HW4_solutions_fa11

# Ok so does it solve the s e plug in and see here are

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that makes it easier? OK, so does it solve the S. E.? Plug in and see. Here are the derivatives: 2 2 2 2 2 2 2 2 2 2 2 3 2 3 2 2 2 3 5 1 2 2 3 2 2 x x x x d x x A e A e dx d x x A e A e dx   The S. E. then becomes: 2 2 2 2 2 2 2 2 2 2 2 2 3 3 2 2 0 2 2 2 2 0 2 2 2 3 5 1 3 2 2 2 2 2 2 2 x x x x m d x x x mE x m x A e A e A e A e m dx     or 2 2 2 2 0 2 4 2 2 3 2 m x m x E Treat this result as we did above: Choose the width of the Gaussian to remove the spatial dependence of this result. You get exactly the same width as for the ground state. Then substitute in the width to find: 1 0 3 2 E What do you know! We found the correct first excited state. 3. First example of quantum tunneling. (5 points) For the ground state of the harmonic oscillator, calculate the probability of finding the particle outside the well. To find the probability that a particle in t he SHO ground state is outside the well, we’d need to integrate the probability density in the region outside the well and see what things add up to give. A good starting point is to decide what you mean by ‘outside the well’, and the natural thing to use is the classical turning points. Once the energy is below the potential, at least on the basis of classical reasoning, the particle is outside (and should also have a classical probability of being found there of exactly zero).

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Physics 3220 Homework 4 Solutions Physics 3220 HW4_solutions.6 Fall 2011 For the ground state, the energy is given above, so you look for the region where: 2 2 0 0 1 1 2 2 E m x Or 0 x m  There’s that same length again! Now you can also see that the Gaussian wave function you used in Problem 2 is chosen to have a natural width that is just equal to the distance between the classical turning points. OK, so the probability of being outside the well, after inserting the normalization factor and noticing that the range of integration is symmetric, is: 2 2 0 0 * 2 2 0.16 x outside m m P dx e dx      The final evaluation is done by recognizing that the lower limit is also equal to the width of the Gaussian, and is evaluated numerically in Mathematica in terms of error functions. You have a roughly 16% chance of finding the particle outside! 4. The half harmonic oscillator. (15 points) Imagine the ‘Half Harmonic Oscillator’, that has the following potential:   2 2 0 0 1 0 2 x V x m x x a) Find the allowed eigen energies. (5 points) The full game here is to find solutions to the S. E. that solve the equation, are consistent with the boundary conditions, and then see what energies you predict. You knowledge of the simple harmonic oscillator functions should allow you to see that the antisymmetric solutions are just the ticket here: They are zero at x=0, so they match the infinite potential boundary, and they satisfy the harmonic oscillator S. E. in the remaining space. Therefore,
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