# Hamiltonian Structure of Chern Simons Electrodynamics

I am reading the review paper "Aspects of Chern-Simons Theory" by Gerald Dunne

Starting from p. 17, Dunne works on the Hamiltonian structure of the CS electromagnetism. When there is no Maxwell term, the CS action is given by

$$L = \frac{1}{2} \epsilon^{ij} \dot{A}_i A_j + A_0 B\tag{70}$$

where I set $$\kappa = 1$$, and this is given in his equation 70. The conjugate momenta is then

$$\Pi^i = \frac{\partial L}{\partial \dot{A}_i} = \frac{1}{2} \epsilon^{ij} A_j\tag{73}$$

which can also be found in his eq. (66) given that $$e \rightarrow \infty$$. The equal-time canonical commutation relations is then given by

$$\left[A_{i} (x) , \Pi^j (x) \right] = i \delta^j_i \delta^{2} (x-y)\tag{68}$$

which is given in his eq. (68). Then, he uses the definition of the conjugate momenta and finds that

$$\left[A_{i} (x) , A_j (x) \right] = i \epsilon_{ij} \delta^{2} (x-y).\tag{72}$$

I do not know how to get this result. Now let me write down what I got

$$$$\left[A_{i} (x) , \Pi^j (x) \right] = \frac{1}{2} \epsilon^{jk} \left[A_{i} (x) , A_k (x) \right]$$$$

On the other hand, since $$\left[A_{i} (x) , \Pi^j (x) \right] = i \delta^j_i \delta^{2} (x-y)$$, we have

$$$$i \delta^j_i \delta^{2} (x-y) = \frac{1}{2} \epsilon^{jk} \left[A_{i} (x) , A_k (x) \right]$$$$

Multiplying each side by $$2 \epsilon_{jm}$$ and using $$\epsilon^{jm} \epsilon_{jn} = \delta^m_n$$, i obtain $$$$\left[A_{i} (x) , A_j (x) \right] = 2 i \epsilon_{ij} \delta^{2} (x-y)$$$$

Apparently i am missing a factor of 2, but i have no idea what i do wrong.

• I used Dirac bracket but i still got the same result. It is actually just plugging in some definition into an equation, isn't it. What could i possibly i do wrong? does he have a weird definition of $\epsilon$ such that $\epsilon^{ab} \epsilon_{bc} = 2 \delta^a_c$? Oct 8, 2018 at 14:37
• my struggle is simpler than the degree of freedom counting. it is just going from one equation to the next one by plugging in some definition. i cannot even pass that point. How to get rid of the factor of 2 in A,A commutator? Oct 8, 2018 at 14:46

There is no factor of 2. The Dirac-Bergmann analysis$$^1$$ goes as follows. The second-class constraints are $$\chi^i ~=~\pi^i - \frac{1}{2}\epsilon^{ij} A_j, \qquad i~\in~\{1,2\}.$$ The matrix of Poisson brackets$$^2$$ of second-class constraints is $$\Delta^{ij}(x,y)~:=~\{\chi^i(x),\chi^j(y)\}~=~-\epsilon^{ij}\delta^2(x-y),$$ so the inverse matrix is $$(\Delta^{-1})_{ij}(x,y)~=~-(\epsilon^{-1})_{ij}\delta^2(x-y).$$ The Dirac bracket becomes $$\{A_i(x),A_j(y)\}_D~=~-(\epsilon^{-1})_{ij}\delta^2(x-y).$$
$$^1$$ Ref. 1 implicitly mentions between eqs. (70)-(71) a shortcut via the Faddeev-Jackiw method.
$$^2$$ To go from brackets to commutators, multiply with $$i\hbar$$.