I am currently reading Srednicki's Quantum field theory Book and am having some troubles with evaluating the trace of some gamma matrix expressions.

For instance in Equation 59.19 Srednicki defines $$\newcommand{\fsl}[1]{#1\kern-0.4em\raise0.22ex\hbox{/}}\langle\Phi_{tu}\rangle= \frac{1}{4}{\rm Tr} [\gamma_\nu(-\fsl{p}_1+\fsl{k}_1'+m)\gamma_\mu(-\fsl{p}_1+m)\gamma^\nu(-\fsl{p}_1+\fsl{k}_2'+m)\gamma^\mu(-\fsl{p}_2-m)]. \tag{59.19c} $$

All the $p$'s and $k$'s are slashed (If you know the syntax for this please help!)

The book has developed expressions such as $\gamma^\mu \fsl{p}\gamma_\mu = 2p$, I am having trouble reconciling $\gamma_\nu(-\fsl{p}_1+\fsl{k}_1'+m)\gamma_\mu$ where the indices of the two gamma matrices are different. Could someone help me explicitly evaluate an expression of this form.


I tried to solve this as follows;

consider $\gamma^\nu((-\fsl{p}_1+\fsl{k}_2'+m)\gamma^\mu(-\fsl{p}_2-m)\gamma_\nu$ to be of the form $abc$ (everything is slashed)

I simplified and obtained this expression = $Tr((-\fsl{p}_2-m)(-4m^2-1/2(s+t+u)+2m(-2\fsl{p}_1+\fsl{k}_1'+\fsl{k}_2')(-\fsl{p}_1+m))$ everything except the mandelstam variables are slashed.

Is there any way to simplify this further? Srednicki has a beautiful expression reducing to just $2m^2(s-4m^2)$


As a first step one can reduce the complexity of the expression by expanding it in terms of powers of $m$. This should give three terms: one without $m$'s, an $m^2$-term and an $m^4$-term. The terms with odd powers of $m$'s would be zero, because they contain the trace of an odd number of $\gamma$-matrics, which is zero.

Let's look at each in turn. The one without $m$'s is the most complicated. It has the form $${\rm Tr}\{\gamma_{\nu} \fsl{p}_1 \gamma_{\mu} \fsl{p}_2 \gamma^{\nu} \fsl{p}_3 \gamma^{\mu} \fsl{p}_4 \} . $$ Now, we can use the identity $ \gamma^{\mu} \gamma^{\rho} \gamma^{\nu} \gamma^{\sigma} \gamma_{\mu} = -2 \gamma^{\sigma} \gamma^{\nu} \gamma^{\rho} , $ to write the trace as $$ -2 {\rm Tr}\{\gamma_{\nu} \fsl{p}_1 \fsl{p}_3 \gamma^{\nu} \fsl{p}_2 \fsl{p}_4 \} . $$ Then we use the identity $ \gamma^{\mu} \gamma^{\rho} \gamma^{\sigma} \gamma_{\mu} = 4 g^{\rho\nu} {\cal I}$ to write it as $$ -8 ({p}_1\cdot {p}_3){\rm Tr}\{\fsl{p}_2 \fsl{p}_4 \} . $$ The remaining trace over two $\gamma$ matrices then gives $$ -32 ({p}_1\cdot {p}_3)({p}_2\cdot {p}_4) . $$

There would be 6 terms with $m^2$. They would either be of the form $m^2 {\rm Tr}\{\gamma_{\nu} \gamma_{\mu} \gamma^{\nu} \fsl{p}_3 \gamma^{\mu} \fsl{p}_4 \}$ or be of the form $m^2 {\rm Tr}\{\gamma_{\nu} \gamma_{\mu} \fsl{p}_2 \gamma^{\nu} \gamma^{\mu} \fsl{p}_4 \}$. For the first case one can use $ \gamma^{\mu} \gamma^{\nu} \gamma_{\mu} = -2 \gamma^{\nu}$ to obtain $$-2 m^2 {\rm Tr}\{\gamma_{\mu} \fsl{p}_3 \gamma^{\mu} \fsl{p}_4 \} $$ and then $$4 m^2 {\rm Tr}\{\fsl{p}_3 \fsl{p}_4 \} , $$ which gives $$16 m^2 ({p}_3\cdot {p}_4) . $$ For the second case we use one of the previously mentioned identities to obtain $$4 m^2 {\rm Tr}\{\fsl{p}_2 \fsl{p}_4 \} , $$ which becomes $$16 m^2 ({p}_2\cdot {p}_4) . $$

The final term with $m^4$ has the trace $$m^4 {\rm Tr}\{\gamma_{\nu} \gamma_{\mu} \gamma^{\nu} \gamma^{\mu} \} . $$ From one of the identities we obtain $$-2 m^4 {\rm Tr}\{\gamma_{\mu} \gamma^{\mu}\} . $$ The last trace is just 4 times the identity. Hence, it becomes $$-8 m^4 . $$


This link show a lots of properties : https://en.wikipedia.org/wiki/Gamma_matrices Be careful, I think you forgot a minus sign in one of your equation (-2p if I'm not mistaken) But I'm not sure here about the indices of p1 and p2 with the Feynman slash. Can you explicitly gives the indices for those gamma matrices ? I hope the trace identities might help you.

  • $\begingroup$ Thank you very much, the trace identities were helpful but now I encounter the problem of finding the trace of terms like Tr{p1(slash)p1(slash)p2(slash)} $\endgroup$ – Aditya Varna Iyer May 12 '16 at 9:31
  • $\begingroup$ For each slashed variable you have a gamma matrices. The trace of an odd number of gamma is zero. I think those terms simply disappear. $\endgroup$ – JSFDude May 12 '16 at 9:34
  • $\begingroup$ Yes I thought so too and managed to ignore those terms but when I try to simplify the expression I got in my edited post it's nasty because I have to expand everything out. I was wondering if there was a more elegant way. $\endgroup$ – Aditya Varna Iyer May 12 '16 at 9:36

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