Phonon Dispersion of the Diatomic Chain

Derivation

Phonon Dispersion of the Diatomic Chain

D-255 Home PU-303 Threads waves · matter · energy Depends on Bloch's Theorem from Translational Symmetry, coupled-oscillator-normal-modes

Statement

For a one-dimensional chain with a two-atom basis (alternating masses \(M\) and \(m\), identical nearest-neighbour springs of stiffness \(K\), lattice constant \(a\)), diagonalising the harmonic equations of motion under a Bloch ansatz yields two dispersion branches, \[ \omega_\pm^2(q) = K\left(\frac{1}{M}+\frac{1}{m}\right) \pm K\sqrt{\left(\frac{1}{M}+\frac{1}{m}\right)^2 - \frac{4\sin^2(qa/2)}{Mm}}, \] an acoustic branch (\(\omega_-\), vanishing linearly as \(q\to 0\)) and an optical branch (\(\omega_+\)), separated at the zone boundary \(q=\pi/a\) by a forbidden frequency gap running from \(\sqrt{2K/M}\) to \(\sqrt{2K/m}\).

Why it matters

The diatomic chain is the minimal model that produces an optical phonon branch and a phononic band gap — features absent from the monatomic chain but generic to every crystal with more than one atom per primitive cell. It explains why ionic solids absorb infrared light in the reststrahlen band, why heat-carrying acoustic modes are kinematically distinct from the high-frequency optical modes probed by Raman and neutron scattering, and how a basis folds the Brillouin zone.

It is also the cleanest illustration that the number of phonon branches equals the number of degrees of freedom per cell: two atoms give two branches (one acoustic, one optical) in 1D, generalising to \(3p\) branches for \(p\) atoms in three dimensions.

Assumptions

Harmonic (small-amplitude) potential: forces linear in displacement.Cubic and higher terms would couple the normal modes, giving phonon-phonon scattering, finite lifetimes and temperature-dependent frequencies — the modes would no longer be exact eigenstates.

Nearest-neighbour springs only, all of stiffness \(K\).Second-neighbour or asymmetric springs add terms in \(\cos(qa)\) and \(\cos(2qa)\), reshaping the branches and shifting the gap edges.

Perfect periodicity, so Bloch's theorem applies and \(q\) is a good quantum number.Mass or spring disorder destroys translational symmetry, delocalised plane-wave modes give way to (Anderson-)localised states, and a sharp dispersion \(\omega(q)\) ceases to exist.

Short-range (spring) forces only — no long-range Coulomb interaction.In a real ionic crystal the macroscopic electric field of a longitudinal optical mode adds a non-analytic term, splitting the zone-centre optical frequency into distinct LO and TO values; the neutral spring model misses this LO-TO splitting entirely.

Purely one-dimensional, longitudinal motion.In 3D each branch splits into one longitudinal and two transverse polarisations with generally different frequencies.

Derivation

\[ M\,\ddot{u}_n = K\,(v_n - u_n) + K\,(v_{n-1} - u_n) = K\,(v_n + v_{n-1} - 2u_n) \] \[ m\,\ddot{v}_n = K\,(u_{n+1} - v_n) + K\,(u_n - v_n) = K\,(u_{n+1} + u_n - 2v_n) \]

Newton's second law for the two atoms of cell \(n\): \(u_n\) is the displacement of mass \(M\), \(v_n\) that of mass \(m\); each atom feels a Hooke restoring force from its two spring-connected neighbours. Note the factor \(2K\), not \(K\). A

\[ u_n = A\,e^{i(qna-\omega t)}, \qquad v_n = B\,e^{i(qna-\omega t)} \]

Bloch ansatz: periodicity guarantees eigenmodes are plane waves labelled by a single wavevector \(q\), with per-sublattice amplitudes \(A,B\). This is exactly Bloch's theorem applied to the lattice; it reduces the infinite set of coupled ODEs to a single cell. B

\[ -M\omega^2 A = K\!\left(B + B\,e^{-iqa} - 2A\right) \] \[ -m\omega^2 B = K\!\left(A\,e^{iqa} + A - 2B\right) \]

Substitute the ansatz and cancel the common factor \(e^{i(qna-\omega t)}\); \(\ddot{u}_n \to -\omega^2 u_n\), and \(u_{n\pm1}\to e^{\pm iqa}u_n\) shifts the cell index. A

\[ \begin{pmatrix} 2K - M\omega^2 & -K\!\left(1+e^{-iqa}\right) \\[2pt] -K\!\left(1+e^{iqa}\right) & 2K - m\omega^2 \end{pmatrix}\begin{pmatrix} A \\ B \end{pmatrix} = 0 \]

Collect terms into a \(2\times2\) dynamical matrix acting on the amplitude vector \((A,B)^{\mathsf T}\); the eigenvalue problem is now algebraic. B

\[ \left(2K - M\omega^2\right)\left(2K - m\omega^2\right) - K^2\left(1+e^{iqa}\right)\left(1+e^{-iqa}\right) = 0 \]

A non-trivial amplitude vector exists only when the coefficient matrix is singular, i.e. its determinant vanishes. This is the secular (characteristic) equation. B

\[ \left(1+e^{iqa}\right)\left(1+e^{-iqa}\right) = 2 + 2\cos qa = 4\cos^2\!\left(\tfrac{qa}{2}\right) \]

Expand the modulus-squared of \(1+e^{iqa}\) and apply the half-angle identity \(1+\cos\theta = 2\cos^2(\theta/2)\). A

\[ Mm\,\omega^4 - 2K(M+m)\,\omega^2 + 4K^2\left[\,1-\cos^2\!\left(\tfrac{qa}{2}\right)\right] = 0 \] \[ Mm\,\omega^4 - 2K(M+m)\,\omega^2 + 4K^2\sin^2\!\left(\tfrac{qa}{2}\right) = 0 \]

Multiply out the determinant, collect powers of \(\omega^2\), and use \(1-\cos^2 = \sin^2\). The result is a quadratic in \(\omega^2\). A

\[ \omega^2 = \frac{2K(M+m) \pm \sqrt{4K^2(M+m)^2 - 16\,MmK^2\sin^2(qa/2)}}{2Mm} \] \[ \omega_\pm^2 = K\left(\frac{1}{M}+\frac{1}{m}\right) \pm K\sqrt{\left(\frac{1}{M}+\frac{1}{m}\right)^2 - \frac{4\sin^2(qa/2)}{Mm}} \]

Solve the quadratic with the standard formula and divide through by \(Mm\), using \((M+m)/Mm = 1/M+1/m\). The two roots are the two branches. A

Result

\[ \omega_\pm^2(q) = K\left(\frac{1}{M}+\frac{1}{m}\right) \pm K\sqrt{\left(\frac{1}{M}+\frac{1}{m}\right)^2 - \frac{4\sin^2(qa/2)}{Mm}} \]

Reading. The lower root \(\omega_-\) is the acoustic branch: it starts at \(\omega=0\) at \(q=0\), where both sublattices oscillate in phase (\(A=B\)) like a long-wavelength sound wave, and rises to \(\sqrt{2K/M}\) at the zone boundary. The upper root \(\omega_+\) is the optical branch: at \(q=0\) it sits at \(\omega=\sqrt{2K(1/M+1/m)}=\sqrt{2K/\mu}\) (with reduced mass \(\mu=Mm/(M+m)\)), where the two sublattices move exactly out of phase (\(MA=-mB\), centre of mass fixed), and it falls to \(\sqrt{2K/m}\) at the boundary. Between \(\sqrt{2K/M}\) and \(\sqrt{2K/m}\) no propagating mode exists — the zone-boundary gap, which is finite precisely when \(M\neq m\).

Units check. \(K\) has units \(\mathrm{N\,m^{-1}=kg\,s^{-2}}\), so \(K/M\) has units \(\mathrm{s^{-2}}\) and \(\sqrt{K/M}\) has units \(\mathrm{s^{-1}}=\mathrm{rad\,s^{-1}}\), correctly a frequency. Every term under the square root scales as \((\mathrm{kg\,s^{-2}}/\mathrm{kg})^2=\mathrm{s^{-4}}\), matching \(\omega^4\); \(\sin^2(qa/2)\) is dimensionless since \(qa\) is (\([q]=\mathrm{m^{-1}}\), \([a]=\mathrm{m}\)).

Limiting cases

Long wavelength (\(q\to0\)): acoustic branch \(\omega_-\simeq q\,a\sqrt{K/[2(M+m)]}\), a linear dispersion with sound speed \(v_s = a\sqrt{K/[2(M+m)]}\); optical branch flat at \(\omega_+\to\sqrt{2K/\mu}\).
Zone boundary (\(q=\pi/a\)): \(\omega_-=\sqrt{2K/M}\) (heavy atoms alone move) and \(\omega_+=\sqrt{2K/m}\) (light atoms alone move); both branches have zero group velocity here.
Equal masses (\(M=m\)): the discriminant becomes \((1/M-1/m)^2\to0\) at the boundary, the gap closes, and the two branches merge into the folded monatomic dispersion \(\omega=2\sqrt{K/M}\,|\sin(qa/4)|\) of a chain with spacing \(a/2\).
Very light second atom (\(m\ll M\)): optical branch pushed up to \(\omega_+\approx\sqrt{2K/m}\) (a nearly dispersionless high-frequency mode) with a wide gap; acoustic branch governed by the heavy mass.

Breaks when

Anharmonic / large amplitude. When displacements are not small, cubic terms in the potential matter: the normal modes couple (phonon-phonon scattering), frequencies shift with temperature, thermal expansion appears, and the sharp \(\omega(q)\) is replaced by broadened, finite-lifetime resonances.
Long-range Coulomb forces (ionic crystals). The macroscopic depolarising field of a longitudinal optical mode makes \(\omega_+(q\to0)\) non-analytic and splits it into distinct LO and TO frequencies (the Lyddane-Sachs-Teller relation) — physics entirely outside a nearest-neighbour spring model.
Disorder. Random masses or spring constants break periodicity; Bloch's theorem no longer holds, \(q\) is not conserved, and high-frequency optical modes can Anderson-localise, so no dispersion curve exists.
Strong wavevectors outside the first zone / continuum limit misuse. The result is periodic in \(q\) with period \(2\pi/a\); treating \(\omega(q)\) as if it grew without bound at large \(q\) (as in an elastic continuum) is wrong — the lattice caps frequencies at the zone-boundary values.

Failure modes

Dropping the factor of two in the restoring force: each atom has two springs, so the diagonal term is \(2K-M\omega^2\), not \(K-M\omega^2\). This misplaces every gap edge.
Wrong cell length. Using the atom spacing \(a/2\) as the lattice constant, or putting the zone boundary at \(\pi/(a/2)\) instead of \(\pi/a\), doubles or halves the zone and corrupts the folding.
Mislabelling the branches at the boundary. At \(q=\pi/a\) the optical mode is the light-atom frequency \(\sqrt{2K/m}\) and the acoustic mode is the heavy-atom \(\sqrt{2K/M}\); students often swap them.
Taking the \(+\) root as acoustic. The lower (\(-\)) root is acoustic; choosing \(+\) gives a branch that does not vanish at \(q=0\).
Forgetting \(|1/M-1/m|\). When simplifying \(\sqrt{(1/M-1/m)^2}\) at the zone boundary, dropping the absolute value can flip a sign and invert the gap ordering.
Confusing phase and group velocity. The sound speed is the group velocity \(d\omega/dq\) at \(q\to0\); using \(\omega/q\) evaluated at some finite \(q\) gives the wrong number.

Discussion

The two branches embody a division of labour that survives into real crystals. In the acoustic branch neighbouring atoms move nearly together, so long-wavelength acoustic phonons are just sound: they carry momentum and, being low in energy, dominate heat transport and the low-temperature specific heat. In the optical branch the two sublattices move against each other; in an ionic crystal, where the two atoms carry opposite charge, this out-of-phase motion is an oscillating electric dipole that couples resonantly to electromagnetic radiation — hence "optical," and hence the strong infrared absorption of salts.

The zone-boundary gap is the one-dimensional ancestor of every band gap produced by a basis. Because the discriminant collapses to \((1/M-1/m)^2\) at \(q=\pi/a\), the gap is controlled entirely by the mass contrast: identical atoms give no gap (the "gap" is an artefact of using a doubled cell for a monatomic chain, and the branches simply reconnect when you unfold the zone), while a large mass ratio opens a wide forbidden band. This is the same mechanism — a periodic modulation opening a gap at the zone edge where forward and backward waves Bragg-reflect into a standing wave — that gives electrons their band gaps in the nearly-free-electron picture.

Both branches flatten at the zone boundary, \(d\omega/dq\to0\), so the phonons there are standing waves, not travelling ones. These flat regions pile up states into van Hove singularities in the phonon density of states, which show up as peaks in the specific heat and in inelastic-scattering spectra.

More deeply, the model is a two-band tight-binding problem in disguise: the dynamical matrix is a \(2\times2\) Bloch Hamiltonian, and the acoustic/optical splitting is a level repulsion between the two sublattice "orbitals." Reading \(q\) as a parameter that traverses the Brillouin zone, one can ask about the topology of the eigenvector as \(q\) runs from \(0\) to \(2\pi/a\); with a suitable choice of springs the diatomic chain becomes the mechanical Su-Schrieffer-Heeger model, whose gap can carry a non-trivial winding number and host protected edge modes at the ends of a finite chain — the phononic analogue of a topological insulator.

Common misconceptions. The optical branch is not "made of light" — it is a mechanical vibration that merely couples to light in charged crystals; a neutral crystal (e.g. solid Ar with an artificial two-mass basis) still has an optical branch that is infrared-silent. The gap does not mean atoms "cannot vibrate" at those frequencies — an external driver at a gap frequency produces an evanescent, spatially decaying response, not a propagating wave. And the acoustic branch being "acoustic" refers to its linear low-\(q\) dispersion, not to audible sound: its frequencies reach terahertz at the zone boundary.

Worked examples

Chain with \(M=2.0\times10^{-26}\,\mathrm{kg}\), \(m=1.0\times10^{-26}\,\mathrm{kg}\), \(K=20\,\mathrm{N\,m^{-1}}\), \(a=3.0\times10^{-10}\,\mathrm{m}\). Find the zone-centre optical frequency, the sound speed, and the zone-boundary gap.

State the target quantities before substituting numbers. A

\[ \omega_+(0)=\sqrt{2K\!\left(\tfrac1M+\tfrac1m\right)} = \sqrt{2(20)\!\left(\tfrac{1}{2.0\times10^{-26}}+\tfrac{1}{1.0\times10^{-26}}\right)} \]

Insert the zone-centre optical formula (\(\sin=0\) leaves \(\omega_+^2=2K(1/M+1/m)\)). A

\[ =\sqrt{40\times1.5\times10^{26}} = \sqrt{6.0\times10^{27}} = 7.75\times10^{13}\ \mathrm{rad\,s^{-1}} \]

Arithmetic: \(1/M+1/m = (0.5+1.0)\times10^{26}=1.5\times10^{26}\,\mathrm{kg^{-1}}\). A

\[ v_s = a\sqrt{\frac{K}{2(M+m)}} = 3.0\times10^{-10}\sqrt{\frac{20}{2(3.0\times10^{-26})}} = 5.48\times10^{3}\ \mathrm{m\,s^{-1}} \]

Small-\(q\) acoustic slope; \(K/[2(M+m)]=20/(6.0\times10^{-26})=3.33\times10^{26}\,\mathrm{s^{-2}}\). B

\[ \omega_-^{\mathrm{ZB}}=\sqrt{2K/M}=\sqrt{40/2.0\times10^{-26}}=4.47\times10^{13},\quad \omega_+^{\mathrm{ZB}}=\sqrt{2K/m}=\sqrt{40/1.0\times10^{-26}}=6.32\times10^{13}\ \mathrm{rad\,s^{-1}} \]

Zone-boundary limits from \(\sin(qa/2)=1\). A

\[ \omega_+(0)=7.75\times10^{13}\,\mathrm{rad\,s^{-1}},\quad v_s=5.48\ \mathrm{km\,s^{-1}},\quad \text{gap: } 4.47\text{–}6.32\times10^{13}\,\mathrm{rad\,s^{-1}}\ (\Delta\omega=1.85\times10^{13}) \]

Reading. A realistic optical mode in the tens-of-THz range, a sound speed of a few km/s typical of solids, and a clear terahertz-wide forbidden band set by the 2:1 mass ratio.

Units check. \(20/(10^{-26}\,\mathrm{kg})=2\times10^{27}\,\mathrm{s^{-2}}\Rightarrow\sqrt{}=4.5\times10^{13}\,\mathrm{s^{-1}}\); \(\mathrm{m}\times\mathrm{s^{-1}}=\mathrm{m\,s^{-1}}\) for \(v_s\). All consistent.

Same chain. Test the linear (sound-wave) approximation for the acoustic branch at \(q = 0.1\,\pi/a\) by comparing the exact dispersion with \(\omega\approx v_s q\).

Choose a small but finite \(q\); state both the exact and approximate expressions. B

\[ qa = 0.1\pi = 0.3142,\quad \sin^2\!\left(\tfrac{qa}{2}\right)=\sin^2(0.1571)=2.447\times10^{-2} \]

Evaluate the dimensionless phase. A

\[ \omega_-^2 = K S - K\sqrt{S^2 - \tfrac{4\sin^2(qa/2)}{Mm}},\quad S\equiv\tfrac1M+\tfrac1m=1.5\times10^{26},\ Mm=2.0\times10^{-52} \] \[ = 20\left(1.5\times10^{26} - \sqrt{(1.5\times10^{26})^2 - \tfrac{4(2.447\times10^{-2})}{2.0\times10^{-52}}}\right) \]

Substitute into the exact lower root. \(4\sin^2/(Mm)=4.89\times10^{50}\); \(S^2=2.25\times10^{52}\). B

\[ \omega_-^2 = 20\left(1.5\times10^{26}-1.4836\times10^{26}\right)=3.28\times10^{25}\ \Rightarrow\ \omega_-^{\mathrm{exact}}=5.73\times10^{12}\ \mathrm{rad\,s^{-1}} \]

Take the square root of the difference. A

\[ \omega^{\mathrm{lin}} = v_s q = (5.48\times10^{3})\!\left(\frac{0.1\pi}{3.0\times10^{-10}}\right)=(5.48\times10^{3})(1.047\times10^{9})=5.74\times10^{12}\ \mathrm{rad\,s^{-1}} \]

Linear estimate using the sound speed from Example 1. A

\[ \omega_-^{\mathrm{exact}}=5.73\times10^{12},\quad \omega^{\mathrm{lin}}=5.74\times10^{12}\ \mathrm{rad\,s^{-1}}\quad(\text{difference }\approx0.1\%) \]

Reading. At one-tenth of the way to the zone boundary the acoustic branch is still linear to about \(0.1\%\); the continuum sound picture is excellent at long wavelengths and only fails as \(q\) approaches \(\pi/a\), where the branch flattens.

Units check. \(v_s q\): \(\mathrm{(m\,s^{-1})(m^{-1})}=\mathrm{s^{-1}}\), a frequency, matching \(\omega_-\).

Problems

A diatomic chain has \(M=3.0\times10^{-26}\,\mathrm{kg}\), \(m=1.0\times10^{-26}\,\mathrm{kg}\), \(K=15\,\mathrm{N\,m^{-1}}\), \(a=3.0\times10^{-10}\,\mathrm{m}\). Find (a) the zone-centre optical frequency, (b) the two zone-boundary frequencies, and (c) the gap width.
Solution
(a) \(\omega_+(0)=\sqrt{2K(1/M+1/m)}=\sqrt{30\times(0.333+1.0)\times10^{26}}=\sqrt{30\times1.333\times10^{26}}=\sqrt{4.0\times10^{27}}=6.32\times10^{13}\,\mathrm{rad\,s^{-1}}\). (b) \(\omega_-^{\mathrm{ZB}}=\sqrt{2K/M}=\sqrt{30/3.0\times10^{-26}}=\sqrt{1.0\times10^{27}}=3.16\times10^{13}\); \(\omega_+^{\mathrm{ZB}}=\sqrt{2K/m}=\sqrt{30/1.0\times10^{-26}}=\sqrt{3.0\times10^{27}}=5.48\times10^{13}\,\mathrm{rad\,s^{-1}}\). (c) Gap runs \(3.16\text{–}5.48\times10^{13}\), width \(\Delta\omega=2.31\times10^{13}\,\mathrm{rad\,s^{-1}}\).
For the chain of Problem 1, compute the long-wavelength sound speed \(v_s\) and the acoustic-branch group velocity at \(q=\pi/a\). Comment on the ratio.
Solution
\(v_s = a\sqrt{K/[2(M+m)]}=3.0\times10^{-10}\sqrt{15/(2\times4.0\times10^{-26})}=3.0\times10^{-10}\sqrt{1.875\times10^{26}}=3.0\times10^{-10}\times1.369\times10^{13}=4.11\times10^{3}\,\mathrm{m\,s^{-1}}\). At the zone boundary \(d\omega/dq=0\) exactly (the acoustic branch is flat there, a standing wave), so the group velocity is \(0\). The ratio \(v_s/0\) is formally infinite: energy transport is efficient at long wavelength but the shortest-wavelength acoustic phonons do not propagate.
Show that at the zone boundary \(\omega_+^{\mathrm{ZB}}/\omega_-^{\mathrm{ZB}}=\sqrt{M/m}\). A neutron measurement on some crystal gives this ratio as \(1.50\). Find the mass ratio \(M/m\).
Solution
\(\omega_+^{\mathrm{ZB}}=\sqrt{2K/m}\) and \(\omega_-^{\mathrm{ZB}}=\sqrt{2K/M}\), so \(\omega_+^{\mathrm{ZB}}/\omega_-^{\mathrm{ZB}}=\sqrt{(2K/m)/(2K/M)}=\sqrt{M/m}\). Setting \(\sqrt{M/m}=1.50\) gives \(M/m=1.50^2=2.25\). The heavier atom is \(2.25\times\) the lighter one; equivalently the gap edges are fixed purely by the mass contrast, independent of \(K\).
Starting from the exact acoustic root, derive the sound speed \(v_s=a\sqrt{K/[2(M+m)]}\) by expanding to lowest order in \(q\).
Solution
Write \(S=1/M+1/m\) and \(P=4\sin^2(qa/2)/(Mm)\), so \(\omega_-^2 = KS - KS\sqrt{1-P/S^2}\). For small \(q\), \(P/S^2\ll1\), so \(\sqrt{1-P/S^2}\approx1-\tfrac{P}{2S^2}\), giving \(\omega_-^2\approx KP/(2S)=\dfrac{K}{2S}\cdot\dfrac{4\sin^2(qa/2)}{Mm}\). Since \(S=(M+m)/Mm\), \(\dfrac{1}{S\,Mm}=\dfrac{1}{M+m}\), so \(\omega_-^2\approx\dfrac{2K\sin^2(qa/2)}{M+m}\). Using \(\sin(qa/2)\approx qa/2\), \(\omega_-^2\approx\dfrac{2K}{M+m}\cdot\dfrac{q^2a^2}{4}=\dfrac{Ka^2q^2}{2(M+m)}\), hence \(\omega_-\approx qa\sqrt{K/[2(M+m)]}\) and \(v_s=d\omega_-/dq=a\sqrt{K/[2(M+m)]}\).
At \(q=0\) show that the optical eigenvector satisfies \(MA=-mB\) (centre of mass stationary), while the acoustic eigenvector has \(A=B\). Interpret physically.
Solution
From row 1 of the dynamical matrix at \(q=0\): \((2K-M\omega^2)A = K(1+e^{0})B=2KB\). Acoustic (\(\omega_-^2=0\)): \(2KA=2KB\Rightarrow A=B\) — both sublattices move identically, a rigid long-wavelength translation (sound). Optical (\(\omega_+^2=2K(1/M+1/m)\)): \(2K-M\omega_+^2 = 2K-2KM(1/M+1/m)=2K-2K-2KM/m=-2KM/m\), so \((-2KM/m)A=2KB\Rightarrow B=-(M/m)A\), i.e. \(MA=-mB\). The mass-weighted displacement sums to zero, \(MA+mB=0\): the centre of mass is fixed and the two sublattices oscillate in antiphase. In an ionic crystal this is an oscillating dipole, the origin of infrared activity.