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When Yang-Mills field theory was introduced, a problem is that the gauge invariance can not allow mass for the gauge field. Later people invented spontaneous symmetry breaking and Higgs mechanism to give the gauge field mass. The Higgs particle is almost confirmed at LHC.

My question is, since there is symmetry non-conservation (P/CP) in nature, why not simply put a mass on Yang-Mill's theory directly, got a non-abelian Proca action, say gauge symmetry breaking (although gauge may not be a symmetry actually Gauge symmetry is not a symmetry?; actually this point is more subtle, one can also take a Stueckelberg action then fixing the gauge, it leads to the same Lagrangian)? Is there any theoretical reason for not adding mass by hand on Yang-Mills theory? Or just because Higgs particle was found, it works, that's it.

My friend has a guess, that gauge invariance implies BRST symmetry, which restricts the possible form of Lagrangian. If one did a renormalization flow transformation to lower energy scale without BRST symmetry, there will be other coupling in the effective Lagrangian at lower energy scale. I am not sure about this reasoning, because BRST symmetry can restrict the possible counter terms, can it also restrict the possible terms in the effective Lagrangian?

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Related: – Qmechanic Jul 25 '14 at 15:25

3 Answers 3

In a quantum theory, gauge symmetry is an inevitable consequence of Poincare invariance and long range interactions at the classical level (the weak and strong interactions aren't long range because of quantum effects, such as confinement and the Higgs mechanism). If one "breaks" a gauge symmetry (what it doesn't have much since since gauge symmetries are mathematical ambiguities rather than physical symmetries), the one has to give up either:

  1. Poincare invariance.
  2. Existence of a normalizable vacuum state (or existence of states with negative norm). This prevents the probabilistic interpretation of quantum mechanics.

Note that breaking a gauge symmetry is different from formulating a theory without gauge invariance. For example, classical electrodynamics in terms of the electric and magnetic field doesn't have a gauge symmetry, but it doesn't break it.

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About 2, is there any problem with quantizing Proca field? – user26143 Jul 18 '13 at 23:17
You can not say that "In a quantum theory, gauge symmetry is an inevitable consequence of Poincare invariance", since gauge symmetry are internal symmetries independent from spacetime coordinates and may exist in addition to the Poincare-invariance of a theory, independent to whether the theory relavisitically spacetime invariant or not! The Proca field is massive and thus not that difficult to quantize as the Maxwell field. The book "Field quantization" from Walter Greiner shows for example how to quantize Maxwell and Proca field. – Hansenet Jul 18 '13 at 23:38
The objects on which gauge symmetry have a defined transformation property under Lorentz transformations. The crucial property required in QFT Lagrangians is Lorentz-invariance and renormalizability. Gauge invariance is a possible additive feature of a theory. – Hansenet Jul 18 '13 at 23:41
Hello @Hansenet . I can say that. Note that I'm talking about an interacting theory with long range interactions in 3+1 dimension, etc. Note that one needs a massless vectorial field. If one starts with a massless vectorial field that contains additional (more than 2) non-physical polarizations, then one needs gauge invariance to decouple them (otherwise the theory has pathologies). If, however, one prefers to begin with a field with only the 2 physical polarizations, then one needs gauge invariance to preserve Poincare invariance. – drake Jul 19 '13 at 0:06
@user26143 Almost any introductory book to QFT that deals with QED will explain that one needs gauge invariance to decouple unphysical polarizations to avoid negative norm states or absence of a vacuum state. For instance, Greiner's book, the one mentioned by Hansenet. You can read a couple of papers by Weinberg in the 60s where he shows that if one starts with just the 2 physical polarizations, one needs gauge invariance to preserve Lorentz invariance. In his QFT book, it is also sketched. – drake Jul 19 '13 at 0:12

In general Gauge Theories, abelian or non-abelian, mass terms are not by construction forbidden. If one has a chiral gauge symmetry, that is a gauge symmetry in which left and right handed particles transform differently under the gauge transformation, then mass terms will inevitably destroy the gauge symmetry and thus are forbidden. The most famous example is the SU(2)xU(1) symmetry of the electroweak symmetry, where on invokes the mechanism of spontaneous symmetry breaking (called Higgs Mechanism induced by the Higgs field) to allow mass terms for the fermions. Any other way to introduce mass terms in this chiral theory destroys the gauge symmetry! In QCD like QED there is a left-right symmetry under gauge transformation, so mass terms are allowed at least at what concerns the gauge symmetry.

In order to quantize a not spontaneously broken gauge symmetry where the gauge bosons remain massless, one is forced to fix the gauge of the lagrangian to get a physical and sensitive theory. This gauge fixed lagrangian, however has still the BRST-symmetry, which is, if one looks at the infinitesmal transformation properties of BRST-transformation, a special gauge transformation with a nilpotent transformation parameter.

What this BRST-symmetry actually does in pertubative calculations is to ensure that only physical degrees of freedom appear in asymptotical particle states, i.e. particles that are created in in-and out-states of the S-matrix in some scattering process. The nilpotent BRST-symmetry operator sorts the states on which it acts into different state spaces, depending on whether they are physical or not.

If you are particlarly interested in BRST-symmetry, I can recommend you the textbooks by 1.Peskin & Schroeder (An Introduction to Quantum Field Theory). 2.Steven Weinberg (The quantum theory of fields, Volume II), 3.Mark Sredenicki (Quantum Field Theory, a more readible introduction than the books by Peskin and Weinberg in my opinion).

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My question is, why do we need to protect gauge symmetry – user26143 Jul 18 '13 at 22:33

Because the basic concept of a gauge theory is its invariance under a gauge symmetry, which is realized in nature. Without a gauge symmetry gauge bosons would be useless and dispensible. But the crucial fact is that these gauge bosons are observed in nature and the way they interact with matter content build out of fermions has been experimentally proved throughoutly and confirmed with great accuracy. The great experimental success of the application of gauge theories to the strong and the electroweak interactions has established these gauge theories. In physics you may always choose out of different model that one that describe the phenomenological observations the best. And in the interactions between fundamental particle the model of interactions based of gauge theories one has found the best theory to fit the experimental data.

The reason why we protect gauge symmetry is thus that it seems to be a fundamental ingredient in the building plan of nature which restricts together with renormalizability the possible terms in lagrangians describing the nature the way that we observe it experimentally. So nature itself gives you the most powerful alibi, which gauge symmetry should be preserved.

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No that's not it at all. If you put a mass term in Yang-Mills by hand you end up violating unitarity in scattering at high energy. You can get away with it in an abelian theory by the Stueckelberg trick because the longitudinal mode becomes weakly coupled. But in the nonabelian case the longitudinal modes become strongly coupled and you either have to break Lorentz invariance or unitarity. – Michael Brown Jul 19 '13 at 0:14
I meant mass terms for the fermions not the bosons. Of course I know that mass terms for gauge bosons break the gauge symmetry and thus are not allowed. – Hansenet Jul 19 '13 at 0:20
@Michael Brown, I heard the problem of non-abelian Stueckelberg action is either breaking renormalizability or unitarity… although for effective theory, renormalizability is not crucial. Has non-abelian Stueckelberg action already been ruled out by LHC? – user26143 Jul 19 '13 at 0:27
This is exactly what I was looking for! Unfortunately, that URL doesn't work any more. For future visitors of this page: The correct URL is now – Noiralef Sep 26 at 12:22

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