# Are Feynman rules unique?

I am new here and I got a question that I never read in any textbook. Are Feynman rules actually unique or can one derive different rules, which are equivalent?

Of course, we have a freedom to choose a gauge. So my question concerns whether or not Feynman rules are unique after we have chosen such a gauge and if there is still the possibility to gain another Feynman rule, which is equivalent to the other one.

• There is some freedom you have certainly, since for example you can use different choices of gauge in QED which result in a different expression for the propagator but give the same physical answers. Mar 22, 2018 at 10:22
• Okay, of course there is a gauge freedom we have. I forgot to mention that I am aware of this. But how is it besides the gauge that we can choose? After we have chosen a gauge, are the derived Feynman rules unique then? Mar 22, 2018 at 10:29
• Can you try to clarify the question? Mar 22, 2018 at 10:35
• Sure. :) After we derived a particular Feynman rule and have chosen a particular gauge: is that derived Feynman rule then unique or is it possible, that still another equivalent Feynman rule can be derived? I hope, I could clarify my question. (Maybe I also should edit my original post to prevent confusion.) Mar 22, 2018 at 10:39
• Hi and welcome to PSE. There is no need to put the "edit" word when you edit. The question should be as concise as possible. Mar 22, 2018 at 10:46

• The vertices depend on how you split the Lagrangian into free and interacting parts. For example, the propagator for a massive scalar field is usually $1/(p^2 + m^2)$, but one could also say the free part is the kinetic term alone, giving a propagator $1/p^2$. Then we gain a two point self-interaction from the mass term, which gives a factor of $m^2$. This point of view is useful when the Higgs mechanism is involved, because every 'free' particle really is massless.
• Similarly, the allowed vertices depend on the basis we choose for our propagating particles. For example, quarks/neutrinos in the mass basis can change generations upon interacting with a $W$ boson, but can't in the flavor basis. Which you use is up to convention and context.
• "Integrating out" some fields will give new Feynman rules from the effective interactions generated. For example, for scales much lower than the $W$-boson mass we can ignore the $W$ boson, at the cost of having a new, effective four-fermion interaction.
• The adjoint representation for gluons is "almost" the same as the fundamental times and antifundamental, as $$N \times \bar{N} = (N^2 - 1) + 1.$$ Therefore we can write Feynman rules for gluons as if they're made of a fictitious charged particle and antiparticle pair, even though the Lagrangian says no such thing. This is called t'Hooft double line notation and is especially useful for large $N$.