# How can spontaneous $P$ or $CP$ violation arise? Examples needed

In quantum field theory, there can be a spontaneous breakdown of discrete and continuous symmetries. Can there be a spontaneous breakdown of P and CP violation where P represents parity and C represents charge conjugation? Can we think about such possibilities?

• See e.g. here for an example from the huge literature on this topic. – user178876 Jul 30 '18 at 2:25

In the Standard Model (SM), with three generations of quarks and leptons, the CP violation is a consequence of the complex Yukawa couplings. Hence, CP is not the symmetry of the Lagrangian. However, if you can come up with a field theory where CP is a symmetry of the Lagrangian but the vacuum breaks the CP symmetry, it is quite possible to have spontaneous CP violation. Later I would like to expand on it.

For further details, please considers looking at the following references.

The second reference explains why spontaneous CP violation (SPCV) is not a possibility in the SM because you can always choose a vacuum which is invariant under CP. However, with more than one scalars such as with one doublet and one singlet you can achieve SCPV.

There are some restrictions, most notably the theorem of Vafa and Witten. These restrictions can be avoided at finite chemical potential and non-zero theta angle.

The standard examples are CP violation at $\theta=\pi$, see here, and P violation at finite isospin density (pion condensation).

Consider pion condensation: Take QCD and add an isospin chemical potential $${\cal L } = \mu_I\psi^\dagger \tau_3\psi$$ where $\psi=(u,d)$ is a two-component spinor in isopsin space. This operator conserves $P$, so there is no explicit symmetry breaking here. It is also relevant to the real world, because finite nuclei and neutron stars have an excess of neutron over protons.

The partition function involves $H-\mu_I Q_I$, where $Q_I$ is the isospin charge. This means that the dispersion relation for negatively/positively charged pions (with isopsin $\pm 1$) is $$E_\pi = \sqrt{m_\pi^2+\vec{p}^2} \pm \mu_I$$ and negatively/positively charged pions Bose condense for $|\mu_I|>m_\pi$. Pions have negative intrinsic parity, and the Bose condensate breaks $P$.

• Your answer is very technical for my purposes. – mithusengupta123 Jul 25 '18 at 7:58
• @mithusengupta123 added a few remarks – Thomas Jul 25 '18 at 16:35
• How should one interpret the statement "These restrictions can be avoided at finite chemical potential and non-zero theta angle."? A non-zero theta angle, more precisely $\theta\ne 0,\pi$ is known to violate P, and likewise a chemical potential C. – user178876 Jul 30 '18 at 2:26
• @marmot That's the point, $\theta=\pi$ does not violate $CP$, but sponatenous $CP$ violation may occur, as explained in the paper. The claim about $C$ in the case of pion condensation was indeed not right. This is just an example of spontaneous $P$ violation. I fixed this. – Thomas Aug 14 '18 at 15:20