Pauli-Lubanski Vector: Proving Eigenvector of s_pW

[neworder1]

Homework Statement

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 $$s_{\vec{p}}W$$.

Homework Equations

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.

 

[Jhero]

To prove that \vert \vec{p}, \sigma \rangle is an eigenvector of s_{\vec{p}}W, we need to show that s_{\vec{p}}W \vert \vec{p}, \sigma \rangle = \lambda \vert \vec{p}, \sigma \rangle, where \lambda is some eigenvalue. Let's start by expanding s_{\vec{p}}W as s_{\vec{p}}W = L_{\vec{p}}(s)W = L_{\vec{p}}(s) \cdot \frac{1}{2}\epsilon_{\mu\nu\alpha\beta}J^{\nu\alpha}P^{\beta}.

Now, using the definition of L_{\vec{p}}, we can rewrite this as s_{\vec{p}}W = L_{\vec{p}}(s) \cdot \frac{1}{2}\epsilon_{\mu\nu\alpha\beta}J^{\nu\alpha}L_{\vec{p}}(P)^{\beta}.

Next, we can use the fact that L_{\vec{p}}(P) = P + \vec{p}, since the Lorentz transform only affects the spatial components of the momentum. This gives us s_{\vec{p}}W = L_{\vec{p}}(s) \cdot \frac{1}{2}\epsilon_{\mu\nu\alpha\beta}J^{\nu\alpha}(P^{\beta} + p^{\beta}).

Since \vert \vec{p}, \sigma \rangle is an eigenvector of P^{\beta} with eigenvalue \vec{p}, we can rewrite this as s_{\vec{p}}W = L_{\vec{p}}(s) \cdot \frac{1}{2}\epsilon_{\mu\nu\alpha\beta}J^{\nu\alpha}(p^{\beta}) \vert \vec{p}, \sigma \rangle.

Now, we can use the fact that \vec{s} \cdot \vec{J}\vert 0, \sigma \rangle = \sigma \vert 0, \sigma \rangle, and the definition of the Pauli-Lubianski vector, to rewrite this as s_{\vec{p}}W =