HW #4 (221b)

1)

a)  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  (γ=100 still), we see the expected resonances at k a  ≈  n π, corresponding to the almost-bound  states inside the shell.

Plot[(Subscript[σ, 0])[k, 100],  {k,  2,  10},  
      PlotRange -> {0, 2}]

(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],  {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}]

b)  Lecture notes `Scattering Theory III.'

c)  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}]

We can show analytically that the decrease in probability intside the shell is equal to the probability flux flowing out through the shell.  With ρ = \(TraditionalForm\`\(\(|\)\(ψ\)\( | \^2\)\)\), j = \(\/\(2  m\ i\)\ \ \((\ \(ψ\^*\)\)\)▽ψ-\(ψ▽ψ\^*\)), we want to show ∫ d(ρ/dt) dV+∫ j·da = 0.  In particular, we integrate d(ρ/dt)over the volume inside r < a integrate j over the sphere at r = a.  For the j integral, we can use the expression for the wavefunction just inside or just outside r = a.  The problem asks that we use the latter.

Using the above wavefunction with all constants restored, 

\(TraditionalForm\`dρ\/dt = \ \(-Γ\)\/\ ρ\) = \(TraditionalForm\`\(-Γ\)\/\(2  \ r\^2\)\ \((cosh\ 2  κ\ r - \ 
            cos\ 2\ \(k\_0\) r)\)\ e\^\(\(-Γt\)/\)\),   

so that

∫ d(ρ/dt) dV  =  ((∫_0)^a) \(TraditionalForm\`\(\(dρ\/dt\)\(\ \)\(4\)  \(π\)\(\ \)\(r\^2\)\(\ \)\)\)=  \(TraditionalForm\`\(\(-2\) π\ Γ\)\/\(\(\)\(\ \)\)\ \((\(sinh\ 2  κ\ a\)\/\(2  κ\) - \ \(sin\ 2\ \(k\_0\) a\)\/\(2  k\_0\))\)\ e\^\(\(-Γt\)/\)\).

A similar computation for j using the r > a wavefunction gives

j = \(TraditionalForm\`\/\(2  m\ i\)\ \ \((\ i\ k\ r\&^\)\) + \(TraditionalForm\`\(\(i\ \(\(\(k\)\(\ \)\)\^*\) r\&^\)\()\)\)\) ρ  =  \(TraditionalForm\`\(\ k\_0\)\/\(2  m\)\ \(r\/r\^2\)\&^\ \((cosh\ 2  κ\ r - \ 
            cos\ 2\ \(k\_0\) r)\)\ e\^\(\(-Γt\)/\)\), 

and

(∫_(r = a)) j·da  =  4π (a^2)j  =  \(TraditionalForm\`\(2  π\ \ k\_0\)\/m\ \((cosh\ 2  κ\ a - \ cos\ 2\ \(k\_0\) a)\)\ e\^\(\(-Γt\)/\)\).

Plugging in Γ = \(TraditionalForm\`\(2 \( \^2\) \(k\_0\) κ\)\/m\), 

∫ d(ρ/dt) dV+∫ j·da = 0   
	⟺  
\(TraditionalForm\`  \)	\(TraditionalForm\`\(\(\(-\(\(4\ π\ \(\^2\) \(k\_0\) κ\)\/m\)\)\ \((\(sinh\ 2  κ\ a\)\/\(2  κ\) - \ \(sin\ 2\ \(k\_0\) a\)\/\(2  k\_0\))\)\ e\^\(\(-Γt\)/\)\)\(\ \)\(+\)\(\ \)\)\)\(TraditionalForm\`\(2  π\ \ k\_0\)\/m\ \((cosh\ 2  κ\ a - \ cos\ 2\ \(k\_0\) a)\)\ e\^\(\(-Γt\)/\)\  = \ 0\)\(TraditionalForm\`\)
	⟺  
	sinh 2 κ a - (κ/(Subscript[k, 0]))sin 2 (Subscript[k, 0]) a = cosh 2  κ a -  cos 2 (Subscript[k, 0]) a .

To demonstrate this last equality, we must use the condition we developed when finding this solution to the Schrodinger equation, i.e. the condition from matching the discontinuity in derivatives at r = a with the delta function singularity.  Equivalently, this is the condition for having a pole in the scattering amplitude:  \(TraditionalForm\`e\^\(2\ i\ k\ a\)\) =  1 - 2  i \(\(\^2\ k\)\/\(2  mγ\)\).  Plugging in k  =  (Subscript[k, 0])  -  i κ , the real and imaginary parts of this equation read

(e^(2  κ a)) cos 2 (Subscript[k, 0]) a =  1 - \(TraditionalForm\`\(\^2\ κ\)\/mγ\) 	  and 	   (e^(2  κ a))sin 2 (Subscript[k, 0]) a = - \(TraditionalForm\`\(\^2\ k\_0\)\/mγ\).

Plugging these expressions into our condition  

sinh 2 κ a - (κ/(Subscript[k, 0]))sin 2 (Subscript[k, 0]) a = cosh 2  κ a -  cos 2 (Subscript[k, 0]) a 

gives 

sinh 2 κ a  + \(\(\^2\ κ\)\/mγ\)(e^(-2 κ a))= cosh 2  κ a- (1 - \(TraditionalForm\`\(\^2\ κ\)\/mγ\)) (e^(-2 κ a))

or 

sinh 2 κ a  = cosh 2  κ a - (e^(-2 κ a)), 

which does in fact hold.