Traces in Compton Scattering

[JD_PM]

TL;DR Summary

I want to understand how to get (as an example)

$X_{aa} = \operatorname{Tr}\{\gamma^{\beta} (\not{\!p} + \not{\!k} +m) \gamma^{\alpha} (\not{\!p} + m) \gamma_{\alpha} (\not{\!p} + \not{\!k} +m) \gamma_{\beta} (\not{\!p'} +m)\}$

Given the Feynman amplitude (Compton Scattering by electrons) $\mathscr{M}=\mathscr{M}_a+\mathscr{M}_b$, where

$$\mathscr{M}_a = -i e^2 \frac{\bar u' \not{\!\epsilon'} (\not{\!p}+\not{\!k}+m) \not{\!\epsilon} u}{2(pk)}, \ \ \ \ \mathscr{M}_b = i e^2 \frac{\bar u' \not{\!\epsilon} (\not{\!p}-\not{\!k'}+m) \not{\!\epsilon'} u}{2(pk')} \tag 1$$

Given also the formula to sum over spin and polarization states.

$$Y:= \frac 1 4 \sum_{r} \sum_{s} |\mathscr{M}|^2=\frac 1 4 \sum_{r} \sum_{s} \Big\{|\mathscr{M_a}|^2+|\mathscr{M_b}|^2+\mathscr{M_a}\mathscr{M_b}^{\dagger}+\mathscr{M_b}\mathscr{M_a}^{\dagger}\Big\}$$ $$=\frac{e^4}{64 m^2} \Big\{ \frac{X_{aa}}{(pk)^2} + \frac{X_{bb}}{(pk')^2} - \frac{X_{ab}+X_{ba}}{(pk)(pk')}\Big\} \tag 2$$

Where $r$ represents polarization states and $s$ spin states.

My issue is understanding how to get the $X_{..}$ expressions, which are traces.

My idea is work out $X_{aa}$ as an example, so that I understand the process. The answer is

$$X_{aa} = \operatorname{Tr}\{\gamma^{\beta} (\not{\!p} + \not{\!k} +m) \gamma^{\alpha} (\not{\!p} + m) \gamma_{\alpha} (\not{\!p} + \not{\!k} +m) \gamma_{\beta} (\not{\!p'} +m)\} \tag 3$$Useful trick: setting $\mathscr{M}:=\epsilon_{\alpha}\epsilon_{\beta}'\mathscr{M}^{\alpha \beta}$ allows us to write $Y=\frac 1 4 \sum_{r} \sum_{s} |\mathscr{M}|^2=\frac 1 4 \sum_{s} \mathscr{M^{\alpha \beta}}\mathscr{M_{\alpha \beta}}^{\dagger}$

• My attempt to get $X_{aa} = \operatorname{Tr}\{\gamma^{\beta} (\not{\!p} + \not{\!k} +m) \gamma^{\alpha} (\not{\!p} + m) \gamma_{\alpha} (\not{\!p} + \not{\!k} +m) \gamma_{\beta} (\not{\!p'} +m)\}$

Note: from here on $\mathscr{M}$ really means $\mathscr{M_a}$

$$\frac 1 4 \sum_{r} \sum_{s} |\mathscr{M}|^2=\frac 1 4 \sum_{r} \sum_{s}
(\bar{u_s} (\vec p') \Gamma u_r (\vec p))( \bar{u_r} (\vec p) \tilde \Gamma u_s (\vec p')) \tag 4$$

Where I've defined $\Gamma:=\not{\!p}+\not{\!k}+m$, $\tilde \Gamma = \gamma^0 \Gamma^{\dagger} \gamma^0$

But this leads to a trace of the form

$$\frac{1}{4} \operatorname{Tr}\Big\{\frac{\not{\!p'}+m}{2m}\Gamma\frac{\not{\!p}+m}{2m}\tilde \Gamma\Big\} \tag 5$$

Which is not the provided answer.

Source: Mandl & Shaw second edition, pages 143 & 144.

Any help is appreciated.

 

[Dr.AbeNikIanEdL]

I did not check everything else, but from the "useful trick" to Eq. (4) the indices $\alpha$ and $\beta$ seem to go missing (at least with the $\Gamma$ as you define it), and with them the corresponding gamma matrices.

 

[JD_PM]

I did not check everything else, but from the "useful trick" to Eq. (4) the indices $\alpha$ and $\beta$ seem to go missing (at least with the $\Gamma$ as you define it), and with them the corresponding gamma matrices.

True, my attempt was actually a broad idea.

Let me provide you with some context. You may remember we discussed Compton Scattering assuming photons and electron were definite states. This is explained below up to Eq. 8.63.

From Eq. 8.63. on definite states are not assumed anymore and average sums begin; what I am trying to do now is understand how to get the $X_{..}$ trace-expressions presented below. To do so I am aimed at getting $X_{aa}$

Thank you.

 

[Dr.AbeNikIanEdL]

Yes, I understood what you are trying to do. As I said, you seem to be missing the gamma matrices that are contracting the $\epsilon$s in your (1). I also don't understand where the $\gamma^0$s are coming from in $\tilde{\Gamma}$? Apart from that you appear to be getting the right structure?

 

[nrqed]

Summary:: I want to understand how to get (as an example)

$X_{aa} = \operatorname{Tr}\{\gamma^{\beta} (\not{\!p} + \not{\!k} +m) \gamma^{\alpha} (\not{\!p} + m) \gamma_{\alpha} (\not{\!p} + \not{\!k} +m) \gamma_{\beta} (\not{\!p'} +m)\}$

Given the Feynman amplitude (Compton Scattering by electrons) $\mathscr{M}=\mathscr{M}_a+\mathscr{M}_b$, where

$$\mathscr{M}_a = -i e^2 \frac{\bar u' \not{\!\epsilon'} (\not{\!p}+\not{\!k}+m) \not{\!\epsilon} u}{2(pk)}, \ $$

As Dr Abe pointed out, this can be written as

$$ \mathscr{M}_a = -i e^2 \epsilon'_\alpha~\epsilon_\beta ~ \frac{\bar u' \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u}{2(pk)}, \ $$

If you square this and sum over the spins and polarization, you should directly get the answer.

 

[JD_PM]

Thank you Dr.AbeNikIanEdL and nrqed, I indeed missed the gamma matrices. I am now closer to get the answer

As Dr Abe pointed out, this can be written as

$$ \mathscr{M}_a = -i e^2 \epsilon'_\alpha~\epsilon_\beta ~ \frac{\bar u' \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u}{2(pk)}, \ $$

If you square this and sum over the spins and polarization, you should directly get the answer.

I'll include as many details as possible.

The trick $Y=\frac 1 4 \sum_{r} \sum_{s} |\mathscr{M}|^2=\frac 1 4 \sum_{s} \mathscr{M^{\alpha \beta}}\mathscr{M_{\alpha \beta}}^{\dagger}$ allows us to only have to worry about summing over spin states; let me label them $s_1$ and $s_2$.

Note that I use Einstein's notation with polarization indices (that's why there are only two sum signs instead of 4).

$$Y=\frac 1 4 \sum_{s_1} \sum_{s_2} \mathscr{M^{\alpha \beta}}\mathscr{M_{\alpha \beta}}^{\dagger}=\frac{e^4}{16(pk)^2}\sum_{s_1} \sum_{s_2}(\bar u'_{s_1} \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u_{s_2})(\bar u'_{s_1} \gamma_{\alpha}(\not{\!p}+\not{\!k}+m) \gamma_\beta \ u_{s_2})^{\dagger}$$ $$=\frac{e^4}{16(pk)^2}\sum_{s_1} \sum_{s_2}(\bar u'_{s_1} \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u_{s_2})(u_{s_2}^{\dagger}\gamma_\beta (\not{\!p}+\not{\!k}+m) \gamma_{\alpha} \bar u_{s_1}^{'\dagger})$$ $$=\frac{e^4}{16(pk)^2}\sum_{s_1} \sum_{s_2} (\bar u'_{s_1} \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u_{s_2})(\bar u_{s_2} \gamma^0 \gamma_\beta (\not{\!p}+\not{\!k}+m) \gamma_\alpha \gamma^0 u_{s_1}')$$

Where I assumed that $\Big(\gamma_{\alpha, \beta}\Big)^{\dagger}=\gamma_{\alpha, \beta}$, used $u^{\dagger} = \bar u \gamma^0$ (this actually holds due to $\Big(\gamma^0\Big)^2 = 1$, which is proven at #6 here) and used $\bar u := u^{\dagger} \gamma^0$

Now it's time to write out the matrix indices explicitly. We want the indices to be contracted so that we end up getting the trace (i.e. we want $A_{ij}B_{ji} = (AB)_{ii} =\operatorname{Tr}(AB)$).

Let us label the matrix indices as follows

$$(\bar u'_{s_1 \color{red}{\sigma}} \gamma_{\color{\green}{\rho} \color{grey}{\theta}}^\alpha(\not{\!p}+\not{\!k}+m)_{\color{blue}{\pi} \color{\green}{\rho}} \gamma_{\color{red}{\sigma}\color{blue}{\pi}}^\beta \ u_{s_2\color{grey}{\theta}})(\bar u_{s_2 \color{yellow}{\nu}} \gamma_{\beta, \color{violet}{\eta} \color{brown}{\kappa}} (\not{\!p}+\not{\!k}+m)_{\color{pink}{\lambda} \color{violet}{\eta}} \gamma_{\alpha, \color{yellow}{\nu} \color{pink}{\lambda}} u_{s_1\color{brown}{\kappa}}')$$

When writing out matrix indices explicitly, we can treat them as numbers. That means we can rearrange them at will. I choose this particular way:

$$Y = \frac{e^4}{16(pk)^2}\gamma_{\color{red}{\sigma}\color{blue}{\pi}}^\beta (\not{\!p}+\not{\!k}+m)_{\color{blue}{\pi} \color{\green}{\rho}}\gamma_{\color{\green}{\rho} \color{grey}{\theta}}^\alpha \Big(\sum_{s_2} u_{s_2\color{grey}{\theta}}\bar u_{s_2 \color{yellow}{\nu}}\Big)\gamma_{\alpha, \color{yellow}{\nu} \color{pink}{\lambda}} (\not{\!p}+\not{\!k}+m)_{\color{pink}{\lambda} \color{violet}{\eta}}\gamma_{\beta, \color{violet}{\eta} \color{brown}{\kappa}}\Big(\sum_{s_1} u_{s_1\color{brown}{\kappa}}' \bar u'_{s_1 \color{red}{\sigma}}\Big)$$ $$=\frac{e^4}{16(pk)^2}\gamma_{\color{red}{\sigma}\color{blue}{\pi}}^\beta (\not{\!p}+\not{\!k}+m)_{\color{blue}{\pi} \color{\green}{\rho}}\gamma_{\color{\green}{\rho} \color{grey}{\theta}}^\alpha \Lambda_{\color{grey}{\theta}\color{yellow}{\nu}}^+\gamma_{\alpha, \color{yellow}{\nu} \color{pink}{\lambda}} (\not{\!p}+\not{\!k}+m)_{\color{pink}{\lambda} \color{violet}{\eta}}\gamma_{\beta, \color{violet}{\eta} \color{brown}{\kappa}}\Lambda_{\color{brown}{\kappa}\color{red}{\sigma}}^{'+}$$ $$=\frac{e^4}{16(pk)^2}\operatorname{Tr}\{\gamma^\beta (\not{\!p}+\not{\!k}+m) \gamma^\alpha \Lambda^+\gamma_{\alpha} (\not{\!p}+\not{\!k}+m) \gamma_{\beta}\Lambda^{'+}\}$$ $$=\frac{e^4}{64(pk)^2m^2} \operatorname{Tr}\{\gamma^\beta (\not{\!p}+\not{\!k}+m) \gamma^\alpha (\not{\!p}+m)\gamma_{\alpha} (\not{\!p}+\not{\!k}+m) \gamma_{\beta}(\not{\!p'}+m)\}$$

I learned a lot so thank you ! But I do not want to celebrate anything yet... I have two questions:

1) What happened to $\epsilon'_{\alpha} \epsilon_{\beta}$?

Note I ignored $\epsilon'_{\alpha} \epsilon_{\beta}$ since the very beginning; if they are not supposed to appear I do not know why not (I simply did it because I noticed I'd get the answer anyway).

2) Am I cheating?

Honestly, it feels like I am. Note I labelled the indices in that specific way only to match the provided answer. What if I were not given the answer? I'd get the same terms but simply disordered. Is that OK? Maybe I am missing something...

 

[Dr.AbeNikIanEdL]

1) What happened to ϵ′αϵβϵα′ϵβ\epsilon'_{\alpha} \epsilon_{\beta}?

Note I ignored ϵ′αϵβϵα′ϵβ\epsilon'_{\alpha} \epsilon_{\beta} since the very beginning; if they are not supposed to appear I do not know why not (I simply did it because I noticed I'd get the answer anyway).

You also sum over the photon polarizations (this is why Eq. 8.65 in your book works). You should have relations like $\sum_{\lambda} \epsilon_\lambda^\mu \epsilon_\lambda^\nu = -g^{\mu\nu}$ (depending on gauge choices etc.), where $\lambda$ denotes the polarization states. Note how the lorentz indices in $\mathcal{M}$ are completely contracted, so if writing out $|\mathcal{M}|^2$ the indices in $\mathcal{M}$ and $\mathcal{M}^\dagger$ should really be given different names. Only after summing over polarization states is the contraction actually between gammas pairwise from $\mathcal{M}$ and $\mathcal{M}^\dagger$.

2) Am I cheating?

Honestly, it feels like I am. Note I labelled the indices in that specific way only to match the provided answer. What if I were not given the answer? I'd get the same terms but simply disordered. Is that OK? Maybe I am missing something...

The labels on the indices should correspond to the original matrix product, i.e. $\bar{u}\Gamma u = \bar{u}_i\Gamma_{ij}u_j$. So your step in between there looks wrong to me indeed.

 

[JD_PM]

You should have relations like $\sum_{\lambda} \epsilon_\lambda^\mu \epsilon_\lambda^\nu = -g^{\mu\nu}$ (depending on gauge choices etc.), where $\lambda$ denotes the polarization states.

Oh I see. All I've found in Mandl & Shaw about such an expression is this https://ia800108.us.archive.org/32/items/FranzMandlGrahamShawQuantumFieldTheoryWiley2010/Franz%20Mandl%2C%20Graham%20Shaw-Quantum%20Field%20Theory-Wiley%20%282010%29.pdf

$$\sum_{\lambda=1}^2 \epsilon_{\lambda}^{\alpha} (\vec k) \epsilon_{\lambda}^{\beta} (\vec k) = -g^{\mu \nu}-\frac{1}{(kn)^2} \Big[ k^{\mu} k^{\nu} -(kn)( k^{\mu}n^{\nu}+k^{\nu} n^{\mu}) \Big]$$

I have to say I know little on this but here's my guess: I'd assume that in the gauge we're working with the second term in the above equation vanishes (I was going to use the on-shell condition $k^2=0=k'^2$ as justification but I think we potentially get $\frac 0 0$ indeterminacy so I indeed do not know the reason). Thus we end up getting

$$\sum_{\lambda=1}^2 \epsilon_{\lambda}^{\mu} (\vec k) \epsilon_{\lambda}^{\nu} (\vec k) = -g^{\mu \nu}$$

Which is exactly what you suggested.

Note how the lorentz indices in $\mathcal{M}$ are completely contracted, so if writing out $|\mathcal{M}|^2$ the indices in $\mathcal{M}$ and $\mathcal{M}^\dagger$ should really be given different names. Only after summing over polarization states is the contraction actually between gammas pairwise from $\mathcal{M}$ and $\mathcal{M}^\dagger$.

So if I am not mistaken you suggest to relabel the indices of $\mathscr{M}$ and $\mathscr{M}^{\dagger}$. Thus we start from an equation:$$Y=\frac{e^4 (\epsilon_{\lambda}^{\mu} \epsilon_{\lambda}^{\nu})}{16(pk)^2} \Big(\sum_{s_1} \sum_{s_2}(\bar u'_{s_1} \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u_{s_2})(\bar u'_{s_1} \gamma_{\alpha}(\not{\!p}+\not{\!k}+m) \gamma_\beta \ u_{s_2})^{\dagger}\Big)_{\mu \nu}$$

And I actually have to show that

$$Y=\frac{e^4 (\epsilon_{\lambda}^{\mu} \epsilon_{\lambda}^{\nu})}{16(pk)^2} \Big(\sum_{s_1} \sum_{s_2}(\bar u'_{s_1} \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u_{s_2})(\bar u'_{s_1} \gamma_{\alpha}(\not{\!p}+\not{\!k}+m) \gamma_\beta \ u_{s_2})^{\dagger}\Big)_{\mu \nu}$$ $$=
\frac{e^4}{16(pk)^2}\sum_{s_1} \sum_{s_2} (\bar u'_{s_1} \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u_{s_2})(\bar u_{s_2} \gamma^0 \gamma_\beta (\not{\!p}+\not{\!k}+m) \gamma_\alpha \gamma^0 u_{s_1}')$$

Do you agree?

 

[nrqed]

Oh I see. All I've found in Mandl & Shaw about such an expression is this https://ia800108.us.archive.org/32/items/FranzMandlGrahamShawQuantumFieldTheoryWiley2010/Franz%20Mandl%2C%20Graham%20Shaw-Quantum%20Field%20Theory-Wiley%20%282010%29.pdf

$$\sum_{\lambda=1}^2 \epsilon_{\lambda}^{\alpha} (\vec k) \epsilon_{\lambda}^{\beta} (\vec k) = -g^{\mu \nu}-\frac{1}{(kn)^2} \Big[ k^{\mu} k^{\nu} -(kn)( k^{\mu}n^{\nu}+k^{\nu} n^{\mu}) \Big]$$

I have to say I know little on this but here's my guess: I'd assume that in the gauge we're working with the second term in the above equation vanishes (I was going to use the on-shell condition $k^2=0=k'^2$ as justification but I think we potentially get $\frac 0 0$ indeterminacy so I indeed do not know the reason). Thus we end up getting

$$\sum_{\lambda=1}^2 \epsilon_{\lambda}^{\mu} (\vec k) \epsilon_{\lambda}^{\nu} (\vec k) = -g^{\mu \nu}$$

Which is exactly what you suggested.
So if I am not mistaken you suggest to relabel the indices of $\mathscr{M}$ and $\mathscr{M}^{\dagger}$. Thus we start from an equation:$$Y=\frac{e^4 (\epsilon_{\lambda}^{\mu} \epsilon_{\lambda}^{\nu})}{16(pk)^2} \Big(\sum_{s_1} \sum_{s_2}(\bar u'_{s_1} \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u_{s_2})(\bar u'_{s_1} \gamma_{\alpha}(\not{\!p}+\not{\!k}+m) \gamma_\beta \ u_{s_2})^{\dagger}\Big)_{\mu \nu}$$

This is not correct. First, it does not make sense to have indices $\mu \nu$ at the end, all the Lorentz indices of what you wrote inside the expression are already shown.

And you did not really square because you would then have four polarization vectors, not two. First, write $\mathscr{M}$ and $\mathscr{M}^{\dagger}$ in detail. The key point is that you must use different Lorenzt indices in each, for example use $\alpha,\beta$ in the first and $\mu,\nu$ in the second. Then multiply them and do the sums and averages over polarizations and spins.

 

[Dr.AbeNikIanEdL]

Oh I see. All I've found in Mandl & Shaw about such an expression is this https://ia800108.us.archive.org/32/items/FranzMandlGrahamShawQuantumFieldTheoryWiley2010/Franz%20Mandl%2C%20Graham%20Shaw-Quantum%20Field%20Theory-Wiley%20%282010%29.pdf

2∑λ=1ϵαλ(→k)ϵβλ(→k)=−gμν−1(kn)2[kμkν−(kn)(kμnν+kνnμ)]∑λ=12ϵλα(k→)ϵλβ(k→)=−gμν−1(kn)2[kμkν−(kn)(kμnν+kνnμ)]​

\sum_{\lambda=1}^2 \epsilon_{\lambda}^{\alpha} (\vec k) \epsilon_{\lambda}^{\beta} (\vec k) = -g^{\mu \nu}-\frac{1}{(kn)^2} \Big[ k^{\mu} k^{\nu} -(kn)( k^{\mu}n^{\nu}+k^{\nu} n^{\mu}) \Big]
I have to say I know little on this but here's my guess: I'd assume that in the gauge we're working with the second term in the above equation vanishes (I was going to use the on-shell condition k2=0=k′2k2=0=k′2k^2=0=k'^2 as justification but I think we potentially get 0000\frac 0 0 indeterminacy so I indeed do not know the reason). Thus we end up getting

2∑λ=1ϵμλ(→k)ϵνλ(→k)=−gμν∑λ=12ϵλμ(k→)ϵλν(k→)=−gμν​

\sum_{\lambda=1}^2 \epsilon_{\lambda}^{\mu} (\vec k) \epsilon_{\lambda}^{\nu} (\vec k) = -g^{\mu \nu}
Which is exactly what you suggested.

It is a little more complicated then just the second term vanishing. Basically using only $-g_{\mu\nu}$ in as the polarization sum is only effectively correct if everything is correctly plugged into a calculation of a gauge invariant amplitude. The terms proportional to $k$ can be dropped due to a Ward identity. Mandl and Shaw appear to give more details in deriving (8.36), their (8.33) is the Ward identity in this case.

Do you agree?

No, see @nrqed ’s post.

 

[JD_PM]

Alright, let's go step by step.

First, write $\mathscr{M}$ and $\mathscr{M}^{\dagger}$ in detail. The key point is that you must use different Lorenzt indices in each, for example use $\alpha,\beta$ in the first and $\mu,\nu$ in the second.

$$\mathscr{M}= -i e^2 \epsilon'_\alpha~\epsilon_\beta ~ \frac{\bar u' \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u}{2(pk)}, \tag 6 $$

$$\mathscr{M}^{\dagger}= i e^2 \epsilon_\nu~\epsilon'_\mu ~ \frac{\bar u \gamma^\nu(\not{\!p}+\not{\!k}+m) \gamma^\mu \ u'}{2(pk)}, \tag 7 $$

Then multiply them and do the sums and averages over polarizations and spins.

So we have (where $\lambda$ denotes polarization states)

$$Y=\frac{e^4}{16(pk)^2}\sum_{\lambda} (\epsilon_{\alpha}^{'\lambda} \epsilon_{\beta}^{\lambda} \epsilon_{\nu}^{\lambda} \epsilon_{\mu}^{'\lambda}) \sum_{s_1} \sum_{s_2}(\bar u'_{s_1} \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u_{s_2})(\bar u_{s_2} \gamma^\nu (\not{\!p}+\not{\!k}+m) \gamma^\mu u_{s_1}') \tag 7$$

Time to sum over polarization states; we know that $\sum_{\lambda=1}^2 \epsilon_{\lambda}^{\mu} (\vec k) \epsilon_{\lambda}^{\nu} (\vec k) = -g^{\mu \nu}$ so (I think) we get

$$Y=-\frac{e^4}{16(pk)^2} g_{\alpha \beta \nu \mu} \sum_{s_1} \sum_{s_2}(\bar u'_{s_1} \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u_{s_2})(\bar u_{s_2} \gamma^\nu (\not{\!p}+\not{\!k}+m) \gamma^\mu u_{s_1}') \tag 8$$

Before summing over spin states I have to contract indices correctly.I was thinking of using the following rule

$$x_{\mu} = g_{\mu \nu} x^{\nu}$$

But I guess I've done something wrong, because I expect to get an answer with the indices of the $\gamma$ matrices summed over (i.e. $\gamma^{\alpha}...\gamma_{\alpha}$). However, using the above rule I end up with all indices in a covariant way (i.e. $\gamma_{\alpha}...\gamma_{\beta}$):

$$Y=-\frac{e^4}{16(pk)^2} \sum_{s_1} \sum_{s_2}(\bar u'_{s_1} \gamma_\alpha(\not{\!p}+\not{\!k}+m) \gamma_\beta \ u_{s_2})(\bar u_{s_2} \gamma_\nu (\not{\!p}+\not{\!k}+m) \gamma_\mu u_{s_1}') \tag 9$$

I know that Eq. 9 is wrong. Besides, I do not get rid of the -ive sign...

Is Eq. 8 OK or neither?

 

[Dr.AbeNikIanEdL]

Ok, matrix elements look good to me now. The problem however already starts in your (7) (note that you have two (7)s, I will ignore the first tag at the conjugate matrix element). There really are two photons for which you have to sum the polarization states, one incoming and one outgoing, you might want to call them $\lambda$ and $\lambda^\prime$ matching the $\epsilon$ and $\epsilon^\prime$, and then have separate sums over them (just as you have different sums for $s_1$ and $s_2$ and not just one spin sum).

Your (8) has an object $g_{\alpha\beta\mu\nu}$ that does not make sense (might be a typo but it is impossible to tell wether you mean the right thing). I assume this and (9) should be resolved by what I wrote above.

 

[JD_PM]

Ok, matrix elements look good to me now. The problem however already starts in your (7) (note that you have two (7)s, I will ignore the first tag at the conjugate matrix element).

Ups my bad.

There really are two photons for which you have to sum the polarization states, one incoming and one outgoing, you might want to call them $\lambda$ and $\lambda^\prime$ matching the $\epsilon$ and $\epsilon^\prime$, and then have separate sums over them (just as you have different sums for $s_1$ and $s_2$ and not just one spin sum).

Oh I see, thank you very much. So Eq. 7 becomes

$$Y=\frac{e^4}{16(pk)^2}\sum_{\lambda} \sum_{'\lambda} (\epsilon_{\alpha}^{'\lambda} \epsilon_{\beta}^{\lambda} \epsilon_{\nu}^{\lambda} \epsilon_{\mu}^{'\lambda}) \sum_{s_1} \sum_{s_2}(\bar u'_{s_1} \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u_{s_2})(\bar u_{s_2} \gamma^\nu (\not{\!p}+\not{\!k}+m) \gamma^\mu u_{s_1}') \tag 7$$

Then using $\sum_{\lambda=1}^2 \epsilon_{\lambda}^{\mu} (\vec k) \epsilon_{\lambda}^{\nu} (\vec k) = -g^{\mu \nu}$ we get$$Y=\frac{e^4}{16(pk)^2}g_{\beta \nu}g_{\alpha \mu} \sum_{s_1} \sum_{s_2}(\bar u'_{s_1} \gamma^\alpha(\not{\!p}+\not{\!k}+m) \gamma^\beta \ u_{s_2})(\bar u_{s_2} \gamma^\nu (\not{\!p}+\not{\!k}+m) \gamma^\mu u_{s_1}') \tag{10}$$

Time to contract indices using $x_{\mu} = g_{\mu \nu} x^{\nu}$. We get

$$Y=\frac{e^4}{16(pk)^2} \sum_{s_2}(\bar u'_{s_1} \gamma_\mu(\not{\!p}+\not{\!k}+m) \gamma_\nu \ u_{s_2})(\bar u_{s_2} \gamma^\nu (\not{\!p}+\not{\!k}+m) \gamma^\mu u_{s_1}') \tag{11}$$

Eq. 11 makes sense to me.

 

[JD_PM]

Now we have to write out the matrix indices explicitly.

Let's supose we're not given the answer, so that selecting the spinor indices in the following convinient way

$$(\bar u'_{s_1 \color{red}{\sigma}} \gamma_{\color{\green}{\rho} \color{grey}{\theta}}^\alpha(\not{\!p}+\not{\!k}+m)_{\color{blue}{\pi} \color{\green}{\rho}} \gamma_{\color{red}{\sigma}\color{blue}{\pi}}^\beta \ u_{s_2\color{grey}{\theta}})(\bar u_{s_2 \color{yellow}{\nu}} \gamma_{\beta, \color{violet}{\eta} \color{brown}{\kappa}} (\not{\!p}+\not{\!k}+m)_{\color{pink}{\lambda} \color{violet}{\eta}} \gamma_{\alpha, \color{yellow}{\nu} \color{pink}{\lambda}} u_{s_1\color{brown}{\kappa}}')$$

to get the answer is not an option.

How could we proceed to write out the spinor indices then?

 

[nrqed]

Now we have to write out the matrix indices explicitly.

Let's supose we're not given the answer, so that selecting the spinor indices in the following convinient way

$$(\bar u'_{s_1 \color{red}{\sigma}} \gamma_{\color{\green}{\rho} \color{grey}{\theta}}^\alpha(\not{\!p}+\not{\!k}+m)_{\color{blue}{\pi} \color{\green}{\rho}} \gamma_{\color{red}{\sigma}\color{blue}{\pi}}^\beta \ u_{s_2\color{grey}{\theta}})(\bar u_{s_2 \color{yellow}{\nu}} \gamma_{\beta, \color{violet}{\eta} \color{brown}{\kappa}} (\not{\!p}+\not{\!k}+m)_{\color{pink}{\lambda} \color{violet}{\eta}} \gamma_{\alpha, \color{yellow}{\nu} \color{pink}{\lambda}} u_{s_1\color{brown}{\kappa}}')$$

to get the answer is not an option.

How could we proceed to write out the spinor indices then?

I am puzzled by the way you assign the indices. If you have, say, $\bar u' A B C u$, the indices should belike this (it is conventional to use Greek letters for Lorentz indices and Latin letters for the matrix indices):

$$ \bar u'_i \, A_{ij} \, B_{jk} \, C_{kl} \, u_l $$

Note the order.

 

[JD_PM]

Thank you nrqed and Dr.AbeNikIanEdL, I think I got it

I am puzzled by the way you assign the indices. If you have, say, $\bar u' A B C u$, the indices should belike this (it is conventional to use Greek letters for Lorentz indices and Latin letters for the matrix indices):

$$ \bar u'_i \, A_{ij} \, B_{jk} \, C_{kl} \, u_l $$

Note the order.

And I actually missed that Dr.AbeNikIanEdL pointed it out at #7. My apologies.

The labels on the indices should correspond to the original matrix product, i.e. $\bar{u}\Gamma u = \bar{u}_i\Gamma_{ij}u_j$. So your step in between there looks wrong to me indeed.

So writing out the matrix indices of Eq. 11 yields

$$Y=\frac{e^4}{16(pk)^2} \sum_{s_1}\sum_{s_2}(\bar u'_{s_1 \color{red}{i}} \gamma_{\mu, \color{red}{i}\color{green}{j}}(\not{\!p}+\not{\!k}+m)_{\color{green}{j}\color{grey}{k}} \gamma_{\nu, \color{grey}{k}\color{blue}{l}} \ u_{s_2 \color{blue}{l}} \bar u_{s_2 \color{orange}{m}} \gamma^\nu_{\color{orange}{m} \color{brown}{n}} (\not{\!p}+\not{\!k}+m)_{\color{brown}{n}\color{pink}{ñ}} \gamma^\mu_{\color{pink}{ñ}\color{aqua}{o}} u_{s_1\color{aqua}{o}}') \tag{12}$$ $$=\frac{e^4}{16(pk)^2} \sum_{s_1} (\bar u'_{s_1 \color{red}{i}} \gamma_{\mu, \color{red}{i}\color{green}{j}}(\not{\!p}+\not{\!k}+m)_{\color{green}{j}\color{grey}{k}} \gamma_{\nu, \color{grey}{k}\color{blue}{l}} \Lambda_{\color{blue}{l}\color{orange}{m}}^+ \gamma^\nu_{\color{orange}{m} \color{brown}{n}} (\not{\!p}+\not{\!k}+m)_{\color{brown}{n}\color{pink}{ñ}} \gamma^\mu_{\color{pink}{ñ}\color{aqua}{o}} u_{s_1\color{aqua}{o}}')$$

We are allowed to rearrange it as follows and get

$$Y=\frac{e^4}{16(pk)^2} (\gamma_{\mu, \color{red}{i}\color{green}{j}}(\not{\!p}+\not{\!k}+m)_{\color{green}{j}\color{grey}{k}} \gamma_{\nu, \color{grey}{k}\color{blue}{l}} \Lambda_{\color{blue}{l}\color{orange}{m}}^+ \gamma^\nu_{\color{orange}{m} \color{brown}{n}} (\not{\!p}+\not{\!k}+m)_{\color{brown}{n}\color{pink}{ñ}} \gamma^\mu_{\color{pink}{ñ}\color{aqua}{o}} \sum_{s_1} u_{s_1\color{aqua}{o}}'\bar u'_{s_1 \color{red}{i}})$$ $$=\frac{e^4}{16(pk)^2} (\gamma_{\mu, \color{red}{i}\color{green}{j}}(\not{\!p}+\not{\!k}+m)_{\color{green}{j}\color{grey}{k}} \gamma_{\nu, \color{grey}{k}\color{blue}{l}} \Lambda_{\color{blue}{l}\color{orange}{m}}^+ \gamma^\nu_{\color{orange}{m} \color{brown}{n}} (\not{\!p}+\not{\!k}+m)_{\color{brown}{n}\color{pink}{ñ}} \gamma^\mu_{\color{pink}{ñ}\color{aqua}{o}} \Lambda_{\color{aqua}{o}\color{red}{i}}^{'+})$$ $$=\frac{e^4}{64 m^2(pk)^2} (\gamma_{\mu, \color{red}{i}\color{green}{j}}(\not{\!p}+\not{\!k}+m)_{\color{green}{j}\color{grey}{k}} \gamma_{\nu, \color{grey}{k}\color{blue}{l}} (\not{\!p}+m)_{\color{blue}{l}\color{orange}{m}} \gamma^\nu_{\color{orange}{m} \color{brown}{n}} (\not{\!p}+\not{\!k}+m)_{\color{brown}{n}\color{pink}{ñ}} \gamma^\mu_{\color{pink}{ñ}\color{aqua}{o}} (\not{\!p'}+m)_{\color{aqua}{o}\color{red}{i}}) \tag{13}$$

We use $A_{ij}B_{ji} = (AB)_{ii} =\operatorname{Tr}(AB)$ on Eq. 13 to get the desired answer:

$$Y= \frac{e^4}{64 m^2(pk)^2} (\gamma_{\mu, \color{red}{i}\color{green}{j}}(\not{\!p}+\not{\!k}+m)_{\color{green}{j}\color{grey}{k}} \gamma_{\nu, \color{grey}{k}\color{blue}{l}} (\not{\!p}+m)_{\color{blue}{l}\color{orange}{m}} \gamma^\nu_{\color{orange}{m} \color{brown}{n}} (\not{\!p}+\not{\!k}+m)_{\color{brown}{n}\color{pink}{ñ}} \gamma^\mu_{\color{pink}{ñ}\color{aqua}{o}} (\not{\!p'}+m)_{\color{aqua}{o}\color{red}{i}})$$ $$=\frac{e^4}{64 m^2(pk)^2} \operatorname{Tr}\{\gamma_{\mu}(\not{\!p}+\not{\!k}+m) \gamma_{\nu} (\not{\!p}+m) \gamma^\nu (\not{\!p}+\not{\!k}+m) \gamma^\mu (\not{\!p'}+m)\}$$