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Definition of Delta as a Sum

  1. Jul 31, 2008 #1
    Is it true that [tex] \sum_x e^{i(k-k')x} = \delta_{k-k'} [/tex], where [tex]\delta[/tex] is the Kronecker delta? I've come across a similar relation for the Dirac Delta (when the sum is an integral). I do not understand why [tex]k-k' \neq 0 [/tex] implies the sum is zero.

    Edit: In fact, I'm really confused, since it seems that when the [tex]x=0...\inf[/tex] and k=k' the sum is infinite. So is it a Dirac delta?
    Last edited: Jul 31, 2008
  2. jcsd
  3. Jul 31, 2008 #2


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    In that sum, x should be an integer, so it is usually written as n.
    The sum should be from n=-N to n=+N, and divided by 2N+1.
    Then in the limit N-->infinity, it is the Kronecker delta.
  4. Jul 31, 2008 #3


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    There are really two different formulas here. For [itex]x[/itex] real (not necessarily an integer), we have

    \sum_{k=-\infty}^{+\infty}e^{ikx}=2\pi\sum_{n=-\infty}^{+\infty}\delta(x-2\pi n)

    where [itex]\delta(x)[/itex] is the Dirac delta function.

    For [itex]n[/itex] an integer,

    \sum_{k=1}^{N}e^{2\pi ikn/N}=N\delta_{n0}

    where [itex]\delta_{nm}[/itex] is the Kronecker delta. To get an idea of why this sums to zero when [itex]n\ne 0[/itex], consider the case [itex]N=4[/itex]; then the four numbers being summed are [itex]i^n, (-1)^n, (-i)^n, 1^n[/itex].
    Last edited: Jul 31, 2008
  5. Aug 5, 2008 #4
    Thank you both, I've written these handy formulas for my future reference.
  6. Aug 5, 2008 #5


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    [tex]N\delta_{n0}[/tex] should have been [tex]N\delta_{n\mathop{\rm mod}N,0}[/tex]
    That is, [itex]N[/itex] if [itex]n=0,\pm N,\pm 2N, \ldots,[/itex] and zero otherwise.
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