10.34: Numerical Solution for the Feshbach Potential
- Page ID
- 136990
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)
Parameters go here: xmax = 5 m = 1 V0 = 2.5 \( \mu\) = 0 d = .5
Potential energy:
\[ V(x) = V_{0} tanh \left( \frac{x}{d}\right)^{2} \nonumber \]
Given:
\[ \frac{-1}{2m} \left( \frac{d^2}{dx^2} \psi (x) \right) + V(x) \psi (x) = E \psi (x) \nonumber \]
\[ \psi (-x_{max} = 0~~ \psi \left( -x_{max} \right) = 0.1 \nonumber \]
\[ \psi = Odesolve (x, x_{max}) \nonumber \]
Normalize wavefunction:
\[ \psi (x) = \frac{ \psi (x)}{ \sqrt { \int_{0}^{x_{max}} \psi (x)^2 dx}} \nonumber \]
Enter energy guess: E = 1.44949