# Question on the 2-dim representation of the Lorentz group

1. Mar 6, 2014

### Kontilera

Hello! I'm currently reading some QFT and have passed the concept of Weyl spinors 2-4 times but this time it didn't make that much sense..
We can identify the Lorentz algebra as two su(2)'s. Hence from QM I'm convinced that the representation of the Lorentz algebra can be of dimension (2s_1 + 1)(2s_2 + 1).
The Weyl spinor is two dimensional so it's either a (s_1, s_2) = (1/2, 0) or a (0,1/2) representation (i.e. left or right handed).

But it then seems (since one representation of the su(2)s is the trivial) as if I only need to specify three parameters when lorentz transforming my Weyl spinors.. What happend to my choice of three rotations and three boosts?

Thanks! :)

2. Mar 6, 2014

### Bill_K

But they are three complex parameters. The SU(2) factorization of the Lorentz group involves the complex generators J ± iK. A Weyl spinor remains invariant under one of these subgroups.

Specifically, the infinitesimal transformation of a left-handed spinor is (Peskin & Schroeder, p44)

ψL → (1 - iθ·σ/2 - β·σ/2) ψL

where θ is a rotation and β is a Lorentz boost. Individually they change ψL, and any real combination of them changes ψL, but a complex combination of them leaves ψL invariant.

Last edited: Mar 6, 2014
3. Mar 6, 2014

### Avodyne

Another way to say the same thing is to let $\vec A$ be the generators for one of the SU(2)'s and $\vec B$ be the generators for the other SU(2). Then the angular momentum generators are $\vec J=\vec A +\vec B$ and the boost generators are $\vec K=i(\vec A-\vec B)$. For the (1/2,0) rep, $\vec A=\vec\sigma/2$ and $\vec B=0$. Then $\vec J =\vec\sigma/2$ and $\vec K=i\vec\sigma/2$.

4. Mar 7, 2014

### Kontilera

Yeah, I with you. The su(3) has three generators... but the relation above gives me six generators and that is what I can work with. :)
Or equivalently, I work with three generators but they now span a complex vectorspace, hence my coefficients amount to six degrees of freedom.

Thanks.

5. Mar 7, 2014

### Kontilera

Does it exists a Loretnz transformation such as (1, 1/2) ? In that case, have do we make A and B to same dimension?

6. Mar 7, 2014

### dextercioby

Of course there is. It's 'half' of a Rarita-Schwinger spinor. The dimension of the spinor space is (2x1+1)(2x1/2+1) = 6.

7. Mar 8, 2014

### Kontilera

Haha. But what about the dimension? I guess we use a reducible rep for the lower s_i? Does it matter which one we choose?