Jacobian for CoM coordinates

  • #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 :)
 

Answers and Replies

  • #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!)
 

Suggested for: Jacobian for CoM coordinates

Replies
2
Views
742
Replies
3
Views
420
Replies
3
Views
2K
Replies
3
Views
641
Replies
7
Views
2K
Replies
4
Views
2K
Back
Top