# Disconnected diagram Feynman rules momentum space

1. Aug 6, 2008

### Sunset

Hi!

I wonder if the Feynman rules in momentum space can also be applied to disconnected diagrams. So aussume I have a disconnected Feynman diagram 1 , i.e. the product of two connected Feynman diagrams 2 and 3.
I can translate my diagrams with the position space Feynman rules to explicit mathematical expressions:
1->D
2->C_2
3->C_3

D=C_2 * C_3

I can compute the Fourier transforms F(C_2) and F(C_3) by using the Feynman rules in momentum space.

Now if the FR in moentum space would be applicable for disconnected diagrams, this would mean
F(D)=F(C_2)*F(C_3)

This means the inverse Fourier transform yields
D= F^-1 [ F(C_2)*F(C_3) ]

=> C_2 * C_3 = F^-1 [ F(C_2)*F(C_3) ]

How can that be?? This rule is not generally true for Fourier transformations...

Best regards, Martin