I need some clarifications about Poisson brackets.

I know canonical brackets and the properties of Poisson Brackets and I also know something about Levi-Civita symbol (definition and basic properties), but I have some doubts.

I don't know how I could apply Poisson brackets properties if I have a summation, for example if in the case of a system of N particles I have to solve $[L_i, x_{\alpha j}]$. I know that a generical component of total angular momentum is given by $L_a=\sum_{\alpha=1} ^N l_{\alpha a}$ and also know that the components of algular momentum af a particle is given by $l_a= \epsilon_{aij} x_i p_j$. Now, if I have to calculate $[L_i, x_{\alpha j}]$, I have these doubts:

1) how can I decide the indices of Levi-Civita symbol that I'm going to use for solving the problem?

2) how can I use the property of linearity of Poisson brackets in this case?

and an other (general) question:

3) If I have a Levi-Civita symbol that multiplyes a sum of two terms and each term is mulplied for a Kronecker delta, I have to follow these steps:

a) multiply the Levi-Civita symbol for each term

b) impose the condition thanks to each Kronecker delta isn't equal to zero

c) eventually, substitute these conditions in the two Levi-Civita symbol, but I have to substitute in each Levi-Civita symbol the condition that I found for that Kronecker delta that at the step a) was multipling just that one Levi-Civita symbol

Is it correct this way to go on?


1 Answer 1


1.) Always choose indices in such a way that the free indices on both sides of the equation match. Furthermore, make sure that you don't mix up summation indices of different summations.
2.) Linearity implies that if you enter a sum as the argument of a Poisson brackets, you get a sum of Poisson brackets, with each of them having a single term your original sum as the argument, i.e. \begin{equation}[\Sigma^N_{\alpha=1}B_\alpha,A]=[B_1,A]+[B_2,A]+[B_3,A]+\ldots+[B_N,A] \end{equation} 3.) If you mean something like \begin{equation}\epsilon_{ijk}(A_{jl}\delta_{lk}+B_{kl}\delta_{jl})=\epsilon_{ijk}A_{jl}\delta_{lk}+\epsilon_{ijk}B_{kl}\delta_{jl}=\epsilon_{ijk}A_{jk}+\epsilon_{ijk}B_{kj},\end{equation} the answer is yes.


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