Results 1 to 6 of 6

Math Help - Quantum Physics

  1. #1
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,927
    Thanks
    332
    Awards
    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.
    __________________________________________________ _________________________________________________
    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

    __________________________________________________ _________________________________________________
    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,927
    Thanks
    332
    Awards
    1
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    538
    Thanks
    11
    Well....

    Substitute <br />
\Psi^*_{\alpha} (x)=\sum_{n=1}^{\infty} c_n^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right )<br />
into the expression <br />
\left | \int_0^L dx \, \Psi_{\alpha}^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right ) \right| <br />
, 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 <br />
\left | \int_0^L dx \, \Psi_{\alpha}^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right ) \right|=\ldots=|c_n|. <br />



    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.
    Last edited by Rebesques; March 28th 2006 at 01:16 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,927
    Thanks
    332
    Awards
    1
    Quote Originally Posted by Rebesques
    Well....

    Substitute <br />
\Psi^*_{\alpha} (x)=\sum_{n=1}^{\infty} c_n^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right )<br />
into the expression <br />
\left | \int_0^L dx \, \Psi_{\alpha}^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right ) \right| <br />
, 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 <br />
\left | \int_0^L dx \, \Psi_{\alpha}^* \sqrt{\frac{2}{L}}\sin \left ( \frac {n \pi x}{L} \right ) \right|=\ldots=|c_n|. <br />



    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    538
    Thanks
    11
    Hmmm...

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

    You probably cannot tell more than the expression
    <br />
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.<br />

    Let us ponder on that...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,927
    Thanks
    332
    Awards
    1
    Quote 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
    <br />
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.<br />

    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quantum Physics
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: November 16th 2009, 07:19 PM
  2. A Physics [Quantum Theory] Problem... help please...
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: September 12th 2008, 07:15 AM
  3. how do you solve quantum physics equations example is included
    Posted in the Advanced Applied Math Forum
    Replies: 12
    Last Post: March 27th 2008, 07:05 PM
  4. how do solve quantum physics or physics equations
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: March 26th 2008, 04:33 AM
  5. Quantum Physics
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: February 24th 2008, 07:43 PM

Search Tags


/mathhelpforum @mathhelpforum