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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!
     
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