Problem Set Solutions for Phys 3810, Spring 2009 - Hint-o-licious Hints, Assignments of Quantum Mechanics

Download Problem Set Solutions for Phys 3810, Spring 2009 - Hint-o-licious Hints and more Assignments Quantum Mechanics in PDF only on Docsity! Phys 3810, Spring 2009 Problem Set #4, Hint-o-licious Hints 1. Griffiths, 4.2 Show that the energies are given by E(nx.ny ,nz) = π2h̄2 2ma2 (n2x + n 2 y + n 2 z) The latter part of the problem is a math–puzzle sort of thing where you figure out how many ways one can get the same E with different sets of integers. You can find the values and the degeneracies from E1 up to E14, but if you run out of patience, I’ll tell you that E14 = π2h̄2 2ma2 (27) So what’s different about this value from the first few? 2. Griffiths, 4.38 The separated solution is ψ(r) = X(x)Y (y)Z(z) Note that when you write out the Schrödinger equation in cartesian coordinates for this potential with the separated solution (substitute and then divide by ψ = XY Z) it becomes a sum of three terms each of which depends only on x y or z. This means that you can write three separate Schrödinger equations, i.e. − h̄2 2m 1 X d2X dx2 + 1 2 mω2 = Ex , etc. which we know how to solve. (Be as clear as you can about this part.) The total energy is just sum of energies from each separate Schrödinger equation: E = Ex + Ey + Ez = h̄ω(nx + ny + nz + 3 2 ) The second part involves more thought. The energy of the oscillator state just depends on n ≡ nx + ny + nz for nx = 0, 1, 2, . . . etc. so the question is how many ways can you get n from the three separate indices. That’s a sort of puzzle–math problem. First, see if you can spot the pattern for the lowest n’s. (To give you one: n = 3 has 10 possible states.) 3. Griffiths, 4.8 Test that the u(r) radial solution u1(r) = Arj1(kr) really does solve Eq. (4.41). 1