# Higgs Boson

1. Jun 2, 2010

### roflwaffle

how do we "know" it must exist?

btw im new here, dont flame me too hard

2. Jun 2, 2010

### xlines

Why should we flame you?! On the question, I don't "know", that is for sure. Maybe you should check out:

http://math.ucr.edu/home/baez/constants.html

Baez is making plausible argument that a piece of puzzle is missing, so there is a good reason, among several others, to believe there is something there. But we do not *know* until particle is detected.

3. Jun 2, 2010

### DJsTeLF

The best (in the sense of most accurately tested and confirmed by comparison to experiment) theory we have of particles is something called the Standard Model of Particle Physics. In a nutshell it is a quantum field theory with a group structure SU(3)*SU(2)*U(1) that respects special relativity.

Within it there is a fundamental field associated with each type of fundamental particle. The particles themselves are then the quanta (or excitations) of the fields.

The group structure of the theory has many qualities that allows us to derive a lot of testable physics from it. The trouble is (should say 'was' here really) if we were to simple put the known masses of the particles in by hand we would break many of the symmetries of the theory and it loses its predictive power.

The Standard Model gets around this by suggesting 'how' previously massless fields can acquire mass using something called the Higgs Mechanism. (Check it out on Wiki or similar for more info.) The mechanism involves the addition of at least one (in its simplest form) additional field, namely the Higgs field. The qaunta (particle) of this field is the Higgs boson. Something like the Higgs mechanism must exist in nature so as to explain why some particles have mass.

A nice way for a layperson to picture this is with the following analogy. Imagine a popular person entering a busy room of uniformally distributed people. As he / she walks through through room the people nearby flock round and thus add to effective mass of the person trying to get across the room. As he progresses the people at the back of the huddle back off and return to uniform distribution but more continue to gather from infront of he / she. In the analogy the popular person is a massive particle, say for example neutron. The room is the Higgs field and the uniformaly distributed people are the Higgs bosons. A photon could be imagined as a wholey unpopular / invisible person that passes through the room unaffected by the crouds. (In physics speak we say the Higgs field couples to others with a magnitude proportional to their mass.)

Last edited: Jun 2, 2010
4. Jun 2, 2010

### tom.stoer

The question is not if there is something, but what it is exactly. Perhaps the Higgs is only a kind of effective degree of freedom, a quasi-particle like in condensed matter physics. A scalar particle is a stranger in the SM. Alternatives are wellcome

5. Jun 5, 2010

### DJsTeLF

Heya tom, your post isn't very clear- I wonder if you could elaborate. Also, the Standard Model is based on the Higgs (scalar), without which nothing has mass...

6. Jun 5, 2010

### tom.stoer

My idea is the following: the concept of sponaneous symmetry breaking and a condensate (= vacuum expectation value) is borrowed from condensed matter physics. But in condensed matter physics a condensate is never an effect from an elementary particle. There is a similar effect in QCD: the quark condensate $$<\bar{q}q>$$ serves as order parameter of the chiral symmetry (spontaneous breaking of the chiral symmetry goes hand in hand with the nearly massless pions, the corresponding Goldstone bosons).

But the Higgs is totally different, a stranger in the SM. It is the only scalar; its mass is affected by huge quantum corrections and it is by no means clear why it should not run towards the unification scale; it has a rather strange self-interaction; and last but not least there is no reason for a particle at all (all that is needed for the spontaneous symmetry breaking is a condensate that coupes to the other particles).

So why shouldn't it be possible that the Higgs is an effective degree of freedom, like a phonon or something like that? That would explain why it has such a strange potential which could be created by quantum fluctuations of the condensate.

Last edited: Jun 5, 2010
7. Jun 5, 2010

### Kevin_Axion

I thought the Higgs Boson breaks SU(2)xU(1) symmetry giving mass to three bosons namely: W+- Boson and leaves one boson massless - the Photon. And then there are other mechanisms that give mass to the other particles. Correct me if I'm wrong, does it give mass to every elementary particle?

8. Jun 5, 2010

### tom.stoer

Yes, its coupling to the elementary fermions is responsible for their mass.

9. Jun 5, 2010

### humanino

To insist, the QCD quark condensate contributes to the W and Z masses by breaking electroweak symmetry as well, just not enough (more than 3 order of magnitudes too little) to account for the entire mass (besides, as such it does not generate fermion masses).

The Higgs mechanism could stem from an analog condensate of a stronger coupled unbroken Yang-Mills interaction (technicolor models).

Chris Quigg has a nice review article which mentions those.

10. Jun 5, 2010

### Parlyne

In addition to the condensate, you also need three degrees of freedom which can be absorbed by the $W^+$, $W^-$, and $Z$ to become their longitudinal polarization states.

Whether or not the excitations responsible for the symmetry breaking are fundamental, it's pretty hard to come up with a set of fields that couples correctly to make the $Z$ massive, but not the $\gamma$ without there being at least one physical state left over.

11. Jun 6, 2010

### tom.stoer

I agree that its very hard to figure out something like that.

Technicolor as mentioned above is rather obscure. Strong binding (in order to hide the techni-quarks from us) requires high energy and would usually generate large masses - which is not what we observe.

I think we need something new.

Has anybody thought about a W- and Z-boson self-interaction?

12. Jun 6, 2010

### Parlyne

I'm pretty sure that any self-interaction not already present in the SU(2)xU(1) gauge interactions would break the gauge symmetry sufficiently badly that we wouldn't expect the observed relationship among the coupling strengths and W and Z masses. I believe it would also violate unitarity in much the same way that the Standard Model does when implemented without the Higgs.

13. Jun 6, 2010

### tom.stoer

Perhaps the interactiosn already present are sufficient. Afaik non-perturbative aspects haven't been investigated in much detail for SU(2)*U(1). Für QCD we know all these instantons etc., so perhaps we miss something in el.-weak theory.

14. Jun 6, 2010

### Parlyne

Non-perturbative solutions of the electroweak theory have definitely been investigated. See, for example, sphaelerons, which break baryon and lepton number conservation.

It's my understanding, however, that a gauge theory's instantons can't break the gauge symmetry.

15. Jun 7, 2010

### tom.stoer

OK, I agree.

An instanton certainly not.

I do not say that it's obvious, otherwise it would have been investigated already. I am only saying that my gut feeling is that the Higgs looks more like an effective degree of freedom. In that case there is no additional particle to be detected, only the vev is required.

16. Jun 7, 2010

Staff Emeritus
The W and Z charges - and thus their interactions - are fixed by the SU(2)xU(1) theory. That's what SU(2)xU(1) means. If you want to have a different self-coupling, you need to add a new interaction - and now you have Technicolor.

17. Jun 7, 2010

### tom.stoer

Charges and coupling are not fixed; they are subject to running according to renormalization.

At low energies they are fixed by Fermi's theory, i.e. phenomenology, not by GSW. At high energies they may run, but of course this would not help as the masses we are talking about are low-enbergy phenomena.

18. Jun 7, 2010

### Parlyne

Charges do not run. Only the the coupling strength does. More importantly, though, the running of the couplings cannot, by itself, break the gauge symmetry.

Vanadium's point was that the symmetry structure of the theory fully determines the coupling structure of the gauge bosons. That is, what terms are present and how they depend on the gauge couplings. None of this is changed by the running of those couplings. The running means nothing more than that there is an energy-dependent change in the strength of each interaction term, which is entirely determined by the structure of the theory already present.

19. Jun 7, 2010

### tom.stoer

I agree, charges do not run.

My point was that renormalization reflects the existence of quantum corrections. I do not know if e.g. the mass of the vector bosons could be created by purely by renormalization effects. Of course this must not be mass renormalization which would break gauge symmetry (Slavnov-Taylor identities), but something different. One (strange) possibility could be that renormalization creates non-minimal coupling terms.

Again: it is certainly not something obvious, otherwise it would have been found already.

20. Jun 15, 2010

### Trenton

Forget spin, symmetry and all that for one second. What I want to understand is how the W and Z bosons with something like 90 times the mass of a proton, come to exist. I know they only exist for trillionths of a trillionth of a second but they still exist.

What is going on here with specific reference to mass-energy equivalence?