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Starting from a model with two complex Higgs doublets (as e.g. in the MSSM) we arrive at 5 physical Higgs bosons (instead of 1 as in the Standard Model), 2 of which are charged and 3 are neutral. One of the neutral Higgs bosons is a pseudoscalar (CP-odd), the other two are "proper" scalars (CP-even).

In Drees' "Sparticles", it says

...imply [...] a neutral scalar which is CP odd on being a linear combination of the imaginary components of the neutral Higgs fields.

The corresponding equation is

$\frac{A}{\sqrt{2}} = \Im h_1^0 \sin\beta + \Im h_2^0\cos\beta$

The statement suggests that it should be obvious (and it's likely more general than the case at hand), but it's not, to me. Can someone explain this reasoning and / or make explicit how this can be seen from the equation?

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  • $\begingroup$ simply because when applying C you use the complex conjugate. Then the imaginary part of the fields gets an extra minus sign (hence CP odd). $\endgroup$
    – Paganini
    Mar 9, 2016 at 8:22
  • $\begingroup$ Sounds reasonable, but: I know the form of $C$ for Dirac spinors, but what does it look like for scalar fields? How do I know that once I get a minus from applying $C$ that applying $P$ to the field doesn't give another? And: I thought $\Im$ would give me something real? $\endgroup$ Mar 9, 2016 at 10:52

1 Answer 1

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Consider terms in the lagrangian of the form \begin{eqnarray} &\bar{\psi_L}\psi_R \phi + \bar{\psi_R}\psi_L\phi^{\ast}\nonumber\\ &=\bar{\psi}P_R\psi\phi+\bar{\psi}P_L\psi\phi^{\ast}\nonumber \end{eqnarray} Now if you consider $\phi$ to be complex, i.e. $\phi=\phi_1+i\phi_2$, these terms become, \begin{eqnarray} &\bar{\psi}\left(P_R+P_L\right)\psi\phi_1+i\bar{\psi}\left(P_R-P_L\right)\psi\phi_2\nonumber\\ &=\bar{\psi}\psi\phi_1+i\bar{\psi}\gamma^5\psi\phi_2.\nonumber \end{eqnarray} Clearly, looking at just the fermionic parts of both terms, the first one is CP even and the second CP odd. Now if you want your theory to be CP invariant, as MSSM is, you need the scalar part coupling to these terms to be even or odd respectively. That is, $\phi_1$ needs to be CP even while $\phi_2$ needs to be CP odd.

I guess it is obvious to you now why a linear combination of the imaginary parts of the Higgs doublet components would be CP odd.

Why this unique feature in MSSM and not SM? That would be because we have just one Higgs doublet in SM (four components) as opposed to two doublets in the MSSM (eight components). Three components are gobbled up by the $W^{\pm}$ and Z Bosons to get mass. Hence, in the SM all we have got left is a neutral scalar, CP even, massive physical Higgs (three eaten up from four components). Whereas in the MSSM we still have five of the components left, including imaginary parts.

Hope that helps.

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