HW #2

1. 1D scattering

(a)

The delta function potential requires a discontinuity of the wave function, ψ' (+0) - ψ' (-0) = ((2  
                m μ)/(ℏ^2)) ψ(0), while the wavefunction itself is continuous.

Solve[{1 + R == T, 
        I k T - I k (1 - R) == ((2  m μ)/(ℏ^2)) 
            T}, {R, T}]

Simplify[%]

Simplify[(-((m μ)/(m μ - I k (ℏ^2)))) (-((m μ)/(m μ + I k (ℏ^2)))) + (((k (ℏ^2))/(I m μ + k (ℏ^2)))) (((k (ℏ^2))/(-I m μ + k (ℏ^2))))]

It is worth noting that the reflection probabiliy \(TraditionalForm\`\(\(|\)\(R\)\( | \^2\)\)\) is 100% at the low momentum limit k = 0, while it monotonically decreases to zero at higher momenta.

(b)

We assume T and R to be approximately constant.  Hence the Gaussian wave packet is given by the integrals

Integrate[(E^(I q x)) (E^(-(((q - k))^2)/
                2/(σ^2))) 
        (E^(-I ((ℏ^2) (q^2)/2/m) t/ℏ)), {q, -∞, ∞}, 
      Assumptions -> {σ ∈ Reals && 
            m ∈ Reals && ℏ ∈ Reals && 
            t ∈ Reals}]

Integrate[(E^(-I q x)) (E^(-(((q - k))^2)/
                2/(σ^2))) 
        (E^(-I ((ℏ^2) (q^2)/2/m) t/ℏ)), {q, -∞, ∞}, 
      Assumptions -> {σ ∈ Reals && 
            m ∈ Reals && ℏ ∈ Reals && 
            t ∈ Reals}]

ψ = 
      If[x < 0, (((E^(((2 I k m x - m (x^2) (σ^2) - I (k^2) t ℏ)/(2 m + 2 I t (σ^2) ℏ)))) (Sqrt[2 π]))/(Sqrt[(1/(σ^2)) + ((I t ℏ)/m)])) + 
          R (( (E^(-((2 I k m x + m (x^2) (σ^2) + I (k^2) t ℏ)/(2 m + 2 I t (σ^2) ℏ)))) (Sqrt[2 π]))/(Sqrt[(1/(σ^2)) + ((I t ℏ)/m)])), 
        T (( (E^(((2 I k m x - m (x^2) (σ^2) - I (k^2) t ℏ)/(2 m + 2 I t (σ^2) ℏ)))) (Sqrt[2 π]))/(Sqrt[(1/(σ^2)) + ((I t ℏ)/m)]))]

Table[
      Plot[Abs[ψ]^2 /. {R -> -((m μ)/(m μ - I k (ℏ^2))), 
              T -> ((k (ℏ^2))/(I m μ + k (ℏ^2)))} /. {k -> 100, 
            m -> 1, μ -> 30, σ -> 
              1, ℏ -> 1}, {x, -10, 10}, 
        PlotRange -> {0, 10}], {t, -0.1, 0.1, 0.01}]

Select the graphics and go to "Animate Selected Graphics."

(c)

Rewriting Eq. (1) into the form Eq. (3) gives f(π) = -i R, f(0) = i (1 - T).  Therefore, the unitarity relation is rewritten as
 \(TraditionalForm\`1 = \(\(\(|\)\(R\)\( | \^2\)\(+\(\(|\)\(T\)\( | \^2\)\)\)\)
 = \(\(\(|\)\(i\ \(f(π)\)\)\( | \^2\)\(+\(\(|\)\(1 + 
                    i\ \(f(0)\)\)\( | \^2\)\)\)\)
 = \(\(\(|\)\(f(π)\)\( | \^2\)\(\(+1\) + i\ \(f(0)\) - 
                  i\ \(\(f\^*\)(0)\) + \)\(|\)\(f(
                  0)\)\( | \^2\)\)
 = σ + 1 - 
                2  Im\ \(\(f(0)\)\(.\)\)\)\)\)\)
 Therefore, σ = 2  Im f(0), the one-dimensional version of the optical theorem.

2. Probability currents

The plane wave ψ = 
        (e^(i (Overscript[k,  ->]) ·(Overscript[x,  ->]) )) gives simply \(TraditionalForm\`\(j\& -> \) = \(ℏ\ \(k\& -> \)\)\/m\), while the spherical wave ψ = (e^(i k r))/r gives \(TraditionalForm\`\(j\& -> \) = \(\(\(ℏ\ k\)\/m\) \(x\& -> \)\/r\^3 = \(-\(\(ℏ\ k\)\/m\)\) \(∇\& -> \)\(1\/r\)\)\). For the former, (Overscript[j,  ->])  is constant and hence its divergence simply vanishes.  For the latter, \(TraditionalForm\`\(\(\(∇\& -> \)\(\(·\)\(j\& -> \)\)\)\(=\)\)\)-((ℏ k)/m) Δ 
           (1/r) = ((ℏ k)/m) 
          4  π δ((Overscript[x,  ->]) ) indeed.  Therefore, the spherical wave is a wave that emanates from the origin. 

For fun, I plot below the probability current on the x-y plane at z=0 for both cases.

<<Graphics`PlotField`

PlotVectorField[{1, 0}, {x, -1, 1}, {y, -1, 1}]

PlotVectorField[{x, y}/((((x^2) + (y^2)))^(3/2)), {x, -1, 
        1}, {y, -1, 1}]

3. Classical Hard Sphere Scattering

For the impact parameter b > a, the point-like particle will not get scattered by the hard sphere and merely passes it by.  On the other hand, if b < a, it gets scattered into the angle θ = π - 2  (sin^(-1)) (b/a).  Solving it, we find b = a cos (θ/2).  Because the original beam is assumed to have a uniform area density, the cross section is
\(TraditionalForm\`d\ σ = \(b\ \ d\ b\ \ d\ ϕ = \(\(-a\)\ cos\ \(θ\/2\) a\ sin\ \(θ\/2\) 
              d\ θ\ d\ ϕ = \(\(-\(a\^2\/4\)\) 
                sin\ θ\ d\ θ\ d\ ϕ = \(a\^2\/4\) 
                d\ cos\ θ\ d\ ϕ\)\)\)\).
Hence the differential cross section is
((d σ)/(d Ω)) = 
        ((a^2)/4), and the total cross section is σ = ∫((a^2)/4) 
              d Ω = π (a^2), nothing but the geometric cross section of the hard sphere.