Strong $CP$ problem and fine tuning

Question

I have worked my way up to and through Srednicki's Quantum Field Theory chapter 94 and was also doing some reading on the internet about the strong $CP$ problem.

Wikipedia's entry on the strong $CP$ problem describes it as an example of a "Fine Tuning" problem.

Since I've been working on QFT mainly on my own through the book and other internet sources, perhaps I've missed some things and wondered if my thinking is correct and/or if anyone could offer some context.

The only other place I encountered a fine tuning problem was in Ch 29 of Srednicki. When I re-read that in light of the proposed solutions to the Strong CP problem, I wonder if the reason why Wikipedia lists the Strong CP problem as an example of fine tuning is because of the precise cancellations in higher-order calculations of $\theta$'s value required to get it to be zero (one of the proposed solutions to Strong CP).

Or it is equally probable that I don't know what I'm talking about.

Is my thinking close? Is there a better way to think of it?

Answer 1

The Wikipedia entry you linked explains it quite nicely. It is a fine-tuning problem because in principle the dimensionless variable $\bar \theta$ can take on any value between $0$ and $2\pi$ (physical observables are periodic in $\bar \theta$ with period $2\pi$). The fact that it seems to be almost exactly $0$ is therefore suspicious - maybe there is some unknown mechanism that forces $\bar \theta$ to take on such a small value.

This can seem like a bit of pseudoscience, e.g. there is nothing inherently contradictory with $\bar \theta$ being exactly $0$. However thinking in these terms of 'naturalness' and 'fine-tuning' has led to various breakthroughs in physics. For example, in the standard model there are an infinite amount of operators with mass-dimension $d>4$ which could in principle be added, but somehow are not present, i.e. all of their coefficients are nearly zero. A strict logician may see no issue with this and be perfectly fine setting all coefficients equal to zero, but a scientist understands that this is very 'unnatural'. The ultimate reason for this is that under renormalization, when we restrict our attention to mass/energy scales $E$ much below the renormalization scale $\Lambda$, all these operators become multiplied by a factor $(E/\Lambda)^{4-d}$ and therefore vanish when we take $\Lambda \gg E$.