3
$\begingroup$

So in Peskin and Schoreder, when computing the amplitude of $e^+e^-\rightarrow\mu^+\mu^-$, summing up over spin they write

\begin{align}\sum_{s,s'}\bar{v}^{s'}_a(p_2)\gamma^\mu_{ab}u^s_b(p_1)\bar{u}^{s}_c(p_1)\gamma^\nu_{cd}v^s_d(p_2) & =(\not{p}_2-m)_{da}\gamma^\mu_{ab}(\not{p}_1+m)_{bc}\gamma_{cd}^\nu\\ &= \textrm{tr}[(\not{p}_2-m)\gamma^\mu(\not{p}_1+m)\gamma^\nu] \end{align}

Why is that a trace?

As I understand that $a$, $b$, $c$, $d$ indexes are the matrix indexes and this is the combination to make a trace. Can someone please clarify, I can't find the answer anywhere.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
8
$\begingroup$

An $A_{ab}B_{bc}$ yields a $C_{ac}$. Contracting all indices, but the outer ones of your expression yields a $[(\not{p}_2-m)\gamma^\mu(\not{p}_1+m)\gamma^\nu]_{dd}$. Now executing the $dd$ contraction is just the trace.

Improve this answer
$\endgroup$
5
$\begingroup$

I was puzzled by this same thing when I took QFT classes several years back. After thinking about it, the reason is so trivial as to not merit an explanation in the literature, especially Peskin and Schroeder.

Look at th LHS of your first equation:

$\begin{align}\sum_{s,s'}\bar{v}^{s'}_a(p_2)\gamma^\mu_{ab}u^s_b(p_1)\bar{u}^{s}_c(p_1)\gamma^\nu_{cd}v^s_d(p_2) & =(\not{p}_2-m)_{da}\gamma^\mu_{ab}(\not{p}_1+m)_{bc}\gamma_{cd}^\nu\\ &= \textrm{tr}[(\not{p}_2-m)\gamma^\mu(\not{p}_1+m)\gamma^\nu] \end{align}$

$\bar{v}$ is a 1 X 4 matrix, $\gamma$ is a 4 X 4 matrix and $u$ is a 4 X 1 matrix. The product of these three matrices in this order is a 1 X 1 matrix. But the trace of a 1 X 1 matrix is equal to the matrix element itself! And because of the property of the trace, tr$(A*B*C)$ = tr$(C*A*B)$ = tr$(B*C*A)$. The same reasoning applies to the next three terms.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I should have thought of that, thank you for showing me. $\endgroup$ Commented Jun 12, 2016 at 10:10
  • $\begingroup$ @user3218736 You're welcome. I like to read and answer questions if I can from those who's interest coincide with some of mine. $\endgroup$ Commented Jun 12, 2016 at 19:27

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.