Deriving the commutation relations in Maxwell-Chern-Simons theory

I am learning quantization of MCS theory. $$L_{MCS}=-\frac{1}{4}F^{\mu \nu}F_{\mu\nu}+\frac{g}{2} \epsilon^{\mu \nu \rho}A_\mu\partial_\nu A_\rho$$ I have reached the commutation relation $$[A_i(\vec x),\pi^j(\vec y)] = i \delta^j_i \delta^2(\vec x- \vec y).$$ Now I don't know how to reach the commutation relation between Electric fields and electric and magnetic fields. The relations are as follows. $$[E_i(\vec x),E_j(\vec y)] = - i g \epsilon_{ij} \delta^2(\vec x- \vec y)$$ $$[E_i(\vec x),B_j(\vec y)] = i \delta_{ij}\partial_j \delta^2(\vec x- \vec y)$$ Note: I have tried some conventional 3+1 QFT methods to get these relations but I get stuck on polarization sums and also this $\epsilon_{ij}$ term can not appear there. I got $$\pi^i = \dot A_i + \frac{\mu}{2}\epsilon^{ij}A_j$$

• What's the expression for the electric field? $E=-i\partial_{A_i} -\frac{g}{2}\epsilon_{ij}A_j$? – Demosthene Jun 20 '17 at 7:50
• I dont know if the expression for $E$ is modified here. I think $E=-\nabla A_0-\frac{\partial \vec A}{\partial t}$. How can we see whether the expression for $E$ is changed or not? – Sami Khan Jun 20 '17 at 9:24
• In Weyl guage it simply becomes $E^i= \dot A_i$. – Sami Khan Jun 20 '17 at 9:28

I have done this and here are main steps. $$\pi^i = \dot A_i + \frac{\mu}{2}\epsilon^{ij}A_j$$ In above expression we have $E^i = \dot A_i$ so we have $$\pi^i = E^i + \frac{\mu}{2}\epsilon^{ij}A_j$$ or $$E^i= \pi^i - \frac{\mu}{2}\epsilon^{ij}A_j.$$
we can compute $[E^i(\vec x),E^j(\vec y)]$ as $$[(\pi^i - \frac{\mu}{2}\epsilon^{ij}A_j),(\pi^i - \frac{\mu}{2}\epsilon^{ij}A_j)]$$expand above commutator and use following known commutation relations
$$[A_i(\vec x),\pi^j(\vec y)] = i \delta^j_i \delta^2(\vec x- \vec y)$$ $$[A_i(\vec x),A_j(\vec y)] =0$$ $$[\pi^i(\vec x),\pi^j(\vec y)] =0$$ and you will get your result. similarly we can use the relation $B=\epsilon^{ij}\partial_iA_j$ to compute the second commutation relation.