When can we ignore the delta function in th Feynman rules?

[Vereinsamt]

in peskin-schroeder and http://www.hep.phy.cam.ac.uk/batley/particles/handout_04.pdf" the amplitude for $$e^-e^+\rightarrow \mu^- \mu^+$$ is written using feynman rules as follows
$$-iM=[\bar{v}(p_2)(-ie\gamma^\mu )u(p_1)] \frac{-ig_{\mu\nu}}{q^2}[\bar{u}(k_1)(-ie\gamma^\nu )v(k_2)]$$

but what about the delta function integeration? is it already done here?

thanks in advanced!

 

[SporadicSmile]

I may well be wrong here, but that looks like the M matrix part of the S matrix, and it is the S matrix that has the delta function in it, so you shouldn't be expecting a delta function.

again i DID only just do this, so i may be wrong.

 

[denian]

The delta function in the Feynman rules is used to represent the conservation of four-momentum at each vertex in the diagram. In the specific example you provided, the delta function integration is already accounted for in the expression for the amplitude. This is because the delta function in this case is represented by the factor of -ig_{\mu\nu}/q^2, which is the propagator for the photon. This factor ensures that the four-momentum is conserved at that vertex. Therefore, there is no need to explicitly include the delta function integration in this case. However, there may be cases where the delta function integration needs to be included separately, such as when there are multiple particles involved in the interaction. It is important to carefully follow the Feynman rules and pay attention to the conservation of four-momentum at each vertex to determine when the delta function integration needs to be included.