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Homework Help: QFT: Srendicki Problem 2.4

  1. Sep 11, 2010 #1

    G01

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    1. The problem statement, all variables and given/known data

    Derive the following commutation relations from the general commutation relation for the Lorentz generators:

    [tex] [J_i,J_j]=i\hbar\epsilon_{ijk}J_k[/tex]

    [tex][J_i,K_j]=i\hbar\epsilon_{ijk}K_k[/tex]

    [tex][K_i,K_j]=-i\hbar\epsilon_{ijk}J_k[/tex]


    2. Relevant equations

    The commutator for the Lorentz generators:

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


    [tex]J_i=\frac{1}{2}\epsilon_{ijk}M^{jk}[/tex]

    [tex]K_i=M^{i0}[/tex]

    3. The attempt at a solution

    I've got the first one.

    The second two I'm having slight problems and just need help finding my mistake.

    For the second commutator, I have an extra factor of 1/2 on the RHS. I start from:

    [tex][M^{jk},M^{j0}]= i\hbar ((g^{jj}M^{k0} - (\mu \doublearrow \nu))-(\rho \doublearrow \sigma))[/tex]

    Only the first term on the right have side is non zero since all off diagonal g are 0.

    Now this implies:

    [tex][J_{i},K_j]= \frac{1}{2}\epsilon_{ijk}i\hbar M^{k0}[/tex]

    How do I get rid of that pesky 1/2?

    Similarly on the third commutator:

    I start from the same place and get to the line:

    [tex][K_i,K_j]=[M^{i0},M^{k0}]=-i\hbar M^{ij}[/tex]

    I can't figure out how to put the RHS in terms of J_k without getting a factor of 2!

    Any help will be appreciated. I'm sure its just stupid errors. Thanks!
     
    Last edited: Sep 11, 2010
  2. jcsd
  3. Sep 11, 2010 #2

    diazona

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    For part b, I don't think you should be getting to [itex][M^{jk},M^{j0}[/itex] at all. It looks like you may be mixing up two different indices: the j in
    [tex][J_i,K_j] = i\hbar\epsilon_{ijk}K_{k}[/tex]
    is not the same as the j in
    [tex]J_i = \frac{1}{2}\epsilon_{ijk}M^{jk}[/tex]
    It might help to rewrite the definitions of J and K as
    [tex]J_a = \frac{1}{2}\epsilon_{abc}M^{bc}[/tex]
    [tex]K_d = M^{d0}[/tex]
    to keep the different indices straight.

    Then again, I tried to do the problem myself and I'm getting the same extra factor of 1/2 that you are...
     
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