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I’m studying the generators of Lorentz group in QFT. How can I prove the algebra of the rotation generators $J_i$ $$\left [ J_i,J_j\right ] =i\epsilon_{ijk}J_k $$ from the algebra of the Lorentz group generators $M^{\mu\nu}$?

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    $\begingroup$ By giving particular values to $\mu, \nu$ from from 1,2,3. $\endgroup$
    – DanielC
    Commented Dec 8, 2020 at 16:09

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Write $J_i$ in terms of $M_{\mu \nu}$, namely, $J_i=\tfrac 1 2\varepsilon_{ijk}M_{jk}$ and compute the commutator.

Let's use only the spatial part \begin{eqnarray} [M_{ij},M_{kl}]=i \left ( M_{ik}\delta_{jl}-M_{jk}\delta_{il}-M_{il}\delta_{jk}+M_{jl}\delta_{ik}\right), \end{eqnarray} then \begin{eqnarray} [J_p,J_q]& = &\frac{1}{4}\varepsilon_{ijp}\varepsilon_{klq}[M_{ij},M_{kl}]= \frac{i}{4}\varepsilon_{ijp}\varepsilon_{klq} \left ( M_{ik}\delta_{jl}-M_{jk}\delta_{il}-M_{il}\delta_{jk}+M_{jl}\delta_{ik}\right)\\ & = & \frac{i}{4}\left ( M_{ik}\varepsilon_{ijp}\varepsilon_{kjq}-M_{jk}\varepsilon_{ijp}\varepsilon_{kiq}-M_{il}\varepsilon_{ijp}\varepsilon_{jlq}+M_{jl}\varepsilon_{ijp}\varepsilon_{ilq}\right)=i M_{ik}\varepsilon_{jip}\varepsilon_{jkq} \\ & = & i M_{ik}(\delta_{ik}\delta_{pq}-\delta_{iq}\delta_{pk})=iM_{pq}=i\varepsilon_{pq k} J_k. \end{eqnarray} The way it is supposed to be. Do not forget that there is no $\frac{1}{2}$ for the inverse relation between $J$ and $M$, namely, that $M_{jk}=\varepsilon_{jki}J_i$. To see that multiply $J_i=\tfrac 1 2\varepsilon_{ijk}M_{jk}$ by $\varepsilon_{ipq}$, that will lead to \begin{equation} \varepsilon_{pqi} J_i=\frac{1}{2}\varepsilon_{ipq}\varepsilon_{ijk}M_{jk}=\frac{1}{2}(\delta_{pj}\delta_{qk}-\delta_{pk}\delta_{qj})M_{jk}=\frac{1}{2}(M_{pq}-M_{qp})=M_{pq}. \end{equation}

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  • $\begingroup$ Done that, but I get a $1/2$ factor since we’ve defined $J_i=\frac{1}{2}\epsilon_{ijk}M^{jk}$. $\endgroup$
    – john
    Commented Dec 8, 2020 at 13:17
  • $\begingroup$ my bad, of course it should be defined with $\frac{1}{2}$ $\endgroup$
    – nwolijin
    Commented Dec 8, 2020 at 13:23
  • $\begingroup$ And I get the $1/2$ factor also on the commutator between $J_i$ and $K_j$ and between $K_i$ and $K_j$ $\endgroup$
    – john
    Commented Dec 8, 2020 at 13:51
  • $\begingroup$ Are you saying that with $\frac{1}{2}$ you do not get the result? $\endgroup$
    – nwolijin
    Commented Dec 8, 2020 at 16:11
  • $\begingroup$ Thank you very much! Could you also explain why there is no $\frac{1}{2}$ in the inverse relation? $\endgroup$
    – john
    Commented Dec 8, 2020 at 17:16

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