Poisson Brackets / Levi-Civita Expansion

[Bismar]

Hi,

I am stumped by how to expand/prove the following identity,

$$\{L_i ,L_j\}=\epsilon_{ijk} L_k$$

I am feeling that my knowledge on how to manipulate the Levi-Civita is not up to scratch.

Am i correct in assuming,

$$L_i=\epsilon_{ijk} r_j p_k$$
$$L_j=\epsilon_{jki} r_k p_i$$

Which follows on to,

$$\{L_i ,L_j\}=\{\epsilon_{ijk} r_j p_k,\epsilon_{jki} r_k p_i\}$$

And then I'm stuck. I'm assuming the Kronecker Delta identities with the Levi-Civita work into it in some way, but i do not understand how/why.

I can work it out if i expanded the Levi-Civita to such,

$$L_1=r_2 p_3 - r_3 p_2$$
$$L_2=r_3 p_1 - r_1 p_3$$
$$L_3=r_1 p_2 - r_2 p_1$$

But then that's trivial... :(

 

[dextercioby]

Please, use different summation indices.

$$\left\{\epsilon_{ilk} r_{l}p_{k}, \epsilon_{jmn}r_{m}p_{n}\right\}$$

Then, of course,

$$\left\{r_i, p_j\right\} = \delta_{ij}$$

 

[Bismar]

Sorry, I'm afraid i do not understand, where does that get you?

$$\epsilon_{ikl} \epsilon_{jmn}(\frac{dr_k p_l}{dr_o} \frac{dr_m p_n}{dp_o} - \frac{dr_m p_n}{dr_o} \frac{dr_k p_l}{dp_o})$$

$$= \epsilon_{ikl} \epsilon_{jmn}(\delta_{ko}\delta_{no}p_l r_m -\delta_{mo}\delta_{lo} p_n r_k)$$

If that's even right, which I'm sure isn't, I'm stuck again.

 

[vanhees71]

$$\begin{split} \{L_a,L_b\} &=\epsilon_{acd} \epsilon_{bef} \{x_c p_d,x_e p_f\} \\ &= \epsilon_{acd} \epsilon_{bef} (\{x_c,x_e p_f \} p_d+x_c \{p_d,x_e p_f \}) \\ &= \epsilon_{acd} \epsilon_{bef}(x_e p_d\delta_{cf} -x_c p_f \delta_{de}) \\ &= \epsilon_{acd} \epsilon_{bec} x_e p_d - \epsilon_{acd} \epsilon_{bdf} x_c p_f \\ &=[(-\delta_{ab} \delta_{de}+\delta_{ae} \delta_{db}) x_e p_d+ (\delta_{ab} \delta_{cf} - \delta_{af} \delta_{cb}) x_c p_f] \\ &= -\delta_{ab} \vec{x} \cdot \vec{p} + x_a p_b+\delta_{ab} \vec{x} \cdot \vec{p} -x_b p_a \\ &=x_a p_b-x_b p_a=\epsilon_{abc} L_c \end{split}$$

 

[PJC01]

Hi,

Thank you for reaching out with your question. The Poisson bracket and Levi-Civita expansion are both important concepts in mathematics and physics, and understanding their relationship can be challenging. Let's break it down step by step to see if we can make it clearer for you.

First, it is important to note that the Poisson bracket is a mathematical operation used to describe the dynamics of a system, while the Levi-Civita symbol is a mathematical object used to represent the cross product in three-dimensional space. The two are related, but they serve different purposes.

Now, let's start with your initial assumption that L_i=\epsilon_{ijk} r_j p_k and L_j=\epsilon_{jki} r_k p_i. These expressions are correct, as they represent the angular momentum components in terms of position and momentum variables. However, it is important to note that the Levi-Civita symbol is not a vector, but rather a tensor, so the order of indices matters.

Next, let's look at the Poisson bracket of these two expressions, \{L_i ,L_j\}. Using the definition of the Poisson bracket, we can expand this expression as follows:

\{L_i ,L_j\}=\frac{\partial L_i}{\partial r_k}\frac{\partial L_j}{\partial p_k}-\frac{\partial L_i}{\partial p_k}\frac{\partial L_j}{\partial r_k}

Substituting in the expressions for L_i and L_j that you provided, we get:

\{L_i ,L_j\}=\epsilon_{ikl} p_l \epsilon_{jkm} r_m - \epsilon_{ikl} r_l \epsilon_{jkm} p_m

Using the properties of the Levi-Civita symbol, we can rewrite this as:

\{L_i ,L_j\}=\epsilon_{ikl} \epsilon_{jkm} (p_l r_m - r_l p_m)

Now, using the well-known identity \epsilon_{ijk}\epsilon_{lmn}=\delta_{il}\delta_{jm}\delta_{kn}-\delta_{im}\delta_{jl}\delta_{kn}+\delta_{in}\delta_{jl}\delta_{km}, we can simplify this expression to:

\{L_i ,L_j\}=\delta_{ij} L_k - \delta_{ik} L_j

We can further