- #1

tannhaus

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- TL;DR Summary
- I need to calculate the square of the Pauli-Lubanski pseudovector in a rest frame such that the results is proportional to the square of the spin operator.

Hello there, recently I've been trying to demonstrate that, $$\textbf{W}^2 = -m^2\textbf{S}^2$$ in a rest frame, with ##W_{\mu}## defined as $$W_{\mu} = \dfrac{1}{2}\varepsilon_{\mu\alpha\beta\gamma}M^{\alpha\beta}p^{\gamma}$$ such that ##M^{\mu\nu}## is an operator of the form $$ M^{\mu\nu}=x^{\mu}p^{\nu} - x^{\nu}p^{\mu} + \frac{i}{2}\Sigma^{\mu\nu}$$ and ##S^i## defined as $$S_i = \varepsilon^{ijk}\frac{i}{2}\Sigma^{jk}$$ Where ##\Sigma^{\mu\nu} = [\beta^{\mu}, \beta^{\nu}]##. I've managed to show that ##\textbf{S}^2 = \dfrac{1}{2}\Sigma^{ij}\Sigma_{ij}## but I can't for my life work out the necessary result. Any sort of light towards this is very welcome!