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MathGroup Archive 2005

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Re: Mathematica can't calculate Fourier transform (Dirac mean position eigenfunction)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54988] Re: Mathematica can't calculate Fourier transform (Dirac mean position eigenfunction)
  • From: dh <dh at metrohm.ch>
  • Date: Wed, 9 Mar 2005 06:34:15 -0500 (EST)
  • References: <d0juvv$n6i$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hallo Jacob,
in order that a FourierTransform exists, the function must be absolute 
integrable. Now try to integrate ( (1 + k^2 + (1 + k^2)^(1/2) )^(-1/2) 
from zero to infinity and you will see that it is diverging.
Sincerely, Daniel

Jacob wrote:
> Hi, I'm attempting to use Mathematica to calculate a mean-position
> eigenfunction of the Dirac equation. To do so I need to evaluate
> Fourier transforms (from k-space to r-space) of wavefunctions dependent
> on:
> 
> ( (1 + k^2 + (1 + k^2)^(1/2) )^(-1/2)
> 
> where k is in units of the Compton wavevector.
> 
> Cell expression:
> 
> Cell[BoxData[
>     FractionBox["1",
>       SqrtBox[
>         RowBox[{"1", "+",
>           SuperscriptBox["k", "2"], "+",
>           SqrtBox[
>             RowBox[{"1", "+",
>               SuperscriptBox["k", "2"]}]]}]]]], "Output"]
> 
> 
> Mathematica is unable to evaluate the FT of the above (either Fourier
> sine transform or normal FT). Can anyone give any suggestions as to how
> I could evaluate it?
> 
> More specifically, I am making a reverse Foldy-Wouthuysen
> transformation of a mean-position eigenfunction in p-space, then
> transforming the result into r-space assuming spherical symmetry. The
> first component of the r-space eigenfunction is given by the Fourier
> sine transform of:
> 
> k ( 1 + (1 + k^2)^(-1/2) )^(1/2)
> 
> Cell[BoxData[
>     RowBox[{"k", " ",
>       SqrtBox[
>         RowBox[{"1", "+",
>           FractionBox["1",
>             SqrtBox[
>               RowBox[{"1", "+",
>                 SuperscriptBox["k", "2"]}]]]}]]}]], "Output"]
> 
> 
> Thanks for any help.
> 


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