1 Introduction 2 Resolvents and Green's Functions

4. If H is self-adjoint, then G(x, y; z) is, in fact, our earlier discussed
Green’s function. It is given by Eqn. 59 for all z (including real z),
not in the spectrum of H. The discrete eigenvalues of H correspond to
poles of G as a function of z. Thus, the bound states can be found by
searching for the poles of G, in a suitably cut complex plane (if H has
a continuous spectrum, we have a branch cut).
5.2 Example: Force-free Motion
Let us evaluate the Green’s function for force-free motion, V (x) = 0, in
x ∈ (−∞, ∞) configuration space:
d 2
Hu(x; z) = − 1 u(x; z) = zu(x; z). (64)
2m dx2 We must have u L → 0 as x → −∞, and u R → 0 as x → ∞. Then we have
solutions of the form:
where
u L (x; z) = e −iρx , u R (x; z) = e iρx , (65)
ρ = √ 2mz. (66)
We select the branch of the square root so that the imaginary part of ρ is
positive, as long as z is not along the non-negative real axis. We know that
the spectrum of H is the non-negative real axis. Thus, we cut the z-plane
along the positive real axis.
To obtain the Wronskian, note that u ′ L = −iρu L and u ′ R = iρu R . Evaluate
at x = 0 for convenience: u L (0) = u R (0) = 1. Hence,
W (z) = (−iρ) − (iρ) = −2iρ = −2i √ 2mz. (67)
Thus, we obtain the Green’s function:
G(x, y; z) =
2m
−2i √ 2mz
[
e −iρx e iρy θ(y − x) + e −iρy e iρx θ(x − y) ]
= i√ m
2z eiρ|x−y| . (68)
We could have noticed from the start that G had to be a function of (x − y)
only, by the translational invariance of the problem.
Let us continue, and obtain the time development transformation U(x, y; t):
U(x, y; t) = − 1 ∫
dze −itz G(x, y; z), (69)
2πi C ∞
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