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Dirac equation and friends

  1. Feb 28, 2008 #1
    Dirac equation and friends :)

    I was playing with Dirac equations and deriving some usefull details,
    Note sure for a calculation, is all the math right?

    we require for a pure Lorentz trasf that the spinor field trasform linearly as:

    [tex]\psi'(x')=S(\Lambda)\psi(x)[/tex] (1)

    where [tex]x'=\Lambda x[/tex] and [tex]S(\Lambda)[/tex] is a linear operator that we can write follow:

    [tex]S(\Lambda)=exp(-\frac{i}{a}\sigma_{ab}\omega^{ab})[/tex] (2)

    If we take the [tex]\dagger[/tex] and use Dirac gammas on (1) we obtain the transformation law for the dagger spinor:


    and using gamma zero we have:


    Now what i want to show is that:


    correct me in the following equalities if i make something wrong:


    Now using the properties of gamma zero on the matrices:

    [tex]\gamma_{0}^2=Id[/tex]; [tex]\gamma_{0}\sigma_{ab}\gamma_{0}=(\sigma_{ab})^{\dagger}[/tex]

    we get to:

    [tex]\gamma^{0}exp(\frac{i}{a}(\omega^{ab})^{\dagger}(\sigma_{ab})^{\dagger})\gamma^{0}=exp(\gamma^{0}\frac{i}{a}(\omega^{ab})^{\dagger}\gamma^{0}\gamma^{0}(\sigma_{ab})^{\dagger}\gamma^{0})=exp(\frac{i}{a}\sigma_{ab}\omega^{ab})\equiv S^{-1}(\Lambda)[/tex]

    The last follow from (2).

    Am i correct??

    thanks in advance.

  2. jcsd
  3. Feb 28, 2008 #2
    Sorry i wrote a instead of 4 in the S operator underneath the i...
  4. Feb 28, 2008 #3


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    It looks right to me. And you can edit your posts here for 24 hours so you can fix things like the a versus 4 detail.
  5. Mar 1, 2008 #4
    thanx because my prof....Im sure will be very miticoulous on this "Not so important" calculus....
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