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Tree-level diagram Moller scattering

  • Thread starter kelly0303
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  • #1
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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.

Homework Equations




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!
 

Answers and Replies

  • #2
nrqed
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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.

Homework Equations




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^-##).
 
  • #3
king vitamin
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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 [itex]e^+[/itex] particle.
 
  • #4
nrqed
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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 [itex]e^+[/itex] 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).
 

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