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Suppose you know everything about Quantum Field Theory, but nothing about the very specific interactions and particles which exist in our real word.

Which physical or mathematical principles could be used to reduce the infinite number of possibilities of QFT, to finish with some class of theories, in which the Standard Model, for instance, would belong.

[If the question is too "extreme", one may soften it by considering having some partial information, to be detailed, about our real world]

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The first step might be to realize that most of the observed particles have spin-0, spin-1/2 or spin-1. Knowing this we would like to write down Lagrangians for these particles. Renormalizability is of great help here, since it essentially brings down the set of allowed theories to a small number.

For purely scalar field theory, you can only have $\phi^4$ interactions. If you now add spinors, the only allowed interaction is the Yukawa coupling. Finally, if you add vector bosons, the allowed interactions is that of QED or scalar QED. Another kind of interaction allowed by renormalizability of gauge fields includes interactions of the type $A^2(\partial_\mu A^\mu)$ or $A^4$. Gauge symmetry restricts the coupling constants of these interactions a lot, and the only allowed theory is Yang-Mills. And, we're done! Nothing else is allowed.

So, we have already reduced the potentially infinite number of theories to

  1. $\phi^4$
  2. Yukawa ${\bar \psi} \psi \phi$
  3. QED ${\bar \psi}\gamma^\mu \psi A_\mu$
  4. Scalar QED $A_\mu \left( \phi \partial^\mu \phi^* - \phi^* \partial^\mu \phi\right)$
  5. Yang-Mills

One could go further from here, and reduce it a little bit more. This is still an infinite class of theories. For example, the Lie group of Yang-Mills is unknown at this point.

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Do you need Poincare invariance to show spin (Pauli-Lubanski) is a Casimir operator for Poincare group? –  user26143 Sep 27 '13 at 14:56
I've never really understood the argument that renormalizability limits the number of theories. If we know everything about QFT, we know that QFT's are only useful as effective theories, in which case we can't really rule out irrelevant terms in the lagrangian, we just don't need them for IR calculations. Given that a UV complete theory won't be a local QFT at all, it seems bizarre to rule out these terms. If you define a renormalizable theory in the UV, the rest of the terms will be generated under the renormalization group anyway, they just aren't as important. Can you clarify my confusion? –  Dan Sep 27 '13 at 17:42
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