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Let the non-renormalizable $\phi^6$ theory behaves as a low-energy, effective field theory, and works perfectly well below a finite energy (or momentum) scale $\Lambda$ for a system.

In this theory, all the loop diagrams will be finite if the loop momenta are carefully integrated up to $\Lambda$. This implies that there will be no divergences in scattering amplitudes at any order.

  1. Does this theory still require renormalization? If yes, why?

  2. If yes, then instead of saying $\phi^6$ to be a non-renormalizable theory, shouldn't we say that (i) it is renormalizable at low energy but (ii) non-renormalizable at high energies?

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There's a bit of confusion here.

Whenever you work with a QFT you have to able to define it first. It is not enough to just write down diverging integrals and assert that they correspond to transition amplitudes. This doesn't make sense.

So I assume that what you mean is: define a theory with an explicit momentum-scale cutoff $\Lambda$ such that all the integrals are finite. Consider $\Lambda$ just as physical as other constants like mass $m$ and the coupling constant $\lambda$.

There's a whole bunch of issues with this definition. Like, for example,

  • It isn't clear how to consistently impose momentum constraints on the loop integrals, as we can pass to different loop momenta integration variables for which different constraints have to apply. Note that you don't care about this subtlety when $\Lambda \gg p$.
  • The theory with a momentum cut-off isn't Lorentz invariant. It simply isn't. You can, however, neglect this Lorentz violations when $\Lambda \gg p$.

The list can go on, but I think I've made my point already. But say that we somehow found relatively satisfactory answers to all of the questions above. What now?

1. Does this theory still require renormalization? If yes, why?

Yes, it does! Renormalization isn't about getting rid of infinities, and it isn't about getting rid of the unphysical $\Lambda$ (though it achieves both of these goals in the mean time). It is about making sense of the results that your theory gives.

Like, for example, you would want to give a particle interpretation to your theory, with an $S$-matrix corresponding to particle scattering. What do you require to give your theory a particle interpretation? One of the requirements is that the 2-point function has a pole when $p^2 = M^2$ with residue 1. This follows directly from normalization of states, and it allows you to talk about interacting particles of mass $M$ which your theory is supposed to describe.

If you calculate this 2-point function to some loop order you will find that both the location and the value of the pole are not $m$ and $1$ as you would've expect naively, but depend on $\Lambda$. But what's it mean? It means that your particles are of mass $M = M(m, \lambda, \Lambda)$ and are generated by a Fock-space operator associated to the physical observable $Z(m, \lambda, \Lambda) \cdot \phi$, not just the field operator $\phi$.

Again, renormalization is about reinterpreting predictions in terms of interacting particles.

2. If yes, then instead of saying $\boldsymbol{\phi^6}$ to be a non-renormalizable theory, shouldn't we say that (i) it is renormalizable at low energy but (ii) non-renormalizable at high energies?

What happens is: the higher-valency correlation functions become highly dependent on $\Lambda$ even in the $\Lambda \gg p$ regime. Physically, your theory becomes pathologically sensible to short-scale fluctuations.

With renormalizable theories we can say that $\Lambda$ is very large and corresponds to the boundary of the domain of applicability of our theory. But the exact details of this boundary aren't relevant for the long-range physics: we can just adopt the limiting value for the higher-valency correlations.

In case of $\phi^6$ in $4d$ though, this is not the case. Instead, we have the following nonrenormalizable behaviour:

The long-range properties of your theory depend explicitly on the details of the cut-off procedure. It can be the value of $\Lambda$, the way you resolve the cutoff ambiguity in the loop momenta integration variables, masses of the Pauli-Villars regularizer fields, etc. The key fact is that - your results depend on something for which you can't really say how it works and if it is physical or not. That's what is bad about nonrenormalizability.

With nonrenormalizable theories you can tweak the theory to give whatever predictions you want by simply changing the cut-off mechanism a bit. Not a lot of predictive power there.

UPDATE: allright, I admit that it is not 100% true. I was trying to make a point in the context of HEP, but once you let go of your ambitions to describe arbitrary high-energy processes, you can actually do something useful with nonrenormalizable theories as well.

For instance, you can fix the perturbation theory order $k$ prior to renormalization, and then simply determine the values of counterterms from experiments. With renormalizable theories this could be done once, i.e. using a fixed finite number of counterterms independent of the perturbation theory order. But nonrenormalizable theories require more and more tweaking and adjustment with increasing order.

Of course one can claim that this is fine, since perturbation expansion is only an assymptoitic series and thus even renormalizable theories can't be solved with an a-priori arbitrary precision using perturbation theory. And it is probably true.

There's another property of nonrenormalizable theories which has to do with Wilsonian renormalization group flow. Effective couplings used in perturbation theory blow up in the ultraviolet regime thus rendering the whole concept of perturbation theory meaningless. Thus we end up with phase transitions in the high-energy regime which we can't describe with perturbation theory.

It is also worth mentioning that such phase transitions aren't specific to nonrenormalizable theories. As a useful example, QED (Quantum Electrodynamics), though renormalizable, has an ultraviolet phase transition (the Landau Pole problem). Renormalizable theories without these phase transition are called asymptotically free.

And asymptotic freedom, together with renormalizability, is enough to ascertain that a HEP theory can be used to make sensible predictions up to whatever energy is associated with the boundary of its domain of validity. This is because the further into the UV you go, the less becomes the coupling, making the asymptotic expansion a better approximation for an even further order in perturbation theory (remember how the closer the coupling is to zero – the more orders of perturbation theory we can trust without worrying about asymptotic expansion blowing up?).

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  • $\begingroup$ "Not a lot of predictive power there." I disagree. EFTs need not to be renormalizable, and yet one can compute quite a lot of things with them, Indeed, at each order of the perturbative expansion (in $p/\Lambda$) one just need to fix a finite number of parameters, which allows to make quite a few predictions (to this order in $p/\Lambda$). Furthermore, cut-offs can be physical, not everything is about HEP ;-) $\endgroup$ – Adam May 31 '17 at 8:16
  • $\begingroup$ I'm also confused by the last part; I thought non renormalizable theories could still be used at low energies, while you seem to be saying that they're not predictive. $\endgroup$ – Javier Jun 1 '17 at 21:48
  • $\begingroup$ @Javier what happens is, to each new order of the perturbative expansion, you have to tweak more parameters. You can still use them as approximations to some order $k$. Breaking down at high energy is another property of nobrebornalizable theories related to Wilsonian renormalization group flow. I decided not to write about it in my answer for reasons of keeping things simple. $\endgroup$ – Solenodon Paradoxus Jun 2 '17 at 1:39
  • $\begingroup$ @Adam please see my update $\endgroup$ – Solenodon Paradoxus Jun 2 '17 at 5:53
  • $\begingroup$ @Javier please see my update $\endgroup$ – Solenodon Paradoxus Jun 2 '17 at 5:53

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