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

$$\Lambda^+_{\color{red}{\delta}\color{blue}{\alpha}}(\vec p_2')= \sum_{s_1} u_{s_1 \color{red}{\delta}}(\vec p_2') \bar u_{s_1 \color{blue}{\alpha}}(\vec p_2')=\Big(\frac{\not{\!p_2'}+m_l}{2m_l}\Big)_{\color{red}{\delta}\color{blue}{\alpha}} \ \ \ \ (4)$$

$$\Lambda^-_{\color{green}{\nu}\color{gray}{\beta}}(\vec p_1')= -\sum_{s_2}v_{s_2 \color{green}{\nu}}(\vec p_1') \bar v_{s_2 \color{gray}{\beta}}(\vec p_1')=-\Big(\frac{\not{\!p_1'}-m_l}{2m_l}\Big)_{\color{green}{\nu}\color{gray}{\beta}} \ \ \ \ (5)$$

Thus we get

$$A_l = \Lambda^+_{\color{red}{\delta}\color{blue}{\alpha}}(\vec p_2')\gamma_{\color{blue}{\alpha}\color{green}{\nu}}\Lambda^-_{\color{green}{\nu}\color{gray}{\beta}}(\vec p_1')\gamma_{\color{gray}{\beta}\color{red}{\delta}}=-\operatorname{Tr}\Big[\frac{\not{\!p_2'}+m_l}{2m_l}\gamma_{\alpha}\frac{\not{\!p_1'}-m_l}{2m_l}\gamma_{\beta}\Big] \ \ \ \ (6)$$

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 :)


1 Answer 1


The trace of a matrix is equal to the sum of the diagonal elements, so in index notation, $\mathrm{Tr} (A) = A_{\alpha \alpha}$ (sum over $\alpha$). In your equation (2), it's easy to see that you have this structure, since

$$ \Lambda_{\delta \alpha}^+ (\vec p') \Gamma _{\alpha \beta} \Lambda_{\beta \gamma}^+ (\vec p) \tilde \Gamma _{\gamma \delta} = \left(\Lambda^+(\vec{p}') \Gamma \Lambda^+ \tilde{\Gamma}\right)_{\delta \delta} = \mathrm{Tr} \left(\Lambda^+(\vec{p}') \Gamma \Lambda^+ \tilde{\Gamma}\right). $$

In equation (1), it's a bit more difficult to see, because the indices of the spinors are surpressed. If I use $a,b,c,d,\dots$ for the spinor indices, then you see that:

$$ \sum_{s_1,s_2} \bar u_{s_2}^a (\vec p_2') \gamma_{\alpha}^{ab} v_{s_1}^b (\vec p_1'))(\bar v_{s_1}^c(\vec p_1')\gamma_{\beta}^{cd} u_{s_2}^d (\vec p_2')) = \sum_{s_1,s_2} (v_{s_1}^b (\vec{p}_1' ) \bar{v}^c_{s_1}(\vec{p}_1')) \gamma_\beta^{cd} (u_{s_2}^d (\vec{p}_2') \bar{u}^a_{s_2}(\vec{p}_2'))\gamma_\alpha^{ab}. $$

Now, I can take the sum over all polarizations, giving me the energy projectors:

$$ A_{\alpha \beta} = \Big( \frac{ \not{\!p}-m}{2m} \Big)^{bc} \gamma_\beta^{cd} \Big( \frac{ \not{\!p}+m}{2m} \Big)^{da} \gamma^{ab}_\alpha. $$ This time, the sum over spinor indices is what gives you the trace.

  • $\begingroup$ Wizact thank you for your answer! Could you please check out my edit? Why does the negative sign cancel out? $\endgroup$
    – JD_PM
    Jun 1, 2020 at 10:13
  • 1
    $\begingroup$ Which negative sign do you mean? Could you point it out? $\endgroup$
    – Wizact
    Jun 1, 2020 at 10:30
  • $\begingroup$ Please note equation (5). When I use such a property of the (negative) energy projection operator, I pick a negative sign. I guess we pick another negative sign and that's why we get a positive trace, but where? $\endgroup$
    – JD_PM
    Jun 1, 2020 at 10:41
  • 1
    $\begingroup$ Equation (5) has two minus signs. In the expression for the amplitude, you don't have any minus signs, hence the resulting energy projector will not have a minus sign either. $\endgroup$
    – Wizact
    Jun 1, 2020 at 10:46
  • 1
    $\begingroup$ When you write the spinor indices, something like $u_{s_1}^d$ is just a number, so you can place it wherever you want. Then, it's just rearranging the sum in a convenient way. $\endgroup$
    – Wizact
    Jun 1, 2020 at 11:00

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