# Poincare-representation of majorana spinor

1. Jun 16, 2009

### torus

Hi,
one labels the Weyl- and Vector-representations of the Poincare group by (0,1/2), (1/2,0), (1/2,1/2) etc., where does the Majorana spinor fit into this?
Or can you say it belongs somehow to the real part of the (0,1/2)+(1/2,0) rep, although this sounds pretty unfamiliar.

Thanks for every response.
torus

2. Jun 19, 2009

### RedX

I think you can say that a Majorana spinor is (0,1/2).

As for saying that it belongs to the real part of (0,1/2)+(1/2,0), I guess you can say that, if by real part you mean charge conjugation which turns (0,1/2) to (1/2,0) and vice versa (i.e., swaps chirality).

3. Jun 19, 2009

### torus

Thanks for your response!
Isn't (0,1/2) the Weyl spinor, I don't think they are quite the same, since the Majorana is supposed to be a real representation, it can't be in the complex (0,1/2).
One text I read sort of implied that you could get the Majorana by a unitary transformation of (0,1/2)+(1/2,0), but I've never seen that.
Unfortunatly I can't seem to find a paper or book about this, so any literature would be greatly appreciated.

4. Jun 19, 2009

### RedX

SO(N) has only real representations (the generators are antisymmetric and Hermitian), but maybe this doesn't include spinors?

A Dirac 4-component spinor is in the representation (1/2,0)+(0,1/2). You can call the fields $$(\xi , \eta^{\dagger})$$ where the dagger indicates (0,1/2).

A Majorana spinor is special because the (1/2,0) and the (0,1/2) are the same field, so you'd call the 4-component spinor $$(\xi , \xi^{\dagger})$$.

In other words a Dirac spinor is two Weyl fields, and a Majorana spinor is only one Weyl field.

So you can say in a sense that if you swap the the position of the two fields, a Majorana spinor is the same, while a Dirac isn't:

$$(\xi , \eta^{\dagger})$$ ---> $$(\eta , \xi^{\dagger})$$

$$(\xi , \xi^{\dagger})$$ ---> $$(\xi , \xi^{\dagger})$$

So the bottom line is the only one that's the same under swap of fields. Swap of fields is done by charge conjugation operation, so a Majorana spinor is its own charge conjugate. I'm not sure about the terminology of real as applied to spinor representations of SO(4), but this is one way to specify the difference between Majorana and Dirac spinors.

Last edited: Jun 19, 2009
5. Jun 20, 2009

### torus

Yes, I'm aware of that, I was just wondering how to express this through the usual representation-notation.

6. Jun 20, 2009

### RedX

Well, call a spinor representation of SO(4) the letter 'R'. Then $$R^{\dagger}=R$$ is the condition for Majorana spinor? The adjoint operation swaps SU(2)s. If you swap SU(2)s and it's the same representation, then that's a Majorana representation?

I don't know how to express it mathematically, but if I was asked on an exam, that's what I'd put.

7. Jun 20, 2009

### Avodyne

This is discussed in various places, including (IIRC) the textbooks by Weinberg, Polchinski, and Brown.

8. Jun 21, 2009

### torus

Thanks, I'll try to get a hold of those books.