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

https://arxiv.org/abs/hep-th/9902115

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

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

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

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

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

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

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  • $\begingroup$ 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 $\epsilon^{ab} \epsilon_{bc} = 2 \delta^a_c$? $\endgroup$
    – SprCsm
    Oct 8, 2018 at 14:37
  • $\begingroup$ 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? $\endgroup$
    – SprCsm
    Oct 8, 2018 at 14:46

1 Answer 1

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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 $$ \{A_i(x),A_j(y)\}_D~=~-(\epsilon^{-1})_{ij}\delta^2(x-y). $$

References:

  1. G.V. Dunne, arXiv:hep-th/9902115.

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$^1$ Ref. 1 implicitly mentions between eqs. (70)-(71) a shortcut via the Faddeev-Jackiw method.

$^2$ To go from brackets to commutators, multiply with $i\hbar$.

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