HW #3 (221b)

I thank Ed Boyda for the solution.

1)

d)  Set =m=a=1.

(Subscript[δ, 0])[k_, γ_] := (1/(2  I)) 
        Log[((1 + ((2 γ)/k) (E^(-I k)) Sin[k])/(1 + ((2 γ)/k) (E^(I k)) Sin[k]))]

(Subscript[σ, 0])[k_, γ_] := ((4  π)/(k^2)) 
        Sin[(Subscript[δ, 0])[k, γ]]^2

(γ=100):  Large γ emulates hard sphere scattering except at resonances.  The graph below essentially matches that of the hardsphere cross section ((4  π)/(k^2)) (sin^2)ka; in particular, (Subscript[σ, 0]) -> 4  π as k  ->  0.

Plot[(Subscript[σ, 0])[k, 100],  {k,  0,  10},  
      PlotRange -> {0, 15}]

When we zoom in and improve the resolution (γ=100 still), we see the expected resonances at k a  ≈  n π, corresponding to the almost-bound  states inside the shell. we see that the unitarity bound is saturatred at the resonance.

Plot[{(Subscript[σ, 0])[k, 100], 4  π/(k^2)},  {k,  2,  10},  
      PlotRange -> {0, 2}, PlotPoints -> 100, 
      PlotDivision -> 100]

(γ=10):  For middling  γ, the resonances are shifted towards smaller k  values according to equation (III.54) of the lecture notes.  The peakes are higher and wider than the γ=100 resonances because of the increased width Γ.  (See equation III.63, which is valid for poles close to the real axis.)

Plot[{(Subscript[σ, 0])[k, 10], 4  π/(k^2)},  {k, 2,  10},  
      PlotRange -> {0, 2}]

(γ=1):  For small γ, we can't distinguish the resonances from the oscillatory background.  By III.52, small γ means poles far from the real axis, so the disappearance of the resonances confirms that the particle doesn't `feel' poles far from its real momentum.

Plot[(Subscript[σ, 0])[k, 1],  {k, 2,  10},  
      PlotRange -> {0,  .5}]

f)  The wavefunctions are derived in the lecture notes III.  Following the conventions there but setting =m=a=1,  k = (Subscript[k, 0]) - i κ,  E = (Subscript[E, 0]) - i Γ/2 = 
          ((Subscript[k, 0])^2)/2 - i (Subscript[k, 0]) κ  +  O((κ^2)).  I choose the approximate value of k at the second resonance and leave γ arbitrary for now.

(Subscript[E, 0]) = (k0^2); Γ = 2 k0 κ;  
    k0 = ((2  π)/(1 + (1/(2  γ))));  κ = (((((2  π)/(4  γ))))^2);

The wavefunction, defined piecewise on the intervals r=(0,1), r=(1,∞), is

rR[r_,  t_] := 
        If[r < 1, 
          Sin[(k0 - I κ) r] (E^(-I (Subscript[E, 0]) t)) 
            (E^(-Γ t/2)), 
          Sin[(k0 - 
                  I κ)] (E^(I (k0 - I κ) (r - 
                      1))) (E^(-I (Subscript[E, 0]) t)) 
            (E^(-Γ t/2))];

Double click on the graph to watch the animation (γ=5):

Do[Plot[
        Abs[rR[r, i]]^2 /. {γ -> 5, t -> 2  i}, {r, 0, 
          20},  PlotRange -> {0, 2}],  {i, 0, 20}]