Abstract

CHAPTER 1. OVERVIEW 14
But because limy→∞ ρ(x + 1 1 y, x − 2
2y) = 0 and limy→−∞ ρ(x + 1
2
y, x − 1
2 y)=0since
for any wavefunction ψ(q) wehavethatlimq→∞ ψ(q) = limq→−∞ ψ(q) = 0, we also
know that
∞
−∞
dy ∂
∂y
Therefore, we can conclude that
∞
−∞
dy[ ∂
∂y
Placing this into A(ρ), we have
[ρ(x + 1
2
ρ(x + 1
2
1 −ipy
y, x − y)e ]=0
2
1 −ipy
y, x − y)]e
2
= ip
f(x, p)
A(ρ) =−
im∗ ∂
∂x [ip
p
f(q, p)] = −
m∗ ∂f(x, p)
∂x
Now we want to analyze B(ρ). First, by Fourier transforms, we have that
So we have that
B(ρ) = 1
∞
i
−∞
∞
= 1
i
= 1
2iπ2 ρ(x + 1
2
dy[U(x + 1
2
−∞
∞
dp
−∞
′ f(x, p ′ ∞
)
−∞
∞
1 1
y, x − y)= dpf(x, p)e
2 2π −∞
ipy
1 1 1 −ipy
y) − U(x − y)]ρ(x + y, x − y)e
2 2 2
dye −ipy
[U(x + 1
∞
1 1
y) − U(x − y)](
2 2 2π −∞
p−p′
−iy
dye [U(x + 1
1
y) − U(x −
2 2 y)]
dp ′ f(x, p ′ )e ip′ y
)