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Is global gauge symmetry really a symmetry transformation and local conserved current in gauge theories

+ 3 like - 0 dislike
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One way to define a gauge theory is that whenever the Lagrangian is invariant under some local transformations, we say these local transformations are local gauge transformations and the theory is a gauge theory.  

However, I think a better way to understand gauge theory is not about the local symmetry of the Lagrangian, but about the structure of the Hilbert space: a gauge theory is a theory where all states in the Hilbert space are invariant under the local transformations. This constraint on Hilbert space already encodes that the Hamiltonian or Lagrangian should be gauge invariant, otherwise under time evolution a gauge invariant state may become a gauge non-invariant state. Also, I think in principle one can have a theory which does have a local symmetry but it is not a gauge theory. For example, we can take the electromagnetism Lagrangian without imposing gauge equivalence, namely, $A_\mu$ and $A_\mu+\partial_\mu\chi$ label different states. This theory has a Lagrangian that is invariant under a local transformation, but it is not a gauge theory. Of course, this theory does not describe relativistic massless spin-1 particle since it has more degrees of freedom than required.

Now here is a question: is the global gauge transformation a symmetry transformation (a transformation acting on states and mapping one state to another)? Suppose the global gauge transformation is a special case of local gauge transformation, as it seems to be, it should not be a symmetry transformation.

On the other hand, according to Noether's theorem, a symmetry of the Lagrangian, which may or may not be a symmetry of the theory, leads to local conserved current. However, Weinberg-Witten theorem states that there is no gauge invariant conserved current in non-Abelian gauge theory, although the total charge is conserved. Is this related to the observation that a global gauge transfomation is also not a symmetry transformation? And why does this theorem only apply to non-Abelian gauge theories?
asked Nov 26, 2014 in Theoretical Physics by Mr. Gentleman (255 points) [ no revision ]
I don't quite agree with your premises.

1. Given an abstract Hilbert space, what do you even mean by "local transformations" on states?

2. Are you suggesting there are no classical gauge theories?
@JiaYiyang

Thank you for your comments and sorry for any confusion. The two questions you asked are related, so let me answer them once.

 

The canonical way of thinking about the Hilbert space in a quantum field theory is to first solve the classical equation of motion and identify the eigenmodes, after that, we construct one-particle and multi-particle states in the eigenmodes. Remarkably, for a translational invariant system, momentum eigenstates are always a set of eigenmodes of the equation of motion. Then doing a Fourier transformation, we see that position eigenstates are also such eigenmodes. However, if we want to construct one-particle and multi-particle states in a gauge theory, we have to keep in mind that different modes connected by a gauge transformation are identified, this makes the Hilbert space of a gauge theory very non-local, in the sense that it is not a tensor product of local Hilbert spaces, instead, we need to modulo gauge equivalence classes. So by local transformations on states, I presume a prescription of labeling states by modes and I do the transformation on modes. Since it is assumed that the modes label the states, I am also doing a transformation on states. For classical gauge theory, I just do the transformation on a mode without saying anything about the quantum states associated with it.

 

Gauge theory with continuous gauge group in continuous space may be confusing to think about, to have a concrete example, one can just think about a lattice $Z_2$ gauge theory. For detailed discussion, please see Wen's book Quantum Field Theory of Many-Body Systems. One may also want to think about toric code, where the ground state of the deconfined lattice $Z_2$ gauge theory is written in a gauge invariant way (meaning without doing a gauge fixing).

 

In that sense, a global gauge symmetry is also not a symmetry transformation. Then this bring me to the question of local conserved current.
I see, I wrote a partial answer.

1 Answer

+ 5 like - 0 dislike

I see(after your clarification in comments), then your description is indeed a valid way of phrasing gauge theory and it highlights the saying "a gauge symmetry is actually a redundancy in description of the theory", but I'm afraid your rejection to the local symmetry vocabulary is unfair. As long as your local symmetry is not too trivial, it almost always gives rise to redundancy. I touched the issue briefly in my exposition article "Semantics in Gauge theory". A more complete account can be found in the lucidly written textbook "Classical and Quantum Dynamics of Constrained Hamiltonian Systems".

In particular, a global gauge transformation is not a special case of local gauge transformations. A local gauge transformation--to qualify as a "change of redundant labelings"--must preserve initial or boundary conditions of the system, while a global gauge transformation almost always breaks initial/boundary conditions.(Of course they both preserve the Lagrangian hence the equation of motions).

I don't know anything about Weinberg-Witten, I'll leave it to others.  

answered Nov 27, 2014 by Jia Yiyang (2,465 points) [ revision history ]
edited Nov 27, 2014 by Jia Yiyang
Thank you for your answer. I looked up your notes and I like the point that a global gauge transformation is not a gauge transformation because it almost always (not really always, for example, the "global" $Z_2$ gauge transformation in lattice $Z_2$ gauge theory preserves boundary conditions) changes boundary conditions, thank you for pointing it out.

 

However, I do not quite see why local symmetries always induce redundancy. Taking lattice $Z_2$ gauge theory from Wen's book as an example, there we identified different link configurations which are related by gauge transformations. With this in mind, we reach the Hamiltonian. But we do not have to identify those link configurations while still write the same Hamiltonian to describe a system (which is not really $Z_2$ gauge theory). Here we still have the local symmetry of the Hamiltonian but there is no redundancy, so in this sense we can say that local symmetry is not a local gauge symmetry, which is in fact not a symmetry. I think the crucial difference between these two ways of defining a gauge theory is, in a real (fake) gauge theory, determined by the nature of the system to be described, the Hamiltonian have to (can but have no responsibility to) be gauge invariant.

 

In fact the relation from this discussion to Weinberg-Witten theorem is just my "random" thought, and I was just trying to understand the theorem intuitively.
@Mr.Gentleman, I don't know anything about $Z_2$ lattice gauge theory. But I did use the qualification "almost" for "it almost always gives rise to redundancy". In fact some not so trivial-looking local transformation may become trivial after the equations of motion are imposed, such local transformation does not lead to redundancy. But I suspect it is not the issue with $Z_2$ lattice gauge theory, and I don't understand what a fake gauge theory is.

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