Pair production in boson-boson collision

[member 251684]

Homework Statement

I'm given this interaction Lagrangian:

$$L_I (x) = -g \overline{\psi}(x) \psi (x) \phi (x)$$

where $$g$$ is the coupling constant, $$\psi$$ is an electron and $$\phi$$ is a neutral scalar field. I have to calculate the amplitude for process
$$\psi + \psi \rightarrow e^- + e^+$$ that is to say that I have to calculate the matrix element

$$\langle e^+ (\vec{p}_3) e^- (\vec{p}_4) |S| \phi (\vec{p}_1 ) \phi (\vec{p}_2)\rangle$$

Homework Equations

I'm stuck determining the relative sign of the contributions. I know that there are two Feynman diagrams contributing to the matrix element and that the final fermions are swapped in the two diagrams, so there's a - (minus) sign in the relative contributions.

On the other side, I find that in the first graph the propagator is given by

$$\langle 0 | T [\overline{\psi}(x) \psi(y)] |0 \rangle$$

while in the second graph the propagator is given by

$$\langle 0 | T [\psi (x) \overline{\psi}(y)] |0 \rangle$$

I just need to know if this change in the propagator adds a new - (minus) sign, so that the relative contributions end up having the same sign or if they ends up having opposite sign.

The Attempt at a Solution

I know thath this is a second order process in g and I know also that the final state is produced by a fermionic propagator between the neutral bosons. I've calculated many similar processes but this question on relative signs of the two contributions is really blocking me.

Sorry for my English, I'm not a native speaker. Thank you in advance for you help.

 

[Avodyne]

There is no relative minus sign. Physically, this is because the two outgoing fermions are not identical. Or, you can see the two diagrams as related by swapping the two incoming particles, which are identical; but these are bosons, so there is no relative sign.

 

[member 251684]

Oh well. I've to think about it. I just tought that "swapping" two fermions always lead to a minus sign...
But, about the propagator, it stays the same in both cases?

Thank you for your help