- #1
maverick280857
- 1,789
- 4
Hi,
How is
[tex]\frac{1}{\displaystyle{\not}{P}-m+i\epsilon}-\frac{1}{\displaystyle{\not}{P}-m-i\epsilon} = \frac{2\pi}{i}(\displaystyle{\not}{P}+m)\delta(P^2-m^2)[/tex]
? This is equation (4-91) of Itzykson and Zuber (page 189). I know that
[tex]\frac{1}{x\mp i\epsilon} = \mathcal{P}\left(\frac{1}{x}\right) \pm i\pi\delta(x)[/tex]
But this doesn't seem to give the right hand side of the first equation above. What am I missing?
Thanks in advance!
How is
[tex]\frac{1}{\displaystyle{\not}{P}-m+i\epsilon}-\frac{1}{\displaystyle{\not}{P}-m-i\epsilon} = \frac{2\pi}{i}(\displaystyle{\not}{P}+m)\delta(P^2-m^2)[/tex]
? This is equation (4-91) of Itzykson and Zuber (page 189). I know that
[tex]\frac{1}{x\mp i\epsilon} = \mathcal{P}\left(\frac{1}{x}\right) \pm i\pi\delta(x)[/tex]
But this doesn't seem to give the right hand side of the first equation above. What am I missing?
Thanks in advance!