# Why amplitudes are rational functions?

In Bootstrap and Amplitudes: A Hike in the Landscape of Quantum Field Theory there are few statements about analytical structure of amplitudes.

I want to understand statement:

Tree amplitudes must be rational functions of Mandelstam variables.

What is the reason for such a property of tree level amplitudes? What is the simplest explanation of such a fact?

• The tree level feynman rules contain two elements: the vertex factors and propagators. Locality implies that vertex factors are always polynomials in momenta (since we are allowed to keep only a finite number of derivates in the action). For the same reason, propagators are also polynomials in momenta. This implies that amplitudes(vertex factor/propogators) are rational functions in mandelstam variables? Are you looking for a deeper explanation? Oct 8 '20 at 16:44
• I would like to add the it is also agreed for the same reasons, that loop integrands are rational functions of momenta. This is why BCFW formalism is useful in finding loop integrands. You should explain if there is a specific kind of explanation you are looking for? Oct 8 '20 at 16:51
• @Anonjohn, could you explain how we obtain simple poles in amplitude? You logic fails in prediction of such terms.. Oct 8 '20 at 17:30

After the OP explained in the comments what exactly they're looking for, I will attempt an answer. There are a few separate facts that need explanation:

• Tree amplitudes are rational functions of Mandelstam invariants: This is an obvious statement, that I have explained in the comments, and I will repeat here to preserve: The tree level feynman rules contain two elements: the vertex factors and propagators. Locality implies that vertex factors are always polynomials in momenta (since we are allowed to keep only a finite number of derivates in the action). For the same reason, propagators are also polynomials in momenta. This implies that amplitudes(vertex factor/propogators) are rational functions in mandelstam variables.

• The poles are simple poles in Mandelstam invariants provided we have a massless, local theory: This requires a little explanation, so I will explicitly write down the line of reasoning that leads to this conclusion.

If we have a local, massless theory, the denominator is always quadratic in momenta. This is a totally nontrivial point since naively, we could think of interactions of the form $$\mathcal{L}_{int} \sim g_n \phi \partial ^n \phi$$. But of course, massless theories do not admit such terms bilinear in fields since generic values of the $$\{g_i\}$$ introduces new poles in the two point function, that is to be interpreted as a new massive particle. Therefore all propagators are of the form $$\frac{i}{(k_{i_1}+k_{i_2}+k_{i_3}...+k_{i_m})^2}$$. No two propagators of a tree diagram carry the same momenta(for generic values of the external momenta), and hence when this propagator goes on shell, we find a simple pole in the Mandelstam variable $$s_{{i_1 i_2....i_m}}=(k_{i_1}+k_{i_2}+k_{i_3}...+k_{i_m})^2$$.

@Anonjohn has given a good answer. I would like to add a bit more. I think, that the author of statement tacitly assumes dealing with massless theory with a linear dispersion relation (kinetic term of form $$\phi \Box \phi$$) : Why does nature favour the Laplacian?. So each propagator is of the form: $$\frac{1}{(\sum_{i \in I} k_i)^2}$$ The analytical properties are crucial in determining the exact expression for tree amplitudes in massless theories. For instance, the proof of famous Parke-Taylor formula by BCFW https://arxiv.org/pdf/hep-th/0501052.pdf - is based on them. For a good introduction I recommend these notes, written by the same author as the article you cite - https://arxiv.org/pdf/1308.1697.pdf.

• Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. Oct 8 '20 at 20:39
• Author work with particle with linear dispersion law Oct 8 '20 at 23:36
• @Nikita, sorry, misguiding notation Oct 9 '20 at 5:37