Image Processing Reference

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This would imply that

−

e

iK y

,

gKy

(,)

=

(2.37)

aaiK

+

(

)

+

+

aiK

(

)

α

0

1

1

nn

n

n

Now taking the Fourier transform of Equation 2.37, we have

1

2

e

iK

,(

xy

−

)

d

K

∫

Gxy

(, )

=−

(2.38)

()

π

n

aaiK

+

(

)

+

+

aiK

(

)

α

0

1

1

nn

n

n

This is the particular solution. The homogeneous equation can be written

as
L
x
U
(
x
) = 0, giving the total solution
G
(
x
,
y
) +
U
(
x
) subject to

+
∫

ϕ

()

xUxGxy

=

()

(, )( )

ρ

yy

d

(2.39)

2.2.2 the Integral equation of Scattering

We can apply a Green's function approach to all wave and particle scattering

problems, for example, Schrödinger's equation for a single particle, subject to

a potential energy
V
(
r
) (e.g., to describe an electron coming into an atom). And

for the time-independent case, this reduces to the inhomogeneous Helmholtz

equation to

2

(2.40)

∇+

2

ψ

()

rVrr Er

() ()

ψ

=

ψ

()

2
m

where
E
= (
p
2
/2
m
) is the positive energy of the incoming particle and
m
is its

mass in a center of mass coordinate frame. Therefore, we now have

(

∇+

2

2
Kr

)

ψ

()

=

Ur

() ()

ψ

r

=

ρ

()

r

(2.41)

∫

ψϕ

()

r

=

()

r

+

Grrrdr

(, )( )

′

ρ

′

′

(2.42)

Using Fourier transforms and following the earlier analysis, we can write

∫

∫

gK r

(

′′

,

)(

∇+

2

K

2

)

e

iK

′

,

r

d

K

′ =−

e

−

iKr iK r

′

,

′

e

,

d

K

′

(2.43)

∇= −

2
e

iK

′

,

r

()

K

′

e

iKr

,

(2.44)

′

∫

∫

gK rK

(,)

′′

(

−

(

Ke K

′

)

)

d

′

= −
−

e

e

d

K

′

2

2

iK

′

,

r

iKr iK

′

,

′

,

r

(2.45)

′

e

KK

−

−

iK

′

,

r

(2.46)

gK r

(,)

′′

=

2

−

()

′

2

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