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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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  • @AccidentalFourierTransform Finite-dimensional representations are not the representations on the Hilbert space of the theory (precisely because they are not unitary), they just live on the target space of the fields, hence Wigner's theorem doesn't hold. When we speak of "generators" or "Noether charge" in quantum mechanics, we generally mean an operator on the space of states, not on the target space of fields. – ACuriousMind Apr 08 '16 at 17:09
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    Wigner's theorem is a statement about symmetries acting on the underlying Hilbert space. Noether's theorem concerns symmetries internal to the Lagrangian. I'm not too sure how the two are related. In particular, I'm not entirely convinced that every symmetry for the Lagrangian needs to be unitary. – EuYu Apr 08 '16 at 17:27
  • @EuYu: That's why I linked what I linked. In the Hamiltonian framework, the Noether charge generates the symmetry. After (canonical) quantization, the Hamiltonian symmetries and generators correspond to operators on the Hilbert space, and then Wigner's theorem hits and tells us the symmetry operator must be unitary. I don't see where you see a gap in this argument. – ACuriousMind Apr 08 '16 at 17:33
  • Oh, yes, that is correct. See also this post, you're looking at the "wrong thing" to decide hermiticity. We would have to take the adjoint of the $\sigma$ in their representation on the Hilbert space, not on the non-unitary finite-dim rep to conclude whether or not $J$ is Hermitian. Whether or not it is Hermitian in the representation on the target space of the fields is a physically irrelevant information, at least it has no meaning I could see. – ACuriousMind Apr 08 '16 at 17:37
  • I actually have to go offline now; if you want to discuss this further you could drop me a line in chat. – ACuriousMind Apr 08 '16 at 17:40
  • Sorry for being obtuse, but the argument still seems circular to me. Let me see if I understand you correctly. Let's use a complex scalar field as example. The Lagrangian is invariant under $\phi \mapsto e^{i\theta}\phi$, which generates a charge $Q$. This charge acts on the Hilbert space, with the symmetry implemented via $V(\theta) = e^{i\theta Q}$. So far, I agree. But if we do not know a priori that $Q$ is Hermitian, how do we know that we have $|\langle\xi|V^{\dagger} V|\psi\rangle| = |\langle\xi|\psi\rangle|$, which is the condition for Wigner's theorem to apply? – EuYu Apr 08 '16 at 17:54
  • @EuYu: Uh-oh, you're right, that is a gap. I'll think about that, if I can't fix it, I'll delete this answer. – ACuriousMind Apr 08 '16 at 18:33
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    Thanks for the updated answer. I agree for the most part, but it still doesn't seem satisfying. If we have a symmetry of the Lagrangian, we don't get to pick the charge that comes out. The associated symmetry $V(\theta)=e^{i\theta Q}$ will either be unitary or not, and this is not a condition we can set. Of course, we can reject any symmetries of the Lagrangian which generates such a non-unitary $V$ as non-physical, which is what I think you're suggesting. But the underlying mathematical problem is whether such non-physical symmetries exist in the first place, and that hasn't been resolved. – EuYu Apr 08 '16 at 19:39
  • @EuYu: Well, I think the problem is that there is e.g. the ordering ambiguity in the choice of quantization procedure that makes it impossible to decide from the classical expression for $Q$ whether $Q$ will be Hermitian in the quantum theory or not, so you cannot really phrase this as a purely mathematical problem because the quantization procedure is "physically" chosen to make all generators of symmetries come out Hermitian. For instance, if I choose Weyl quantization, all phase space functions will be sent to Hermitian operators due ot the symmetrization, so Q is trivially Hermitian. – ACuriousMind Apr 08 '16 at 19:43
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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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  • Might the "fermionic Lorentz generators" be "non-Hermitian" for the same reason the Boosts are not unitary? I.e. are you taking the adjoint as operators on the target space of the field, and not on the Hilbert space of the theory? This is not a true counterexample to generators being unitary, since the relevant notion of "unitary/self-adjoint" is that of operators on the Hilbert space of states, not on the target space of the fields. – ACuriousMind Apr 11 '16 at 16:20
  • First of all, the initial question was about Hermitian Noether charges, and nothing about unitarity. I just answered that. Second, the supercharges acting on a state does not change the "unitarity" of the system (I think this answer the first part of your question.) Second, to "create" states in the Hilbert space of the theory (which is the second part of your question) you have to construct, out of the supercharges, the $a$ and $a^\dagger$, which, as in the harmonic oscillator case, are not Hermitians. – CGH Apr 12 '16 at 14:04