This preview shows pages 1–2. Sign up to view the full content.
Solutions to problem set 4 (141A Sp07)
1
1. From ISSP page 91:

m
n
ω
2
u
n
=
C
(
u
n
+
1

2
u
n
+
u
n

1
)
For
n
>
0 we have

m
ω
2
e

kn
=
C
(
e

k
(
n
+
1)

2
e

kn
+
e

k
(
n

1)
)
,
ie.

m
ω
2
=
C
(
e

k

2
+
e
k
)
.
For
n
=
0, we have

M
ω
2
=
C
(
e

k
(1)

2
+
e

k

1

)
=
C
(2
e

k
(1)

2)
=
2
C
(
e

k

1)
Therefore
2
C
M
(
e

k

1)
=
C
m
(
e

k

2
+
e
k
)
2
C
M
(1

e
k
)
=
C
m
(1

2
e
k
+
e
2
k
)
Let
x
=
e
k
; then
1

2
x
+
x
2

(1

x
)
2
m
M
=
0
x
2

2
±
1

m
M
²
x
+
±
1

2
m
M
²
=
0
So
x
=
±
1

m
M
²
±
r
±
1

m
M
²
2

±
1

2
m
M
²
=
±
1

m
M
²
±
m
M
.
There is a trivial solution with
x
=
1, ie
k
=
0 and
ω
=
0 (this corresponds to a bulk motion of
the lattice.) The nontrivial solution has
x
=
1

2
m
/
M
, ie
k
=
ln(1

2
m
/
M
)
.
Remember that the displacement function looks like
u
n
=
e

kn
for
n
>
0. For
M
>
2
m
,
k
will
be negative and so the displacement will blow up. Therefore there is no localized mode for
M
>
2
m
.
For
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 08/01/2008 for the course PHYSICS 141A taught by Professor Souza during the Spring '08 term at University of California, Berkeley.
 Spring '08
 SOUZA
 Solid State Physics

Click to edit the document details