PHY 140A: Solid State Physics. Solution to Homework #4

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1 PHY 140A: Solid State Physics Solution to Homework #4 TA: Xun Jia 1 November 5, jiaxun@physics.ucla.edu

2 Fall 006 Physics 140A c Xun Jia (November 5, 006 Problem #1 Problem. in Ashcroft and ermin: Diatomic Linear Chain. Consider a linear chain in which alternate ions have mass 1 and, and only nearest neighbors interact. (a. Show that the dispersion relation for the normal modes is: ω = C ( 1 + ± cos ka 1 (1 (b. Discuss the form of the dipersion relation and the nature of the normal modes when 1. (c. Compare the dispersion relation with that of the monatomic linear chain when 1 =. Solution: (a. As in Fig. 1, let the coordinates of two atoms in the n th unit cell be u n,1 and u n, then the equation of motions are: a 1 C u n-1,1 u n-1, u n,1 u n, u n+1,1 u n+1, Figure 1: The diatomic linear chain with alternating masses. 1 ü n,1 = C(u n, u n,1 C(u n,1 u n 1, = C(u n, + u n 1, u n,1 ü n, = C(u n+1,1 u n, C(u n, u n,1 = C(u n+1,1 + u n,1 u n, ( For the normal modes, they are in the form: u n,1 = A 1 e i(kna ωt (3 u n, = A e i(kna ωt substitute (3 in to (, we have: A 1 (ω 1 C + A C(1 + e ika = 0 A 1 C(1 + e ika + A (ω C = 0 (4 1

3 Fall 006 Physics 140A c Xun Jia (November 5, 006 to have non trivial solutions, it requires: ( ω det 1 C C(1 + e ika C(1 + e ika ω C = 0 (5 this gives the dispersion relations of: ω = C ( 1 + ± cos ka 1 (6 (b. when 1, then / 1 1, expand (6: ω = C ( 1 + ( ± cos ka [ ( = C 1 + ( ± 1 + ( ] (7 cos ka + o( For + in (7, it follows that: C [ 1 + o( ] 1 (8 plug in to (4, we get: A 1 A 0 (9 this gives a normal mode in which the atom with mass 1 does not oscillate, while the atom with mass oscillate with frequency given in (9. For - in (7, it follows that: [ C ka sin 1 + o( ] 1 plug in to (4, we get: 1 (10 A 1 A 1 (11 this gives a normal mode in which the two atoms within a same primitive cell move in phase, just a whole unit, since 1, the frequency is determined by the lager one, say 1. (c. When 1 = =, (6 gives: ω = C ( ± + cos ka (1

4 Fall 006 Physics 140A c Xun Jia (November 5, 006 Optical branch Acoustic branch (/C 1/ ka=ka' Figure : The dispersion relation for the diatomic linear chain when 1 = =. Shaded area gives the Brillouin zone for the diatomic chain. this dispersion relation is shown as in Fig., where for the first Brillouin zone for the diatomic chain is ka [ π, π]. In terms of the new lattice constant a = a/, the dispersion relation is then: ω = C (1 ± cos ka, ka [ π, π ] (13 For the optical branch, consider the part with ka [ π, 0], then: ω = C (1 + cos ka = C [1 cos(ka + π] (14 notice that ka +π [ π, π], above expression indicates that the optical branch in dispersion relation for the diatomic chain in [ π, 0] is equivalent to the acoustic branch in [ π, π]. Similarly, the optical branch in dispersion relation for the diatomic chain in [0, π] is equivalent to the acoustic branch in [ π, π ]. Thus, instead of describing the dispersion relation with both optical and acoustic branch in ka [ π, π ], it is equivalent to just consider the acoustic branch in ka [ π, π], which is just the dispersion relation for the monatomic chain. Therefore, when setting 1 = =, the dispersion relation for monatomic chain is recovered. 3

5 Fall 006 Physics 140A c Xun Jia (November 5, 006 Problem # Problem 4.3 in Kittel. For the problem treated by (18 to (6, find the amplitude ratio u/v for the two branches at K max = π/a. Show that at this value of K the two lattices act as if decoupled: one lattice remains at rest while the other lattice moves. Solution: From Eqn. (1 in Kittel, let K = K max = π/a, we get: ( C 1 ω det 0 o C ω = 0 (15 The solution to this equation gives the normal modes frequencies at this K value: ω = C/ 1 ; ω = C/ (16 Substitute this expression of ω in to (0 in Kittel, we found: When ω = C/ 1, it follows that: which gives a mode that u is finite, and v is zero. When ω = C/, it follows that: u v (17 u v = 0 (18 which gives a mode that v is finite and u is zero. In any cases, the two lattices act as if decoupled: one lattice remains at rest while the other lattice moves. Problem #3 Problem 4.5 in Kittel. To assist solving the following problem, solve this with coupling constants C 1 and C, then set C 1 = C, and C = 10C. Consider the normal modes of a linear chain in which the force constants between nearest-neighbor atoms are alternately C and 10C. Let the masses be equal, and let the nearest-neighbor separation be a/. Find ω(k at k = 0 and k = π/a. Sketch in the dispersion relation by eye. This problem simulates a crystal of diatomic such as H. Solution: Now as in Fig. 3, let the coupling constants be in general C 1, and C, then the equations of motions are: ü n,1 = C 1 (u n, u n,1 C (u n,1 u n 1, = C 1 u n, + C u n 1, (C 1 + C u n,1 ü n, = C (u n+1,1 u n, C 1 (u n, u n,1 (19 = C u n+1,1 + C 1 u n,1 (C 1 + C u n, 4

6 Fall 006 Physics 140A c Xun Jia (November 5, 006 a C 1 C C 1 C C 1 C u n-1,1 u n-1, u n,1 u n, u n+1,1 u n+1, Figure 3: The diatomic linear chain with alternating coupling constants. For the normal modes, they are in the form: substitute (0 in to (19, we have: u n,1 = A 1 e i(kna ωt u n, = A e i(kna ωt (0 A 1 [ω (C 1 + C ] + A (C 1 + C e ika = 0 A 1 (C 1 + C e ika + A [ω (C 1 + C ] = 0 to have non trivial solutions, it requires: ( ω det (C 1 + C C 1 + C e ika C 1 + C e ika ω = 0 ( (C 1 + C this gives the dispersion relations of: ω = 1 ( C 1 + C ± C1 + C + C 1 C cos ka (1 (3 set C 1 = C, and C = 10C, we found the dispersion relation is: ω = C ( 11 ± cos ka and at k = 0, and k = π/a, the normal mode frequencies are: at k = 0, we have: 0 C : for optical branch : for acoustic branch (4 (5 at k = π/a, we have: 0C C The dispersion relation is drawn in Fig. 4. : for optical branch : for acoustic branch (6 5

7 Fall 006 Physics 140A c Xun Jia (November 5, Optical branch Acoustic branch (/C 1/ 0-0 ka Figure 4: The dispersion relation for diatomic linear chain with alternating coupling constants C 1 = C, and C = 10C. Problem #4 Problem.3 in Ashcroft and ermin: Lattice with a Basis Viewed as a Weakly Perturbed onatomic Bravais Lattice. It is instructive to examine the dispersion relation for the one-dimensional latttice with a basis and unequal coupling constants (see previous problem, in the limit in which the coupling constants C 1 and C become very close: C 1 = C +, C = C, C. (a. Show that when = 0, the dispersion relation reduces to that for a monatomic linear chain with nearest-neighbor coupling. (Warning: If the length of the unit cell in the diatomic chain is a, then when C 1 = C it will reduce to a monatomic chain with lattice constant a/; furthermore, the Brillouin zone ( π/a < k < π/a for the diatomic chain will only be half the Brillouin zone ( π/(a/ < k < π/(a/ of the monatomic chain. You must therefore explain how two branches (acoustic and optical in half the zone reduce back to one branch in the full zone. To demonstrate the reduction convincingly you should examine the behavior of the amplitude ratio u/v when = 0. (b. Show that when 0, but C, then the dispersion relation differs from that of the monatomic chain only by terms of order ( /C, except when π ka is of order /C. Show that when this happens the distortion of the dispersion relation relation for the monatomic chain is linear in /C. Solution: 6

8 Fall 006 Physics 140A c Xun Jia (November 5, 006 (a. Same as in last problem in a linear chain with alternating coupling constants, set C 1 = C +, C = C, we get the dispersion relation of: ω = 1 ( C ± C (1 + cos ka + (1 cos ka (7 at the case = 0, above equation reduces to: ω = C 1 + cos ka 1 ± (8 the equivalence between this expression and the dispersion relation of a monatomic chain is the same as discussed in problem #1 part (c. (b. when C, namely, /C 1, we can Taylor expand (7 in terms of /C as: ω(k = C ± (1 + cos ka + C(1 cos ka = C ± (1 + cos ka 1 + ka tan C [ = C ± ( ] (1 + cos ka 1 + ka tan + o( 4 C C 4 = ω 0 (k ± Cf(k + o( 4 C 4 (9 where ω 0 (k is the dispersion relation for the case = 0, i.e. the monatomic chain, and f(k is defined to be: f(k = C 1 + cos ka tan ka (30 Now, take the square root of (9 and Taylor expand again, we get: ω(k = ω 0 (k ± Cf(k + o( 4 C 4 = ω 0 (k ± C f(k ω 0 (k + o( 4 C 4 (31 that is, in general, the dispersion relation ω(k differs from that of the monatomic chain ω 0 (k only by terms of order ( /C. 7

9 Fall 006 Physics 140A c Xun Jia (November 5, 006 However, to ensure the Taylor expansion in (9 valid, from the first line of that equation, it requires that: 1 + cos ka C(1 cos ka (3 This condition is in general satisfied for almost all k value, except for those k such that (1 + cos ka is of same order as (1 cos ka/c, and this indeed happens at when π ka is of order /C, since then 1 + cos ka = 1 + cos π + 1 (π ka +... A C C(1 cos ka C where A is a constant. Now from (9, we have: ω(k C (1 ± A C + C C ± C 1 + A C and then: ω(k ω 0 (k ± C ω 0(k (33 ( A + o( C (35 where ω 0 (k ω 0 (k = π/a = C/, since when π ka is of order /C, ka is close to π. From above expression, we found that when π ka is of order /C, the distortion of the dispersion relation for the monatomic chain is linear in /C. Problem #5 Problem 4.6 in Kittel. Consider point ions of mass and charge e immersed in a uniform sea of conduction electrons. The ions are imagined to be in stable equilibrium when at regular lattice points. If one ion is displaced a small distance r from its equilibrium position, the restoring force is largely due to the electric charge within the sphere of radius r centered at the equilibrium position. Take the number density of ions (or conduction electrons as 3/4πR 3, which defines R. (a Show that the frequency of a single ion set into oscillation is (e /R 3 1/. (b Estimate the value of this frequency from sodium, roughly. (c From (a, (b, and some common sense, estimate the order of magnitude of the velocity of sound in the metal. Solution: 8

10 Fall 006 Physics 140A c Xun Jia (November 5, 006 (a. Since the number density of ions is 3/4πR 3, the charge density will be 3e/4πR 3, thus when an ion is displaced by a small distance r from its equilibrium position, the total charge contributing to the restoring force is: q = 3e 4πr 3 4πR 3 3 thus the equation of motion (in Gauss unit is = er3 R 3 (36 r = qe r = e r R 3 (37 which gives an oscillation frequency of: e (38 R 3 (b. To estimate the value of this frequency for sodium, which is a bcc lattice, we find these values: 3 R = 3 8π a = 3 3 8π 49.06pm = m (39 = kg = kg e = C = C since above numbers are all in SI unit, the value for the frequency ω can be computed by (refprob5:eqn in its SI unit version: e 4πε 0 R = rad/s (40 (c. If we estimate the wave length to be of order of lattice constant, say λ m, then v = ω k = ωλ π 104 m/s (41 9