Physics 411 (Emanuel Gull ): Midterm III

Relaxation Method

Consider the potential in a square domain [0, 1] × [0, 1], where the boundary at x = 0 is held at constant potential Φ = 0, the boundary at x = 1 is held at constant potential Φ = 1, the boundary y = 0 follows the functional form Φ = x², and the boundary y = 1 the form Φ = x⁴ − x² + x.

1. Solve the Laplace equation ∇²Φ = 0 for this problem via relaxation. Choose a discretization ∆x = 0.1 and iterate Φ until convergence.

2. Plot the potential in [0, 1] × [0, 1] as a contour plot. Label some of the contours.

3. Compute the electric field at (0.5, 0.5) with centered finite differences.

Bound states in a potential well

A potential is given by V (x) = 0 for x < 0 and x > 1. For 0 < x < 1, the potential is given by

V (x) = c(x² − x),

where c characterizes the depth of the potential.

        Find the solutions of the 1D Schrodinger equation with E < 0 as a function of c, with m = 1 and h = 1. Note that for c = 1 you recover last week’s homework problem. As you increase c, additional bound states will appear. Plot the energy of the lowest four bound states as a function of c, from c = 1 until you have at least four bound states.

        Additional helpful information:

1. Start from your implementation of the Numerov method and the bisection from last time.

2. In order to start the Numerov IVP, use the analytically known exponential decay in the ‘forbidden’ regions x < 0 and x > 1

3. The ground state will have no node. The first excited state will have one node, the second excited state two nodes, etc. The ground state will have the lowest energy, the 1st excited state the next higher eigen-energy, etc.

4. It is easier to choose c large enough that four bound states exist, and reduce c towards 1, than to increase c from 1 and find the place where additional bound states start to appear.