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Unitary Operators and Lorentz Transformations

  1. Aug 25, 2007 #1
    1. The problem statement, all variables and given/known data
    I am trying to learn from Srednicki's QFT book. I am in chapter 2 stuck in problem 2 and 3. This is mainly because I don't know what the unitary operator does - what the details are.

    Starting from:
    [tex]U(\Lambda)^{-1}U(\Lambda')U(\Lambda)=U(\Lambda^{-1}\Lambda'\Lambda)[/tex]
    How does one arrive at:
    [tex]\delta\omega_{\mu\nu}U(\Lambda)^{-1}M^{\mu\nu}U(\Lambda)=\delta\omega_{\mu\nu}\Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\sigma}M^{\mu\nu}[/tex]




    2. Relevant equations
    Given that:
    [tex]\Lambda '=1+\delta\omega [/tex]

    [tex]U(1+\delta\omega)=I + \frac{i}{2 \hbar}\delta\omega_{\mu\nu}M^{\mu\nu}[/tex]

    3. The attempt at a solution
    Working out the left hand side from the given, I end up with
    [tex]I+\frac{i}{2 \hbar}\delta\omega_{\mu\nu}U(\Lambda)^{-1}M^{\mu\nu}U(\Lambda)[/tex]

    As for the RHS, I dont know the details. Why do I end up with those contractions of two Lorentz transformations with the generator of the Lorentz group M?

    Also, since I'm already here, I would also like to ask what is meant by,
    [tex][M^{\mu\nu},M^{\rho\sigma}]=i\hbar (g^{\mu\rho}M^{\mu\nu} - (\mu \leftrightarrow \nu) ) - ( \rho \leftrightarrow \sigma ) [/tex]

    specifically the notation with the double arrow. It seems like an index replacement?
     
    Last edited: Aug 25, 2007
  2. jcsd
  3. Aug 25, 2007 #2

    dextercioby

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    Yes, it's a shorthand notation for an index replacement. And correct the line in which you wrote what you're trying to prove. See how the summations are done. I'll be doing the calculations for myself an see if i get to the result you're supposed to prove.
     
    Last edited: Aug 25, 2007
  4. Aug 25, 2007 #3

    dextercioby

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    Here's what is get

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

    Now where does the result you quote follow from ?? The result you're quoting appears also in the books by Weinberg and Gross. But i don't see how my expression translates to what they all are saying, namely

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

    Hmmm.
     
  5. Aug 25, 2007 #4
    Starting from:
    [tex]U(\Lambda)^{-1}U(\Lambda')U(\Lambda)=U(\Lambda^{-1}\Lambda'\Lambda)[/tex]

    It says to plug in [tex] \Lambda '=1+\delta\omega[/tex], to first order in [tex]\delta\omega[/tex]

    Working out the left hand side from the given, I end up with
    [tex]I+\frac{i}{2 \hbar}\delta\omega_{\mu\nu}U(\Lambda)^{-1}M^{\mu\nu}U(\Lambda)[/tex]

    My guess is that the right hand side would give me
    [tex]I+\frac{i}{2 \hbar}\delta\omega_{\mu\nu}\Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\sigma}M^{\rho\sigma}[/tex]

    So that i get the result. I have no idea how to get the contractions. My thinking is that [tex]\Lambda^{-1}\Lambda'\Lambda[/tex] is a matrix representation for a change of basis that is why the M comes out in [tex]\rho[/tex] and [tex]\sigma[/tex]. But why does delta stay in [tex]\mu[/tex] and [tex]\nu[/tex]?

    The result we want is:
    [tex]\delta\omega_{\mu\nu}U(\Lambda)^{-1}M^{\mu\nu}U(\Lambda)
    =\delta\omega_{\mu\nu}\Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\sigma}\underline{M^{\rho\sigma}}[/tex]

    Note the typo I made in my first post (underlined).
     
    Last edited: Aug 25, 2007
  6. Aug 28, 2007 #5

    Avodyne

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    According to Srednicki's eq.(2.5), [tex](\Lambda ^{-1}) _\rho{}^\mu = \Lambda^\mu{}_\rho[/tex], so you're done.
     
  7. Aug 29, 2007 #6

    dextercioby

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    Yes, you're right. The inverse of a Lorentz matrix is the transposed matrix and then the indices can be shuffled.
     
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