We already know that the generators of Lorentz group are three boosts and three rotations. We have the relation $$ \left[ M_{\mu \nu},M_{\sigma \lambda} \right] = ig_{\mu \sigma} M_{\nu\lambda} - ig_{\mu \lambda} M_{\nu\sigma} + ig_{\mu \nu} M_{\mu\sigma} - ig_{\nu \sigma} M_{\mu \lambda} $$ and from this we can obtain the commutation relations of $J_i$ and $K_i$ using $$ M_{ij}=\varepsilon_{ijk}J_k, \qquad M_{0i}=K_i $$

I want to make the inverse way: how can we obtain $ \left[ M_{\mu \nu},M_{\sigma \lambda} \right]$ knowing the commutation relations of $J_i$ and $K_i$?

  • $\begingroup$ Insert the definitions of $M_{\mu\nu}$ in terms of $J$ and $M$ into the commutator and expand? $\endgroup$
    – TimRias
    Oct 18, 2017 at 14:57
  • $\begingroup$ How would you know the commutation relations for boosts in the first place? $\endgroup$
    – DanielC
    Oct 18, 2017 at 15:15

1 Answer 1


Let us start from

$$ M_{ij}=\epsilon_{ijk}J_{k}\ ,\quad M_{0i}=-K_i $$

(the minus sign is my convention, sorry!). If you know, a priori, that

\begin{align*} [J_i,J_j]&=i\epsilon_{ijk}J_k\ ,\\ [K_i,K_j]&=-i\epsilon_{ijk}J_k\ ,\\ [J_i,K_j]&=i\epsilon_{ijk}K_j\ , \end{align*}

then you can just do a brute force computation for appropriate indices and figure out the pattern for the general case. For example

\begin{align*} [M_{0i},M_{0j}]&=[K_i,K_j]=-i\epsilon_{ijk}J_k=-i\,M_{ij}\ , \end{align*}

and so on. Alternatively, you can write down $J$ explicitly in terms of $M$ by doing

$$ \epsilon_{ijq}M_{ij}=\epsilon_{ijq}\epsilon_{ijk}J_k=2\delta_{qk}J_k\Rightarrow J_k=\frac{1}{2}\epsilon_{kij}M_{ij}\ . $$

Then we can substitute this directly on the commutation relation for $J$'s

$$ \frac{1}{4}[\epsilon_{mni}M_{mn},\epsilon_{pqj}M_{pq}]=i\epsilon_{ijk}\epsilon_{uvk}M_{uv}=i\left(\delta_{iu}\delta_{jv}-\delta_{iv}\delta_{ju}\right)M_{uv}=-2iM_{ij}\ .\quad (\Box) $$

Then you must replace the product of Levi-Civita symbols on the LHS of $(\Box)$ by their representation of the product of $\delta$'s to get

$$ [M_{ik},M_{jk}]=-2iM_{ij}\ . $$

You then carry on substituting $J$ on the commutation relation for $J$ with $K$, but in the end you still must guess the general pattern.

Unfortunately none of these approaches are very practical. If all you want is a way to find the commutation relations for the generators of the Lorentz group, I suggest writing them down on a particular representation, e.g.

$$ (M_{\mu\nu})^{\sigma}_{\ \ \rho}=i\left(\eta_{\mu\rho}\delta^\sigma_{\ \ \nu}-\eta_{\nu\rho}\delta^\sigma_{\ \ \mu}\right)\ . $$

On the above expression, $\sigma$ and $\rho$ represent matrix indices sort of speak. Then one knows how $M_{\mu\nu}$ transforms under a Lorentz transformation

$$ \Lambda M_{\mu\nu}\Lambda^{-1}=M_{\lambda\sigma}\Lambda^{\lambda}_{\ \ \mu}\Lambda^{\sigma}_{\ \ \nu}.\quad (\star) $$

Considering infinitesimal transformations $\Lambda=1-\frac{i}{2}\xi^{\lambda\sigma}M_{\lambda\sigma}$ (for some antisymmetric parameter $\xi$), the LHS of $(\star)$ can be written as

$$ \Lambda M_{\mu\nu}\Lambda^{-1}=M_{\mu\nu}+\frac{i}{2}\xi^{\lambda\sigma}[M_{\mu\nu},M_{\lambda\sigma}]+\mathcal{O}(\xi^2)\ , $$

while the RHS of $(\star)$ can be written as

$$ M_{\lambda\sigma}\Lambda^{\lambda}_{\ \mu}\Lambda^{\sigma}_{\ \nu}=M_{\mu\nu}-\frac{1}{2}\xi^{\lambda\sigma}\left(M_{\mu\lambda}\eta_{\nu\sigma}-M_{\mu\sigma}\eta_{\nu\lambda}+M_{\lambda\nu}\eta_{\mu\sigma}-M_{\sigma\nu}\eta_{\mu\lambda}\right)+\mathcal{O}(\xi^2)\ , $$

thus implying

$$ [M_{\mu\nu},M_{\lambda\sigma}]=i\left(M_{\mu\lambda}\eta_{\nu\sigma}-M_{\mu\sigma}\eta_{\nu\lambda}+M_{\lambda\nu}\eta_{\mu\sigma}-M_{\sigma\nu}\eta_{\mu\lambda}\right)\ . $$

Hope this helps! Cheers.


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