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Wave Function Solution

  1. Feb 23, 2008 #1
    [SOLVED] Wave Function Solution

    1. The problem statement, all variables and given/known data

    An electron is found to be in a state given by the wave function [​IMG]

    Find the value of A.

    2. Relevant equations

    The normalization of the wave function: [​IMG]

    where [tex]\psi[/tex]* is found by replacing i with -i.

    3. The attempt at a solution

    Since the wave function is completely real, the normalization is simply the wavefunction squared, then [​IMG]

    I know that the integral of the normalization from negative infinity to infinity is equal to one, when integrated with respect to dx, but I have no idea as to how to solve this integral and find A. Is there a simplification that I am missing, or is it just a really difficult integral?

    This is my first time posting with an equation editor, so if the programming does not come out I will repost with pictures. Thanks so much for your help.
     
    Last edited: Feb 23, 2008
  2. jcsd
  3. Feb 23, 2008 #2

    cepheid

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    No need for pictures. A simple correction to your syntax (you need to surround equations with [ tex ] ... [ /tex ] does the trick.

     
  4. Feb 23, 2008 #3

    cepheid

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    Just a correction to your terminology:

    [tex] |\psi|^2 [/tex]

    is not the "normalization" of the wavefunction. It is the modulus squared of the wavefunction. The normalization is not a "thing". It's a condition. When this condition is satisfied

    [tex] \int |\psi|^2 \, dx = 1 [/tex]

    we say that the wavefunction is normalized. Hence, the above equation is called the normalization condition. Stated in words, the normalization condition says that the integral of the modulus squared of the wavefunction over all space is unity.
     
  5. Feb 23, 2008 #4

    cepheid

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    Also remember that

    [tex] (e^a)^2 = e^{2a} [/tex]

    So your squared wavefunction should be

    [tex] |\psi(x)|^2 = A^2 e^{-2[(x-a)/2\epsilon]^2} [/tex]
     
  6. Feb 23, 2008 #5

    cepheid

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    A hint for solving the integral is that is you substitute y = x - a, it looks like you end up with a standard Gaussian integral.
     
  7. Feb 23, 2008 #6
    Thanks, I forgot to square the wavefunction psi of my last equation, so with simplification I get

    [tex] |\psi(x)|^2 = A^2 e^{-(x-a)^2/2\epsilon^2} [/tex]
     
  8. Feb 23, 2008 #7
    The standard Gaussian integral which cepheid stated is

    [tex]\int_{-\infty}^\infty e^{-y^2}\,d\,y=\sqrt\pi[/tex]
     
  9. Mar 24, 2008 #8
    Hello! It did turn out to be Gaussian integration, which we hadn't been taught yet. :) Thanks for your help!
     
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