Why is the Higgs boson considered a boson?

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Bosons are described as force carrier particles and, as I understand it, the Higgs mechanism explains why photons are massless, just as all the carriers of the strong nuclear force, the gluons, whereas the weak force bosons have mass. A peculiar type of field is postulated, the ‘Higgs field’. It is believed that this Higgs field is a sort of background field that fills the whole Universe. The SM predicts that via a Higgs mechanism it breaks the symmetry of the electroweak interaction and lets the electroweak bosons acquire a mass. The same field and mechanism explains also why all the particles we know have a mass. Initially all particles are massless, but most of them interact with this background Higgs field and they do so by acquiring mass.

I don't know if the above sketchy description is accurate but do not see in it anything that suggests the Higgs particle as being a 'boson' as a force carrier particle. Why is it considered a boson?
 

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  • #2
Orodruin
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Bosons are described as force carrier particles and, as I understand it
This is incorrect. This is not the defining property of a boson. A boson is a particle which follows Bose-Einstein statistics. According to the spin-statistics theorem, any particle with integer spin will be a boson.
 
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  • #3
Bosons are described as force carrier particles and, as I understand it, the Higgs mechanism explains why photons are massless, just as all the carriers of the strong nuclear force, the gluons, whereas the weak force bosons have mass.
Actually, all SM particles, except Higgs, are formally "massless": there is no explicit mass term in SM Lagrangian. IOW:
Higgs mechanism does not explain why photons are massless, it explains why fermions and W, Z have mass.

A peculiar type of field is postulated, the ‘Higgs field’. It is believed that this Higgs field is a sort of background field that fills the whole Universe.
Every field fills the entire Universe, not only Higgs field. Electron field is everywhere. It's just equals to zero in the regions of space where there are no electrons. Higgs is peculiar by the fact that it is NOT zero in empty space, unlike all other fields.

do not see in it anything that suggests the Higgs particle as being ... a force carrier particle.
Well, what is "force"? If you define "interaction" as any term in Lagrangian which multiplies more than one field, and the bosonic part of this term to be a "force carrier", then SM has fermion-Higgs interactions where Higgs is a force carrier.
 
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  • #4
haushofer
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The Higgsparticle is a so-called Goldstone-particle, which arises when continuous symmetries in a quantum field are spontaneously broken. These symmetries are (part of) the gauge symmetries of the standard model, which are generated by "bosonic generators" (the corresponding conserved charges obey Bose-statistics). As such the corresponding particle is a boson.

You can also spontaneously break symmetries generated by fermionic generators, which would lead to fermionic Goldstone particles. One example (the only one I can think of, actually) of that would be supersymmetry.


Edit: this is wrong, see post #7
 
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Orodruin
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The Higgsparticle is a so-called Goldstone-particle, which arises when continuous symmetries in a quantum field are spontaneously broken. These symmetries are (part of) the gauge symmetries of the standard model, which are generated by "bosonic generators" (the corresponding conserved charges obey Bose-statistics). As such the corresponding particle is a boson.

You can also spontaneously break symmetries generated by fermionic generators, which would lead to fermionic Goldstone particles. One example (the only one I can think of, actually) of that would be supersymmetry.
This is not entirely accurate. The Goldstone bosons of the symmetry breaking correspond to the generators that were broken. In the Standard Model, the three Goldstone bosons that appear upon EW symmetry breaking are "eaten" by the gauge fields (making the Ws and Z massive). The Higgs particle itself is not a Goldstone boson as it corresponds to the unbroken direction.

Generally, Goldstone particles are massless (as long as they are Goldstone and not pseudo-Goldstone), which is certainly not true for the Higgs particle.
 
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vanhees71
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The Higgs boson is not a Goldstone boson of a spontaneously broken symmetry. In the electroweak Standard Model there is no spontaneously broken symmetry. It's unfortunately almost allways called a broken symmetry, despite the fact that a local gauge symmetry cannot be spontaneously broken.

This is very important since it's the whole point of introducing the Higgs sector in the first place. Suppose you'd have a model with the symmetry of the electroweak model but as a global symmetry, i.e., you have the quarks and leptons with weak isospin and hypercharge coupled to the Higgs bosons (in the usually used minimal model you just have a Higgs doublet). Then you'd have indeed three massless Goldstone modes by breaking the (chiral) symmetry ##\mathrm{SU}(2)_{\text{WISO}} \times \mathrm{U}(1)_{\text{WY}}## spontaneously to ##\mathrm{U}(1)_{\text{em}}##. Indeed the original group has four real parameters, the symmetry group of the ground state only one. Thus of the four real field-degrees of freedom used in the Higgs doublet, you have three massless Goldstone modes and one massive particle, the Higgs boson.

However, that's not the Higgs mechanism. In the Higgs mechanism the ##\mathrm{SU}(2)_{\text{WISO}} \times \mathrm{U}(1)_{\text{WY}}## is a local gauge symmetry, and thus you can absorb the would-be Goldstone modes by choosing a particular gauge, called the "unitary gauge" completely into the gaug-boson fields, which makes them massive, and the absorbed would-be Goldstone modes provide the necessary additional field-degrees of freedom for a massive vector particle (a massive vector particle has three polarization states (or spin states), a massless one only two (e.g., the helicity states)). Thus there are no massless scalar particles left in the physical spectrum but only the massive Higgs boson.
 
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haushofer
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Yes, you're right of course, the Higgs particle does not correspond to the "Goldstone particle"; this degree of freedom becomes the longitudinal degree of freedom for the massive particle after the breaking. That's beyond sloppiness of me :P

But the question in the OP is answered by the nature of the gauge groups which are spontaneously broken, I'd say.
 
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Orodruin
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Bosons are described as particles with intergal spin.
You mean "integer" spin. As was mentioned already in post #2, this is not the definition of what a boson is. Bosons are particles that follow Bose-Einstein statistics and the spin-statistics theorem shows that this is equivalent to having integer spin.
 
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  • #10
vanhees71
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Yes, you're right of course, the Higgs particle does not correspond to the "Goldstone particle"; this degree of freedom becomes the longitudinal degree of freedom for the massive particle after the breaking. That's beyond sloppiness of me :P

But the question in the OP is answered by the nature of the gauge groups which are spontaneously broken, I'd say.
The important point of my posting was to stress that local gauge symmetries cannot be spontaneously broken:

https://doi.org/10.1103/PhysRevD.12.3978
 

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