Question: A one-dimensional double-well potential has V = for x < -1/2 l, V = 0 for -1/2 l x -1/4 l, V

A one-dimensional double-well potential has V = ∞ for x < -1/2 l, V = 0 for -1/2 l ≤ x ≤ -1/4 l, V = V0 for -1/4 l < x < 1/4 l, V = 0 for 1/4 l ≤ x ≤ 1/2 l, and V = ∞ for x > 12 l, where l and V0 are positive constants. Sketch V. Use the Numerov method to find the lowest four eigenvalues and the corresponding unnormalized eigenfunctions for the following values of V0 / (h2/ml2):
(a) 1;
(b) 100;
(c) 1000. Compare the wave functions and energies for (a) with those of a particle in a box of length l, and those for (c) with those of a particle in a box of length 1/4 l.
Use either a program similar to that in Table 4.1, a spreadsheet, or a computer-algebra system such as Mathcad. If negative eigenvalues are being sought using Excel 2010, you must uncheck the Make Unconstrained Variables Non-Negative box in the Solver Parameters box.

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a The potentialenergy function is As in the particle in a box Prob 430 E r Eh 2 ml 2 x r xl x r goes ... View full answer

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