E291: Cross section and optical theorem in 2D
Submitted by: Noam Tal
The problem:
In this problem you have to ”generalize” the phase shift method to 2D
(1) Define the channel functions: χn (ϕ).
(2) Write down the free wave functions: φE,n (r, ϕ). Keep it normalized.
A plane wave eikx strikes a ”spherical”(round) target.
(3) Write down the incident wave as a sum of ”spherical” waves, as was defined in (2).
(4) What is the incident flux in channel n?
Given a phase shift δn in channel n.
(5) Find an expression for the partial cross section σn .
i(kE r− π
4)
The asimptotic behaviour of the wave function can be expressed by: Ψ(x) = eikx + if (ϕ) e
(6) Write down an expression for the scattering function f (ϕ).
(7) Find the the proportion coefficient in σtotal ∝ Im[f (0)].
Given:
eikx
=
eikr cos(ϕ)
∞
X
=
in Jn (kr)einϕ
n=−∞
Jn (z) ∼
q
2
πz
cos(z − π4 (2n + 1))
The solution:
Item (2) confuses a different approach see Ex0292
(1) In 2D the spacial function Ylm is reduced to the normalized function χ(ϕ) =
√1 einϕ
2π
(2) Let’s move to the basis of |E, ϕi. The completeness demand:
Z
Z
Z
d2 k
kE dkdϕ
kE dEdϕ
1 = |ki
hk|
=
|E,
ϕi
hE,
ϕ|
=
|E, ϕi
hE, ϕ|
(2π)2
(2π)2
vE (2π)2
When hE, ϕ|E 0 , ϕ0 i = 2πδ(E − E 0 )δ(ϕ − ϕ0 )
q
Hence: |E, ϕi = kvEE |ki. Keeping the normalization gives:
E,n
|φ
i=
√
r
2π
kE
Jn (kr)χn (ϕ)
vE
(3) We shell put the spherical wave function in the given plane wave:
ikx
e
=
∞
X
n=−∞
n
inϕ
i Jn (kr)e
=
∞
X
n=−∞
n
i
r
vE E,n
φ (r, ϕ)
kE
(4) The incident flux in channel n is the square of the amplitude of the plane wave n term:
1
√
r
iincident = |An |2 =
vE
kE
(5) In analogy to the T matrix (Ψ = φ + ψscatt ) we will find ψscatt from the S matrix. We know
that the ingoing and outgoing amplitudes are:
r
vE
An = in
Bn = Snn0 An
KE
Using the relation: S = 1 − iT and the fact that for a spherical symmetrical target the S matrix is
diagonal, we get Tnn0 = −δnn0 eiδn 2 sin(δn ). We can now derive the outgoing flux:
r
vE 2
vE
2
iδn
n+1
in = | − iTnn0 An | = |δnn0 e 2 sin(δn )i
| = 4 sin2 (δn )
≡ σn vE
kE
kE
Hence: σn =
4
kE
sin2 (δn )
(6) We use again the same method to find ψscatt only now we will use the given terms of Jn (z) and
eikx :
r
r
1 i(kr− π (2n+1))
1 i(kr− π ) −i π n
4
4 e
2
Jn (kr) ∼
e
=
e
2πkr
2πkr
r
π
π
∞
X
1 ei(kr− 4 ) inϕ
ei(kr− 4 )
n − π2 n
ikx
i2δn
√
− 1)i e
ψscatt = −iTnn0 e = −i
(e
e
= if (ϕ) √
2πk
r
r
n=−∞
Hence: f (ϕ) =
∞
X
−i
√
(ei2δn − 1)einϕ
2πk
n=−∞
(7) We shall use the above result for f (ϕ):
∞
X
∞
X
−i
−i
0
0
inϕ
i2δn
√
√
|f (ϕ)| =
− 1)e
(e
(ei2δn − 1)ein ϕ
2πk
2πk
n=−∞
n0 =−∞
2
1 X 2i(δn −δn0 )
0
0
[e
− e−2iδn − e2iδn + 1]ei(n−n )ϕ
2πk 0
nn
Z 2π
1 X 2i(δn −δn0 )
2X
σtotal =
|f (ϕ)|2 dϕ =
[e
− e−2iδn − e2iδn0 + 1]2πδ(n − n0 ) =
[1 − cos(2δn )]
2πk 0
k n
0
=
nn
Now let’s calculate Im[f (0)]:
∞
X
∞
X
−i
1
i2δn
√
√
(e
− 1)] =
[1 − cos(2δn )]
Im[f (0)] = Im[
2πk
2πk
n=−∞
n=−∞
We can see that:
r
σtotal = 2
2π
Im[f (0)]
k
We can derive this result using the optical theorem:
X
|hφE,Ω |T |φE,ϕ0 i|2 = −2ImhφE,ϕ0 |T |φE,ϕ0 i
ϕ
q
Since for 2D |E, ϕi = kvEE |ki and G(r|r0 ) = −i m
2 H0 (kr), we can calculate the total cross section
to obtain the same result.
2