Massless representations of the Poincare group

[nrqed]

Never mind, I answered my own question...

 

[meopemuk]

By definition

W_0 = \mathbf{P} \cdot \mathbf{J}

Applying this operator to | p \rangle we obtain

W_0 | p \rangle = P_3 J_3 | p \rangle

J_3 is a generator of the "little group" which leaves this vector invariant (up to a constant factor), so (see eq. (2.5.39) in Weinberg's "The quantum theory of fields")

W_0 | p \rangle = P_3 \lambda| p \rangle = p_0 \lambda| p \rangle

where \lambda is helicity.

 

[nrqed]

By definition

W_0 = \mathbf{P} \cdot \mathbf{J}

Applying this operator to | p \rangle we obtain

W_0 | p \rangle = P_3 J_3 | p \rangle

J_3 is a generator of the "little group" which leaves this vector invariant (up to a constant factor), so (see eq. (2.5.39) in Weinberg's "The quantum theory of fields")

W_0 | p \rangle = P_3 \lambda| p \rangle = p_0 \lambda| p \rangle

where \lambda is helicity.

Thank you. My problem was in convincing myself that J_3 leaves the state invariant. I will look at Weinberg when I can

Thanks again