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Jacobian for CoM coordinates

  1. Nov 18, 2014 #1
    I am computing matrix elements of a two body quantum-mechanical potential, which take the form

    [tex] V_{k l m n} = \int d^3 r_1 d^3 r_2 e^{-i k \cdot r_1} e^{-i l \cdot r_2} V( | r_1-r_2 | ) e^{i m \cdot r_1} e^{i n \cdot r_2} [/tex]

    To do this integral, I make the change of coordinates
    [tex] \overset{\rightarrow}{r} \equiv ( \overset{\rightarrow}{r}_1 - \overset{\rightarrow}{r}_2 ) / 2, \overset{\rightarrow}{R} \equiv ( \overset{\rightarrow}{r}_1 + \overset{\rightarrow}{r}_2 ) / 2 ,[/tex]

    which gives a momentum conserving delta function times the Fourier transform of the potential. This is exactly what as expected, but I am concerned that I am missing an overall Jacobian factor when I make the swap

    [tex] d^3 r_1 d^3 r_2 \rightarrow d^3 R d^3 r [/tex]

    I know how to get Jacobians for a single particle's coordinates, but for some reason I can't think straight about two particles. Can anyone provide guidance on this issue?

    thanks :)
     
  2. jcsd
  3. Nov 18, 2014 #2
    I figured it out...

    Suppose [itex] X \equiv (x + y)/2 , Y \equiv (x - y)/2 [/itex]. Then [itex] dx \wedge dy = (dX + dY) \wedge (dX - dY) = dY \wedge dX - dX \wedge dY = -2 dX \wedge dY [/itex]

    I am being sloppy, because I am planning on throwing out that minus sign; perhaps someone can enlighten me (I think it has to do with orientation; in any case area elements can't be negative -- that would just be nonsense!)
     
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