advanced29 min studyLesson 495 in Physics

Monoatomic and diatomic chain phonons

Derive harmonic chain dispersions with explicit cell-length and mass conventions, check acoustic limits, and identify the origin of a diatomic optical branch.

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What you’ll understand

  • Derive the monoatomic nearest-neighbor dispersion from a plane-wave displacement.
  • Compute diatomic acoustic and optical frequencies at symmetry points.
  • Distinguish primitive-cell length, neighbor spacing, and zone folding.

A lattice vibration is a collective displacement pattern, so its frequency depends on the phase relation between neighboring atoms. A monoatomic chain gives one branch for one scalar displacement per cell. Alternating masses introduce a second independent displacement and a second branch. We will declare cell lengths carefully: in the diatomic chain a denotes the full two-atom cell length, while nearest neighbors are separated by a divided by two. This avoids a frequent factor-of-two error in the dispersion and zone edge.

Nearest-neighbor forces produce a sine dispersion

First consider one mass m per primitive interval a, with identical springs of stiffness K joining neighbors. Expanding the bond potential to quadratic order gives force K(un+1+un−1−2un)K(u_{n+1}+u_{n-1}-2u_n) on displacement un. Substituting un=uei(qna−ωt)u_n=u e^{i(qna-\omega t)} diagonalizes translation and yields a frequency determined by the phase difference qa. The resulting branch is periodic in q with period two pi over a. Frequencies are nonnegative; the sine must therefore appear with an absolute value when taking the square root.

Near zero wavevector the frequency is proportional to the magnitude of q, with sound speed aK/ma\sqrt{K/m}. At the zone boundary it saturates, while the group velocity vanishes in this ideal nearest-neighbor model. The phase velocity and group velocity need not coincide away from the long-wavelength limit. These conclusions follow from the chosen force law. Longer-range forces can reshape the dispersion, and the continuum sound approximation cannot be extended blindly to wavelengths comparable with the lattice spacing.

ω(q)=2Km∣sin⁡qa2∣,−πa≤q<πa,cs=aKm\omega(q)=2\sqrt{\frac Km}\left|\sin\frac{qa}{2}\right|,\qquad -\frac\pi a\le q<\frac\pi a,\qquad c_s=a\sqrt{\frac Km}
Here a is the monoatomic primitive spacing, m is mass in kg, and K is stiffness in N/m. Omega is angular frequency in s⁻¹ and q is angular wavevector in m⁻¹.↔ Scroll wide equations horizontally.

Two masses give two eigenfrequencies

For the diatomic chain, place mass m at na and mass M at na plus a over two, with the same spring K on every nearest-neighbor bond. A cell therefore has two displacement amplitudes and a two-by-two generalized eigenproblem. Its determinant gives the two squared frequencies below. The minus branch goes to zero at q equal to zero; the plus branch remains finite. The primitive zone boundary is now pi over the two-atom cell length a, not pi over the neighbor spacing.

At zero wavevector the acoustic displacement moves the two atoms together. The optical displacement moves them oppositely with zero cell centre-of-mass displacement. When M exceeds m, the boundary frequencies are the square roots of two K over M and two K over m. Thus a mass contrast opens a gap between the two branches in this simple chain. Equal masses close that gap; the true primitive interval then becomes a over two, and the two branches can be understood as a folded monoatomic dispersion rather than two chemically distinct vibrations.

ω±2(q)=K(1m+1M)±K(1m+1M)2−4mMsin⁡2qa2\omega_\pm^2(q)=K\left(\frac1m+\frac1M\right)\pm K\sqrt{\left(\frac1m+\frac1M\right)^2-\frac{4}{mM}\sin^2\frac{qa}{2}}
In this diatomic formula a is the full two-atom primitive-cell length and nearest-neighbor spacing is a/2. Equal nearest-neighbor spring constants and scalar longitudinal displacements are assumed; the minus and plus signs label acoustic and optical branches.↔ Scroll wide equations horizontally.

A phonon is a quantum of a normal mode

Classical normal-mode frequencies are obtained before quantization. Quantizing each independent harmonic mode makes its energy levels equally spaced by hbar omega, and one excitation is called a phonon. A phonon is not an individual atom moving along a sine curve; it is a quantum of the collective displacement coordinate. Counting branches begins with the number of independent coordinates per primitive cell. These scalar chains consequently have one or two branches, while an actual three-dimensional crystal also has displacement polarizations.

The harmonic approximation assumes small excursions around a mechanically stable periodic equilibrium and neglects damping, defects, and anharmonic scattering. It predicts perfectly sharp mode frequencies but no intrinsic finite lifetime. A simple wave-superposition lab cannot generate this dispersion by changing its phase slider: that instrument fixes its component frequencies. To test a chain model quantitatively, use the equations with declared masses, stiffnesses, cell length, and wavevector, then check zero-wavevector translation, boundary values, and the equal-mass folding limit independently.

Diatomic frequencies in a mechanical chain

A one-dimensional mechanical analogue has masses m = 1 kg and M = 4 kg with every spring K = 2 N/m. Using a as its two-atom cell length, find both angular frequencies at q = 0 and q = pi/a.

  1. At q = 0 the sine term vanishes. The acoustic squared frequency is zero, while the optical value is 2K(1/m+1/M)=4(1+1/4)=5 s−22K(1/m+1/M)=4(1+1/4)=5\,\mathrm{s^{-2}}.

  2. At q = pi/a the sine squared is one. The square-root term becomes ∣1/m−1/M∣=3/4|1/m-1/M|=3/4 in inverse kilograms, so the two squared frequencies are 2(1.25∓0.75)=12(1.25\mp0.75)=1 and 4 s−24\,\mathrm{s^{-2}}.

  3. Taking nonnegative square roots gives boundary frequencies 1 and 2 radians per second. The zero-wavevector pair is zero and square root of five radians per second. The lower boundary mode is associated with the heavier mass scale, consistent with slower oscillation.

The result

The acoustic and optical frequencies are (0, square root of 5) at the zone centre and (1, 2) radians per second at the zone edge.

Explore the model

Interactive lab

Interactive lab · oscillator

Mass–stiffness frequency scaling in a single mode

Reset and set BOTH Damping ratio ζ = 0 and Natural frequency ω₀ = 1; predict period 2π. For a fresh run, Reset and set ζ = 0 and ω₀ = 2; predict period π. Play or Step each run from its reset initial position 1 and initial velocity 0.

Mass–stiffness frequency scaling in a single modeUnit 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 forThis unit-mass single oscillator illustrates a mode's frequency and period. It has no coupled chain, masses, wavevector, acoustic branch, optical branch, or adjustable phonon dispersion.

Check your understanding

In the stated diatomic chain, where is the first-zone boundary?

Try it yourself

A monoatomic mechanical chain has m = 2 kg, K = 8 N/m, and primitive spacing a = 0.10 m. Find omega and positive-q group velocity at q=π/(2a)q=\pi/(2a), and compare the group velocity with the sound speed.

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 sine and cosine of qa/2 are both one over square root of two. Differentiate omega with respect to q on the positive branch.

Compare with the solution

The sound speed is cs=aK/m=0.10(2)=0.20 m s−1c_s=a\sqrt{K/m}=0.10(2)=0.20\,\mathrm{m\,s^{-1}}. At the chosen q, ω=2(2)/2=22 s−1\omega=2(2)/\sqrt2=2\sqrt2\,\mathrm{s^{-1}}. The positive-q derivative is vg=aK/mcos⁡(qa/2)=0.20/2 m s−1v_g=a\sqrt{K/m}\cos(qa/2)=0.20/\sqrt2\,\mathrm{m\,s^{-1}}. It is smaller than the sound speed, showing that the linear acoustic approximation already overestimates group velocity at this finite fraction of the zone.

Take it with you

  • The cell translation fixes the reciprocal zone independently of the number of atoms it contains.
  • Two displacement amplitudes produce two branches with distinct zero-wavevector behavior.
  • Harmonic frequencies describe collective modes, while quantization assigns phonon energies.

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