Quantum Physics

**topsquark** · March 25th 2006, 05:04 AM

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

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

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

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

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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 $c_n$ 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 $c_n$ 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 $\Psi_{\alpha} (x)$ 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.

**topsquark** · March 27th 2006, 03:48 AM

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

**Rebesques** · March 28th 2006, 12:50 AM

Well....

Substitute $ \Psi^*_{\alpha} (x)=\sum_{n=1}^{\infty} c_n^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right ) $ into the expression $ \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 $\left\{ \left(\frac{2}{L}\right)^{\frac{1}{2}}\sin \left ( \frac {n \pi x}{L} \right )\right\}$ in [0,L].

A few calculations later, we get $ \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.

**topsquark** · March 28th 2006, 03:27 AM

Originally Posted by Rebesques

Well....

Substitute $ \Psi^*_{\alpha} (x)=\sum_{n=1}^{\infty} c_n^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right ) $ into the expression $ \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 $\left\{ \left(\frac{2}{L}\right)^{\frac{1}{2}}\sin \left ( \frac {n \pi x}{L} \right )\right\}$ in [0,L].

A few calculations later, we get $ \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

**Rebesques** · March 28th 2006, 05:12 AM

Hmmm...

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

You probably cannot tell more than the expression
$ 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...

**topsquark** · March 28th 2006, 07:17 AM

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
$ 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

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