# Derivation of Maxwell-Chern-Simon Hamiltonian

I am trying to derive famous relations of MCS theory. Starting from $$L_{MCS}=-\frac{1}{4}F^{\mu \nu}F_{\mu\nu}+\frac{g}{2} \epsilon^{\mu \nu \rho}A_\mu\partial_\nu A_\rho$$ I am ok with following results $$\mathcal{L}=\frac{1}{2e^2}E^2_i - \frac{1}{2e^2}B^2_i + \frac{\kappa}{2}\epsilon^{ij} \dot{A}_i A_j + \kappa A_0 B$$ and $$\Pi^i=\frac{1}{e^2}\dot{A}_i + \frac{\kappa}{2}\epsilon^{ij}A_j$$ but when I go for hamiltonian every things becomes a mess. $$\mathcal{H}=\Pi^i \dot{A}_i - \mathcal{L}= \frac{e^2}{2} \left( \Pi^i-\frac{\kappa}{2}\epsilon^{ij}A_j\right)^2+\frac{1}{2e^2}B^2+ + A_0(\partial_i \Pi^i + \kappa B)$$ Here I show what I have tried. $$\mathcal{H}= \Pi_i(\Pi^i - \frac{\kappa}{2}\epsilon^{ij}A_j)+ \frac{1}{4}F^{\mu \nu}F_{\mu\nu}-\frac{g}{2} \epsilon^{\mu \nu \rho}A_\mu\partial_\nu A_\rho$$ $$\mathcal{H}= \Pi_i(\Pi^i - \frac{\kappa}{2}\epsilon^{ij}A_j)+ \frac{1}{2}\partial_\mu A_\nu \partial^\mu A^\nu-\frac{1}{2}\partial_\mu A_\nu \partial^\nu A^\mu-\frac{g}{2} \epsilon^{\mu \nu \rho}A_\mu\partial_\nu A_\rho$$ Next I expanded indices from $0 i j$ and created mess. Somebody tell me a decent method please.

• Comment to the post (v1): Consider to add references to fix conventions & notation. – Qmechanic May 14 '17 at 5:41

You have done a lot well. Just follow these steps. Take $$\mathcal{H}= \Pi_i(\Pi^i - \frac{\kappa}{2}\epsilon^{ij}A_j)-[\frac{1}{2e^2}E^2_i - \frac{1}{2e^2}B^2_i + \frac{\kappa}{2}\epsilon^{ij} \dot{A}_i A_j + \kappa A_0 B]$$ instead of $$\mathcal{H}= \Pi_i(\Pi^i - \frac{\kappa}{2}\epsilon^{ij}A_j)+ \frac{1}{4}F^{\mu \nu}F_{\mu\nu}-\frac{g}{2} \epsilon^{\mu \nu \rho}A_\mu\partial_\nu A_\rho$$ write $E_i$ as $F_{i0}=\partial_i A_0-\partial_0 A_i$ open the square and you will get $\dot A_i$ terms. Replace all $\dot A_i$ with conjugate momenta. Solve each step explicitly and completely. At the end you have to rearrange the things and here it is.