Feynman diagrams in effective theories

Question

I've been seeing many feynman diagrams lately that I can't quite interpret yet. I've heard a basic Quantum Field Theory lecture and so to me, a Feynman diagram is simply a mnemonic picture to quickly write down (and remember) all of the possible terms in the perturbation series of the matrix element in scattering.

But they seem to be much more than that.

Consider for example this diagram of proton-neutron scattering via pion exchange. The picture seems to have an intuitive meaning, but how can it be a valid Feynman diagram? Wouldn't that imply that there is an underlying theory with a proper Lagrangian from which Feynman rules can be derived that would assign a number to this diagram?

Question: When describing nucleon-nucleon interactions in this way, do people actually write down a Lagrangian and derive those rules? (I am confused because I've seen diagrams like that in several books and they usually write down the cross sections simply from analogy arguments to other theories.)

Any clarification will be greatly appreciated,

Answer 1

For nucleon-nucleon interaction please keep in mind that in this low-energy regime pertubative QCD breaks down and reactions are not really calculable. For the specific pion exchange you mention have a look at

http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html

as to why this QCD-process can be seen as an exchange of a pion.

In general you can get Lagrangians for effective (i.e. low-energy) theories by integrating out the high momentum degrees of freedom. For example the W-boson propagator

$\frac{-i(g_{\mu\nu}-\frac{q_\mu q_\nu}{m_W})}{q^2-M^2_W}$ in the limit of small $q^2$ becomes $\frac{ig_{\mu\nu}}{M^2_W}$

so instead of the full electro-weak symmetry there would be a new lagrangian with a four point interaction term $\frac{G_F}{\sqrt{2}}J_\mu J^\mu$ where $J_\mu$ is a left-handed Dirac current.

In more general terms this integrating out of high momentum degrees of freedom is the point of view on renormalization taken by Wilson. In this view our current theories are effective theories that result from an unknown fundamental Lagrangian from which all the high momentum d.o.f. have been integrated out. (c.f. renormalization group)

Answer 2

Well, it need not be a Lagrangian, you can work in Hamiltonian formalism too (which is often possible outside of particle physics, e.g. in condensed matter theory). But except for this caveat the answer is a resounding yes: the theory comes first and diagrams are just mnemonics. Diagram by itself doesn't make any sense if you can't reconstruct the corresponding term in the expansion from it. And you can only do this if you know precisely what theory you are dealing with (so you need to know what types of particles there are and in particular their Lorentz structure and Dirac structure, etc.).

Of course, if you are familiar enough with the diagrammatic approach and comfortable with most of QFT and particle physics (so that you can write down Lagrangian of the most common types of interactions and also "see" what interaction vertices it corresponds to) then there is no problem working directly with Feynman diagrams and never mention any Lagrangian at all.

Let me make (by now perhaps superfluous) analogy to the differential calculus. You can learn your ${{\rm d} x^n \over {\rm d} x} = n x^{n-1}$ formulas and product and chain rules and comfortably differentiate all kinds of functions. Yet, at some point you should learn that these are just mnemonics and differentiation is defined in terms of limits (which is what you will have to return to anytime you encounter a function for which you have no rule). Still, most of the time the standard rules are all you need.

Answer 3

The effective Lagrangian upon which this Feynman diagram is based is the Yukawa effective theory. See, for example, the following exposition by: Christof Wetterich. The kinetic terms are given in equations 1.3 and 1.4 and the intercation term (Yukawa interaction) is given in equation 1.5 for a charged pion doublet and by 1.19 for a pion triplet including the neutral Pion. Please notice that the difference between the proton and neutron masses explicitely breaks the isospin invariance.

This is an old theory that was used before the standard model was discovered. Although, this is a perturbatively renormalizable theory, there is no profit in performing computation beyond the tree level, because the proton, neutron and pions are not elementary. Also both the Yukawa coupling and the particle masses are external parameters.