HW #5

1. Not-so-hard sphere

(a)

We solve the problem exactly.  The wave function is (Subscript[R, l])(
          r) = (Subscript[j, l])(k r) cos (Subscript[δ, l]) + (Subscript[n, l])(k r) 
            sin (Subscript[δ, l])for r > a, and (Subscript[R, l])(r) = (Subscript[j, l])((Sqrt[(k^2) - (K^2)]) r) for r < a at high energies k > K = (Sqrt[2  m V])/ℏ, .  Requiring the logarithmic derivatives to match,

Simplify[
      1/((Sqrt[(π/(2  k r))]) BesselJ[(((2  l + 1))/2), k r] 
                Cos[(Subscript[δ, l])] + (Sqrt[(π/(2  k r))]) 
                BesselY[(((2  l + 1))/2), k r] 
                Sin[(Subscript[δ, l])])
D[(Sqrt[(π/(2  
                        k r))]) BesselJ[(((2  l + 1))/2), k r] 
              Cos[(Subscript[δ, l])] + (Sqrt[(π/(2  k r))]) 
              BesselY[(((2  l + 1))/2), k r] Sin[(Subscript[δ, l])], 
          r]]

Simplify[
      D[(Sqrt[(π/(2 (Sqrt[ (k^2) - (K^2)]) r))]) BesselJ[(((2  l + 1))/2), (Sqrt[(k^2) - (K^2)]) r], r]/((Sqrt[(π/(2 (Sqrt[ (k^2) - (K^2)]) r))]) BesselJ[(((2  l + 1))/2), (Sqrt[(k^2) - (K^2)]) r])]

sol = 
      Solve[-(-k r BesselJ[-(1/2) + l, k r] cot + 
                    BesselJ[(1/2) + l, k r] cot + 
                    k r BesselJ[(3/2) + l, k r] cot - 
                    k r BesselY[-(1/2) + l, k r]  + 
                    BesselY[(1/2) + l, k r] + 
                    k r BesselY[(3/2) + l, k r] )/(2 r BesselJ[
                      (1/2) + l, k r] cot + 
                  2 r BesselY[(1/2) + l, 
                      k r] ) == -(-(Sqrt[(k^2) - (K^2)]) r BesselJ[-(1/2) + l, (Sqrt[(k^2) - (K^2)]) r] + 
                    BesselJ[
                      (1/2) + l, (Sqrt[(k^2) - (K^2)]) r] + (Sqrt[(k^2) - (K^2)]) r BesselJ[(3/2) + l, (Sqrt[(k^2) - (K^2)]) r])/(2 r BesselJ[
                    (1/2) + l, (Sqrt[(k^2) - (K^2)]) r]) /. {r -> a}, 
        cot]

sin2deltal = Simplify[(1/(1 + (cot^2))) /. sol[[1]]]

For K a = 3,

table1 = 
      Table[{l, 
          sin2deltal /. {a -> 1, k -> 30. , K -> 3. }}, {l, 0, 
          50}]

plot1 = ListPlot[table1]

sigma1 = 
      Sum[((4  π (2  l + 1))/(k^2)) 
            table1[[l + 1, 2]] /. {k -> 30. }, {l, 0, 
          50}]

table2 = 
      Table[{l, 
          sin2deltal /. {a -> 1, k -> 30. , K -> 10. }}, {l, 
          0, 50}]

plot2 = ListPlot[table2]

sigma2 = 
      Sum[((4  π (2  l + 1))/(k^2)) 
            table2[[l + 1, 2]] /. {k -> 30. }, {l, 0, 
          50}]

(b)

In the semi-classical formula, the term with the potential is obtained from that without the potential by the replacement k -> (Sqrt[(k^2) - (K^2)]).  Therefore, \(TraditionalForm\`δ\_l = \((\@\(\((k\^2 - K\^2)\) a\^2 - l\^2\) - 2  l\ arctan\ \@\(\((k\^2 - K\^2)\) a\^2 - l\^2\)\/\(\(\@\(k\^2 - K\^2\)\) a + l\))\) - \((\@\(\(k\^2\) a\^2 - l\^2\) - 
              2  l\ arctan\ \@\(\(k\^2\) a\^2 - l\^2\)\/\(k\ a + l\))\)\)
Note that the term should be dropped when the argument of the square root is negative because it implies there is no integration region.

table3 = 
      Table[Sin[ ((Sqrt[Max])[(a^2) ((k^2) - (K^2)) - (l^2), 0] - 2  l ArcTan[(Sqrt[Max])[(a^2) ((k^2) - (K^2)) - (l^2), 0]/(a (Sqrt[(k^2) - (K^2)]) + l)] - (Sqrt[Max])[(k^2) (a^2) - (l^2), 0] + 2  l ArcTan[(Sqrt[Max])[(k^2) (a^2) - (l^2), 0]/(a k + l)])]^2 /. {a -> 1, k -> 30. , K -> 3. }, {l, 
          0, 50}]

plot3 = 
      ListPlot[table3, PlotJoined -> True, 
        PlotStyle -> {RGBColor[1, 0, 0]}]

Compared to the exact result, it is quite close.

Show[plot1, plot3]

For the total cross section, compared to the exact result 0.140557,it is agian quite close, with about 5% error.

sigma3 = 
      Sum[((4  π (2  l + 1))/(k^2)) 
            table3[[l + 1]] /. {k -> 30. }, {l, 0, 50}]

((sigma3 - sigma1)/sigma1)

Now for K a = 10,

table4 = 
      Table[Sin[ ((Sqrt[Max])[(a^2) ((k^2) - (K^2)) - (l^2), 0] - 2  l ArcTan[(Sqrt[Max])[(a^2) ((k^2) - (K^2)) - (l^2), 0]/(a (Sqrt[(k^2) - (K^2)]) + l)] - (Sqrt[Max])[(k^2) (a^2) - (l^2), 0] + 2  l ArcTan[(Sqrt[Max])[(k^2) (a^2) - (l^2), 0]/(a k + l)])]^2 /. {a -> 1, k -> 30. , 
            K -> 10. }, {l, 0, 50}]

plot4 = 
      ListPlot[table4, PlotJoined -> True, 
        PlotStyle -> {RGBColor[1, 0, 0]}]

Compared to the exact result, it is quite close

Show[plot2, plot4]

For the total cross section, compared to the exact result 8.74545,it is agian quite close, with about 5% error.

sigma4 = 
      Sum[((4  π (2  l + 1))/(k^2)) 
            table4[[l + 1]] /. {k -> 30. }, {l, 0, 50}]

((sigma4 - sigma2)/sigma2)

(c)

The eikonal approximation in Sakurai relates the phase shift to \(TraditionalForm\`δ\_l = \(\(Δ(
            b)\)\( | \_\(b = l/k\)\)\)\) (7.6.24) where \(TraditionalForm\`Δ(
          b) = \(-\(m\/\(2  
                  k\ ℏ\^2\)\)\) \(∫\_\(-∞\)\%∞\( V(\@\(b\^2 + z\^2\))\) dz\)\).  In our case, we only need to know the distance the straight line with impact parameter b in Figure 7.5 (page 393) goes through the radius r, which is 2 (Sqrt[ (a^2) - (b^2)]).  Therefore, Δ(
          b) = -(m/(2  k (ℏ^2))) 
            V 2 (Sqrt[ (a^2) - (b^2)]) = -((K^2)/(2  
                    k)) (Sqrt[(a^2) - (b^2)]).  The result is hence (Subscript[δ, l]) = -((K^2)/(2  
                    k)) (Sqrt[(a^2) - (((l/k))^2)]) = -((K^2)/(2 
                     (k^2))) (Sqrt[(k^2) (a^2) - (l^2)]).

Expand the semi-classical formula,

Simplify[
      Series[((Sqrt[((k^2) - (K^2)) (a^2) - (l^2)]) - 
              2  l ArcTan[((Sqrt[((k^2) - (K^2)) (a^2) - (l^2)])/((Sqrt[(k^2) - (K^2)]) a + l))]) - ((Sqrt[(k^2) (a^2) - (l^2)]) - 
              2  l ArcTan[((Sqrt[(k^2) (a^2) - (l^2)])/(k a + l))]), {K, 
          0, 2}]]

PowerExpand[%]

This is precisely the result from the eikonal approximation.  

In general, the eikonal approximation is a simplified formula of the semi-classical result when the potential is weak compared to the kinetic energy.

(d)

The Born approximation says 
(f^((1))) = -((2  
                  m)/(ℏ^2)) (1/q) (∫_0^∞) 
              r V(r) sin q r dr
\(TraditionalForm\`\(\(=\)\(\(-\(\(2  
                    m\)\/ℏ\^2\)\) \(1\/q\) \(∫\_0\%a 
                r\ \(\(\(ℏ\^2\) K\^2\)\/\(2  m\)\) 
                sin\ q\ r dr\) = \(-\(K\^2\/q\)\) \(sin\ q\ a - q\ a\ cos\ q\ a\)\/q\^2\)\)\)

Integrate[r Sin[q r], {r, 0, a}]

Note that (q^2) = 2  (k^2)(1 - cos θ) and hence d cos θ = (1/(2  (k^2))) 
            d (q^2) = (q/(k^2)) d q.  The cross section is then

sigmaBorn = 
      Simplify[Integrate[
          2  π ((((K^2) ((-a q Cos[a q] + Sin[a q])/(q^3))))^2) (q/(k^2)), {q, 0, 2  k}]]

sigmaBorn /. {k -> 30, a -> 1, K -> 3. }

((% - sigma1)/sigma1)

This is very close, as good as 0.6%.

sigmaBorn /. {k -> 30, a -> 1, K -> 10. }

((% - sigma2)/sigma2)

This one has 100% error!  As K is increased, it goes beyond the validity of the Born approximation.  In comparison, the semi-classical formula remained very good.

2. Gaussian Wave Packet with Resonance

(a)

\(TraditionalForm\`\(d\/\@\(2  π\)\) \(∫\_0\%∞\(\( 
                  e\^\(\(-\((q - k)\)\^2\) d\^2/2\)\)(\(e\^\(i\ q\ r\)\) 
                  e\^\(2  i\ δ\) - \(\((\(-1\))\)\^l\) 
                  e\^\(\(-i\)\ q\ r\))\) \(e\^\(\(-i\)\ ℏ\ \(q\^2\) t/2 
                   m\)\) d\ q\)\)
Without the scattering, it is
\(TraditionalForm\`\(d\/\@\(2  π\)\) \(∫\_0\%∞\(\( e\^\(\(-\((q - k)\)\^2\) d\^2/2\)\)(
              e\^\(i\ q\ r\) - \(\((\(-1\))\)\^l\) 
                  e\^\(\(-i\)\ q\ r\))\) \(e\^\(\(-i\)\ ℏ\ \(q\^2\) t/2 
                   m\)\) d\ q\)\)
Assuming that the Gaussian is sufficiently narrow, we can extend the integration to -∞ to ∞.  The integrand is dominated where q = k, and we expand the exponent
±i q r - 
          i ((ℏ (q^2))/(2  m)) t = ±i k r - 
          i ((ℏ (k^2))/(2  m)) 
            t + i(± r - ((ℏ k)/m) t) (q - 
                k) + ((O(q - k))^2).
We use the notation v = ℏ k/m, the classical velocity.

Within this approximation, the incoming piece is 
(d/(Sqrt[2  π])) (e^(-i k r - 
              i ℏ (k^2) t/2  
                m)) (∫_(-∞)^∞) 
                (e^(-(((q - k))^2) 
                  (d^2)/2)) (e^(i(-r - v t) (q - k))) 
            d q
=(e^(-i k r - 
                i ℏ (k^2) t/2  m)) 
          (e^(-(((r + v t))^2)/2  (d^2)))
which is appreciable only if t < 0, while the outgoing piece is
(d/(Sqrt[2  π])) (e^(i k r - 
              i ℏ (k^2) t/2  
                m)) (∫_(-∞)^∞) 
                (e^(-(((q - k))^2) 
                  (d^2)/2)) (e^(i(r - v t) (q - k))) d q
=(e^(i k r - 
                i ℏ (k^2) t/2  m)) 
          (e^(-(((r - v t))^2)/2  (d^2)))
which is appreciable only if t > 0.

(b)

Around the resonance, the phase shift is well approximated by \(TraditionalForm\`e\^\(2  i\ \(\(δ\_l\)(q)\)\) = \(q - k\_0 - i\ κ\)\/\(q - k\_0 + i\ κ\)\), and hence the scattered wave is
\(TraditionalForm\`\(d\/\@\(2  π\)\) \(∫\_0\%∞\( 
                e\^\(\(-\((q - k)\)\^2\) 
                  d\^2/2\)\) \(e\^\(i\ q\ r\)\) \(\(\(-2\) 
                  i\ κ\)\/\(q - k\_0 + 
                  i\ κ\)\) \(e\^\(\(-i\)\ ℏ\ \(q\^2\) t/2  m\)\) 
            d\ q\)\).
First of all, assuming that the Gaussian is wider than the resonance, we substitute (Subscript[k, 0]) into q, and we extend the integral from -∞ to ∞,
(d/(Sqrt[2  π])) (e^(-((((Subscript[k, 0]) - k))^2) 
              (d^2)/4)) (∫_(-∞)^∞) 
                (e^(i q r)) ((-2 i κ)/(q - (Subscript[k, 0]) + 
                  i κ)) (e^(-i ℏ (q^2) t/2  m)) 
            d q.
We expand the phase factor around (Subscript[k, 0]) up to the first order,
q r  -  ((ℏ (q^2))/(2  m)) 
            t = (Subscript[k, 0]) r - ((ℏ ((Subscript[k, 0])^2))/(2  m)) 
            t + (r - ((ℏ (Subscript[k, 0]))/m) t) (q - 
                (Subscript[k, 0])) + ((O(q - (Subscript[k, 0])))^2),
\(TraditionalForm\`\(d\/\@\(2  π\)\) \(e\^\(\(-\((k\_0 - k)\)\^2\) 
              d\^2/4\)\) \(e\^\(i\ \(k\_0\) r - 
              i \(\( ℏ\ k\_0\^2\)\/\(2  m\)\) 
                t\)\) \(∫\_\(-∞\)\%∞\(\(\(-2\) 
                  i\ κ\)\/\(q - k\_0 + i\ κ\)\) 
            e\^\(\(i(r - ℏ\ \(k\_0\) t/m)\) \((q - k\_0)\)\)\ d\ q\)\)
The integral can be extended to the lower half plane if r - ((ℏ (Subscript[k, 0]))/m) t < 0, and we find
θ(
          v t - r) (d/(Sqrt[2  π])) (e^(-((((Subscript[k, 0]) - k))^2) 
              (d^2)/2)) (e^(i (Subscript[k, 0]) r - 
                i (( ℏ ((Subscript[k, 0])^2))/(2  m)) 
                  t))(-2 π i) (-2 i κ) 
        (e^(i(r - ℏ (Subscript[k, 0]) t/m) (-i κ)))
\(TraditionalForm\`\(\(=\)\(\(θ(
            v\ t - r)\) \(d\/\@\(2  π\)\) \(e\^\(\(-\((k\_0 - k)\)\^2\) 
                d\^2/2\)\) \(\(e\^\(i\ \(k\_0\) r - 
                  i \(\( ℏ\ k\_0\^2\)\/\(2  m\)\) 
                    t\)\)(\(-4\) π\ κ)\)\ \(e\^\(κ\ r\)\) \(\(e\^\(\(-ℏ\)\ \(k\_0\) κ\ t/m\)\)\(.\)\)\)\)\)

(c)

For plotting, I use k = (Subscript[k, 0]) = 1, κ = 0.1, ℏ = 1, m = 1, and d = 3.

rR = (E^(I k r - I ℏ (k^2) 
                  t/(2  m))) 
          (E^(-(((r - ℏ k t/m))^2)/(2  (d^2)))) - (((-1))^l) (E^(-I k r - I ℏ (k^2) t/(2  m))) 
          (E^(-(((r + ℏ k t/m))^2)/(2  (d^2)))) + 
        If[r < ((ℏ (Subscript[k, 0]))/m) 
              t, (d/(Sqrt[2  π])) (E^(-((((Subscript[k, 0]) - k))^2) 
                  (d^2)/2)) (E^(I (Subscript[k, 0]) 
                    r - I ℏ ((Subscript[k, 0])^2) 
                    t/(2  
                          m))) (-4 π κ) (E^(κ r)) (E^(-ℏ (Subscript[k, 0]) κ t/m)), 0]

Table[
      Plot[Abs[rR]^2 /. {k -> 1, 
            (Subscript[k, 0]) -> 1, κ -> 0.2, ℏ -> 1, 
            m -> 1, d -> 0.5, l -> 1}, {r, 0, 10}, 
        PlotRange -> {{0, 10}, {0, 1}}], {t, -8, 8, 0.5}]

I see the "delayed" piece with an exponential profile.  It is also interesting to see that the probability is scraped off from the "prompt" Gaussian peak by a destructive interference, which is used to create the delayed piece consistent with the probability conservation, with a prominent dip right after the prompt peak.

(d)

The exponential tail has the time dependence (e^(-ℏ (Subscript[k, 0]) κ t/m)) = 
        (e^(-t/2  τ)) with τ = m/(2  ℏ (Subscript[k, 0]) κ).  Here, there is a factor of two in the exponent because the probability is the square of the wave function (e^(-t/τ)).

On the other hand, the imaginary part of the energy at the pole is \(E = \(\(\(ℏ\^2\)(k\_0 - i\ κ)\)\^2\/\(2  m\) = \(\(ℏ\^2\)(k\_0\^2 - κ\^2)\)\/\(2  m\) - 
                  i\ \(\(ℏ\^2\) \(k\_0\) κ\)\/m\)\)=(Subscript[E, 0]) - i (Γ/2).  Therefore, \(TraditionalForm\`Γ = \(2 \( ℏ\^2\) \(k\_0\) κ\)\/m\).

I find Γ τ = ℏ.