Evaluating Feynman diagram in QFT without external fermion lines

[Amanheis]

...like in Photon-Photon-scattering. I know this doesn't make sense physically but the value for the diagram should still be computeable. If I want to put the expression for the matrix element together, I get a matrix, but it should be a scalar, right? Since the spinor bi-product is missing...
Do I have to take the trace?

 

[Vanadium 50]

I'm not quite sure what you are doing, but it sounds like you are missing the polarization indices for the external photon lines. Could that be it?

 

[Amanheis]

No, contracting e_\mu with gamma^\mu gives still a matrix, since e_\mu is just a 4-vector. Or am I wrong on this part?

 

[Avodyne]

For every closed fermion loop, there is a trace over the spin indices.

 

[Amanheis]

Thanks, that helped