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Pauli-Lubanski vector

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data

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

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

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

    2. Relevant equations

    3. The attempt at a solution

    I tried the following way:
    [tex]s_{\vec{p}}W L_{\vec{p}} \vert 0, \sigma \rangle= L_{\vec{p}} s_{\vec{p}}W \vert 0, \sigma \rangle[/tex], since [tex]s_{\vec{p}}W[/tex] is a four-scalar, so commutes with Lorentz transforms. Now I can use the fact that [tex]P^{\mu} \vert 0, \sigma \rangle \neq 0[/tex] only for [tex]\mu = 0[/tex], but what then? The resulting expression is easily seen to be [tex]L_{\vec{p}}\vec{s_{\vec{p}}} \cdot \vec{J}\vert 0, \sigma \rangle[/tex], but it doesn't help.
     
    Last edited: Oct 17, 2009
  2. jcsd
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