Bhabha Scattering: Why Only Two Leading Order Diagrams?

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SUMMARY

Bhabha scattering, represented by the process e+ e- → e+ e-, involves only two leading order Feynman diagrams: the s-channel and t-channel diagrams. The s-channel diagram corresponds to the reaction AB → CD, while the t-channel diagram represents AC → BD. The u-channel diagram, which involves Moller scattering (e+ e+ → e+ e+), does not contribute to Bhabha scattering. The concept of "crossing" in Feynman diagrams refers to the rotation of the diagram rather than the literal crossing of lines.

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I don't understand why in Bhabha scattering (e+ e- -> e+ e-) why there are only two leading order Feynman diagrams. It seems to me like there should be s, t, and u-channel diagrams. Could someone explain why I am wrong?
 
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Actually there is only one diagram. If the reaction in the s-channel is AB → CD (e+ e- → e+ e-) then the reaction in the t-channel is ACBD (e+ e- → e+ e-) and both of these contribute to Bhabha scattering. But the reaction in the u-channel is Moller scattering: AD → CB (e+ e+ → e+ e+).
 
Sorry if I'm being slow, but the attached Feynman diagram seems to be a u-channel diagram for the reaction e+ e- -> e+ e-.

(Apologies -- I don't know how to make Feynman diagrams in latex)
 

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! "Crossing" does not mean that the lines literally cross over one another. The diagram is just rotated, so that one or more legs move from a past direction to a future direction, or vice versa.
 
Oh my. This is embarrassing -- I thought for a minute the diagram I just drew was distinct from the other two!

Thank you!
 
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