Prove the following commutation relations

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SUMMARY

The discussion centers on proving the commutation relations involving the Pauli-Lubanski vector and the generators of the little group for massive representations, specifically the SO(3) group. The initial approach involves the expression $$[J^k, J^i] = \frac{1}{M^2}\ (L^{-1})^k_{\mu}\ (L^{-1})^i_{\nu}\ [W^{\mu}, W^{\nu}]$$, leading to a calculation that incorporates the structure constants of the Lorentz group. Participants emphasize the importance of working in the rest frame of the particle, where the momentum vector simplifies to $$P^{\mu}/M=(1,0,0,0)$$, and the Pauli-Lubanski vector is represented as $$W^{\mu}=(0,\vec{W})$$.

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  • Understanding of quantum field theory (QFT) principles
  • Familiarity with the Pauli-Lubanski vector and its role in representation theory
  • Knowledge of the Lorentz group and its algebraic structures
  • Experience with commutation relations in quantum mechanics
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  • Study the derivation of the Pauli-Lubanski vector in quantum field theory
  • Learn about the representations of the Lorentz group and their implications in particle physics
  • Research the mathematical framework of Lie algebras and their applications in QFT
  • Explore the relationship between SO(3) and SU(2) representations in quantum mechanics
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The discussion is beneficial for theoretical physicists, particularly those specializing in quantum field theory, representation theory, and particle physics, as well as advanced students seeking to deepen their understanding of the mathematical foundations of these concepts.

Hibarikyoya
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Homework Statement
I have to prove that:

$$[J^k, J^i] = i \epsilon^{kij} J^j$$

.
Relevant Equations
where $$J^k = \frac{1}{M} (L^{-1})^k_{\mu} W^{\mu},$$ in which M is a real number (with the meaning of a mass), $L(\vec{P},M)$ is a Lorentz transformation (in particular a boost in the direction identified by the momentum $\vec{P}$). I can also provide the explicit form of this transformation, but I think is not needed for this exercise. $W^{\mu}$ is the Pauli-Lubanski four vector. Moreover k, i and j run from 1 to 3 and they are spatial indices, while the greek indices run from 0 to 3
I tried in this way:
$$[J^k, J^i] = \frac{1}{M^2}\ (L^{-1})^k_{\mu}\ (L^{-1})^i_{\nu}\ [W^{\mu}, W^{\nu}] $$
$$ = \frac{1}{M^2}\ (L^{-1})^k_{\mu}\ (L^{-1})^i_{\nu}\ (-i) \epsilon^{\mu \nu \rho \sigma} W_{\rho} P_{\sigma}.$$
At this point I had no idea how to going on with the calculation. Can anyone help me?
 
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Dealing with
\mathbf{J}=\mathbf{r}\times\mathbf{p}
is a primitive way. Do you like to solve the problem in an advanced way ?
 
Hint: What you are supposed to prove is that the Pauli-Lubanski vector provides the generators (Lie-algebra basis elements) of the little group for the massive representations, where indeed the little group is the SO(3) (represented by representations of the SU(2), i.e., its covering group in QT).

Obviously the idea is to consider the Pauli-Lubanski-vector operators on the subspace of fixed three-momentum ##\vec{P}##. The boost transforms to the rest frame of the particle, where ##P^{\mu}/M=(1,0,0,0)##, and where ##W^{\mu}=(0,\vec{W})##.

In other words: For your prove you can simply work in the restframe of the particle and the three spatial components of ##\vec{W}## wrt. this frame.
 
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vanhees71 said:
Hint: What you are supposed to prove is that the Pauli-Lubanski vector provides the generators (Lie-algebra basis elements) of the little group for the massive representations, where indeed the little group is the SO(3) (represented by representations of the SU(2), i.e., its covering group in QT).

Obviously the idea is to consider the Pauli-Lubanski-vector operators on the subspace of fixed three-momentum ##\vec{P}##. The boost transforms to the rest frame of the particle, where ##P^{\mu}/M=(1,0,0,0)##, and where ##W^{\mu}=(0,\vec{W})##.

In other words: For your prove you can simply work in the restframe of the particle and the three spatial components of ##\vec{W}## wrt. this frame.
Very nice idea. The professor showed us his solution and it was a mess
 

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