# Reps of lorentz group and pauli and gamma matrices

1. Apr 2, 2009

I'm very confused

By performing a lorentz transformation on a spinor $$\psi\rightarrow S(\Lambda)\psi(\Lambda x)$$ and imposing covariance on the Dirac equation $$i\gamma^{\mu}\partial_{\mu}\psi=0$$ we deduce that the gamma matrices transform as

$$S(\Lambda)\gamma^{\mu} S^{-1}(\Lambda)=\Lambda^{\mu}_{\nu}\gamma^{\nu}$$

I understand that.

Now the Gamma matrices can be given by
$$\gamma^{\mu}=\left[ \begin{array}{cccc} 0&\sigma^{\mu}\\ \bar{\sigma}^{\mu} & 0\end{array} \right]$$

with $$\sigma^{\mu}=(1,\sigma^1,\sigma^2,\sigma^3)$$ and $$\bar{\sigma}^{\mu}=(-1,\sigma^1,\sigma^2,\sigma^3)$$

and the dirac equation is reducible into the weyl equations.

$$i\sigma^{\mu}\partial_{\mu}\psi_L=0$$ and $$i\bar{\sigma}^{\mu}\partial_{\mu}\psi_R=0$$

What is the way to write the lorentz transformations in this case, and how to the pauli matrices transform.

2. Apr 2, 2009

### Ben Niehoff

The gamma matrices are constants! They do not transform. Neither do the Pauli matrices.

It is the spinor (either Dirac or Weyl) that transforms. Try going over it again, carefully.

3. Apr 2, 2009

Then now I really do not understand. In order to maintain covariance the following expression is true.

$$S_{-1}(\Lambda)\gamma^{\mu} S(\Lambda)=\Lambda^{\mu}_{\nu}\gamma^{\nu}$$

if we write a lorentz transformation on a spinor as $$\psi'(x')=S(\Lambda)\psi(x)=S(\Lambda)\psi(\Lambda_{-1}x')$$

then say $$i\gamma^{\mu}\partial_{\mu}\psi(x)=0$$ then

$$i\gamma_{\mu}\Lambda^{\rho}_{\mu}\partial_{\rho}'\psi(\lambda_{-1}x)=0$$

so $$i\gamma_{\mu}\Lambda^{\rho}_{\mu}\partial_{\rho}'S^{-1}\psi(x')=0$$

Then multiply on the left with S

$$iS\gamma^{\mu}S_{-1}\Lambda^{\rho}_{\mu}\partial_{\rho}'\phi'(x')$$

and then impose covariance so the form of the equation is unaltered in the transformed frame so that $$S\gamma^{\mu}S_{-1}\Lambda^{\rho}_{\mu}=\gamma^{\rho}$$

That's in every text book and my lecture notes lol. What happens with 2 component spinors though?

Last edited: Apr 2, 2009
4. Apr 2, 2009

### Avodyne

This is true in the sense that if you act with a transformation matrix on every index (both spinor indices and the vector index), the gamma matrices are unchanged. This is equivalent to saying (as the OP did) that transforming the two spinor indices gives the same result as transforming only the vector index (actually by the inverse transformation).

The Lorentz transformation properties of Weyl fields are treated in mind-numbing detail in the QFT book by Srednicki, available free online.

5. Apr 2, 2009

lol thanks. I was worried I was being completely thick then. dont suppose you have a link do you for where I can see it for free? can't find it.

Im generally confused on this whole matter. Ultimately I want to perform a super poincare transformation on superspace. I.e a susy transform then a lorentz transform which I know how to do on the super space cos I have seen it in a book. Then I want to invert the transformation, so inverse lorentz then inverse susy.

6. Apr 2, 2009

### George Jones

Staff Emeritus
7. Apr 2, 2009

oh cheers. Yeah I came across that page but didn't see the bit where it said 'click here' :lol:

Sorry that just made me look lazy.

8. Apr 2, 2009

### RedX

Let me first say that you have a lot of mix-ups where inverse transformations should be transformations and transformations should be inverse transformations. I'm not just talking about using a different convention than what I'm used to - what you have is not internally consistent.

Anyways, if you're confused about how the 2-component Weyl spinors inside your 4-component Dirac spinors transform, try calculating the 4x4 spin-matrices. If you do that (in the Weyl basis), you should find that resulting matrix is block-diagonal, and that 2x2 Pauli matrices transform each 2-component Weyl spinor (the top should transform with opposite sign of the bottom - this is the result of left-handed and right-handed Weyl fields being related by the adjoint operation)

9. Apr 3, 2009