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How to fourier invert a plane wave

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Ok so got a solution to the Klein-Gordon equation and need to solve for a(k)

    [tex] \varphi(x) = \int \tilde{dk} \left[ a(\bold{k}) e^{ikx} + a^{\dagger}(\bold{k} ) e^{-ikx} \right] [/tex]

    [tex] \tilde{dk} = \frac{d^{3}k}{(2 \pi)^{3} 2 \omega} [/tex]

    The way it's done in Srednicki p.26 has me confused when taking the fourier transform of [itex] \varphi [/itex]

    [tex] \int d^3x e^{-ikx} \varphi(x) = \frac{1}{2\omega} a(\bold{k}) + \frac{1}{2\omega} e^{2i\omega t} a^{\dagger}(\bold{-k} ) [/tex]

    2. Relevant equations

    [tex] kx = \bold{k} \cdot \bold{x} - \omega t [/tex]


    3. The attempt at a solution


    [tex] \int d^3x e^{-ikx} \varphi(x) =\int d^3x e^{-ikx} \int \tilde{dk} \left[ a(\bold{k}) e^{ikx} + a^{\dagger}(\bold{k} ) e^{-ikx} \right] [/tex]


    [tex] = \int d^3x \int \tilde{dk} a(\bold{k}) + \int d^3x \int \tilde{dk} e^{-2kx} a^{\dagger}(\bold{k} ) [/tex]


    so the problem is how do these integrals with respect to dx and dk disappear?
     
  2. jcsd
  3. Mar 29, 2009 #2

    tiny-tim

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    dummy!

    Hi waht! :smile:

    It's a different k …

    you have ∫∫ eikx eik'x … for a dummy k' :wink:
     
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