Talk:Higgs mechanism/Archive 1

Archive 1

Archive 2

applied to photons

Why does the Higgs mechanism not give mass to photons? --63.26.56.28 17:43, 5 May 2006 (UTC)

Actually under certain circumstances it does. (At least an absolutely similar mechanism can explain the meißner-effect in superconductors.) But in the standard model (so not in a solid but in the vacuum) you can just measure masses for the weak gauge bosons. Perhaps, there is a good article in en:wikipedia on the electroweak theory. This ought to be the right place to turn to. :) -- 84.61.164.214 20:56, 30 June 2006 (UTC)

A simple question: what is the $v^{2}$ in this equation?

$V(x,y,z)=\left(|H(x,y,z)|^{2}-v^{2}\right)^{2}$

Mass generation

I think the sentence "It also gives mass to all the other elementary particles in the standard model" is a bit misleading. Someone could think that fermions get their mass from this mechanism exclusively. —Preceding unsigned comment added by 134.2.78.97 (talk) 11:51, 29 October 2007 (UTC) Saying that the Higgs particle or Higgs mechanism "gives mass" to other particles is completely incomprehensible to anyone except a theoretical physicist. Some type of explanation for students and professors in other fields would be greatly appreciated. --Bob Goldstein MIT Physics 1970. —Preceding unsigned comment added by Bob124c41 (talk • contribs) 02:52, 11 February 2008 (UTC)

According to what I have read, yes, all fermions derive their mass from the Higgs field. In other words all the particles in the Standard Model (theory) are massless. Measured values have to be plugged into the theory to make it work. This is according to the mathematical theories of the Standard Model. The only thing missing is the discovery of the Higgs - and hopefully that will be happening soon at CERN with the LHC, which comes back to life in September of 2009. You can read about the connection of mass to the Higgs field in several books that I know about. Below I have listed three books. The first one has an article about it here at wikipedia, just click on the title.

The God Particle: If the Universe Is the Answer, What Is the Question?

Author Leon M. Lederman, with Dick Teresi

Longing for the Harmonies, Themes & Variations from Modern Physics

by Frank & DEVINE, Betsy WILCZEK

The Particle Garden: Our Universe As Understood By Particle Physicists

by Gordon Kane Ti-30X (talk) 05:34, 18 May 2009 (UTC)

By this statement people mean that all the particles in the standard model are necessarily massless--- they cannot get a mass because their different helicity states do not have the same charges.

A massive particle with spin 1/2 can be thought of as two different helicity states (not counting antiparticles) which flip into each other constantly. But in the standard model, the different helicities of particles have different charges. One helicity of the electron is part of an SU(2) doublet with one U(1) charge and the other is a singlet with a different value of the U(1) charge. So you can't have a massive electron because if you flip helicity you violate charge conservation. For the same reason, you can't have massive quarks.

But the weak SU(2) and U(1) are Higgsed, so the electron helicity can flip by absorbing a particle from the condensate to make up the missing charge. This effect can be written down simply by just replacing the condensate with its classical value, and then the electron in this classical field is massive.

The neutrino cannot get a mass because it cannot flip helicity by absorbing one Higgs. In order for the neutrino to get a mass it must flip helicity, and turn into an anti-neutrino, and the easiest way for this to happen is for the neutrino to absorb two Higgs bosons at once. This effect is suppressed naturally by the improbability of two nearly simultaneous Higgs absorptions, but it can happen if there are intermediate states available for a neutrino + 1 higgs.

This is important for several reasons:

1. it tells you that all the particle states with definite helicity have to have different charges, so that they cannot have a mass if not for the Higgs. The idea is that if they could have a mass, you would expect it to be of order 1 unit, where the unit is the Planck mass which is enormous. We would never see particles that heavy.
2. it tells you that the mechanism of SU(2)/U(1) breaking also sets the scale of particle masses--- so that all particles (electrons/quarks) have the same mass up to factors of order unity.
3. it tells you that the neutrinos must stay massless, except for interactions which are not renormlizable. This puts severe constraints on extensions to the standard model.
4. it explains the CKM matrix.

This is the standard model. But it is only peripherally related to the Higgs mechanism itself, so it might not belong on this page.Likebox (talk) 00:50, 12 February 2008 (UTC)

Likebox, Not true - The Higgs boson and Higgs field are very much connected to the Standard Model. The Higgs is seen as another step toward completion of the Standard Model. It's kind like a missing piece of a picture puzzle. Ti-30X (talk) 05:42, 18 May 2009 (UTC)

"Higgs mechanism" has two meanings. The narrow meaning is the Higgs boson and the Higgs field in the standard model, the wider meaning is any charged scalar with a VEV, in any quantum field theory, related to nature or not. The wider meaning is more illuminating as an explanation, because the special case of the standard model is not particularly interesting, except for the reason that it is true of our world.Likebox (talk) 17:56, 18 May 2009 (UTC)

citations

where are they? 69.112.164.135 (talk) 18:46, 26 April 2008 (UTC)

eh? --Michael C. Price ^talk 09:42, 27 April 2008 (UTC)

History of Science

The way in which history is written is a strange one, especially when it comes to science... The "official" story that you can read everywhere (including here) is that the Higgs mechanism was (as the name suggests) 'originally proposed' by Peter Higgs. Slightly better-informed sources sometimes mention that the mechanism was "independently discovered by Brout and Englert as well as Hagen, Guralnik, and Kibble". In reality, and as Higgs himself has several times pointed out: "they were clearly ahead of me",see e.g. http://www.cpa.ed.ac.uk/bulletinarchive/1996-1997/11/news/26.html and http://physicsworld.com/cws/article/print/19750

The chronological truth is indeed that the so-called Higgs mechanism was 'originally proposed' by Peter Higgs , François Englert and Robert Brout , and by G. S. Guralnik, C. R. Hagen, and T. W. B. Kibble .

The above is the publishing order but not who was really first. Guralnik at Brown Univ. recalls his part in Higgs history here: http://chep.het.brown.edu/stlouis-v4.pdf. The GHK team had the results figured out by spring of 1963 and presented this prior to the other teams. They delayed their PRL article in hope to figure out another related calculation.—Preceding unsigned comment added by Moose-32 (talk • contribs) 15:56, 15 September 2008

I always wonder why the anglo-saxon scientific world is always so imperialistic, not to say revisionistic. Either it's because anglosaxons just can't remember any non-english sounding name, or they really want people to believe that every single thing in modern science has been invented either by an American or a British.

I just left a note on the talk page of Higgs on the German wikipedia bringing up this issue. Hopefully this can be straightened out fairly and civilly.-Richard L. Peterson130.86.14.12 (talk) 19:37, 17 March 2008 (UTC)Rich (talk) 19:38, 17 March 2008 (UTC)

It's perfect just the way it's written now in the main article: Brout and Englert were first, Higgs a couple of months later, Kibble & co a few months later again, so they should indeed be stated in this order. The mechanism is usually referred to as the Higgs mechanism (probably mainly because Higgs was the one who made most publicity for it and maybe also because his derivation was easier to understand than Brout and Englert's) - so the main title shouldn't be changed, although it might be nice to have separate entries created for Brout-Englert-Higgs mechanism etc and link these to this entry. (Imtg5102 (talk) 15:11, 18 April 2008 (UTC))

I think the reason for the name is that 'tHooft read Higgs, and 'tHooft is Dutch so it's not an anglo-saxon conspiracy. I don't think that 'tHooft and Veltman really cared very much about proper attribution--- they called some of their original relations "Bell Trieman identities" as a joke, to see if Bell and Trieman would get credit for them. I think Slavnov and Taylor got the credit in the end.Likebox (talk) 20:45, 14 May 2008 (UTC)

Ironically, 't Hooft is now probably one of the people outside of Belgium (though IMHO correctly) calling it the "Brout-Englert-Higgs mechanism", see http://arxiv.org/abs/0708.3184. In any case, interesting anecdote, though I don't know if it's correct, Higgs seems to give another story for why only his name got attached to it, see his article "prehistory of the higgs boson" (end of par.3). Maybe we should delete the reference to 't Hooft coining the name, ar on the contrary add a reference? (I guess we could also just stop discussing this, wait if anything shows up at the LHC, see who gets the Nobel prize and then start arguing again...) Imtg5102 (talk) 15:56, 15 May 2008 (UTC)

'tHooft and Veltman care very much about attribution - and retribution. In the particle physics world, it is thought by the older theorists that t'Hooft and Veltman started referencing this mechanism or boson without the GHK team. Additionally, t'Hooft feels very strongly that many Europeans have been cut out of their share of Nobel prizes and he is trying to help correct this. The Wolf Prize is one of the results of this effort.

I think that is right – everyone will just have to wait. All six will not get the Nobel Prize as only three can win it. With six people (plus the experimentalists) claiming (and deserving) credit, the Nobel Prize will have to wait until there are only three alive or the academy can expect quite a controversy on its hands.

Ginzburg and Landau (with their "superconductivity theory" of 1950) were also clearly ahead, but what about Stueckelberg (1948), who was directly related to the 'particle community', whereas Ginzburg and Landau were not! (So sometimes it is useful to look beyond the border of one's scientific "carpet", although usually this does not help, except "afterwards".) On the other hand, the London brothers (1935!) were clearly ahead of Ginzburg and Landau, not to forget the experimentalists themselves (Meissner and Ochsenfeld, 1933). Of course, they were completely off from any particle physics in the conventional sense. Consequence? Such is life, especially history! Also in mathematics the famous "theorems" and "lemmas", and whatsoever, are typically not named after those people who first invented them. Who cares?

Thus, although it seems always worthwile to look at different fields: who knew of the superconductivity analogon, when the "Higgs" theory was derived by Brout and Englert, Higgs, and Guralnik, Hagen, Kibble? One should remember that superconductivity remained a riddle almost half a century. With the Higgs mechanism it does not seem to be different, especially after the recent incidents at the LHC. - With regards, user 87.160.124.101 (talk) 17:36, 28 September 2008 (UTC)

Except in this case, I got a chance to ask Brout, and he gave great credit to Nambu. Nambu studied superconductivity, and was directly inspired to postulate a quark condensate. Brout was inspired to study a charged condensate by this, and Anderson was arguing for a charge condensate indepently. So in this case, history is that the Higgs mechanism came directly from superconductivity. Since this is also the clearest way to explain the effect, I think it should be worded as it was before.Likebox (talk) 20:28, 4 November 2008 (UTC)

Guralnik, Hagen Kibble (GHK) team was really the first to the party and made the largest contribution to the mass boson work. In fact it was solved in 1963 by them and starting giving talks on this prior to the others. Too bad (from a Nobel Prize standpoint) GHK has the worst publicists of the three teams - in that area GHK team clearly behind Higgs and Brout / Englert.

Here is a recent paper by Guralnik that highlights their work and the history of the mass boson. Interesting from a history standpoint as well as the evaluation of the three papers by GHK, BE, and Higgs. The History of the Guralnik, Hagen and Kibble development of the Theory of Spontaneous Symmetry Breaking and Gauge Particles —Preceding unsigned comment added by 65.42.208.133 (talk) 16:37, 24 July 2009 (UTC)

Your comments are appreciated, but I personally would prefer that you either sign your wikipedia username or your real life name to the comments. Regards, Rich (talk) 04:40, 1 August 2009 (UTC)

"A form of superconductivity" vs "spontaneous symmetry breaking"

In my opinion the phrase

"a form of superconductivity in the vacuum"

is vague, uninformative, and not suitable as the first sentence of an article in Wikipedia. Indeed many different things can be "a form of superconductivity", from nuclei to solid states. On the other hand

"spontaneous symmetry breaking in a gauge theory"

is a concise and precise definition of what the Higgs mechanism is. The first sentence is all about concise and precise definition, and not at all about historical development.

Look at other places in the internet -- Britannica, PhysicsWorld -- everywhere the Higgs mechanism is the symmetry breaking, while in Wikipedia it is "a form of superconductivity". I find it ridiculous.

Take an arbitrary physicist and ask them,

What is "a form of superconductivity in the vacuum"?

and they would not be able to answer anything particular. Then ask them

What is "a spontaneous symmetry breaking in a gauge theory"?

and they will immediately answer -- well, that's the Higgs mechanism.

The superconductivity may have its place in the history section, but I strongly believe it does not belong to introduction. Bakken (talk) 23:15, 4 November 2008 (UTC)

Except that in this case the phrase "spontaneous symmetry breaking" is not 100% accurate. The "symmetry" is gauged, and is exact even in the Higgs phase. The difference is that the gauge-transformed states have a different phase for the condensate. So some people quibble that it isn't SSB. I don't mind the phrase spontaneous breaking, personally, because the conserved charge is no longer conserved just like a normal SSB. But I think it is more accurate to say "Superconductivity in the vacuum state", because that captures the essential point that the vacuum is filled with stuff, that the stuff is charged, and that it can flow from place to place without resistance to neutralize any excess charge.Likebox (talk) 23:23, 4 November 2008 (UTC)

by the way, in German, French, and Spanish Wikipedia articles the Higgs mechanism is "spontaneous symmetry breaking". You must be joking, folks, -- "a form of superconductivity"... :)) Bakken (talk) 23:30, 4 November 2008 (UTC)

"Superconductivity in the vacuum" is also more historically accurate. Nambu was directly inspired by superconductivity for the chiral quark condensate, and what Brout, Higgs, etc, did is to realize that the relativistic generalization of the Meissner effect is massive gauge bosons. I don't think that what other encyclopedias say on stuff like this is important, because non-internet media has always been obtuse about physics. I think we should focus only on the professional literature, with special focus on the original papers.Likebox (talk) 23:36, 4 November 2008 (UTC)

the Russian Wikipedia also says that Higgs mechanism is "a spontaneous symmetry breaking". Could you please specify who are those people who "quibble that Higgs mechanism is not SSB"? Then how about such an introductory sentence: The Higgs mechanism is a spontaneous symmetry breaking [references to all possible places], however some people believe that it is not symmetry breaking [references to "some people"] but rather a form of superconductivity. I for one would accept it. Bakken (talk) 23:51, 4 November 2008 (UTC)

And I disagree -- Wikipedia is not a cult club of initiated members, it is for everybody -- I believe it does matter how other encyclopaedias define the term because that reflects the language people speak and understand.

In this case, you are talking about language that many people speak and pretend to understand. The people that really understand it think of it as superconductivity. To answer your question about who objects to calling it SSB (because I personally don't, I just think superconductivity is more descriptive and accurate)--- Sidney Coleman used to object to calling it this, because the generators of the symmetry don't push you into different states, but into different descriptions of the same state.Likebox (talk) 23:59, 4 November 2008 (UTC)

Also Gerard 't Hooft objects to this language (I think--- from memory) and a bunch of other people. As I said, I don't object, because the global symmetry associated with a gauge symmetry can still be thought of as a global symmetry, sort of. If you have two Higgs fields for one Gauge symmetry, and both have a VEV, there will be Goldstone bosons, and I wouldn't be able to say which field is Higgsing and which field is doing SSB, so the two concepts meld together to a large extent. But it's a little tough to see what "spontaneous breaking of a gauge symmetry" means as a physics statement because a gauge transformation does not make a physically distinct state. So it's a judgement call whether you want to use SSB language.

I remember that Coleman used to illustrate this point as follows: he would take a piece of chalk, put it on his desk, and say "I will translate this chalk", and then he would move it. Then he would say "Now, I will rotate this chalk" and he would rotate the chalk. Then he would say "Finally, I will gauge transform this chalk", and then he would wiggle his fingers over the chalk (and people would laugh). The gauge transformation is not a physical transformation, it's just a redundancy.Likebox (talk) 00:21, 5 November 2008 (UTC)

Just one person in private communication!? How about these texts:

Michael Edward Peskin, Daniel V. Schroeder, An Introduction to Quantum Field Theory, page 690: "The Higgs Mechanism. In this section we analyze some simple examples of gauge theories with spontaneous symmetry breaking..."

Jagdish Mehra, Helmut Rechenberg, The Historical Development of Quantum Theory, page 1127: "Spontaneous Symmetry-Breaking and the Higgs Mechanism: The idea of spontaneous symmetry-breaking went back to the 19th-century..."

K. Zuber, Neutrino Physics, page 44: "The concept of spontaneous symmetry breaking is then used for particles to receive mass through the so-called Higgs mechanism..."

F. Strocchi, Symmetry breaking, page 193: "...the so called Higgs mechanism, by which the breaking of a gauge symmetry is not accompanied by Goldstone bosons ..."

:) Are you going to tell me that all these people are stupid and have no clue in Higgs mechanism? Folk, this "form of a superconductivity" is a complete joke... What was your textbook in quantum field theory? —Preceding unsigned comment added by Bakken (talk • contribs) 00:35, 5 November 2008 (UTC)

Look, it's a judgement call. It's not clearly wrong to say SSB, and as I said, I don't personally object to calling it SSB, even thoug it isn't 100% accurate. SSB in a gauge theory just means Higgsing to most people. I don't like it because it isn't as physical a way to describe the phenomenon as "superconductivity", but I don't want to argue about terminology.Likebox (talk) 00:49, 5 November 2008 (UTC)

Isn't it a very clear call? There are plenty of references to refereed books where the Higgs mechanism is firmly associated with spontaneous symmetry breaking. You, so far, only referred to some private communications "from memory". Is there a single textbook in quantum field theory where the Higgs mechanism is not associated with spontaneous symmetry breaking? I understand now why some people are so wary of wikipedia... "a form of superconductivity" -- this phrase has very little information in it. First, there is no fundamental theory of superconductivity, only phenomenology, so superconductivity is itself not well defined. Then "a form of" means not precisely, but somewhat similar. So "a form of superconductivity" means "not completely dissimilar to something we don't really know what..." :(

Good night, sleep well, and have nice dreams. Bakken (talk) 01:10, 5 November 2008 (UTC)

(deindent) Of course I don't think it's a clear call, or I wouldn't waste time on it. You shouldn't put so much stock in textbooks. Textbooks are written and selected by a political process which has little to do with whether their contents are technically correct. The criteria for textbooks is how quickly they generate the illusion of understanding in the majority of students, measured by test performance and student evaluations. This criterion is incapable of selecting textbooks that actually communicate knowledge correctly and clearly. In order to write an encyclopedia, textbooks need to be ignored, and you need to pay attention only to the professional literature, especially the original papers.

You say you think people should be wary of Wikipedia. Actually, what you are seeing is that you need to be wary of textbooks.

A form of superconductivity is saying that the Higgs mechanism is the obvious relativistic version of the Landau theory of superconductivity. That contains the complete description of the phenomenon, and a competent physicist can reconstruct the whole thing from this description. This is also true of the description "The Higgs mechanism is where you start with an model with a spontaneously broken global symmetry and then gauge the symmetry that is broken in the model". What is incorrect is to say "The Higgs mechanism is where you start with a gauge theory, and then you break the Gauge symmetry", that's not right because Gauge symmetry can't be broken.Likebox (talk) 02:08, 6 November 2008 (UTC)

"Superconductivity" does not communicate the core idea of the Higgs mechanism and so should not appear in the lead. That it gives masses to particles is much more relevant and should be what we lead off with. --Michael C. Price ^talk 16:29, 8 November 2008 (UTC)

Looking over this thread I see that plenty of references have been supplied for the description of the Higgs mechanism as a form of SSB and none for describing it as a form of superconductivity. That should really conclude the matter. Superconductivity is not appropriate for the lead of the article.--Michael C. Price ^talk 06:38, 27 April 2009 (UTC)

Standard Model vs. General Mechanism

There is the Higgs mechanism in the standard model, which is a special case of the general Higgs mechanism. It has the virtue of existing in our universe, and that makes it special, but otherwise it just a not particularly interesting example. I believe that the article should stick to describing the general Higgs mechanism. Maybe this should fork an article "Higgs mechanism in the standard model", with a list of the different Higgs models out there.Likebox (talk) 02:25, 6 November 2008 (UTC)

The definition of the Higgs mechanism (as EWSB) can be found in a number of books on quantum field theory (citations from which I have showed above) as well as in the review of the Particle Data Group which I refer to (PDG is an accepted authority in this field of physics). It seems logical to me that the definition in Wikipedia reflects the definition from the modern textbooks and reviews. Bakken (talk) 09:38, 6 November 2008 (UTC)

However, some textbooks discuss the general mechanism in first, before moving onto specifics. --Michael C. Price ^talk 09:43, 6 November 2008 (UTC)

I've added that it also explains the masses of the leptons (in E-W). And quarks. --Michael C. Price ^talk 09:56, 6 November 2008 (UTC)

Also--- the edits here are making the intro less clear, and more jargonny. That's not reasonable, since all the additional information is provided much more explicitly later in this article and in the standard model article.Likebox (talk) 20:51, 7 November 2008 (UTC)

Clarity is obviously a subjective thing, because I don't think explaining the concept, in the lead, by analogy with superconductivity is helpful. As for jargon, what's a "quantum fluid"? --Michael C. Price ^talk 22:02, 7 November 2008 (UTC)

Upon reflection I think the point about duplication is correct. We should shorten the over-long lead and leave the explanation of the details to the body of the article. I suggest removal of paragraphs 3, 4 & 6 -- as Ron says, all this is covered later in the article.

Also the 2nd paragraph implies (to the casual reader) that all the bosons acquire a mass; in fact some of the bosons remain massless (e.g. photon). We should be explicit about this (or remove this paragraph as as well). --Michael C. Price ^talk 09:59, 8 November 2008 (UTC)

Please No Plagiarizing

Don't copy the standard model specific, jargonny description of the Higgs mechanism in the particle data book. Especially not when you are replacing a thought out, accurate, clear description.Likebox (talk) 04:02, 8 November 2008 (UTC)

"thought out, accurate and clear" -- that's your POV. In my POV it's clumsy, unspecified and misleading. What is "superconductivity"? What is "quantum liquid"? These are no fundamental theories for these things. They are just phenomenological words. The accepted definition from Particle Data Group -- that is a thought out, accurate and clear. I did not plagiarized it (have you read the review?). I rephrased it. And then Michael C. Price rephrased it even further.

It is *the definition* of the Higgs mechanism. It is what most of physicists define as Higgs mechanism. If you believe it's broader than that, make a chapter in the article and write it down there. But the first paragraph must give the most accepted definition used by most of the textbooks.

And you can't even give a reference. Bakken (talk) 08:35, 8 November 2008 (UTC)

This is not a dispute on accuracy or content-- it is a dispute on how to communicate the idea best. That's a somewhat subjective call, and references are not going to help. We should just discuss it until we all agree.

The reason that I don't like the particle data stuff is because what most physicists call the "Higgs mechanism" is not that. When physicists say "we higgs SU(5) to the standard model", they don't mean the specific higgs in the standard model. They just mean what it says--- that there is a charged field with a vacuum expectation value which breaks the symmetry to a subgroup.Likebox (talk) 16:04, 8 November 2008 (UTC)

I agree this entirely a communication issue, not a factual dispute. And I agree that Higgs is broader than its application to the standard model. For instance we believe the Higgs will give mass to the neutrinos, yet this is well outside the standard model. Some of my changes have tried to make this point already, but no doubt they could be further improved. --Michael C. Price ^talk 16:10, 8 November 2008 (UTC)

PS Only one of Higgs components is charged, the other is neutral, so perhaps we shouldn't describe it as a "charged field". --Michael C. Price ^talk 16:13, 8 November 2008 (UTC)

when he says "charged" he probably does not mean "electric charge" but rather a generator of the gauge symmetry.

and he again refuses to give a reference to a textbook in quantum field theory where the Higgs mechanism is *not* a spontaneous symmetry breaking. I can agree on not stressing "electroweak symmetry breaking" -- that is indeed standard model. But a symmetry is always broken in the Higgs mechanism, this is how the very Higgs condensate appears. Without broken symmetry the condensate would be zero. Bakken (talk) 16:47, 8 November 2008 (UTC)

Yeah--- I was using the word "charged" loosely to mean gauge charged, or under a nontrivial representation of the gauge group, but that usage is common. People often say "SU(2) neutral" or "SU(2) charged" to mean a trivial or nontrivial representation.

It's not that I refuse to give a reference, I can dig up some references, but they're going to be few in number, because the issue is not that important. The point is that Higgs and SSB are closely related, but they are treated differently in a path integral. In an SSB model you integrate over different configurations that are linked by an infinitesimal local symmetry rotation, and the integral over the configurations which are linked by infinitesimal rotation gives the Goldstone boson contribution to the partition function. In a Higgs mechanism, you don't integrate over these different configurations, because they are gauge equivalent. So you don't get Goldstone bosons, and you get massive vector mesons instead. Its a point thats belabored in the body of the article, and I don't know if it should be mentioned so prominently at the beginning, because it doesn't make any difference in writing down the Lagrangian. It only makes a difference when you go about a path integral. So if the wording in the beginning is a little different, I'll live.Likebox (talk) 20:28, 8 November 2008 (UTC)

Fermions Masses and Higgs Mechanism

The Fermions in the standard model get masses from the interaction with the Higgs field. That is a different effect from the superconductivity that gives masses to the gauge bosons, its a Yukawa interaction, and it would be there in an SSB model too.Likebox (talk) 18:47, 9 November 2008 (UTC)

I showed you references to verifiable sources (e.g. PDG review) where Higgs mechanism is firmly connected to spontaneous symmetry breaking. These sources also attribute masses of fermions in the Standard Model to the Higgs mechanism. No texbook in quantum field theory defines the Higgs mechanism as superconductivity. There is no fundamental theory of superconductivity. Superconductivity is a very complicated many-body phenomenon, not fully understood. It does not help to define an easy, clear and obvious Higgs mechanism through a lot more complicated and involved superconductivity. Bakken (talk) 19:09, 9 November 2008 (UTC)

I agree that it is firmly connected to SSB, both are caused by stuff in the vacuum. The Fermion masses are also due to Higgs interactions, but that's not the mechanism itself. Please stop relying on textbooks, they are useless. Understand the effect. Then you won't need a textbook.Likebox (talk) 19:14, 9 November 2008 (UTC)

Well, in my opinion wikipedia has to reflect the generally accepted point of view on the subject reflected in the modern textbooks and review articles. If you insist, you can add a section at the end of the article with your interpretation of the concept, provided you have reputable and verifiable sources to refer to. —Preceding unsigned comment added by Bakken (talk • contribs) 19:22, 9 November 2008 (UTC)

What I'm telling you is that the generally accepted viewpoint is what I wrote. It's not incompatible with the textbooks, just pictorial. You won't ever get me to cite a source. It just won't happen. Not for stuff this elementary.Likebox (talk) 19:26, 9 November 2008 (UTC)

This is degenerating into an edit war. I don't understand why. Please act in good faith. The language you are inserting is inferior.Likebox (talk) 19:28, 9 November 2008 (UTC)

Please, can't you see the lack of objectivity in that statement? It's an edit war because you refuse to see both sides of the argument. --Michael C. Price ^talk 19:45, 9 November 2008 (UTC)

Nothing is fully objective. I give up for now. Hopefully you folks will change your mind in a year or two.Likebox (talk) 19:56, 9 November 2008 (UTC)

What I mean is that there is no point in telling someone that their language is "inferior" without explaining why it is inferior. --Michael C. Price ^talk 20:22, 9 November 2008 (UTC)

(deindent) Sorry--- I just meant the following. There is the high-energy physics notion of SSB and Higgs, and there is the condensed matter version, of SSB and superconductivity. Historically, the idea began in the condensed matter community, and migrated over to high energy physics with the work of Nambu.

The idea is that there is a field in the vacuum, and it is charged. This means that the spin 1 bosons become short-ranged, which is the Meissner effect in condensed matter. This was the historical inspiration for the Higgs mechanism, and it is not possible to understand it fully without seeing it in this context.

Unfortunately, most textbooks and treatments outside of the original literature leave out the condensed matter part, making it difficult to understand the Higgs mechanism. I tried to fix it here, by writing a clear article which is based on history and which emphasizes the historical point of view. But now it is getting undone--- and the article is turning into the same a-historical, overly formal treatment which you can find everywhere else. Of course, this is just an opinion.Likebox (talk) 20:45, 9 November 2008 (UTC)

The condensed matter approach is a point that can be explained in the body of the article, but is inappropriate for the lead, which is intended just to introduce the subject, not explain it. In fact the lead is way too long. The manual of style says that the maximum number of paragraphs should be between 2-4 (depending the total size of the article -- not sure where we are on that score), yet we have 7, so we are clearly over the limit no matter how we judge it. No wonder we can't agree about the lead's contents.

Personally, I'm happy with the Feynman pictorial notion of bosons and fermions acquiring mass by scattering off a background VEV field, but at the same time I recognise that there are many equally valid and different ways of understanding physical processes; superconductivity is an interesting and insightful analogy and the article covers the subject. But, to repeat, the purpose of the lead is not to explain the mechanism, just to introduce it, so we shouldn't really be getting our knickers in a twist over the subject.

One subject which has already been raised and which is legitimate to scope in the lead is the scope of the mechanism itself. Do we include fermion masses? Originally the Higgs mechanism just described the acquisition of mass by the gauge bosons, but nowadays the term is some times used to describe the masses of the fermions as well. Why don't we just say that?--Michael C. Price ^talk 06:54, 10 November 2008 (UTC)

Abelian Higgs and superconductivity are two words for the same thing. The reason people keep objecting, I think, is because they don't like the concept that it was the condensed matter people who discovered something so fundamental. Sorry, that's the history of the thing.

What people used to call high energy physics (before string theory) is just as phenomenological as studying a carbon crystal. It's the study of the low energy effective theory of the background we live in. It's full of fields and condensates, and it's very interesting, but it is no different than studying any other collection of matter of medium complexity.

The fermion masses are not part of the "higgs mechanism" itself. They are a side-effect, the same as any other SSB. While the "Feynman picturial notion" is sufficient to understand the Fermion masses, it is not sufficient for understanding the Higgs mechanism proper, because massless gauge bosons scattering off massless goldstone bosons don't make massive vector mesons. To understand how they mix together, you need to understand gauge invariance and the condensate, which is harder to see in a Feynman perspective.Likebox (talk) 15:06, 16 November 2008 (UTC)

I don't agree; once you understand that a massive particle propagates as a massless particle scattering off a fixed value background (= -i x mass^2 for scalar bosons; = -i x mass for fermions) then whole Higgs thing is completely clear, since $i/(p-m)=\Sigma (i/p)^{n}(-im)^{n-1}\,$. --Michael C. Price ^talk 19:49, 16 November 2008 (UTC)

Yes--- that's exactly the picture for the lepton masses (I mean, you need a condensate to get the perturbative scattering, but you have that). But how is it supposed to work for the gauge bosons? The scattering of zero mass gauge bosons can't produce a mass, because you need a longitudinal component. In the Higgs model, that's provided by the goldstone modes that aren't there. I can't quite figure out how you find this picture by summing a series. But that might be a failure of the imagination on my part.Likebox (talk) 20:46, 16 November 2008 (UTC)

With the gauge fields, isn't it exactly the same picture as with the fermions, but after you've made a linear transformation of the Higgs field to the VEV minimum?--Michael C. Price ^talk 09:41, 17 November 2008 (UTC)

In the Lagrangian, yes, but in terms of resummation, I'm not sure. There's an additional thing for the gauge fields--- when you have a charged condensate, you have to deal with Gauss's law, the condensate produces a long range field. This can be thought of as emitting zero-mass bosons, but depending on the gauge it can also be an additional thing, the Coulomb law can be separate from the photons. I got confused on this. You might be right that the diagrams add up to a gauge boson mass, but I didn't see exactly how it works.Likebox (talk) 21:21, 17 November 2008 (UTC)

The Higgs mechanism consists of:

• transforming to centralise the broken vacuum, so that the old Higgs field is now parameterised by the longitudinal massless Nambu-Goldstone bosons (along the bottom of the Mexican hat potential), and parameterised by massive boson fields in the tranverse direction (i.e. up the hill). You ignore the Goldstone bosons since they can be gauged away. But there are still a lot of extra Lagrangian terms to consider.
• application of $i/(p-m)=\Sigma (i/p)^{n}(-im)^{n-1}\,$ for the remaining fermion terms which couple quadratically with the VEV.
• application of $i/(p^{2}-m^{2})=\Sigma (i/p^{2})^{n}(-im^{2})^{n-1}\,$ for the remaining boson terms which couple quadratically with the VEV. This includes the original gauge bosons alongside the transformed Higgs field itself.

Viewed this way the acquisition of masses by the gauge bosons is by the same mechanism as the acquisition of masses by everything else. I think this is the point that Bakken was trying to make.--Michael C. Price ^talk 09:56, 19 November 2008 (UTC)

Except that this is not quite right for the gauge bosons. The massive spin 1 propagator is not g_mu nu /k^2-m^2, like the Feynman gauge zero mass propagator with a shifted pole, it's got an extra term from longitudinal projection: (g_munu - k_mu k_nu/M^2) /(k^2 - m^2). It doesn't have an expansion the way you want, at least not naively, and it represents three propagating degrees of freedom, not two.

The reason that you get a difference can be understood in "Dirac gauge", that's where you just have photons and an instantaneous coulomb interaction. The (analog of) plasma waves in the condensate are the massive photon, and you would think that plasma oscillations would have an have zero frequency at infinite wavelength, because the plasma is translationally invariant. That conclusion is wrong, plasma oscillations have a nonzero frequency at long wavelength. The reason is that the coulomb law is long ranged.

So when you are sucking up the goldstone modes into the photon to make a massive vector meson, you can either do it in the lagrangian formulation, where it is clear that you are doing it right, or you can do it by summing a series, which is like separating out the photon into a longitudinal part which is the Higgs goldstone oscillation together with the transverse part which is the original photon. You are probably right--- it probably can be done---but it's more complicated and I don't see how to do it exactly.Likebox (talk) 19:19, 19 November 2008 (UTC)

Is that the Landau gauge you're using? Surely the propagator has the form $g_{\mu \nu }/(k^{2}-m^{2})$ in the Feynman gauge? --Michael C. Price ^talk 20:20, 19 November 2008 (UTC)

There is no gauge fixing for massive gauge bosons--- the propagator is always (g_mu nu - k_mu k_nu/m^2)/(k^2-m^2). The propagator you give is no good--- it has a wrong-sign in the 0-0 component. The proper massive propagator projects out that part of the metric tensor, and has only positive sign residues for the poles corresponding to the propagating fields. The Feynman propagator only works for massless gauge fields, and then only because of magic (discovered by Feynman).

The question here is how do you get this massive spin-1 propagator by resumming diagrams for massless gauge boson diagrams and a scalar Higgs in a way that is independent of the gauge choice for the original massless field. If you want to think of this as the massless gauge field bouncing off the condensate ( that's the picture for the fermions), you have to add up diagrams in whatever gauge you choose for the original massless propagator and recover the massive propagator. I couldn't see how to do that at first.

The textbook way to get the massive propagator from the Higgs model in the Lagrangian formalism is by using the condensate gauge, where you fix the gauge by making the phase of the condensate constant everywhere. Then the propagator comes out like that automatically. But you can't use that gauge if you want to think of the condensate perturbatively as particles, because you should be starting with a gauge choice which depends on the condensate being there. The point is that you should get the same massive propagator answer in Feynman gauge, or Landau gauge, or any other gauge by resumming diagrams, and if you do that, this would give a detailed particle picture for how the mass for the gauge field comes about.

I sorted it out this afternoon, but I didn't internalize it. You do get the massive propagator by resumming, but it takes a little effort because the interaction is derivative. The quadratic interaction between the massless gauge boson and the condensed field phi is from the terms A_mu D_mu phi, where D_mu is the gradient. Writing phi as the condensate times a phase factor theta(x) gives the equation for the phase factor, but you have the phase as an independent variable now, because you didn't set it to zero by your gauge choice--- you're in Feynman gauge. The equation for the phase makes the phase a function of A and the gauge fixing function F, which is the longitudinal part of A, which is not surprising--- the phase is mixed up with the A by gauge invariance. Then if you sum over all photon-phi mixed diagrams, you recover the massive propagator, like you should, because that's what you get in the textbook theta(x)=0 gauge. The answer has to be the same by gauge invariance, and so it had to work out, in hindsight. I did it out roughly, convinced myself it does work out, but I wasn't very careful with signs and factors. The diagrams which mix up the A and theta to make the massive propagator have more powers of k than usual in the numerator, because of the derivative coupling--- it's not theta times A, it's grad theta that multiplies A.

Compare this type of interaction with the Higgs induced fermion mass terms, which are just yukawa-term: Higgs times psi-bar psi. This term just describes scattering of fermions off the condensate in a simple-minded way--- there's no weird mixing of components or swallowing of components. Even if you had a scalar field A and you gave it mass using a Higgs mechanism, that would come from a scalar-scalar term Higgs^2 times A^2. Again, this term can be interpreted as simple scattering off the condensate, and the adjusted mass of the scalar comes out from summing diagrams in a very straightforward way. The same mechanism would work in a non-higgs SSB model, where you turn off the gauge field, and it would work in the exact same way (except now there would be some extra Goldstone bosons).

But for the gauge field, everything is more complicated in terms of diagrams because the interaction is gradient type. The reason is because the condensate is charged and shaking, and shaking the condensate is what gives longitudinal component of A. This picture is hard to see by summing diagrams. This is the essential magic of the Higgs mechanism--- the only reason I can see that it would be obvious physically that the A gets massive is because you know that this is what is happening in a superconductor.23:58, 19 November 2008 (UTC)

Just to respond to your opening point; the wrong 00 sign business doesn't bother me -- if it drops out of the maths then we have to live with it. I agree it is precisely because the gauge bosons are initially massless that they are gauge invariant and therefore have to be fixed with Feynman's magical gauge breaking term to get sensible propagators (i.e.just $ig_{\mu \nu }/k^{2}$) . Once we apply the Higgs translation to centralise the VEV then we hide the gauge invariance and gain masses upon resummation. The choice of the Feynman gauge breaker is justified physically (in my view) by the neat cancellation it gives between all the ghost loops and some of the gauge loops; the cancellation of the Feynman gauge breaker with the Faddeev-Popov determinant is one of the most amazing things in physics; it saves the particle viewpoint. --Michael C. Price ^talk 10:39, 20 November 2008 (UTC)

All I was trying to say in the long-winded passage above is that the summation of diagrams is much less trivial for the spin-1 particles than for the spin-1/2 or spin-0 particles interacting with the Higgs. The "Higgs translation that centralizes the VEV" is a nontrivial operation in terms of diagrams for spin-1 fields, while for spin-1/2 and spin-0 it is pretty straightforward. The reason is that the translation involves sucking up part of the Higgs into the spin-1 field to make a massive spin-1 field. The physics of the resummation is also different.

Tacking another way on the same issue--- why is glass transparent? Why doesn't the photon acquire a mass by bouncing around inside a dielectric medium? If you take the point of view that the mass of the photon is just the photon bouncing off a background, then a photon bouncing around in glass would be massive--- that would make glass be as opaque and shiny as a superconductor. The reason that glass is transparent is that it is not enough to have scattering to give the photon a mass--- you need a charged condensate. If you just have a dipole density then you get a dielectric correction which attenuates photon modes, and changes the speed of the waves, but the interaction is still long-ranged. In order to make a short ranged photon, you have to do something nontrivial--- you need to have modes of oscillation of a condensate that can mix up with the modes of oscillation of the photon to make a massive spin-1 particle.Likebox (talk) 23:23, 20 November 2008 (UTC)

I don't see why the diagram summation is significantly more difficult for spin 1 particles over spin 0 or 1/2. In the Feynman-t'Hooft gauge the only difference is Lorentz metric in each term - but they all collapse down anyway. It certainly is a tedious operation collecting together all the terms that come flying out of the transformed Lagrangian, but conceptually it is the same process; I don't see why you say it is different.

I don't get the "charged condensate" stuff, but that is due to my lack of insight into superconductivity. Something I must look into, but I still think is a distraction in the article's lead.--Michael C. Price ^talk 07:53, 21 November 2008 (UTC)

The reason it is much more difficult is because the propagator for the massive vector meson is a combination of the propagator for the massless vector meson and the propagator for a scalar field. For spin-1/2 or spin-0 particles, bouncing off the Higgs just moves the mass pole around. But that's not true for vectors. The massive vector diagrams are not the diagrams for the massless vector with the mass changed from 0 and keeping g_mu_nu on top.Likebox (talk) 22:52, 21 November 2008 (UTC)

This makes no sense at all, since the transformations are defined in such a way as make the combinations all sort themselves out. (The transformations are a combination of rotations of the vector gauge fields (to prevent mixing of the kinetic terms) and gauge transformations of everything which stops the massless Goldstone bosons mixing with the massive vector particles. Plus a trivial shift of origin.) When the dust settles we have the usual kinetic free terms and (to lowest order) self interaction (=mass) terms such as $-m^{2}Z_{\mu }Z^{\mu }=-m^{2}\eta ^{\mu \nu }Z_{\mu }Z_{\nu }$, which lead to straight forward pole shifts

$i\eta _{\mu \nu }/(k^{2}-m^{2})=i\eta _{\mu \nu }/k^{2}+(i\eta _{\mu \alpha }/k^{2})(-im^{2}\eta ^{\alpha \beta })(i\eta _{\beta \nu }/k^{2})+...\,$.

I.e. the metric assigned to the vertices cancels the additional propagator metrics, so the vector summation is the same as the scalar summation. --Michael C. Price ^talk 08:20, 22 November 2008 (UTC)

I see why there is confusion--- there is mixing of the kinetic terms. It comes from the term multiplying the derivative of the Higgs and the vector potential, and that comes from the convariant derivative of the scalar field.Likebox (talk) 17:26, 22 November 2008 (UTC)

But in terms of the new fields, there is no mixing. The forms of the vector kinetic terms (by which I mean just $-tr(F^{2})/4$) remain unchanged (they're gauge invariant and the rotation between the vector fields to isolate the massless U(1) boson leaves their form unchanged as well). The terms you're referring to sound like interaction terms that don't affect the expression of mass (which is purely a quadratic self interaction; e.g. $-m^{2}Z_{\mu }Z^{\mu }$). --Michael C. Price ^talk 17:47, 22 November 2008 (UTC)

Cheng and Li (Gauge theory and elementary particle physics, ISBN 0198519613) appendix B pg 307 are explicit about the gauge transformation canceling the mixing terms that arise exactly as you describe. --Michael C. Price ^talk 01:15, 23 November 2008 (UTC)

Look, this is the whole point--- you wanted to do the change of variables by summing diagrams. I already know how to do it by changing gauge, that's the standard way. You made a new additional claim that you can do it by summing diagrams. That's not completely trivial because if you don't use the standard gauge and use something like Feynman gauge, the summation is completely not obvious.

The term you are stuffing in by hand "m^2 Z_mu Z_mu" comes from the Higgs interaction in the case that you choose the gauge where the phase of the condensate is fixed. That's a different gauge than Feynman gauge. It doesn't work for the case where there is no condensate. If you are following a textbook, you are not summing diagrams, you are doing Lagrangian manipulations. Since all the terms are quadratic, it should be equivalent to summing diagrams. I'm repeating myself here.

But no matter--- I hope you have come to agree that the mechanism for the vector mass is completely different than the mechanism for the fermion masses.Likebox (talk) 05:49, 23 November 2008 (UTC)

No, I absolutely do not agree; the mechanism is the same. Somewhere we are getting our wires seriously crossed, because the Feynman gauge is what makes the summation so easy. Yes, the Feynman gauge is different from the group gauge. So what? They are applied separately. I don't know why you think I'm stuffing the Z^2 term in "by hand" -- of course it comes from the Higgs D^2 term, where else could it come from? I'm not proposing anything new here, just pointing out that there are some equivalent approaches; namely that the mechanism of mass acquisition by the Higgs boson, fermions and vector gauge bosons is the same. This is why everyone (except you, it seems) accepts the term "Higgs mechanism" to cover all these cases. You actually say Since all the terms are quadratic, it should be equivalent to summing diagrams. Precisely so! -Michael C. Price ^talk 10:43, 23 November 2008 (UTC)

(deindent) Ok. We do have a minor disagreement here. I am trying to convince you that the mechanism of mass acquisition by the vector mesons is physically different from the mechanism of mass acquisition for the fermions, and you disagree.

spin-1/2: no mixing of kinetic terms: the Higgs degrees of freedom stay completely separate from the fermions, and everything is the same as in ordinary SSB. To convince yourself of this, imagine taking the U(1) and SU(2) gauge fields out of the standard model so that the U(1) and SU(2) become global symmetries. Keep what used to be the Higgs field with the same potential, except now it's not a Higgs field anymore because there are no gauge fields, it's just a scalar undergoing SSB. There are going to be three Goldstone bosons, three massless particles, in addition to the standard model fermionic matter. The global SU(2) and U(1) are broken down by the SSB to a global U(1) subgroup, what used to be EM, but now it's just a global symmetry, there's no photon. Most importantly, the Fermions acquire a mass just the same as they do in the standard model.

spin-1: mixing of kinetic terms: the Higgs degrees of freedom are partially combined with the vector degrees of freedom to make massive vector degrees of freedom. By adding a gauge field with two propagating degrees of freedom to a scalar with one goldstone massless goldstone mode, you get a massive gauge field with three propagating degrees of freedom.

The second case is much more interesting, because the Goldstone bosons get "swallowed" by the gauge field. There is nothing analogous in the Fermion case. The Fermions leave the goldstone bosons alone. The swallowing up is what makes the spin-1 mass generation different than spin-1/2.

By the way, the reason people call all mass generation in the standard model "the Higgs mechanism" is not because the physics is the same. It's because in the standard model the Higgs field is what gives mass to everything. But this last point is just a quibble about terminology. I am trying to make a point about physics.

Take the abelian Higgs model for simplicity. The kinetic terms for the vectors are |F|^2, and there is no mass term. The kinetic term for the scalar is |D\phi|^2, which includes three types of terms:

$|\nabla \phi |^{2}-2i\nabla \phi \cdot A+|\phi |^{2}A^{2}$

There is also an additional potential for \phi with a minimum away from zero:

$V(\phi )\,$

Writing $\phi$ as $\scriptstyle Re^{i\theta }$, where $A$ and $\theta$ are the magnitude and phase, $A$ has a vacuum expecation value $H$, while $\theta$ is free to take any value. Expanding for small fluctuations, $A\rightarrow H+r$, where $r$ is small, and $\theta$ is small,

$\phi =(H+r)e^{i\theta }\,$

The scalar lagrangian splits up as follows:

$|\nabla \phi |^{2}$ splits into $|\nabla r|^{2}$ and $|\nabla \theta |^{2}$, in the quadratic part, in addition to higher order derivative interactions.

the potential for the scalar gives a mass term for the "r" fluctuations and no mass term for the $\theta$ fluctuations:

$V(\phi )=m^{2}r^{2}\,$

That's it for the purely scalar Lagrangian. Then there are the terms which involve the vector potential A. The quadratic part of the third term is:

$|\phi ^{2}|A^{2}=H^{2}A^{2}\,$

this part is what gives a mass to the vector meson, but it doesn't come alone. It can't appear just like that because it isn't gauge invariant all by itself. There is also the middle term:

$-2i\nabla \phi \cdot A$ which becomes $2H\nabla \theta \cdot A$

This term mixes up $\theta$ and A with a derivative interaction which has no counterpart in the spin-1/2 case. This term is completely absent in normal SSB models. It's what makes the physics different, in a diagram point of view. This term is what I am talking about.

To figure out the effect of this middle term, there are two roads you can take:

1. Choose a gauge where $\theta =0$ everywhere. That fixes the gauge completely, and the troublesome term proportional to grad $\theta$ drops out in this gauge. The A Lagrangian now just becomes the massive vector Lagrangian F^2 + m^2 A^2, but there's no gauge freedom left, that's been take care of. This gauge shows you that the theta and A degrees of freedom make 1 massive vector, and nothing else, no Goldstone bosons, nothing.
2. Choose Feynman gauge. Now you can't make $\theta =0$, because you've already chosen a gauge. In this gauge, you still have a naively massless \theta running around, and, if you put the vector potential mass in the propagator, you also have a massive vector potential with a bogus propagator, and the troublesome middle term leads to derivative interactions between the two. When you take all the interactions into account, you ought to still get the same thing as in the other gauge, a regular ghost-free massive vector, but it's not completely obvious how it happens.

Option 1 makes everything obvious, and is done in many textbooks. However, if you take option 1, then you can't say "I'm using Feynman gauge", because you're not. You're using a gauge where the propagator is transverse: (g_mu_nu - k_mu k_nu/m^2)/(k^2 - m^2), and the gauge condition is $\theta =0$. Feynman gauge is the condition $\scriptstyle \nabla \cdot A=f$ where f is a gaussian random field, and that's a different condition. You can't use both at once.

Option 2 should give the same answer, but how? There are naively massless $\theta$ scalars, and they have to somehow conspire with massive vector with the bogus Feynman style propagator to make a good massive vector with the usual propagator. The way in which this happens is that the kinetic terms of the A in Feynman gauge and the kinetic terms of \theta mix up to make a good massive vector. It works out. To see it, you have to sum all the diagrams that mix $A$ and $\theta$ propagators.

So the physics is more complicated, it involves nontrivial mixing up of components. You are right that it is the same as summing diagrams, but it would be hard to get the summation straight without a physical picture to guide you.Likebox (talk) 19:37, 23 November 2008 (UTC)

I choose option 2 (the Feynman gauge). The physical picture that guides me is the Mexican hat potential. We don't see the massless Goldstone oscillations along the ring-minima because they are in some SU(n) gauge group dimension, ergo non-physical ("eaten"). Only the orthogonal oscillations that climb up the sides of the hat survive -- and because they are climbing away from a minima they give masses to the vector fields. The physical and non-physical (gauge) dimensions are orthogonal to each other, hence the lack of mixing terms. Hence when you finally get to the summations it's all rather trivial and the same as the summations for the fermions and scalar Higgs.--Michael C. Price ^talk 22:52, 23 November 2008 (UTC)

The transverse oscillations are eaten, but it is precisely the transverse massless oscillations, the Goldstone oscillations, that are responsible for the vector mass. The orthogonal massive oscillations are responsible for the spin-1/2 masses, but not the vector masses. This is why the Higgs mass (which is the mass of the orthogonal oscillation) is larger than the vector mass (which is also the mass of the transverse oscillations,since they have been eaten).

To make this completely obvious, there's a nice limit which is the "Stuckelberg model" (which is called the affine Higgs mechanism in the literature). Start with a U(1) gauge group and a mexican hat Higgs, and take the limit that the Higgs vacuum expectation value goes to infinity, but the Higgs charge goes to zero (you can do that with a U(1) gauge group--- the charges don't have to be quantized) in such a way that the product of the two eH is constant. In this limit, the orthogonal Higgs oscillations, the ones that go up the potential well, become infinitely massive, and drop out of the theory. Still, the vector meson has a finite mass because the limit is taken to keep the mass finite.

In the complete mathematical limit, the Higgs field is all transverse, there are no orthogonal oscillations. The only field that survives in this limit is $\theta$, and the U(1) group acts affinely on $\theta$, meaning that a Gauge transformation adds to theta instead of multiplying by a phase. This mechanism gives a mass to a vector meson without introducing a Higgs at all, and it's the reason that the theory of massive electrodynamics is renormalizable, even without a Higgs. Historically, this accident prevented Schwinger and Feynman from understanding that masses are not compatible with renormalizability, and it took 't Hooft and Veltman to figure that out.Likebox (talk) 03:47, 24 November 2008 (UTC)

Surely the bare masses of electron, W and Higgs are independent? I don't see how you can relate one to the other, since they contain different coupling constants from the original Lagrangian. --Michael C. Price ^talk 16:20, 24 November 2008 (UTC)

Yes, the masses of the electron, the W and the orthogonal Higgs (the one going up the side of the well, what's usually called the Higgs) are different. But the mass of two components of the transverse Higgs, the Goldstone modes, are exactly equal to the W mass, and the mass of one other transverse Higgs component is equal to the Z mass. Those transverse components are not there if you choose option 1 for the gauge, but you wanted option 2. In option 2, they are separate fields and the equality of their mass with the vector mass looks like a coincidence (but it's a consequence of gauge invariance and rotational invariance).

I think the whole thing will become completely clear if you think about the case of the Affine Higgs mechanism. In this case, there is only a single real field $\theta$ which has the following gauge transformation rule:

$\theta \rightarrow \theta +\alpha \,$

$A\rightarrow A+\nabla \alpha$

The covariant derivative is $(\nabla \theta -A)$, and you can check that this is a gauge covariant quantity (it is actually completely gauge invariant). Note that there's no "i", this is a shift-translation, and theta is a phase variable.

Now write the Lagrangian for the Abelian gauge field A coupled to theta: it's

${1 \over e^{2}}F^{2}+|D\theta |^{2}$

Now if you want to see what's going on, make a gauge transformation with $\alpha =-\theta$ to set $\theta =0$ (option 1). So the theta terms disappear along with the gauge freedom, and you get the action for a massive vector meson:

${1 \over e^{2}}F^{2}+m^{2}A^{2}$

The propagator for this thing is (you have no gauge freedom--- option 1)

$\langle A_{\mu }A_{\nu }\rangle (k)={g_{\mu \nu }-{k_{\mu }k_{\nu } \over m^{2}} \over k^{2}-m^{2}}$

Which is the normal massive vector meson propagator. To understand what is going on, realize that this is the limit of e goes to zero and H goes to infinity in the usual mexican hat Abelian Higgs model.

The condensate is infinitesimally charged in this limit, and so does not allow violations of electric charge conservation in perturbation theory (this is a surprise--- charge is still conserved in this model). The second term in the propagator does not contribute to scattering, because in any process current conservation implies that:

$k_{\mu }J^{\mu }=0\,$

and the correct massive vector propagator is equivalent (only in the case of massive U(1), only for the affine Higgs mechanism) to the bogus Feynman-style propagator:

${g_{\mu \nu } \over k^{2}-m^{2}}$

Because the external fields don't couple to the longitudinal component. So that you can give a mass to electromagnetism in the most naive way, and current conservation works out and the diagrams work out by just shifting the pole in the Feynman propagator. But this is a complete accident. It only happens because there is this crazy affine Higgs mechanism for a non-compact U(1).

This accidental property utterly fails for non-U(1) gauge groups. But it was historically very confusing, because it means that the most naive massive extension of electromagnetism to the massive case turns out the be renormalizable. Schwinger and Feynman knew that massive electrodynamics was renormalizable, and I believe that they thought it was because the mass perturbation is a super-renormalizable term (I think they thought that--- it's hard to guess the wrong thoughts that people think. I think they assumed that any superrenormalizable perturbation of a renormalizable theory is renormalizable, which is incorrect). So they, along with everybody else, thought that massive nonabelian gauge theory would be renormalizable, and it's not. You always need a Higgs mechanism to have a massive vector meson, even in the U(1) case, even though in the U(1) you can take the affine limit, where it doesn't look like a Higgs field at all.Likebox (talk) 21:36, 24 November 2008 (UTC)

Higgs Mechanism is exactly 100% the same thing as superconductivity

And this fact is not only well known, it was the motivation for discovering the phenomenon. For sociological reasons, this is usually kept secret from high energy students.Likebox (talk) 15:54, 24 April 2009 (UTC)

Ok, the term "superconductivity" itself is only used for electromagnetic U(1). But the phenomenon is identical.Likebox (talk) 03:59, 25 April 2009 (UTC)

Superconductivity is not usually kept secret from high energy students. We were taught the analogy, but it just isn't very useful unless you already understand superconductivity, which I don't. I am still unhappy with the phrase charged condensate appearing in the article, especially the lead. Unless this term is explained it is seriously misleading since the VEV is explicitly chargeless.--Michael C. Price ^talk 09:48, 26 April 2009 (UTC)

The VEV is explicitly charged. This is not an analogy, it's the same exact phenomenon. Analogy is to say "higgs is like SSB". It is like SSB in many ways. But it isn't an analogy to say Higgs is like superconductivity, superconductivity is just what you call a Higgs mechanism for electromagnetism and a charged condensate made out of ordinary matter, like cooper pairs.Likebox (talk) 18:06, 26 April 2009 (UTC)

But we are not talking about ordinary matter and EM, we are talking about the HIGGS mechanism, not cooper pairs. --Michael C. Price ^talk 19:14, 26 April 2009 (UTC)

Or, if the matter is bosonic nuclei, it could be a condensate of nuclei. It could be a new charged scalar field. The only thing that matters is that there is a charged field, and that it has a VEV. Everything else is details. The usual Higgs mechanism for a U(1) field is superconductivity for condensate made out of a relativistic elementary boson.Likebox (talk) 02:43, 27 April 2009 (UTC)

Yeah, but why are focussing on U(1) in the lead? --Michael C. Price ^talk 05:51, 27 April 2009 (UTC)

The reason is that, at least for me, U(1) gauge theory is completely obvious. Nonabelian physics is one level harder to visualize, because there are lots of fields, and they all mix together. But the principle of the Higgs mechanism is exactly the same whether it's U(1) or SU(3000).

I think that the reason we are arguing so much is because we are using too much jargon. It is possible to say this in a way that has no jargon at all, and that appeals to the interested non-specialist who is the target audience.

For example, something like this: How do we know that a gauge boson is always massless? The reason is that a pure gauge field, in the U(1) case, an A which is a gradient of a scalar, has exactly zero energy. A pure gauge field is physically exactly the same as the vacuum. A constant A field of magnitude C pointing say in the x-direction is always pure gauge, it is the gradient of Cx, so a constant A field has zero energy. This is just as true for nonabelian gauge fields.

This means that an A field which is nearly constant must have have nearly zero energy, which means that a long wavelength A field has an energy that goes to zero as the wavelength gets longer. So there's no mass gap, the A field excitation is always massless. If the A field was massive, the energy per particle in a long wavelength A field would have to go to a nonzero constant, equal to the mass.

So how can you have massive gauge fields? You can't do it by simple scattering, because the argument is too general. Scattering of A particles can't give a mass, because an A which is constant would still have exactly zero energy. You need to somehow make it that a constant A field has a nonzero energy.

The way you do this is by having a condensate around. Now a gauge transformation shifts both A and the rotates the phase of the condensate. If you just shift A, but don't rotate the condensate, it's not a gauge transformation. If you just rotate the condensate, but you don't shift A, that's not a gauge transformation. But if you shift A and rotate the condensate at the same time, it's a gauge transformation and has zero energy. Notice that gauge invariance is still exactly true.

But this implies that the energy change from shifting A is exactly the same as the energy change from rotating the condensate, because they cancel each other. The energy change from rotating the phase of the condensate is easy to understand: the condensate gets a momentum. So the energy of a constant gauge field is just the kinetic energy of a relativistic Bose-Einstein gas. In terms of the A field, the energy density per-wavelength (the energy density per-particle) goes to a constant, which means that the gauge field in the presence of a condensate is massive. In classical field theory terms, which is all you need for understanding the Higgs mechanism, this just means that the frequency of a long wavelength mode is nonzero.

The effect is just caused by the kinetic energy of a charged superfluid (aka, a superconductor). The long-wavelength gauge field inside a charged superfluid gets a mass (aka becomes short ranged). So the mass of a photon inside a superconductor (the inverse electromagnetic screening length) is nonzero, and this is the Meissner effect. The nonabelian case introduces no new phenomenon, it just changes the type of vortices you can have inside the Higgs condensate.Likebox (talk) 18:29, 27 April 2009 (UTC)

Mass prediction

One should not say "The mass of the leptons is acquired by Higgs mechanism". This is wrong. The mechanism is different physically for the leptons and for the gauge bosons (see discussion above). The Higgs condensate gives mass to the gauge bosons, the Higgs boson gives mass to the leptons and quarks. Also, the absolute magnitude of the gauge bosons were predicted, with input the coupling constants, not just the ratio.Likebox (talk) 16:05, 24 April 2009 (UTC)

What I said above is not quite right. The Higgs condensate is necessary, to allow the leptons and quarks to violate charge conservation. The right thing to say is that the interactions with the fermions involve the trasverse oscillations in a direct way.Likebox (talk) 16:12, 24 April 2009 (UTC)

When I read the "discussion above", I come to the opposite conclusion, namely that the mechanism is essentially the same for the fermion mass acquisition as it is for the gauge boson mass acquisition. It just depends on what you mean by the Higgs mechanism, and it means different things to different people at different times.--Michael C. Price ^talk 22:53, 24 April 2009 (UTC)

The vectors eat the massless goldstone modes, the fermions acquire mass by higgs-fermi direct coupling. Both involve interaction with a condensate, but there is no "eating" for the fermions. The "eating" is what makes it interesting for the vectors.Likebox (talk) 04:02, 25 April 2009 (UTC)

Higgs mechanism vs Higgs-Kibble mechanism

Kaku seems to make the distinction between the Higgs mechanism generally (all mass acquisition, i.e. by fermions and bosons) and the Higgs-Kibble mechanism specifically (eating the massless gauge bosons and giving them mass). Is this a distinction we can make here? It might make things clearer and save a lot of the endless debate about names we're having. --Michael C. Price ^talk 10:16, 26 April 2009 (UTC)

I agree that the debate is almost entirely about names, but there is a reason for the quibbling. There is a simple reason that the Goldstone bosons don't appear--- the Goldstone modes are oscillations which are related by a pure gauge transformation, they aren't physical. This is the essential point, and it is obscured by making the interaction look like normal scattering.

As far as terms go: the Higgs mechanism (as far as I know, in theorist's usage) is always the gauge boson mass business. The fermion mass is interaction with the Higgs, but not usually Higgs mechanism par se. Higgs-Kibble vs. Higgs should not be distinguished, because Higgs and Kibble discovered the same thing. It would be historically inaccurate.Likebox (talk) 17:43, 26 April 2009 (UTC)

I disagree that the issue is obscured by making the interaction look like normal scattering. The mechanism by which the gauge bosons acquire mass is the same as the mechanism by which the fermions acquire mass. What is different is way the massless gauge bosons vanish and that is what you are referring to as the Higgs mechanism. The question is, does the Higgs mechanism refer to just the gauge bosons being eaten, or does it also refer to the mass acquisition (by scattering off the Higgs) as well?

Look, I explained it before, simple scattering does not lead to mass acquisition for gauge bosons. If it did, the Higgs mechanism wouldn't have a name, it would have been just one of the many anonymous Lagrangians discovered in the 1950s. The reason the phenomenon has a name is because it involves new physics--- in order to get a mass, the gauge bosons must mix up, or "eat", the transverse oscillations of the scalars. It is this "eating" that is called the Higgs mechanism.Likebox (talk) 02:46, 27 April 2009 (UTC)

I see no reason to start with a dismissive Look : your reasoning is flawed. As I explained above, the scattering is the same for all mass acquisition, it's the eating that different: the gauge bosons get eaten, the non-gauge fermions don't. What you choose to call that is one thing, but the physics is clear. --Michael C. Price ^talk 05:44, 27 April 2009 (UTC)

Sorry, that sounded bad. I just meant that the physics is not as clear as you are making it out to be. Without a condensate, there is no scattering mechanism that can give a gauge boson a mass. If you don't have a nontrivial vacuum, gauge bosons are always exactly massless. But they still scatter off stuff. If you make a neutral condensate, all the scattering in the world won't give the gauge bosons mass. But that's also true for the fermions, with a neutral condensate, you won't get Fermion masses. So maybe it is wrong to make such a big distinction. But there is a little distinction, because of the eating.Likebox (talk) 18:36, 27 April 2009 (UTC)

Not SSB, but Almost

There's a difference between two procedures:

1. "break a symmetry, then gauge it"
2. "gauge a symmetry, then break it"

The second concept doesn't make any sense. If a symmetry is gauged, it can't be broken, it can only be Higgsed. If a symmetry is broken, it can still be gauged, but then it becomes Higgsed. The difference in terminology is real: a broken symmetry means that the symmetry no longer works in the path integral.Likebox (talk) 17:47, 26 April 2009 (UTC)

This doesn't make sense to me. We start with a gauge symmetry. The symmetry is broken, and we are left with a residual symmetry. It's the residual symmetry that does the eating. But the original symmetry is still present, it's just the vacuum is not invariant under the full symmetry, but it's presence is still important.--Michael C. Price ^talk 19:37, 26 April 2009 (UTC)

That's the whole point. If you start with a gauge symmetry, the symmetry is never broken. It's still there even when you have a condensate, it's always there. It just can't be broken. The vacuum is invariant under the symmetry, because there wasn't a "symmetry" to begin with. Two states that differ by a gauge transformation are the same exact state.Likebox (talk) 02:41, 27 April 2009 (UTC)

I'm not going to argue the point. You start by saying the there is a gauge symmetry, then that it was never there. Suit yourself. --Michael C. Price ^talk 05:53, 27 April 2009 (UTC)

I'm sorry for being unclear and argumentative. I just meant something simple, that gauge invariance can't be broken, it's a symmetry of the states. Two states which differ by a gauge transformation are the same exact state, whether you have Higgs mechanism or not. Its a stupid quibble, it's not that deep as physics. The global part of the gauge symmetry, the constant gauge transfomations, do stuff at infinity, and those are symmetries which can be broken. So if you mean "at infinity", maybe it's OK to say "broken gauge symmetry".Likebox (talk) 18:42, 27 April 2009 (UTC)

I want to say, my thinking about this was faulty before I read the page on infraparticle. I used to think that the local gauge transformation don't have any physical meaning, because they don't give you new conservation laws. But infraparticle explains that they do, but only when they go out to infinity, and do nontrivial stuff there.Likebox (talk) 18:49, 27 April 2009 (UTC)

Particle Language/Field Language

The duality between particles and fields allows you to state any quantum field theory effect in two ways, and it is up to the author to decide which way is clearer, and write accordingly. In the case of the Higgs mechanism, both pictures give insight, and I think that the lead can include both with a minimum of text. The mathematics of the thing is the same, but I think that talking about particles and condensates makes it more intuitive for most readers.Likebox (talk) 13:55, 29 April 2009 (UTC)

I disagree -- particles are nothing more but quanta of fields. Particles are smallest possible excitations from ground states. Particles are "normal modes" of fields, nothing more. Particle language is approximate, perturbative, phenomenological, often misleading and falsely intuitive. Tell me -- what are those mysterious "particles" the Higgs condensate is allegedly made of? Certainly not Higgs bosons, since they are small fluctuations from Higgs condensate... --Bakken (talk) 14:13, 29 April 2009 (UTC)

That's a common point of view, and it leads many people who talk about quantum field theory to unnecessarily abandon the intuitive particle pictures. But the "only fields are fundamental, particles are small oscillations", point of view is not complete and leaves out useful stuff. Obviously, you can do everything in field theory in terms of fields, so I'm not saying your picture is incorrect, it's just not the only picture.

The particle point of view for field theory is the one originally due to Feynman (and also to Schwinger), and it is not just a perturbative expansion of the S-matrix, meaning that it doesn't just describe long-distance asymptotic particles. Sure, that's part of it, but it has a much more important role in describing short-distance particles which are not long-distance objects. The Feynman diagrams give you a description of short-distance field correlation functions and operator product expansions, including correlation functions that don't have an obvious interpretation as S-matrix elements. This is especially true for QCD, where the particles called "quarks" and "gluons" are not small oscillations of any field which is defined at long distances, they have no S-matrix interpretation. The original S-matrix-y idea that the only definition of "particle" is a quantum of well defined long wavelength modes with finite mass is gone. Nobody argues that quarks and gluons aren't particles anymore.

The important thing about the short-distance particle picture is that it allows you talk about particles even when the long-distance reference vacuum state is not completely understood. In particular, if I want, I can treat the Higgs mechanism as a pure quartic potential, with no negative unstable mass squared term, and define the Higgs particles as the objects created by the Higgs field in this model. Then there are four species of Higgs, which interact with the bosons by Feynman diagrams. If I then choose to add the negative mass-squared term as a perturbation, the path integral for the Higgs field, or for the Higgs particles, don't change at all at short distances, except for a small amount of new scattering from the negative m^2 term, but the new vacuum is totally different, because of the Higgs condensate forms. The Higgs condensate, in this definition of particles, is a coherent state of the small oscillations around the symmetric vacuum. This is the short-distance-particle picture of the Higgs.

You have a choice: you can describe this condensate in terms of the symmetric charged Higgs particles you see at short distances, or in terms of the single long-wavelength Higgs particles and massive vector mesons, the particles in the long-distance S-matrix. It's a choice, because the short distance description doesn't care at all about the long-distance physical spectrum. This was a major paradigm shift of the seventies.

The claim that particle pictures are "perturbative" is also slightly misleading. The question is what you mean by "perturbative". Yes, the expansion of short-distance correlators as particle paths expands the interactions in the coupling, but it isn't an S-matrix expansion, it's an expansion for correlation functions. In short, the particle picture is nowadays fine, but somebody forgot to tell the people who write textbooks. So particle interpretations are often left out in an overzealous attempt to not offend S-matrix practitioners.Likebox (talk) 15:07, 29 April 2009 (UTC)

If you want to talk in your convoluted language "from seventies", just do it, but don't impose on everybody else. Please, first find a modern textbook with your "particle" language and make a reference to it. —Preceding unsigned comment added by Bakken (talk • contribs) 15:30, 29 April 2009 (UTC)

You totally misunderstood. The language you are using is from the seventies, and it is also correct. The particle language is another point of view. And I do intend to impose it on everybody else. To do otherwise would be unfair to the next generation. There's references out there, don't make me dig them up. It's not that controversial anymore.Likebox (talk) 15:35, 29 April 2009 (UTC)

I want to add that I don't mean to demean the S-matrix approach, which is absolutely correct when applied to quantum gravity. There, you can't separate out short-distance from long distances, and say "I define my objects at short distances only". I mean, it's fine in field theory, but in string theory, the short distance behavior is inextricably tied up with the long-wavelength behavior, and this is a fundamental infrared/ultraviolet property. It's closely tied up with holography, but it was discovered long before, and it's the reason people believe string perturbation theory is finite even though this hasn't been fully proven to all orders. The reason is that the short-distance divergences in string theory have an interpretation as long-wavelength massless modes running amok, because high-energy states in quantum gravity are big floppy black holes, not small concentrated bullets.Likebox (talk) 15:35, 29 April 2009 (UTC)

Here's the first source I found, there are a million more: An introduction to quantum field theory By Michael Edward Peskin, Daniel V. Schroeder p692.Likebox (talk) 15:39, 29 April 2009 (UTC)

Sorry, but they also identify the Higgs mechanism with "spontaneous symmetry breaking in a gauge theory": Michael Edward Peskin, Daniel V. Schroeder, An Introduction to Quantum Field Theory, page 690: "The Higgs Mechanism. In this section we analyze some simple examples of gauge theories with spontaneous symmetry breaking...". Another textbook, perhaps? :) --Bakken (talk) 00:16, 30 April 2009 (UTC)

"Gauge theories with spontaneous symmetry breaking" is not a phrase that is all that wrong. The reason is that they don't imply that the gauge symmetry itself is broken by using that phrase. If they said "Spontaneous breaking of a gauge symmetry" then it would be closer to wrong. It's all stupid terminology quibbles that hang up on precise definition of words, and it is annoying to haggle over.Likebox (talk) 23:51, 30 April 2009 (UTC)

I'm with Likebox on this, fields and particles are equivalent, which is why Feynman diagrams are useful. Yes, they're perturbative, but so what? They work. And there are plenty of textbook references for this. --Michael C. Price ^talk 16:04, 29 April 2009 (UTC)

All right, some people believe in God, others in particles. I have left untouched the superconductivity analogy in the lead. If you believe in particles with imaginary mass nobody has ever seen... :) Dear Michael, please, look at the Lagrangian of the Standard Model -- there are fields and only fields there, no particles. It is only after the theory is quantized that the particles appear as field quanta. So fields and particles are not equivalent. Not in the standard model in any case. --Bakken (talk) 17:02, 29 April 2009 (UTC)

Since the theory is quantised, I don't see the relevance of the fact that only fields appear in the classical formulation; classical physics false, quantum physics true. --Michael C. Price ^talk 21:05, 29 April 2009 (UTC)

"classical physics false, quantum physics true". Aha... General relativity is wrong? Maxwell equations are wrong? Newtonian mechanics is wrong? Thermodynamics is wrong? But anyway, if you want to call fields "particles" -- just be my guest. I personally will continue to call them "interacting quantum fields". —Preceding unsigned comment added by Bakken (talk • contribs) 22:00, 29 April 2009 (UTC)

Actually, yes, Newtonian mechanics etc is wrong. I'm sure this isn't news to you, so I'll leave the discussion here. --Michael C. Price ^talk 05:48, 30 April 2009 (UTC)

I didn't mean to sound so defensive and cranky above. But quarks are particles with an imaginary mass that nobody has ever seen, so yes, modern field theorists do believe that particles include "imaginary" entities divorced from asymptotic S-matrix states.

But this is not an act of faith, it is an act of operator product expansion. The notion of a local particle in the modern formulation is tied up with the algebra of field operators at short distances, not anymore to their asymptotic states at infinity. The Feynman expansion for asymptotic states has a counterpart in the Feynman expansion for short-distance correlators, and the notion of a particle is contained in the propagator for fields which have nearly free correlators at short distances. While this does not quite include the Higgs, which becomes strongly interacting at ultra-short distances in the simplest models, it is a good enough approximation to Higgs correlation functions that nobody worries about the asymptotic ill-definedness anymore. The notion of particle provided by this picture is essential for understanding the further advances in physics, especially string theory.

This is why it is important to explain the particle picture properly. If it is not done, then string theory is incomprehensible gibberish.Likebox (talk) 18:06, 29 April 2009 (UTC)

One should also point out that it is also possible to give a classical particle formulation of the standard model, where there are no fields at all. Then the classical fields appear only after quantization. The point is that there is a quantum duality between particle pictures and field pictures, which is not there classically.Likebox (talk) 18:22, 29 April 2009 (UTC)

It might be possible, but nobody does it. Probably for a reason, ah? :) --Bakken (talk) 00:19, 30 April 2009 (UTC)

String theorists do it. String actions are usually entirely defined in terms of the particle spectrum, and the field formulation of string theory is either non-existent or really muddy, and it is defined in terms of very non-local fields. This is why getting comfortable with this terminology is important. It gives a person learning field theory a handle on how to translate field theory results to string theory.Likebox (talk) 23:56, 30 April 2009 (UTC)

"But quarks are particles with an imaginary mass that nobody has ever seen, so yes, modern field theorists do believe that particles include "imaginary" entities..." -- exactly. Particles are the low-energy spectrum of interacting quantum fields. In the case of QCD the low-energy spectrum is represented not by quarks, but by e.g. pions and nucleons which are the observed particles. You are again starting an undo-war. Well, let's see where we end up. I will not compromise on Spontaneous symmetry breaking and on Particle Data Group. However, I will accept your particle formulation provided it is placed after the formulation of the Particle Data Group. --Bakken (talk) 19:01, 29 April 2009 (UTC)

No edit war--- I am trying to convince you to go along with this. Spontaneous symmetry breaking is not good language, in 'tHooft's opinion. I am personally not dead set against it like he is, because the global part of the symmetry is indeed broken, but I see the point he is making. The gauge symmetry, the local part, can't possibly break.Likebox (talk) 23:10, 29 April 2009 (UTC)

In my opinion the field formulation has to go before the particle formulation if only for historical reasons. I do not think you are constructive when you try to delete perfectly valid and generally accepted formulations (backed by references to internationally accepted authorities, like PDG) just because you personally believe the formulations are "outdated". —Preceding unsigned comment added by Bakken (talk • contribs) 23:14, 29 April 2009 (UTC)

That's not what I am doing. I am deleting perfectly valid and generally accepted formulations because they are not the subject of this page. This page is about the Higgs mechanism in general, without only a minor focus on the specific Higgs mechanism in the standard model. That case is the only one that is for sure experimentally relevant, so of course PDG focuses on this case. The question is, should we?Likebox (talk) 23:21, 29 April 2009 (UTC)

Quibble--- I did not put the particle interpretation first, I put it second, just as it is now.Likebox (talk) 23:25, 29 April 2009 (UTC)

Surely the Higgs condensate / VEV is an example where the particle picture makes more sense than the field. Consider (for example) an electron or quark scattering off the condensate. At every scattering event it flips chirality (left->right->left etc), and it does so it changes weak hypercharge and I₃, which are both conserved quantities. The particle interpretation explains this by saying that an appropriately Higgs has been absorbed from / supplied to the condensate. We don't see the individual Higgs particles because they form a superfluid, but we need them to maintain the various conservation laws. --Michael C. Price ^talk 15:10, 19 June 2009 (UTC)

Form of the Potential in the Lead

The mexican hat potential is a property of Higgs mechanism with the simplest possible structure--- pointlike renormalizable scalar fields. But I believe that the term "Higgs Mechanism" is used for technicolor models too, where the Higgs is a fermion condensate made out of some techniquarks. Then the potnetial would not necessarily be quartic. Does "Mexican hat potential" mean arbitrary symmetry breaking potential?Likebox (talk) 14:01, 29 April 2009 (UTC)

I believe "Mexican hat" signifies a generic potential with a symmetry and a smooth minimum at a non-zero value. But then again -- there is probably no standard definition of that. --Bakken (talk) 14:16, 29 April 2009 (UTC)

Particle Data Group Again

The particle data group has a pro forma description of the Higgs mechanism which is a complete aside. Their main mission is to explain the experimental data on particles. So the description they give is annoyingly terse, and inappropriate to a more discoursive article like this one. Plus I don't think it sounds good in the lead (it was inserted before).Likebox (talk) 18:08, 29 April 2009 (UTC)

Well, Likebox, the particle data group is a group of internationally recognised experts. I am sorry, but I tend to believe more in their opinion, and their formulation than in yours. In any case, their opinion has the same right to be in the lead paragraph, as your opinion.--Bakken (talk) 18:30, 29 April 2009 (UTC)

This stuff is well known, and completely objective, and everybody can check every statement without any references. There is no need to defer to anybody's opinion, except for insight about pedagogy and clarity. These people are not concerned with the actual details of the Higgs mechanism proper, they are concerned with tabulating experimental data. Their point of view is specific to the Higgs mechanism in the standard model, and this is not satisfactory for a discussion of the general phenomenon. We had this exact same debate before, about the exact same source, with different contributors, and at that time people decided to leave the particle data group summary out.Likebox (talk) 22:42, 29 April 2009 (UTC)

Sorry, that was you then too. What's up with this PDG business? Why should we put it up front as the definition?Likebox (talk) 23:12, 29 April 2009 (UTC)

Because they are a group of internationally recognised experts in the field of particle physics. Because their definitions are thought out and well formulated. Because their reviews with their definitions are regularly published in refereed international journals. Because the editors and the referees of those journals agree with their formulations. That's why we should put their formulation first, and your personal formulation second. And moreover, I believe your definition, "...a form of superconductivity in the vacuum...", is just not very meaningful and certainly a lot worse that PDG's "...spontaneous symmetry breaking in a gauge theory...".--Bakken (talk) 23:56, 29 April 2009 (UTC)

Yeah, yeah. I don't think so, and I don't think you'll get anyone who isn't in the group to agree with you.Likebox (talk) 00:09, 30 April 2009 (UTC)

Well, judging from the quotations I gave you, the authors Peskin, Schroeder, Mehra, Rechenberg, Zuber, Strocchi, apparently agree with me. Or rather I agree with them. :) --Bakken (talk) 00:27, 30 April 2009 (UTC)

The quotes are referring to the phrase "spontaneous symmetry breaking" in a gauge theory.

You could get all the members of the APS to sign an affidavit that they agree with you, what difference does it make? We're trying to figure out what is the clearest language for this page. Unless they are reading this page and thinking about it, their opinion is not very apropos.

These authors are right that it is a "spontaneous symmetry breaking model" where "the symmetry that is broken is gauged", it's just not breaking the local part of the symmetry, the part that is usually thought of as the "gauge symmetry". That's all. It's not that deep a point. So I try to rephrase "spontaneous symmetry breaking" to "spontaneous breaking of the symmetries of the potential, which include the gauge symmetry" (that's absolutely true), or "spontaneous symmetry breaking of the global part of the gauge symmetry" (that's also absolutely true), to avoid the potential confusion caused by the phrase "spontaneous breaking of the local gauge symmetry", because the local symmetry is in the kinematics, in the definition of the states, and it just can't be broken. It's not a complicated point, nor a very deep point. I personally don't have such a big complaint with this, because I haven't seen anybody ever get confused on this point. But 'tHooft doesn't like it.Likebox (talk) 16:16, 30 April 2009 (UTC)

Lead

Ok, so people like the phrasing from before. The changes I made were cosmetic mostly. The Higgs mechanism is not controversial, so it shouldn't say "according to the particle data group". That's according to everyone else too. Uncontrovertial statements should not be attributed, otherwise they seem to be disputed. The remaining changes were reordering of words to sound nicer, and removing the "expert" tag, which is completely unnecessary because there is nothing here that is disputed on accuracy. It's all style issues.Likebox (talk) 08:04, 16 May 2009 (UTC)

interactions

No, I don't get the math. Are scattering and coupling the only possible interactions? Do they have different functions or must they occur together to get Higgs working? Etc. 4.249.3.140 (talk) 18:14, 22 May 2009 (UTC)

Scattering and coupling are the same thing. --Michael C. Price ^talk 09:59, 19 June 2009 (UTC)

Not charged

The article's lead says

In particle language, the constant Higgs field is a superfluid of charged particles,

However any adjoint Higgs field with a diagonalised VEV (such as in minimal SU(5)) must be completely chargeless. It is true in that the electro-weak higgs VEV possesses I₃ and weak hypercharge, but this is evidently specific to E-W and is not a generic property. It seems to imply that all Higgs condensates are superfluids, but not necessarily superconductors. --Michael C. Price ^talk 09:59, 19 June 2009 (UTC)

All Higgses are charged under the big group. Once you diagonalize the VEV, all Higgs look neutral under the small group, that's the definition of the small group. The unbroken subgroup is the subgroup that fixes the Higgs vev.

So in your example, and in every other example, the Higgs ends up neutral after breaking. The point is that it is a superconductor for the original big group.Likebox (talk) 12:30, 19 June 2009 (UTC)

That's not the way I would express it. Both the vector and adjoint Higgs are variously charged before SSB. After SSB the diagonalised adjoint VEV is neutral with respect to all charges, whereas the vector Higgs' VEV carries the hidden charges (mediated by massive bosons), but is neutral with respect to the unbroken charges (mediated by massive bosons).--Michael C. Price ^talk 15:00, 19 June 2009 (UTC)

That's the same thing, said slightly differently. The SU(5) higgs is coupled to the SU(5) gauge bosons that don't become the SU(3)xU(2) gauge bosons. So the SU(5) condensate is charged under some of the SU(5), not the top 3 by 3 block, nor the bottom 2 by 2, nor the leftover diagonal guy. Just the off diagonal 3 by 2 part.Likebox (talk) 16:47, 19 June 2009 (UTC)

But there is no off-diagonal condensate. The SU(5) / adjoint Higgs condensate is completely diagonal. With regards to the vector Higgs, note that there could be/was a period (between the adjoint SSB and the vector SSB) when there was only the adjoint Higgs condensate around (ignoring more speculative larger groups). So for this period there was a Higgs superfluid / condensate, happily giving masses to the X & Y bosons, but no superconductivity. --Michael C. Price ^talk 18:53, 19 June 2009 (UTC)

No, no, you can diagonalize the SU(5) condensate, so that the SU(5) Higgs field is all diagonal, but it's still charged under the off diagonal gauge bosons. The off diagonal 3 by 2 block that's broken consists of SU(5) matrices act by rotating the top 3 components into the bottom two components. The condensate is not invariant under these transformations, by construction, because that's what it means to break the gauge symmetry. This also means that the condensate is charged under these off diagonal elements, which makes them a superconductor for the broken part of the group.Likebox (talk) 21:45, 19 June 2009 (UTC)

No, the diagonal condensate is not charged, since the charges, Q, of any adjoint field are Q_i^j = Q_i - Q_j. Hence the diagonal elements, i=j, have zero charge. True, the off diagonal gauge bosons are charged, but that is irrelevant; it's the chargeless state of the diagonal adjoint Higgs condensate that is relevant. The gauge transformation business you raise is a red herring, since GTs don't change any measurable quantities (cf Smolin's chalk); since the adjoint Higgs condensate is chargeless / diagonalised in one frame it is effectively chargeless in all frames. A GT won't generate a superconductor from nowhere. --Michael C. Price ^talk 07:13, 20 June 2009 (UTC)

Smolin's chalk?? He must have used that old bit from Sidney Coleman's lectures.

It's still ok. The "charged" part you are talking about, has nothing to do with the "charge" I am talking about.

You are calculating the charge under the diagonal subgroup of SU(5). That's the charge under diagonal gauge transformations. Indeed, any diagonal adjoint higgs vacuum expectation value has zero charge under the diagonal subgroup.

That's exactly why this diagonal subgroup is unbroken. It's the low energy U(1), plus a subgroup of SU(3) which only does diagonal rotations, and a subgroup of SU(2) that only does diagonal rotations. This is part of the standard model gauge group. So this is a red herring. The "charged" you need to be thinking about is the part of the group that is broken.

When I say a scalar field is "charged", I mean it "couples to the vector bosons". When a gauge transformation leaves a field alone, then this field does not couple to the vector bosons. When a gauge transformation changes a field, then this field does couple to the vector bosons. Always. Without exception. This is the definition of a gauge theory.

• charged = changes under gauge transformation
• neutral = invariant under gauge transformation

The SU(5) higgs does not change under:

1. The diagonal U(1) subgroup which rotates the top three components one way and the bottom two components the other way
2. the upper 3 by 3 SU(3) matrices
3. the lower 2 by 2 SU(2) matrices

That's 12 generators in all, which are the standard model U(1), SU(3), and SU(2) respectively. The remaining 12 generators are broken, and these are the ones that the Higgs is charged under. The generators of these transformations are

1. 3 by 2 (anti)hermitian blocks, which mix the top 3 components with the bottom 2

That's 12 generators, giving 24 in all (that's all of SU(5)). These are the 12 gauge bosons which the Higgs couples to. These are the generators that are broken. These are the charges which the condensate has.Likebox (talk) 15:03, 20 June 2009 (UTC)

Yes, I meant Coleman's chalk.

Okay, I see why we were talking past eachother; we meant different things by "charge". I was using it to mean the eigenvalues of the mutually commuting Cartan generators, which I think corresponds to what you would call conserved charges. Your definition is broader (and more standard now I think about it :-) , so I apologise for the confusion). Where that leaves the superconductor business I'm not sure, since electric charge is conserved, of course.

BTW you know any good online sources for Pati-Salam model? It seems more viable than SU(5), as far as I can see. Both come from SO(10), but less constraint on the (unobserved) proton decay rate with P-S. --Michael C. Price ^talk 20:38, 20 June 2009 (UTC)

The Cartan business is also a legitimate way of looking at it. The notion of "charge" is sort of intuitive. If you think of charges as a maximal set of commuting U(1)'s only, not as coupling to gauge bosons, then what I said would be awfully confusing. Maybe we should clarify that, say "charged--- meaning nontrivial rep".

I never studied Pati Salam in any way. I only know the gauge group is SU(4) times something, which now I looked at the page and it says SU(2) cross SU(2). Pati Salam came before SU(5), and it's not simple, it's at least two groups with two independent coupling constants. So it's much less elegant than SU(5), and I think much less viable. But some people like it.

I don't know how it comes from SO(10). I only know the reduction from SO(10) to SU(5). That's actually slightly tricky, you have to figure out how the 16 components of a ten dimensional spinor mix up with each other under "holomorphic" rotations. It's an interesting exercise, because it shows you the relation between holomorphicity and spinors very clearly. That comes up a lot.

I think that the SO(10) models are still the most plausible. The SO(10) matter rep is simple, and it unifies the couplings. SU(5) is no worse. If Pati Salam is true, then SU(5) would just be a cruel joke. Like Einstein used to say, God is subtle, not malicious.Likebox (talk) 21:31, 20 June 2009 (UTC)

I think the breakdown for Salam-Pati goes:

SO(10) --> SO(6) x SO(4), with

SO(6) = SU(4)_C and SO(4) = SU(2)_L x SU(2)_R

with a further breakdown

SU(4)_C --> SU(3)_C x U(1)_B-L

giving:

Q = I_3L + I_3R + (B-L)/2

and then

SU(2)_R x U(1)_B-L --> U(1)_Y

so that we finally get to the standard model

SU(3)_C x SU(2)_L x U(1)_Y

Yes, nature may have played a cruel joke on us...--Michael C. Price ^talk 04:16, 21 June 2009 (UTC)

Superfluid?

As far as I can tell there are one or two authors on here who are trying to push an explanation in terms of a superfluid analogy. I'm not sure if this is even correct, but in the very least it is not the way that the Higgs mechanism is usually explained. In fact if I search arxiv I only found one HEP Higgs paper that references superfluid. So all I ask for is a single citation which supports this explanation. If not, then this analogy should be deleted or moved much further down in the article. —Preceding unsigned comment added by 24.186.196.7 (talk) 04:57, 10 September 2009 (UTC)

The more I read the article, the more I am losing confidence in this article's credibility. I've never heard the Meissner Effect in conjuction with any Higgs dicsussion. No link supplied discusses any superconductivity. I admit that I may be seriously wrong, but all I ask is someone go through and put in some REFERENCES. With google books and arxiv it should be very easy to supply the much-needed references to support these explainations. Until this occurs I warn any passer-by that there is a chance that this explaination is not orthodox HEP theory. --24.186.196.7 (talk) 05:08, 10 September 2009 (UTC)

When we were taught about the Higgs (physics BSc), the Meissner effect was mentioned.

I agree that more refs would be nice, although this is hardly a problem confined to this article.--Michael C. Price ^talk 06:03, 10 September 2009 (UTC)

To anon: There is no direct analogy with superfluidity, it is superconductivity that is the correct counterpart as is mentioned in the article. The explanation in the article is self contained, and can be checked by any competent physicist.Likebox (talk) 13:27, 10 September 2009 (UTC)

All superconductors are superfluids, hence the lack of a Higgs resistive drag to motion. --Michael C. Price ^talk 16:45, 10 September 2009 (UTC)

The difference between them is that the no-energy-gap superflow doesn't exist in a superconductor. Because the superfluid is charged, the lowest lying modes are plasma oscillations with a finite frequency.Likebox (talk) 19:22, 10 September 2009 (UTC)

Don't the Cooper pairs superflow?--Michael C. Price ^{[ [User talk:Michael C Price|talk]]} 19:56, 10 September 2009 (UTC)

They do superflow, sort of, except that the phase of the superflow oscillates with a finite frequency even at infinite wavelengths. That's exactly the "eating" of the charged superflow by the photon to make a short ranged electromagnetic field.

This is in contrast to a slow long-wavelength superflow in liquid helium, where the phase of the wavefunction slowly changes from place to place, and which has nearly zero energy. If you take out the overall phase variation, and ignore density variations, then superfluid He4 has an energy gap too. The low energy flows are all incompressible potential flows, which just means that the flow velocity is everywhere the gradient of the wavefunction phase. This type of flow does not cause friction.

The flows that do lead to friction is the production of closed vortex lines in the Helium. The vortex lines exist in the superconductor too. The potential flow, however, doesn't.Likebox (talk) 21:35, 11 September 2009 (UTC)

I had the same reaction, and a search on Google Scholar (e.g., "Higgs" together with superconductive, superfluid, etc.) turns up no obvious evidence of this connection in the current literature. I see no strong arguments in the discussion above for keeping this statement, it isn't referenced, and it seems contrary to general understanding, so I have removed it from the introduction. I have also moved the associated section on superconductivity to the end of the article, in light of its questionable relevance. Eric Drexler (talk) 04:34, 18 May 2010 (UTC)

Your removal of the link to superfluidity is ill-advised. As I said earlier the analogy to the Meissner effect is made in the literature (e.g. Kaku's QFT) and in undergraduate courses which cover the Higgs mechanism. --Michael C. Price ^talk 02:49, 21 May 2010 (UTC)

Should there be a link

to Absolute time and space here? Might a non-zero Higgs field provide a 'thing' relative to which all other 'things' may move, diferentiating rest from constant motion and providing a solution to Newton's Bucket? I am fully aware that I may have simply misunderstood the reasoning and theories here, particularly given the current lack of links between the articles.--94.195.167.7 (talk) 20:59, 12 October 2009 (UTC)

The Higgs field is relativistically invariant--- it doesn't define a rest frame. If you are moving, you see the exact same Higgs as when you are stationary. For Newton's bucket, that's a statement of absoluteness of rotation, and that isn't going to be adressed by a special relativistic theory.Likebox (talk) 22:01, 12 October 2009 (UTC)

Thanks - I remember reading something about it somewhere - I have bought a new book to brush up.--94.195.167.7 (talk) 20:34, 13 October 2009 (UTC)

Utterly meaningless/invalid math

the equations in the article are utterly meaningless without definition of the variables/constants, found neither in the article or the links in the sections containing the equations. Would be nice if somebody could rectify this. —Preceding unsigned comment added by 121.79.6.107 (talk) 00:28, 5 November 2009 (UTC)

The symbols are all standard: A is the vector potential, q is the charge of the particle described by the quantum field psi/phi, "i" is the imaginary unit, t is time. I think that's all the symbols. If there are more mystery symbols, just list them below.Likebox (talk) 00:47, 5 November 2009 (UTC)

Yes, they're standard, but that doesn't help the general reader. We should place explicit links at the very least.--Michael C. Price ^talk 01:33, 5 November 2009 (UTC)

They may be standard to the field, but one should not require a degree in science majoring in Quantum Mechanics to understand a wikipedia article. A is commonly used to represent just about any matrix/system of simultaneous equations, i is also used to represent instantaneous current in electrical systems, phi represents phase angle difference in AC systems. I assume lambda is some sort of eigenvalue for the system, but that could be made a lot clearer too. —Preceding unsigned comment added by 121.79.6.107 (talk) 07:04, 5 November 2009 (UTC)

By saying "standard" I didn't mean you shouldn't clarify the symbols, I just meant that I don't know which ones are confusing. The constant "lambda" is a dimensionless number that measures how strongly the phi-particles bounce off each other. It is not an eigenvalue, it is called a "coupling constant".Likebox (talk) 21:18, 5 November 2009 (UTC)

the easy way

For those of you which still could not grasp the concept of the higgs field, I found a very kind video on CERN TV (a youtube channel edited by the CERN) : http://www.youtube.com/user/CERNTV#p/a/f/0/1cHW1FJ45z4 (for review) —Preceding unsigned comment added by 88.179.19.69 (talk) 23:45, 27 November 2009 (UTC)

Overwhelming?

The evidence for the Higgs mechanism is comparable to the pre-COBE evidence for the big-bang. The Higgs mechanism is both theoretically required, and it predicts the quantitative masses of the W and Z bosons to 5 decimal places. The assumption that the W and Z are gauge bosons with breaking pattern as in the standard model predicts their interaction/decays to several decimal places. Is this overwhelming?

Compare to Big Bang. The big bang was supported by a 1-2 decimal place Hubble law, and by big-bang nucleosynthesis data which predicted results for element abundances to 1-2 decimal place for 5 elements. That's about 10 decimal places worth of data in both cases. The situation for both the Higgs mechanism and Big Bang are now much more precise, but lets say its 10 decimal places of agreement. To realize how good evidence this is, consider that the answer could have been anything, a-priori. So 10 decimal places of agreement is a probability of 1 part in 10 billion. This means that if the Higgs mechanism is wrong, you would need a one part in 10 billion coincidence to explain the data. That's almost 8 sigma of evidence, where 5 sigma would be sufficient for scientific certainty. That's pretty overwhelming.Likebox (talk) 07:03, 16 December 2009 (UTC)

The above is original research. If you want to claim in the article that the evidence is overwhelming then it will suffice to quote a recognised authority who says just that. Xxanthippe (talk) 07:10, 16 December 2009 (UTC).

Bullshit.Likebox (talk) 07:13, 16 December 2009 (UTC)

See, inter alia, the 1979, 1984 and 1999 Nobel Prizes in Physics, all awarded for application of the Higgs mechanism (spontaneous "breaking" of gauge symmetry) to the specific case of electroweak interactions. -David Schaich ^Talk/_Cont 09:42, 17 December 2009 (UTC)

Agreement with experiment is a necessary but not sufficient condition for a physical theory to be considered "overwhelming" and arguments that enumerate Nobel Prize-recipients can cut both ways. For example, perturbative QED, which has almost as many Nobel Prize-recipients as electroweak theory, predicts the magnetic moment of the electron also to a large number of decimal places. However, there is yet no proof that the perturbation expansion for QED converges, and the presence of the Landau pole indicates that the theory has serious difficulties at high energy. QED is now considered to be a phenomenological low energy approximation to some more fundamental theory - string theory or whatever (see Steven Weinberg's books on field theory). The Higgs mechanism is no less flimsy and has the extra embarrassment of the non-appearance of the Higgs boson. Maybe some qualification is needed in the article but a dispute over whether the status of the Higgs mechanism is to be described by the words "strong" or "overwhelming" has a chance of being seen as one of Wikipedia's Lamest edit wars. Xxanthippe (talk) 06:14, 18 December 2009 (UTC).

Irrelevant quibble. QED is still one of the most successfull physical theories of all time. --Michael C. Price ^talk 07:16, 18 December 2009 (UTC)

Yes, That is what is so disturbing. (I think, though, that I would rate classical gravity higher). Xxanthippe (talk) 08:39, 18 December 2009 (UTC).

(deindent) I think the essence of this dispute is that you don't like quantum field theory. Yes, QED blows up at distances 100 orders of magnitude smaller than the Planck length. So what? The standard model "blows up" (meaning reaches a unified coupling) at a few orders of magnitude below the Planck length. Who cares? String theory will fix the ultraviolet problem for these, and QCD is fine at all distances, no matter how short.

The Higgs mechanism is the only way that a fundamental gauge boson can get a mass, and that's the W and Z boson. These are established to be gauge bosons with experimental certainty, and the assumption that they are fundamental carriers of the weak interaction predicted their existence and their masses and interactions to many decimal places of accuracy. So there's no way that this is wrong. The Higgs mechanism is a part of this prediction.

The Higgs boson is another story. In the Stuckelberg mechanism, you can take the limit that there is no Higgs boson, in that the Higgs boson becomes infinitely massive, but the mass of the vector stays finite. This can't be done with the standard model scalar Higgs: there's an argument by Weinberg--- you want the self-interaction of the Higgs to not blow up until the Planck scale. But if the Higgs is composite, then pretty much anything goes. The only thing we know for sure is that the W and Z get a mass from some sort of Higgs mechanism, and we know the charge and representation of the Higgs condensate. We have no certainty about the boson itself, it could still be some convoluted pion-like state of some new "technicolor" interaction. That's why we need the LHC.Likebox (talk) 09:51, 18 December 2009 (UTC)

By the way, the perturbative expansion of QED is known to diverge. It's no big deal.Likebox (talk) 09:53, 18 December 2009 (UTC)

Higgs could reveal itself in Dark Matter collisions?

This is mainly to David Schaich, but not exclusively to him or her :- According to a remarkable preprint of Taoso et al., commented as "important physics news" e.g. by http://physicsworld.com/cws/article/news/41218 , there may be a measurable, and thus important, relation between our "well-known" dark matter and the "equally well-known" Higgs field. To my opinion, this relation is very speculative, but not totally, and somehow even natural, and perhaps the paper itself is not the best (I am not the person to referee that). But inspite of this caveat, the ideas in the paper, already commented in the above-mentioned physicsworld article, deserve citation in the Wikipedia. Otherwise one might become accused sometimes to have overlooked, or even suppressed or reverted, an essential piece of progress, which in principle is already in the public. - Best regards, Meier99 (talk) 15:45, 17 December 2009 (UTC)

I agree that it's an interesting paper, (positively reviewed in New Scientist, IIRC), but would it not be more appropriate to the Higgs particle article? --Michael C. Price ^talk 16:05, 17 December 2009 (UTC)

Yes, I agree with your better alternative. In the last-mentioned article, there are also some suitable Feynman diagrams. - Regards, Meier99 (talk) 17:45, 17 December 2009 (UTC)

Can we solve the problem of Higgs mechanism by proving the hypothesized Higgs boson does not exist at all?

Yes, we can. The brief answer is in http://mechanism-revealedphysics-bcz37.blogspot.com/

BCZ§ —Preceding unsigned comment added by BingchengZMRP (talk • contribs) 18:40, 23 March 2010 (UTC)

Mathematical parts

The sections with equations in them have the following structure--- first explain the nonrelativistic case of the Landau model of superconductivity, a charged nonrelativistic field with a vacuum expectation value, then explain the relativistic case, which is a charged scalar field with a vacuum expectation value (the only difference is the relativistic invariance of the condensate--- the calculations are exactly the same), finally do the full-blown nonabelian Higgs mechanism. This is the structure of the article, and recent edits have made it unreadable.

This is more or less the historical order of development, the Landau model of 1961 and the Stueckelberg analysis of 1957 is followed by the Brount-Englert 1964 paper. There is nothing controversial here, and anyone who is confused about the accuracy of any of the statements here is too ignorant to work on this article.69.86.66.128 (talk) 05:22, 27 August 2010 (UTC)

Local symmetries CANNOT be spontaneously broken

I think there are two problems with this article

1) It is a common misconception among students and even young researchers that SSB of local symmetries is possible, this primarily originates from simplified (and sometimes wrong) explanations of the Higgs mechanism in introductory QFT books (Peskin & Schroeder, and many other similar books). As the famous Elitzur theorem says, spontaneously breaking of local symmetries is impossible! See >>Phys. Rev. D 12, 3978–3982 (1975)<<. What is really happening is that the local symmetry is EXPLICITLY broken by a gauge fixing procedure and THEN it is possible to break the remaining GLOBAL symmetry.

This is also mentioned on scholarpedia (Kibble, one of the fathers of the mechanism, is the Curator): >>This mechanism is often said to exhibit "spontaneously broken gauge symmetry". That is a convenient shorthand description, but the terminology is potentially somewhat misleading. The process of quantization requires a choice of gauge, i.e., an explicit breaking of the gauge symmetry. However, the resulting theory does retain a global phase symmetry that is broken spontaneously by the choice of the phase of $\langle \phi \rangle$.<<

2) The introduction indicates that the Higgs mechanism is only connected to particle physics, this is of course not true. It is widely known that the mechanism originates from superconductivity (and first noticed by Anderson), but it is today used in many different Condensed Matter systems (Condensed Matter physicists call it the Anderson-Higgs mechanism). My point is that the mechanism is also widely used in modern theoretical Condensed Matter physics (also in non-superconductors) and the article should also equally discuss this aspect.

A last comment. In this comment section, the user Bakken has written tons of inaccurate, misleading and even wrong statements. Let me correct one of these: Bakken repeatedly claim that there is no fundamental/microscopic theory of superconductivity. This is utterly wrong, for conventional superconductors a very successful theory has been known since 1957 (BCS-theory) and was awarded the noble prize in 1972. For high Tc-superconductors a microscopic theory is not fully developed (although the ingredients are well known), but this is irrelevant wrt. Higgs mechanism.

Element4element4 (talk) 22:16, 21 September 2010 (UTC)

Dear Elemnet4element4, thank you for reading my statements, I am truly flattered :). Although it is not quite clear to me what exactly your point is, but please note that the term spontaneous symmetry breaking refers to the symmetry of the ground state of the system, not the symmetry of the Lagrangian. It can happen, that the Lagrangian of a system has certain symmetry, but its ground state has not. That's precisely what the term "spontaneous symmetry breaking" signifies: the Lagrangian is symmetric, but the ground state is not. Again: it is not the symmetry of the Lagrangian that is broken (you are right here -- it can not be broken), it is the symmetry of the ground state that is broken. Note also that *that* article of Higgs is called "Broken Symmetries and the Masses of Gauge Bosons" :) Bakken (talk) 23:04, 22 September 2010 (UTC)

About the theory of superconductivity: BCS is not an a-priori microscopic solution of quantum-mechanical many-electron problem in a metal. BCS is only a crude phenomenological approximation to this complicated solution. For example, BCS can not predict from the first principles the critical temperature for any real metal. BCS is simply a generic effect in mean-field models of fermionic systems (understood with the Bogoliubov transformation). I personally do not call a mean-field model "a fundamental theory" ;) A Nobel prize for a theory does not necessarily mean that the theory is fundamental. For example, Lev Landau received a Nobel prize for a phenomenological theory of super-fluidity. Bakken (talk) 23:04, 22 September 2010 (UTC)

charged particles

In classical physics, all charged particles have mass as a result of the electromagnetic field surrounding them. (Granted, you have to add general relativity in order to keep that mass finite!) Does this not happen in quantum field theory? Harryjohnston (talk) 04:10, 18 May 2011 (UTC)

Yes, it does. Look up mass renormalization. Dauto (talk) 00:43, 19 May 2011 (UTC)

OK, so why do the W particles need the Higgs mechanism to have mass? (The same question may apply to, e.g., the electron, except that there seem to be conflicting opinions about whether the Higgs mechanism is responsible for fermion mass or not.) Harryjohnston (talk) 01:50, 23 May 2011 (UTC)

In quantum field theory a particle's mass is generated by a specific term in the Lagrangian. The W is a gauge boson which means the Lagrangian has a symmetry called gauge symmetry. It so happens that the mass term for a vector boson does not obey the gauge symmetry so by force a gauge boson must be massless unless the gauge symmetry is broken somehow. The Higgs mechanism provides a method for the gauge symmetry to be (dynamically) broken allowing for a mass term for the Z, the two Ws but not for the photon since the higgs particle does not interact with the photon which therefore remains massless. Dauto (talk) 04:39, 23 May 2011 (UTC)

Sorry, I'm still missing something. The mass that a particle acquires by virtue of being charged is surely due to the term in the Lagrangian describing the interaction between the charge and the electromagnetic field. If this term doesn't have gauge symmetry, how can a gauge boson be charged in the first place? Or does that mean that the Higgs mechanism is necessary to allow the W particles to be charged? Harryjohnston (talk) 20:35, 23 May 2011 (UTC)

At high energies the Lagrangian has a symmetry described by the symmetry group $U(1)_{Y}\times SU(2)_{L}\times SU(3)_{C}$. The first factor is the weak hypercharge gauge symmetry group, the second factor is the weak isospin gauge symmetry group, and the last factor is the strong force gauge symmetry group which is not important for what I'm about to explain. Don't confuse those groups with the flavor isospin and flavor hypercharge which are not gauge groups. The latter are mere accidental symmetries with no dynamics in themselves. Note the absence of the electromagnetic symmetry group. At lower energies through the Higgs mechanism this symmetry is reduced to the smaller group $U(1)_{em}\times SU(3)_{C}$ where both weak isospin and weak hypercharge symmetries have been broken being replaced by the electromagnetic gauge symmetry group as a residual symmetry. Clearly electromagnetism is not independent from hypercharge and isospin which is shown by the Gell-Mann–Nishijima formula: $Q=I_{3}+{\frac {1}{2}}Y.\$, where Q is the charge, I3 is the third component of the isospin and Y is the hypercharge. The point is that eventhough the W+, W-, and Z particles have different charges, seemly violating the isospin symmetry, they all in fact have the same hypercharge but have different values for the third component of their isospin, as would be expected from members of a isospin triplet. Dauto (talk) 16:37, 24 May 2011 (UTC)

Thanks. I think with a bit more thought I'll almost understand that, though I'll probably need to brush up on isospin first. At any rate now I at least know that there definitely is an answer. :-) Harryjohnston (talk) 23:22, 24 May 2011 (UTC)

Lay explanation

Would it be possible to give the gist of the mechanism for the informed lay person? For instance, I know something about physics, but I'm not a physicist; I understand the gist of spontaneous symmetry breaking, quantum probability amplitudes, spacetime curvature etc. But I can't make any sense of this article. --Doradus (talk) 15:15, 27 July 2011 (UTC)

Yes, it would, but physicists are notorious for their failure to communicate through the common language. This points out the actual problem with this article. The article is composed of a series of sentences whose interconnection is disjoint. Literally, each sentence could be written as a stand alone paragraph. A physics article can get technical but only by building upon simpler, if brief, previously stated sentences in topical paragraphs. It isn't necessary to cover the entire field of physics by leaning on principles or words which could be mentioned and highlighted as referential links. For example,

Abelian Higgs Mechanism

"Gauge invariance means that certain transformations of the gauge field do not change the energy at all".

Gauge invariance hasn't been defined nor has it been linked to a definition. The above pseudo definition is vague or misleading. It says, "transformations (do not) change". If you're experienced with these concepts, what is omitted is understood. In wiki we seek to provide explanations that are complete up to concise. That only can be achieved by hyperlink to stay within rigor and brevity, but if a reader has the patience to link out for substance, then the most abstruse concepts can be reified in a short section.

The next sentence which should be a separate paragraph even though it attempts to qualify the pseudo definition is also unclear.

"If an arbitrary gradient is added to A, the energy of the field is exactly the same".

What is "A"? Gradient of what? A physicist knows that it's the EM vector potential, so that if A is replaced by A + d(potential) the resulting equation has the same form (invariance), but the intelligent lay reader may not know.

However, the above may be relegated to mere nitpicks against the below:

"This makes it difficult to add a mass term,"

Add a mass term? What? Weren't we talking about ethereal gauge fields and their invariance? For that matter gauge field hadn't been defined nor linked either. Why do I want to add a mass term? What is a mass term? Indeed, this whole program of seeking to add mass in some electromagnetic form may be specious! Staying within the standard model we would like to present the vogue prevailing view, but we must do so with foundation. Doing so is critical for clarity about the purpose that Higgs, et al, had in creating a mechanism to explain why certain particles which act like massless quantum fields(link required) can suddenly have mass(no link on "mass" because of the ambiguity lurking in the vogue prevailing view).

It gets worse:

"because a mass term tends to push the field toward the value zero".

Push? And, what is a "value zero"? A "value, zero". Different objects. The naive reader may not understand the elision!

Coup de grace:

"But the zero value of the vector potential is not a gauge invariant idea".

Is this the origin of mass? A trick played by nature? Or, is it merely a poorly worded sentence?

"What is zero in one gauge is nonzero in another".

How is this sentence connected with the previous sentence? That has to be explicated.

The article then opens the next paragraph with,

"So in order to give mass to a gauge theory, the gauge invariance must be broken by a condensate".

The word, "condensate" is ambiguous unless the entire article is understood to be referring to the Higgs model form as applied to solid state physics, or that "condensate" refers abstrusely to a vacuum, a ground state, otherwise. These are profoundly different arenas, or are they? Is matter prior to vacuum or vice versa or are these notions categorically disjoint? These are issues of foundations which hardly can be touched in a survey. In any case this article should reside in a section dedicated to solid state physics while a companion form would reside in foundations of the standard model.

Ok. Well, those are some specific criticisms. I guess what we need are some reworded paragraphs that are more accessible to the informed layman. I offer myself as the guinea pig, if someone wants to try to explain this to me. --Doradus (talk) 02:08, 15 August 2011 (UTC)

I'm going to chime in and ask for a non-technical lay description. I don't have a math background, I don't understand the terminology being used, the equations, or sometimes even the context. Considering what's happening in this field right now this article needs some serious attention. --JSleeper — Preceding unsigned comment added by 209.203.71.2 (talk) 03:23, 14 December 2011 (UTC)

I will agree with the previous comments; as it is, the article is basically incomprehensible to a layman: I read it, and I don't know any more about the Higgs mechanism than I did before.

How does the Higgs mechanism give a particle mass? The article not only does not answer this question, it doesn't seem to even address the question. Guy who reads a lot (talk) 04:29, 4 January 2012 (UTC)

Lead Section Cleanup

Only citation in the first paragraph isn't actually a reference. Why is mass in quotes?

The second paragraph uses a lot of Expressions of Doubt, and the second half is vague. Why are the only mention of fermions electrons and quarks? According to to WP's Manual of Style, lead sections should be well cited.

The lead is also kinda lengthy, maybe paragraphs 5-7 could be moved into the history of research section. Thoughts?

Jzampardi (talk) 09:52, 1 February 2014 (UTC)

Discovery of Weak Force

Article states weak force was discovered in 1967, I am pretty sure it was discovered decades earlier than that

Spope3 (talk) 19:07, 6 December 2014 (UTC)

Higgs field

I tried to create a lay article on the Higgs field but was told by an editor that I should add the material here. I think this is wrong, but given the demand for lay material, I have added it here 'for the moment'. If you think there should be a separate Higgs field article please mention this to the editor to try and get him to change his decision [field] Mal (talk) 12:32, 15 January 2012 (UTC)

That's just one editor. There are lots of introductory articles for lay readers in Wikipedia. http://en.wikipedia.org/wiki/Introduction_to_genetics Ostensibly, these articles are supposed to make it easier for a non-technical reader. In reality, all Wikipedia articles are supposed to be understandable by a non-technical reader, but they're often not. WP:AUDIENCE The introductory articles keep editors from arguing with each other over whether an article is too technical or too simple.

The rule that an article should be understandable by a lay reader is strongest for the introduction to the article. If you want to add an understandable introduction, please give it a try. If other editors don't like it, you can fork it over to a non-technical introductory article.

But the introduction here should be much easier for the lay reader. I took a semester of quantum mechanics and a semester of relativity in college (decades ago), and I can't understand anything after "More precisely...." --Nbauman (talk) 19:45, 4 July 2012 (UTC)

To be fair, the introduction is also supposed to explain the actual topic, as well as be a set-up for the rest of the article. Without the necessary background knowledge, it's only reasonable that 'More precisely...' is the place where you'd lose track of things. It is, when being precise, an exceedingly complex subject. Anyways, maybe the article they meant to send you to was Introduction to the Higgs field. Darryl from Mars (talk) 23:51, 6 July 2012 (UTC)

then perhaps someone more daring than I should insert that very link somewhere near the top of this article. as it is, the article is impenetrable to a lay-reader, & contains no suggestions for an introductory read.

duncanrmi (talk) 12:56, 28 February 2017 (UTC)