research40 min studyLesson 63 in Physics

Phonons and collective modes

Find normal modes of a harmonic chain and explain why quantizing their collective motion produces phonons.

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What you’ll understand

  • Derive the acoustic dispersion of a monatomic spring chain.
  • Distinguish atom displacement, normal-mode amplitude, and phonon occupation.
  • Identify the approximations behind harmonic phonons and their finite lifetimes.

A displacement at one point in a crystal pulls on neighboring atoms, so the natural oscillations belong to the whole structure. A phonon is a quantum of one such collective normal mode. It is not a tagged atom traveling through the solid. Starting with a monatomic spring chain makes the distinction explicit: first diagonalize coupled classical motion, then quantize the resulting independent oscillators. The same logic motivates collective excitations in systems whose microscopic ingredients look very different.

Let the coupling choose the coordinates

Let identical atoms of mass mm sit at equilibrium positions jaja in a one-dimensional periodic chain. Write uju_j for the small displacement from equilibrium. Nearest-neighbor harmonic springs contribute energy proportional to (uj+1−uj)2(u_{j+1}-u_j)^2. A uniform displacement costs no spring energy because all relative separations remain unchanged. Newton’s equation therefore contains a discrete second derivative. This starting model assumes a stable equilibrium, small displacements, identical masses, and interactions truncated to nearest-neighbor quadratic terms. It does not include bond breaking, thermal expansion, or arbitrary nonlinear motion.

Insert a plane-wave displacement with wavevector qq and frequency ω\omega. Translational symmetry makes these waves normal modes, so each allowed wavevector evolves independently in the harmonic approximation. The resulting acoustic branch approaches zero frequency at long wavelength: neighboring atoms move almost together, producing little restoring force. A finite chain has discrete allowed wavevectors, while an infinite-chain approximation replaces their sum by an integral. The exact zero mode in an unpinned chain is rigid translation of the center of mass; it is handled separately from an ordinary finite-frequency oscillator.

mu¨j=K(uj+1+uj−1−2uj),ω(q)=2K/m ∣sin⁡qa2∣m\ddot u_j=K(u_{j+1}+u_{j-1}-2u_j),\qquad \omega(q)=2\sqrt{K/m}\,\left|\sin\frac{qa}{2}\right|
The spring constant K is positive. The dispersion is periodic in reciprocal space, and q can be restricted to one Brillouin zone.↔ Scroll wide equations horizontally.

Quantize each independent oscillator

The normal-mode transformation diagonalizes the coupled harmonic Hamiltonian. Each nonzero-frequency real normal coordinate has the energy spectrum of a quantum harmonic oscillator. Its excitation number is the phonon occupation. Increasing that number adds one quantum of collective energy without adding an atom to the crystal. The zero-point term survives even at zero occupation. When using complex Fourier coordinates, the reality of atom displacements relates opposite wavevectors, so count independent degrees of freedom consistently rather than accidentally doubling the number of oscillators.

At small wavevector, the sine dispersion becomes approximately linear, with sound speed c=aK/mc=a\sqrt{K/m}. A wavepacket assembled from nearby modes travels with group velocity dω/dqd\omega/dq, which falls to zero at the zone boundary in this particular chain. A diatomic unit cell introduces additional branches, including an optical branch in which the two sublattices can move oppositely. Optical here labels the branch structure and possible electromagnetic coupling; it does not mean that every mode must visibly emit light. Real three-dimensional crystals have several polarizations as well.

Hharm=∑νℏων(bν†bν+12),[bν,bμ†]=δνμ,ω(q)≃c∣q∣H_{\mathrm{harm}}=\sum_{\nu}\hbar\omega_\nu\left(b_\nu^\dagger b_\nu+\frac12\right),\qquad [b_\nu,b_\mu^\dagger]=\delta_{\nu\mu},\qquad \omega(q)\simeq c|q|
The sum runs over independent nonzero-frequency normal oscillators; rigid translations are excluded. The linear approximation requires |qa| much smaller than one.↔ Scroll wide equations horizontally.

Separate frequency from lifetime

An isolated exactly harmonic crystal has no intrinsic phonon-phonon scattering: its normal occupations are constants of motion. Finite lifetimes appear when anharmonic couplings, defects, boundaries, electrons, or other degrees of freedom are included. A damped classical oscillator can model the decay of one measured coherent displacement over a suitable regime, but damping is not the same as assigning an exact phonon occupation. Even the distinction between well-defined quasiparticles and strongly broadened vibrations depends on comparing decay rates with oscillation frequencies and on the observable being measured.

The oscillator lab lets you inspect one classical unit-mass coordinate with initial displacement one and initial velocity zero. Change its natural frequency, then increase damping to compare oscillation and envelope decay. It does not contain a chain, wavevector-dependent dispersion, Bose occupation statistics, or quantum zero-point fluctuations. The honest bridge is that a harmonic normal coordinate obeys the same classical differential equation before quantization. To study heat capacity or thermal conductivity, one must additionally count modes, assign quantum populations, and, for transport, model how energy moves and relaxes.

Frequency and speed in a model chain

Take a=0.30 nma=0.30\,\mathrm{nm} and K/m=1.0×1013 s−1\sqrt{K/m}=1.0\times10^{13}\,\mathrm{s}^{-1}. Find the long-wavelength sound speed and the angular frequency at q=π/(3a)q=\pi/(3a). Then find the energy added by one phonon of that mode.

  1. The sound speed is c=aK/m=(3.0×10−10)(1.0×1013)=3.0×103 m s−1c=a\sqrt{K/m}=(3.0\times10^{-10})(1.0\times10^{13})=3.0\times10^3\,\mathrm{m\,s}^{-1}. This is the slope near q = 0, not the phase velocity at every wavevector.

  2. At qa=π/3qa=\pi/3, the exact harmonic-chain dispersion gives ω=2(1.0×1013)sin⁡(π/6)=1.0×1013 s−1\omega=2(1.0\times10^{13})\sin(\pi/6)=1.0\times10^{13}\,\mathrm{s}^{-1}. It is an angular frequency; dividing by 2π2\pi would give cycles per second.

  3. Using ℏ=1.0546×10−34 J s\hbar=1.0546\times10^{-34}\,\mathrm{J\,s} gives ΔE=ℏω=1.0546×10−21 J\Delta E=\hbar\omega=1.0546\times10^{-21}\,\mathrm{J}. Dividing by 1.6022×10−22 J/meV1.6022\times10^{-22}\,\mathrm{J/meV} gives approximately 6.58 meV6.58\,\mathrm{meV}.

The result

The chain supports long-wavelength sound at 3000 m/s, and one quantum of the selected mode adds 6.58 meV above its previous occupation energy.

Explore the model

Interactive lab

Interactive lab · oscillator

One classical normal coordinate

Begin with zero damping and compare natural frequencies for the analytic oscillator with initial displacement one and velocity zero. Then increase the damping ratio and distinguish a decaying displacement from an oscillation frequency. This single classical coordinate does not compute a crystal dispersion or a quantum phonon population.

One classical normal coordinateUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forDiagonalizing a harmonic crystal produces independent coordinates; quantization adds discrete occupation energies beyond this classical lab.

Check your understanding

What does adding one phonon to a harmonic normal mode do?

Try it yourself

Compare the exact chain dispersion with its linear approximation at qa=πqa=\pi. Express both frequencies in units of K/m\sqrt{K/m}. Which is larger, and why is this not a failure of the harmonic approximation?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The exact answer contains sin(qa/2), while the long-wavelength expression contains |qa|. Small displacement and small wavevector are different assumptions.

Compare with the solution

At the zone boundary, the exact frequency is 2K/m2\sqrt{K/m}, whereas the linear approximation predicts πK/m\pi\sqrt{K/m}. The linear result is larger by a factor π/2≈1.57\pi/2\approx1.57. The harmonic model can remain valid for sufficiently small displacements at this wavevector; it is the additional long-wavelength expansion of the sine that fails. A lattice is not a continuum at distances comparable to a cell.

Take it with you

  • Phonons count excitations of collective normal coordinates, not atoms.
  • A monatomic unpinned chain has an acoustic branch because uniform translation has no restoring force.
  • Harmonic frequencies, quantum populations, and relaxation mechanisms are separate pieces of a physical prediction.

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