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In the last page, Daniel Faraday asked about the spring potential, specifically why we can look at the limit of x → infinity and have it mean anything (see the bottom of page 51). I have what I think is a good answer, but I'd like a second opinion on it.

The reason that we normally look at SHO potentials only with small displacements x is that the potential is always an approximation to something, whether that means a physical spring deforming at large x or looking at a small segment of a complicated potential. However, there is still a solution to the SHO potential going to very large x; it isn't the Schrödinger equation failing at large x or the spring potential failing at large x. Rather, the approximation that the potential we're working with is a SHO potential fails at large x. Therefore, observing the behavior of solutions to an SHO potential in the limit of large x can still be useful. As long as one doesn't assume that the system behaves the same way outside that limit (which Griffiths doesn't), there's no problem.

I rambled a bit, but hopefully, that was clear enough.

My understanding of Griffiths reasoning was that he was looking at the limit of x→ infinity only in order to get the basic form of the differential equation, which he used to work backwards and get the exact solution.

The only major issue I have with the SHO analytical solution is a seemingly-logical contradiction in the step from equation [2.72] to [2.74], culminating in equation [2.77] in conjunction with footnote 23: “Note that although we invoked some approximations to motivate Equation 2.77, what follows is exact.” –How can an approximation used to remove complexity in an equation then be solved to an “exact” solution of the original equation? Doesn't an “approximation” to the DE, by definition, require that the exact solution to the approximate DE be an approximate solution to the exact DE? Or is there something very obvious or very tricky I'm missing here?

Yuichi Even though 2.76 is approximate and a very good approximation only when <math>\xi\approx\pm\infty</math>, when you insert <math>h(\xi)</math> as in 2.77, we are looking for exact solutions again. The solution in the form of 2.77 was subbed in 2.72, not 2.74, to figure our what equation <math>h(\xi)</math> must satisfy (resulting in 2.78 - that's why “K” is back in the business). So when one finds solutions, <math>h(\xi)</math>, and when you combine them with 2.76, they are exact solutions of 2.72, not 2.74.

I was looking at problem 2.15, and I was wondering how would you go about solving this. I haven't even been able to come up with a way to start it. Any ideas?

The problem is asking about the ground state for the harmonic oscillator. So I would start with Equation 2.59. Also, since the problem is asking what the probability of finding the particle outside the classically allowed region I would try modifying 2.59 with the dimmensionless variable from Equation 2.71.

I would start by writing <math>E_{0} = \hbar\omega/2</math> and then substitute this for <math>E</math> in the given equation and solve for x. Given this and the probability density function the problem should be pretty straight forward.

I was just wondering about the discussion problem. When we extend the barrier from L to 2L does it matter where the particle is along the ground path?

The barrier is assumed to move fast enough that the wave function is unable to respond immediately. Since the wave function is a description of the probability of where the particle may be, it does not matter where the particle is specifically.