# [Physics] The square of Pauli-Lubanski operator

operatorspoincare-symmetryquantum-spin

Let's have Pauli-Lubanski operator:

$$\hat {W}^{\alpha} = \frac{1}{2}\varepsilon^{\alpha \beta \gamma \delta}\hat {J}_{\beta \gamma}\hat {P}_{\delta} = \frac{1}{2}\varepsilon^{\alpha \beta \gamma \delta}\hat {S}_{\beta \gamma}\hat {P}_{\delta},$$

where $\hat {J}_{\beta \gamma} = \hat {L}_{\beta \gamma} + \hat {S}_{\beta \gamma}$ and $\hat {S}_{\beta \gamma}$ is the spin-tensor.

In some books there is a following expression for square of this operator:

$$\hat {W}_{\lambda}\hat {W}^{\lambda } = \hat {P}_{\alpha}\hat {P}_{\beta }\hat {S}^{\alpha \mu}\hat {S}_{\beta \mu} -\frac{1}{2}\hat {P}_{\alpha}\hat {P}^{\alpha}\hat {S}_{\alpha \beta}\hat {S}^{\alpha \beta }.$$
But I don't understand how to get it, because I can't claim, that $\hat {S}_{\alpha \beta}$ is antisymmetrical. So, I only get, using expression
$$\varepsilon_{\alpha \beta \gamma \lambda}\varepsilon^{\lambda \mu \nu \sigma} = \delta^{\mu}_{\alpha}\varepsilon^{\nu \sigma}_{\beta \gamma} + \delta^{\mu}_{\gamma}\varepsilon^{\nu \sigma}_{\beta \alpha} + \delta^{\mu}_{\beta}\varepsilon^{\nu \sigma}_{\gamma \alpha}=\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}\delta^{\sigma}_{\gamma} – \delta^{\mu}_{\alpha}\delta^{\nu}_{\gamma}\delta^{\sigma}_{\beta} + \delta^{\nu}_{\alpha}\delta^{\sigma}_{\beta}\delta^{\mu}_{\gamma} – \delta^{\nu}_{\alpha}\delta^{\mu}_{\beta}\delta^{\sigma}_{\gamma} + \delta^{\sigma}_{\alpha}\delta^{\mu}_{\beta}\delta^{\nu}_{\gamma} – \delta^{\sigma}_{\alpha}\delta^{\nu}_{\beta}\delta^{\mu}_{\gamma},$$
following:
$$\hat {W}_{\lambda}\hat {W}^{\lambda} = \frac{1}{4}\varepsilon_{\lambda \alpha \beta \gamma }\hat {J}_{\alpha \beta}\hat {P}_{\gamma }\varepsilon^{\lambda \mu \nu \sigma}\hat {J}_{\mu \nu}\hat {P}_{\sigma} = -\varepsilon_{\alpha \beta \gamma \lambda}\varepsilon^{\lambda \mu \nu \sigma}\hat {J}_{\alpha \beta}\hat {P}_{\gamma }\hat {J}_{\mu \nu}\hat {P}_{\sigma} =$$

$$= \frac{1}{4}\left(\delta^{\mu}_{\alpha}\varepsilon^{\nu \sigma}_{\beta \gamma} + \delta^{\mu}_{\gamma}\varepsilon^{\nu \sigma}_{\beta \alpha} + \delta^{\mu}_{\beta}\varepsilon^{\nu \sigma}_{\gamma \alpha}\right)\hat {J}_{\alpha \beta}\hat {P}_{\gamma }\hat {J}_{\mu \nu}\hat {P}_{\sigma} =$$

$$=-\frac{1}{4}\left( \hat {J}_{\alpha \beta}\hat {P}_{\gamma}\hat {J}^{\alpha \beta}\hat {P}^{\gamma} – \hat {J}_{\alpha \beta}\hat {P}_{\nu}\hat {J}^{\alpha \nu}\hat {P}^{\beta} + \hat {J}_{\alpha \beta}\hat {P}_{\gamma}\hat {J}^{\gamma \alpha}\hat {P}^{\beta} – \hat {J}_{\alpha \beta}\hat {P}_{\gamma}\hat {J}^{\beta \alpha}\hat {P}^{\gamma} + \hat {J}_{\alpha \beta}\hat {P}_{\gamma}\hat {J}^{\beta \gamma }\hat {P}^{\alpha} – \hat {J}_{\alpha \beta}\hat {P}_{\gamma}\hat {J}^{\gamma \beta}\hat {P}^{\alpha}\right).$$
So, if spin-tensor operator is antisymmetrical, I'll get the answer easily using Poincare algebra. But if not, I can't continue.

Can you help me to get the answer? Maybe, I can get the answer, because there is the identity of spin-tensor operator and orbital momentum tensor operator?

$$\hat {W}^{\alpha} = \frac{1}{2}\varepsilon^{\alpha \beta \gamma \delta}\hat {J}_{\beta \gamma}\hat {P}_{\delta}.$$
Where ${J}_{\beta \gamma}$ and ${P}_{\delta}$ correspond to the generators of the PoincarĂ© group. (respectively total angular moment, and momentum)
So, yes, ${J}_{\beta \gamma}$ is of corse antisymmetric, as the orbital angular moment $L_{\beta \gamma}$, and as the spin operator $S_{\beta \gamma}$.