# When does a “constant of the motion” imply a Noether current in a quantum field theory?

Assume we are given a quantum field theory described by some functional. If $J$ is a Noether current, i.e. it is associated with a symmetry of the functional and satisfies $\partial_s J^s=0$ (Noether theorem), we can always obtain a conserved quantity as $Q(x^0) = \int J^0(x^0, x^1,...,x^n)d^{n}x$. What is the converse to the previous statement? I would like to know what assumptions are necessary to build a conserved current $J$ given $Q$ swith $\dot Q = [H,Q]=0$ where $H$ is the Hamiltonian of the (quantum field) theory.

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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. – Igor Khavkine May 28 '13 at 22:10
@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? – Daniel May 28 '13 at 23:10
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…). – The User May 29 '13 at 0:02
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"). – Igor Khavkine May 29 '13 at 0:03
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. – The User May 29 '13 at 0:20

The answer to the question in your title is "no" in general. Noether current and conserved charge only go hand in hand if the symmetry that gives rise to them is a continuous symmetry, such as translation (conserved momentum) or a $U(1)$ symmetry (conserved charge). A discrete symmetry will, in general, give rise to a conserved quantity that can only take on discrete values, so it cannot "flow" and there is no current associated with that conservation law. An example is inversion symmetry, with parity as a conserved quantity. There is no current associated with parity.