Dirac Equation and Spinor Field Transformations: Understanding the Basics

[Marco_84]

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:

\psi'(x')=S(\Lambda)\psi(x) (1)

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

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

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

\psi'(x')^{\dagger}=\psi(x)^{\dagger}S(\Lambda)^{\dagger}

and using gamma zero we have:

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

Now what i want to show is that:

\gamma^{0}S(\Lambda)^{\dagger}\gamma^{0}=S^{-1}(\Lambda)

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

\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})

Now using the properties of gamma zero on the matrices:

\gamma_{0}^2=Id; \gamma_{0}\sigma_{ab}\gamma_{0}=(\sigma_{ab})^{\dagger}

we get to:

\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)

The last follow from (2).

Am i correct??

thanks in advance.

marco

 

[Marco_84]

Sorry i wrote a instead of 4 in the S operator underneath the i...
regards
marco

 

[CarlB]

Am i correct??

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.

 

[Marco_84]

thanx because my prof...Im sure will be very miticoulous on this "Not so important" calculus...
regards
marco