I'm trying to find $[M_i, M_j]$ Poisson brackets.

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

I know that:

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

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

and so:

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

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

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 give me some hints to continue?

  • $\begingroup$ The Poisson bracket is clearly $ij$-antisymmetric – at every step. It means that the $i\leftrightarrow j$ permutation changes its overall sign. So it must be true for the final result, too. Why don't you just evaluate the last step properly, using $\epsilon_{abc}\epsilon_{ade} = \delta_{bd}\delta_{ce} - \delta_{be}\delta_{cd}$? Note that even this right hand side of mine is $bc$ and $de$ antisymmetric $\endgroup$ Commented Dec 27, 2012 at 10:12
  • $\begingroup$ try to factorize the $e_{ijk} $ and see if you obtain something that looks like the $k$ component of a cross product $\endgroup$
    – J L
    Commented Dec 27, 2012 at 10:14
  • $\begingroup$ @LubošMotl If I haven't done any calculus mistakes, I have obtained $\sum_l (\delta_{in}\delta_{km}-\delta_{kn}\delta_{im})(-p_k q_n \delta_{jl} \delta_{ml}+q_j p_m \delta_{kl} \delta_{nl})$. But now what can I do? Thanks a lot! :) $\endgroup$
    – sunrise
    Commented Dec 27, 2012 at 10:46
  • $\begingroup$ @burzum: how can I factorize $\epsilon_{ijk}$? $\endgroup$
    – sunrise
    Commented Dec 27, 2012 at 10:48

1 Answer 1


You are confusing in the index, such calculations must be carried out very carefully. I would start with your difention. $$M_i=\epsilon _{ijk} q_j p_k$$

$$M_p=\epsilon _{pnm} q_n p_m$$ $$\{M_i, M_p\}=\sum_l \left(\frac{\partial M_i}{\partial q_l}\frac{\partial M_p}{\partial p_l}-\frac{\partial M_i}{\partial p_l}\frac{\partial M_p}{\partial q_l}\right)$$

First term

$=\epsilon _{ijk}p_k\delta_{jl}\epsilon _{pnm}q_n\delta_{ml}=\epsilon _{ilk}p_k\epsilon _{pnl}q_n=(-1)\epsilon _{lik}p_k(-1)^2\epsilon _{lpn}q_n=-\epsilon _{lik}p_k\epsilon _{lpn}q_n=-\left(\delta_{ip}\delta_{kn}-\delta_{in}\delta_{kp}\right)p_kq_n$

Here I used the antisymmetry of $\epsilon _{lik}$ and equation $\epsilon_{ijk}\epsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}$

Second term

Absolutely the same calculations. $=\epsilon _{ijk}q_j\delta_{kl}\epsilon _{pnm}p_m\delta_{nl}=\epsilon _{ijl}q_j\epsilon _{plm}p_m=\epsilon _{plm}p_m\epsilon _{ijl}q_j=-\epsilon _{lpm}p_m\epsilon _{lij}q_j=-\left(\delta_{pi}\delta_{mj}-\delta_{pj}\delta_{mi}\right)p_mq_j=$

Make the change $m=k,j=n$. Then


All together

$\{M_i, M_p\}=-\left(\delta_{ip}\delta_{kn}-\delta_{in}\delta_{kp}\right)p_kq_n+\left(\delta_{pi}\delta_{kn}-\delta_{pn}\delta_{ki}\right)p_kq_n=\delta_{in}\delta_{kp}p_kq_n-\delta_{pn}\delta_{ki}p_kq_n=p_pq_i-p_iq_p=q_ip_p-p_iq_p$

  • 1
    $\begingroup$ What a wonderful answer!! :) And so, when I write the components of angular momentum, all indices of Levi-Civita symbols mustn't be equal, must they? thanks again! $\endgroup$
    – sunrise
    Commented Dec 27, 2012 at 17:07
  • $\begingroup$ For example, if you have $\epsilon_{112}$, since Levi-Civita symbol is antisymmetric, then $\epsilon_{112}=-\epsilon_{112}$ (swap the first two indexes), so $\epsilon_{112}=0$. $\endgroup$
    – Oiale
    Commented Dec 27, 2012 at 19:07
  • $\begingroup$ I'm sorry, I have explained in a wrong way my last doubt.. I try again: in the original problem I set $M_i=\epsilon _{ijk} q_j p_k$ and $M_j=\epsilon _{jnm} q_n p_m$. So I had $j$ both in $M_i$ and in $M_j$. You set $M_i=\epsilon _{ijk} q_j p_k$ and $M_p=\epsilon _{pnm} q_n p_m$. So you had only one j: my mistake is having $j$ twice, isn't it? thanks!! $\endgroup$
    – sunrise
    Commented Dec 27, 2012 at 21:34
  • 2
    $\begingroup$ Yes, that is your original mistake. $\endgroup$ Commented Dec 28, 2012 at 1:20

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