Why was it always assumed that the Higgs boson is a CP even particle?

I understand that experimentally, it just is so but I am under the impression that before its discovery people took it to be CP even and I do not see why.

  • $\begingroup$ Near duplicate. $\endgroup$ – Cosmas Zachos Feb 5 '19 at 20:18
  • $\begingroup$ Hint: it is in the same piece of the complex Higgs doubllet field that the v.e.v. thereof is, with quantum numbers of the vacuum. So its coupling to fermions has the same symmetries as their mass term. $\endgroup$ – Cosmas Zachos Feb 5 '19 at 20:35

A Higgs boson can be endowed with a VEV entailing both CP even (scalar) and CP odd (pseudscalor) sectors, reflected as a "complex" fermion mass term as: $$ m\bar{\psi} e^{\theta i\gamma_5} \psi = m\cos\theta \bar{\psi} \psi + m\sin\theta \bar{\psi} i\gamma_5\psi. $$

The fun fact is that after a global rotation of the fermion field $$ \psi \rightarrow e^{-\frac{1}{2}\theta i\gamma_5} \psi. $$ the "complex" mass term can be transformed into a scalar mass term: $$ m\bar{\psi} e^{\theta i\gamma_5} \psi \rightarrow m\bar{\psi} \psi. $$ In other words, via redefining the fermion field, the CP odd part of the Higgs boson can be effectively rotated away. That being said, it's achievable if there is only one Higgs boson. If you fancy a fancy-schmancy beyond standard model involving multiple Higgs bosons, the said rotation can only make one of the Higgs bosons CP even.

Added note: Dirac arrived at the Dirac equation via the "square root" of Klein–Gordon equation. Let's double check whether the Dirac equation with "complex" mass term (in Planck units $c= \hbar = 1$) $$ i\gamma^\mu\partial_\mu\psi = me^{\theta i\gamma_5} \psi $$ can get us back to the Klein-Gordon equation $$ (\partial^\mu\partial_\mu + m^2)\psi = 0. $$ Let's get cracking on the nitty-gritty: $$ m^2\psi \\ = (me^{-\theta i\gamma_5}) (me^{\theta i\gamma_5})\psi \\ = (me^{-\theta i\gamma_5}) (i\gamma^\mu\partial_\mu)\psi \\ = (i\gamma^\mu\partial_\mu) (me^{\theta i\gamma_5}) \psi \\ =(i\gamma^\mu\partial_\mu)(i\gamma^\nu\partial_\nu) \psi \\ = -\partial^\mu\partial_\mu\psi. $$ Voila! We indeed recover the classic Klein-Gordon equation, without any funny "complex" factor. Note that in the 4th line we leveraged the crucial anti-commuting property between $\gamma_5$ and $\gamma^\mu$.

In a nutshell, the most general "square root" counterpart of the Klein–Gordon equation is the Dirac equation with a "complex" mass term $m e^{\theta i\gamma_5} \psi$. The "real" mass Dirac equation is merely a special case of $\theta = 0$.

  • $\begingroup$ Let me check my understanding with you: the first term in the first line is a scalar and does not change under CP however the second term picks up a minus sign meaning that they are odd and even respectively. I could always have coupled the Higgs to fermions in this way had I wanted and then by a redefinition I could always remove the odd part (in the standard model). $\endgroup$ – Kris Feb 7 '19 at 13:14
  • $\begingroup$ In what what is the exponential factor you introduce related to the more usual $\phi$ field? $\endgroup$ – Kris Feb 7 '19 at 13:15
  • $\begingroup$ @Kris, the exponential factor $\theta$ is the axial transformation angle, usually linked to the axion, which is a pseudo Nambu-Goldston boson with a mass subject to the ABJ quantum anomaly effects. $\endgroup$ – MadMax Feb 7 '19 at 15:53
  • $\begingroup$ @Kris, the usual electroweak $\phi$ doublet is a scalar Higgs, which is CP even. The CP odd $\phi `$ is an independent electroweak doublet. The exponential factor $\theta$ is the axial transformation angle linking these two fields, so that they in aggregation form a quadruplet, if you will. $\endgroup$ – MadMax Feb 7 '19 at 16:04
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    $\begingroup$ @SRS, nothing, I mean NOTHING, can prevent a fermion from having a "complex" mass term $m\bar{\psi}e^{i\theta\gamma_5}\psi$. It meets all the usual requirements: electromagnetic $U(1)_{EM}$ gauge symmetry, Lorentz symmetry, and renormalizability. Of course, it breaks the electroweak and chiral symmetries, in the same way as a typical "real" mass term $m\bar{\psi}\psi$. $\endgroup$ – MadMax Oct 17 '19 at 13:51

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