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

• I
• JD_PM
In summary, the trace in the equation for the Bhabha scattering shows up because the positive energy projection operator in Eq. 5 satisfies the following equation.
JD_PM
TL;DR Summary
I want to understand why the trace shows up while dealing with unpolarized cross sections.
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

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

More details

vanhees71

## 1. Why does the trace show up in unpolarized cross sections?

The trace appears in unpolarized cross sections because it is a mathematical operation used to calculate the total probability of a scattering event. It takes into account all possible spin states of the particles involved in the scattering process.

## 2. What is the significance of the trace in computing unpolarized cross sections?

The trace is important because it allows us to calculate the total probability of a scattering event, taking into account all possible spin states. This is essential in accurately predicting the outcome of experiments and understanding the behavior of particles at the subatomic level.

## 3. How is the trace calculated in unpolarized cross sections?

The trace is calculated by summing the diagonal elements of the scattering matrix. The scattering matrix is a mathematical representation of the interaction between particles and contains information about the probabilities of different spin states.

## 4. Can the trace be used to calculate polarized cross sections?

Yes, the trace can also be used to calculate polarized cross sections. However, in this case, it is necessary to consider the off-diagonal elements of the scattering matrix, which represent the interference between different spin states.

## 5. Is the trace always present in calculations of unpolarized cross sections?

Yes, the trace is always present in calculations of unpolarized cross sections. It is a fundamental part of the mathematical framework used to describe and understand particle interactions and cannot be omitted without compromising the accuracy of the results.

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