Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Poisson brackets of angular momentum components

  1. Dec 27, 2012 #1
    I want to find [M_i, M_j] Poisson brackets.

    $$[M_i, M_j]=\sum_l (\frac{\partial M_i}{\partial q_l}\frac{\partial M_j}{\partial p_l}-\frac{\partial M_i}{\partial p_l}\frac{\partial M_j}{\partial q_l})$$

    I know that:

    $$M_i=\epsilon _{ijk} q_j p_k$$

    $$M_j=\epsilon _{jnm} q_n p_m$$

    and so:

    $$[M_i, M_j]=\sum_l (\frac{\partial \epsilon _{ijk} q_j p_k}{\partial q_l}\frac{\partial \epsilon _{jnm} q_n p_m}{\partial p_l}-\frac{\partial \epsilon _{ijk} q_j p_k}{\partial p_l}\frac{\partial \epsilon _{jnm} q_n p_m}{\partial q_l})$$

    $$= \sum_l \epsilon _{ijk} p_k \delta_{jl} \cdot \epsilon_{jnm} q_n \delta_{ml}- \sum_l \epsilon_{ijk}q_j \delta_{kl} \cdot \epsilon_{jnm} p_m \delta_{nl}$$

    Then I have thought that values that nullify deltas don't add any informations in the summations. And so, $$m=l, j=l$$ but so I obtain $$m=j$$. But if $$m=l$$, the second Levi-Civita symbol in the first summation is zero... And if I go on, I obtain $$[M_i, M_j]=-p_iq_j$$ instead of $$[M_i, M_j]=q_ip_j-p_iq_j$$

    Where am I wrong? :| Could you say to me how to go on? Thanks a lot!
  2. jcsd
  3. Dec 27, 2012 #2
    You have 3 j's in the same term. Make sure your dummy indices (i.e. the ones that are summed over) are different from the variable indices. Use a different letter for each dummy index to avoid confusion.
  4. Dec 30, 2012 #3
    Thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook