Image Processing Reference
In-Depth Information
In general:
1
4π
e
rr
iK rr
− ′
Grr
(, )
′
→
Grr
( ,)
′
=
+
(2.57)
− ′
We note that an equation of the form ∇− ∂∂=
ψ
(
1
/
v
)(
ψ
/
t
)
0
(where
2
2
2
2
∇=∂∂
ψψ
(
/
x
))
is solved by setting
x
−
vt
=
p
and
x
+
vt
=
q
; therefore
2
2
2
∂∂∂=
ψ/
pq
0
. Having the solution ψ = ψ
1
(
p
) + ψ
2
(
q
) = ψ
1
(
x
−
vt
) + ψ
2
(
x
+
vt
) =
2
+
+ , we see that we have, as expected, an ingoing and an outgoing wave.
From this it can be written
ψψ
e
±−
iK rr
′
′
∫
ψ
()
r
=
ϕ
()
r
−
rr
Ur
() ()
′
ψ
r
′
d
3
r
′
±
±
(2.58)
−
and this is an inhomogeneous Fredholm type of equation of the second kind.
If
U
(
r
′)ψ
+
(
r
′) decreases rapidly as |
r
′| → ∞ (e.g., if it vanishes for |
r
′| >
R
), then
letting |
r
| → ∞(which is equivalent to |
r
′| → 0) leads to
1
4
e
iKr
∫
ψ
+
()
r
→−
ϕ
()
r
Ur
() ()
′
ψ
+
r
′
d
3
r
′
π
r
constant
1
4π
e
iKr
(2.59)
→+
ϕ
()
r
C
′
r
rC
e
r
iKr
→+
ϕ
()
where
C
is the scattering amplitude.
Similarly, it follows that
C
e
−
iKr
(2.60)
ψ
−
()
r
→+
ϕ
()
r
r
At |
r
| → ∞, by looking at the time dependence and locus of the phase, it
can be determined whether we have an ingoing or outgoing wave; for example,
with Schrödinger's equation,
ψ
(,)
rt
=
ψ
( )
re
−
iE
(/)
t
=
ψ
( )
re
−
i
ω
t
Therefore
Ce
iKr
(
−
ω
t
)
(2.61)
ψ
+
(,)
rt
=
e
iKr
(
−
ω
t
)
+
r
For ψ
+
one can see that
Kr
increases as ω
t
increases, corresponding to an
outgoing wave (i.e., its constant phase front moves out). This would typically
be written as
e
iKr
(2.62)
ψψ
+
=
()
r
=
e
fiKz
+
f
(, )
θϕ
r
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