Scattering (Homework 10-325)
a) What is the scattering amplitude f(θ) for a Gaussian potential of range a, V (r) =
α exp(−r2/a2).
b) Make a plot of the differential cross-section dσ/d
The scattered wave is always:
$ψ_{scattered} (r,θ)=f(θ)*(e^{i k r}/r)$
It can be show (very difficult) that for a spherically symmetrical potential (a potential where
$V(r ⃗ )=V(r)$) the function (equation [11.88])
$f(θ)=-2m/(ℏ^2 k) ∫_0^∞ r V(r) sin(k r)d r$ where $k=2K*sin θ/2$
$K=\sqrt{2mE}/ℏ$ (E is energy of incident particle-wave) and θ is the scattering angle
For the given potential
$V(r)=α*exp(-r^2/a^2 )$
$f(θ)=-2m/(ℏ^2 k) α∫_0^∞ r*exp(-r^2/a^2 ) sin(k r)d r=$
$=α (√π/4)*a^3*exp(-(a^2 k^2)/4) =(αa^3 √π)/4*exp(-(a^2 K^2 sin^2 θ/2)/2)$
b)
$dσ/dΩ=|f(θ) |^2=(α^2*πa^6)/16*exp (-a^2 K^2*sin^2 (θ/2)$
c)
In the limit of weak of low energy $(k→0)$ it can be shown that
$f(θ)=-m/(2πℏ^2 )*∫ V(r ⃗ ) d^3 r=-m/(2πℏ^2 ) α ∫_0^∞ exp (-r^2/a^2 )*(r^2 d r)∬dΩ$
$f(θ)=-m/(2πℏ^2 ) α*(a^3 √π)/4*4π=-(m*α*a^3 √π)/(2ℏ^2 )$
$dσ/dΩ=|f|^2=(m^2*α^2 a^6 π)/(4ℏ^4)$
and
$σ=(m^2*α^2 a^6 π)/(4ℏ^4 )*∬ dΩ=(m^2*α^2 a^6 π)/(4ℏ^4 )*4π=(m^2*α^2 a^6 π^2)/ℏ^4$
