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Math Help - Fourier Integral and the Dirac Delta

  1. #1
    Forum Admin topsquark's Avatar
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    Fourier Integral and the Dirac Delta

    This is going to end up being a simple question with a simple answer, but I just can't tweak the answer out of my Math Methods text.

    I have the following statement:
    \dot{\Delta}(\vec{x},0) = i \int \frac{d^3k}{2(2 \pi )^3} \left [ e^{i \vec{k} \cdot \vec{x}} + e^{-i \vec{k} \cdot \vec{x}} \right ] ( \vec{k} and \vec{x} are real 3-vectors and \vec{k} \cdot \vec{x} is the usual dot product.)

    I'm supposed to get that
    \dot{\Delta}(\vec{x},0) = -i \delta ^3 (\vec{x})

    But looking at the integral I'm thinking it ought to be:
    \dot{\Delta}(\vec{x},0) = \frac{i}{2} \left ( \delta ^3 (\vec{x}) + \delta ^3 (-\vec{x}) \right )
    which would give me 0, which obviously isn't true. So (sigh) what am I doing wrong?

    Thanks!
    -Dan
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    This is going to end up being a simple question with a simple answer, but I just can't tweak the answer out of my Math Methods text.

    I have the following statement:
    \dot{\Delta}(\vec{x},0) = i \int \frac{d^3k}{2(2 \pi )^3} \left [ e^{i \vec{k} \cdot \vec{x}} + e^{-i \vec{k} \cdot \vec{x}} \right ] ( \vec{k} and \vec{x} are real 3-vectors and \vec{k} \cdot \vec{x} is the usual dot product.)

    I'm supposed to get that
    \dot{\Delta}(\vec{x},0) = -i \delta ^3 (\vec{x})

    But looking at the integral I'm thinking it ought to be:
    \dot{\Delta}(\vec{x},0) = \frac{i}{2} \left ( \delta ^3 (\vec{x}) + \delta ^3 (-\vec{x}) \right )
    which would give me 0, which obviously isn't true. So (sigh) what am I doing wrong?

    Thanks!
    -Dan
    Do you think this might be related to:

    <br />
\delta(x)=\delta(-x)\ ?<br />

    and that you are using the other sense convention for what is the forward FT

    RonL
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Do you think this might be related to:

    <br />
\delta(x)=\delta(-x)\ ?<br />

    and that you are using the other sense convention for what is the forward FT

    RonL
    Okay, I see what you are saying about the convention. And (duh!) I forgot that \delta is even. I knew it was something simple! Thanks.

    -Dan
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