Question

Problem 5: Find the differential and total cross sections in the first Born approximation for elastic scattering of a particle of mass m, which is initially traveling along the z-axis, from a nonspherical, double-delta potential V (~ r) = V0δ(~ r − aˆ k)+V0δ(~ r+aˆ k), where ˆ k is the unit vector along the z-axis.

Answer 1

Dσ/ dΩ = |f(θ, φ)| ^2 = µ^ 2/ 4π 2¯h^ 4     | e^ ^{i~q^-1~r^-4} V (r^-4 )d ^3 r \^2

Here V (~r) is not symmetric but because it is in the form of a δ-function the integration is direct but first we express the δ function in the cartesian system because k being along Oz it will be easier to integrate.

dσ /dΩ = m^2 4π^ 2¯h ^4      Z V0 h [(~r − a~k) + δ(~r + a~k) i e^ iq·r0 d ^3 r^1 ]^ 2

= m^2V0 / 4π^ 2 h ^4 I ^2

First we write δ(~r ± ak) = δ(x)δ(y)δ(z ± a)

which then allow us to express the integral as:

I = Z δ(x)e ^ixqx dx       Z δ(y)e ^{iyqy dy} Z [δ(z − a) + δ(z + a)] e ^{izqz dx}

= 1 × 1 × [e ^iaqz + e −^iaqz ]

= 2 cos (aq_z)

Now we need to evaluate the magnitude of qz. Since the magnitude of ~ki and ~kf is the same, we denoted k (elastic scattering) the angle of ~q and Oz is half that of the angle between ~k and ~k0. Note that ~k0 lays on the Oz axis.

Thus we have qz = q sin θ/2 = 2k sin2 θ/2 leading to

I = 2 cos (aqz) = 2 cos 2ak sin2 (θ/2)

replacing I into equation we obtain

dσ /dΩ = m^2V0 / π ²h ⁴ cos² ( 2ak sin² θ/2 )---answer