# Nonabelian global symmetries, $SO(N)$ charges in terms of creation and annihilation operators

Consider an $SO(N)$ symmetric theory of $N$ real scalar fields,$$\mathcal{L} = {1\over2} \partial_\mu \Phi^a \partial^\mu \Phi^a - {1\over2} m^2 \Phi^a \Phi^a - {1\over4} \lambda (\Phi^a \Phi^a)^2.$$The Noether charge is$$Q_{ab} = \int_{\Omega^3} d^3x\,J_{ab}^0,$$where $\Omega^3$ is all space. $Q$ is constant in time. We can express $J^0$ in terms of $\pi$ and $\Phi$ as$$J_{ab}^0 = \partial^0 \Phi_a \epsilon_{ab}\Phi_b = \pi_a \epsilon_{ab} \Phi_b.$$Define$$\epsilon^{ij} = \begin{cases} 1 & \text{if }(i, j) = (a, b) \\ 0 & \text{otherwise.}\end{cases}\Big\} = -\epsilon^{ji},$$so $\epsilon^{ab} = 1 = -\epsilon^{ba}$ and all other entries are $0$. Then the Noether charge becomes$$Q_{ab} = \int d^3x\,J_{ab}^0 = \int d^3x\,\pi_a \epsilon_{ab} \Phi_b = {1\over2} \int d^3x(\pi_a \Phi_b - \pi_b \Phi_a).$$While summing over $SO(N)$ indices, we pick up a factor of ${1\over2}$ since the skew-symmetry of $\epsilon$ causes us to double-count.

My question is, what are the $SO(N)$ charges $Q_{ab}$ here in terms of bosonic creation and annihilation operators?

• $J_{ab}^0 = \pi^a \epsilon^{ab} \Phi^b$ is not a valid expression, since there is a summation on the right hand side of the equation, but a and b are free indices on the left hand side. Your definition of $\epsilon$ is a bit weird, too. What you mean is $$J_{ab}^0 = \pi^i \epsilon_{ab}^{ij} \Phi^j$$ where the matrices $\epsilon_{ab}$ are the generators of the Lie algebra $so(N)$, d.h. $$\epsilon_{ab}^{ij} = \begin{cases} 1 & \text{if } (a,b)=(i,j) \\ -1 & \text{if } (a,b)=(j,i) \\ 0 & \text{otherwise} \end{cases}$$
• Actually, $\pi^i \epsilon_{ab}^{ij} \Phi^j = \pi^a \Phi^b - \pi^b \Phi^a$, so the $\frac 1 2$ is too much in the last line.
As to your question, have you tried it yourself? You have to insert $$\Phi^a = \int \frac{d^3 p}{(2\pi)^3 \sqrt{2E}} \left( a^a(p) e^{ipx} + (a^a)^\dagger(p) e^{-ipx} \right)$$ (plus the similar expression for $\pi^a$) and use $\int d^3x e^{i(p-p')x} = (2\pi)^3 \delta(p-p')$, then you should get a result.