Abstract

CHAPTER 1. OVERVIEW 15
Now we will look at 0
p−p′
dye−iy
−∞
coordinates s = −y, wehavethat
0
−∞
[U(x + 1
p−p′
−iy
dye [U(x + 1
1
y) − U(x − y)] =
2 2
So we can conclude that
0
−∞
p−p′
−ir
dye [U(x + 1
1
y) − U(x − y)] = −
2 2
2
1
y) − U(x − y)]. If we use the change of
2
∞
0
∞
0
p−p′
is
dse [U(x − 1
1
s) − U(x +
2 2 s)]
p−p′
ir
dye [U(x + 1
1
y) − U(x −
2 2 y)]
Therefore, for the inner integral in B(ρ), we get that ∞
p−p′
dye−iy [U(x + −∞ 1y)
− 2
U(x − 1
2y)] equals ∞
∞
−∞
p−p′
−iy
dye
Finally, we have
0
p−p′
p−p′
iy dy[e−iy − e
[U(x + 1
1
y) − U(x − y)] =
2 2
B(ρ) = 1
2iπ2 ∞
−∞
= − 1
π2 ∞
dp ′ f(x, p ′ ∞
)
0
dp
−∞
′ f(x, p ′ ∞
)
0
][U(x + 1
∞
0
dy[−2i sin(y
2
dy[2i sin(y
1
y) − U(x − y)]. So
2
p − p′
p − p′
)][U(x + 1
1
y) − U(x −
2 2 y)]
1
1
)][U(x + y) − U(x −
2 2 y)]
dy[U(x + 1
1 p − p′
y) − U(x − y)][sin(y
2 2 )]
Recall p = k, wherek is the wave number. So if we change from the momentum
variable p in these equations to the wave number variable k and use the fact that