450: Numerical Solution for the Feshbach Potential

Parameters go here: x_max = 5 m = 1 V₀ = 2.5 \( \mu\) = 0 d = .5

Potential energy:

\[ V(x) = V_{0} tanh \left( \frac{x}{d}\right)^{2}\]

Given:

\[ \frac{-1}{2m} \left( \frac{d^2}{dx^2} \psi (x) \right) + V(x) \psi (x) = E \psi (x)\]

\[ \psi (-x_{max} = 0~~ \psi \left( -x_{max} \right) = 0.1\]

\[ \psi = Odesolve (x, x_{max})\]

Normalize wavefunction:

\[ \psi (x) = \frac{ \psi (x)}{ \sqrt { \int_{0}^{x_{max}} \psi (x)^2 dx}}\]

Enter energy guess: E = 1.44949