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Math Help - simple substitution problem

  1. #1
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    simple substitution problem

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  2. #2
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    Quote Originally Posted by transgalactic View Post
    Your link does not work...
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  3. #3
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    my link works kust press it
    maybe you have a popup blocker
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  4. #4
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    i used this property

    \frac{\partial}{\partial t}[ X(x)T(t) ] = X(x)[\frac{\partial}{\partial t}T(t)] because X(x) won't vary with t

    i applied it and still didnt get the needed expresion

    http://i50.tinypic.com/67uyc7.jpg
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  5. #5
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    your RHS is wrong,.. it should be - T(t) \frac{h^2}{2m}\frac{\partial^2}{\partial x^2}X(x)

    rearrange it and you'll get the result.
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  6. #6
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    Your equation is originally,
    i h\frac{\partial \Psi}{\partial t}= \frac{h^2}{2m}\frac{\partial^2\Psi}{\partial x^2}
    If you let \Psi= X(x)T(t), you have \frac{\partial\Psi}{\partial t}= X(x)\frac{dT}{dt} and \frac{dX}{dx}T(t) (since X and T are functions of a single variable you should not have " \partial ") so your equation becomes
    i h X(x)\frac{dT}{dt}= \frac{h^2}{2m}T\frac{d^2X}{dx^2}
    which is what you have. Now divide both sides by XT:
    \frac{i h}{T}\frac{dT}{dt}= \frac{h^2}{2m X}\frac{d^2X}{dx^2}
    the equation you want to get to.

    The importance of this, by the way, is the fact that the right side now depends only of x while the left side depends only on t. In order to be equal for all x and t, each must be a constant! That allows you to separate the partial differential equation into two ordinary differential equations,
    \frac{ih}{T}\frac{dT}{dt}= \lambda
    and
    \frac{h^2}{2m X}\frac{d^2X}{dx^2}= \lambda
    which are the same as
    ih \frac{dT}{dt}= \lambda T
    and
    \frac{h^2}{2m}\frac{d^2X}{dx^2}= \lambda X.

    Of course, you have to determine possible values for \lambda and you usually do that by looking at boundary conditions on X. Notice that \lambda is an "eigenvalue" for both differential operators.
    Last edited by HallsofIvy; December 9th 2009 at 08:44 AM.
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  7. #7
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    wow super
    THANKS
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