Lorentz commutator

  1. I am trying to derive the algebra and I get a factor of 2 wrong...
    Consider the Lorentz group elements near the identity
    \Lambda_1^\mu\,_\nu = \delta^\mu\,_\nu + \omega_1^\mu\,_\nu, \quad \Lambda_2^\mu\,_\nu = \delta^\mu\,_\nu + \omega_2^\mu\,_\nu
    and write a representation as
    U(\Lambda)=U(1 +\omega)=1-\frac{1}{2}i\omega^{\mu \nu} M_{\mu \nu}
    where [itex]M[/itex] is a generator. Now the term [itex]\Lambda \equiv (\Lambda_2^{-1}\Lambda_1^{-1}\Lambda_2 \Lambda_1) [/itex] belongs to the group and up to 2nd order is [itex] 1+[\omega_2,\omega_1] [/itex].
    So its representation is
    U(\Lambda)=U(1+[\omega_2,\omega_1])=1-\frac{1}{2}i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}
    I know this is wrong and I am supposed to get
    1-i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}
  2. jcsd
  3. samalkhaiat

    samalkhaiat 1,194
    Science Advisor

    What algebra? Is it Lorentz algebra?
  4. yes.
  5. I am not planning to make the calculation, but did you use the fact that for [tex] \Lambda_{\mu}^{\nu} = \delta_{\mu}^{\nu} + \omega_{\mu}^{\nu}[/tex]
    then its inverse matrix is: [tex] (\Lambda^{-1})_{\mu}^{\nu} = \delta_{\mu}^{\nu} - \omega_{\mu}^{\nu} +O(\omega^2)[/tex]?
  6. Yes. I used it, and I am pretty sure it leads to [itex]\Lambda = 1+ [\omega_2,\omega_1] [/itex]. I am not sure then how it would lead to
    U(1+[\omega_2,\omega_1])=1-i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}
    (there is a missing factor 1/2 according to my convention)
  7. Yes, you are, your calculation is right.

    Do you use a specific reference where this equation is given there?

    I must say that I myself find QFT quite hard as you can follow my posts in the forum in the last month or so.
  8. Yes. Sorry, I am using the following reference:
    and trying to fill in the gaps. This is at the end of page 41, and on page 42. My problem is the top of eq. (4.30) given the definition (4.29).
  9. samalkhaiat

    samalkhaiat 1,194
    Science Advisor

    No, in fact that is the correct form.

    This is wrong. If you use it, you will have a factor (2) multiplying the RHS of the Lorentz algebra.
  10. Thank you. you are right, it works out! I stopped when I didn't understand this line, which is a typo, and should have continued in the first place
    Last edited: Jul 26, 2014
  11. samalkhaiat

    samalkhaiat 1,194
    Science Advisor

    The point of having (1/2) in the [itex]U( \Lambda )[/itex] is to get "mormalized" algebra. There is nothing wrong in choosing [itex]U = 1 + i \omega^{ \mu \nu } M_{ \mu \nu }[/itex], but your algebra, in this case becomes [itex][M , M] = (1/2) ( g M + \cdots )[/itex]. So, when you mormalize [itex]M \rightarrow (1/2) M[/itex] you get the "nice" algebra [itex][M , M] = g M + \cdots [/itex].
    The point to remember is this, once you made a choice, you have to use it for all group elements: [itex] ( \Lambda ), ( \Lambda_{1} \Lambda_{2} ) , \cdots , ( \Lambda_{1} \cdots \Lambda_{n} )[/itex].

    I think it is lazyness not a "typo", because he repeats the same mistake when he "derives" the Poincare' algebra.

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