1
$\begingroup$

There is a coupling in the standard model of the form $\theta F^{\mu\nu}\tilde{F}_{\mu\nu}$ where $F^{\mu\nu}$ is the QCD field strength. I've read that $\theta<10^{-8}$, which is speculated to be related to Peccei-Quinn symmetry.

  1. What would be the consequences of this term if $\theta$ were not small? In particular, I know that this term is CP-violating. How will this CP-violation be manifested in experiments?

  2. How do we know $\theta<10^{-8}$?

$\endgroup$
5
$\begingroup$

To answer point $2$, $\theta$ is derived from the neutron elecric dipole moment that is measured to be like $ <10^{-18}\,\text{e}\cdot\text{m}$. You can prove that the dipole moment is proportional to $\theta$ and from that you can get your upper limit for $\theta$. See Donoghue, Golowich, Holstein - "Dynamics of the Standard Model", pages 44-45 for more details.

As far as your first question is concerned, a value of $\theta$ different from $0$ would make $CP$ not a symmetry for QCD, but I don't really know what effect this could have in detail.

$\endgroup$
0
$\begingroup$

The term $\theta F^{\mu\nu}\tilde{F}_{\mu\nu}$ gives the chromoelectric and magnetic fields $\theta \vec E\cdot\vec B$. To compensate for this angle $\theta$ a scalar field is introduced that has the wave equation $$ \square\phi~=~\phi\vec E\cdot\vec B. $$ If this term includes the electromagnetic field $SU(3)\times U(1)$ this scalar field couples to electric and magnetic fields. This scalar field is called the axion. It may have in the above sourced equation physics with the QED field.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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