r
′
d
3
r
′′
e
−
ikvectore
r
(
θ,φ
)
·
vector
r
′
T
(
vector
r
′
,vector
r
′′
)
e
i
vector
k
·
vector
r
′′
=
1
(2
π
)
3
/
2
bracketleftbigg
e
i
vector
k
·
vector
r
+
e
ikr
r
f
(
θ,φ

vector
k
)
bracketrightbigg
,
(65)
where we have introduced the scattering amplitude,
f
(
θ,φ

vector
k
),
2
defined as
f
(
θ,φ

vector
k
) =
−
M
2
π
planckover2pi1
2
integraldisplay
d
3
r
′
d
3
r
′′
e
−
ikvectore
r
(
θ,φ
)
·
vector
r
′
T
(
vector
r
′
,vector
r
′′
)
e
i
vector
k
·
vector
r
′′
.
(66)
With the definitions
vector
k
′
=
kvector
e
r
(
θ,φ
)
,
(67)
and
f
(
vector
k
′

vector
k
) =
f
(
θφ

vector
k
)
,
(68)
the scattering amplitude becomes
f
(
vector
k
′

vector
k
)
=
−
M
2
π
planckover2pi1
2
integraldisplay
d
3
r
′
d
3
r
′′
e
−
i
vector
k
′
·
vector
r
′
T
(
vector
r
′
,vector
r
′′
)
e
i
vector
k
·
vector
r
′′
=
−
(2
π
)
2
M
planckover2pi1
2
integraldisplay
d
3
r
′
d
3
r
′′
(
vector
k
′

vector
r
′
)(
vector
r
′

T

vector
r
′′
)(
vector
r
′′

vector
k
)
=
−
(2
π
)
2
M
planckover2pi1
2
(
vector
k
′

T

vector
k
)
.
(69)
This shows that the scattering amplitude is the Fourier transform of the Tmatrix.
In optics it is well
known that the image in the farfield is the Fourier transform of the nearfield image.
This is because
propagation over a large distance allows the Fourier components of the field to spatially separate. This is
exactly what we are seeing here. The Tmatrix describes the exact scattered field over all space, including
inside the interaction region. The Fourier transform of the Tmatrix therefore describes the same scattered
field in momentum space. The scattering amplitude, on the other hand, describes the scattered field only
in the farfield region (
r
→ ∞
), hence it follows that the different momentum components would spatially
separate, with the result that the scattering amplitude should reflect the Fourier transform of the scattered
field.
6
The Scattering CrossSection
Up to now we have taken the incident wave to be a planewave

ψ
0
)
=

vector
k
)
, where
(
vector
r

vector
k
)
=
e
i
vector
k
·
vector
r
(2
π
)
3
/
2
.
(70)
This state is deltanormalized so that
(
vector
k

vector
k
′
)
=
δ
3
(
vector
k
−
vector
k
′
). We note that this choice of normalization has
the drawback that that

ψ
0
(
vector
r
)

2
does not have the right units to be a probability density. The units of a
probability density are 1
/
[
volume
], while Eq. (
??
) shows that

ψ
0
(
r
)

2
=
(
vector
r

vector
k
)
2
is clearly dimensionless.
One possible way to give a probabilistic interpretation to

ψ
0
(
vector
r
)

2
would be to assume a finite quantiza
tion volume
V
, so that the probability density would be defined as
ρ
(
vector
r
) =

ψ
0
(
vector
r
)

2
(2
π
)
3
V
, chosen so that
2
Note that
f
(
x, y

a
) should be read as “f of x and y given a”.
9
integraltext
V
d
3
rρ
(
vector
r
) = 1. For a planewave, the probability density is uniform, giving
ρ
(
vector
r
) =
ρ
= 1
/V
. A plane wave
corresponds to a uniform probability flow at speed
v
=
planckover2pi1
v/M
, so that the current density is
j
=
ρv
=
v/V
.
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 Spring '10
 MichaelMoore
 mechanics, Angular Momentum, SCATTERING, Fundamental physics concepts, plane wave, scattering amplitude