# Understanding how to get the unpolarized cross-sections

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• JD_PM
In summary: Gamma u_s (\vec p') \Big)= \frac 1 2 \Big( \sum_s u_{s \delta} (\vec p') \bar u_{s \alpha} (\vec p') \Big) \Gamma_{\alpha \beta} \Big( \sum_r u_{r \beta} (\vec p) \bar u_{r \gamma}(\vec p)\Big) \tilde \Gamma_{\gamma \delta} \ \ \ \ (5)\$In summary, we discussed how to obtain the cross-sections from ##S_{fi}## elements, assuming they are fully polarized. However, in reality
JD_PM
TL;DR Summary
I want to understand the beautiful technique to sum over spin states.
The following material comes from the book Quantum Field Theory by Mandl & Shaw, chapter 8, section 8.2

Here we discussed how to obtain the cross-sections from ##S_{fi}## elements. These cross-sections were assumed to be fully polarized (i.e. the particles involved in the collision have definite initial and final polarization states). However, the beams of colliding particles are usually unpolarized, and the polarizations of the particles produced in the collision cannot be measured. Thus, we must average and sum over polarization states of initial and final particles, respectively. This procedure implies summing over spin and and photon polarization states.

In this post I am aimed at discussing the technique to obtain the unpolarized cross-section involving lepton-spin states only.

As an example Mandl & Shaw worked out Compton Scattering for electrons and took the following Feynman Amplitude
$$\mathscr{M}=\bar{u_s} (\mathbf p') \Gamma u_r (\mathbf p) \ \ \ \ (1)$$

Where the four-component spinors ##u_r( \mathbf p)## and ##\bar u_s(\mathbf p')## completely specify the momentum and spins of the electron in the initial and final states, and the operator ##\Gamma## is a ##4 \times 4## matrix built up out of ##\gamma##-matrices. Eq. (1) gives rise to an unpolarized cross-section of the form

$$X:= \frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2 |\mathscr{M}|^2 \ \ \ \ (2)$$

Where we've averaged over initial spins (i.e ##\sum_r##) and summed over final spins (i.e ##\sum_s##)

Defining:

$$\tilde \Gamma := \gamma^0 \Gamma^{\dagger} \gamma^0 \ \ \ \ (3)$$

Eq. (2) can be rewritten as follows

$$X:= \frac 1 2 \sum_{r=1} \sum_{s=1} \Big( \bar{u_s} (\mathbf p') \Gamma u_r (\mathbf p))( \bar{u_r} (\mathbf p) \tilde \Gamma u_s (\mathbf p') \Big) \ \ \ \ (4)$$

Writing out the spinor indices explicitly, (4) can be written as

$$X= \frac 1 2 \Big( \sum_s u_{s \delta} (\mathbf p') \bar u_{s \alpha} (\mathbf p') \Big) \Gamma_{\alpha \beta} \Big( \sum_r u_{r \beta} (\mathbf p) \bar u_{r \gamma}(\mathbf p)\Big) \tilde \Gamma_{\alpha \delta} \ \ \ \ (5)$$

The positive energy projection operator satisfies the following equation

$$\Lambda_{\alpha \beta}^+ (\mathbf p) = \Big( \frac{ \not{\!p}+m}{2m} \Big)_{\alpha \beta} = \sum_{r=1}^2 u_{r \alpha} (\mathbf p) \bar u_{r \beta} (\mathbf p) \ \ \ \ (6)$$

Eq. (6) allows to eliminate the sums over positive energy states, leading to the final result

$$X= \frac 1 2 \Lambda_{\delta \alpha}^+ (\mathbf p') \Gamma _{\alpha \beta} \Lambda_{\beta \gamma}^+ (\mathbf p) \tilde \Gamma _{\gamma \delta}=\frac 1 2 Tr \Big[\Lambda^+ (\mathbf p') \Gamma \Lambda^+ (\mathbf p) \tilde \Gamma \Big] \ \ \ \ (7)$$So my questions are:1) How to get Eq. (4)

I am aware of the following definition

$$\bar u_{s,r} (\mathbf p) := u^{\dagger}_{s,r} (\mathbf p) \gamma^0$$

But I do not know what property of ##\gamma##-Dirac matrices should I use to get (4).2) How to get Eq. (5).

Again, there has to be a ##\gamma##-Dirac matrix property I am missing...3) Why does the Trace show up in (7)?Any help is appreciated.

Thank you.

JD_PM said:
3) Why does the Trace show up in (7)?Any help is appreciated.

Thank you.

This is why

JD_PM said:
2) How to get Eq. (5).

Again, there has to be a ##\gamma##-Dirac matrix property I am missing...

Ahhh so it has nothing to do with Dirac matrix properties.

Apparently, it turns out that when we write out matrix indices explicitly we can treat spinors and ##\Gamma## matrices as numbers!

That means that we can rearrange ##(5)## at will (note that at #1 I made a typo; it is actually ##\tilde \Gamma_{\gamma \delta}##). For instance

$$X= \frac 1 2 \Big( \sum_s u_{s \delta} (\mathbf p') \bar u_{s \alpha} (\mathbf p') \Big) \Gamma_{\alpha \beta} \tilde \Gamma_{\gamma \delta} \Big( \sum_r u_{r \beta} (\mathbf p) \bar u_{r \gamma}(\mathbf p)\Big) \ \ \ \ (5)$$

JD_PM said:
1) How to get Eq. (4)

Alright, I got a proof for Eq. (4) (I know it'll be trivial for most of you but I am pretty excited I have to say ). Please feel free to check it out.

I assumed that

$$X:= \frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2 |\mathscr{M}|^2=\frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2\mathscr{M}\mathscr{M}^{\dagger}$$

Where by ##\dagger## I mean the conjugate transpose. Thus explicitly we get

$$X:= \frac 1 2 \sum_{r=1}^2 \sum_{s=1}^2\mathscr{M}\mathscr{M}^{\dagger}=\frac 1 2 \sum_{r=1} \sum_{s=1} ( \bar{u_s} (\vec p') \Gamma u_r (\vec p))\Big( \bar{u_s} (\vec p) \Gamma u_r (\vec p')\Big)^{\dagger}=\frac 1 2 \sum_{r=1} \sum_{s=1} ( \bar{u_s} (\vec p') \Gamma u_r (\vec p))\Big( u_r^{\dagger} (\vec p') \Gamma^{\dagger} \bar{u_s}^{\dagger} (\vec p)\Big) \ \ \ \ (6)$$

Now it is about working out ##\bar{u_s}^{\dagger} (\vec p)## and ##u_r^{\dagger} (\vec p')##

We know that the adjoint is, by definition

$$\bar{u_s} (\vec p) := u_s^{\dagger} \gamma^0$$

Taking ##\dagger## on both sides of such equation we get

$$\bar{u_s}^{\dagger} (\vec p) := (u_s^{\dagger} \gamma^0)^{\dagger}=\gamma^{0\dagger} u_s \ \ \ \ (7)$$

Where

$$\gamma^{0\dagger}=\gamma^{0}$$

And here comes the key step: I assumed that ##\Big(\gamma^{0}\Big)^{-1}=\gamma^{0}## Thus we get$$u_r^{\dagger} (\vec p') = \bar u_r (\vec p') \gamma^{0} \ \ \ \ (8)$$

Plugging ##(7)## and ##(8)## into ##(6)## we get the desired ##(4)##

$$X=\frac 1 2 \sum_{r=1} \sum_{s=1} ( \bar{u_s} (\vec p') \Gamma u_r (\vec p))\Big( u_r^{\dagger} (\vec p') \Gamma^{\dagger} \bar{u_s}^{\dagger} (\vec p)\Big)=\frac 1 2 \sum_{r=1} \sum_{s=1} ( \bar{u_s} (\vec p') \Gamma u_r (\vec p))\bar u_r (\vec p') \gamma^{0}\Gamma^{\dagger}\gamma^{0} u_s(\vec p')=\frac 1 2 \sum_{r=1} \sum_{s=1} \Big( \bar{u_s} (\vec p') \Gamma u_r (\vec p))( \bar{u_r} (\vec p) \tilde \Gamma u_s (\vec p') \Big)$$

Note that in my proof I assumed ##\Big(\gamma^{0}\Big)^{-1}=\gamma^{0}##. If this is correct, why is this true?

JD_PM said:
Note that in my proof I assumed ##\Big(\gamma^{0}\Big)^{-1}=\gamma^{0}##. If this is correct, why is this true?

I think I got it!

##\Big(\gamma^{0}\Big)^{-1}=\gamma^{0}## is true and we can prove it based on the anticommutation relation (which can be found in page 452; Mandl & Shaw) ##\left\{\gamma^\mu, \gamma^{\nu} \right\}=2g^{\mu \nu}##.

Using the convention ##(+, -, -, -)## for the metric, we get for ##\mu=0, \nu=0##

$$\Big(\gamma^0\Big)^2=1$$

Which indeed proofs my assumption.

## 1. What is the purpose of understanding unpolarized cross-sections?

Understanding unpolarized cross-sections allows scientists to accurately measure the probability of a particle interacting with other particles. This is crucial in fields such as particle physics and nuclear physics.

## 2. How are unpolarized cross-sections measured?

Unpolarized cross-sections are typically measured by colliding particles and measuring the number of interactions that occur. The cross-section is then calculated by dividing the number of interactions by the intensity of the particle beam.

## 3. What factors can affect unpolarized cross-sections?

The size and shape of the particles, the energy of the particle beam, and the type of interaction being measured can all affect unpolarized cross-sections. Other factors such as temperature and pressure can also play a role.

## 4. How do unpolarized cross-sections differ from polarized cross-sections?

Unpolarized cross-sections measure the probability of a particle interacting with other particles without taking into account the spin or orientation of the particles. In contrast, polarized cross-sections measure the probability of interaction while taking into account the spin or orientation of the particles.

## 5. What are some real-world applications of understanding unpolarized cross-sections?

Understanding unpolarized cross-sections is crucial in many fields, such as nuclear power, medical imaging, and astrophysics. It allows scientists to accurately predict and control particle interactions, which has practical applications in energy production, medical treatments, and understanding the behavior of the universe.

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