9.6: Particle in a Semi-infinite Potential Well

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Numerical Solutions to Schrödinger's Equation for the Particle in the Semi-infinite Box

Parameters go here: \( x_{min} = 0\) \( x_{max} = 5\) \( m = 1\) \( lb = 2\)

Potential energy \( V(x) = if[(x \geq lb), V_{0}, 0]\)

Given \( \frac{d^{2}}{dx^{2}} \psi (x) = 2 m (V(x) -E) \psi (x)\) \( \psi (x_{min}) = 0\) \( \psi '(0) = 0.1\)

\( \psi := Odesolve (x, x_{max})\) \( \psi = \frac{ \psi (x)}{ \sqrt{ \int_{x_{min}}^{x_{max}}} \psi(x)^{2} dx}\)

Enter energy guess: E = 0.766

Screen Shot 2019-02-06 at 1.39.31 PM.png

Calculate the probability that the particle is in the barrier: \(\int_{2}^{5} \psi (x)^{2}dx = 0.092\)

Calculate the probability that the particle is not in the barrier: \(\int_{0}^{2} \psi (x)^{2}dx = 0.908\)

Calculate and display the momentum distribution:

Fourier transform: p = -10,-9.9 .. 10 \( \Phi (p) = \int_{ -x_{min}}^{ x_{max}} exp(-1 p x) \psi (x)~dx\)

Screen Shot 2019-02-06 at 1.39.36 PM.png