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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