# Dirac equation and friends

1. Feb 28, 2008

### 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

2. Feb 28, 2008

### Marco_84

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

3. Feb 28, 2008

### CarlB

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.

4. Mar 1, 2008

### Marco_84

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

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