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When doing a computation of scattering cross sections of particles with spin, one usually averages over the initial spins and sums over the final ones. I'm a bit puzzled as to how to do the calculation for a longitudinally polarized cross section of Moller scattering though. The reference I'm using is Scattering of longitudinally polarized fermions by A.M. Bincer, where he does exactly what I'm after, but I think the notation's a bit outdated and he omits some steps, so I've tried re-doing it for clarity.

For Moller scattering, there's two tree-level diagrams for the process, the $t$ and $u$ channel, however I am unable to complete the derivation to obtain the square of the amplitude $\left|\mathcal{M}_{fi}\right|^2$. The probability of course consists of a sum of three terms, the $t$ channel, the $u$ channel (same as $t$, but with the final momenta exchanged), and an interference term with a relative minus sign. For instance, the $\left|\mathcal{M}_{fi}\right|^2$ for the $t$ channel is:

$\frac{e^4}{(p_3-p_1)^4}\bar u^s(p_1)\gamma^\mu u^{s'}(p_2)\bar u^r(p_2)\gamma_{\mu}u^{r'}(p_4)\bar u^{s'}(p_3)\gamma^\nu u^s(p_1)\bar u^{r'}(p_4)\gamma_\nu u^r(p_2)$

where $p_1,p_2$ and $p_3,p_4$ are initial and final momenta, and upper unprimed (primed) indices denote spins of initial (final) particles. If we put spinor indices as well, we can rearrange some terms and use the standard $\Sigma u(p)\bar u(p)=\gamma\cdot{p}+m$ trick, but this doesn't work for the initial spinors as we're assuming they're eigenstates of helicity. In the reference it states that we can write the polarized spinors as: $u^s(p)=\frac{1}{2}\sum\limits_{\epsilon=\pm 1}(1+sh)u^\epsilon(p)$, where $h$ is the helicity operator, and $s=\pm 1$, however I can't for the life of me figure out how to complete the derivation. The author doesn't really show all of the steps in the calculation, and when I write out terms like $u^s_a(p_1)\bar u^s_b(p_1)$ using the said formula, it doesn't look like anything helpful as there's now a sum of the form (with spinor indices summed over):

$u^s_a(p_1)\bar u^s_b(p_1)=\frac{1}{4}\sum\limits_{\epsilon,\epsilon'}(\delta_{aq}+sh_{aq})(\delta_{rb}+sh_{rb})u^\epsilon_q(p_1)\bar u^{\epsilon'}_r(p_1)$,

which I can't seem to simplify a lot as there's now two different spin indices, and a whole bunch of other terms. There's a reference to an identity from W. Heitler - The quantum theory of radiation, but I don't have access to the book so I'm pretty much stuck. Any help or hints would be greatly appreciated.

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  • $\begingroup$ You just write $u_s \bar u_s$ for the particle solution with the particular helicity $s$. It's $(p\cdot \gamma +m)$ times an additional operator involving $p,\gamma$ and perhaps $\gamma_5$. You should be able to find this extra gamma-matrix-like factor from the condition that the helicity operator acting on this $u\bar u$ gives you what it should. (Helicity plus minus one annihilates it both from the right and the left). At the end, even if you compute the specific-helicity cross sections (without summing over polarizations), you will sum traces of products of gamma matrices, just harder ones $\endgroup$ – Luboš Motl May 29 '16 at 17:55
  • $\begingroup$ So, if I understand you correctly, I should write $u^s(p)\bar u^s(p)=(p\cdot \gamma+m)A^s$, where $A^s$ depends on which spin state I take, $s=\pm1$, and then simply invert the relation to get $A^s$ (which I'm assuming depends on the basis, but won't matter for the end result), and then simply plug that back into $\left|\mathcal{M}\right|^2$ and do the traces? $\endgroup$ – blueshift May 29 '16 at 18:43
  • $\begingroup$ Yup, I forgot what $A^s$ really is right now, it's some $(1+\gamma_5 \gamma_i\cdot p_i / |\vec p|)$, or something like that, try it or find it somewhere. When you have this form, the calculation of the cross section boils down to similar operations as in the polarization-blind case, you just need to know the traces of longer products of gamma matrices, aside from a larger number of terms etc. $\endgroup$ – Luboš Motl May 29 '16 at 18:54
  • $\begingroup$ Okay, I've tried doing the calculation, but the result turns out to be too complicated to make anything out of it, can I just pick my coordinate system to be along the $z$ axis to make the eigenspinors take the form $\begin{pmatrix}\sqrt{E-p^3}\begin{pmatrix}1\\0\end{pmatrix}\\ \sqrt{E+p^3}\begin{pmatrix}1\\ 0\end{pmatrix}\end{pmatrix}$ for $s=1$, and similar for the other one? In this case, I can apparently write $A^s$ as a linear combination of the 16 matrices ($\gamma^\mu,\sigma^{\mu\nu}$ etc.) spanning the space, so I can compute the traces easily. Would that work? $\endgroup$ – blueshift May 30 '16 at 17:16
  • $\begingroup$ If you picked particular directions of the momenta and a particular basis for the spinors, you could have written the explicit forms of all the gamma matrices at all other places, too. That's not what I wanted to recommend you but it's possible, too. If you have the explicit form of the spinors, you don't need to waste your time with finding $A^s$, do you? Just calculate things like $\bar u \gamma v$ from the Feynman diagrams directly in components. $\endgroup$ – Luboš Motl May 30 '16 at 17:20

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