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One-Dimensional Quantum Scattering; Unbound Particles

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The Problem

The table below shows the dependence of the reflection coefficients on scattering energy. Determine the one-dimensional scattering potential. Use $m=1=\hbar$.

$R$	$E$
$1.00000$	$1.8$
$1.00000$	$1.9$
$1.00000$	$2.0$
$0.52319$	$2.1$
$0.39900$	$2.2$
$0.32312$	$2.3$
$\vdots$	$\vdots$
$0.11591$	$3.0$
$0.10260 $	$3.1$
$0.09107$	$3.2$
$\vdots$	$\vdots$
$0.01626$	$4.8$
$0.01469$	$4.9$
$0.01328$	$5.0$

The Solution

Using the energies and reflection coefficients above, the potentials $V$ corresponding to each $R$ and $E$ can be found using the step by step computational procedure below. \begin{align} &k_1 = \sqrt{ \frac{2mE}{\hbar} } \\\ &k_2 = k_1 \left( \frac{1-\sqrt{R}}{1+\sqrt{R}} \right) \\\ &V = E - \frac{\left(\hbar k_2\right)^2}{2m} \end{align}

After all the potentials have been found, assume equal position spacing between measurements; plot $V$ vs $x$.

It appears to be some polynomial function. Try fitting a polynomial; I used Python's curve_fit method from the SciPy optimize library. This method uses non-linear least squares to fit a function [1].

def func(x, a, b, c, d, l, m, n):
    return a*x**6 + b*x**5 + c*x**4 + d*x**3 + l*x**2 + m*x + n

popt, pcov = curve_fit(func, x, V)

func(x, *popt) is then the ydata to plot against $x$.

So a proper approximate potential to fit the experimental data is: \begin{equation} V(x) = \begin{cases} ax^6 +bx^5+cx^4 +dx^3 +lx^2 + mx + n \qquad 0 \leq x \leq 1 \\\ \!{1.8} \hphantom{l+bx^5+cx^4 +dx^3 +lx^2 + mx + n} \qquad \text{otherwise} \end{cases} \end{equation}

References

$\hphantom{++}$[1] https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.curve_fit.html.