delta$, whose leading value is $\pi /k$. Keeping the slow phase $\Delta t$ is essential: expanding $\sin (wt)$ at fixed $t$ and then using $t=O(1/k)$ would lose the energy exchange. Although angles are retained only to order $k$, their leading energies are of order $k^2$.

In the harmonic approximation, each in-plane displacement coordinate contributes one quadratic kinetic energy term and one quadratic restoring potential energy term. The equipartition theorem assigns $k_{B}T/2$ to each. There are two such coordinates per chemical atom, henceHere the fixed-size heat capacity is at constant area in this two-dimensional model. On cooling, phonon excitation freezes out and the heat capacity decreases. For each nonzero angular frequency, the quantum harmonic oscillator contributeswhich is below $k_B$ at finite $x$ and approaches zero as $T\to 0$. Two linearly dispersing in-plane acoustic phonon branches give the familiar two-dimensional Debye model scaling $C_v\propto T^2$ at sufficiently low temperature. Out-of-plane modes are excluded by the specified model.

A phonon is one quantum of a normal vibrational mode of a crystal. In the harmonic approximation, each mode with wavevector $\mathbf{q}$, branch index $s$ and angular frequency ${\omega }_s(\mathbf{q})$ is an oscillator with energies $\hslash {\omega }_s(n+1/2)$. A phonon changes the occupation $n$ by one; it is not a particular atom moving through the solid.

For a three-dimensional primitive cell containing $r$ atoms there are $3r$ branches. Three acoustic branches approach zero frequency as $\mathbf{q}\to 0$: at zero wavevector they represent rigid translations, with all basis atoms displaced together. Their small-wavevector frequencies usually grow linearly with wavevector. The remaining $3r-3$ optical branches have internal relative motion of basis atoms and generally nonzero zone-centre frequencies. The names reflect their coupling to radiation in suitable polar crystals, not a rule that every optical mode must be infrared active. Acoustic and optical branches can each have longitudinal or transverse polarizations.

Let $a$ be a stable equilibrium separation: $V^{\prime }(a)=0$, $K=V^{\prime \prime }(a)>0$. With bond extension $\delta =x-a$, the Taylor expansion isThe harmonic approximation applies when the typical thermal and zero-point bond fluctuations satisfy $|\delta |\ll a$ and the higher terms are negligible, for example $|V^{\prime \prime \prime }(a)\delta |\ll 3K$ and $|V^{\prime \prime \prime \prime }(a){\delta }^2|\ll 12K$. The relevant small quantity is the relative displacement of neighbouring atoms. Small bond excursions about a stable minimum give an amplitude-independent spring constant $K=V^{\prime \prime }(a)$. Large thermal excursions, bond breaking or appreciable anharmonicity invalidate the approximation.

The Debye model replaces the crystal by a harmonic elastic continuum with acoustic dispersion $\omega =v_{s}k$. It counts three polarizations and introduces a cutoff so that the total number of vibrational modes is $3N$, where $N$ is the atom count. Periodic-box counting givesFor an isotropic average speed, $3/v_D^3=1/v_L^3+2/v_T^3$. With cutoff frequency ${\omega }_D$ fixed by the mode count,This density of states is quadratic because a three-dimensional shell has volume proportional to $k^{2}dk$ and the acoustic dispersion is linear. The cutoff is a mode-counting device; it is not a claim that every actual branch stays linear to the zone boundary.

Each phonon mode has mean thermal energy $\hslash \omega /(e^{\beta \hslash \omega }-1)$, using Bose-Einstein statistics with zero phonon chemical potential. The fixed harmonic zero-point contribution has zero temperature derivative. With $x=\hslash \omega /(k_{B}T)$,and differentiating at fixed volume givesFor $T\ll {\Theta }_D$, extend the upper limit to infinity. Expanding $1/(e^x-1)={\sum }_{j\ge 1}{e}^{-jx}$ gives ${\int }_0^{\infty }{x}^3/(e^x-1)dx=6\sum j^{-4}={\pi }^4/15$. integration by parts makes the heat-capacity integral $4{\pi }^4/15$, soOnly modes with $\hslash \omega \lesssim k_{B}T$ are appreciably excited; their number scales as $T^3$, explaining the power law physically. At high temperature every mode contributes $k_B$, giving the Dulong-Petit limit $C_V\to 3N{k}_B$.

The model assumes the harmonic approximation, a macroscopic three-dimensional solid and acoustic continuum behaviour for the thermally populated modes. Anisotropic sound speeds change the low-temperature coefficient rather than its cubic power. Real optical branches and nonlinear acoustic dispersion limit the full-temperature approximation; electronic excitations can add a linear term in a metal, and anharmonicity produces thermal expansion and distinctions between $C_P$ and $C_V$. At temperatures below the lowest finite-sample mode, the bulk continuum law also fails. Thus the $T^3$ result is the low-temperature lattice contribution, not a universal total heat capacity for every solid.

A phonon is one quantum of a crystal's vibrational normal modes. In the harmonic approximation, each mode is a quantum harmonic oscillator with energy levels $(n+1/2)\hslash \omega$. A phonon carries excitation energy $\hslash \omega$ and crystal momentum $\hslash q$ modulo a reciprocal lattice vector. It is a collective lattice excitation, not one atom travelling along the chain. Phonons obey Bose-Einstein statistics; their number need not be conserved in thermal equilibrium.

For central potential $V(r)=\beta r^k$ and nonzero angular momentum $J$, a circular orbit satisfies $\beta k{r}_0^{k+2}=J^2/m$, so attraction requires $\beta k>0$. Twice differentiating the radial effective potential $V+J^2/(2m{r}^2)$ gives $V_{\mathrm{eff}}^{\prime \prime }(r_0)=(k+2)J^2/(m{r}_0^4)$. Its harmonic approximation therefore has radial angular frequency ${\omega }_r=\sqrt{k+2}|J|/(m{r}_0^2)$, whereas $|{\omega }_{\theta }|=|J|/(m{r}_0^2)$. This proves the ratio and shows stability precisely when $k>-2$. At $k=-2$ the tuned effective potential is flat; at $k<-2$ the radial disturbance grows exponentially. Rationality of this local frequency ratio is only a necessary condition for all nearby orbits to close; Bertrand's theorem makes a stronger finite-amplitude assertion.

Excitation of one quantum of a normal mode from the vibrational ground state. For an infrared-active mode in the linear-dipole harmonic approximation, this is the allowed absorption underlying the usual fundamental band. A vibrational overtone excites two or more quanta from the ground state.

A vibrational spectrum resolves transitions associated with molecular vibrations. In the harmonic approximation, a normal mode of angular frequency $\omega$ has levelsFor a diatomic molecule, $\omega \simeq \sqrt{k/\mu }$, where $k$ is the effective bond stiffness and $\mu$ the reduced mass. Replacing an atom by a heavier isotope therefore usually lowers the associated vibrational frequency without changing the basic electronic bonding.

Infrared spectroscopy detects a vibration when it couples to the electric dipole moment; Raman spectroscopy instead involves changing polarizability. Their selection rules can make complementary modes visible. Anharmonicity shifts frequencies and permits weaker transitions absent from the ideal harmonic picture.