# Higgs Mechanism, how does it work?

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1. Aug 17, 2013

### friend

I'm not sure this thread belongs in this forum since we have now discovered a Higgs like boson.

But I'm trying to get a better understanding of how the Higgs mechanism works. Let me share what I think I know, and you tell me how far off I am.

Particles gain mass as they travel through the Higgs field. The Higgs field exists everywhere; it is a scalar, spin-0 field, and it has a non-zero value everywhere. There is a coupling constant between the Higgs field and particle fields. The larger the coupling the larger the mass.

In another contexts, IIRC, I'm told that mass is acquired through a symmetry breaking process since, for example, muons have more mass than the electrons but are otherwise the same.

So one question I have is does the Higgs field break particle symmetries to produce mass? For example, if the Higgs field had a changing value, I could understand how this would break symmetries and produce mass. Thanks.

2. Aug 17, 2013

### Staff: Mentor

Even before its discovery, the Higgs boson was an (undiscovered) part of the Standard Model, I moved the thread.

The symmetry breaking is a broken symmetry of the Higgs field. It leads to the non-zero vacuum expectation value (->broken symmetry) you mentioned.

The coupling strengths of fermions to the Higgs field are arbitrary, there is no reason why they should be equal (but we don't understand why they have this structure).

3. Aug 18, 2013

### friend

I appreciate your help. But your answers were kind of sparse. So I have some follow up questions that you or anyone else might want to answer.

When you say, "broken symmetry of the Higgs field", I assume that means that the Higgs field has different values at different spacial coordinates, right? And is it this changing value of the scalar Higgs field breaking the U(1)SU(2)SU(3) symmetries of the SM particles and giving mass to those particles?

That would seem to confirm what I'm understanding in the lecture, Demystifying the Higgs Boson with Leonard Susskind. He calls the Higgs field a "condensate" so that absorbing or emitting a higgs boson (or is it a ziltch boson) does not change the condensate in any way. And if the Higgs condensate does not change during boson emittion or absorbtion, then that is why particles don't change momentum during absorbtion or emittion, right? I could use a better understanding of what condensate means in more general terms in particle physics if anyone is game.

At 0:57:00, Susskind states, "higgs boson is an oscillation in the potential". In another place he says it is "a soundwave propagating through the condensate". Is this accepted language, or is he simplifying for his audience? This leads to the question of if there are higgs bosons of any wavelength like the photon? Or is there only the 126GeV Higgs boson? Another question is whether the Higgs "potential" is the same as the Higgs "condensate", just to be clear about terminology.

By "It" you mean the changing Higgs field (higgs bosons)? And by "non-zero vacuum expectation value" you mean of the particles derived from the U(1)SU(2)SU(3) symmetry (and NOT the vacuum expectation of the higgs field itself), right?

Is it true that we really don't know where the Higgs field itself comes from or why it has the non-zero value everywhere, it just seems to help give mass to particles, right?

Thanks.

Last edited: Aug 18, 2013
4. Aug 18, 2013

### Staff: Mentor

No, the field could have a different expectation value everywhere (for example: 0, with an unbroken symmetry).

??
Particles do change their momentum if they emit or absorb particles.

The whole lecture is an attempt to describe the result of formulas (=the actual physics) without using those formulas.

There is only the 126GeV Higgs boson as there is only the 0 eV photon. Those numbers are rest masses. Moving (in your reference frame) Higgs bosons have an energy of more than 126 GeV.
"It" means symmetry breaking.

"non-zero vacuum expectation value" is the non-zero vacuum expectation value of the Higgs field.

We have no idea why the universe has all those fields we can observe.

5. Aug 18, 2013

### Bill_K

No. The symmetry that's broken is described by Susskind and his sombrero. The original electroweak symmetry [STRIKE]SU(3)[/STRIKE] SU(2) x U(1) requires that photon, W and Z all be massless. This symmetry is broken by the existence of the Higgs condensate. The only massless particle remaining is the photon, and the only symmetry remaining is the electromagnetic gauge symmetry (around the periphery of the sombrero).

It's unusually colorful, but not incorrect.

126 GeV is the rest mass of the Higgs boson. If the Higgs boson is moving its energy will increase, just as for other particles.

The condensate is the Higgs ground state, which occurs at the value where the Higgs potential is a minimum.

We don't know where ANY of the particles "come from", or why they exist, or why they have the properties they have.

Last edited: Aug 18, 2013
6. Aug 18, 2013

### friend

I assume that the particle's interaction with the Higgs field could come from any direction. So the question would be why doesn't the Higgs interaction change the momentum of the SM particle as would happen with any other type of particle interaction. I'm thinking that it may be because the Higgs interaction causes no change in the condensate, and so there is no corresponding change in the momentum of the SM particle.

7. Aug 18, 2013

### Staff: Mentor

I think you are using some model way beyond its range of applicability.

8. Aug 18, 2013

### Bill_K

Because the Higgs condensate is uniform in space and time, it does not "come from any direction". It conveys no momentum to the particles that interact with it.

9. Aug 18, 2013

### RGevo

Rather technical points, but fairly important when you come to run through the full details.

The symmetry breaking is of an su(2)u(1), an su(3) example is more complicated and not what is generally discussed.

Before symmetry breaking the fields are w1, w2, w3 and b corresponding the the 3 degrees of freedom of the su(2) an the u(1) - the su(3) example would have more complicated extended Higgs sector. The massive charged w and neutral z and massless photon are combinations of these.

10. Aug 18, 2013

### Bill_K

Oops, sorry! Corrected, with thanks.

11. Aug 18, 2013

### friend

You answer "No", but you seem to be affirming the implication in my question - that it is the U(1)SU(2)SU(3)=SM symmetry that is broken by the Higgs. (I was otherwise concerned that the previous poster may be referring to breaking the symmetry of the Higgs field itself)

Maybe I'm trying to make a distinction between the Goldstone bosons of the mexican hat potential and whatever else about the Higgs field that might be interacting with the SM particles to give them mass.

I was listening to Lesson39: Higgs Boson basics. At time 28:00, he draws the "mexican hat" potential and says, "mass is related to the curvature at that point". I assume he's talking about the curvature of the U($\phi$) curve at the point of the minimum at M/L (See figure at 28:00). Just because there is curvature there does not mean there is a Goldstone boson oscillating in the troph. But perhaps the existence of the curvature alone has something to do with breaking the SM symmetries to give particles mass. Is he trying to imply that the greater the curvature of U($\phi$) at M/L, then the more the mass?

I'm trying to reconcile the Higgs mechanism of breaking the SM symmetries with my conception of "breaking" the commutator relations between the generators associated with that symmetry if some field multiplies differently on each of the generators to destroy the symmetry. If that field were a constant multiplying the generators, then the symmetry remains. But if the field has different values for each generator, then the commutation relation and thus the symmetry would be broken. I remember elsewhere where Susskind says that it is changing potential that breaks symmetry. I'd have to look it up.

Last edited: Aug 18, 2013
12. Aug 19, 2013

### Bill_K

The SU(2) x U(1) electroweak symmetry is broken by the nonzero value of the Higgs field. "No" referred to the statement I quoted. The value of the Higgs field is constant and does not vary from point to point:

Later,

The mass of the Higgs boson comes from the second derivative of U(φ) at the minimum, but all the other masses (W, Z, and fermions) do not, they arise from the value of the Higgs field.

An interesting feature of the Higgs mechanism is that the symmetry group and its commutators are unaffected. When we say the symmetry is broken, we mean the vacuum state does not respect the symmetry.

13. Aug 19, 2013

### friend

I really appreciate your help in all this. So just to clearify terminology, it sounds like the Higgs boson is a Goldstone boson of the mexican hat potential of the higgs field, is this right?

Something seems a little inconsistent here. I could use a little help. Maybe I need a paradigm shift in my (mis)conceptions. Isn't every continuous symmetry expressible in terms of a commutation relation between generators? If so, then how can symmetry breaking have any meaning without breaking the commutator relations as well? And how else can you break the commutation relations without destroying the multiplication properties by introducing some non-constant field that multiplies differently each factor in the multiplication of the commutator? What do you mean "the vacuum state does not respect the symmetry"; that sounds a bit anthropomorphic?

Last edited: Aug 19, 2013
14. Aug 19, 2013

### Bill_K

The vacuum state is not invariant under the symmetry. It lies in the valley of the mexican hat potential. There are many such states, and they transform into one another under the gauge transformations, but you have to choose one of them. And it will not be invariant under the gauge, because a gauge transformation will take it into another one of the states. Thus you are forced to break the symmetry by choosing one.

This is starting to get a bit technical, and maybe I should refer you back to Susskind's lectures at this point. A Goldstone boson is a massless particle. There is a Goldstone boson corresponding to each broken symmetry. In the electroweak case the full symmetry group SU(2) x U(1) has four generators, three of which are broken. So - three Goldstone bosons. But this is where the Higgs mechanism comes in. When we rewrite the theory in terms of the true vacuum state, the three Goldstone bosons are no longer there. They have been absorbed by the W and Z bosons, providing the third component that a massive vector particle requires. The result is that the gauge bosons have mass, and the Goldstone bosons are no more.

15. Aug 19, 2013

### friend

This would confirm my suspicions, if my terminology is correct. Please correct me here: a goldstone boson oscillates at the minimum of the mexican hat potential, towards the center and away from it, rolling up and down the potential around the minimum (not around in the rim), right?

What book would you recomment that best explains these things?

16. Aug 20, 2013

### Bill_K

No, that's the Higgs boson. Repeat after me:

An argument against early electroweak theories was that broken symmetries implied the existence of massless Goldstone bosons, and no such particles were observed. Higgs showed how to construct a theory in which Goldstone bosons were not present.

17. Aug 20, 2013

### bapowell

The Goldstone modes are associated with the degrees of freedom along the trough of the potential, where $dV/d\phi = 0$.

18. Aug 20, 2013

### Bill_K

There are no Goldstone modes. The degrees of freedom along the trough of the potential are gauge transformations.

19. Sep 2, 2013

### friend

Is it true that interaction with the Higgs field causes the spin of the electron (and other massive particles, I assume) to oscillate? How does that work, I wonder?

20. Sep 2, 2013

### RGevo

I am guessing what is being referred to is the higgs interactions with fermions. i.e.

y {f_{L}}^{\dagger} phi f_{R}

i.e. the sandwich of the higgs doublet phi with an SU(2) doublet of left handed fermions and right handed fermion singlet (y is the yukawa coupling). The interaction is between the left and right handed fields.. which is probably what is meant from the 'spin of the electron (and other massive particles, I assume) to oscillate?'

Perhaps someone else has a better explanation..

Last edited: Sep 2, 2013
21. Sep 3, 2013

### Bill_K

It's not the spin that's involved, it's a quantity called the chirality. The chirality operator is γ5, with eigenvalues ±1 and projection operators ½(1 ± γ5). Given any fermion field ψ, we split it into right- and left-handed parts, ψR = ½(1 + γ5) ψ and ψL = ½(1 - γ5) ψ.

For a massless fermion, chirality is a good quantum number, but for a massive one it is not. In fact the mass term in the Hamiltonian can be written

mψψ = m(ψLψR + ψRψL)

showing that it couples ψL and ψR. Hence the energy eigenstates of the fermion do not have unique chirality.

Don't blame it on the Higgs! This holds equally true whether the mass crossterm is put in by hand, or generated by coupling to the Higgs field.

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