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A simple equality of Generalized Lorentz Operators

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi

    we have Lorentz operators
    [itex]J^{\mu\nu} = i(x^{\mu}\partial^{\nu} - x^{\nu}\partial^{\mu})[/itex]

    and these have
    [itex][J^{\mu\nu}, J^{\rho\sigma}] = i(\eta^{\nu\rho}J^{\mu\sigma} + \eta^{\mu\sigma}J^{\nu\rho} - \eta^{\mu\rho}J^{\nu\sigma} - \eta^{\nu\sigma}J^{\mu\rho})[/itex]

    Now define generalized rotation operators, for i, j, k space coordinates
    [itex]M^{i} = \epsilon_{ijk}J^{jk}[/itex]

    Show that [itex]M^{i}[/itex] have the SU(2) algebra. i.e.
    [itex][M^{i}, M^{j}] = i\epsilon_{ijk}M^{k}[/itex]

    2. Relevant equations
    (all is above)

    3. The attempt at a solution
    I've done a few attempts and failed. so I tried taking an example
    [itex]M^{1}= \epsilon_{123}J^{23}+\epsilon_{132}J^{32} = J^{23} - J^{32}[/itex]
    [itex]M^{2}= \dots = J^{31} - J^{13}[/itex]

    and now
    [itex][M^{1},M^{2}] = [J^{23}, J^{31}] - [J^{32}, J^{31}] + [J^{23}, J^{13}] - [J^{32}, J^{13}][/itex]

    well [itex]J^{ij} = -J^{ji}[/itex] so the second term negates the first one ([itex]J^{32} for J^{23}[/itex]) and like wise the fourth and third term. So all in all I get zero. and that's no SU(2) :(

    where did I go wrong?

    thanks a lot for reading this
    Tamir
     
  2. jcsd
  3. Nov 30, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello, tamiry.

    Did you leave out a factor of 1/2 here?

    Those terms don't cancel. Watch the signs carefully.
     
  4. Nov 30, 2013 #3
    but of course...
    thanks!
     
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