Changing to polar coordinates in an exponential

  • Thread starter physichu
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  • #1
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Hello :)
I dont get this integral (Peskin & Schroeder P. 27 )

##\int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{1 \over {{E_{\bf{p}}}}}{e^{i{\bf{p}} \cdot {\bf{r}}}}} = {{2\pi } \over {{{\left( {2\pi } \right)}^3}}}\int\limits_0^\infty {dp{{{p^2}} \over {2{E_{\bf{p}}}}}{{{e^{ipr}} - {e^{ - ipr}}} \over {ipr}}} ##.

Actualy I dont get the next stage iethr :(

## = {{ - i} \over {2{{\left( {2\pi } \right)}^2}r}}\int\limits_{ - \infty }^\infty {dp{{p{e^{ipr}}} \over {\sqrt {{p^2} + {m^2}} }}} ##.

Thanx in advence :)
 

Answers and Replies

  • #2
Dr. Courtney
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Between the first two, it appears as if they have integrated over the angular coordinates in a spherical coordinate sytem, leaving only the radial integral. There needs to be some justification about spherical symmetry to simply yield a factor of 2 pi by integrating over the angular coordinates.

Between the second and third expressions, they seem to be using E = p^2/2m and simplifying.
 
  • #3
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where do the other exponential and the ##ipr## denominator come from?
 
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