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I am currently reading Thomson's "Modern particle physics", and I have trouble understanding a concept that he somewhat gives for granted. Using Feynman rules, I am able to compute the matrix element of a Feynman diagram (let's restrict ourselves to QED here), given specific initial and final spin states.

Now, I think the general way of computing the total matrix element is to sum all the matrix elements for possible helicity configurations, then take the modulus of this number. Indeed, if I understand correctly, the matrix elements may interfere. However, in the book, he says that since all the helicities eigenstates are orthogonal, we can just add the moduli, instead of summing before taking the modulus.

Let me give and example to be clearer : Suppose we have an electron quark scattering. Suppose also both the incoming electron and quark are right handed. Then, in the general way we should do : $$|M_\textrm{tot}|^2 = |M_{RR\rightarrow RR}+M_{RR\rightarrow RL}+M_{RR\rightarrow LR}+M_{RR\rightarrow LL}|^2.$$

In the book, invoking orthogonality, he does instead : $$|M_\textrm{tot}|^2 = |M_{RR\rightarrow RR}|^2+|M_{RR\rightarrow RL}|^2+|M_{RR\rightarrow LR}|^2+|M_{RR\rightarrow LL}|^2$$

I tried proving that they both yield the same thing with trace techniques, but I am unable to simplify the extra sums that appear.

Another question is in the case where the initial quark and electron do not have determinate helicities. In the book, he makes the mean of the matrix elements for each possible initial helicity configuration. Now, is this a general case, or can it be that initial state configurations interfere ? If the answer is yes, how would we go about making the mean in that case ?

If it can help, my questions revolve around page 130 of Thomson's book.

Here's what I tried to do. Let's first write the matrix element for know initial $(s,r)$ and final $(s',r')$ helicity configurations : $$M_{sr\rightarrow s'r'}=\alpha \overline{u_e^{s'}}(p_3)\gamma^{\mu}u_e^s(p_1)\times \overline{u_q^{r'}}(p_4)\gamma_{\mu}u_q^r(p_2)$$

Where $\alpha$ is just a constant in front that we don't really care about for what I'm doing.

Now, in the case of the book, what we would do is compute the modulus of this matrix, and then sum on $s,r,s',r'$. This will make appear terms like $\sum_s u^s_i\overline{u^s}_j = (p_{mu}\gamma^{mu}+m)_{ij}$ which can then simplified into traces. However, in my (supposedly more general case), we sum before and then take the modulus.

Let's choose $s,r=1$ to simplify, otherwise we will be having 8 sums, but it's easily generalizable. We would thus have something like that for the "total" M :

$$M_\textrm{tot} = \sum_{s',r'}\alpha \overline{u_e^{s'}}(p_3)\gamma^{\mu}u_e^s(p_1)\times \overline{u_q^{r'}}(p_4)\gamma_{\mu}u_q^r(p_2)$$

Taking $|M_\textrm{tot}|^2 = M_\textrm{tot}M_\textrm{tot}^{\dagger}$ $$|M_\textrm{tot}|^2=\alpha^2 \sum_{a,b}\sum_{c,d}\overline{u_e^{a}}(p_3)\gamma^{\mu}u_e^s(p_1)\overline{u_e^{s}}(p_1)\gamma^{\nu}u_e^c(p_3)\times \overline{u_q^{b}}(p_4)\gamma_{\mu}u_q^r(p_2)\overline{u_q^{r}}(p_2)\gamma_{\nu}u_q^d(p_4)$$

So what I need to show in short, is that somehow the orthogonality relations make a $\delta_{ac}$ and $\delta_{bd}$ appear. In that case, everything becomes like he does in the book. I cannot however figure out if there is a way to do that. (Preferably without choosing a representation, but I can't be too picky really!)

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    $\begingroup$ "I tried proving that they both yield the same thing with trace techniques, but I am unable to simplify the extra sums that appear. " Can you show what you did? $\endgroup$ – Brian Moths Dec 20 '16 at 16:27

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