1 Introduction 2 Resolvents and Green's Functions

The integral is now in the form of the integral of a Gaussian, and has the
value √ 2πσ. Therefore,
U(x, y; t − iτ) = 1 √ 1 2π √
2π
1
(τ + it)ea2
2m
√
[
]
m
=
2π(τ + it) exp m(x − y)2
− , (75)
2(τ + it)
with Re √ τ + it > 0.
It is interesting to look at this result for t = 0:
√ [
]
m
U(x, y, −iτ) =
2πτ exp m(x − y)2
− . (76)
2τ
Referring back to our earlier discussion, we see that this gives the solution
to the diffusion equation, or, for example, to the heat conduction problem
(for a homogeneous medium) with an initial heat distribution proportional to
δ(x − y). As time τ increases, the heat propagates out from x = y, spreading
according to a broadening Gaussian.
In quantum mechanics, we are more interested in the limit τ → 0 + . The
only subtle issue is the phase in √ τ + it. Let
√
τ + it =
√
Re
iθ
= √ Re iθ/2 , (77)
→
where R = √ τ 2 + t 2 |t|. Referring to Fig. 4, for t > 0 we have 0 <
τ→0 +
θ < π/2, approaching θ = π/2 as τ → 0 + . Similarly, for t < 0 we have
−π/2 < θ < 0, approaching θ = −π/2 as τ → 0 + . Hence,
⎧
√ √ ⎨ e iπ/4 = √ 1
τ + it →τ→0 + |t|
2
(1 + i), t > 0,
⎩ e −iπ/4 = √ 1
(78)
2
(1 − i), t < 0.
Thus,
U(x, y; t) = 1 2
(
1 − i t ) √ [ ]
m im(x − y)
2
|t| 2π|t| exp . (79)
2t
This is the time development transformation for the free particle Schrödinger
equation in one dimension, where we have kept proper track of the phase for
all times.
We may check that the behavior of this transformation is as expected
when we transform by time t, followed by transforming by time −t. The
result ought to be what we started with, i.e., this product should be the
identity. Thus, we consider the product
U(y 2 , x; −t)U(x, y 1 ; t) =
m
2π|t| exp { im
2t
[
(x − y1 ) 2 − (x − y 2 ) 2]} . (80)
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