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In supersymmetry, for each particle (boson/fermion), there is a symmetric particle which is a fermion/boson.

The MSSM predicts five Higgs bosons: two neutral scalar ones (H and h), a pseudo-scalar (A) and two charged scalars.

Does this mean that there are five higgsinos?

My question arises because I read:

What is the difference between the SM Higgs and the supersymmetric H and h?

Nothing.

So I'm confused, is it saying that H and h are the supersymmetrical parters of the Higss boson?

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...because the particle that we discovered is, in fact, a boson. So, as you correctly state, the Higgsino would have to be a fermion. –  Dmitry Brant Oct 25 '13 at 16:13
    
@DmitryBrant So the answer is just physics.stackexchange.com/q/44618 ? –  jinawee Oct 25 '13 at 16:16
    
"Have we found the Higgsino?", no. –  user29727 Oct 25 '13 at 17:17
    
The spin is 0 for a Higgs, since it' is a boson. –  Dimensio1n0 Nov 1 '13 at 15:05
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2 Answers 2

The Higgs field of the standard model is a complex SU(2) doublet. This means it has two complex-number-valued components which interact with the weak-force gauge bosons. Two complex-number-valued degrees of freedom, equals four real-valued degrees of freedom. Three of these degrees of freedom are absorbed into W+,W-,Z bosons (giving them each a third, spin-0 polarization, and thus making them massive rather than massless), and the fourth real degree of freedom is "the Higgs boson". It is the left-over part of the Higgs field.

In the minimal supersymmetric standard model, there are two Higgs superfields (they differ from each other by having opposite "hypercharge"). One will give mass to up-type quarks, the other to down-type quarks and to the charged leptons. In the standard model, it's the same unique Higgs field which gives mass to all these fermions, but supersymmetry forbids certain couplings, so the MSSM is more complicated.

Each Higgs superfield has a bosonic part that is a scalar field with two complex-number-valued components, as above, and a fermionic part consisting of two two-component spinors. This fermionic part is the higgsino.

If we add up everything, the MSSM then has two higgsinos and eight real degrees of freedom in the bosonic part of its Higgs superfields. As before, three of the real degrees of freedom are absorbed by W+,W-,Z bosons, but now there are five real degrees of freedom left over, and these are the five Higgs bosons of the MSSM. They are conventionally denoted h, H, A, H+, H-, and have various charges and symmetry properties. h and H are neutral and "CP-even" like the lone Higgs boson of the nonsupersymmetric standard model. That is the sense in which they are the same as it.

You can read about all this in "A Supersymmetric Primer".

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Although I agree with everything that Mitchell Porter wrote in his nice answer, I would like to add a few remarks.

With supersymmetry, there are equal numbers of bosonic and fermionic degrees of freedom for each type of field.

Prior to electroweak symmetry breaking, in the simplest supersymmetric models (e.g., the so-called MSSM), the Higgs scalars have $8$ real degrees of freedom. There must be (at least) two Higgs-doublets in supersymmetric models.

After electroweak symmetry breaking, there are indeed $5$ Higgs bosons ($h$, $H$, $A$, $H^\pm$), as you correctly write. The missing $3$ degrees of freedom are eaten by the $W^\pm$ and the $Z$-bosons, when they acquire masses.

There are $4$ higgsinos, labeled by their charge and by whether their scalar superpartner helps give mass to up-type or down-type quarks: $\tilde h_u^0$, $\tilde h_d^0$, $\tilde h_u^+$, $\tilde h_d^-$. Each higgsino has $2$ degrees of freedom - so in total higgsinos have $4\times2=8$ degrees of freedom, matching those of the Higgs scalars.

Nobody has observed direct evidence for higgsinos, or any other supersymmetric particle. Though there are strong theoretical hints and indirect evidence for their existence.

That is not the end of the story. The higgsinos mix with fermions with identical quantum numbers. The neutral higgsinos mix with the photino and zino, resulting in four neutral particles called neutralinos, and labeled $\chi_{i=1,2,3,4}^0$, where $\chi_1^0$ is the lightest neutralino etc. Similarly, the charged higgsinos mix with the charged wino, forming two charginos, $\chi_{i=1,2}^\pm$.

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