- #1
pellman
- 684
- 5
I always tended to think that we ought to use formulae which explicitly remind us that position and momentum are on equal footing in quantum theory (even though this may not be ultimately true) and write my transforms symmetrically
[tex]f(x)=(2\pi)^{-3/2}\int{F(p)e^{ix\cdot p}d^3p[/tex]
[tex]F(p)=(2\pi)^{-3/2}\int{f(x)e^{-ix\cdot p}d^3x[/tex]
However, I frequently see the convention of putting the 2-pi's with the momentum volume element, that is,
[tex]f(x)=\int{F(p)e^{ix\cdot p}\frac{d^3p}{(2\pi)^3}[/tex]
[tex]F(p)=\int{f(x)e^{-ix\cdot p}d^3x[/tex]
Here is an example. Folland here is acquainting mathematicians with physics conventions and explicitly states this convention.
http://books.google.com/books?id=3J...O8gN0J&sa=X&oi=book_result&ct=result&resnum=1
but is the there any reason behind this convention?
[tex]f(x)=(2\pi)^{-3/2}\int{F(p)e^{ix\cdot p}d^3p[/tex]
[tex]F(p)=(2\pi)^{-3/2}\int{f(x)e^{-ix\cdot p}d^3x[/tex]
However, I frequently see the convention of putting the 2-pi's with the momentum volume element, that is,
[tex]f(x)=\int{F(p)e^{ix\cdot p}\frac{d^3p}{(2\pi)^3}[/tex]
[tex]F(p)=\int{f(x)e^{-ix\cdot p}d^3x[/tex]
Here is an example. Folland here is acquainting mathematicians with physics conventions and explicitly states this convention.
http://books.google.com/books?id=3J...O8gN0J&sa=X&oi=book_result&ct=result&resnum=1
but is the there any reason behind this convention?