Gauge symmetry is not a symmetry?

Question

I have read before in one of Seiberg's articles something like, that gauge symmetry is not a symmetry but a redundancy in our description, by introducing fake degrees of freedom to facilitate calculations.

Regarding this I have a few questions:

1. Why is it called a symmetry if it is not a symmetry? what about Noether theorem in this case? and the gauge groups U(1)...etc?
2. Does that mean, in principle, that one can gauge any theory (just by introducing the proper fake degrees of freedom)?
3. Are there analogs or other examples to this idea, of introducing fake degrees of freedom to facilitate the calculations or to build interactions, in classical physics? Is it like introducing the fictitious force if one insists on using Newton's 2nd law in a noninertial frame of reference?

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Revo

Comments

As it was mentioned, I just recommend to pay more attention to the phrase "This implies for example the conservation of the electric charge irrespective of the equation of motion." in David Bar Moshe answer.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Misha

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This is a great question, but the answers are misleading. There is always a global part to the gauge symmetry which is a real symmetry. The Noether theorem gives you a current which is conserved due to the equations of motion, and there are conserved quantities associated to boundary transformations.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Ron Maimon

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While gauge symmetry is, of course, classical and seems no quantum content, gauge symmetry breaking is purely quantum. This "correction" (or breaking) is a profound quantum phenomenon.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user user15692

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Duplicate: physics.stackexchange.com/questions/83735/…

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Prahar

Answer 1

In order:

1. Because the term "gauge symmetry" pre-dates QFT. It was coined by Weyl, in an attempt to extend general relativity. In setting up GR, one could start with the idea that one cannot compare tangent vectors at different spacetime points without specifying a parallel transport/connection; Weyl tried to extend this to include size, thus the name "gauge". In modern parlance, he created a classical field theory of a $\mathbb{R}$-gauge theory. Because $\mathbb{R}$ is locally the same as $U(1)$ this gave the correct classical equations of motion for electrodynamics (i.e. Maxwell's equations). As we will go into below, at the classical level, there is no difference between gauge symmetry and "real" symmetries.

2. Yes. In fact, a frequently used trick is to introduce such a symmetry to deal with constraints. Especially in subjects like condensed matter theory, where nothing is so special as to be believed to be fundamental, one often introduces more degrees of freedom and then "glue" them together with gauge fields. In particular, in the strong-coupling/Hubbard model theory of high-$T_c$ superconductors, one way to deal with the constraint that there be no more than one electron per site (no matter the spin) is to introduce spinons (fermions) and holons (bosons) and a non-Abelian gauge field, such that really the low energy dynamics is confined --- thus reproducing the physical electron; but one can then go and look for deconfined phases and ask whether those are helpful. This is a whole other review paper in and of itself. (Google terms: "patrick lee gauge theory high tc".)

3. You need to distinguish between forces and fields/degrees of freedom. Forces are, at best, an illusion anyway. Degrees of freedom really matter however. In quantum mechanics, one can be very precise about the difference. Two states $\left|a\right\rangle$ and $\left|b\right\rangle$ are "symmetric" if there is a unitary operator $U$ s.t. $$U\left|a\right\rangle = \left|b\right\rangle$$ and $$\left\langle a|A|a\right\rangle =\left\langle b|A|b\right\rangle $$ where $A$ is any physical observable. "Gauge" symmetries are those where we decide to label the same state $\left|\psi\right\rangle$ as both $a$ and $b$. In classical mechanics, both are represented the same way as symmetries (discrete or otherwise) of a symplectic manifold. Thus in classical mechanics these are not separate, because both real and gauge symmetries lead to the same equations of motion; put another way, in a path-integral formalism you only notice the difference with "large" transformations, and locally the action is the same. A good example of this is the Gibbs paradox of working out the entropy of mixing identical particles -- one has to introduce by hand a factor of $N!$ to avoid overcounting --- this is because at the quantum level, swapping two particles is a gauge symmetry. This symmetry makes no difference to the local structure (in differential geometry speak) so one cannot observe it classically.

A general thing -- when people say "gauge theory" they often mean a much more restricted version of what this whole discussion has been about. For the most part, they mean a theory where the configuration variable includes a connection on some manifold. These are a vastly restricted version, but covers the kind that people tend to work with, and that's where terms like "local symmetry" tend to come from. Speaking as a condensed matter physicist, I tend to think of those as theories of closed loops (because the holonomy around a loop is "gauge invariant") or if fermions are involved, open loops. Various phases are then condensations of these loops, etc. (For references, look at "string-net condensation" on Google.)

Finally, the discussion would be amiss without some words about "breaking" gauge symmetry. As with real symmetry breaking, this is a polite but useful fiction, and really refers to the fact that the ground state is not the naive vacuum. The key is commuting of limits --- if (correctly) takes the large system limit last (both IR and UV) then no breaking of any symmetry can occur. However, it is useful to put in by hand the fact that different real symmetric ground states are separately into different superselection sectors and so work with a reduced Hilbert space of only one of them; for gauge symmetries one can again do the same (carefully) commuting superselection with gauge fixing.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user genneth

Comments

+1 Great answer!

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Heidar

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when i try to browse your personal blog, i get a "Unknown control sequence '\Gam'"

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Larry Harson

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+1; really clear and to the point, thanks!

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Slaviks

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I didn't ask why it is called gauge symmetry. I was asking about how if gauge symmetry is not a symmetry, then how the gauge groups are not a symmetry group either! That is what I do not understand

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Revo

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@Revo: in classical field theory, they are symmetries. David Bar Moshe below explains how Noether's theorem works in this case. This is not the case in a quantum theory. People kept the terminology even though now we understand better how things work.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user genneth

Answer 2

The (big) difference between a gauge theory and a theory with only rigid symmetry is precisely expressed by the Noether first and second theorems:

While in the case of a rigid symmetry, the currents corresponding to the group generators are conserved only as a consequence of the equations of motion. This is called that they are conserved "on-shell", in the case of a continuous gauge symmetry, the conservation laws become valid "off-shell", that is independently of the equations of motion. This implies for example the conservation of the electric charge irrespective of the equation of motion.

Now, the conservation law equations can be used in principle to reduce the number of fields.

The procedure is as follows:

1. Work on the subspace of the field configurations satisfying the conservation laws. However, there will still be residual gauge symmetries on this subspace. In order to get rid of those:

2. Select a gauge fixing condition for each conservation law.

This will reduce the "number of field components" by two for every gauge symmetry. The implementation of this procedure however is very difficult, because it actually requires to solve the conservation laws, and moreover, the reduced space of field configurations is very complicated. This is the reason why this procedure is rarely implemented and other techniques like BRST are used.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user David Bar Moshe

Comments

Can you give a reference for such a calculation where by a physically conserved quantity is derived from local gauge symmetries? I would think that is impossible since after all gauges can be fixed and there would be no remnant symmetry but nothing physical would have changed either! I would have thought that all conservation laws needs the variation of the action (w.r.t the deformation parameters) to be evaluated on the solutions and hence conservation is always on-shell. That is my understanding of what happens even for non-Abelian gauge field theory.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user user6818

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@Anirbit, Sorry for the late response. The following reference discussing Noether's second theorem: nd.edu/~kbrading/Research/WhichSymmetryStudiesJuly01.pdf Let's consider for definiteness a gauged Klein-Gordon field theory. The equation of motion of the gauge field is $\partial_{\nu}F_{\mu \nu} = J_{\mu}$, where $J_{\mu}$ is the Klein-Gordon field current: $i(\bar{\phi}\partial_{\mu}\phi - \phi\partial_{\mu}\bar{\phi})$.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user David Bar Moshe

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Cont. Thus this current is conserved when the gauge field satisfies its equation of motion, the matter field needs not satisfy its equation of motion for the conservation. Thus, one may say that the current conservation requires only the gauge fields to be on-shell. But this is not the whole story; the time component of the gauge field equations of motion is the Bianchi identity (or the Gauss law).

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user David Bar Moshe

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Cont. The Lagrangian doesn't contain a time derivative for the time component of the gauge field. This component appears as a Lagrange multiplier times the Gauss law, thus its equation of motion is not dynamical, it just describes a constraint surface in the phase space expressing the redundancy of the field components. Thus the conservation of the time component of the Klein-Gordon current i.e., the charge (after integration over the 3-volume) is not dependent on any equation of motion of the "true" degrees of freedom.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user David Bar Moshe

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Dear @DavidBarMoshe: Minor thing. It seems to me that the Klein-Gordon field current should depend on the gauge potential, cf. this Phys.SE answer.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Qmechanic

Answer 3

1) Why is it called a symmetry if it is not a symmetry? what about Noether theorem in this case? and the gauge groups U(1)...etc?

Gauge symmetry is a local symmetry in CLASSICAL field theory. This may be why people call gauge symmetry a local symmetry. But we know that our world is quantum. In quantum systems, gauge symmetry is not a symmetry, in the sense that the gauge transformation does not change any quantum state and is a do-nothing transformation. Noether's theorem is a notion of classical theory. Quantum gauge theory (when described by the physical Hilbert space and Hamiltonian) has no Noether's theorem. Since the gauge symmetry is not a symmetry, the gauge group does not mean too much, in the sense that two different gauge groups can sometimes describe the same gauge theory. For example, the $Z_2$ gauge theory is equivalent to the following $U(1)\times U(1)$ Chern-Simons gauge theory:

$$\frac{K_{IJ}}{4\pi}a_{I,\mu} \partial_\nu a_{J,\lambda} \epsilon^{\mu\nu\lambda}$$ with $$K= \left(\begin{array}[cc]\\ 0& 2\\ 2& 0\\ \end{array}\right)$$ in (2+1)D.

Since the gauge transformation is a do-nothing transformation and the gauge group is unphysical, it is better to describe gauge theory without using gauge group and the related gauge transformation. This has been achieved by string-net theory. Although the string-net theory is developed to describe topological order, it can also be viewed as a description of gauge theory without using gauge group.

The study of topological order (or long-range entanglements) shows that if a bosonic model has a long-range entangled ground state, then the low energy effective theory must be some kind of gauge theory. So the low energy effective gauge theory is actually a reflection of the long-range entanglements in the ground state.

So in condensed matter physics, gauge theory is not related to geometry or curvature. The gauge theory is directly related to and is a consequence of the long-range entanglements in the ground state. So maybe the gauge theory in our vacuum is also a direct reflection of the long-range entanglements in the vacuum.

2) Does that mean, in principle, that one can gauge any theory (just by introducing the proper fake degrees of freedom)?

Yes, one can rewrite any theory as a gauge theory of any gauge group. However, such a gauge theory is usually in the confined phase and the effective theory at low energy is not a gauge theory.

Also see a related discussion: Impossibility of breaking gauge-symmetry in lattice gauge theories

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Xiao-Gang Wen

Comments

@Xiao-GangWen: Why do you think that a gauge symmetry (that goes to the identity in the boundary) is a true symmetry in classical physics? In my opinion, in neither case it is a true symmetry, but only a redundancy in the description. Thank you in advance.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user drake

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@drake: I think I agree with you: gauge symmetry is not a true symmetry even in classical physics. But it could be viewed as a symmetry (ie local symmetry) in classical physics. Gauge symmetry cannot be viewed as a symmetry in quantum physics.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Xiao-Gang Wen

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@Xiao-GangWen could you describe the terms in the expression that you wrote down for the $U(1)$ x $U(1)$ Cherns-Simons gauge theory? In particular I am unfamiliar with some of the terms, namely $a$ and the matrix $K$.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user sunspots

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@Xiao-GangWen if this notation is from a paper, could you direct me to the paper?

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user sunspots

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@Airwoz: for the U(1) x U(1) Cherns-Simons gauge theory, see arXiv:0803.2300 Mutual Chern-Simons theory for Z_2 topological order Su-Peng Kou, Michael Levin, Xiao-Gang Wen. I also wrote a book which explain gauge symmetry is not a symmetry: Quantum Field Theory of Many-Body Systems ---from the Origin of Sound to an Origin of Light and Electrons

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Xiao-Gang Wen

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Better write questions as own questions/post, than as an answer here - its, not a forum (even though that would have some advantages).

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user NiftyKitty95

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@ Jook: There are three kinds of gauge theories: (1) Classical gauge theory where both gauge field and charged matter are treated classically. (2) fake quantum gauge theory where gauge field is treated classically and charged matter is treated quantum mechanically. (3) real quantum gauge theory where both gauge field and charged matter are treated quantum mechanically. Most papers and books deal with the fake quantum gauge theory, and so does your question/answer it seems. My answer deals with the real quantum gauge theory, which is very different.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Xiao-Gang Wen

Answer 4

When talking about symmetry, one should always indicate: symmetry of what?

If I measure the length of a stick in inches and then in centimeters, i.e. in different gauges, then I get two different answers, although the stick is the same in both cases. Similarly, when I measure the phase of a sine wave with two clocks that have different phases, then I get two different phases, and phase shifts form the group U(1). In the first example the stick is invariant under the change of gauge from centimeters to inches, but this has nothing to do with any physical symmetry of the stick. Noether's theorem has to do with symmetries of the Lagrangian. E.g. if the Lagrangian has spherical symmetry, then total angular momentum is conserved. The Noether theorem obviously also applies to quantum systems. A change of gauge is not a physical transformation, that is all. In quantum field theory one starts with a simple Lagrangian (e.g. Dirac Lagrangian), and then changes it so that it becomes invariant under local gauge changes, i.e. one then changes the derivative in the Dirac equation into a D which has a "gauge field" in it: to make this sound cryptic, one then says that "local gauge invariance has generated a gauge field", although this is not true. Imposing local gauge invariance simply puts a constraint on what sort of Lagrangians can be written. It is similar to demanding that a function F(z) be analytic in the complex plane, this also has serious consequences.

This post imported from StackExchange Physics at 2014-04-04 15:40 (UCT), posted by SE-user Martin