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The Higgs is not detected in the asymptotic data, so it is possible that there is no particle interpretation for the Higgs quantum field. Indeed, the Higgs potential is only positive definite if the quartic term is included --- the quadratic term corresponds to a negative mass term. It would seem, therefore, that the Higgs field does not have an on-shell mass spectrum, so that there is no straightforward particle interpretation.

One can say that there is an effective field theory in which there is a resonance near a given mass that we will call the Higgs resonance, but in the absence of a pure mass shell spectrum (that is, if there is a continuous mass spectrum), it is generally taken in QFT that there is no particle interpretation. The resonance is clearly not a $\delta$-function, so is there some other precise way in which we can call the Higgs a particle?

Of course this doesn't call into question the empirical effectiveness of the Standard Model of Particle Physics, it only asks about its interpretation and about how we put the Mathematics into words.

I was somewhat struck by Rolf Heuer's observation (this morning) that this is the first observation of a scalar particle. Indeed, according to the SM, there are no quantum fields that have non-zero mass terms in the absence of interactions. In the absence of interactions, the Higgs field is a massless scalar field. Should we say that it is the Higgs interaction that gives mass to the standard model? (EDIT: Is it better to say that every term that is not quadratic in the fields contributes towards the effective masses of each of the asymptotically observed fields? Or what alternative phrasing is closer to the Mathematics of the interacting fields?)

EDIT(2, $\scriptstyle\mathsf{see\ below\ for\ the\ comment\ that\ prompted\ this\ possible\ rephrasing}$): Is there any part of the definition of "particle" that is not a matter of convention? Does the Higgs cross that bar?

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Rather than saying that it gives mass, maybe it's more correct to say that the Higgs interaction is mass in the SM. –  Alfred Centauri Jul 4 '12 at 14:43
    
OK, but are "gives" and "is" separated by a precise distinction in the Mathematics? –  Peter Morgan Jul 4 '12 at 15:08
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Regarding the implicit question in paragraph 1: You could by the same reasoning, ask if neutrons or muons or any unstable particle is really a particle. It seems to me that if you don't like the answer for neutrons, you're probably using too rigid a definition of particle. –  user1504 Jul 4 '12 at 15:23
    
@user1504 If we really did experiments in the asymptotic regime, then we would never detect neutrons, we would only ever detect photons, protons, and electrons. AFAIK, it's generally taken that asymptotic completeness is necessary for a particle interpretation, which makes photons, protons, and electrons particles, everything else is quantum fields that cause the particles to behave differently. If we do experiments in the near-field, then asymptotic properties of a theory are irrelevant to how we interpret it. Is any resonant peak in the energy spectrum, however wide, a particle? OK, maybe. –  Peter Morgan Jul 4 '12 at 15:41
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Mesons are scalar particles, observed lng before the Higgs. - Particles are excitations of quantum fields, and it is a matter of convention which excitations are regardfed as a particle. Thus one cannot really answer your question. –  Arnold Neumaier Jul 4 '12 at 16:09
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The Higgs is not detected in the asymptotic data, so it is possible that there is no particle interpretation for the Higgs quantum field.

Possible but not plausible. The particle interpretation comes out of quantum field theory quite naturally, so if you want to get rid of a particle but keep the field, it seems that you will have to replace much of QFT as well. Not an easy task. Note that there are other fields whose particles have not been detected asymptotically --- the top quark is but one example --- so the Higgs is not special in this regard.

Indeed, the Higgs potential is only positive definite if the quartic term is included --- the quadratic term corresponds to a negative mass term. It would seem, therefore, that the Higgs field does not have an on-shell mass spectrum, so that there is no straightforward particle interpretation.

This is a common misconception. To have a theory with a well-defined particle-spectrum, one should consider perturbations near a stable vacuum. At this vaccuum, it is not necessary that scalar fields like the Higgs have a zero expectation value. In the case of the Higgs, at zero expectation value we do not have a stable vacuum, so it is meaningless to discuss the mass spectrum there, as you say. But the correct thing to do is to consider the field near the value where its potential is at a minimum, namely to write $\phi = \phi_0 + \delta\phi$, where the potential $V$ is minimized at $\phi_0$, and $\delta\phi$ is the dynamical field. The field $\delta\phi$ does have a positive mass-squared, and this is what we call the mass of the Higgs particle.

Note that shifting the field is also the realistic thing to do, because $\phi$ really does have a non-zero expectation value in nature.

One can say that there is an effective field theory in which there is a resonance near a given mass that we will call the Higgs resonance, but in the absence of a pure mass shell spectrum (that is, if there is a continuous mass spectrum), it is generally taken in QFT that there is no particle interpretation. The resonance is clearly not a δ-function, so is there some other precise way in which we can call the Higgs a particle?

The fact that the resonance has a width (and is not a $\delta$-function) is true for all particles that can decay. It is a question of semantics whether you would like to call such resonances particles, and physicists have decided that it makes sense to do so. But again note that the Higgs is not unique in having a width: the top, $W$, $Z$, and other fundamental particles also have this property. Would you like to stop calling them particles as well?

I was somewhat struck by Rolf Heuer's observation (this morning) that this is the first observation of a scalar particle.

It is the first observation of a fundamental scalar particle. Pions have been observed a long time ago.

Indeed, according to the SM, there are no quantum fields that have non-zero mass terms in the absence of interactions. In the absence of interactions, the Higgs field is a massless scalar field. Should we say that it is the Higgs interaction that gives mass to the standard model?

Whether to say that it is the Higgs field or its interactions that give mass to other particles is a question of semantics. Certainly both are necessary.

Is there any part of the definition of "particle" that is not a matter of convention? Does the Higgs cross that bar?

Particles are the perturbative excitations of quantum fields. Once you expand around the correct vacuum, as I explained above, this definition applies to the Higgs particle as well.

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The OP is using Wigner's definition of "particle", which is different from the quantum field theory definition of particle, and was standard before 1974. –  Ron Maimon Aug 3 '12 at 18:15
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I understand; this is why I gave two additional definitions that explain why people think of Higgs excitations as particles. Sorry if this did more to confuse than to clarify. –  Guy Gur-Ari Aug 3 '12 at 20:38
    
No problem, +1, just wanted to make sure. –  Ron Maimon Aug 4 '12 at 0:48
    
@GuyGur-Ari Nice (+1). Regarding the compositeness of the Higgs: While it is true that experimental data (almost) rule out technicolor models, I think we cannot yet claim that the Higgs boson is elementary, or rather, we cannot even state up to what scale is elementary in experimental terms. Do you agree? –  drake Sep 9 '12 at 0:55
    
I agree that we cannot really claim at this point that the Higgs is elementary. –  Guy Gur-Ari Sep 10 '12 at 21:04
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What is the difference between a field described by $E=E_{0}\sin(\omega t)$ and a field described by $\phi = \phi_{0}+ \phi_{1}\sin(\omega t)$? The particle is the excitation of the field, not the field itself. The only difference between the Higgs and anything else is that its vibrations are relative to a nonzero background, rather than vacuum. There's nothing in Maxwell's equation that prevents you from quantizing around non-zero backgrounds, btw.

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The Higgs potential is certainly a type of interaction term that other quantum fields do not have. Absent that and the interactions of the Higgs field with other quantum fields, the Higgs is a straightforward zero mass scalar field. –  Peter Morgan Jul 4 '12 at 17:49
    
Well, so what? Zero is a minimum of $\phi^{2}$, and $\phi_{0}$ is the minimum of the Higgs potential. Expand around the minimum, and the behaviour is identical. –  Jerry Schirmer Jul 4 '12 at 17:54
    
Expanding around the minimum plays fast and loose with gauge fixing. –  Peter Morgan Jul 4 '12 at 18:21
    
That's the whole reason for the theory. You give the gauge degrees of freedom of the vector field to the scalar, and you get to have massive vector bosons without killing renormalizability. –  Jerry Schirmer Jul 4 '12 at 18:43
    
Nonetheless, choosing a global value for $\phi_0$ fixes the $SU(2)$ gauge. As far as local degrees of freedom are concerned, sure, what you say. In the QFT context, I will admit to considerable uncertainty about how best to talk about this. I don't understand interacting QFT very well, and non-commutative gauge theories raise the bar to understanding considerably higher. –  Peter Morgan Jul 4 '12 at 18:55
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