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Noether's theorem tells us that to every continuous symmetry of the Lagrangian there corresponds a conserved current $j^\mu$. From the time component of this current, we can then define the Noetherian charge $$Q = \int d^3\mathbf{x}\ j^0(\mathbf{x}),$$ which is a time independent operator. In all examples I've seen, the Noether charge $Q$ is always a Hermitian operator (up to a trivial rescaling by $i$). But no one ever seems to explicitly mention this fact in full generality.

Can we prove that Noether's theorem will always give us a Hermitian charge operator? If not, are there counter-examples?

Qmechanic
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EuYu
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2 Answers2

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The Noether charge is the generator of the symmetry it belongs to, see e.g. this answer by Qmechanic. This relationship is also preserved in the quantum theory, see this question, in the sense that the quantum Noether charge $Q$ must commute with the Hamiltonian $H$, at least in the absence of anomalies and if we do not run into "quantization issues" when using canonical quantization.

Now, if we assume that the classical symmetry transformation must be represented by a unitary transformation upon the Hilbert space (note: I'm not assuming it is a quantum symmetry transformation), then we can directly conclude by Stone's theorem that $Q$ is Hermitian and that the transformation associated to the classical symmetry is indeed a symmetry.

We might ask whether one can drop this assumption, I think one cannot. Dropping the requirement that transformations are represented by unitary operators leads to the normalization of states not being preserved, in particular, it means that the probabilities after the transformation to find one state in other states that form a basis do not sum to 1. This wrecks havoc with the entire structure of the quantum theory; it is a reasonable physical assumption that all physical transformations be represented unitarily upon the Hilbert space of states.

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ACuriousMind
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Supersymmetry generators are not always Hermitian. If you impose SUSY, and then compute de corresponding Noether's currents, and then you calculate the conserved charge, i.e., the fermionic Lorentz generators, you will get two non-Hermitian conserved currents.

(By the way, the relation $Q^\dagger=\bar Q$ is only valid in Lorentzian signature, in Euclidean signature, this relation is not true.)

Actually, a nice computation of this was made by Olive and Witten. They did just what I outlined for $\mathcal N=2$ SYM without matter fields and obtained the central charge of this theory. See section 2.8 of http://arxiv.org/abs/hep-th/9701069 for a detailed computation.

Also, not always you would get a pair $Q$ and $\bar Q$. Just take, for example, $\mathcal N=(n,m)$ 6D theories.

The moral of this story is the following: You will always assume $S=S^\dagger$ (a real action.) If your symmetry generator is also Hermitian, the conserved current, and therefore, the conserved charge, will also be Hermitian. But this might not be the case.

CGH
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