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I am very puzzled by the quintet of Higgs bosons in the MSSM: two charged, two scalars and a pseudoscalar. I wonder if they could be understood better if they were considered jointly with the three "bosons" that are "eaten" by the Higgs mechanism. For instance, if they were two charged bosons and a pseudoscalar one, then the whole set should be four charged, two pseudoscalars and two scalars, which seems more sensible if you wish, for instance, reorder them into chiral and dirac supermultiplets jointly with the higgsinos.

I have been reminded of this question because it seems that now LHC is trying to guess if the discovered Higgs boson is a pseudoscalar or an scalar, and they are doing this by looking how it couples to ZZ, if it is to transversal or to longitudinal modes.

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The MSSM has 2 complex doublets i.e. 8 real components in the Higgs fields. Four of them are electrically neutral (real bosons, antiparticles to themselves), four of them (i.e. two particle-antiparticle pairs) are electrically charged.

One neutral real boson and two charged ones (one charged pair) get eaten by the gauge bosons because 3 generators are broken, going from $SU(2)\times U(1)_Y$ to $U(1)_{\rm em}$.

The electrically charged ones are CP-partners of their oppositely charged antiparticles, of course. We don't learn anything if we consider CP for charged particles. One combination of the positive and negative particle is always CP-even and the orthogonal combination is CP-odd.

The four neutral ones are evenly split to two CP-even and two CP-odd ones, too.

It's the CP-odd ones that are eaten by the gauge bosons.

So one is left with the CP-even charged ones, three CP-even neutral bosons, and one CP-odd neutral Higgs boson. Why are CP-odd bosons eaten? It's because the gauge bosons pick a minus sign under C, simply because they're linked to generators which are C-odd, too. (Antiparticles need to be transformed by the opposite gauge transformation phases.) So under P, the gauge bosons are ordinary vectors, not axial vectors, but the C flips their sign, which is why they're "axial vectors under CP".

The Goldstone bosons that are eaten are CP-odd. Now, are they C-even, P-odd, or vice versa?

This is really a meaningless question because the spectrum of the electroweak theory (the fermions) isn't P-symmetric. Not even the free Lagrangian (not even when mixing is neglected). It's because there are left-handed doublets and right-handed lepton singlets, etc. So it makes no sense to try to assign P parity to fields - because no choice will lead to a P-invariant theory.

If you remove fermions, you may say that the Goldstone bosons are P-even, C-odd. They have the "normal sign" under P because the gauge bosons do, but they have the opposite one under C.

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