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I'm working on trying to solve the following problem:

Using the following expressión for the square of Pauli-Lubanski vector:$$W^2=-\frac{1}{2}M_{\mu\nu}M^{\mu\nu}P_{\alpha}P^{\alpha}+M^{\mu\nu}M_{\lambda\nu}P_{\mu}P^{\lambda}$$ show that $$[W^2,P_{\sigma}]=0$$ I know that there are other ways to prove the same result; for example, first showing that $[W_{\mu},P_{\nu}]=0$ but I have to use the formula that it gives to me, because It was deduced on the previous exercise. My calculations are like following; first of all $$[W^2,P_{\sigma}]=-\frac{1}{2}[M_{\mu\nu}M^{\mu\nu}P_{\alpha}P^{\alpha},P_{\sigma}]+[M^{\mu\nu}M_{\lambda\nu}P_{\mu}P^{\lambda},P_{\sigma}]$$ but, using the properties of conmutator, the conmutation relations of generators of Poincaré Algebra and renaming indexes I obtain, for the first conmutator on the right side, the following: $$[M_{\mu\nu}M^{\mu\nu}P_{\alpha}P^{\alpha},P_{\sigma}]=(M_{\mu\nu}[M^{\mu\nu},P_{\sigma}]+[M_{\mu\nu},P_{\sigma}]M^{\mu\nu})P_{\alpha}P^{\alpha}=i(M_{\mu\nu}\eta_{\sigma\lambda}(\eta^{\nu\lambda}P^{\mu}-\eta^{\nu\lambda}P^{\nu})+(\eta_{\nu\sigma}P_{\mu}-\eta_{\mu\sigma}P_{\nu})M^{\mu\nu})P_{\alpha}P^{\alpha}=2i(M_{\mu\sigma}P^{\mu}+\eta_{\nu\sigma}P_{\mu}M^{\mu\nu})P_{\alpha}P^{\alpha}$$ On the other hand, for the second conmutator on the right side I obtain the following: $$[M^{\mu\nu}M_{\lambda\nu}P_{\mu}P^{\lambda},P_{\sigma}]=[M^{\mu\nu}M_{\lambda\nu},P_{\sigma}]P_{\mu}P^{\lambda}=(M^{\mu\nu}[M_{\lambda\nu},P_{\sigma}]+\eta_{\sigma\epsilon}[M^{\mu\nu},P^{\epsilon}])P_{\mu}P^{\lambda}=(M^{\mu\nu}i(\eta_{\nu\sigma}P_{\lambda}-\eta_{\lambda\sigma}P_{\nu})+i\eta_{\sigma\epsilon}(\eta^{\nu\epsilon}P^{\mu}-\eta^{\mu\epsilon}P^{\nu})M_{\lambda\nu})P_{\mu}P^{\lambda}=i(\eta_{\nu\sigma}M^{\mu\nu}P_{\lambda}P_{\mu}P^{\lambda}+P^{\mu}M_{\lambda\sigma}P_{\mu}P^{\lambda})$$ The problem, accoding to me, comes when I try to compare both expressions using again the conmutation relations of Poincare algebra, because I obtain the following expression: $$\eta_{\nu\sigma}M^{\mu\nu}P_{\lambda}P_{\mu}P^{\lambda}+P^{\mu}M_{\lambda\sigma}P_{\mu}P^{\lambda}=\eta_{\nu\sigma}([M^{\mu\nu},P_{\mu}]+P_{\mu}M^{\mu\nu})P_{\lambda}P^{\lambda}+([P^{\mu},M_{\lambda\sigma}]+M_{\lambda\sigma}P^{\mu})P_{\mu}P^{\lambda}=\eta_{\nu\sigma}(i\eta_{\mu\alpha}(\eta^{\nu\alpha}P^{\mu}-\eta^{\mu\alpha}P^{\nu})+P_{\mu}M^{\mu\nu})P_{\lambda}P^{\lambda}+M_{\lambda\sigma}P_{\mu}P^{\mu}P^{\lambda}-i\eta^{\mu\alpha}(\eta_{\sigma\alpha}P_{\lambda}-\eta_{\lambda\alpha}P_{\sigma})P_{\mu}P^{\lambda}=(M_{\mu\sigma}P^{\mu}+\eta_{\nu\sigma}P_{\mu}M^{\mu\nu}-3iP_{\sigma})P_{\lambda}P^{\lambda}$$ I have tried for three complete days to eliminate the $3iP_{\sigma}$ term, but I couldn't. I have checked all my calculations, and made it many times using other ways, but I couldn't find where the mistake is, or what I'm doing wrong. Could anybody help me, please?. Am I not considering something? Note: In some steps I used that $M_{\mu\nu}=-M_{\nu\mu}$ :D .

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You have a subtle and interesting error. In your second commutator, there are four terms, and you thought that the second and fourth are both zero. Only the second is. The second term contains $M^{\mu\nu}P_\nu P_\mu$, and this is indeed zero because $M^{\mu\nu}$ is antisymmetric while $P_\nu P_\mu$ is symmetric. The fourth term contains $P^\nu M_{\lambda\nu}P^\lambda$, but this is not zero; $M$ and $P$ don't commute, so it isn't the contraction of an antisymmetric tensor with a symmetric tensor. After you use the commutation relation to move the $P$'s together, that part will vanish but the extra terms coming from the commutator won't.

I suggest that your calculation would be a bit nicer if you eliminated $\eta$ by simply using it to raise or lower indices. I would also write the contraction of $P$ with itself as $P^2$ whenever I could. Finally, I would use the commutator of $M$ and $P$ to put each of the two commutators in a “standard” form where $M$ is to the left of $P$. If you do this, you should find that the first commutator is $2i(2M_{\mu\sigma}P^\mu+3iP_\sigma)P^2$ and the second commutator is $i(2M_{\mu\sigma}P^\mu+3iP_\sigma)P^2$. Thus -1/2 times the first, plus the second, is zero.

By the way, in the first line of the second commutator you have a typo where the second $M$ in the second term is missing.

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  • $\begingroup$ Thank you very much for your answer. I really appreciate the time you spent to check my calculations, you helped me a lot!!! $\endgroup$ – Carlos Beltran Aug 31 at 23:50

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