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Spinor map and Lorentz transformations

  1. Aug 3, 2013 #1
    At the moment I'm reading a little bit about the connection between the Lorentz group [itex] SO^{+}(1,3)[/itex] and its covering group [itex]SL(2,\mathbb{C})[/itex].

    On wiki I found this:
    https://en.wikipedia.org/wiki/Lorentz_group#Elliptic

    But I think there is a sign error in the [itex]P_{1}[/itex] matrix. It should be:
    [tex]Q_1 = \left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & \cos(\theta) & \sin(\theta) & 0 \\ 0 & -\sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right][/tex]

    Could anyone confirm (or disporve) this?

    Here my calculation:

    I started with:
    [tex]
    X = x^{\mu} \sigma_{\mu} = \left[ \begin{matrix} t+z & x-iy \\ x+iy & t-z \end{matrix}, \right]
    [/tex]
    where [itex] x^{\mu} = (t, x, y, z) [/itex] and [itex] \sigma_{\mu} = (1_{2 \times 2}, \vec{\sigma}) [/itex]
    and
    [tex]
    P_1 = \left[ \begin{matrix} \exp(i \theta/2) & 0 \\ 0 & \exp(-i \theta/2) \end{matrix} \right]
    [/tex]

    Computing [itex] X' \equiv P_{1} X P_{1}^{\dagger} [/itex] leads to:
    [tex]
    X' = \left[ \begin{matrix} t+z & \exp(i \theta/2) (x-iy) \\ \exp(-i \theta/2) (x+iy) & t-z \end{matrix} \right].
    [/tex]
    and using the relation [itex] (x')^{\mu} = \dfrac{1}{2} \mathrm{tr} \left( \underline{\sigma}^{\mu} X' \right) [/itex], where [itex] \underline{\sigma}^{\mu} = \sigma_{\mu} [/itex] I found:
    [tex]
    (x')^{1} = \dfrac{1}{2} \mathrm{tr} \left( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} X' \right)
    = x \cdot \cos \theta + y \cdot \sin \theta
    [/tex]

    If I check the other matrices like [itex] P_{2} [/itex] everything is ok.
     
  2. jcsd
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