Does $CP$-violation mean $CP$-odd?

Question

This might be a very simple question but I'm doing some reading into the Higgs boson and $CP$-symmetry breaking. I've seen the terms $CP$-even and $CP$-odd terms in the Hamiltonian floated around and have two questions:

1. Does CP-odd interactions mean $CP$-violation?

2. I've seen that the Higgs is $CP$-even in the SM, but that this could be because the $CP$-odd part can be 'rotated away'. What does this mean? Is the Higgs even or odd? does it obey violation?

I am aware that the Higgs is currently being investigated for $CP$-violating properties which could deviate from the SM, so I'm a bit confused.

Answer 1

I adduced the crucial reference for you in your question, and the crucial missing statement, so it makes sense.

1. A theory is CP-invariant if the Lagrangian, and so the Hamiltonian, commutes with the CP operator, $[H,CP]=0$, that is, it is invariant under $$ CP~H ~(CP)^{-1} = H.$$ This means that the eigenvalue of the Hamiltonian under a CP operation is +1, so it is even. By contrast, if the Hamiltonian has an odd piece, so H is not even, so a piece of it changes sign under CP, the theory violates that symmetry.

Your reference is investigating the Yukawa terms \begin{equation} \mathcal{L}_{\mathrm{Y}}=-\frac{m_\tau}{v}h(\kappa_\tau\overline{\tau} \tau +\tilde{\kappa} \overline{\tau} i\gamma_{5} \tau )\tag{1,2}\\ = -\frac{m_\tau}{v}h\kappa_\tau(\overline{\tau} \tau + i\tan (\alpha)~ \overline{\tau} \gamma_{5} \tau ), \end{equation} which extend the standard model result. The first term is the SM Higgs coupling, even under CP, but the second is odd, parameterized by the angle α, so that term is violating CP . The α =0 limit is the SM.

N.B. The SM also violates CP by just a little on account of phases of the CKM matrix comparing the three generations, but that is a separate effect, independent of this Higgs coupling, and, as stressed, very small. Here, the emphasis is on a larger effect due to additional Higgses, not covered here, which result in the above effective term. They are there, however, prejudicing the next subquestion.

2. Under a chiral rotation, the kinetic and gauge-coupling terms for the fermions are invariant, but not the above gamma-less bilinears, scalar and pseudo scalar, respectively. Under such a transformation, the two terms above mix. If there is only one Higgs, as in the SM, even if you had the above (2), a chiral rotation would get rid of α, and absorb it into the definition of the fermions such as this one. The reason you cannot do this in the general extended theories is because the h here is a linear combination of scalar and pseudoscalar Higgses which might be there, and are physical, i.e. a chiral rotation may not eliminate them. No mater how you redefine your fermions, the general form above is there, to be determined by experiment. See the linked questions and especially this one. The above h is thus taken to be a scalar, even under CP, so, manifestly, the second term violates CP for non-vanishing angle.

Answer 2

The term “CP odd” refers to an observable whose sign changes under a CP transformation. The term “CP violating” refers to an observable which is not an eigenstate of the CP transformation.

Consider these statements about electromagnetism:

observable	$C$	$P$	$CP$
charge $q$	$-$	$+$	$-$
displacement $\vec x$, velocity $\vec v$	$+$	$-$	$-$
angular momentum $\vec L\sim \vec x\times\vec v$	$+$	$+$	$+$
current $\vec I\sim q\vec v$	$-$	$-$	$+$
electric field $\vec E\sim q\hat x$	$-$	$-$	$+$
magnetic field $\vec B\sim q\vec v\times\hat x\sim q\vec L$	$-$	$+$	$-$
electric dipole moment $\vec p = q\vec x$	$-$	$-$	$+$
magnetic dipole moment $\vec\mu\sim \vec B$	$-$	$+$	$-$
electric work done $q\int\vec E\cdot\mathrm d\vec x$	$+$	$+$	$+$
electric dipole energy $U_E=\vec p\cdot\vec E$	$+$	$+$	$+$
magnetic dipole energy $U_B=\vec\mu\cdot\vec B$	$+$	$+$	$+$

A system with monopole electric field and dipole magnetic field may be an eigenstate of $CP$, with eigenvalue $-1$. But a system with both magnetic and electric dipole moments cannot be an eigenstate of CP. The statement that “the Standard Model approximately conserves CP” is equivalent to the prediction that Standard Model particles with monopole electric charge and/or magnetic dipole moments will, approximately, have zero electric dipole moment.

The Higgs is a predicted to be a scalar field, with even parity. The Particle Data Group is does not yet assign eigenvalues for C, P, or CP to “the signal that has been discovered in Higgs searches.” Because our universe has a CP-violating matter-antimatter asymmetry, we predict that none of the Standard Model fields are exact CP eigenstates. How much of this mixing leaks into the Higgs sector is the subject of vigorous research.