Problems are listed in approximate order of difficulty. A single dot (•) indicates straigh...

Problems are listed in approximate order of difficulty. A single dot (•) indicates straightforward problems involving just one main concept and sometimes requiring no more than substitution of numbers in the appropriate formula. Two dots (••) identify problems that are slightly more challenging and usually involve more than one concept. Three dots (•••) indicate problems that are distinctly more challenging, either because they are intrinsically difficult or involve lengthy calculations. Needless to say, these distinctions are hard to draw and are only approximate.

••• (Section 7.8) Consider the problem of a particle in a finite square well of depth U0, as described in Section 7.8. To solve this problem, use a coordinate system centered on the box, with the left edge of the well at x = −a/2 and the right edge at x = + a/2. From the symmetry of the potential, one can argue that the stationary states should have symmetric probability distributions, that is, |ψ1(x)|2 = |ψ1(−x)|2. This implies that solutions are either symmetric and satisfy ψ(x) = ψ(−x) or are antisymmetric and satisfy ψ(x) = −ψ(−x). Thus, with our choice of coordinates, the solutions within the well are either of the form cos(kx) (for n = 1, 3, 5, … ) or of the form sin(kx) (for n = 2,4, 6, … ). (a) For the case of the symmetric solutions, derive an equation relating k and α. [Hint: Using the boundary conditions that both ψ(x) and ψ′(x) are continuous at x = −a/2, you can produce two equations that relate k, α, and the arbitrary coefficients A and G in Section 7.8. By dividing these equations, you can eliminate the coefficients. The final equation you produce is a transcendental equation relating E and U0; it cannot be solved for E using elementary methods.] (b) Show that the equation derived in (a) produces the correct stationary-state energies in the limit U0 → ∞. (c) Using numerical techniques, find the ground-state energy E1 of a square well of depth U0 = 3E1(∞) where E1(∞) = π2ℏ2/2ma2 is the lowest energy of the infinite square well. [Hint: We know that E1 will be less than E1(∞), so write E1 = zE1(∞) to give an equation for z and look for a solution around z = 0.5. A simple trial-and-error search for the solution works surprisingly well.] (d) Repeat for U0 = 30E1(∞).