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Does gauge invariance imply charge neutrality? I understand that all physical observables must be gauge invariant. Does this mean that physical observables must be neutral?

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

Is the electric charge of an electron an observable in QED? Is that correct that all observables in QED must be neutral? Are magnetic monopoles observables?

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You are right of course! Physical observables must be gauge invariant. But this does not mean that they must be neutral. They could be charged under the global symmetry and be neutral under the local gauge symmetry.

In particular, a local gauge symmetry is generated by a function $\alpha(x)$ where $\alpha(x) \to 0$ as $|x| \to \infty$. A global symmetry of course has $\alpha(x) = $ constant which does not satisfy the above property. One way to have a charged gauge invariant operator is to connect it to a Wilson line that joins the operator to a point at infinity.

To add a bit more detail, a Wilson line $W_{{\cal P},q}(x_1,x_2)$ is a line operator (defined along a path ${\cal P}$) that under a gauge symmetry transforms as (assuming abelian symmetry for simplicity) $$ W_{{\cal P},q}(x_1,x_2) \to e^{- i q \alpha(x_1) } W_{{\cal P},q}(x_1,x_2) e^{ i q \alpha(x_2) } . $$ A charged local operator transforms under gauge symmetry as $$ {\cal O}(x) \to e^{ - i q \alpha(x) } {\cal O}(x) . $$ where $q$ is the charge of the state. We now construct the operator $$ {\tilde {\cal O}}(x) = W_{{\cal P},q}(\infty,x){\cal O}(x) $$ This transforms as $$ {\tilde {\cal O}}(x) \to e^{ - i q \alpha(\infty)} {\tilde {\cal O}}(x) . $$ Then, ${\tilde {\cal O}}(x)$ is invariant under local gauge transformations but not invariant under global symmetry transformations.

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  • $\begingroup$ Thank you for your answer. Are there any references explaining attaching Wilson lines to an operator? $\endgroup$ – The Last Knight of Silk Road Feb 26 '18 at 0:38
  • $\begingroup$ I suppose any book on QFT should have a discussion on Wilson lines/loops and similar such objects. I have added a bit more detail so you can understand the basics. $\endgroup$ – Prahar Feb 26 '18 at 2:10
  • $\begingroup$ Thank you very much. Could you leave a comment for the magnetic monopole? $\endgroup$ – The Last Knight of Silk Road Feb 26 '18 at 3:20
  • $\begingroup$ The same thing holds except you now use a 't Hooft line. $\endgroup$ – Prahar Feb 26 '18 at 12:42
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A rough answer is that yes, all physical states must be charge neutral on net, but there can be local charge imbalances that are compensated somewhere else in space. A quick and dirty way to see this is to consider periodic spatial boundary conditions, and notice that the integral form of Gauss's law gives that the total charge $Q_\text{total} = \oint_{\partial V} {\bf E} \cdot d{\bf S}$ vanishes trivially because there is no boundary.

More generally, it's convenient in QFT to assume that all fields are localized and die off quickly at spatial infinity, which allows us to freely integrate by parts and ignore boundary terms. We can't do this for a gauge theory with nonzero total charge, because then the integral form of Gauss's law gives than the total flux does not die off with distance, and we always need to carefully include boundary terms whenever we integrate by parts. This makes mathematical formalization of gauge theories with net charge quite difficult.

Yet another way to see the problem is that in QFT we want particle creation operators to be local, so that a particle can be locally created from the vacuum. But a charged particle (e.g. a magnetic monopole) is a topological defect which "makes its presence felt" arbitrarily far away, so individual gauge-charged particles cannot be created locally. But multiple charges can be created locally (e.g. vacuum pair-creation and annihilation) as long as their net charge is zero, because their far fields cancel out and far away they look trivial. This is like how a Dirac magnetic monopole needs to be connected to a magnetic flux tube that either goes out to infinity (and is therefore nonlocal) or ends at an oppositely-charge monopole.

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