# 400: Particle in a Semi-infinite Potential Well

## 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

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 $$\Phi (p) = \int_{ -x_{min}}^{ x_{max}} exp(-1 p x) \psi (x)~dx$$