3
$\begingroup$

I'm having some trouble wrapping my head around the strong $CP$ problem. I know that the non-trivial vacuum structure of QCD induces the topological theta term in the strong sector of the SM, which is also consistent with the chiral/ABJ anomaly and the resolution of the $U(1)_A$ problem. The quark mass matrix is in general complex (since it arises from Yukawa couplings) and one can perform a chiral rotation to get to a physical mass basis, which via the chiral anomaly induces a change in the theta term given by $$\bar{\theta} = \theta + Arg det M.$$ Then the strong $CP$ problem is the question of why this $\bar{\theta}$ is so small.

Now my question is based on a couple of quotes taken from Helen's Quinn lectures on the Strong $CP$ Problem, found here: https://arxiv.org/abs/hep-ph/0110050. The relevant quotes are:

In our current Standard Model theory (and in any extension of it which maintains the well-established QCD theory for the strong interactions) any breaking of $CP$ symmetry in the weak interactions can, and generally will, induce a breaking of that symmetry in the strong interactions.

and

Thus the $θ$-dependent term in the action induces $CP$-violating effects. Conversely, any $CP$-violating interaction, even in the weak interaction sector of the theory, will, in general, induce a non-zero value of $θ$ via loop-effects.

As far as I know, the complex phase in the CKM matrix (which cannot be eliminated) is the one responsible for $CP$ violation in the weak sector, and this is different from the complex angle we eliminate by a chiral transformation above (which changed the value of $\theta$ via the anomaly).

If this is the case, why can't we simply insist that the strong sector must respect $CP$ and set $\bar{\theta} = 0$? How does the presence of the complex phase in the CKM matrix not allow us to do this, and which loop effects are being talked about in the quotes given here? Or in essence, how exactly does $CP$ violation in the weak sector necessitate it in the strong sector?

$\endgroup$

1 Answer 1

1
$\begingroup$

The answer is pretty simple. To make $$ \bar{\theta} = \theta + Arg det M = 0 $$ you have to chirally rotate the quarks so that $$ Arg det M = -\theta $$ The chiral rotation of quarks such as $$ \psi \rightarrow e^{\alpha i\gamma_5}\psi $$ will change the phase of the quark mass term from $$ m\bar{\psi}{\psi} $$ to $$ m\bar{\psi}e^{2\alpha i\gamma_5}{\psi} $$ Therefore you have limited freedom in chirally rotating the quarks, if you want to keep the quark mass $m$ as real, in stead of as non-real $me^{2\alpha i\gamma_5}$.

If the quark mass is zero (which is of course not true) $$ m=0 $$ you would have unlimited freedom in chirally rotating the quarks. And that is the actually the first solution proposed to "solve" the strong CP problem.


Added note: The very existence of the strong CP problem is called into question in this new paper Absence of strong CP violation by Schierholz, which basically says that the measurements of the electric dipole moment of the neutron can NOT confirm that $\bar{\theta}$ is small. See more details here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.