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Chemistry LibreTexts

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:

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

    Screen Shot 2019-02-15 at 12.10.51 PM.png

    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

    Screen Shot 2019-02-15 at 12.10.56 PM.png