quantum field theory - How to tell the order of a Feynman diagram?

How can we know the order of a Feynman diagram just from the pictorial representation?

Is it the number of vertices divided by 2?

For example, I know that electnro-positron annihilaiton is first order:

So what's the order of, for instance, a radiative penguin diagram (below) ?

Answer

Using elementary graph theory identities one can show that the number of loops in a connected diagram is related to the number of external lines and the number of vertices of type $i$ each of which has $n_i$ lines attached to it, is related by

$$\sum \left(\frac{n_i}{2}-1\right) V_i -\tfrac{1}{2}E +1= L$$ So you can see that for a fixed process (fixed $E$), knowing the number of vertices of each type is equivalent to knowing the number of loops (which can correspond to a multitude of diagrams in the same "order").

In the standard model we have two classes of vertices; those with three or four lines. So as you can see specifying the total number of vertices (equivalently the order with respect to the sum of powers of all coupling counstants), isn't going to uniquely fix the number of loops, however specifying the number of vertices of each class, you can get a one to one correspondence between loop order and coupling constant power order, in which case both are equivalent to the quantum mechanical expansion in powers of $\hbar$.

Derivation:

To derive this formula you can treat each external line as type of vertex with only one line attached to it. That is $E\equiv V_1$ and corresponding to it $n_1=1$. Then we can rewrite

$$\sum \left(\frac{n_i}{2}-1\right) V_i +1= L$$ This formula can be understood by recursion, first we prove it's true for zero vertices, but this is obvious since for zero vertices we do have $L=1$, just draw a circle!

Now to prove by recursion we assume the formula is correct, and prove that if we add one vertex of type $i$, we must introduce $(n_i/2-1)$ new loops. This can be easily seen by taking your diagram and putting a vertex anywhere on an internal line (notice that we no longer distinguish between internal and external lines because $E$ is just another type of vertex now).

When you insert this vertex, two of it's legs are already eaten automatically, so we need to connect the remaining $n_i-2$ legs, note that we must connect them with each other, because all other vertices are already saturated, and leaving a leg hanging is equivalent to introducing an external vertex which we are not doing by assumption. Now this is only possible if $n_i$ is even, in which case we get $(n_i-2)/2$ new loops, which proves the recursion for even vertices. If the vertex is odd, we must introduce them in pairs and the same discussion ensues.