Image Processing Reference
In-Depth Information
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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