2
$\begingroup$

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.

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

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

Where:

  • $\not{\!A} := \gamma^{\alpha} A_{\alpha}$
  • $\Gamma$ 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.

EDIT

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 :)

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 at 10:13
  • 1
    $\begingroup$ Which negative sign do you mean? Could you point it out? $\endgroup$ – Wizact Jun 1 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 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 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 at 11:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.