Timeline for When does a "constant of the motion" imply a Noether current in a quantum field theory?
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| May 29, 2013 at 0:20 | comment | added | The User | Conversely, in general it is not possible to translate a quantum observable into a classical observable (consider parity). Thus you should not expect that there is even an associated current (from which you would get a classical observable, the charge). I think your question makes sense in classical field theory (are you interested in that case?): conserved currents induce conserved charges and you can ask whether a conserved quantity is given as a charge associated to a current. | |
| May 29, 2013 at 0:03 | comment | added | Igor Khavkine | Locality is easier to understand in the classical theory. Any conserved quantity $Q$ is a functional of the field, say $\psi(x)$. The functional $Q$ is local if its variational derivative $\delta Q/\delta \psi(x)$ depends only on $\psi(x)$ and finitely many derivatives at the same point $x$. Using this definition, you can check that your $Q$ is a local functional while $Q^2$ is not (though it could be said to be "bilocal", $Q^3$ would be "trilocal" and any polynomial in $Q$ would be "multilocal"). | |
| May 29, 2013 at 0:02 | comment | added | The User | If your question is actually about quantum field theory, then even the direction you stated is not true in general. The $Q(x^0)$ is the classical conserved Noether chargeIn quantum field theory it is a heuristic to transfer these charges into an operator (quantisation), but you still have to prove that it is actually a conserved quantity. There are a lot of “anomalies” where this is not the case (conformal anomaly, chiral anomaly…). | |
| May 28, 2013 at 23:11 | history | edited | Yaniel Cabrera | CC BY-SA 3.0 |
made title more precise
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| May 28, 2013 at 23:10 | comment | added | Yaniel Cabrera | @Igor: Your comment is pointing in what I think it is a right direction. I have seen a Noether current built from a conserved operator which is assumed to be local. My failure is understanding what is meant by locality in this case. Could you elaborate or give some references? | |
| May 28, 2013 at 22:10 | comment | added | Igor Khavkine | In addition to Carlo's answer below consider the conserved operator $Q$ from your question and consider $Q^2$. $Q^2$ is an equally good conserved operator, but it is no local given by a single integral of a local functional. So local conserved operator are rather special among all conserved operators. | |
| May 28, 2013 at 21:10 | answer | added | Carlo Beenakker | timeline score: 3 | |
| May 28, 2013 at 18:21 | history | asked | Yaniel Cabrera | CC BY-SA 3.0 |