Bhabha Scattering

[gamma5772]

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?

[Bill_K]

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 AC → BD (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+).

[gamma5772]

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)

[Bill_K]

! "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.

[gamma5772]

Oh my. This is embarrassing -- I thought for a minute the diagram I just drew was distinct from the other two!

Thank you!