# 401: Particle in a Slanted Well Potential

## Numerical Solutions for Schrödinger's Equation for the Particle in the Slanted Box

Parameters go here: $$x_{max} = 1$$ $$\mu = 1$$ $$V_{0} = 2$$

Potential energy $$V(x) = V_{0} x$$

Given

$\frac{-1}{2 \mu} \frac{d^{2}}{dx^{2}} \psi (x) + V(x) \psi (x) = E \psi (x)$

with these boundary conditions: $$\psi (0) = 0$$ and $$\psi '(0) = 0.1$$

$$\psi = Odesolve(x, x_{max})$$ Normalize wavefunction: $$\psi (x) = \frac{ \psi (x)}{ \sqrt{ \int_{0}^{x_{max}} \psi (x)^{2} dx}}$$

Enter energy guess: E = 5.925

Calculate most probably position: x = 0.5 Given $$\frac{d}{dx} \psi (x) = 0$$ Find (x) = 0.485

Calculate average position: $$X_{avg} = \int_{0}^{1} \psi (x) (x) \psi (x) dx$$ $$X_{avg} = 0.491$$

Calculate potential and kinetic energy:

$$V_{avg} = V_{0} X_{avg}$$ $$V_{avg} = 0.983$$

$$T_{avg} = E - V_{avg}$$ $$T_{avg} = 4.942$$