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Lorentz generators

  1. Mar 27, 2009 #1
    I know the representations of Lorentz generators M[tex]\mu\nu[/tex] as 4X4 matrices.

    This matrices satisfy the commutation relation(Lie algebra of O(3,1))

    However I think these 4X4 matrix representations are not unique.

    Is there any other representations satisfying the commutation relation? 2X2 matrix or

    another?
     
  2. jcsd
  3. Mar 27, 2009 #2

    Ben Niehoff

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    Yes, there are infinitely many representations of the Lorentz algebra. One trivial example is to form a set of 8x8 matrices by taking the Kronecker product of M with the 2x2 identity.
     
  4. Mar 27, 2009 #3

    George Jones

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    Because [itex]SL \left(2 , \mathbb{C} \right)[/itex] is the universal cover of the restricted Lorentz group, they have isomorphic Lie algebras.
     
  5. Mar 27, 2009 #4
    This means can I get the 2x2 representation of M[tex]\mu\nu[/tex] from SL(2,C)?

    But how? What I know is [tex]\sigma_{\mu}A^{\mu} _{\phantom{\mu}\nu}}[/tex]=[tex]L\sigma_{\nu}L^{\dagger}[/tex]
     
  6. Mar 28, 2009 #5

    samalkhaiat

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    Take

    [tex]
    M^{0i} = \frac{i}{2} \sigma^{i}, \ \ M^{ij} = \frac{1}{2} \epsilon^{ijk}\sigma^{k}
    [/tex]

    Also, if you define

    [tex]\omega_{i0} = 2 Re(\alpha_{i}), \ \ \omega_{ij} = 2 \epsilon^{ijk} Im(\alpha_{k})[/tex]

    then

    [tex]
    1 + \frac{i}{2}\omega_{\mu\nu}M^{\mu\nu} = \left( \begin {array}{rr} (1 + \alpha_{3}) & (\alpha_{1} - i \alpha_{2}) \\ (\alpha_{1} + i \alpha_{2}) & (1 - \alpha_{3}) \end {array} \right)
    [/tex]

    represents elements of SL(2,C) infinitesimally close to the identity element.

    regards

    sam
     
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