Problem 2.56 Find the first three excited state energies (to five significant digits) for the harmonic oscillator, by wagging the dog (Problem 2.55). For the first (and third) excited state you will need to set uſ[0] == 0, u'[0] == 1.) Problem 2.55 Find the ground state energy of the harmonic oscillator, to five significant digits, by the "wag-the-dog" method. That is, solve Equation 2.73 numerically, varying K until you get a wave function that goes to zero at large E. In Mathematica, appropriate input code would be Plot[ Evaluate[ u[x] /. NDSolve[ {w/[x] -(x² - K)*u[x] == 0, u[0] == 1, w[0] == 0}, u[x], {x, 0, b} ] ], {x, a, b}, PlotRange -> {c, d} (Here (a, b) is the horizontal range of the graph, and (c, d) is the vertical range-start with a = 0, b = 10, c = – 10, d = 10.) We know that the 2n + 1, so you might start with a “guess" of K = 0.9. Notice what the "tail" of the wave function does. Now try K = 1.1, and note correct solution is K = %3D that the tail flips over. Somewhere in between those values lies the correct solution. Zero in on it by bracketing K tighter and tighter. As you do so, you may want to adjust a, b, c, and d, to zero in on the cross-over point.

Problem 2.56 Find the first three excited state energies (to five significant digits) for the harmonic oscillator, by wagging the dog (Problem 2.55). For the first (and third) excited state you will need to set uſ[0] == 0, u'[0] == 1.) Problem 2.55 Find the ground state energy of the harmonic oscillator, to five significant digits, by the "wag-the-dog" method. That is, solve Equation 2.73 numerically, varying K until you get a wave function that goes to zero at large E. In Mathematica, appropriate input code would be Plot[ Evaluate[ u[x] /. NDSolve[ {w/[x] -(x² - K)*u[x] == 0, u[0] == 1, w[0] == 0}, u[x], {x, 0, b} ] ], {x, a, b}, PlotRange -> {c, d} (Here (a, b) is the horizontal range of the graph, and (c, d) is the vertical range-start with a = 0, b = 10, c = – 10, d = 10.) We know that the 2n + 1, so you might start with a “guess" of K = 0.9. Notice what the "tail" of the wave function does. Now try K = 1.1, and note correct solution is K = %3D that the tail flips over. Somewhere in between those values lies the correct solution. Zero in on it by bracketing K tighter and tighter. As you do so, you may want to adjust a, b, c, and d, to zero in on the cross-over point.