# $SO(N)$ symmetric theory of $N$ real scalar fields, why do charges have correct commutation relations of generators?

Consider an $SO(N)$ symmetric theory of $N$ real scalar fields,$$\mathcal{L} = {1\over2} \partial_\mu \Phi^a \partial^\mu \Phi^a - {1\over2} m^2 \Phi^a \Phi^a - {1\over4} \lambda(\Phi^a \Phi^a)^2.$$For the quantum theory, consider the $SO(N)$ charges $\hat{Q}_{ab} = \int d^3 \textbf{x}\,\hat{J}_{ab}^0$. Why do the charges $\hat{Q}_{ab}$ have correct commutation relations of the generators of the $SO(N)$ symmetry?

• Please give a definition of $\hat{J}^0_{ab}$. For instance, if one defines them in the "right" way, that is as the conserved currents under the obvious $\mathrm{SO}(N)$ symmetry, the question becomes rather trivial because the conserved charges of a symmetry generically are the generators of that symmetry. – ACuriousMind Nov 15 '15 at 4:57

1. Let us consider the corresponding Hamiltonian theory, so that we have a notion of a commutator that we can use to form a Lie algebra bracket. Moreover, let us consider the classical theory for simplicity. Then the Poisson bracket $$\tag{1} \{\Phi^a({\bf x}),\Pi_b({\bf y})\}_{PB}~=~\delta^a_b~\delta^3({\bf x}-{\bf y}), \qquad \text{etc},$$ plays the role of commutator.

2. The Hamiltonian Lagrangian density ${\cal L}_H$ for OP's theory is of the form $$\tag{2} {\cal L}_H~=~ \Pi_a \dot{\Phi}^a -{\cal H}, \qquad {\cal H} ~=~ \frac{1}{2} \Pi^2 +{\cal V}(\Phi^2, (\nabla\Phi)^2) ,$$ where we have introduced the notation $$\tag{3} \Phi^2 :=\Phi^a g_{ab} \Phi^b, \qquad \Pi^2 :=\Pi^a g_{ab} \Pi^b, \qquad \text{etc}.$$

3. Here the $o(N)$-indices $a,b,\ldots,$ are raised and lowered with a constant metric $g_{ab}$. The $O(N)$ symmetry $$\tag{4} \delta \Phi^a ~=~ \varepsilon \{\Phi^a, Q[\omega]\}_{PB}~=~\varepsilon\omega^{a}{}_{b}\Phi^b, \quad \delta \Pi^a ~=~ \varepsilon \{\Pi^a, Q[\omega]\}_{PB}~=~\varepsilon\omega^{a}{}_{b}\Pi^b, \qquad$$ has generators $$\tag{5} Q[\omega]~:= \frac{1}{2}\omega_{ab} Q^{ab} ~=~\Pi^a\omega_{ab}\Phi^b, \qquad Q^{ab}~:=~\int\! d^3x\left( \Pi^a\Phi^b-\Phi^a\Pi^b\right),$$ where $$\tag{6} \omega_{ab}~=~-\omega_{ba}$$ are anti-symmetric matrices.

4. It is not hard to check that the generators (5) equipped with the Poisson bracket (1) form an $o(N)$ Lie-algebra. See also this related Phys.SE post.

5. The Noether charges are the generator of the symmetry, as is always the case for Hamiltonian theories, cf. e.g. this Phys.SE post. This can also be explicitly confirmed in the above case.

6. We conclude that the Noether charges (5) form an $o(N)$ Lie algebra.

7. Finally, it seems that OP is also asking a more general question of whether any symmetry of any action lifts to a symmetry of the corresponding Noether charges? This is a good question and discussed in e.g. this Phys.SE post. The answer is: Not always! There can be either classical or quantum obstructions/anomalies, such as e.g. central charges and 2-cocycles.