# Tree-level diagram Moller scattering

## Homework Statement

If the electron and photon were spinless, in the non-relativistic limit we can write the Lagrangian as: $$L=-\frac{1}{2}\phi_e(\Box+m_e^2)\phi_e-\frac{1}{2}A_0\Box A_0+em_eA_0\phi_e\phi_e$$ For Moller scattering $$e^-e^- \to e^-e^-$$ Draw the three tree level diagrams following from the Lagrangian.

## The Attempt at a Solution

So the interaction part must couple 2 electrons with a photon. 2 tree level diagrams are normal scattering (like in QED) in t and u channel. What is the 3rd one? Is there another vertex I am missing from the Lagrangian? Thank you!

Related Advanced Physics Homework Help News on Phys.org
nrqed
Homework Helper
Gold Member

## Homework Statement

If the electron and photon were spinless, in the non-relativistic limit we can write the Lagrangian as: $$L=-\frac{1}{2}\phi_e(\Box+m_e^2)\phi_e-\frac{1}{2}A_0\Box A_0+em_eA_0\phi_e\phi_e$$ For Moller scattering $$e^-e^- \to e^-e^-$$ Draw the three tree level diagrams following from the Lagrangian.

## The Attempt at a Solution

So the interaction part must couple 2 electrons with a photon. 2 tree level diagrams are normal scattering (like in QED) in t and u channel. What is the 3rd one? Is there another vertex I am missing from the Lagrangian? Thank you!
There is no third diagram, indeed. It seems to be a mistake in the question (they probably were thinking of $e^-e^+ \to e^+ e^-$).

king vitamin
Gold Member
I don't see why you can't have an s-channel diagram. The problem is just asking for the scattering between scalars right? There's no $e^+$ particle.

nrqed
I don't see why you can't have an s-channel diagram. The problem is just asking for the scattering between scalars right? There's no $e^+$ particle.
A scalar can still be charged. If the scalar electron is charged, there cannot be an s-channel for the process $e^-e^- \to e^-e^-$.
But now I notice that the Lagrangian contains only $\phi_e$, not $\phi_e^*$ so it seems that you are correct and I was wrong, and that their electron is its own antiparticle (in which case the notation $e^-$ is terribly misleading).