Are Photons Goldstone Bosons?

Question

I'm not interested in more speculative ideas, like the one of [Bjorken et. al.][1] that photons are Goldstones of broken Lorentz symmetry. 

Instead, I want to understand if photons are simply the Goldstones of the spontaneously broken large gauge symmetry? (Large gauge symmetry here simply means those gauge transformations that do not become trivial at infinity.)

I recently read Stromingers "[Lectures on the Infrared Structure of Gravity and Gauge Theory"][2], where he argues that

 large gauge symmetry is spontaneously broken, resulting in an
infinite vacuum degeneracy with soft photons as the Goldstone bosons.

Moreover I discovered that already in the 70s there were quite a few papers that argued that photons are simply the Goldstones of the broken asymptotic gauge symmetry.

For example:

•  [Gauge invariance and mass][3] by Richard A. Brandt and Ng Wing-Chiu Phys. Rev. D 10, 4198; who argued that "the physical photon can be interpreted as a Goldstone boson arising from the spontaneous breakdown of the R -transformation invariance." (R-transformations is simply another name for gauge transformations that are non-trivial at infinity)
• [Spontaneous breakdown in quantum electrodynamics][4] by R.Ferrari L.E.Picasso Nuclear Physics B Volume 31, Issue 2, 1 September 1971, Pages 316-330: "In the context of quantum electrodynamics we discuss the spontaneous breakdown of the symmetry associated to gauge transformations of the second kind, with gauge functions linear in the coordinates. We show that the photons (both physical and unphysical) can be considered as the Goldstone particles of this symmetry, and that the Ward identity and, in general, all self-photon theorems, are dynamical consequences of the spontaneous breakdown of the symmetry considered."

This seems like a well established fact. For example, in a [recent paper][5] by Yuta Hamada, Sotaro Sugishita they noted in passing:

 The statement that photons and gravitons are NG bosons is not new and it is discussed in [20, 21, 22, 23].

I really like this perspective. However, I am a bit confused, because no textbook and almost no paper mentions this although the papers quoted above are 40+ years old. 

  [1]: https://arxiv.org/abs/hep-th/0111196
  [2]: https://arxiv.org/abs/1703.05448
  [3]: https://journals.aps.org/prd/abstract/10.1103/PhysRevD.10.4198
  [4]: http://www.sciencedirect.com/science/article/pii/0550321371902355?via%3Dihub
  [5]: https://arxiv.org/abs/1709.05018

Comments

If it has been proven many times, then why do you ask again this question? Have you found (a) loop hole(s) in the previous proofs and arguments?

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@Vladimir Kalitvianski I asked, because I was not sure if this is an established fact or not. (I'm pretty sure that most theoretical physicists do not know this, for whatever reason.) Is there any textbook that discusses this?

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I do not know any textbook that discusses this (maybe in Russian there are some, following N.N. Bogolubov's 1/q-theorem in quantum statistical physics).

I did not get why Strominger calls them "soft photons" - all photons, soft and hard ones, are massless. And I do not understand why this perspective is so attractive to you?

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Thanks for your comment!

Okay that's strange. Now I'm really curious why this is not "textbook knowledge".

I was confused by this specification to "soft photons", too... 

I like this perspective, because it illustrates nicely why photons are massless and yields insights into what the Higgs mechanism is really doing. Usually people claim that "local $U(1)$ gauge invariance" is the reason that photons are massless. However, local gauge invariance is simply a redundancy in our description and can be "modded out". In contrast large gauge transformations are physical (c.f. strong CP problem) and only the spontaneous breaking of this physical symmetry leads to massless particles, i.e. the photons as Goldstone bosons.@VladimirKalitvianski 

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@VladimirKalitvianski: Soft photons are those that go into the coherent state photon cloud of an electron, while hard photons are those that are detectable as photons.

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The main reason why this is not discussed in textbooks is that few have an extensive treatment of the infrared problem. Strominger's research brought it again into the foreground.

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I belong to an old generation of physicists and in my opinion the masslessness of photons follows from the Maxwell equations for the propagating to infinity electric and magnetic fields (strengths). Any other explanation is a secondary - like some property of a massless theory which cannot be a true reason, but just a property. Besides, photons are always coupled to charges, either to the "sourcing charges" or to the "sucking ones". In the latter case the photons are just external fields for probe charges, which scatter or absorb external photons, depending on their being free or coupled to some other charges. For some reason we forget it and think of photons as of a gas/flux of massless particles existing per se.

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@ArnoldNeumaier Thanks for the clarification. So, only soft photons are Goldstones of the asymptotic gauge symmetry, and hard photons aren't?

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@VladimirKalitvianski: Hard photons (produced by a laser, say) are not coupled to charges, in the sense that they are not external fields of any charge.

The photon field is the Goldstone field of  the asymptotic gauge symmetry, independent of how or whether it is interpreted in a particle language.

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@ArnoldNeumaier: I am surprised to see that you call the near field (for example, that with the strength $E_{{\rm{near}}}\propto 1/R^2$) a "cloud" of "coherent states of photons", and the propagating to infinity field ($E_{{\rm{prop}}}\propto 1/R$) are "hard" ones as they are "detectable". All frequencies may propagate to infinity, so photons are of all frequencies and are detectable (long radio waves is an example). All frequencies may be present in the near field too and the sum of them gets into the equations of motion of detecting device charges. Being sourced means being coupled to the sourcing charges. Being detected means being coupled to the probe charges ("absorbers"). There is no other meaning of fields/photons and their properties. Otherwise (i.e., when non detectable) they are void of physical meaning.

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@VladimirKalitvianski:

1. The e/m field corresponding to soft photons extends (in vacuum) to infinity, too. This has nothing to do with frequency. Hard photons may have any frequency where they are detectable.

2. After production, laser light is no longer coupled to the source. Neither is light from distant stars when we perceive it. 

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@ArnoldNeumaier: No, the Coulomb, or more generally the retarded near field is not detectable at infinity: it is just too weak (zero). It has no effect at all at infinity. The propagating photons, on the contrary, may be felt at any distance: they may cause excitations, chemical reaction, heating, etc.

After production, the laser pulse has no meaning if it does not get into a measuring device. You even cannot judge about your laser efficiency without measuring the distribution the radiated energy and the lost energy in the total energy supplied to the laser. Measuring is always implied explicitly or implicitly.

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@VladimirKalitvianski: The mass of the moon has a physical meaning even when nobody measures it. So has the energy of a laser pulse. 

A near field is unmeasurable at infinity by definition. But a soft photon cloud is not a near field but the stuff that modifies a bare electron as asymptotic (scattering) state, and hence at infinity. 

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@ArnoldNeumaier: Just tell clearly whether your soft photon cloud (by the way, why "cloud" and around what?) is measurable at all. Do not replace the electron state measurement with the soft photon measurement. If "at infinity" one measures the electron field "at infinity", where is the cloud? In vicinity of the electron or far away from the electron?

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@ArnoldNeumaier  a side issue but this  MIT   video I think contradicts  the statement "Hard photons (produced by a laser, say) are not coupled to charges," . Energy conservation forces   (8'15") the conclusion that the photons are connected to the source, the whole experimental system is a lasing set up.

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@ArnoldNeumaier: I order to speak of Moon, one has first to establish experimentally what it is from many points of view and only after that extrapolate its properties to the future, when one does not look at the Moon. Such extrapolations do not always work, just because the Moon is not a mathematical unchanged point, but a real physical object observable precisely due to its changes (radiation and absorption of photons). Such changes are relatively small energetically w.r.t. the kinetic energy of the Moon, but for small objects like electron such changes are important. That is why we distinguish (and define) experimentally elastic, inelastic, and inclusive pictures of the observed electron. In nature there is no bare electron, so in the theory we must proceed from this fact and do not impose our imperfect ideas about interactions as the true ones.

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@annav: Photons are produced by a laser and of course in that moment interacting with it, but a few moments later they are independent of the source. Light received from a distant star is certainly not interacting with its source. 

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@VladimirKalitvianski: The photon cloud is an intrinsic part of the physical electron (like for any infraparticle). It is only a figure of speech. The physical electron states are the asymptotic states formed by the product of a free electron wave function and an asymptotic coherent state of the electromagnetic field. The latter can be formally decomposed into soft virtual photon modes which traditionally gave rise to the visualization as a soft photon cloud. 

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@ArnoldNeumaier: Just tell me whether the "free electron" and "the coherent EMF" you are speaking of are asymptotically far away from each other or not. If they are far away physically, there is no background for the notion of "cloud". If they come together, they cannot obey free equations except for the case of my electronium construction with its specific physics.

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@VladimirKalitvianski: They are concentrated along the asymptotic trajectory of the physical electron - the latter carries with it a Coulomb-like field. Note that virtual photons generally don't obey any equations of motion; they are just products of the approximation scheme used. 

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@ArnoldNeumaier: Thank you for your clarifying answer. But your statement that EMF does not have any equation of motion contradicts the Maxwell achievements.

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@VladimirKalitvianski: Maxwell achieved nothing for quantum fields. Moreover, I didn't claim that the electromagnetic field has no equations of motions - only that virtual particles (and hence the photon cloud) don't.

The interacting electron/positron and electromagnetic fields are not asymptotic quantities and obey operator product expansions, which are the relativistic QFT analogue of the equation of motion. And the coherent state factor in the asymptotic electron wave function is represented in the interaction picture, and hence moves freely with the center of mass of the electron. 

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@ArnoldNeumaier: If the virtual photons are a synonymous to the retarder near field, then they have an explicit analytical expression obeying some equation of motion. Coherent photon states obey the classical equation of motion since they are a solution of a classical sourcing current J.

I must quit this discussion as I am fed up with it.

Answer 1

Check out this paper: https://arxiv.org/abs/1412.5148 It is more or less a simple application of the Goldstone/Noether procedure. If you apply the charge operator ($\star F$) to a Coulomb ground state you produce a photon.

By the way it is not good to call these "large gauge transformations" because it is really a global symmetry, but with connections as parameters.

Comments

@RyanThorngren One more thing: you wrote "gauge transformation is a redundancy in the mathematical description of the physics. It is not physical." I think this is not correct, unless you use a different definition of gauge transformation than I do. (Which I'm pretty sure you do). In my understanding, only local gauge transformations, i.e. only those which are non-trivial within a finite volume are non-physical and mere redundancies. The other gauge symmetries I quoted above are physical:

• The global gauge symmetry is responsible e.g. for the conservation of electrical charge.
• The spontaneously broken asymptotic gauge symmetry is responsible for the massless excitations we call photons. 
• Large gauge transformations are responsible for the non-trivial vacuum of QFTs like QCD. ($\theta$-term, instantons, etc.)

I'm pretty sure that you know and agree that these kind of "gauge" transformations are physical. I'm really interested in how you call these transformations, i.e. a map from my terminology to yours. (Especially, because I tried to read the papers you quoted, but unfortunately had problems understanding the terminology. In particular the physical motivation of these higher q-form symmetries is unclear to me. Although now I suspect these are simply what I call asymptotic gauge symmetries etc. above. ) 

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@JacobS you are right, there is some care required with the definitions, and some of the popular but blunt statements about the nature of gauge transformations need further qualification.

Standard accounts, e.g. Henneaux-Teitelboim's "Quantization of gauge systems" give the definition of gauge transformations as parameterized variations that leave the Lagrangin invariant up to a totdal spacetime derivative. For compactly supported choice of gauge parameters this happens to be examples of what in other texts are called "gauge transformations", namely compactly supported symmetries of the Lagrangian, up to total derivatives.

But for non-compactly supported gauge parameters, we still get something that needs to be called "gauge transformation".

In my A first Idea of Quantum Field Theory I speak of "gauge symmetry" for the compactly supported symmetries, and of "gauge-parameterized gauge symmetry" for the other case. (Keeping in mind that, despite the terminology, neigher definition in general subsumes the other, but that's just how it is).

Another point to notice is that, with either of the two definition, one has to be careful with the ever popular slogan "gauge symmetry is just a redundancy". This is in general too naive, namely when there are more that one gauge symmetries connecting any two gauge equivalent configuration.

If one strictly belives that all gauge symmetry is redundancy, then for instance locality of field theory breaks (exposition pdf)

There is related discussion on PhysicsForums right now, around here

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@VladimirKalitvianski I have to admit that I am at a loss seeing how your comment is at all related to anything I wrote in my comment, or, for that matter, to anything at all.

Your first sentence has the words "some classifications according to some criteria": apart from me not having mentioned any classification of anything, what could be more vague of you than speaking of "some classification according to some criteria" without specifying any context? I have not the least of a clue what you are referring to here.

Next you say "even in Classical Filed Theory the exact physically reasonable solutions do not exist". This seems content-free to me I just can't see what you mean to be saying. What is "the" even referring to?

Before getting to the point that I could agree or disagree with you on anything, I have trouble parsing your message to anything of semantic content.

Why is that? It's not the first time, I have this problem with many of your messages. I am wondering if there is a way that you could improve your communication method, for it would help this forums here, given your high rate of activity.

My suspicion is that you are internally captured in some old arguments that you have once been involved in, with some other people, and are providing counter-arguments here in reply to statements that once were made elsewhere, by other people. Could this be?

Maybe @ArnoldNeumaier can help? In other places he seems to have been able to understand what you have in mind.

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@UrsSchreiber Thanks for your comment. I've read your recent introductions (the PhysicsForums Insights, and in the nLab) and liked them. I think it's a real problem that the few people who do try to get these things like gauge symmetries, global symmetries etc. precise do not talk to each other and therefore do not seem to speak a common language. Thus it's no wonder that almost everyone seems extremely confused.  So, to come back to my original question: are you able to translate the approaches by Strominger or also the approach by Kapustin, Seiberg, et. al. into your language? Would you agree with the statement that gauge bosons are Goldstones? (And if yes, in your language, of which broken symmetry?)

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@VladimirKalitvianski you write " I am clearly speaking of existence or non existence of physical solutions of a Theory equations/Lagrangian you make some transformations of. I suppose you imply existence of such solutions. Can you confirm my supposition, please?"

Sure, we are talking about examples like the Yang-Mills Lagrangian and the Chern-Simons Lagrangian, whose equations of motion are well known and whose solution spaces are well studied.

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@JakobS I still have to study these articles to answer these questions. Maybe later.

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Dear Urs, it was you who taught me that the "gauge transformations between gauge transformations" are redundancy in the redundancy (and so on). One imagines something like a chain complex of gauge symmetries, and the global symmetries are its cohomology.

JakobS, I do not know the precise relationship between 1-form symmetries and asymptotic symmetries but one observation is that the conserved charges are the same, namely surface integrals of F and \star F. We can leverage this observation to propose a thought experiment. We know the 1-form symmetries are spontaneously broken. This implies a (rather large) ground state degeneracy. A natural question is "which of these ground states are coherent?" in the sense "which are classically observed?". Take spontaneous 0-form symmetry breaking as an example. Let Q be the charge operator (a volume integral of j^0 over all space). In a finite size system, there is an exponentially small splitting between energy eigenstates of definite Q, with the neutral Q as the lowest energy state. In the infinite size limit, this degeneracy goes quickly to zero and we see degenerate ground states that are characteristic of spontaneous symmetry breaking. However, these degenerate ground states are distinguished by a global operator Q which is impossible to measure. Instead, there are *local* order parameters f which [Q,f] not 0 which we can measure, and when we do, because of this noncommutation, we completely decohere the charge in the observed vacuum state. The states of definite Q are called "cat states" in the literature by a colorful analogy with Shroedinger's famous feline.

So in the case of our 1-form symmetries, Q is now a surface integral and f is something like a Wilson line. In the usual classical gauge theory basis (where we expect to see photons), the Wilson line is a diagonal operator e^i \int A and so our observed states have indefinite Q. However, suppose we only have experimental access to a finite region of spacetime (a causal diamond, say). Then only the Q and f which intersect can noncommute, and we can only measure f in our small region. Therefore, we can imagine a preparation of the system where we observe states with both gauge bosons and definite Q *at infinity*. I think these are the states Strominger et al are talking about. See also this paper https://arxiv.org/abs/1709.08632 which shows that indeed the asymptotic charges are precisely those which are inaccesible to our experiments.

Besides being incredibly impressionistic, this discussion still surffers from the problem that on a contractible space there are no nontrivial 1-form symmetry generators. Maybe one needs to think about a causal diamond inside a torus or something? I don't know. At best it seems like the two approaches are clearly related but not the same. I would guess that Strominger's approach is the shadow of the 1-form symmetry approach in contractible space. I'm happy to discuss more, over email, say.