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The Pauli-Lubanski pseudo-vector is defined as:

$$W_{\mu}=\frac{1}{2}\epsilon_{\mu \nu \lambda \rho}J^{\nu \lambda}P^{\rho}$$

Where the rotation and translation operators transform as:

\begin{align} U(\Lambda,a)P^{\mu}U^{-1}(\Lambda,a)&=\Lambda^{\mu}_{\nu}P^{\nu}\\ U(\Lambda,a)J^{\mu \nu}U^{-1}(\Lambda,a)&=(\Lambda^{-1})^{\mu}_{\lambda}(\Lambda^{-1})^{\nu}_{\rho}(J^{\lambda \rho}-a^{\lambda}P^{\rho}+a^{\rho}P^{\lambda}) \end{align}

I'm working on how the Pauli-Lubanski transforms under Lorentz, to show that it's in fact Lorentz covariant. I saw a solution in one of the answers of this post Calculating the commutator of Pauli-Lubanski operator and generators of Lorentz group the following:

\begin{align} U(\Lambda,a)W_{\mu}U^{-1}(\Lambda,a)&=\frac{1}{2}\epsilon_{\mu \nu \lambda \rho}(\Lambda^{\nu}_{\alpha}\Lambda^{\rho}_{\beta}J^{\alpha \beta}+a^{\nu}\Lambda^{\rho}_{\alpha}P^{\alpha}-a^{\rho}\Lambda^{\nu}_{\rho}P^{\rho})\Lambda^{\sigma}_{\delta}P^{\delta}\\ &=\frac{1}{2}\Lambda^{\alpha}_{\mu}\epsilon_{\alpha \nu \rho \sigma}J^{\nu \rho}P^{\sigma}\end{align}

However, I'm having trouble handling the $\Lambda's$ in order to get from step one to step two. Could you offer me some guidance?

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  • $\begingroup$ Hint: are you comfortable with the transformation properties of $\epsilon$? $\endgroup$ – Cosmas Zachos Sep 6 '18 at 5:36
  • $\begingroup$ I only know that it is an antisymmetric tensor (equal to +1 for even permutations and -1 for odd permutations). I also read it is Lorentz invariant, so applying a Lorentz transformation doesn't affect it. $\endgroup$ – Charlie Sep 6 '18 at 7:01
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Since you're interested in the transformation law of $W^{\mu}$ under the Lorentz group, let us set $a^{\mu}=0$ in your equations. Also, notice that you've got the first one wrong; it should read

$$ U(\Lambda)P^{\mu}U(\Lambda)^{-1}=(\Lambda^{-1})^{\mu}_{\nu}P^{\nu} $$

Using the latter together with the transformation law for $J^{\mu\nu}$, we write

$$ U(\Lambda)W_{\mu}U(\Lambda)^{-1}=\frac{1}{2}\epsilon_{\mu\nu\lambda\rho}U(\Lambda)J^{\nu\lambda}P^{\rho}U(\Lambda)^{-1}=\frac{1}{2}\epsilon_{\mu\nu\lambda\rho}U(\Lambda)J^{\nu\lambda}U(\Lambda)^{-1}U(\Lambda)P^{\rho}U(\Lambda)^{-1}=\frac{1}{2}\epsilon_{\mu\nu\lambda\rho}(\Lambda^{-1})^{\nu}_{\tau}(\Lambda^{-1})^{\lambda}_{\sigma}(\Lambda^{-1})^{\rho}_{\alpha}\ J^{\tau\sigma}P^{\alpha} $$

Now, observe that from the definition of the determinant

$$ \epsilon_{\mu\nu\lambda\rho}(\Lambda^{-1})^{\mu}_{\beta}(\Lambda^{-1})^{\nu}_{\tau}(\Lambda^{-1})^{\lambda}_{\sigma}(\Lambda^{-1})^{\rho}_{\alpha}=\det(\Lambda^{-1})\ \epsilon_{\beta\tau\sigma\alpha} $$

follows the equality

$$\epsilon_{\mu\nu\lambda\rho}(\Lambda^{-1})^{\nu}_{\tau}(\Lambda^{-1})^{\lambda}_{\sigma}(\Lambda^{-1})^{\rho}_{\alpha}=\det(\Lambda^{-1})\ \Lambda_{\mu}^{\beta}\ \epsilon_{\beta\tau\sigma\alpha}$$

Therefore

$$ U(\Lambda)W_{\mu}U(\Lambda)^{-1}=\det(\Lambda^{-1})\Lambda_{\mu}^{\beta}\ \frac{1}{2}\ \epsilon_{\beta\tau\sigma\alpha}\ J^{\tau\sigma}P^{\alpha}=\det(\Lambda^{-1})\Lambda^{\beta}_{\mu}W_{\beta} $$

Rearranging the indices and using the defining property of the $\Lambda$'s, $\Lambda^{T}\eta\Lambda=\eta$, we find

$$ U(\Lambda)W^{\mu}U(\Lambda)^{-1}=\det(\Lambda^{-1})(\Lambda^{-1})_{\beta}^{\mu}W^{\beta} $$

Since $\det(\Lambda^{-1})=\pm 1$ depending on the Lorentz transformation, we find

$$ U(\Lambda)W^{\mu}U(\Lambda)^{-1}=\pm(\Lambda^{-1})_{\beta}^{\mu}W^{\beta} $$

which is the transformation law for a pseudo-vector.

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  • $\begingroup$ I see, then I can still then rearrange the components in the desired order (I also see there was a mistake in my lecture notes regarding the transformation of $P^{\mu}$). Also, I didn't think about using the definition of the determinant but I clearly see now that's the transformation property I was missing about the Levi-Civita tensor, and that's the part where one can find the pseudo-vector character of $W^{\mu}$ appears. I actually had to compute the general Lorentz transform $U(\Lambda,a)$ but I can work it from here now that I understand better how this tensor transforms. Thank you. $\endgroup$ – Charlie Sep 6 '18 at 11:51
  • $\begingroup$ You totally got the point. You're welcome. P.S.: If you think my answer was enough, you may want to accept it. Just saying! Have a good day! $\endgroup$ – Giorgio Comitini Sep 7 '18 at 12:08

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