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  Question about Charge and Gauge Transformation

+ 1 like - 0 dislike
101 views

If a quark is in red, a gauge transformation can transform it into blue. But gauge transformation cannot change any observable. Thus, the colour of the quark cannot be an observable. 

Is that true that observables in gauge theories (QED and QCD) must be neutral?

Is magnetic monopole an observable?

asked Feb 26 in Q&A by New Student (165 points) [ no revision ]

The origin of your confusion, I guess, is in your admitting a too brave abstraction: a quark "itself". In fact you should never forget its permanent coupling to the gluons because in a gauge theory they always come and are observed together.

Vladimir Kalitvianski Thank you. It wasn't a correct example. 

2 Answers

+ 2 like - 0 dislike

It is true that in a more modern perspective only gauge invariant operators can be observables, however even in the more traditional way of thinking about gauge theory (say via gauge fixing), the appropriate definition of charges is those which are eigenvalues of the generators of global gauge transformations, in this sense $SU(3)$ quarks have eight (nonabelian) charges, and you can only specify two of them at a time because the Cartan subalgebra has dimension two. Global gauge transformations can still change charges but this is not strange, since global transformations are physical.

On the other hand, the three colors do nothing more than labeling the entries of the $3\times 1$ column vectors, and can't be interpreted as charges.

answered Feb 28 by Jia Yiyang (2,635 points) [ revision history ]
edited Feb 28 by Jia Yiyang

Thank you.

+ 1 like - 0 dislike

The Hilbert space of a gauge theory is formed as the global-symmetry-invariant part of the tensor product Hilbert space $\mathcal{H}_{matter}\otimes \mathcal{H}_{gauge field}$ (the symmetry operators act diagonally in this space). So if $\mathcal{O}(x)$ is a point operator in some non-trivial representation, to get it to act on the invariant vectors of this Hilbert space, one must attach to it the end of a Wilson line in that same representation.

So yes $\mathcal{O}(x)$ could be something like the creation operator for a quark and we would need to attach a Wilson line to it. Note this means that at the other end of the Wilson line there needs to be an annihilation operator! (or something else with opposite quantum numbers) This is something you know, that gauge-charged particles are created in particle-antiparticle pairs. In QCD, this Wilson line has tension, and so the quarks are confined.

answered Feb 27 by Ryan Thorngren (1,895 points) [ no revision ]

Thank you.

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