9.7: Particle in a Slanted Well Potential

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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

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

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

Screen Shot 2019-02-06 at 7.03.20 PM.png

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:

\( V_{avg} = V_{0} X_{avg}\) \( V_{avg} = 0.983\)

\( T_{avg} = E - V_{avg}\) \( T_{avg} = 4.942\)