I've been trying this one by the "brute force" method and have gotten essentially nowhere, so I was wondering if there is a more elegant way to approach it.
A particle is in a 1-D infinite potential well. It is a free particle over the interval [0,L]. Presume that we know the particle is in a state such that we know it is at position L/2 with certainty.
Given the structure of the potential we can expand this wavefunction using the orthonormal basis states: where n spans the positive integers. So:
where the are complex in general.
The x=L/2 supposition gives us the following conditions for all integer k:
From basic Quantum Mechanics we have the following condition as well:
I need to find either the coefficients , or more generally:
Now, the problem does go on to say that we can ignore such "trivial" issues such as convergence of the series, but I suspect there is only one solution for the anyway. (That could easily be wrong!) I can say, with certainty, that there are an infinite number of non-zero coefficients, so the series doesn't terminate (making the problem simple, if long.) I suspect as well that there are no non-zero coefficients. I'm pretty sure about this one.
I have tried to do this by coming up with conditions on the by working out the integrals for various values of k, but that approach doesn't appear to help much. I can't seem to separate the terms of the various infinite series such that they are of any use.
As Sherlock Holmes would say, it's quite a "three-piper!"
By the way. One of the deficiencies of my education is that I know only one type of (continuous) function that would produce a state of certain position x = L/2: a Gaussian normal centered on L/2. However we need the additional condition that the wavefunction goes to 0 at x = 0 and x = L, which a Gaussian won't do. Perhaps a tactic for solving the problem would be to apply various functions that DO have this property and see what happens with the expansion into the orthonormal basis, but I don't know how to construct such functions in general.