PHY 140A: Solid State Physics
Solution to Homework #4
TA: Xun Jia1
November 5, 2006
1 Email:
jiaxun@physics.ucla.edu
Fall 2006
c Xun Jia (November 5, 2006)
°
Physics 140A
Problem #1
Problem 22.2 in Ashcroft and Mermin: Diatomic Linear Chain. Consider a linear
chain in which alternate ions have mass M1 and M2 , and only nearest neighbors
interact.
(a). Show that the dispersion relation for the normal modes is:
µ
¶
q
C
2
ω =
M1 + M2 ± M12 + M22 + 2M1 M2 cos ka
M1 M2
(1)
(b). Discuss the form of the dipersion relation and the nature of the normal modes
when M1 À M2 .
(c). Compare the dispersion relation with that of the monatomic linear chain when
M1 = M2 .
Solution:
(a). As in Fig. 1, let the coordinates of two atoms in the nth unit cell be un,1 and
un2 , then the equation of motions are:
a
un-1,1 un-1,2
M1
M2
un,1
un,2
C
un+1,1 un+1,2
Figure 1: The diatomic linear chain with alternating masses.
M1 ün,1 = C(un,2 − un,1 ) − C(un,1 − un−1,2 ) = C(un,2 + un−1,2 − 2un,1 )
M2 ün,2 = C(un+1,1 − un,2 ) − C(un,2 − un,1 ) = C(un+1,1 + un,1 − 2un,2 )
(2)
For the normal modes, they are in the form:
un,1 = A1 ei(kna−ωt)
un,2 = A2 ei(kna−ωt)
(3)
substitute (3) in to (2), we have:
A1 (ω 2 M1 − 2C) + A2 C(1 + e−ika ) = 0
A1 C(1 + eika ) + A2 (ω 2 M2 − 2C) = 0
1
(4)
Fall 2006
c Xun Jia (November 5, 2006)
°
Physics 140A
to have non trivial solutions, it requires:
µ 2
¶
ω M1 − 2C C(1 + e−ika )
det
=0
C(1 + eika ) ω 2 M2 − 2C
this gives the dispersion relations of:
µ
¶
q
C
2
2
2
ω =
M1 + M2 ± M1 + M2 + 2M1 M2 cos ka
M1 M2
(b). when M1 À M2 , then M2 /M1 ¿ 1, expand (6):


µ
µ
¶ s
¶2
C 
M2
M2
M2
ω2 =
1+
± 1+
+2
cos ka
M2
M1
M1
M1
"µ
¶ µ
¶
µ
¶2 #
C
M2
M2
M2
=
1+
± 1+
cos ka + o(
)
M2
M1
M1
M1
• For ”+” in (7), it follows that:
r
ω=
¸
·
2C
M2
)
1 + o(
M2
M1
(5)
(6)
(7)
(8)
plug in to (4), we get:
A1
→0
(9)
A2
this gives a normal mode in which the atom with mass M1 does not oscillate, while the atom with mass M2 oscillate with frequency given in
(9).
• For ”-” in (7), it follows that:
r
2C
ω=
M1
¯
¯·
¸
¯ ka ¯
¯sin ¯ 1 + o( M2 )
¯
2¯
M1
(10)
plug in to (4), we get:
A1
→1
(11)
A2
this gives a normal mode in which the two atoms within a same primitive
cell move in phase, just a whole unit, since M1 À M2 , the frequency is
determined by the lager one, say M1 .
(c). When M1 = M2 = M , (6) gives:
ω2 =
√
C
(2 ± 2 + 2 cos ka)
M
2
(12)
Fall 2006
c Xun Jia (November 5, 2006)
°
Physics 140A
Optical branch
Acoustic branch
(M/C)
1/2
2
1
0
-6
-4
-2
0
2
4
6
ka=2ka'
Figure 2: The dispersion relation for the diatomic linear chain when
M1 = M2 = M . Shaded area gives the Brillouin zone for the
diatomic chain.
this dispersion relation is shown as in Fig. 2, where for the first Brillouin zone
for the diatomic chain is ka ∈ [−π, π]. In terms of the new lattice constant
a0 = a/2, the dispersion relation is then:
ω2 =
2C
(1 ± cos ka0 ),
M
π π
ka0 ∈ [− , ]
2 2
(13)
For the optical branch, consider the part with ka0 ∈ [− π2 , 0], then:
ω2 =
2C
2C
(1 + cos ka0 ) =
[1 − cos(ka0 + π)]
M
M
(14)
notice that ka0 +π ∈ [ π2 , π], above expression indicates that the optical branch in
dispersion relation for the diatomic chain in [− π2 , 0] is equivalent to the acoustic
branch in [ π2 , π]. Similarly, the optical branch in dispersion relation for the
diatomic chain in [0, π2 ] is equivalent to the acoustic branch in [−π, − π2 ]. Thus,
instead of describing the dispersion relation with both optical and acoustic
branch in ka0 ∈ [− π2 , π2 ], it is equivalent to just consider the acoustic branch
in ka0 ∈ [−π, π], which is just the dispersion relation for the monatomic chain.
Therefore, when setting M1 = M2 = M , the dispersion relation for monatomic
chain is recovered.
3
Fall 2006
c Xun Jia (November 5, 2006)
°
Physics 140A
Problem #2
Problem 4.3 in Kittel. For the problem treated by (18) to (26), find the amplitude
ratio u/v for the two branches at Kmax = 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. (21) in Kittel, let K = Kmax = π/a, we get:
µ
¶
2C − M1 ω 2
0
det
=0
o
2C − M2 ω 2
(15)
The solution to this equation gives the normal modes frequencies at this K value:
ω 2 = 2C/M1 ;
ω 2 = 2C/M2
(16)
Substitute this expression of ω 2 in to (20) in Kittel, we found:
• When ω 2 = 2C/M1 , it follows that:
u
→∞
v
(17)
which gives a mode that u is finite, and v is zero.
• When ω 2 = 2C/M2 , it follows that:
u
=0
v
(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 C1 and C2 , then set C1 = C, and C2 = 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/2. 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 H2 .
Solution:
Now as in Fig. 3, let the coupling constants be in general C1 , and C2 , then the
equations of motions are:
M ün,1 = C1 (un,2 − un,1 ) − C2 (un,1 − un−1,2 )
= C1 un,2 + C2 un−1,2 − (C1 + C2 )un,1
M ün,2 = C2 (un+1,1 − un,2 ) − C1 (un,2 − un,1 )
= C2 un+1,1 + C1 un,1 − (C1 + C2 )un,2
4
(19)
Fall 2006
c Xun Jia (November 5, 2006)
°
Physics 140A
a
C1
C1
C2
un-1,1 un-1,2
un,1
C2
un,2
C1
C2 M
un+1,1 un+1,2
Figure 3: The diatomic linear chain with alternating coupling constants.
For the normal modes, they are in the form:
un,1 = A1 ei(kna−ωt)
un,2 = A2 ei(kna−ωt)
(20)
substitute (20) in to (19), we have:
A1 [ω 2 M − (C1 + C2 )] + A2 (C1 + C2 e−ika ) = 0
A1 (C1 + C2 eika ) + A2 [ω 2 M − (C1 + C2 )] = 0
to have non trivial solutions, it requires:
µ 2
¶
ω M − (C1 + C2 )
C1 + C2 e−ika
det
=0
C1 + C2 e−ika
ω 2 M − (C1 + C2 )
this gives the dispersion relations of:
¶
µ
q
1
2
2
2
ω =
C1 + C2 ± C1 + C2 + 2C1 C2 cos ka
M
set C1 = C, and C2 = 10C, we found the dispersion relation is:
´
√
C ³
ω2 =
11 ± 101 + 20 cos ka
M
and at k = 0, and k = π/a, the normal mode frequencies are:
• at k = 0, we have:
r
ω=
ω=0
• at k = π/a, we have:
22C
M
: for optical branch
(21)
(22)
(23)
(24)
(25)
: for acoustic branch
r
20C
r M
2C
ω=
M
ω=
: for optical branch
: for acoustic branch
The dispersion relation is drawn in Fig. 4.
5
(26)
Fall 2006
c Xun Jia (November 5, 2006)
°
Physics 140A
4
Optical branch
(M/C)
1/2
Acoustic branch
2
0
-2
0
2
ka
Figure 4: The dispersion relation for diatomic linear chain with alternating coupling constants C1 = C, and C2 = 10C.
Problem #4
Problem 22.3 in Ashcroft and Mermin: Lattice with a Basis Viewed as a Weakly
Perturbed Monatomic 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 C1 and C2 become
very close: C1 = C + ∆, C2 = 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 C1 = C2 it will reduce to a
monatomic chain with lattice constant a/2; furthermore, the Brillouin zone
(−π/a < k < π/a) for the diatomic chain will only be half the Brillouin zone
(−π/(a/2) < k < π/(a/2)) 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 ∆ 6= 0, but ∆ ¿ C, then the dispersion relation differs from
that of the monatomic chain only by terms of order (∆/C)2 , 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
Fall 2006
c Xun Jia (November 5, 2006)
°
Physics 140A
(a). Same as in last problem in a linear chain with alternating coupling constants,
set C1 = C + ∆, C2 = C − ∆, we get the dispersion relation of:
´
p
1 ³
2
2
2
2C ± 2C (1 + cos ka) + 2∆ (1 − cos ka)
ω =
(27)
M
at the case ∆ = 0, above equation reduces to:

s

1 + cos ka
2C 

ω2 =
1±
M
2
(28)
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 (27) in terms of ∆/C
as:

s

2
∆
C 
2 ± 2(1 + cos ka) + 2 2 (1 − cos ka)
ω(k)2 =
M
C


s
2
p
∆
ka
C 
=
2 ± 2(1 + cos ka) 1 + 2 tan2 
M
C
2
(29)
"
!#
Ã
p
ka
∆4
∆2
C
2
=
2 ± 2(1 + cos ka) 1 +
tan
+ o( 4 )
M
2C 2
2
C
= ω0 (k)2 ±
∆2
∆4
f
(k)
+
o(
)
C2
C4
where ω0 (k) is the dispersion relation for the case ∆ = 0, i.e. the monatomic
chain, and f (k) is defined to be:
s
ka
C 1 + cos ka
f (k) =
tan2
(30)
M
2
2
Now, take the square root of (29) and Taylor expand again, we get:
s
∆2
∆4
ω(k) = ω0 (k)2 ± 2 f (k) + o( 4 )
C
C
2
∆4
∆ f (k)
+ o( 4 )
= ω0 (k) ±
2C 2 ω0 (k)
C
(31)
that is, in general, the dispersion relation ω(k) differs from that of the monatomic
chain ω0 (k) only by terms of order (∆/C)2 .
7
Fall 2006
c Xun Jia (November 5, 2006)
°
Physics 140A
However, to ensure the Taylor expansion in (29) valid, from the first line of that
equation, it requires that:
|1 + cos ka| À |
∆2
(1 − cos ka)|
C2
(32)
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 ∆2 (1 − cos ka)/C 2 , and this indeed
happens at when |π − ka| is of order ∆/C, since then
1
∆2
1 + cos ka = 1 + cos π + (π − ka)2 + . . . ∼ A 2
2
C
2
2
∆
∆
(1 − cos ka) ∼ 2 2
2
C
C
where A is a constant. Now from (29), we have:
s
2C
A ∆2 ∆2
(1 ±
+
)
ω(k)2 ≈
M
2 C2 C2
r
2C 2C ∆
A
±
1+
≈
M
MC
2
(33)
(34)
and then:
r
∆
A
∆2
ω(k) ≈ ω0 (k) ± ω0 (k) 1 + + o( 2 )
(35)
C
2
C
p
where ω0 (k) ≈ ω0 (k = π/a) = 2C/M , 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 M 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 3/4πR3 , which defines R. (a) Show that the
frequency of a single ion set into oscillation is ω = (e2 /M R3 )1/2 . (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
Fall 2006
c Xun Jia (November 5, 2006)
°
Physics 140A
(a). Since the number density of ions is 3/4πR3 , the charge density will be 3e/4πR3 ,
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 4πr3
er3
=
4πR3 3
R3
(36)
thus the equation of motion (in Gauss unit) is
e2 r
qe
M r̈ = − 2 = − 3
r
R
which gives an oscillation frequency of:
r
ω=
e2
M R3
(37)
(38)
(b). To estimate the value of this frequency for sodium, which is a bcc lattice, we
find these values:
s
s
3
3
3
3
R=
a=
429.06pm = 2.1126 × 10−10 m
8π
8π
(39)
M = 22.99 × 1.660 × 10−27 kg = 3.818 × 10−26 kg
e = 11 × 1.602 × 10−19 C = 1.762 × 10−18 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:
s
e2
= 2.8 × 1014 rad/s
(40)
ω=
4πε0 M R3
(c). If we estimate the wave length to be of order of lattice constant, say λ ∼ 1010 m,
then
ω
ωλ
v= =
∼ 104 m/s
(41)
k
2π
9