1. Normal Modes of Vibration One dimensional model # 1: The Monatomic Chain Consider a Monatomic Chain of Identical Atoms with nearest-neighbor, “Hooke’s Law” type forces (F = - kx) between the atoms. This is equivalent to a force-spring model, with masses m & spring constants K.

2. First, consider just two neighboring atoms. Assume that they interact with a known potential V(r). Expand that potential in a Taylor’s series about their equilibrium separation: r R V(R) 0 a Repulsive Attractive min This potential energy is the same as that associated with a spring with spring constant:

3. This one-dimensional chain of identical atoms is the simplest possible solid. Assume that the chain contains a very large number (N   ) of atoms with identical masses m. Let the atomic separation be a distance a. Assume that the atoms move only in a direction parallel to the chain. Assume that only nearest-neighbors interact with each other (the forces are short-ranged). aaaaaa U n-2 U n-1 U n U n+1 U n+2

4. The simple case of a monatomic linear chain with only nearest neighbor interactions. Expand the energy near the equilibrium point for the n th atom. Then, the Newton’s 2 nd Law equation of motion becomes: a a U n-1 U n U n+1 l

5. This can be seen as follows. The total force on the n th atom is the sum of 2 forces: The force to the right: The force to the left: Total force = Force to the right – Force to the left aa U n-1 U n U n+1 The Equation of motion of each atom is of this form. Only the value of ‘n’ changes. l

6. Assume that all atoms oscillate with the same amplitude A & the same frequency ω. Assume harmonic solutions for the displacements u n of the form: Undisplaced PositionDisplaced Position

7. Put all of this into the equation of motion: Mathematical manipulation gives: After more manipulation, this simplifies to: The maximum allowed frequency is: The physical significance of this result is that, for the monatomic chain, the only allowed vibrational frequencies ω must be related to the wavenumber k = (2π/λ) or the wavelength λ in this way. This result is often called the “Phonon Dispersion Relation” for the chain, even though these are classical lattice vibrations & there are no (quantum mechanical) phonons in the classical theory. Solution to the Normal Mode Eigenvalue Problem for the monatomic chain.

8. The “Phonon Dispersion Relations” or Normal Mode Frequencies or ω versus k relation for the monatomic chain. 0л/a2л/a–л–л/a k Because of BZ periodicity with a period of 2π/a, only the first BZ is needed. Points A, B & C correspond to the same frequency, so they all have the same instantaneous atomic displacements. k  C A B

9. Some Physics Discussion We started from the Newton’s 2 nd Law equations of motion for N coupled harmonic oscillators. If one atom starts vibrating, it does not continue with constant amplitude, but transfers energy to the others in a complicated way. That is, the vibrations of individual atoms are not simple harmonic because of this exchange of energy among them. On the other hand, our solutions represent the oscillations of N UNCOUPLED simple harmonic oscillators. As we already said, these are called the Normal Modes of the system. They are a collective property of the system as a whole & not a property of any of the individual atoms. Each mode represented by ω(k) oscillates independently of the other modes. Also, it can be shown that the number of modes is the same as the original number of equations N. Proof of this follows. Monatomic Chain Dispersion Relation

10. To establish which wavenumbers are possible for the one dimensional chain, reason as follows: Not all values are allowed because of periodicity. In particular, the n th atom is equivalent to the (N+n) th atom. This means that the assumed solution for the displacements: must satisfy the periodic boundary condition: This, in turn requires that there are an integer number of wavelengths in the chain. So, in the first Brillouin Zone, there are only N allowed values of k.

11. What is the physical significance of wave numbers k outside of the First Brillouin Zone [-(π/a)  k  (π/a)]? At the Brillouin Zone edge : This k value corresponds to the maximum frequency. A detailed analysis of the displacements shows that, in that mode, every atom is oscillating π radians out of phase with it’s 2 nearest neighbors. That is, a wave at this value of k is a standing wave. x Black k = π/a Green: k = (0.85)π/a

12. Points A and C have the same frequency & the same atomic displacements. It can be shown that the group velocity v g = (dω/dk) there is negative, so that a wave at that ω & that k moves to the left. The green curve (below) corresponds to point B in the ω(k) diagram. It has the same frequency & displacement as points A and C, but v g = (dω/dk) there is positive, so that a wave at that ω & that k moves to the right. Points A & C are equivalent; adding a multiple of 2π/a to k does not change either ω or v g, so point A contains no physical information that is different from point B. The points k = ±π/a have special significance x Bragg reflection occurs at k= ±nπ/a unun x unun a -π/a-π/a kk  CA B ω(k) (dispersion relation) π/aπ/a 2π/a2π/a

13. Briefly look in more detail at the group velocity, v g. The dispersion relation is: So, the group velocity is: v g  (dω/dk) = a(K/m) ½ cos(½ka) v g = 0 at the BZ edge [k =  (π/a)] –This tells us that a wave with λ corresponding to a zone edge wavenumber k =  (π/a) will not propagate. That is, it must be a standing wave! –At the BZ edge, the displacements have the form (for site n): U n = U o e inka = U o e  i(nπ/a) = U o (-1) n

14. One Dimensional Model # 2: The Diatomic Chain

15. U n-2 U n-1 U n U n+1 U n+2 K KK K MM m M m a) b) (n-2) (n-1) (n) (n+1) (n+2) a This is the simplest possible model of a diatomic crystal. a is the repeat distance, so, the nearest-neighbor separation is (½)a Consider a Diatomic Chain of Two Different Atom Types with nearest-neighbor, Hooke’s Law type forces (F = - kx) between the atoms. This is equivalent to a force-spring model with two different types of atoms of masses, M & m connected by identical springs of spring constant K.

16. M m M m M U n-2 U n-1 U n U n+1 U n+2 This model is complicated due to the presence of 2 different atom types, which, in general, move in opposite directions. As before, the GOAL is to obtain the dispersion relation ω(k) for this model. There are 2 atom types, with masses M & m, so there will be 2 equations of motion, one for M & one for m. Equation of Motion for M Equation of Motion for m

17. M m M m M U n-2 U n-1 U n U n+1 U n+2 As before, assume harmonic (plane wave) solutions for the atomic displacements U n : Displacement for M Displacement for m α = complex number which determines the relative amplitude and phase of the vibrational wave.

18. The equation for m becomes: Put all of this into the two equations of motion. The equation for M becomes: (1) & (2) are two coupled, homogeneous, linear algebraic equations in the 2 unknowns α & ω as functions of k. More algebra gives: (1) (2)

19. Combining (1) & (2) gives a quadratic equation for ω 2 : The two solutions for ω 2 are: Since the chain contains N unit cells, there will be 2N normal modes of vibration, because there are 2N atoms and 2N equations of motion for masses M & m.

20. Solutions to the Normal Mode Eigenvalue Problem ω(k) for the Diatomic Chain There are two solutions for ω 2 for each wavenumber k. That is, there are 2 branches to the “Phonon Dispersion Relation” for each k. From an analysis of the displacements, it can be shown that, at point A, the two atoms are oscillating 180º out of phase, with their center of mass at rest. Also, at point B, the lighter mass m is oscillating & M is at rest, while at point C, M is oscillating & m is at rest. 0л/a2л/a–л–л/a k  A B C ω + = Optic Modes ω - = Acoustic Modes

21. Briefly, examine the limiting solutions at points 0, A, B & C. In long wavelength region near k = 0 (ka«1), sin(ka/2) ≈ ½ka. Taylor expansion for small x 0л/a2л/a–л–л/a k  A B C

22. The negative root gives the minimum value of the acoustic branch: Substituting these values of ω into the expression for the relative amplitude α and using cos(ka/2) ≈1 for ka«1gives the corresponding values of α as; OR 0л/a2л/a–л–л/a k  A B C The positive root gives the maximum value of the optic branch:

23. Substitute into the expression for the relative amplitude α This solution corresponds to long-wavelength vibrations near the center of the BZ at k = 0. In that region, M & m oscillate with same amplitude & phase. Also in that region ω = v s k, where v s is the velocity of sound & has the form: k  A B C Optic Acoustic 0π/a2π/a–π–π/ a

24. This solution corresponds to point A in the figure. This value of α shows that, in that mode, the two atoms are oscillating 180º out of phase with their center of mass at rest. Similarly, substituting into the relative amplitude gives: 0π/a2π/a–π–π/a k  A B C Optic Acoustic

25. The other limiting solutions for ω 2 are for ka = π. In this case sin(ka/2) =1, so OR (C) (B) Point C in the plot, which is the maximum acoustic branch point, M oscillates & m is at rest. By contrast, at point B, which is the minimum optic branch point, m oscillates & M is at rest.

26. The Acoustic Branch has this name because it gives rise to long wavelength vibrations ω = v s k, where v s is the velocity of sound The Optic Branch is at higher energy vibration (higher frequencies) & an optical frequency (energy) is needed to excite these modes. Historically, the term “optic” came from how these modes were discovered. Consider an ionic crystal in which atom 1 has a positive charge & atom 2 has a negative charge. As we’ve seen, in those modes, these atoms are moving in opposite directions. (So, each unit cell contains an oscillating dipole.) These modes can be excited these modes with electromagnetic radiation. Acoustic & Optic Branches

27. Transverse optic mode for the diatomic chain The amplitude of vibration is strongly exaggerated!

28. Transverse acoustic mode for the diatomic chain