1 Introduction 2 Resolvents and Green's Functions

We still have u R (x; z) = e iρx , but now the left boundary condition is u L (0; z) =
0. Hence, a left solution is
We obtain W = ρ, by evaluating at x = 0. Thus,
u L (x; z) = sin(ρx). (83)
G(x, y; z) = 2m [
sin(ρx)e iρy θ(y − x) + sin(ρy)e iρx θ(x − y) ]
ρ
= m [(
e iρ(x+y) − e iρ(y−x)) θ(y − x) + ( e iρ(x+y) − e iρ(x−y)) θ(x − y) ]
iρ
√ m [
= i e iρ|x−y| − e iρ(x+y)] . (84)
2z
This Green’s function is not translation invariant. However, if x → ∞,
y → ∞ such that x − y is finite, then this Green’s function tends toward
our first example (Imρ > 0 is still our branch). This is compatible with the
intuition that the local physics far from the wall at x = 0 should be nearly
independent of the existence of the wall.
5.4 Example: Force-free Motion in Three Dimensions
Consider the Green’s function problem for force-free motion in three dimensions:
H = − 1
2m ∇2 , (85)
where x ∈ R 3 . The resolvent is most easily found in momentum space, since
H = p 2 /2m is just multiplication by a factor there. Hence, the resolvent in
momentum space is:
1
G(z) =
p 2
− z . (86)
2m
The Green’s function in momentum space is, formally:
G(x, y; z) = ∑ k
φ k (x)φ ∗ k(y)
, (87)
ω k − z
where
ω k = p2
2m , (88)
1
φ k (x) = eip·x . (89)
(2π) 3/2
15