1
$\begingroup$

In the textbooks --- for instance "Freedman & van Proeyen" -- it is said that the Majorana Rarita Schwinger spinor is a combination of a Majorana spinor and a vector. In order to the Lorentz group representation one would assume that it is the product of a vector representation and a Bispinor representation.

I guess, a Majorana spinor transforms like a Bispinor (Dirac-spinor), i.e. the corresponding Lorentz group representation would be

$$D(\frac{1}{2},0) \oplus D(0,\frac{1}{2})$$

In product with the vector representation it would be something like

$$ D(\frac{1}{2},\frac{1}{2})\otimes ( D(\frac{1}{2},0) \oplus D(0,\frac{1}{2})) = D(1,\frac{1}{2}) \oplus D(\frac{1}{2},1). $$

Is this correct?

And is there any additional group-representational or spinor-technological detail which can be learnt from this?

EDIT

In the paper https://arxiv.org/abs/hep-th/0404131 the decomposition in irreducible components BTW is (equation (1)):

$$ D(\frac{1}{2},\frac{1}{2})\otimes ( D(\frac{1}{2},0) \oplus D(0,\frac{1}{2})) = D(1,\frac{1}{2}) \oplus D(\frac{1}{2},1) \oplus D(0,\frac{1}{2}) \oplus D(\frac{1}{2}, 0)$$

from which I conclude that the decomposition in irreducible components is more complicated (as originally assumed).

Why do these additional components appear ?

$\endgroup$
1
  • $\begingroup$ Is this correct? See the second paragraph of the Wikipedia article. $\endgroup$
    – Ghoster
    Commented Nov 9 at 6:33

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.