# $\theta$ angle QCD technically natural

Why exactly does the observed smallness of the $$\theta$$-angle in QCD constitute a fine-tuning problem?

I thought that the content of 't Hooft's technical naturalness was exactly that a parameter is allowed to be small if, when taken to zero, it leads to an increased symmetry?

In this case, taking $$\theta$$ to zero leads to CP-invariance, so what is the problem?

This is a somewhat subjective thing and I think you'll get about five different answers if you ask ten different people. My impression of the situation is this:

• All that technical naturalness does is say that if some parameter is small at a low scale, it's also small at a high scale. It doesn't explain why it's small at a high scale. If one is worried about the parameters in your theory taking seemingly unlikely values, then technical naturalness only means that you can delay resolving the question to the next layer in the EFT hierarchy.
• One can always argue that we don't know what the probability distribution of parameters is, so naturalness is meaningless. But the $$\theta$$-angle is a very clean example, because it's literally an angle. Absent any more structure, you would expect some UV theory to pick $$\theta$$ randomly in $$[0, 2 \pi)$$, which makes the observed smallness very unlikely.
• If you say, "well, can't there just be some unknown structure at high energies that prefers small $$\theta$$?", then of course the answer is yes, but identifying such structure is precisely what we mean by doing model building to solve the strong CP problem.

For some more perspectives on this, see here and here.

• I agree with the content of this answer, but I think it does not really address the OP, see my answer below. The objection that within pure QCD there is a simple reason to require for $\theta = 0$ is completely correct. – pppqqq Apr 24 '19 at 16:34
• The second bullet point doesn't seem to make sense as is, right? If an angle was chosen randomly there would be no problem with any value. – kηives Apr 24 '19 at 16:54
• @kηives As I said, it's a deeply subjective thing. You can point to almost anything and say "that was very unlikely! there must be a reason!" (For example, the temperature right now is between, say, 72.2918 and 72.2919 degrees, which is extremely "unlikely".) You can also almost always respond by saying "no, it's not impossible, so there's no problem". (For example, somebody could win the lottery a hundred times in a row, but that's not literally impossible.) – knzhou Apr 25 '19 at 14:25
• When is this reasoning actually correct? It's subjective and guided by experience (how it's turned out in other theories) and intuition (how we think more fundamental theories should behave). Ultimately, all science is subjective in exactly this sense. There is no foolproof objective algorithm for doing science. – knzhou Apr 25 '19 at 14:26
• @knzhou I don't think I was clear. I'm being very pedantic. It seems you say that theta would be chosen from a flat distribution, where every angle has the same likelihood. So no value is unlikely at all, right? – kηives Apr 27 '19 at 13:39

The $$\theta$$-angle poses a naturalness problem in the context of the Standard Model (SM).

In fact, as you correctly remark, when you consider QCD in isolation, sending $$\theta \to 0$$ restores $$CP$$-symmetry; that is, a small $$\theta$$ is technically natural according to 't Hooft. Much more strongly, if QCD was the whole thing, the experimental bounds on $$\theta$$ would simply suggest $$\theta \equiv 0$$.

However, in the SM there are other potential sources of CP violation: sending $$\theta \to 0$$ alone is not enough in order to restore CP; in order to restore CP, you must also to send to zero the CP violating phases in the CKM matrix.

These phases are, however, experimentally known to be non zero, which means that (from a naturalness point of view) you would expect an equally big $$\theta$$ angle.

This is why the strong experimental bounds on $$\theta$$ pose a naturalness problem within the SM.

• Interesting answer, but I'm not sure whether I agree. Even in the context of the full SM, there would still be increased symmetry, no? E.g. the dipole moment of the neutron would still disappear etc. More formally, the QCD part of the Lagrangian would still have increased symmetry? – Michael Angelo Apr 24 '19 at 18:37
• As to the last question: yes, but I would say that requiring only part of an effective lagrangian as the SM one to have a (non-accidental) symmetry is a poorly motivated assumption, for several reasons. First of all, the fact that symmetry is explicitly broken from another sector means that it is not fundamental, so that whenever possible you would expect symmetry violations. Second, if you believe that the parameters defining different sectors like QCD and Weak interactions should be explained in the context of a more fundamental theory [...] – pppqqq Apr 25 '19 at 0:36
• [...] (which I think is the only case where it makes sense to speak about naturalness), then in the absence of any particular mechanism I find quite plausible to expect both sectors to violate the symmetry “by equal amounts”. Finally (but probably less relevant in this specific case), let's recall that, in general, radiative corrections will typically spread symmetry violations across different sectors, which implies that the division of the lagrangian into a symmetric and a non-symmetric part can actually be valid only at a given scale. – pppqqq Apr 25 '19 at 0:53