equation follows using integration by parts for terms involving $\theta_{ij}$ and using the time independence of the charges of the commutative theory for the rest. The charges $Q_0^a, Q_\theta^a$ generate the infinitesimal symmetry transformations: $$\begin{aligned}
[Q^a_0, \phi_{i,0}(x)] &=& \sum_j T^a_{ij}\phi_{j,0}(x), \label{cQ}\\
 {[}Q^a_\theta, \phi_{i,\theta }(x)] &=& \sum_j T^a_{ij} \phi_{j,\theta }(x). \label{ncQ}\end{aligned}$$

Goldstone’s theorem
-------------------

Consider the vacuum expectation value of the commutator of the currents $J^{a,\mu}_\theta(y)$ and the quantum field $\phi_{i,\theta}(x)$: $$\langle 0|[J^{a,\mu}_\theta(y), \phi_{i,\theta}(x)]|0\rangle = \langle
0| [e^{\frac{1}{2}\overrightarrow{\partial_y} \wedge P} J^{a,\mu}_0(y),
\phi_0(x)e^{\frac{1}{2} \overleftarrow{\partial_x} \wedge P}] |0\rangle.
%= \langle 0|[J^{a,\mu}_0(y), \phi_{i,0}(x)]|0\rangle.$$ The first term in the commutator here is $$e^{\frac{i}{2}\partial_y\wedge \theta  \partial_x} \langle
0|J_0^{a,\mu}(y)\phi_0(x)|0 \rangle = 
e^{\frac{i}{2}\partial_y\wedge \theta \partial_x} \langle
0|J_0^{a,\mu}(0)\phi_0(x-y)|0 \rangle$$ where we have used translational invariance and the fact that $$\partial_y\wedge \partial_x \langle 0|J_0^\mu(0)\phi_0(x-y)| 0\rangle ~=~-~
\partial_x \wedge \partial_x \langle 0|J_0^\mu(0)\phi_0(x-y) | 0\rangle~=~0$$ in the second step. The $\theta^{\mu\nu}$ dependence in the second term as well disappears in the same manner so that $$\langle 0|[J^{a,\mu}_\theta(y), \phi_{i,\theta}(x)]|0\rangle = \langle
0|[J^{a,\mu}_0(y), \phi_{i,0}(x)]|0\rangle.$$ This commutator being the same as the one for the corresponding commutative case, the standard arguments using spectral density and Lorentz invariance [@gsw] can be used to argue for the existence of massless bosons in the symmetry-broken phase. Following [@weinberg], 