Tadpole symmetry factor Can someone help me with symmetry factor of one-loop tadpole diagram (one loop correction to one point Green function in phi-3 theory)?
 A: The symmetry factor should be $2$. This comes from the fact that exchanging the derivatives at the vertex is the same symmetry operation as swapping the endpoints of the propagator in the loop. Each of them amounts for a multiplication by 2, but since they are identical we are essentially overcounting. Dividing by $2$ corrects this error.  
A: There are some ways to get the symmetry factors correct. If it is an easy diagram like this one you can actually imagine it. For example here, by flipping the loop you get the identical diagram and that means dividing by two. However, there are more difficult diagrams. Then, you have to follow set of rules which are described in many field theory books. I assume you might have Peskin and Schroeder for example so just take a look. For this diagram, detach vertices and external lines. That would give you a 3 point vertex and one external line. For the three point vertex with three identical particles divide by 3! (In QED , the vertex with photon, fermion, antifermion wouldn't get such a factor since all are distinguishable in the vertex). Now, having to construct your diagram, you have to attach the external line to one of the three lines of the vertex. There are three ways. The rest of the lines of the vertex have to close together to form a loop (there is exactly one way to do that). Together:
$\frac{3.1.1}{3!}=\frac 1 2$ .I hope you got it.
