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Fourier expansion

  1. Oct 31, 2006 #1
    I'm used to use

    [tex]\tilde{f} (x) = a_n|e_n>[/tex]

    where

    [tex]|e_n> = e^{2 \pi inx / L}[/tex]

    and

    [tex]a_n = \frac{1}{L}<e_n|f>[/tex]

    for my Fourier expansions.

    How do I expand a function in 3 dimensions, for example

    [tex]V(\vec{r}) = \frac{e^{-\lambda r}}{r}[/tex]

    ?
     
  2. jcsd
  3. Oct 31, 2006 #2

    quasar987

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    There is still just one variable in there, no?
     
  4. Oct 31, 2006 #3

    quasar987

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    ---I deleted this, it's mostly nonsense and it does't apply to the problem.---

    But in my opinion, mathematically, it makes no difference if you have an r or an x in there; just do the fourier expansion btw r_0 and r_1 as you would a fct of x.
     
    Last edited: Oct 31, 2006
  5. Oct 31, 2006 #4

    OlderDan

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    The basis functions are products whose factors are the 1-D functions in x, y, and z.
     
  6. Oct 31, 2006 #5
    So [tex]|e_n> = e^{2 \pi in \vec{r} /L} = e^{2 \pi in x /L} e^{2 \pi in y /L} e^{2 \pi in z /L}[/tex] and [tex]r = \sqrt{x^2 + y^2 + z^2}[/tex]?
     
  7. Oct 31, 2006 #6

    OlderDan

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    Yes for the last product on the right and the r; the Ls could be different for each dimension.

    [tex]|e_n> = e^{2 \pi in x /L_x} e^{2 \pi in y /L_y} e^{2 \pi in z /L_z}[/tex]
     
    Last edited: Oct 31, 2006
  8. Oct 31, 2006 #7
    So to expand [tex]V(\vec{r})[/tex] I have to rewrite it in terms of x, y ,z or does quasar987 have a point there?
     
  9. Oct 31, 2006 #8

    OlderDan

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    I thought you wanted the expansion for any function in 3-D. If the function is only a function of r, then you could do a 1-D expansion in r. There are other orthogonal functions that are often used in 3-D in cylindrical or spherical coordinates.
     
  10. Oct 31, 2006 #9
    Yes, first of all I want to solve this problem but I also want to learn something from it. ;)
     
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