I am trying to organize my knowledge of Noether's theorem in QFT. There are several questions I would like to have an answer to.

In classical field theory, Noether's theorem states that for each continuous global symmetry of the action there is a corresponding current (Noether current) $j^{\mu}$, which satisfies (classically) the conservation condition:

$$ \partial_{\mu} j^{\,\mu} \simeq 0, $$

where I use the $\simeq$ sign to denote that the equation is only valid on-shell, i.e. on field configurations which are subject to the classical equations of motion.

Conserved currents lead eventually to conserved charges, which are given by

$$ Q(t) = \int d^{n-1}x \; j^{\,0}(x, t) \simeq \text{const}. $$

  1. Is it correct that these charges form an algebra (with an algebraic Lie bracket given by the Poisson bracket), which is exactly the Lie algebra of the symmetry group?

  2. Vector fields on the space-time manifold also have a Lie algebra structure given by the Lie derivative. I did some calculations and it turned out that conserved vector fields are algebraically close and therefore form a subalgebra. My question is: do Noether currents of an arbitrary field theory also form a subalgebra through the Lie derivative, and if they do, does this subalgebra have any physical meaning?

Despite the questions, this part is relatively clear. Now comes the quantum magic. In the path integral formalism, the Ward identity is a formal analogue of the classical Noether's theorem.

Noether currents are considered very important components of the quantum theory, a kind of vicars of symmetries on the physical system. I never understood this completely. For example, they are ought to be well-defined in the quantum sense and therefore are subject to normal-ordering. This sometimes leads to the modification (!) of the symmetry algebra itself, the best example being the Witt algebra of conformal symmetries and its quantum counterpart, the Virasoro algebra of the normal-ordered conformal currents.

  1. So why are currents more fundamental than geometrical symmetries of classical configurations itself? Why are they even have to be well-defined in the quantum theory, where the only observable quantities are correlations?

P.S. I have studied lots of literature, and all explanations seemed to me unclear and speculative. So I am not looking for a reference, but rather for some kind of a paraphrasing which would make things clearer.

@JamalS, I am keeping the string-theory tag because I expect string theorists to know a lot about (mentioned in my question) Witt/Virasoro algebras and therefore to give expanded answers.


1 Answer 1


Comments to the question (v5):

  1. If an action functional $S$ is invariant under a Lie algebra $L$ of symmetries, the corresponding Noether currents & charges do not always form a representation of the Lie algebra $L$. There could be (classical) anomalies. In some cases such (classical) anomalies appear as central extensions, cf. e.g. Ref. 1-3 and this Phys.SE post.

  2. The negative conclusion for point 1 is true even for Hamiltonian formulations where a Poisson bracket is defined.

  3. The 3rd question seems like an analogue of the faulty logic that requires a quantum theory to be explained in a classical language, and not vice-versa.


  1. F. Toppan, On anomalies in classical dynamical systems, J. Nonlin. Math. Phys. 8 (2001) 518, arXiv:math-ph/0105051.

  2. Tomas Brauner, Spontaneous Symmetry Breaking and Nambu-Goldstone Bosons in Quantum Many-Body Systems, Symmetry 2 (2010) 609, arXiv:1001.5212; page 6-7.

  3. J.D. Brown and M. Henneaux, Central charges in the canonical realization of asymptotic symmetries: an example from three-dimensional gravity, Commun. Math. Phys. 104 (1986) 207.

  • $\begingroup$ Quantum theory, in the path integral formulation, requires classical degrees of freedom and classical action. $\endgroup$ Feb 10, 2015 at 12:48
  • $\begingroup$ Also, geometric symmetries (like conformal transformations) do not require any degrees of freedom at all. They are as much classical as they are quantum. But in classical theory, we respect them instead of modifying the algebra with central charges. $\endgroup$ Feb 10, 2015 at 12:51
  • $\begingroup$ Further references: 4. V. Iyer & R.M. Wald, arXiv:gr-qc/9403028. 5. G. Compere & A. Fiorucci, arXiv:1801.07064. $\endgroup$
    – Qmechanic
    Aug 23, 2019 at 21:30

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