I've been studying different scattering processes (from Mandl & Shaw QFT's book, chapter 8) and there's always a purely-mathematical common step I do not understand: the showing-up of the trace. Let me give two specific examples.

$$A_{(l) \alpha \beta}=\sum_{s_1} \sum_{s_2} \Big[ \bar u_{s_2} (\vec p_2') \gamma_{\alpha} v_{s_1} (\vec p_1'))(\bar v_{s_1}(\vec p_1')\gamma_{\beta} u_{s_2} (\vec p_2'))\Big]_{(l)}=Tr\Big[\frac{\not{\!p_2'}+m_l}{2m_l} \gamma_{\alpha} \frac{\not{\!p_1'}-m_l}{2m_l}\gamma_{\beta}\Big] \ \ \ \ (1)$$

$$X= \frac 1 2 \Lambda_{\delta \alpha}^+ (\vec p') \Gamma _{\alpha \beta} \Lambda_{\beta \gamma}^+ (\vec p) \tilde \Gamma _{\gamma \delta}=\frac 1 2 Tr \Big[\Lambda^+ (\vec p') \Gamma \Lambda^+ (\vec p) \tilde \Gamma \Big] \ \ \ \ (2)$$


  • $\not{\!A} := \gamma^{\alpha} A_{\alpha}$
  • $\Gamma$ is a $4 \times 4$ matrix
  • $\tilde \Gamma := \gamma^0 \Gamma^{\dagger} \gamma^0$
  • $u, v$ are the Dirac Spinors (which are $4 \times 1$ matrices)
  • The Dirac-$\gamma$-matrices are $4 \times 4$ matrices and the adjoint is defined as $\bar w := w^{\dagger}\gamma^{0}$
  • $\Lambda^+$ is the positive energy projection operator defined as $$\Lambda_{\alpha \beta}^+ (\vec p) := \Big( \frac{ \not{\!p}+m}{2m} \Big)_{\alpha \beta}$$
  • $\Lambda^+$ has the following property

$$\Lambda_{\alpha \beta}^+ (\vec p) = \sum_{r=1}^2 u_{r \alpha} (\vec p) \bar u_{r \beta} (\vec p)$$

But what I do not understand is why does the Trace show up in $(1)$, $(2)$

Any help is appreciated.


Let's write out the spinor indices explicitly for $(1)$

$$A_l = \Big(\sum_{s_1} u_{s_1 \color{red}{\delta}}(\vec p_2') \bar u_{s_1 \color{blue}{\alpha}}(\vec p_2')\Big) \gamma_{\color{blue}{\alpha}\color{green}{\nu}}\Big(\sum_{s_2}v_{s_2 \color{green}{\nu}}(\vec p_1') \bar v_{s_2 \color{gray}{\beta}}(\vec p_1')\Big)\gamma_{\color{gray}{\beta}\color{red}{\delta}} \ \ \ \ (3)$$

We know the following properties

$$\Lambda^+_{\color{red}{\delta}\color{blue}{\alpha}}(\vec p_2')= \sum_{s_1} u_{s_1 \color{red}{\delta}}(\vec p_2') \bar u_{s_1 \color{blue}{\alpha}}(\vec p_2')=\Big(\frac{\not{\!p_2'}+m_l}{2m_l}\Big)_{\color{red}{\delta}\color{blue}{\alpha}} $$

$$\Lambda^-_{\color{green}{\nu}\color{gray}{\beta}}(\vec p_1')= -\sum_{s_2}v_{s_2 \color{green}{\nu}}(\vec p_1') \bar v_{s_2 \color{gray}{\beta}}(\vec p_1')=-\Big(\frac{\not{\!p_1'}-m_l}{2m_l}\Big)_{\color{green}{\nu}\color{gray}{\beta}}$$

Thus we get

$$A_l = \Lambda^+_{\color{red}{\delta}\color{blue}{\alpha}}(\vec p_2')\gamma_{\color{blue}{\alpha}\color{green}{\nu}}\Lambda^-_{\color{green}{\nu}\color{gray}{\beta}}(\vec p_1')\gamma_{\color{gray}{\beta}\color{red}{\delta}}=-\operatorname{Tr}\Big[\frac{\not{\!p_2'}+m_l}{2m_l}\gamma_{\alpha}\frac{\not{\!p_1'}-m_l}{2m_l}\gamma_{\beta}\Big]$$

Now I have two questions:

1) Why are we allowed to manipulate $\sum_{s_1} \sum_{s_2} \Big[ \bar u_{s_2} (\vec p_2') \gamma_{\alpha} v_{s_1} (\vec p_1'))(\bar v_{s_1}(\vec p_1')\gamma_{\beta} u_{s_2} (\vec p_2'))\Big]_{(l)}$ in such a way to get $(3)$? What I mean is that I do not see what mathematical properties allow us to do so.

2) I get a negative sign. I guess it gets cancelled out due to an antysymmetric swap of certein indices, but what pair specifically?

Thank you :)

  • $\begingroup$ A great way to start is to always write out an expression with the spinor indices explicitly written on all terms $\endgroup$
    – Triatticus
    May 31, 2020 at 17:34
  • 1
    $\begingroup$ Crossposted from math.stackexchange.com/q/3699711/11127 Please don't crosspost. $\endgroup$
    – Qmechanic
    Jun 3, 2020 at 15:39

1 Answer 1


Your equation $(2)$ has the answer to your same question $$X= \frac 1 2 \Lambda_{\delta \alpha}^+ (\vec p') \Gamma _{\alpha \beta} \Lambda_{\beta \gamma}^+ (\vec p) \tilde \Gamma _{\gamma \delta}=\frac 1 2 Tr \Big[\Lambda^+ (\vec p') \Gamma \Lambda^+ (\vec p) \tilde \Gamma \Big] \tag{2}$$ if you see, look very well at the spinorial indices $$\Lambda^+_{\color{red}{\delta}\color{blue}{\alpha}}\Gamma_{\color{blue}{\alpha}\color{orange}{\beta}}\Lambda^+_{\color{orange}{\beta}\color{green}{\gamma}}\tilde{\Gamma}_{\color{green}{\gamma}\color{red}{\delta}}$$ I've highlighted them so you can easily see them. As you can see they are all contracted with each other and since $A_{ij}B_{ji} = (AB)_{ii} = Tr(AB)$ it's easy to see where the result comes from.

If you write down all the spinorial indices for the spinors $u, v$ and so, and for the gamma matrices $\gamma_\alpha$ you'll see that the result easily follows.

  • $\begingroup$ Davide Morgante enlightening answer, thank you! :) Could you check my edit? I wrote out the spinor indices explicitly for $(1)$. Do you agree with what I've done? $\endgroup$
    – JD_PM
    May 31, 2020 at 19:21
  • 1
    $\begingroup$ Yes! You're absolutely right! Good job $\endgroup$ May 31, 2020 at 19:32

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