Abstract

CHAPTER 4. THEORY 80
Combining Equation 1.16 for U(x) and Equation 1.9 for T (x, z), we get that the
explicit dependence of T (x, k − k ′ )onu is given by
T (x, k − k ′ )= q2
Lc
2
2πɛ 0
L Kmax
0
[J(x − y, z) − J(x + y, z)] sin(2y(k − k
−Kmax
′ ))u(z,k ′′ )dk ′′ dzdy
Lc L 2
+ q2
ɛ
0
+
0
Lc
2
0
[J(x + y, z) − J(x − y, z)] sin(2y(k − k ′ ))Nd(z)dzdy
dy sin(2y(k − k ′
)) ∆c(x + y) − ∆c(x − y) − 2Vy
.
L
We note that T (x, k − k ′ ) is a continuous function. To get an upper bound on
||T ||∞, we will look at each term of T (x, k − k ′ ) individually. Define T1(x, k − k ′ )as
T1(x, k−k ′ )= q2
Lc
2
2πɛ 0
So,
|T1(x, k − k)| = q2
2πɛ
≤ q2
2πɛ
L Kmax
Lc
2
0
Lc
2
0
0
≤ q2
2πɛ
−Kmax
[J(x−y, z)−J(x+y, z)] sin(2y(k−k ′ ))u(z,k ′′ )dk ′′ dzdy.
L Kmax
0 −Kmax [J(x − y, z) − J(x + y, z)] sin(2y(k − k′ ))u(z,k ′′ )dk ′′ dzdy
L Kmax
0 −Kmax |[J(x − y, z) − J(x + y, z)] sin(2y(k − k′ ))u(z,k ′′ )|dk ′′ dzdy
Lc
2
0
≤ q2
2πɛ
Lc
2
0
L Kmax
0 −Kmax 2||J||∞|u(z,k ′′ )|dk ′′ dzdy
L Kmax
0 −Kmax 4||J||2∞dk ′′ dz
1
2 L
0
= q2Lc 4πɛ 2J∞
√
2KmaxLu2.
Kmax
−Kmax u2 (z,k ′′ )dk ′′ 1
2
dz