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Math Help - Normalization

  1. #1
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    Normalization

    soooo I need to normalize this

    wavefunction (x) = Ne^[(-x^2)/(2a^2)]

    A is a constant, and x goes from neg. to pos. infinite..

    I keep doing it over and over.. trying, but now I'm here.. sigh. Our professor is not doing a good job and to make matters worse, hes on vacation our first week in school (he lectured us once), lol.
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  2. #2
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    Quote Originally Posted by ryukolink View Post
    soooo I need to normalize this

    wavefunction (x) = Ne^[(-x^2)/(2a^2)]

    A is a constant, and x goes from neg. to pos. infinite..

    I keep doing it over and over.. trying, but now I'm here.. sigh. Our professor is not doing a good job and to make matters worse, hes on vacation our first week in school (he lectured us once), lol.
    \int_{- \infty}^{+ \infty} e^{- \frac{x^2}{2a^2}} \, dx = a \sqrt{2} \int_{- \infty}^{+ \infty} e^{-u^2} \, du = a \sqrt{2} \sqrt{\pi} (see Gaussian integral - Wikipedia, the free encyclopedia) = a \sqrt{2 \pi}.
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  3. #3
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    wow, I wish I had known about the gaussian integral..

    k so u-sub I did that as well, couldnt integrate from there, guess I am retarded lol.

    So this is the answer... (now going beyond the question)

    but to find the normalizing factor I would just Divide one by this answer making it

    N = 1 / [answer^(1/2)], I am correct in this assumption... I hope so cause I read it and took notes on it.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    \int_{- \infty}^{+ \infty} e^{- \frac{x^2}{2a^2}} \, dx = a \sqrt{2} \int_{- \infty}^{+ \infty} e^{-u^2} \, du = a \sqrt{2} \sqrt{\pi} (see Gaussian integral - Wikipedia, the free encyclopedia) = a \sqrt{2 \pi}.

    now wait a minute... we are taught that normalization is the complex conjugate of the wave function times the orig. wave function, why did you just plug in the original wave function and not multiply it out ? (obviously the integration and what not is correct.. but I got a little confused here when looking it over for a second time when I was actually not a zombie.
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  5. #5
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    Quote Originally Posted by ryukolink View Post
    now wait a minute... we are taught that normalization is the complex conjugate of the wave function times the orig. wave function, why did you just plug in the original wave function and not multiply it out ? (obviously the integration and what not is correct.. but I got a little confused here when looking it over for a second time when I was actually not a zombie.
    I asumed that you had already done that in what you posted and that your trouble was dealing with the resulting integral.

    See Normalization of the wave-function

    Otherwise what you posted makes no sense (to me anyway).

    Quote Originally Posted by ryukolink View Post
    wow, I wish I had known about the gaussian integral..

    k so u-sub I did that as well, couldnt integrate from there, guess I am retarded lol. Mr F says: The link I posted showed how to do it. Did you read it? Otherwise it's a result that can be assumed.

    So this is the answer... (now going beyond the question)

    but to find the normalizing factor I would just Divide one by this answer making it

    N = 1 / [answer^(1/2)], I am correct in this assumption... I hope so cause I read it and took notes on it.
    ..
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  6. #6
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    just forget it, I got it heh

    thanks for the help, as always.
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