Why does the trace show up while computing unpolarized cross sections?

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  • Thread starter JD_PM
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Summary:

I want to understand why the trace shows up while dealing with unpolarized cross sections.

Main Question or Discussion Point

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


  • Lepton Pair Production ##( e^+ (\vec p_1, r_1) + e^- (\vec p_2, r_2) \rightarrow l^+ (\vec p_1', s_1)+l^- (\vec p_2', s_2))##


Here the unpolarized cross section is given by

$$X = \frac{e^4}{4\Big[(p_1+p_2)^2\Big]^2} A_{(l) \alpha \beta} B_{(e)}^{\alpha \beta} \ \ \ \ (1)$$

Here both ##A_{(l) \alpha \beta}## and ##B_{(e)}^{\alpha \beta}## end up yielding traces; let's show ##A_{(l) \alpha \beta}## explicitly as an example.

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


  • Bhabha scattering (##e^+ (\vec p_1, r_1)+e^- (\vec p_2, r_2) \rightarrow e^+ (\vec p_1', s_1) + e^+ (\vec p_2', s_2)##)


Let's take only a term of the total unpolarized cross section for the Bhabha scattering as an example

$$X_{ab} = \frac{-e^4}{4(p_1-p_1')^2(p_1+p_2)^2} \sum_{spins} \Big[ (\bar u (\vec p_2') \gamma_{\alpha} u (\vec p_2))(\bar u (\vec p_2)\gamma_{\beta} v (\vec p_1))(\bar v (\vec p_1) \gamma^{\alpha} v (\vec p_1'))(\bar v (\vec p_1')\gamma^{\beta} u (\vec p_2')\Big]=\frac{-e^4}{4(p_1-p_1')^2(p_1+p_2)^2} Tr\Big[\frac{\not{\!p_2'}+m}{2m} \gamma_{\alpha}\frac{\not{\!p_2'}+m}{2m} \gamma_{\beta}\frac{\not{\!p_1}-m}{2m} \gamma^{\alpha}\frac{\not{\!p_1'}-m}{2m} \gamma^{\beta}\Big] \ \ \ \ (3)$$


Thus all boils down to understand why

$$X= \frac 1 2 \Lambda_{\delta \alpha}^+ (\mathbf p') \Gamma _{\alpha \beta} \Lambda_{\beta \gamma}^+ (\mathbf p) \tilde \Gamma _{\gamma \delta}=\frac 1 2 Tr \Big[\Lambda^+ (\mathbf p') \Gamma \Lambda^+ (\mathbf p) \tilde \Gamma \Big] \ \ \ \ (4)$$


Where the positive energy projection operator satisfies the following equation


$$\Lambda_{\alpha \beta}^+ (\mathbf p) = \Big( \frac{ \not{\!p}+m}{2m} \Big)_{\alpha \beta} = \sum_{r=1}^2 u_{r \alpha} (\mathbf p) \bar u_{r \beta} (\mathbf p) \ \ \ \ (5)$$

But I do not see how to show that Eq. 5 is the reason why the trace shows up in Eq. 2, Eq. 3 and Eq. 4



Any help is appreciated.

Thank you
 

Answers and Replies

  • #2
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$$A_{ij}B_{ji} = (AB)_{ii} =\operatorname{Tr}(AB)$$

More details :smile:
 
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