# Pauli-Lubanski vector

1. Oct 17, 2009

### neworder1

1. The problem statement, all variables and given/known data

Let $$L_{\vec{p}}$$ be a Lorentz transform which takes a particle with 0 momentum to a particle with momentum $$\vec{p}$$. Define $$\vert \vec{p}, \sigma \rangle = L_{\vec{p}} \vert 0, \sigma \rangle$$, where $$\sigma$$ is spin.

Let $$\vec{s}$$ be a spatial vector such that $$\vec{s} \cdot \vec{J}\vert 0, \sigma \rangle = \sigma \vert 0, \sigma \rangle$$. Put $$s_{\vec{p}}=L_{\vec{p}}(s)$$ ($$s = (0,\vec{s})$$).

Define the Pauli-Lubianski vector $$W_{\mu} = -\frac{1}{2}\epsilon_{\mu\nu\alpha\beta}J^{\nu\alpha}P^{\beta}$$. Prove that $$\vert \vec{p}, \sigma \rangle$$ is an eigenvector of the operator $$s_{\vec{p}}W$$.

2. Relevant equations

3. The attempt at a solution

I tried the following way:
$$s_{\vec{p}}W L_{\vec{p}} \vert 0, \sigma \rangle= L_{\vec{p}} s_{\vec{p}}W \vert 0, \sigma \rangle$$, since $$s_{\vec{p}}W$$ is a four-scalar, so commutes with Lorentz transforms. Now I can use the fact that $$P^{\mu} \vert 0, \sigma \rangle \neq 0$$ only for $$\mu = 0$$, but what then? The resulting expression is easily seen to be $$L_{\vec{p}}\vec{s_{\vec{p}}} \cdot \vec{J}\vert 0, \sigma \rangle$$, but it doesn't help.

Last edited: Oct 17, 2009