The Pauli-Lubanski vector and angular momentum

[redtree]

TL;DR Summary

I have seen it stated but not proven that in the formula for the Pauli-Lubanski vector, the orbital angular momentum drops out from $M_{\rho \sigma}$ because of the anti-symmetry with the momentum operator, such that it includes only the spin part of the angular momentum. Where can I find an explicit proof of this statement?

Given $$M_{\rho \sigma} = i (x^{\rho} \partial_{\sigma} - x^{\sigma} \partial_{\rho})$$
and $$W^{\mu} = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} P_{\nu} M_{\rho \sigma}$$

Why does $W^{\mu}$ pick up only the spin part of the total angular momentum?

 

[Haborix]

Looks to me like you can just plug in $M_{\rho \sigma}$ and the derivative for $P_{\nu}$ and just compute that $W_\mu$ is zero, as you have defined it (without the spin component). You will just need to appeal to the anti-symmetry of the Levi-Cevita symbol. I see a couple terms with a repeated index in the Levi-Cevita symbol and couple where it gets contracted with symmetric terms. It is about 3 lines of index algebra/calculus.