Elastic and Inelastic Scattering of Charged Particles:
Rationale: Charged and uncharged particles can and will react with matter, resulting in
the production of secondary charged particles as well as a release of energy upon
interaction.
Here, 𝒌𝑖 and 𝒌𝑓 are the incident and scattered wavevectors corresponding to the
incident charged particle on the atom. As the result of the interaction between the
incident charged particle and the atom, there will be a change in momentum of the
charged particle,
Δ𝒑 = ℏ𝒒 = ℏ𝒌𝑓 − ℏ𝒌𝑖
Note that the quantum mechanical momentum of a particle is given by
𝒑 = ℏ𝒌, 𝑝 = ℏ /𝜆
Consider: a particle of mass 𝑚0 , total charge ze, momentum 𝒑 = ℏ𝒌𝑖 and kinetic energy
𝑝2
𝐸 = 2𝑚 =
0
(ℏ𝑘𝑖 )2
2𝑚0
. This charged particle is scattered by an atom of charge Ze through a
Coulomb central potential V(r).
Result: Charged particle emerges with momentum ℏ𝒌𝑓 , and the same energy.
Wave equation to describe scattering:
ℏ2 2
[−
∇ + 𝑉(𝑟)] Ψ(𝒓) = 𝐸Ψ(𝒓)
2𝑚0
Interaction potential: Coulomb potential
𝑧𝑍𝑒 2
𝑉(𝑟) = −
4𝜋𝜀0 𝑟
Boundary conditions:
lim Ψ(𝒓) = Ψ𝑖 (𝒓) + Ψ𝑠 (𝒓)
𝑟→+∞
Here,
Ψ𝑖 (𝒓) = exp(𝑖𝒌𝑖 ⋅ 𝒓)
is the wavefunction representing the incident particle, which is in the form of a plane
wave, and Ψ𝑠 (𝒓) is the wavefunction representing scattered spherical plane waves
moving radially outward.
Solutions to the TISE: Lippmann – Schwinger form
𝑚0
exp(𝑖𝑘|𝒓 − 𝒓′ |)
Ψ(𝒓) = exp(𝑖𝒌𝑖 ⋅ 𝒓) −
∫
𝑉(𝒓′ )Ψ(𝒓′ )𝑑 3 𝑟′
|𝒓 − 𝒓′ |
2𝜋ℏ2
Here, we note that
1
𝐺(𝒓, 𝒓′) =
|𝒓 − 𝒓′|
This is a Green’s function for the Laplacian operator which satisfies the following
relation:
1
∇2 (
) = −4𝜋𝛿 3 (𝒓 − 𝒓′)
|𝒓 − 𝒓′|
Substituting this back into the TISE,
ℏ2 2
[−
∇ + 𝑉(𝑟)] Ψ(𝒓)
2𝑚0
ℏ2 2
𝑚0
exp(𝑖𝑘|𝒓 − 𝒓′ |)
= [−
∇ + 𝑉(𝑟)] (exp(𝑖𝒌𝑖 ⋅ 𝒓) −
∫
𝑉(𝒓′ )Ψ(𝒓′ )𝑑 3 𝑟′)
|𝒓 − 𝒓′ |
2𝑚0
2𝜋ℏ2
ℏ2 2
=−
∇ exp(𝑖𝒌𝒊 ⋅ 𝒓)
2𝑚0
ℏ2
𝑚0
exp(𝑖𝑘|𝒓 − 𝒓′ |)
2
−
(−
)
∫
∇
(
) 𝑉(𝒓′ )Ψ(𝒓′ )𝑑 3 𝒓′ + 𝑉(𝒓)Ψ(𝒓)
|𝒓 − 𝒓′ |
2𝑚0
2𝜋ℏ2
ℏ2
(−𝑘𝑖2 exp(𝑖𝒌𝑖 ⋅ 𝒓))
=−
2𝑚0
1
exp(𝑖𝑘|𝒓 − 𝒓′ |)
+
∫ (−4𝜋𝛿 3 (𝒓 − 𝒓′ ) − 𝑘 2
) 𝑉(𝒓′ )Ψ(𝒓′)𝑑 3 𝑟 ′ + 𝑉(𝒓)Ψ(𝒓)
|𝒓 − 𝒓′ |
4𝜋
ℏ2 𝑘𝑖2
=
exp(𝑖𝒌𝑖 ⋅ 𝒓)
2𝑚0
𝑘 2 exp(𝑖𝑘|𝒓 − 𝒓′ |)
−
∫
𝑉(𝒓′ )Ψ(𝒓′)𝑑 3 𝑟 ′ − 𝑉(𝒓)Ψ(𝒓) + 𝑉(𝒓)Ψ(𝒓)
|𝒓 − 𝒓′ |
4𝜋
Finally, with
𝑘 2 = 𝑘𝑖2
This becomes
ℏ2 2
ℏ2 𝑘𝑖2
𝑚0
exp(𝑖𝑘|𝒓 − 𝒓′ |)
[−
∇ + 𝑉(𝑟)] Ψ(𝒓) =
(exp(𝑖𝒌𝑖 ⋅ 𝒓) −
∫
𝑉(𝒓′ )Ψ(𝒓′)𝑑 3 𝑟 ′ )
2
′
|𝒓
|
2𝑚0
2𝑚0
2𝜋ℏ
−𝒓
ℏ2 𝑘𝑖2
=
Ψ(𝒓) = 𝐸Ψ(𝒓)
2𝑚0
Here,
ℏ2 𝑘𝑖2
𝐸=
2𝑚0
At large distances away from the scatterer (i. e. the atom),
1 ′
𝑟′ 2
2
′2
2
′2
′
√
√
|𝒓 − 𝒓′| = 𝑟 + 𝑟 − 2𝒓 ⋅ 𝒓′ = √𝑟 + 𝑟 − 2𝑟(𝒓 ⋅ 𝒓̂) = 𝑟 1 − 2 ( 𝒓 ⋅ 𝒓̂) + ( )
𝑟
𝑟
1
1
1
≈ 𝑟√1 − 2 ( 𝒓′ ⋅ 𝒓̂) ≈ 𝑟 (1 − (2 ( 𝒓′ ⋅ 𝑟̂ ))) ≈ 𝑟 − 𝒓′ ⋅ 𝒓̂
𝑟
2
𝑟
Hence,
𝑘|𝒓 − 𝒓′| ≈ 𝑘𝑟 − 𝒓′ ⋅ (𝑘𝒓̂) = 𝑘𝑟 − 𝒌𝑓 ⋅ 𝒓′
and
exp(𝑖𝑘|𝒓 − 𝒓′ |)
1
→
exp(𝑖𝑘𝑟) exp(−𝑖𝒌𝑓 ⋅ 𝒓′)
𝑟→+∞ 𝑟
|𝒓 − 𝒓′ |
This allows us to simplify the Lippman-Schwinger form of the wavefunctions to
Ψ(𝒓) = exp(𝑖𝒌𝑖 ⋅ 𝒓)
𝑚0 exp(𝑖𝑘𝑟)
−
∫ exp(−𝑖𝒌𝑓 ⋅ 𝒓′) 𝑉(𝒓′ )Ψ(𝒓′ )𝑑3 𝑟 ′
2𝜋ℏ2
𝑟
exp(𝑖𝑘𝑟)
= exp(𝑖𝒌𝑖 ⋅ 𝒓) +
𝑓(𝒌𝑖 , 𝒌𝑓 )
𝑟
The function
𝑚0
𝑓(𝒌𝑖 , 𝒌𝑓 ) = −
∫ exp(−𝑖𝒌𝑓 ⋅ 𝒓′) 𝑉(𝒓′ )Ψ(𝒓′ )𝑑 3 𝑟 ′
2𝜋ℏ2
Is the scattering amplitude of the emerging wave. The scattering amplitude is used to
determine the differential scattering cross section per unit solid angle of the incident
charged particle:
𝑑𝜎
2
= |𝑓(𝒌𝑖 , 𝒌𝑓 )| = |𝑓(𝜃)|2
𝑑Ω
Consider the following schematic diagram:
The quantities in this figure can be defined as follows:
𝑁̇𝑖𝑛 – Number of particles incident on the target per unit time and per unit area, with the
unit area perpendicular to the direction of the incident beam.
𝑁̇𝑜𝑢𝑡 – Total number of particles that have interacted per unit time with the n T particles in
the target, and which have been scattered in a given direction.
Total scattering cross section:
𝜎𝑡𝑜𝑡 =
𝑁̇𝑜𝑢𝑡
𝑁̇𝑖𝑛 𝑛𝑇
If we define
𝑑2𝜎
𝑁̇𝑑Ω𝑑𝒲 = (
) 𝑁̇ 𝑑Ω𝑑𝒲
𝑑Ω𝑑𝒲 𝑖𝑛
Then the double differential cross section per unit solid angle and per unit energy loss is
𝑑2𝜎
𝑁̇𝑑Ω𝑑𝒲 1
=
𝑑Ω𝑑𝒲
𝑁̇𝑖𝑛 𝑑Ω𝑑𝒲
Differential scattering cross section (DCS) in solid angle:
𝑑𝜎
𝑑2𝜎
=∫
𝑑𝒲
𝑑Ω
𝑑Ω𝑑𝒲
For elastic scattering of a charged particle,
2
𝑑𝜎
𝑚0
𝑚0 2
2
′ )Ψ(𝒓′ )𝑑 3 ′
= |−
∫ exp(−𝑖𝒌𝑓 ⋅ 𝒓′) 𝑉(𝒓
𝑟| =(
) |⟨exp(𝑖𝒌𝑓 ⋅ 𝒓) |𝑉(𝒓)|Ψ(𝒓)⟩|
2
2
𝑑Ω
2𝜋ℏ
2𝜋ℏ
If we assume that the effect of the potential is weak,
Ψ(𝒓) ≈ exp(𝑖𝒌𝑖 ⋅ 𝒓)
Thus, the DCS in solid angle is
2
𝑑𝜎
𝑚0
′ )Ψ(𝒓′ )𝑑 3 ′
= |−
∫ exp(−𝑖𝒌𝑓 ⋅ 𝒓′) 𝑉(𝒓
𝑟|
𝑑Ω
2𝜋ℏ2
2
𝑚0 2
3
=(
)
|∫
exp(−𝑖(𝒌
−
𝒌
)
⋅
𝒓)𝑉(𝒓)𝑑
𝑟|
𝑓
𝑖
2𝜋ℏ2
2
𝑚0 2
3
=(
) |∫ exp(−𝑖𝒒 ⋅ 𝒓)𝑉(𝒓)𝑑 𝑟|
2𝜋ℏ2
Using the screened Coulomb potential of Wentzel type in this DCS,
2
𝑑𝜎
𝑚0 2
3
=(
) |∫ exp(−𝑖𝒒 ⋅ 𝒓)𝑉(𝒓)𝑑 𝑟|
𝑑Ω
2𝜋ℏ2
2
𝑚0 2
𝑧𝑍𝑒 2
3
=(
) |∫ exp(−𝑖𝒒 ⋅ 𝒓) (−
) exp(−𝑟/𝑅𝑇𝐹 ) 𝑑 𝑟|
2𝜋ℏ2
4𝜋𝜀0 𝑟
2
2
𝑚0 𝑧𝑍𝑒 2
1
𝑟
=(
) |∫ exp (− (𝑖𝒒 ⋅ 𝒓 +
))|
2𝜋ℏ2 4𝜋𝜀0
𝑟
𝑅𝑇𝐹
2
2
𝑚0 𝑧𝑍𝑒 2
1
𝑟
3
=(
) |∫ exp (− (𝑖𝑞𝑟 cos 𝜃 +
)) 𝑑 𝑟|
2𝜋ℏ2 4𝜋𝜀0
𝑟
𝑅𝑇𝐹
Evaluating the integral,
1
1
∫ exp (−𝑖 ((𝑞 cos 𝜃 +
) 𝑟)) 𝑑 3 𝑟
𝑟
𝑅𝑇𝐹
1
1
= ∫ exp (− ((𝑖𝑞 cos 𝜃 +
) 𝑟)) 𝑟 2 sin 𝜃𝑑𝑟𝑑𝜃𝑑𝜙
𝑟
𝑅𝑇𝐹
2𝜋
+∞
𝜋
𝑟
= ∫ 𝑑𝜙 ∫ 𝑑𝑟 𝑟 exp (−
) ∫ 𝑑(− cos 𝜃) exp(𝑖𝑞𝑟(− cos 𝜃))
𝑅𝑇𝐹 0
0
0
+∞
1
𝑟
= 2𝜋 ∫ 𝑑𝑟 𝑟 exp (−
) ∫ 𝑑𝑢 exp(𝑖𝑞𝑟𝑢)
𝑅𝑇𝐹 −1
0
+∞
1
𝑟
= 2𝜋 ∫ 𝑑𝑟 𝑟
exp (−
) (exp(𝑖𝑞𝑟) − exp(−𝑖𝑞𝑟))
𝑖𝑞𝑟
𝑅𝑇𝐹
0
4𝜋 +∞
𝑟
4𝜋
1
4𝜋
=
∫ 𝑑𝑟 sin(𝑞𝑟) exp (−
)=
= 2
2
𝑞 0
𝑅𝑇𝐹
𝑞 𝑞+ 1
𝑞 + 1/𝑅𝑇𝐹
2
𝑞𝑅𝑇𝐹
Therefore, the DCS in solid angle is
2
2
2
𝑑𝜎
𝑚0 𝑧𝑍𝑒 2
4𝜋
𝑚0 𝑒 2
=(
) | 2
| =(
) (𝑧𝑍)2
𝑑Ω
2𝜋ℏ2 4𝜋𝜀0
𝑞 + 1/𝑅𝑇𝐹
2𝜋ℏ2 4𝜋𝜀0
Equating the classical and quantum momenta,
𝒑 = ℏ𝒌 = 𝑚0 𝒗 → 𝒌 =
Thus, the DCS can be written as
𝑚0
𝒗
ℏ
16𝜋 2
2
𝜃 2
2
((2𝑘 sin 2) + 1/𝑅𝑇𝐹
)
(
)
2
2
𝑑𝜎
𝑚0 𝑒 2
1
𝑧𝑍𝑒 2
1
=( 2 2
) (𝑧𝑍)2 (
)
=
(
)
2
2
2
𝑑Ω
2ℏ 𝑘 4𝜋𝜀0
2𝑚0 𝑣 4𝜋𝜀0
𝜃
𝜃
2
(sin2 2 + 1/4𝑘 2 𝑅𝑇𝐹
)
(sin2 2 + 0.25𝜒𝑎2 )
Here, we define the screening angle as
1
𝜒𝑎 =
𝑘𝑅𝑇𝐹
Using the expression for the classical electron radius,
1
𝑒2
𝑟𝑒 =
4𝜋𝜀0 𝑚𝑒 𝑐 2
the DCS can be simplified further as
2
𝑑𝜎
𝑚𝑒 𝑐 2 1
1
2
= (𝑧𝑍𝑟𝑒 ) (
)
2
𝑑Ω
𝑚0 𝑣
4 (0.5(1 − cos 𝜃) + 0.5(0.5𝜒 2 ))2
2 2
𝑎
𝑚𝑒 𝑐
1
= (𝑧𝑍𝑟𝑒 )2 (
)
2
2
𝑚0 𝑣
((1 − cos 𝜃) + 0.5𝜒𝑎2 )
Finally, using the Lorentz velocity transform,
𝛽2𝑐
𝑣=
√1 − 𝛽 2
The final form of the DCS, also known as the screened Rutherford DCS, is
2
𝑑𝜎
𝑚𝑒 𝑐 2 1 − 𝛽 2
1
2
= (𝑧𝑍𝑟𝑒 ) (
)
2
2
4
𝑑Ω
𝑚0 𝑐
𝛽
((1 − cos 𝜃) + 0.5𝜒 2 )
𝑎