450: Numerical Solution for the Feshbach Potential
- Page ID
- 136990
Parameters go here: xmax = 5 m = 1 V0 = 2.5 \( \mu\) = 0 d = .5
Potential energy:
\[ V(x) = V_{0} tanh \left( \frac{x}{d}\right)^{2}\]
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