Half-Infinite Well: A half-infinite well has an infinitely high wall at the origin and one...

Half-Infinite Well: A half-infinite well has an infinitely high wall at the origin and one of finite height U0 at x = L. Like the finite well, it cannot be solved exactly, Here we find energies numerically for one whose width L is 2 and whose depth U0 is 5 in the simple system of units discussed in the above guidelines. The half-infinite well potential energy isn't symmetric about x = 0, but part of the symmetric-case discussion still applies. Because the left wall is infinite, ψ(x) must be 0 for x ≤ 0, so ψ(0) must be 0. Assuming a nonzero slope there, we can define ψ(Δx) to be 1. (In essence, we are keeping only the conditions that apply to odd functions, which must be 0 at the origin.) For ψ(x), the function 2.5*sign(x - 2) + 2.5 can be adapted to almost any computer and is correct for positive x, which is all that matters, (a) Using 0.001 for Δx, test different trial values of £by finding i// at all multiples of Δx out to x = 4 and plotting the results. Find two allowed energies, (b) What tells you that an energy is correct? (c) The "unsolvable" energy quantization condition for a half-infinite well is given in Exercise 40. Do your values satisfy it reasonably well?