# Massless representations of the Poincare group

1. Jun 17, 2009

### nrqed

Never mind, I answered my own question...

Last edited: Jun 17, 2009
2. Jun 17, 2009

### 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.

3. Jun 17, 2009

### nrqed

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