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Kronecker Delta

  1. Apr 19, 2007 #1
    the Kronecker Delta function is
    What does it mean when the subscript is not i,j but i+j?
     

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  3. Apr 19, 2007 #2
    where did you see this with i+j?
     
  4. Apr 19, 2007 #3
    It was in an assignment.
    I wanted to post the whole question but have no idea how to use latex that well (only basic).
    If you can get this, it was:
    Show that:
    Integrate[e^(ix(m+n)),{x,0,2pi}] = 2pi*delta(m+n)


    the LHS should vaguely resemble a Mathematica input and the RHS (m+n) should be subscript.
    I thought it might be a mistake in his notes but in the lecture he made no correction.
    We haven't learnt this function, but this is a 2nd year course on Mathematical Methods and the lecturer is in love with Mathematica. If it were a person I reckon he would marry it. so knowledge of the function wasn't rea;;y necessary.
     
  5. Apr 19, 2007 #4
    In this case, it means 1 when (m+n) = 0 and 0 otherwise.
     
  6. Apr 19, 2007 #5

    HallsofIvy

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    Integrate[e^(ix(m+n)),{x,0,2pi}] = 2pi*delta(m+n)
    [tex]\int_0^{2\pi} e^{ix(m+n)} dx= 2\pi \delta_{m+n}[/tex]
    (click on the equation to see the code)

    If m+n is not 0, then the integral is
    [tex]-\frac{i}{m+n}e^{ix(m+n)}[/tex]
    evaluated from 0 to [itex]2\pi[/itex]. But [itex]e^{ix(m+n)}[/itex] is 0 at both 0 and [itex]2\pi[/itex] so the integral is 0.

    If m+n= 0 then the integral is
    [tex]\int_0^{2\pi}dx= 2\pi[tex]

    Yep, it looks like that "delta" should be "1 if m+n= 0, 0 otherwise".
     
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