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Theorem 2


\begin{displaymath}
\textrm{Tr}[I] = 4 .
\end{displaymath} (7.58)


$\displaystyle \textrm{Tr}[\not{a}\not{b}] = \textrm{Tr}[\not{b}\not{a}]$ $\textstyle =$ $\displaystyle \frac{1}{2} \textrm{Tr}[\not{a}\not{b} + \not{b}\not{a}] ,$  
  $\textstyle =$ $\displaystyle \frac{1}{2} \textrm{Tr}
[a_\mu b_\nu(\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu)] ,$  
  $\textstyle =$ $\displaystyle \frac{1}{2} \textrm{Tr}[a_\mu b_\nu(2g^{\mu\nu})] ,$  
  $\textstyle =$ $\displaystyle \textrm{Tr}[a\cdot b] ,$  
  $\textstyle =$ $\displaystyle a\cdot b \textrm{Tr}[I] ,$  
  $\textstyle =$ $\displaystyle 4 a\cdot b .$ (7.59)

This also shows


$\displaystyle \textrm{Tr}[a_\mu\gamma^\mu b_\nu\gamma^\nu]$ $\textstyle =$ $\displaystyle 4a_\mu b^\mu ,$  
$\displaystyle a_\mu b_\nu \textrm{Tr}[\gamma^\mu\gamma^\nu]$ $\textstyle =$ $\displaystyle 4a_\mu b_\nu
g^{\mu\nu} ,$  
$\displaystyle \Rightarrow \textrm{Tr}[\gamma^\mu\gamma^\nu]$ $\textstyle =$ $\displaystyle 4g^{\mu\nu} .$ (7.60)


next up previous contents index
Next: Theorem 3 Up: Trace Theorems Previous: Theorem 1
Douglas M. Gingrich (gingrich@ ualberta.ca)
2004-03-18