HW #2

1.  Free-particle Schrödinger Equation

(1) Plane wave ψ = (e^(i k z))does not depend on xor y.  Therefore, the Schrödinger equation becomes (((∂_z)^2)+(k^2)) ψ = 0.  Obviously this is a solution to the equation.

D[(E^(I k z)), {z, 2}] + (k^2) (E^(I k z))

(2) In polar coordinates, the Laplacian can be rewritten as \(TraditionalForm\`\(∇\& -> \)\^2\(\(=\)\(∂\_r\^2\(+\(2\/r\)\) ∂\_r\(+\(1\/r\^2\)\) ∂\_θ\^2\(+\(\(cos\ θ\)\/\(\(r\^2\) 
                      sin\ θ\)\)\) ∂\_θ\(+\(1\/\(\(r\^2\) \(sin\^2\) θ\)\)\) ∂\_ϕ\^2\)\)\).  The spherical wave ψ = ((e^(i k r))/r)does not depend on θ or ϕ.  Therefore, the Schrödinger equation becomes \(TraditionalForm\`\((∂\_r\^2\(+\(2\/r\)\) ∂\_r\(+k\^2\))\) ψ = 0\).

D[((E^(I k r))/r), {r, 2}] + (2/r) 
        D[((E^(I k r))/r), r] + (k^2) ((E^(I k r))/r)

Simplify[%]

2. Double Pin-hole Experiment

(1) As directed, we assume that the denominators are approximately the same between two waves.  This is justified because the corrections are only of the order of d/L, and we are interested in the case where d ≪ L. We require that the numerators have the same phase, namely \(TraditionalForm\`k\ \(r\_+\) - k\ \(r\_-\) = 2  π\ n\).  We expand the l.h.s. with respect to d,

Series[Sqrt[(x^2) + (((y + (d/2)))^2) + (L^2)], {d, 0, 1}]

Series[Sqrt[(x^2) + (((y - (d/2)))^2) + (L^2)], {d, 0, 1}]

Simplify[Normal[%% - %]]

Therefore, k ((d y)/(Sqrt[(L^2) + (x^2) + (y^2)])) = 
        2  π n and hence (y/(Sqrt[(L^2) + (x^2) + (y^2)])) =  n (λ/d),

(2) Let us choose the unit where k = 1.  Then we pick d = 20, L = 1000.  Here is the interference pattern.  First along the y-axis (x = 0):

\(Plot[\(\(Abs[E\^\(I\ \(r\_+\)\)\/\(r\_+\) + E\^\(I\ \(r\_-\)\)\/\(r\_-\)]\^2 /. {\(r\_+\) -> \@\(x\^2 + \((y - d\/2)\)\^2 + L\^2\), \(r\_-\) -> \@\(x\^2 + \((y + d\/2)\)\^2 + L\^2\)}\) /. {d -> 
              20, L -> 1000}\) /. {x -> 0}, {y, \(-3000\), 3000}, 
      PlotPoints -> 100]\)

(3) Now on the plane:

\(ContourPlot[\(Abs[E\^\(I\ \(r\_+\)\)\/\(r\_+\) + E\^\(I\ \(r\_-\)\)\/\(r\_-\)]\^2 /. {\(r\_+\) -> \@\(x\^2 + \((y - d\/2)\)\^2 + L\^2\), \(r\_-\) -> \@\(x\^2 + \((y + d\/2)\)\^2 + L\^2\)}\) /. {d -> 
            20, L -> 1000}, {x, \(-5000\), 5000}, {y, \(-5000\), 5000}, 
      PlotPoints -> 100]\)

(4) For the same parameters as in (2),  First along the y-axis (x = 0):

\(Plot[\(\(Abs[E\^\(I\ \(r\_+\)\)\/\(r\_+\)]\^2 + 
              Abs[E\^\(I\ \(r\_-\)\)\/\(r\_-\)]\^2 /. {\(r\_+\) -> \@\(x\^2 + \((y + d\/2)\)\^2 + L\^2\), \(r\_-\) -> \@\(x\^2 + \((y - d\/2)\)\^2 + L\^2\)}\) /. {d -> 20, L -> 1000}\) /. {x -> 
            0}, {y, \(-3000\), 3000}, PlotPoints -> 100]\)

Now on the plane:

\(ContourPlot[\(Abs[E\^\(I\ \(r\_+\)\)\/\(r\_+\)]\^2 + 
            Abs[E\^\(I\ \(r\_-\)\)\/\(r\_-\)]\^2 /. {\(r\_+\) -> \@\(x\^2 + \((y + d\/2)\)\^2 + L\^2\), \(r\_-\) -> \@\(x\^2 + \((y - d\/2)\)\^2 + L\^2\)}\) /. {d -> 20, L -> 1000}, {x, \(-5000\), 
        5000}, {y, \(-5000\), 5000}, PlotPoints -> 100]\)

The main difference is the absence of the interference pattern.