# Why doesn’t the Higgs boson have a spin?

This is due to the mathematical framework of the standard model of elementary particle physics.I would like to explain a few terms:

**Lagrange density, oak invariance and lorentzin variance**

The motion equations for the particles of the standard model are formed from a mathematical function called the*Lagrange* density.

According to the principle of *oak invariance* and *lorentzinvariance,* this lagrange density must remain invariant under certain symmetry operations and under Lorentz transformations.

Clearly, *oak invariance*means that the fundamental interactions do not change when a certain size is chosen locally (calibrated).

*Lorentzinvariance* means that the motion equations take the same form in all inertial systems, i.e. they are compatible with the special theory of relativity.

These two concepts greatly limit the possible lagrange density.That is, whether certain elementary particles and exist and what properties they have depends on the symmetries and structure of our universe.

**Higgs Field and Higgs Boson**

First of all, the question arises as to why a Higgs field or field in the first place.the suggestion that Higgs boson is postulated. The reason for this is that mass terms for the eichbosons and fermions in the lagrange density would violate the principle of oak invariance.

Therefore, a complex scalar field, the Higgs field, is postulated to construct oak variant mass terms for the calibration bosons and fermions.

Clearly explained, the weak eichbosons and fermions get their mass by Welchsel effect with the Higgsfeld.

**Why doesn’t the Higgs boson have a spin or spin?**Spin 0?

The reason for this is the degrees of freedom of the fields involved in the Higgs mechanism.

The fundamental interactions are mediated by calibration bosons.In the case of electromagnetism, this is a massless vector boson, the photon. In the case of the weak interaction, these are three massive vector bosons: W+, W- and Z-0.

One now introduces complex Higgs-Doublett H = (H+,H-0) to construct oak variant terms in the lagrange density, which describe the coupling of the Higgs field to the weak calibration bosons.

But this Higgs double has four degrees of freedom.In the course of the Higgs mechanism (Higgs mechanism 鈥?Wikipedia) and the spontaneous symmetry calculation (Spontaneous Symmetry Refraction 鈥?Wikipedia) three of the four degrees of freedom become the longitudinal degrees of freedom of the massive calibration bosons W-+,W-,Z 0.The photon remains massless as desired.

The remaining degree of freedom is then assigned to the Higgs boson.

**There are several (mathematical) reasons why the Higgs boson does not have a spin:**

If the Higgfeld were not described by a complex scalar doublett (spin 0) with four degrees of freedom, but by a vector field (Spin 1) then one would not have the right number of degrees of freedom between the weak calibration bosons and the Higgsboson self-split.

Furthermore, the Higgsdoublett is coupled to the lepton doubletts, this is described by the terms in the Yukawa coupling (see Why does the Higgs field have a selected hypercharge of +1?).For these terms, a certain transformation behavior is required – keyword oak variance. This is in the simplest case fulfilled by a scalar doublett.

**Short address:**

The easiest way to describe eichboson and fermion compounds mathematically consistent works with a spin-0 field, the Higgs field.