# Solution manual - Chapter 22-23

```Solution manual - Chapter 22-23
Jens Zamanian
May 7, 2014
Chapter 22
Problem 22.2 - Diatomic Linear Chain
(a) To obtain the dispersion relation we must generalize the discussion of a lattice with a basis
to the case where the di↵erent atoms in the basis have di↵erent mass. We place atoms of mass
M1 at positions na and atoms with mass M2 at positions na + d = na + a/2, so that the atoms
of mass M2 have their equilibrium positions right in the middle of two M1 atoms, see Fig. 1. In
the harmonic approximation we then have the potential energy
U=
KX
[u1 (na)
2 n
2
u2 (na)] +
KX
[u2 (na)
2 n
2
u1 ((n + 1)a)] ,
(1)
where u1 (na) is the deviation from the equilibrium position of the atom of mass M1 with equilibrium position at point na and u2 (na) is the deviation for an atom of mass M2 whose equilibrium
position is at na + a/2. Here we have assumed that only nearest neighbors interact. In the book
Figure 1: Chain with atoms of mass M1 (black dots) and atoms with mass M2 (red dots)
1
the more or less just write up the evolution equations of the atoms, but here I will show you how
this is done using what you know from analytic mechanics. To do this we must find the lagrangian
L = T U , where T is the kinetic energy and U is the potential energy (given above). The kinetic
energy of the atoms are given by
Mi u̇2i
Ti =
(2)
2
and the lagrangian is hence
M1 u̇21
M2 u̇22
L=
+
U.
(3)
2
2
The Euler-Lagrange equations then states that
✓
◆
@L
d
@L
= 0.
(4)
@ui (na) dt @ u̇i (na)
Note that there are two equations since i = 1, 2. We get
K [u1 (na)
u2 (na)]
K [u2 ((n
1)a)
u1 (na)]
M1 ü1 (na) = 0.
(5)
Note that when taking the derivative of the lagrangian with respect to u1 (na) is must find all
terms of the sum where u1 (na) exists. In the first term of (1) this is no problem, but for the
second term we note that u1 (na) occurs in the sum together with u2 ((n 1)a). Also, the sign of
the second term above comes from the inner derivative. Doing the same for u2 we get
M1 ü1 (na)
=
K [2u1 (na)
u2 (na)
u2 ((n
1)a)]
M2 ü2 (na)
=
K [2u2 (na)
u1 (na)
u1 ((n + 1)a)] .
(6)
(7)
Note that these equations are exactly symmetric as is expected. You can compare this with Eq.
(22.33) in the book to check that we got all the signs right. We now make the ansatz
u1 (na)
u2 (na)
=
=
✏1 ei(kna
✏2 e
!t)
i(kna !t)
(8)
.
(9)
Inserting this into the evolution equations above we get
h
! 2 M1 ✏1 ei(kna !t) =
K 2✏1 ei(kna !t) ✏2 ei(kna
h
! 2 M2 ✏2 ei(kna !t) =
K 2✏2 ei(kna !t) ✏1 ei(kna
Dividing by eikna
i!t
!t)
✏2 ei(k(n
!t)
1)a !t)
✏1 ei(k(n+1)a
we get
⇥
⇤
! 2 M1 ✏1 + K 2✏1 ✏2 ✏2 e ika
⇥
⇤
! 2 M2 ✏2 + K 2✏2 ✏1 ✏1 eika
We may write this equation in matrix form as

! 2 M1 + 2K
K 1 + e ika
ika
K 1+e
! 2 M2 + 2K

!t)
i
i
(10)
.
(11)
=
0
(12)
=
0.
(13)
✏1
✏2
=

0
0
.
(14)
Solutions can be found if the determinant of the equation on the left hand side is zero. From this
we get the dispersion relation
! 2 M1 + 2K
! 2 M2 + 2K
K2 1 + e
ika
1 + eika = 0.
(15)
Simplifying this we get
! 4 M1 M 2
! 2 2K (M1 + M2 ) + 2K 2 [1
2
cos (ka)] = 0.
(16)
This is second degree equation in ! 2 and we may solve it to get

q
K
2
! =
M1 + M2 &plusmn; M12 + M22 + 2M1 M2 cos(ka) .
M1 M2
(b) In the case where M1
!2
=
=
=
(17)
M2 we have M2 /M1 ⌧ 1. Expanding the expression above we have

q
K
M1 + M2 &plusmn; M12 + M22 + 2M1 M2 cos(ka)
M1 M 2
s
&quot;
#
K
M2
M22
M2
1+
&plusmn; 1+ 2 +2
cos(ka)
M2
M1
M1
M2
&quot;
✓
◆
✓
◆2 !#
K
M2
M2
M2
1+
&plusmn; 1+
cos(ka) + O
.
(18)
M2
M1
M2
M1
The solutions are then (to lowest order in M2 /M1 )
8
q
2K
&gt;
&gt;
&lt;
M2
!=
q
q
&gt;
&gt;
K
2K
:
cos(ka)] = M
sin
M1 [1
1
(19)
ka
2
where we in the second row used that cos(2x) = 1 2 sin2 (x). Using this in Eq. (14) we get for
the first solution
✏1
1 + e ika
=
! 0.
(20)
✏1
2 (1 M1 /M2 )
This means that the first of the solutions is just oscillations of the atoms of mass M2 while the
atoms of mass M1 are still.
The second solution yields
✏1
2
e ika/2
=
=
.
(21)
✏2
1 + eika
cos(ka/2)
This gives the solutions
u1 (na)
u2 (na)
=
=
✏1 ei(kna
!t)
✏1 cos(ka/2)e
(22)
i(k(n+1/2)a !t)
.
(23)
Remembering that the equilibrium positions of the atoms with mass M2 is at na + a/2 we see
that this solution corresponds to where the atoms oscillate in phase. Note that the oscillations of
the M2 atoms are smaller in amplitude. Perhaps someone can figure out why they vanish at the
boundary of the Brillouin zone (ka = ⇡)? I cannot!
(c) If M 1 = M2 = M we get
i
i
p
p
K h
K h
! 2 ⇡ 2 2M &plusmn; 2M 2 + 2M 2 cos(ka) =
2 &plusmn; 2 + 2 cos(ka) .
(24)
M
M
Now we have that k should be in the first Brillouin zone which means that ka should have values
between ⇡ and ⇡. In order to compare this with the single atom chain we must first recognize
that we here have the distance d = a/2 between the atoms instead of a. Introducing a new lattice
constant a0 = a/2 and ka0 will then have values between ⇡/2 and ⇡/2 we get the dispersion
relation
i
i 2K
p
p
K h
K h
!2 =
2 &plusmn; 2 + 2 cos(2ka0 ) =
2 &plusmn; 2 + 4 cos2 (ka0 ) 2 =
[1 &plusmn; cos(ka0 )] . (25)
M
M
M
In Fig. 2 we have plotted the two branches of this dispersion relation. We have here extended
the plot range to ka0 2 [ ⇡, ⇡]. Note that the acoustic branch in [⇡/2, ⇡] is equivalent to the
optical branch in [ ⇡/2, ⇡]. The figure illustrates that instead of consider two separate branches
on ka0 2 [ ⇡/2, ⇡/2] we may consider just the acoustic branch on ka0 2 [ ⇡, ⇡]. In Fig. 3 we plot
the result for a chain of single atom types (Eq. (22.29) in the book). We see that the dispersion
relations are indeed the same, which is what should be expected.
3
2.0
K
1.0
w
M
1.5
0.5
0.0
-3
-2
-1
0
1
2
3
ka
Figure 2: Dispersion relation for when the masses are equal with the acoustic branch (purple line)
and the optical branch (orange dashes). The plot range has been extended to go beyond the first
Brillouin zone.
22.3 - Lattice with a Basis Viewed as a Weakly Perturbed Monoatomic
Bravais Lattice
Only problem (a).
Using
K = K0 +
G = K0
(26)
in Eq. (22.37) of the book becomes
!2 =
In the limit
2K0
1 p
&plusmn;
(K0 +
M
M
! 0 it becomes
)2 + (K0
)2 + 2(K0 +
)(K0
) cos(ka).
(27)
2K0
K0 p
&plusmn;
2 + 2 cos(ka).
M
M
(28)
2K0
K0 p
2K0
&plusmn;
2 + 2 cos(2ka0 ) =
[1 &plusmn; cos(ka0 )] .
M
M
M
(29)
!2 =
In this limit the distance between the atoms is a/2. Note that the reason why the values of K and
G di↵ered is that the atoms were situated at di↵erent distance from each other in the general case.
Introducing a new lattice constant a0 = a/2 we have
!2 =
Similar arguments as in the previous problem will now show that this is indeed equivalent to
the monoatomic chain with lattice constant a0 . We may again refer to the plots of the previous
problem.
We see that if we only consider the acoustic branch (the minus sign in the dispersion relation)
and extend the range of ka0 to be [ ⇡, ⇡] we include all the values of ! 2 .
Next we consider the ratio between the two amplitudes (note that we only need to consider
the sign which corresponds to the acoustic branch as we have instead extended the range of ka0
to ensure that we cover all frequencies).
✏2
✏1
0
=
0
=
0
1 + eika
1 + e2ika
1 + e2ika
p
+
=
=
0
|1 + eika |
|1 + e2ika |
2 + 2 cos(2ka0 )
0
0
ika
0
1 + e2ika
+ eika
ika0 e
=
e
= eika .
2 cos(ka0 )
2 cos(ka0 )
4
(30)
2.0
1.0
w
K
M
1.5
0.5
0.0
-3
-2
-1
0
1
2
3
ka
Figure 3: The dispersion relation for a monoatomic chain.
Returning again to the solutions we see that we then have
u1 (na)
u2 (na)
=
=
✏1 ei(2kna
✏1 e
ika
0
e
0
!t)
(31)
i(2kna0 !t)
= ✏1 e
i(k(2n+1)a0 !t)
.
(32)
0
Comparing this with the monoatomic chain u = eikna i!t we see that they are intact the same
(just divided into atoms sitting and even and odd positions).
Vibrational modes in a simple linear chain.
The sound velocity is obtained by considering the part of the spectra where the frequency depends
linearly on the wave number. In the book they derive the expression
r
K
!=a
|k|.
(33)
M
The sound velocity is then equal to the group and the phase velocities (which are both equal). It
is given by
r
!
K
vsound =
=a
.
(34)
|k|
M
For the numerical values given we get
vsound ⇡ 3.1 km/s.
(35)
The full dispersion relation is given by
!2 =
2K
(1
M
cos(ka)).
(36)
We can take the derivative of this to find the maximum, but we already know that the max occurs
at the boundaries of the first Brillouin zone and is given by (see Fig. (22.8 on page 433 in the
book)
r
2K
!max =
⇡ 6.2 ⇥ 1012 Hz = 6.2 THz
(37)
M
5
Heating diamond
The heat needed to heat a substance of mass from temperature Ti to Tf is given by
Q=V
Z
Tf
dT cv (T ).
(38)
Ti
(Note that sometimes specific heat is defined as the energy per unit mass and Kelvin, but in
this case it is per unit volume and Kelvin. Dimensional analysis will tell you which it is). The
3
Debye temperature for diamond is ⇥D = 1860 K and the density 3.5 g/cm . Due to this high
temperature we can use the low-temperature limit of the Debye approximation. In this limit we
may completely forget about the optical modes (which do exist because diamond is a lattice with
a basis). This is because only the low frequency modes will contribute significantly to the heat
capacity. The specific heat in the Debye approximation is given by
◆3
nkB .
(39)
V nkB
Tf4
4⇥3D
Ti4
(40)
cV = 234
✓
T
⇥D
Putting this into Eq. (38) we get
Q = 234
The volume is just given by V = 1/3.5 cm3 . We need to know the number density of diamond
which is 176.2 ⇥ 1021 cm 3 (this can easily be calculated by using the molar mass of diamond or
the atomic mass of carbon). Inserting all the numbers we get
Q ⇡ 15 J.
For an ideal gas we have
cv =
(41)
3nkB
.
2
(42)
The heat required is then
Q = V cv T ⇡ 10 J.
(43)
For the ideal gas, all energy added will give rise to increase kinetic energy. For diamond, the
additional heat needed compensates for the extra potential energy needed between the atoms to
move around in the potential from all the other atoms.
Specific heat of a 1D crystal
Obtain the (a) high and (b) low temperature expressions for cv for a one dimensional
crystal with
p
dispersion relation !k = !0 | sin(ka/2)|. Here a is the lattice constant and !0 = 4K/M in which
K is the spring constant and M the particles’ mass. Start from
cv =
@
@T
Z
⇡/a
⇡/a
dk
~!k
2⇡ e~!k /(kB T )
1
(44)
(which is the on dimensional version of Eq. (23.15) in AM). In what way are the results di↵erent
from the three dimensional case? (c) Derive also the one dimensional version of Debye’s interpolation formula.
(a) The high temperature limit To find the high temperature limit we first evaluate the
derivative
✓
◆2
Z
kB ⇡/a
~!k
e~!k /kB T
cv =
dk
.
(45)
2⇡
kB T
(e~!k /kB T 1)2
⇡/a
6
Now we take the high temperature limit ~!0 /kB T ! 0 (I used !0 as this is the maximum value
of the frequency). Limit can be evaluated using l’Hospital’s theorem
lim
x2
x &gt;0
(ex
= lim
1)2
x
x &gt;0
ex
1
= lim
x &gt;0
1
= 1.
ex
(46)
The integral is the simply
kB 2⇡
kB
=
= nkB
(47)
2⇡ a
a
(since a is the distance between the particles, n = 1/a is the number of particles per unit length).
cv =
(b) Low temperature limit To treat the low temperature limit we first note that the integrand
is an even function and we may change the integration into
✓
◆2
Z ⇡/a
Z
@
dk
~!k
kB ⇡/a
~!k
e~!k /kB T
cv =
=
dk
(48)
@T 0
⇡ e~!k /kB T 1
⇡ 0
kB T
(e~!k /kB T 1)2
Now, with this new range we may remove the absolute value in the dispersion relation leaving
!k = !0 sin(ka/2) Next we make the change of variable in the integral
x
=
dx
=
~!k
kB T
✓ ◆
~ d!k
a~!0
ka
dk =
cos
dk.
kB T dk
2kB T
2
We now use that
cos(x) =
so we get
a~!0
dx =
2kB T
s
1
sin
2
✓
q
p
cos2 (x) = 1
(49)
(50)
sin2 (x)
s✓
◆
◆2
ka
a
~!0
dk =
2
2
kB T
✓
~!0
kB T
(51)
◆2
sin2
✓
◆
ka
dk
2
(52)
If we now define x0 = ~!0 /kB T we finally get
dk =
2
dx
p
a x20 x2
Using this variable change in the integral we get
Z
Z x0
2kB x0
ex
x2
p
cv =
dx x
⌘
dxf (x, x0 ).
a⇡ 0
(e
1)2 x20 x2
0
(53)
(54)
where we need to remember that x0 = x0 (T ) when taking the derivative. This integral is very
similar to an integral that occurs when dealing with Bose-Einstein condensation. The key point
to realise here is that due to the Bose-Einstein distribution 1/(ex 1) we will have a large fraction
of the modes in the lowest oscillation mode ! ⇡ 0 (note that the denominator diverges for this
value). The usual way to treat such a problem is to separate out the part of the integral close to
! = 0. We introduce an arbitrary value 0 &lt; q &lt; 1 and divide the integral into two parts
Z qx0
Z x0
cv =
dxf (x, x0 ) +
dxf (x, x0 )
(55)
0
qx0
In the limit T ! 0 we must take x0 ! 1. The second integral will vanish in this limit since the
integration range remains
finite and the integrand goes to zero. In the first integral, on the other
p
hand, the term 1/ x20 x2 ! 1/x0 (since the x part can be neglected)
Z 1
2kB
x2 ex
cv =
dx x
.
(56)
a⇡x0 0
(e
1)2
7
Now this integral can be solved. I used Wolfram Alpha, but you can perhaps find the correct
analytic way to do this. The result is ⇡ 2 /3 which gives
cv =
2
2
2⇡kB
2⇡kB
T
⇡kB
T
=
=
,
3ax0
3a~!0
3~c
(57)
where c = a!0 /2.
(c) Debye approximation In the Debye approximation we approximate the dispersion relation
with a linear dispersion relation
!k = !0 sin
✓
ka
2
◆
= !0
ka
+O
2
✓
ka
2
◆3
⇡ !0
ka
.
2
(58)
In the 1D case the integration limit is simply a line and we do not make any approximations for this
(in the 3D case the first Brillouin zone might be quite complicated and one usually approximates
it as a sphere, in the 1D case we already have a ”1D sphere”). We then have (after evaluation of
the derivative)
✓
◆2
Z
kB ⇡/a
~!k
e~!k /kB T
cv =
dk
.
(59)
⇡ 0
kB T
(e~!k /kB T 1)2
Making now the variable change
we get
x
=
dx
=
2
2kB
T
cv =
a⇡~!0
~!k
a~!0
=
k
kB T
2kB T
a~!0
dk
2kB T
Z
x0
dx
0
x2 ex
.
(ex 1)2
(60)
(61)
(62)
where x0 = ⇡~!0 /(2kB T ). If we plot the integrand we get and wee see that for sufficiently large
x0 we might just as well extend the limit of integration to infinity as the error in doing so is
exponentially small. Doing this and solving the integral (once again yielding ⇡ 2 /3 we finally get
the same result as in (b).
8
```