How can I calculate the square of the Pauli-Lubanski pseudovector?

  • #1
tannhaus
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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!
 
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  • #2
Your expressions are manifestly covariant. "Spin" for a massive particle is, however, most easily to interpret in the rest frame of the particle. So to have some intuitive picture, it's best to calculate it within this frame, and this is simply defined by ##(p^{\mu})=(m c,0,0,0)##. In this frame you have a pretty intuitive interpretation of "spin" and the Pauli Lubanski vector (the latter one being the only viable definition of spin in relativistic physics, where in general a unique split of total angular momentum into "spin" and "orbital" is not possible). For more on "classical spin" in relativity, see Sect. 1.8 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

That becomes much clearer in the context of relativistic QFT and a detailed analysis of the representation theory of the Poincare group, where the Pauli Lubanski vector is the generator for little-group transformations, and the little group for massive-particle representations is the rotation group (or its covering group SU(2)) as defined in the rest frame of the particle. The quantities in other frames is then given by the (rotation free) Lorentz boosts from the rest frame of the particle to an arbitrary frame, where it's moving. For details see

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

(particularly Appendix B).
 

FAQ: How can I calculate the square of the Pauli-Lubanski pseudovector?

What is the Pauli-Lubanski pseudovector?

The Pauli-Lubanski pseudovector is a four-vector used in the study of the representations of the Poincaré group in quantum field theory. It is defined as \( W^\mu = \frac{1}{2} \epsilon^{\mu\nu\rho\sigma} P_\nu J_{\rho\sigma} \), where \( \epsilon^{\mu\nu\rho\sigma} \) is the Levi-Civita symbol, \( P_\nu \) is the four-momentum, and \( J_{\rho\sigma} \) are the components of the angular momentum tensor.

How is the square of the Pauli-Lubanski pseudovector defined?

The square of the Pauli-Lubanski pseudovector is defined as \( W^\mu W_\mu \). This scalar quantity is Lorentz invariant and is often used to classify the spin states of particles in relativistic quantum field theory.

What is the physical significance of the square of the Pauli-Lubanski pseudovector?

The square of the Pauli-Lubanski pseudovector is related to the Casimir invariant of the Poincaré group. For a particle of mass \( m \) and spin \( s \), \( W^\mu W_\mu = -m^2 s(s+1) \). This invariant helps in identifying the spin of the particle in a relativistic framework.

How do you compute the components of the Pauli-Lubanski pseudovector?

To compute the components of the Pauli-Lubanski pseudovector \( W^\mu \), you need the four-momentum \( P^\mu \) and the angular momentum tensor \( J^{\rho\sigma} \). The components are given by \( W^\mu = \frac{1}{2} \epsilon^{\mu\nu\rho\sigma} P_\nu J_{\rho\sigma} \). This involves summing over the indices using the Levi-Civita symbol.

Can you provide an example calculation of \( W^\mu W_\mu \) for a specific particle?

Consider a massive particle with mass \( m \) and spin \( s \). The four-momentum can be taken as \( P^\mu = (m, 0, 0, 0) \) in the rest frame. The angular momentum tensor \( J^{\rho\sigma} \) can be related to the spin operators. Using these, you can calculate \( W^\mu \) and then find \( W^\mu W_\mu \). For a spin-1/2 particle, \( W^\mu W_\mu = -\frac{3

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