Changing the sign of 4-momenta in Feynman diagrams

[geoduck]

If a Feynman diagram only involves bosons, then can you multiply all external 4-momenta by negative one, and still get the same amplitude? Looking at the Feynman rules this seems true.

If the diagram has a fermion (as real or virtual particle), then it seems this is no longer true.

I just want to verify this. There probably should be a deeper reason for this, but just looking at the Feynman rules, it seems it ought to be true when only bosons are involved, and not true when at least one fermion is involved.

 

[kevinferreira]

I don't see any Feynman diagram having only one fermion... Being a spin-1/2 particle it is impossible to have total momentum conservation with only one real or vitusl fermion and all the rest bosons.
Multiplying the external momenta by -1 is equivalent to a time reversal, on my perspective. Time reversal creates a factor of $(-1)^{2j}$ on the wavefunctions, where j is the associsted spin. Of course this is just a phase transformation, so nothing changes when computing cross-sections and other observables.
But I think you don't even have to worry about this, since from my argument we see that there cannot be an odd number of fermions at a vertex, so at each time possible factors will cancel.
Do you think this is reasonable?

 

[geoduck]

I don't see any Feynman diagram having only one fermion... Being a spin-1/2 particle it is impossible to have total momentum conservation with only one real or vitusl fermion and all the rest bosons.
Multiplying the external momenta by -1 is equivalent to a time reversal, on my perspective. Time reversal creates a factor of $(-1)^{2j}$ on the wavefunctions, where j is the associsted spin. Of course this is just a phase transformation, so nothing changes when computing cross-sections and other observables.
But I think you don't even have to worry about this, since from my argument we see that there cannot be an odd number of fermions at a vertex, so at each time possible factors will cancel.
Do you think this is reasonable?

A fermion propagator is neither even nor odd in the sign of the 4-momentum.

A boson propagator however is even.

Since an amplitude is just the product of propagators, you expect anything with just boson propagators to not care about the sign of the 4-momenta flowing through it.

You might be right about having two fermions and it ends up cancelling or something: but I don't immediately see it. I'm going to go through a book that has all the calculations for typical processes and see if there are symmetries in changing the the sign of all 4-momenta.

I'll have to review discrete symmetries like time reversal.

 

[andrien]

you are thinking in terms of propagators which does comes into picture when one will consider interaction.two fermions will interact by an exchange of some boson and propagator of boson gets involved.In no case you will find any propagator all through space.it is rubbish.you might have seen in a simple feynman diagram for say two electrons interacting with each other by exchange of virtual photon.Any electron can emit it.it is not taken into account which one is emitting and which one is absorbing.There is no preferred direction for it.so whether you write it's momentum in opposite direction.it should not change any physics.