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Math Help - Delta Function Well

  1. #1
    Senior Member bugatti79's Avatar
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    Delta Function Well

    I Have a query regarding the delta function well in quantum physics.

    As part of the process to determine the discontinuity of the derivative of the function psi at x =0 (see attached), the integral needs to be evaluated from -E to +E and then taking the limit E tends to 0.

    What i dont understand is how the RHS of eqn goes to zero. Could anyone explain this, the rest is fine.

    Thanks
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  2. #2
    A Plied Mathematician
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    The integrand on the RHS is continuous, right? So you're taking the area under a continuous curve, and the width of the area is getting narrower and narrower...
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  3. #3
    A Plied Mathematician
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    Another way of thinking about it is that a continuous function achieves its maximum on a closed interval. Hence,

    \displaystyle 0\le\left|\int_{-\epsilon}^{\epsilon}\psi(x)\,dx\right|\le\int_{-\epsilon}^{\epsilon}|\psi(x)|\,dx\le |\psi_{\max}|\int_{-\epsilon}^{\epsilon}dx=|\psi_{\max}|(2\epsilon)\to 0.
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  4. #4
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    As long as \psi has an anti-derivative (which is certainly true if it is continuous), say \phi, which is continuous, so that \int_{-\epsilon}^\epsilon \psi(x)dx= \phi(\epsilon)- \phi(-\epsilon) and then
    \displaytype\lim_{\epsilon\to 0} \int_{-\epsilon}^\epsilon \psi(x)dx= \phi(0)- \phi(0)= 0.
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  5. #5
    Senior Member bugatti79's Avatar
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    Thanks guys,

    I wanted to see how this goes to zero mathematically, see atached. I think it is similar to Hallsofivy definition. Is my approach look ok?

    THanks
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  6. #6
    A Plied Mathematician
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    A few comments:

    That's definitely overkill, but it gets the job done. You can get the last expression on the second-to-last line immediately by noticing that the integrand is even, and you're integrating over a symmetric interval.
    Last edited by Ackbeet; December 4th 2010 at 11:40 AM. Reason: Grammar.
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  7. #7
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    A few comments:

    That's definitely overkill, but it gets the job done. You can the last expression on the second-to-last line immediately by noticing that the integrand is even, and you're integrating over a symmetric interval.
    I see the interval is symmetric because the limits are from -E to +E.
    How is the integrand an even function? since psi (-x) not equal to psi (x).
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by bugatti79 View Post
    I see the interval is symmetric because the limits are from -E to +E.
    How is the integrand an even function? since psi (-x) not equal to psi (x).
    It is an even function and $$\psi(-x) does equal $$\psi(x), suppose $$x positive, \psi(x)=Be^{kx} and \psi(-x)=Be^{-k(-x)}=Be^{kx}

    CB
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  9. #9
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    It is an even function and $$\psi(-x) does equal $$\psi(x), suppose $$x positive, \psi(x)=Be^{kx} and \psi(-x)=Be^{-k(-x)}=Be^{kx}

    CB
    ah, ok. I was focusing on the one expression for for x>0 and substituting x and -x. Clearly I need to use the correct expression for each subsitution as you've shown. I see that now. Based on that I think you meant


    \psi(-x)=Be^{k(-x)}=Be^{-kx} and


    \psi(x)=Be^{-k(x)}=Be^{-kx}


    Thanks alot!
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  10. #10
    A Plied Mathematician
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    You're welcome for my contribution. Have a good one!
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