- #1

Marco_84

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**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?

Beginning:

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:

[tex]\psi'(x')^{\dagger}=\psi(x)^{\dagger}S(\Lambda)^{\dagger}[/tex]

and using gamma zero we have:

[tex]\overline{\psi'(x')}=\overline{\psi(x)}\gamma^{0}S(\Lambda)^{\dagger}\gamma^{0}[/tex]

Now what i want to show is that:

[tex]\gamma^{0}S(\Lambda)^{\dagger}\gamma^{0}=S^{-1}(\Lambda)[/tex]

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

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

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.

marco