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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: enter image description here

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

enter image description here

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    $\begingroup$ What do you mean by "order" of the Feynman diagram? The first diagram is first-order in the fine structure constant $\alpha$, but second-order in the electrical charge $e$ (since $\alpha \propto e^2$). In the second, there are two different couplings (quark-quark-weak boson and weak boson-weak boson-photon), so there's not one natural thing to consider the expansion in. $\endgroup$ – ACuriousMind Apr 10 '15 at 18:07
  • $\begingroup$ you may hear of "leading order" (LO) - the diagram corresponding to the smallest order in couplings at which a process can occur, then next-to-leading (NLO) and N$^2$ LO etc. $\endgroup$ – innisfree Apr 10 '15 at 19:28
  • $\begingroup$ usually one says "this process is suppressed at tree level, so we need to consider higher order diagrams". I wanted to quantify that 'higher order diagrams' $\endgroup$ – SuperCiocia Apr 10 '15 at 22:18
  • $\begingroup$ That "order* is more commonly called the loop level, and is literally just counting the loops. (tree-level is the zeroth order for that, containing diagrams like your first which have no loop). $\endgroup$ – ACuriousMind Apr 10 '15 at 22:30
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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.

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  • $\begingroup$ Can you please comment on how this formula can be derived? For example, consider the case with just a single type of vertex. If I am not missing something, then, the formula must be $V-E+L=1$. The formula that you wrote does not reproduce this. Am I correct? $\endgroup$ – QGravity Sep 5 '17 at 2:37
  • $\begingroup$ please see derivation added. Your formula is incorrect unless you are using some other naming convention which I do not understand. $\endgroup$ – Ali Moh Sep 6 '17 at 3:55
  • $\begingroup$ Thank you for posting the derivation. Also, the formula I wrote is based on Euler formula: #verices-#edges+#faces=$\chi$=$2-2g$. For a planar Feynman diagram #edges=#propagators$\equiv E$ and #faces=#loops+1$\equiv L+1$ (considering the outer region of the diagram), and $g=0$. Therefore, we end up with $V-E+L=1$. $\endgroup$ – QGravity Sep 6 '17 at 21:48
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The order of a quantity in general refers to the exponent of the quantity in an expression, ie $$x^3y^2$$ would be 3rd order in $x$ and 2nd order in $y$. According to the Feynman rules, each vertex in a Feynman diagrams contributes a factor of the coupling constant, so the order of each coupling constant is simply the number of vertices of that interaction. For example, the first diagram is second order in $\alpha$.

In QFT it is common to use natural units where $\hbar = c = 1$. However, if you don't follow this convention, then the number of loops in your diagram is equal to the powerof $\hbar$ in you final quantity. Tree diagrams have no dependence on $\hbar$ at all and in some sense can be considered to be the purely "classical" result, with higher loop diagrams giving you the quantum corrections. When people talk about the order of a diagram without mentioning a coupling constant, this is often, but not always what they mean; it really depends on the context.

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