1. ## Quantum Physics

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.
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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 $\displaystyle \Psi_{\alpha} (x)$ 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: $\displaystyle \sqrt{\frac{2}{L}}sin \left ( \frac {n \pi x}{L} \right )$ where n spans the positive integers. So:
$\displaystyle \Psi_{\alpha} (x)=\sum_{n=1}^{\infty} c_n \sqrt{\frac{2}{L}}sin \left ( \frac {n \pi x}{L} \right )$
where the $\displaystyle c_n$ are complex in general.

The x=L/2 supposition gives us the following conditions for all integer k:
$\displaystyle \left ( \frac{L}{2} \right ) ^k = \int_0^L dx \, x^k \left | \Psi_{\alpha} (x) \right | ^2$

From basic Quantum Mechanics we have the following condition as well:
$\displaystyle \sum_{n=1}^{\infty} \left | c_n \right | ^2 =1$

I need to find either the coefficients $\displaystyle c_n$, or more generally:
$\displaystyle \left | \int_0^L dx \, \Psi_{\alpha}^* \sqrt{\frac{2}{L}}sin \left ( \frac {n \pi x}{L} \right ) \right | ^2$

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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 $\displaystyle c_n$ 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 $\displaystyle c_n$ 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!"

Thanks!
-Dan

By the way. One of the deficiencies of my education is that I know only one type of (continuous) function $\displaystyle \Psi_{\alpha} (x)$ 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.

2. Hmmmm.... It would seem that everyone is either getting nowhere or is too scared to even try it. Well, that makes me feel a bit better about not getting it myself!

-Dan

3. Well....

Substitute $\displaystyle \Psi^*_{\alpha} (x)=\sum_{n=1}^{\infty} c_n^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right )$ into the expression $\displaystyle \left | \int_0^L dx \, \Psi_{\alpha}^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right ) \right|$, the convergence being justified by the orthogonality of the system $\displaystyle \left\{ \left(\frac{2}{L}\right)^{\frac{1}{2}}\sin \left ( \frac {n \pi x}{L} \right )\right\}$ in [0,L].

A few calculations later, we get $\displaystyle \left | \int_0^L dx \, \Psi_{\alpha}^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right ) \right|=\ldots=|c_n|.$

Hope that helps .

By the way. Your function will not be a Gaussian in the usual sence, but something quite like it, and also vanishing at the endpoints.

4. Originally Posted by Rebesques
Well....

Substitute $\displaystyle \Psi^*_{\alpha} (x)=\sum_{n=1}^{\infty} c_n^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right )$ into the expression $\displaystyle \left | \int_0^L dx \, \Psi_{\alpha}^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right ) \right|$, the convergence being justified by the orthogonality of the system $\displaystyle \left\{ \left(\frac{2}{L}\right)^{\frac{1}{2}}\sin \left ( \frac {n \pi x}{L} \right )\right\}$ in [0,L].

A few calculations later, we get $\displaystyle \left | \int_0^L dx \, \Psi_{\alpha}^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right ) \right|=\ldots=|c_n|.$

Hope that helps .

By the way. Your function will not be a Gaussian in the usual sence, but something quite like it, and also vanishing at the endpoints.
Thank you for the attempt. Unfortunately I already knew that. What I need to know is the actual value of the integral and as I don't know what the cn's are I can't leave the result that way.

In other words I need some clever way to evaluate the integral such that the result doesn't depend explicitly on the cn's or a way to find what the cn's are using the x^k integral conditions.

-Dan

5. Hmmm...

I think the integral is definately going to depend on the c's.

You probably cannot tell more than the expression
$\displaystyle c_n=\left(\frac{2}{L}\right)^{\frac{1}{2}}\int_{0} ^{L} \Psi_{\alpha} (x)\sin \left ( \frac {n \pi x}{L} \right )dx.$

Let us ponder on that...

6. Originally Posted by Rebesques
Hmmm...

I think the integral is definately going to depend on the c's.

You probably cannot tell more than the expression
$\displaystyle c_n=\left(\frac{2}{L}\right)^{\frac{1}{2}}\int_{0} ^{L} \Psi_{\alpha} (x)\sin \left ( \frac {n \pi x}{L} \right )dx.$

Let us ponder on that...
Well, the problem is that my series techniques have never been that advanced. I can work out any number of conditions on sums involving the cns using the x^k integrals, but I can't seem to find a way to use them to get an explicit expression for the cn out of them. I was hoping that someone here has that kind of expertise and guide me, or show me a clever way to get around using them. (sigh)

It's only a homework problem, and a self-imposed one at that, so no big deal if I can't find a way to solve it but my curiosity is up about this one.

Thanks!
-Dan