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. d²y (2.73) where Kis the energy, in units of (1/2) hw:

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Question related to Quantum Mechanics : Problem 2.55

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.
Transcribed Image Text: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.
d²y
(2.73)
where Kis the energy, in units of (1/2) hw:
Transcribed Image Text:d²y (2.73) where Kis the energy, in units of (1/2) hw:
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