# 10.34: Numerical Solution for the Feshbach Potential

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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

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