Why is the Higgs $CP$ even?

Question

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.

Answer 1

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.

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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 $\theta = 0$.