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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,

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3 Answers 3

up vote 9 down vote accepted

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


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)

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+1 Wilsonian point of view is precisely how one should approach effective theories; or all theories for that matter. [Also, rereading the question, I guess my answer addresses a topic quite orthogonal to what OP wanted...] –  Marek Jul 19 '11 at 10:39
Marek, luksen, thank you both for your answers. Do you have any good references for learning about renormalization? (Can you recommend Peskin/Schröder Chapters 10/11/12, which would otherwise be my starting point?) –  David M. R. Jul 19 '11 at 13:14
I think Peskin's introductino to renormalization is quite good. I'd definitely recommend it. –  luksen Jul 19 '11 at 13:49
Slightly related to that pion exchange is Is it pions or gluons that mediate the strong force between nucleons? –  b_jonas Jan 7 at 11:08

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.

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is there a tractable machinery to start with a system of (simple) feynman diagram terms and mathematically arrive back at the corresponding lagrangian? (for a system where you know a well-defined one exists) I mean, can you essentially say they are mathematically equally powerful descriptions (not considering the intuitive power etc). –  BjornW Jul 19 '11 at 9:43
@Bjorn: Feynman diagrams in particle physics are strictly weaker than the Lagrangian formulation. They represent just S-matrix expansion (i.e. interaction over infinite amount of time between asymptotic states), so the question would be whether S-matrix is all there is to field theory (once people thought so, nowadays this isn't pursued much I think). Also, the series actually does not converge. Also, one needs renormalization that modifies the original Lagrangian and diagrams. There are so many issues that I don't have a slightest clue how to even begin to answer your question :) –  Marek Jul 19 '11 at 10:30
its ok :) still, it feels like at least some of those issues have to do with how you enumerate, select and use the individual feynman diagrams rather than discard or change the diagrams themselves.. –  BjornW Jul 20 '11 at 0:03
Bjorn Wesen: There is, look for example at chapters 2 and 3 of Cvitanovics Field Theory notes: –  orbifold May 5 '12 at 1:18

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.

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