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$\newcommand{\fsl}[1]{#1\kern-0.4em\raise0.22ex\hbox{/}}$How can I simplify the difference of squares $p^2 - m^2$ in order to obtain $$\frac{p^2 - m^2}{\fsl{p} + m} = \fsl{p} - m~?$$ (where $\fsl{p}=\gamma^\mu p_\mu$ ). It is not a simple squares difference because the denominator involves a gamma matrix in the slash notation, so how to do that?

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    $\begingroup$ Hint: what is $\fsl{p}\;\fsl{p}$? $\endgroup$ – probably_someone May 23 '18 at 8:37
  • $\begingroup$ @probably_someone what is fsl supposed to do? $\endgroup$ – Kyle Kanos May 23 '18 at 11:04
  • $\begingroup$ It is a common short notation for $\gamma^\mu p_\mu$ $\endgroup$ – MariNala May 23 '18 at 12:25
  • $\begingroup$ Is there a better way to do Feynman slashes? I've never tried on MathJax before, and I usually use a package when I'm using normal LaTeX. $\endgroup$ – probably_someone May 23 '18 at 22:41
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Observe that

$$ (\fsl{p}-m)(\fsl{p}+m)=\fsl{p}\,\fsl{p}-m^2=p^2-m^2$$

since $\fsl{p}\,\fsl{p}=p^2$. Therefore, multiplying both members on the right by $(\fsl{p}+m)^{-1}$, you get

$$\fsl{p}-m=(p^2-m^2)(\fsl{p}+m)^{-1}$$

Now, $p^2-m^2$ is a number, therefore it commutes with the matrix $(\fsl{p}+m)^{-1}$ and you can write unambiguously

$$\fsl{p}-m=\frac{p^2-m^2}{\fsl{p}+m}$$

where the fraction is to be understood in the sense of the inverse element. The other way round, you just write

$$\frac{p^2-m^2}{\fsl{p}+m}=(p^2-m^2)(\fsl{p}+m)^{-1}=(\fsl{p}-m)(\fsl{p}+m)(\fsl{p}+m)^{-1}=\fsl{p}-m$$

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