# What's the significance of a dimensionless coupling constant?

In the preface to Mark Srednicki's QFT book (an online draft version can be found here http://web.physics.ucsb.edu/~mark/qft.html), Mark mentions that the $\phi^3$ theory in 6 dimensions would be a great model for pedagogical purposes because the coupling constant is dimensionless.

So why does a dimensionless coupling constant make a difference? Notably $\phi^4$ is also dimensionless in 4 dimensions, but I've barely heard any mention of this.

It all has to do with renormalization.

Quantum Field Theories are typically plagued by ultraviolet divergences. These nasty artifacts of our idealizations arise from our desire to include arbitrary short-scale fluctuations in the picture. In other words, if we trust our QFT at arbitrary short scales (which we probably shouldn't), infinities appear as results of the calculations of correlation amplitudes and the theory loses its meaning.

During the first half of 20th century a certain technique was developed in order to overcome these ultraviolet divergences. It comes in two steps:

1. Regularization: one hopes to modify the original theory in such a way that it would be finite and resemble the original theory in some limit. For example, one could use the momentum cutoff by artificially excluding the Fourier modes with large momentum: $\omega^2 + p^2 > \Lambda^2$, where $\Lambda$ is called the momentum cut-off and has dimension of energy or inverse length. The original theory is restored in the limit $\Lambda \rightarrow \infty$.

2. Renormalization: now that we have a theory which makes sense, a clever idea is deployed. We typically allow several parameters in the Lagrangian, like the coupling constants, to depend on $\Lambda$. This allows to compensate the change in $\Lambda$ by a change in these parameters in such a way that the limit $\Lambda \rightarrow \infty$ becomes nonsingular. The physical theory is then defined as a limit of such theories. If this step can be carried out consistently, the theory is called renormalizable. Otherwise it is called perturbatively nonrenormalizable.

Now comes the important part, which is often misunderstood. We haven't just gotten rid of infinities, we have redefined the theory. A theory defined this way is not equivalent to the (infinite, non existing) theory with original Lagrangian. In particular, it does not possess certain symmetries of the original Lagrangian, in particular, the scale symmetry.

Consider rescalings of the spacetime coordinates. The defined above theory behaves nontrivially under such rescalings, which is called the renormalization group action. In fact, we could classify all QFTs in three categories:

• Relevant couplings have positive mass dimensions, like $\phi^4$ in 3d. These decrease as we approach the ultraviolet regime, since the cutoff $\Lambda$ becomes effectively smaller relative to our increasing energy scale, so the interaction becomes unimportant. Alternatively, the coupling increases when we approach the infrared region (large scale fluctuations). Thus such interactions would manifest themselves on large scales.

• Irrelevant couplings have negative mass dimensions, like $\phi^4$ in 5d. Contrary to case one, these increase as we approach the ultraviolet region, but are irrelevant (thus the name) on large scales. Note that perturbation theory breaks down in the UV, since the coupling blows up and we can no longer consider it small and expand in its powers. Such theories are always nonrenormalizable. Perturbative General Relativity belongs to this category of theories.

• The third category is of "marginal" couplings with zero mass dimension, like $\phi^4$ in 4d. In this case, the behaviour of the coupling in the UV and IR is completely determined by quantum fluctuations and can not be inferred from simple dimensional analysis. For example, QED blows up in the UV (aka the Landau pole problem), whilst non-Abelian gauge theories with compact gauge groups are asymptotically safe. These theories can be renormalizable, as well as nonrenormalizable.

In conclusion: dimensionless couplings are most interesting because they can give rise to renormalizable theories. Also, couplings with negative dimension of mass are irrelevant in the infrared region and always perturbatively nonrenormalizable (though they might sometimes make sense nonperturbatively, the best example probably being General Relativity in 3 spacetime dimensions). I hope this answers your question.

• Correct me if I misunderstood. So, a theory with with positive mass dimension (e.g. $\phi^3$ in 3D, $g\sim[M]$), will certainly be UV divergent, since the coupling constant grows with increasing energy; the contrary occurs for negative mass dimension theory; finally, whether a dimensionless theory has UV divergence is uncertain unless you do detailed QFT calculation. Both QED and QCD belong to the final case, but the former has UV divergence and latter asymptotically free. – JamieBondi Sep 21 '16 at 15:18
• Also, it's now curious to me that we can possibly create a dimensionless theory whose coupling constant is really a constant, i.e. doesn't run with energy scale. Somehow between QED and QCD. – JamieBondi Sep 21 '16 at 15:19
• @ZPrime Quite the opposite. Theories with positive mass dimension will certainly be asymptotically safe, since the coupling constant decreases when approaching the UV. This is because the cutoff becomes closer to the relevant scale, and it is possible to say that it is effectively decreases (despite that actually it does not change, while the relevant scale increases). Thus the coupling decreases also. For a more rigorous explanation, see Peskin-Schreder, chapter 12 (Renormalization group). – Prof. Legolasov Sep 21 '16 at 15:38
• @ZPrime everything else of what you wrote is true. (But your terminology is a bit troubling: QED has a Landau pole problem, not a UV divergence. UV divergences are these nasty problems in the original theory, prior to renormalization). Yes, there are interacting quantum field theories which are truly scale-invariant. These are called conformal field theories. – Prof. Legolasov Sep 21 '16 at 15:42
• @VladimirKalitvianski I have a different notion of "wrong", I guess. It is not wrong to solve an everyday-life problem with Newtonian physics. Just as well, it is not wrong to consider interacting QFTs. In both these cases, we know for sure that what we are doing is an approximation, and an idealization. But this doesn't mean that it is wrong. Is it wrong to model a liquid as incompressible? Hardly so, even taken that it is not true. I guess it makes no sense to argue about notation, but anyway, thank you for following! – Prof. Legolasov Sep 22 '16 at 18:00