# Bhabha scattering in the spinor-helicity formalism

I am trying to calculate the square amplitude for Bhabha scattering $$e^-(p_1)e^+(p_2)\rightarrow e^-(p_3)e^+(p_4)$$ using the spinor-helicity formalism but one of the interference terms just will not fit.

The unpolarised, spin-averaged squared amplitude should be $$\overline{|M|^2}=2e^4\left(\frac{u^2+t^2}{s^2}+\frac{u^2+s^2}{t^2}+\frac{2u^2}{st}\right).$$

The first term comes from the s-channel, the second from the t-channel and the last term is an interference term and there is also where my problem originates.

The matrix elements for the s- and t-channel are $$M_s=\frac{-ie^2}{s}\overline{v}_2\gamma^\mu u_1\times\overline{u}_3\gamma_\mu v_4,\quad M_t=\frac{-ie^2}{s}\overline{v}_2\gamma^\mu v_4\times\overline{u}_3\gamma_\mu u_1.$$

We have four possible helicity combinations for each of the channels, which reduce down to two due to crossing symmetry. The four remaining are the following. $$M^{-++-}_s=(M^{+--+}_s)^\ast=\frac{-ie^2}{s}\langle2\gamma^\mu1]\langle3\gamma_\mu4]=\frac{2ie^2}{s}\langle23\rangle,$$ $$M^{-+-+}_s=(M^{+-+-}_s)^\ast=\frac{-ie^2}{s}\langle2\gamma^\mu1]\langle4\gamma_\mu3]=\frac{2ie^2}{s}\langle24\rangle,$$ $$M^{-++-}_t=(M^{+--+}_t)^\ast=\frac{-ie^2}{t}\langle2\gamma^\mu4]\langle3\gamma_\mu1]=\frac{2ie^2}{t}\langle23\rangle,$$ $$M^{--++}_t=(M^{++--}_t)^\ast=\frac{-ie^2}{t}\langle4\gamma^\mu2]\langle3\gamma_\mu1]=\frac{2ie^2}{t}\langle43\rangle,$$

Only the same helicity configurations can interfere. Thus, I find that $$|M^{-+-+}_s|^2=|M^{+-+-}_s|^2=4e^4\frac{t^2}{s^2},$$ $$|M^{--++}_t|^2=|M^{++--}_t|^2=4e^4\frac{s^2}{t^2},$$ $$|M^{-++-}_s+M^{-++-}_t|^2=4e^4\frac{u^2}{s^2}+4e^4\frac{u^2}{t^2}+\frac{e^4}{st}\Big(\underbrace{\langle23\rangle\langle41\rangle+\langle32\rangle\langle41\rangle}_{=-2u^2}\Big).$$

Hence, summing over all helicity configurations (gives a factor of 2) and spin-averaging by dividing by 4, I arrive at the following result: $$\overline{|M|^2}=2e^4\left(\frac{u^2+t^2}{s^2}+\frac{u^2+s^2}{t^2}-\frac{2u^2}{st}\right),$$ which obviously has the wrong sign in front of the last term.

I do not know where I have made a mistake in my calculation. Here is the bit that leads to the third term: $$M^{-+-+}_s\times(M^{-+-+}_t)^\ast\sim\langle23\rangle\underbrace{\Big(\langle23\rangle\Big)^\ast}_{=\langle14\rangle},$$ $$M^{-+-+}_t\times(M^{-+-+}_t)^\ast\sim\langle23\rangle\underbrace{\Big(\langle23\rangle\Big)^\ast}_{=\langle41\rangle}.$$

As we see, we have a factor $$\langle14\rangle=-\langle41\rangle=-2p_1p_4=-u^2$$ in the first line and equally in the second since $$\langle14\rangle=\langle41\rangle$$.

Does anybody have an idea what I did wrong?

It turns out that this is a case of 'know your Feynman rules'. First of all, the helicity notation in my post is not consistent. We should have the following matrix elements: $$M^{----}_s=\frac{-ie^2}{s}\langle2\gamma^\mu1]\langle3\gamma^\mu4]=\frac{2ie^2}{s}\langle23\rangle,$$ $$M^{--++}_s=\frac{-ie^2}{s}\langle2\gamma^\mu1][3\gamma^\mu4\rangle=\frac{2ie^2}{s}\langle24\rangle,$$ $$M^{----}_t=\frac{-ie^2}{t}\langle3\gamma^\mu1]\langle2\gamma^\mu4]=\frac{2ie^2}{s}\langle32\rangle,$$ $$M^{-+-+}_t=\frac{-ie^2}{t}\langle3\gamma^\mu1][2\gamma^\mu4\rangle=\frac{2ie^2}{s}\langle34\rangle.$$
Now it is clear that $$M^{----}_{s,t}$$ interfere. However, what I did not consider is that there is a relative sign between both diagrams due to the fact that the t-channel diagram is related to the s-channel diagram by switching the positron with the electron. This incurs a minus sign. We find that $$|M^{----}_{s-t}|^2=\underbrace{|M^{----}_s|^2}_{=4e^4u^2/s^2}+\underbrace{|M^{----}_t|^2}_{=4e^4u^2/t^2}-\frac{4e^4}{st}\Big(\underbrace{\langle23\rangle\langle41\rangle+\langle32\rangle\langle14\rangle}_{=-2u^2}\Big)$$ $$=4e^4\left(\frac{u^2}{s^2}+\frac{u^2}{t^2}+\frac{2u^2}{st}\right),$$ which together with $$|M^{--++}_s|^2$$ and $$|M^{-+-+}_t|^2$$ gives the right result.
• Hey, quick clarification. By $M(----)$ do you mean $M(e^-_L e^+_R\to e^-_L e^+_R)$? So the 'signs' in your amplitude $M(----)$ are the helicities of the SPINORS involved(eg right handed positron $e^+_R$ has a left helicity spinor), and not the particle/antiparticles? Jul 8, 2021 at 23:32