6  Lattice Dynamics II

NoteLearning Objectives
  • Derive the dispersion relation of a crystal lattice with a one-atomic basis for longitudinal vibrations.
  • Explain the meaning of group velocity, the long-wavelength sound limit, and the first Brillouin zone.
  • Derive the acoustic and optical branches for a lattice with a two-atomic basis.
  • Interpret acoustic and optical modes in terms of in-phase and out-of-phase atomic motion.
  • Generalize the branch counting from one-dimensional model systems to three-dimensional crystals.

6.1 Summary of Lattice Dynamics

For a crystal with basis atoms labeled by \(\alpha,\beta\) and Cartesian components \(i,j\), the harmonic equations of motion can be written as \[ M_{\alpha}\frac{\partial^2 u_{n\alpha i}}{\partial t^2} + \sum_{m,\beta,j} C_{n\alpha i}^{m\beta j}u_{m\beta j} =0. \]

The plane-wave ansatz is mass normalized: \[ u_{n\alpha i} = \frac{1}{\sqrt{M_{\alpha}}} A_{\alpha i}(\mathbf q) e^{\mathrm i(\mathbf q\cdot \mathbf R_n-\omega t)}. \]

Inserting this into the equations of motion gives \[ -\omega^2 A_{\alpha i}(\mathbf q) + \sum_{\beta,j} D_{\alpha i}^{\beta j}(\mathbf q) A_{\beta j}(\mathbf q) =0, \] where \[ D_{\alpha i}^{\beta j}(\mathbf q) = \sum_m \frac{1}{\sqrt{M_{\alpha}M_{\beta}}} C_{n\alpha i}^{m\beta j} e^{\mathrm i\mathbf q\cdot(\mathbf R_m-\mathbf R_n)}. \]

The normal-mode frequencies follow from \[ \det\left[D_{\alpha i}^{\beta j}(\mathbf q)-\omega^2\mathbf 1\right]=0. \]

For a basis with \(r'\) atoms, there are \(3r'\) branches in a three-dimensional crystal. The following sections show how this general result becomes explicit in simple models.

6.2 Crystal Lattice With a One-Atomic Basis

Note

A crystal with a one-atomic basis is the simplest setting in which acoustic lattice vibrations and Brillouin-zone restriction can be seen explicitly.

We first consider a crystal with a one-atomic basis and a wave propagating along a high-symmetry direction. If entire lattice planes move parallel to their normal direction, symmetry eliminates transverse force components. The problem then reduces to a one-dimensional longitudinal vibration of lattice planes.

Figure 6.1: Figure placeholder. Longitudinal and transverse displacements of lattice planes in a crystal with a one-atomic basis. The figure illustrates why propagation along a high-symmetry direction can be treated as an effective one-dimensional problem.

6.2.1 Force Law for Longitudinal Plane Vibrations

Note

The restoring force is determined by relative displacements between neighboring lattice planes, not by absolute displacements.

Let \(u_n\) denote the displacement of plane \(n\). The force exerted on plane \(n\) by plane \(n+p\) is proportional to the relative displacement \(u_{n+p}-u_n\). Summing over all planes gives \[ F_n = \sum_p \Phi_p(u_{n+p}-u_n). \]

Newton’s equation is therefore \[ M\frac{\partial^2u_n}{\partial t^2} - \sum_p \Phi_p(u_{n+p}-u_n) =0. \]

We see that a uniform translation \(u_n=\mathrm{const.}\) produces no restoring force.

6.2.2 Dispersion Relation for General Couplings

Note

The dispersion relation follows by inserting a plane-wave displacement pattern into the discrete equations of motion.

Use the plane-wave ansatz \[ u_n = A e^{\mathrm i(qna-\omega t)}, \] where \(a\) is the distance between the planes (lattice constant).

Then \[ u_{n+p} = u_n e^{\mathrm i qpa}. \]

Substitution gives \[ -\omega^2 M - \sum_p \Phi_p\left(e^{\mathrm i qpa}-1\right) =0. \]

Using inversion symmetry, \(C_{-p}=\Phi_p\), this becomes \[ -\omega^2 M = \sum_{p=1}^{\infty} \Phi_p \left(e^{\mathrm i qpa}+e^{-\mathrm i qpa}-2\right) = 2\sum_{p=1}^{\infty} \Phi_p\left(\cos qpa-1\right). \]

Hence

\[ \omega^2(q) = \frac{2}{M} \sum_{p=1}^{\infty} \Phi_p\left(1-\cos qpa\right). \]

This is the longitudinal dispersion relation for a one-atomic basis with arbitrary plane-plane couplings \(\Phi_p\).

6.2.3 Nearest-Neighbor Approximation

Note

Keeping only the nearest-neighbor coupling gives the simplest analytic form of the acoustic phonon dispersion.

If only nearest-neighbor planes are coupled, then \(\Phi_1\neq 0\) and all other \(\Phi_p\) vanish. The dispersion reduces to \[ \omega^2(q) = \frac{2\Phi_1}{M}(1-\cos qa) =\frac{4\Phi_1}{M}\sin^2\frac{qa}{2}. \]

For \(0\le q\le \pi/a\),

\[ \omega(q) = 2\sqrt{\frac{\Phi_1}{M}}\sin\frac{qa}{2}. \]

The frequency is periodic in reciprocal space and even in \(q\): \[ \omega(q)=\omega(q+G), \qquad \omega(q)=\omega(-q), \] with \(G=2\pi/a\) in one dimension.

Figure 6.2: Figure placeholder. Dispersion relation of lattice vibrations for a crystal lattice with a one-atomic basis in the nearest-neighbor approximation. The first Brillouin zone is indicated.

6.2.4 Group Velocity

A single phonon mode described by a plane wave extends throughout the entire crystal and therefore cannot represent a localized disturbance. Physical vibrational excitations are instead described by wave packets formed from the superposition of modes with wavevectors in a narrow range around a central wavevector \(\mathbf q\).

The superposition produces a rapidly oscillating carrier wave modulated by a slowly varying envelope. The envelope represents the localized vibrational excitation and propagates through the crystal with the group velocity

\[ \mathbf v_g = \nabla_{\mathbf q}\omega(\mathbf q). \]

In one dimension this reduces to

\[ v_g = \frac{d\omega}{dq}. \]

The group velocity determines the propagation of vibrational energy and information through the crystal. For long-wavelength acoustic phonons, where the dispersion relation is approximately linear,

\[ \omega \approx v_s q, \]

the group velocity becomes

\[ v_g = \frac{d\omega}{dq} = v_s, \]

which is equal to the speed of sound in the crystal.

6.2.5 Limiting Cases

6.2.5.1 Zone boundary

At the zone boundary, \[ q=\frac{\pi}{a}, \] the frequency reaches \[ \omega_{\max} = \sqrt{\frac{4\Phi_1}{M}}. \]

The group velocity is \[ v_g = \nabla_q\omega(q). \]

For the one-dimensional nearest-neighbor dispersion, \[ v_g(q) = \sqrt{\frac{\Phi_1a^2}{M}}\cos\frac{qa}{2} \qquad (0\le q\le \pi/a). \]

Thus \[ v_g\left(\frac{\pi}{a}\right)=0. \]

At the Brillouin-zone boundary the wave is a standing wave, not a propagating wave packet.

6.2.5.2 Long-Wavelength Limit

For \(qa\ll 1\), \[ \sin\frac{qa}{2} \simeq \frac{qa}{2}, \] and therefore \[ \omega(q) \simeq \sqrt{\frac{\Phi_1a^2}{M}} q. \]

This is a linear acoustic dispersion, \[ \omega(q)=v_s q, \] with

\[ v_s = \sqrt{\frac{\Phi_1a^2}{M}}. \]

The identification of \(v_s\) as the speed of sound follows from the long-wavelength condition \(\lambda\gg a\). In this limit neighboring atoms move almost in phase, so the discrete lattice no longer resolves individual atomic spacings and behaves like a continuous elastic medium. The acoustic phonon wave packet therefore propagates like an ordinary sound wave. Since the group velocity is the slope of the dispersion relation, \[ v_g = \frac{d\omega}{dq}, \] the linear dispersion gives \[ v_g = \frac{d}{dq}(v_s q) = v_s. \]

In this limit the discrete lattice behaves like an elastic continuum. More generally, the linear relation \(\omega\propto q\) at small \(q\) does not depend on keeping only nearest-neighbor couplings; longer-range couplings merely change the numerical value of the sound velocity.

6.2.6 Continuum Check

In the long-wavelength limit, write \[ u_n(t)\rightarrow u(x,t), \qquad u_{n\pm1}(t)\rightarrow u(x\pm a,t). \]

A Taylor expansion gives \[ u(x\pm a,t) = u(x,t) \pm a\frac{\partial u}{\partial x} + \frac{a^2}{2}\frac{\partial^2u}{\partial x^2} +\cdots. \]

For nearest-neighbor coupling, \[ M\frac{\partial^2u}{\partial t^2} = \Phi_1\left[u(x+a,t)+u(x-a,t)-2u(x,t)\right], \] so \[ \frac{\partial^2u}{\partial t^2} = \frac{\Phi_1a^2}{M} \frac{\partial^2u}{\partial x^2} = v_s^2 \frac{\partial^2u}{\partial x^2}. \]

The lattice model therefore reproduces the standard one-dimensional sound-wave equation when \(\lambda\gg a\).

6.3 The First Brillouin Zone

Note

Only wavevectors inside the first Brillouin zone are needed because wavevectors differing by a reciprocal-lattice vector describe the same lattice displacement pattern.

The dynamical matrix contains phase factors of the form \[ e^{\mathrm i\mathbf q\cdot(\mathbf R_m-\mathbf R_n)}. \]

Since \(\mathbf R_m-\mathbf R_n\) is a Bravais-lattice vector, adding a reciprocal-lattice vector \(\mathbf G\) leaves the phase factor unchanged: \[ e^{\mathrm i(\mathbf q+\mathbf G)\cdot(\mathbf R_m-\mathbf R_n)} = e^{\mathrm i\mathbf q\cdot(\mathbf R_m-\mathbf R_n)}. \]

Therefore \[ D_{\alpha i}^{\beta j}(\mathbf q) = D_{\alpha i}^{\beta j}(\mathbf q+\mathbf G), \] and the phonon frequencies satisfy \[ \omega(\mathbf q)=\omega(\mathbf q+\mathbf G). \]

Time-reversal symmetry further gives \[ \omega(-\mathbf q)=\omega(\mathbf q). \]

For the one-dimensional lattice it is therefore sufficient to specify the dispersion in

\[ -\frac{\pi}{a} \le q \le +\frac{\pi}{a}. \]

Physically, wavevectors that differ by a reciprocal-lattice vector describe the same displacements at the lattice sites. The apparent shape of the wave between atoms is not an observable of the discrete lattice model.

Figure 6.3: Figure placeholder. Two waves with wavevectors differing by a reciprocal-lattice vector produce identical displacements on the lattice sites. This motivates restricting the dispersion to the first Brillouin zone.

6.3.1 Transverse and Mixed Polarizations

Note

In three-dimensional crystals, lattice vibrations are characterized not only by their wavevector but also by their polarization.

For transverse vibrations of a one-atomic basis, the derivation is analogous, but the coupling constants \(\Phi_p\) are generally different from the longitudinal ones. Purely longitudinal and purely transverse waves occur only for propagation along suitable symmetry directions. In a cubic crystal these include the \([100]\), \([110]\), and \([111]\) directions. For general propagation directions, the atomic displacements need not be exactly parallel or perpendicular to \(\mathbf q\), and the polarization is mixed.

6.4 Crystal Lattice With a Two-Atomic Basis

Note

A two-atomic basis introduces internal motion within the unit cell and therefore produces both acoustic and optical branches.

We now consider a lattice with two atoms per basis, with masses \(M_1\) and \(M_2\). The displacement of the \(M_1\) plane in cell \(n\) is \(u_n\), and the displacement of the \(M_2\) plane is \(v_n\). The distance between equivalent planes is \(a\). Only nearest-neighbor interactions are retained, and their coupling constant is denoted by \(f\).

Figure 6.4: Figure placeholder. Longitudinal lattice vibration in a crystal with a two-atomic basis. The two atom types have masses \(M_1\) and \(M_2\) and displacements \(u_n\) and \(v_n\).

6.4.1 Equations of Motion

Note

The coupled equations describe how each atom in the basis is restored by its neighboring atoms of the other type.

The restoring force on \(u_n\) comes from its two neighboring \(v\) planes: \[ M_1\frac{\partial^2u_n}{\partial t^2} = f(v_n-u_n)+f(v_{n-1}-u_n). \]

Similarly, \[ M_2\frac{\partial^2v_n}{\partial t^2} = f(u_n-v_n)+f(u_{n+1}-v_n). \]

Equivalently, \[ M_1\frac{\partial^2u_n}{\partial t^2} + f(2u_n-v_n-v_{n-1}) =0, \] \[ M_2\frac{\partial^2v_n}{\partial t^2} + f(2v_n-u_n-u_{n+1}) =0. \]

6.4.2 Mass-Normalized Plane-Wave Ansatz

Note

Mass normalization converts the coupled equations into a symmetric dynamical-matrix eigenvalue problem.

As an ansatz for the solutions, we use \[ u_n(q) = \frac{1}{\sqrt{M_1}} A_1(q)e^{\mathrm i(qan-\omega t)}, \] \[ v_n(q) = \frac{1}{\sqrt{M_2}} A_2(q)e^{\mathrm i(qan-\omega t)}. \]

Substitution gives \[ \left(\frac{2f}{M_1}-\omega^2\right)A_1 - \frac{f}{\sqrt{M_1M_2}} \left(1+e^{-\mathrm iqa}\right)A_2 =0, \] \[ - \frac{f}{\sqrt{M_1M_2}} \left(1+e^{+\mathrm iqa}\right)A_1 + \left(\frac{2f}{M_2}-\omega^2\right)A_2 =0. \]

The corresponding dynamical matrix is \[ D(q) = \begin{pmatrix} \dfrac{2f}{M_1} & -\dfrac{f}{\sqrt{M_1M_2}}(1+e^{-\mathrm iqa})\\ -\dfrac{f}{\sqrt{M_1M_2}}(1+e^{+\mathrm iqa}) & \dfrac{2f}{M_2} \end{pmatrix}. \]

The allowed frequencies follow from \[ \det[D(q)-\omega^2\mathbf 1]=0. \]

6.4.3 Eigenfrequencies

Note

The two eigenfrequencies for each wavevector correspond to collective translation of the basis and internal relative motion of the basis.

The determinant condition gives

\[ \omega^2_{\pm}(q) = f\left(\frac{1}{M_1}+\frac{1}{M_2}\right) \pm f \left[ \left(\frac{1}{M_1}+\frac{1}{M_2}\right)^2 - \frac{4}{M_1M_2} \sin^2\frac{qa}{2} \right]^{1/2}. \]

There are two branches because there are two independent longitudinal degrees of freedom per cell. The lower branch \(\omega_-(q)\) is the acoustic branch and the upper branch \(\omega_+(q)\) is the optical branch, as we will below.

Figure 6.5: Figure placeholder. Dispersion relation for a crystal lattice with a two-atomic basis, shown for \(M_1=2M_2\). The lower curve is the acoustic branch, and the upper curve is the optical branch.

6.4.4 Long-Wavelength Limits and Branches (Acoustic and Optical)

Note

The two eigenfrequencies for each wavevector correspond to collective translation of the basis and internal relative motion of the basis.

For \(qa\ll1\), expand the square root in the dispersion relation. The two branches become \[ \omega_+^2(q) \simeq 2f\left(\frac{1}{M_1}+\frac{1}{M_2}\right) - \frac{a^2f}{2(M_1+M_2)}q^2, \] and \[ \omega_-^2(q) \simeq \frac{a^2f}{2(M_1+M_2)}q^2. \]

The lower branch, \(\omega_-(q)\), is linear at small \(q\): \[ \omega_-(q) \simeq v_s q, \] with \[ v_s = \sqrt{\frac{a^2f}{2(M_1+M_2)}}. \] This means \(\lim_{q\to 0} \omega_-(q) = 0\). Hence, we call the lower branch the acoustic branch, because it has the same linear low-\(q\) behavior as an acoustic wave.

The upper branch, \(\omega_+(q)\), is thus the optical branch. In the limit \(q\to 0\), it has a nonzero frequency:

\[ \omega_+(0) = \sqrt{ 2f\left(\frac{1}{M_1}+\frac{1}{M_2}\right) }. \]

6.4.5 Mode Character at the Zone Center

At \(q\to0\), the acoustic branch has neighboring atoms moving in phase: \[ \frac{u}{v}=1 \qquad (\mathrm{for}~~ \omega_-). \]

The optical branch has neighboring atoms moving out of phase: \[ \frac{u}{v} = -\frac{M_2}{M_1} \qquad (\mathrm{for}~~ \omega_+). \]

For the optical mode, \[ M_1u+M_2v=0, \] so the center of mass of the two-atom basis remains at rest.

This distinction is the origin of the terms acoustic and optical. Acoustic modes reduce to ordinary sound waves in the long-wavelength limit. Optical modes describe relative motion of the atoms in the basis; in ionic crystals this motion can produce an oscillating electric dipole moment and therefore couple strongly to light.

Figure 6.6: Figure placeholder. Longitudinal and transverse acoustic and optical lattice vibrations. The figure shows in-phase motion for acoustic modes and out-of-phase motion for optical modes.

6.4.6 Zone-Boundary Limits and Frequency Gap

At the one-dimensional zone boundary, \[ q=\frac{\pi}{a}, \] the two branches have frequencies \[ \omega_+\left(\frac{\pi}{a}\right) = \sqrt{\frac{2f}{M_2}}, \] \[ \omega_-\left(\frac{\pi}{a}\right) = \sqrt{\frac{2f}{M_1}}, \] assuming \(M_1>M_2\).

A frequency gap opens between the upper edge of the acoustic branch and the lower edge of the optical branch. The gap grows as the mass ratio \(M_1/M_2\) becomes larger. In the gap the wavevector becomes imaginary, corresponding to a spatially damped wave rather than a propagating normal mode.

6.4.7 Limiting Mass Ratios

If \(M_1\gg M_2\), the acoustic branch is governed mainly by the heavy atoms: \[ \omega_-^2 \simeq \frac{2f}{M_1} \sin^2\frac{qa}{2}. \]

The optical branch is nearly flat and is controlled mainly by the light atoms: \[ \omega_+^2 \simeq \frac{2f}{M_2} - \frac{2f}{M_1} \sin^2\frac{qa}{2}. \]

If \(M_1=M_2=M\), the chain is physically a one-atomic lattice with half the spacing. The apparent separation into acoustic and optical branches is then a consequence of using a doubled unit cell; the dispersion has been folded into the smaller Brillouin zone.

6.5 Lattice Vibrations in Three Dimensions

Note

The one-dimensional examples generalize to real crystals by adding polarization directions and branch counting.

The one-dimensional models are pedagogical reductions. In a real three-dimensional crystal, each atom has three displacement components, and the polarization structure becomes central.

6.5.1 One-Atomic Basis in Three Dimensions

Note

A three-dimensional crystal with one atom per basis has three acoustic branches, corresponding to one longitudinal and two transverse polarizations.

For a three-dimensional crystal with a one-atomic basis, \(r'=1\), there are \[ 3r'=3 \] branches for each wavevector \(\mathbf q\).

They are classified as:

  • one longitudinal acoustic branch, usually labeled LA;
  • two transverse acoustic branches, usually labeled TA.

In high-symmetry directions, the longitudinal branch has displacement parallel to \(\mathbf q\), while the transverse branches have displacement perpendicular to \(\mathbf q\). In general directions, the eigenvectors need not be exactly longitudinal or transverse; the polarization may be mixed.

Usually the longitudinal branch lies at higher frequency than the transverse branches for the same small wavevector magnitude, reflecting the larger stiffness associated with compression compared with shear.

Figure 6.7: Figure placeholder. Phonon dispersion relations of aluminum, a monoatomic fcc crystal. The curves are shown along high-symmetry directions \(\Gamma X\), \(\Gamma K\), and \(\Gamma L\) and illustrate one longitudinal and two transverse acoustic branches.

The periodicity and evenness of the dispersion remain valid: \[ \omega(\mathbf q)=\omega(\mathbf q+\mathbf G), \] \[ \omega(-\mathbf q)=\omega(\mathbf q). \]

Therefore phonon dispersions are plotted only within the first Brillouin zone, typically along high-symmetry paths.

6.5.2 Multiatomic Basis in Three Dimensions

Note

A multiatomic basis produces additional optical branches because atoms inside the unit cell can move relative to one another.

For a basis with \(r'\) atoms, the determinant equation has \[ 3r' \] solutions for each \(\mathbf q\). In a three-dimensional crystal these separate into

\[ 3\text{ acoustic branches} \qquad\text{and}\qquad 3r'-3\text{ optical branches}. \]

The three acoustic branches correspond to long-wavelength translations of the basis. The optical branches correspond to internal relative motion of atoms within the basis.

For example, silicon has the diamond structure with a two-atomic basis. Thus \(r'=2\), and there are \[ 3r'=6 \] phonon branches: three acoustic and three optical.

Figure 6.8: Figure placeholder. Phonon dispersion relations of silicon. Because silicon has a two-atomic basis, the spectrum contains three acoustic and three optical branches, with degeneracies along selected high-symmetry directions.

6.5.3 Sanity Checks for Branch Counting

  • A one-atomic basis in three dimensions has only acoustic branches.
  • A two-atomic basis in three dimensions has six branches in total.
  • Optical branches require internal degrees of freedom inside the basis.
  • Degeneracies can reduce the number of visibly distinct curves along high-symmetry directions, even though the number of branches is unchanged.

6.6 Connection to the Next Lecture: From Dispersion to a Phonon Density of States

The central output of lattice dynamics is the set of branches \[ \omega_r(\mathbf q), \] where \(r\) labels the branch. The next step is to count how many modes lie in a frequency interval. This leads to the phonon density of states.

For a macroscopic three-dimensional crystal with periodic Born–von Karman boundary conditions, the allowed wavevectors become very dense in reciprocal space. The density of allowed \(\mathbf q\)-points is \[ Z(\mathbf q) = \frac{V}{(2\pi)^3}. \]

Sums over phonon modes can then be replaced by integrals: \[ \sum_{\mathbf q,r}F[\omega_r(\mathbf q)] = \sum_r \int_{\mathrm{1.\ BZ}} d^3q Z(\mathbf q) F[\omega_r(\mathbf q)], \] where \(F[\omega_r(\mathbf q)]\) is an arbitrary function of the frequency.

Equivalently, one defines a frequency-space density of states \[ D(\omega) = \sum_r \int_{\mathrm{1.\ BZ}} d^3q Z(\mathbf q) \delta \left(\omega-\omega_r(\mathbf q)\right). \]

The physical point is simple: a flat dispersion produces many states in a small frequency interval, while a steep dispersion produces fewer. This is why zone-boundary flattening and optical branches are important for thermal properties.

NoteTake-Home Messages
  • Lattice vibrations in a periodic crystal reduce to eigenvalue problems for the dynamical matrix.
  • A one-atomic basis gives acoustic branches because uniform translation costs no restoring energy.
  • The long-wavelength limit of an acoustic branch reproduces continuum sound propagation.
  • The first Brillouin zone contains all independent phonon wavevectors.
  • A two-atomic basis produces acoustic and optical branches.
  • Acoustic modes involve in-phase motion of neighboring basis atoms at small wavevector.
  • Optical modes involve relative motion inside the basis and can couple to light in ionic crystals.
  • In three dimensions, a basis with multiple atoms produces both acoustic and optical branches.

Problem Set

  1. One-Atomic Basis With General Couplings. Starting from \[ M\frac{\partial^2u_n}{\partial t^2} - \sum_p \Phi_p(u_{n+p}-u_n) =0, \] insert \(u_n=Ae^{\mathrm i(qna-\omega t)}\) and derive \[ \omega^2(q) = \frac{2}{M} \sum_{p=1}^{\infty} \Phi_p(1-\cos qpa). \] Then specialize to nearest-neighbor coupling.

  2. Group Velocity and the Sound Limit. For the nearest-neighbor one-atomic basis, \[ \omega(q)=2\sqrt{\frac{\Phi_1}{M}}\sin\frac{qa}{2} \qquad (0\le q\le \pi/a), \] compute \(v_g(q)\). Evaluate the limits \(q\to0\) and \(q\to\pi/a\), and interpret the results.

  3. First Brillouin Zone. Show that the displacement pattern produced by \(q\) is identical to that produced by \(q+2\pi/a\) on the lattice sites. Explain why this allows the dispersion relation to be restricted to \(-\pi/a\le q\le \pi/a\).

  4. Two-Atomic Basis. Starting from \[ M_1\frac{\partial^2u_n}{\partial t^2} + f(2u_n-v_n-v_{n-1})=0, \] \[ M_2\frac{\partial^2v_n}{\partial t^2} + f(2v_n-u_n-u_{n+1})=0, \] derive the \(2\times2\) dynamical matrix using the mass-normalized ansatz and obtain the determinant equation for \(\omega^2\).

  5. Branch Counting. For crystals with \(M=1\), \(M=2\), and \(M=4\) atoms in the primitive cell, count the total number of branches, the number of acoustic branches, and the number of optical branches. Then state whether diamond-structure silicon has optical branches.