# Mølller scattering

I came across Mølller scattering today (which is just a fancy name for electron-electron scattering. I'm confused as to why there are two tree level Feynman diagrams for this process: Check out the wikipedia entry for it.

On the right two different channels are pictured. However I don't see how they are different from each other. If you rotate the outgoing Fermion lines of the electrons with respect to the electron-photon vertices (in the lower diagram) you end up with the exact same diagram above.

Is there really more than one tree level diagram for this process? If so, why are they these two?

(Note: Since charge conservation needs to be satisfied, the two electrons can't annihilate to photon and then be created again in a 2nd vertex, as it is possible for Bhabha scattering)

When drawing Feynman diagrams, it is important to fix the incoming and outgoing particles and their momenta. For the inexperienced, this is ideally done before drawing the rest of the diagram in order to avoid confusions like yours. So, let's assign each state some momentum: Let's give the electron in the upper left corner (4-)momentum $p_1$, lower left $p_2$, upper right $p_3$ and lower right $p_4$. Let us consider all momenta as 'flowing' from left to right (this is an arbitrary choice that makes no difference to the actual calculations).
Then, it immediately becomes clear why the two diagrams are different. For instance, check out the momentum on the photon line (call it $q$, and imagine it flowing top to bottom) that connects both vertices. Conservation of (4-)momentum forces always applies, so we can write it in terms of the momenta flowing into the vertices. For the $t$-channel diagram, it is clear that $q=p_1-p_3=p_2-p_4\equiv \sqrt{t}$. However, for the $u$-channel diagram $q=p_1-p_4=p_2-p_3\equiv \sqrt u$. Thus, the photon carries a different momentum. When carrying through the calculation, it will be seen that this makes a serious difference.