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Commutators of Lorentz generators

  1. May 5, 2009 #1

    nrqed

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    Equation 2.14 of Srednicki is:

    [tex] U(\Lambda)^{-1} M^{\mu \nu} U(\Lambda) = \Lambda^{\mu}_{\,\,\rho} \Lambda^{\nu}_{\, \, \sigma} M^{\rho \sigma} [/tex]

    He says that writing [itex] \Lambda = 1 + \delta \omega [/itex], one obtains the usual commutation relation of the [itex] M_{\mu \nu} [/itex]:

    [tex] [ M^{\mu \nu},M^{\rho \sigma} ] = i ( g^{\mu \rho} M^{\nu \sigma} - (\mu \leftrightarrow \nu}) - (\rho \leftrightarrow \sigma) [/tex]

    I get, starting from the first equation

    [tex] [ M^{\mu \nu},M^{\rho \sigma} ] \delta \omega_{\rho \sigma} = i (\eta^\mu_\rho \delta \omega^\nu_\sigma M^{\rho \sigma} + \eta^\nu_\sigma \delta \omega^\mu_\rho M^{\rho \sigma} ) [/tex]

    I don't see how to extract the commutator because the omega on the right side contain one of the indices that is not summed over. Its' not clear to me how to proceed. Any suggestion?
     
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  3. May 5, 2009 #2

    StatusX

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    I think it should work if you stick in a kronecker delta.
     
  4. May 5, 2009 #3
    Check the right hand side of the equation, you can write


    [tex]

    \delta\omega^\nu_\sigma = \delta\omega_{\rho\sigma} \eta^{\rho \nu}

    [/tex]

    Edit:

    Also M is anti-symmetric, which means you can write it as

    [tex]

    M^{uv} = \frac{1}{2} (M^{uv} - M^{vu})

    [/tex]
     
    Last edited: May 5, 2009
  5. May 5, 2009 #4
    what level of quantum physics is this? grad?
     
  6. May 5, 2009 #5

    nrqed

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    Of course! (Slapping myself right now...I should have seen it). Thanks!!
     
  7. May 6, 2009 #6

    Fredrik

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    This is from a book on quantum field theory. You would normally study two books on quantum mechanics before this.
     
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