Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
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 –  Luboš Motl Dec 27 '12 at 10:12
try to factorize the $e_{ijk} $ and see if you obtain something that looks like the $k$ component of a cross product –  Jorge Dec 27 '12 at 10:14
@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! :) –  sunrise Dec 27 '12 at 10:46
@burzum: how can I factorize $\epsilon_{ijk}$? –  sunrise Dec 27 '12 at 10:48
add comment

1 Answer

up vote 5 down vote accepted

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

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

share|improve this answer
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! –  sunrise Dec 27 '12 at 17:07
For example, if you have $\epsilon_{112}$, since Levi-Civita symbol is antisymmetric, then $\epsilon_{112}=-\epsilon_{112}$ (swap the first two indexes), so $\epsilon_{112}=0$. –  Oiale Dec 27 '12 at 19:07
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!! –  sunrise Dec 27 '12 at 21:34
Yes, that is your original mistake. –  Emilio Pisanty Dec 28 '12 at 1:20
@EmilioPisanty: thank you! :) –  sunrise Dec 28 '12 at 9:40
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.