I am going through the notes on QFT by Srednicki
(which is certainly a worth reading on the subject, and can be found online, see http://web.physics.ucsb.edu/~mark/qft.html).
When describing fermions, from the very begining he introduces the Lorentz Group and its algebra, and proves that it is equivalent to two copies of $SU(2)$, so that a representation is specified by two (semi)integers, say $n,n'$ (see pages 213-214). He writes such a representation as $(2n+1,2n'+1)$.
For example, some important representations are $(1,1)$: scalar, $(2,1)$: left-handed spinor and so on. Some pages later (p. 217), he writes the relation $2\otimes 2=1\oplus 3$, which is just the usual result from addition of agular momentum. My problem is, some pages later (p. 219) he writes the following "group theoretic relation" $$(2,2)\otimes (2,2)=(1,1)\oplus(1,3)\oplus(3,1)\oplus(3,3).$$ I am having a hard time trying to understand this relation.
At a first look, it seems that we have to add four spin one-half momenta, that is, $$(2,2)\otimes(2,2)=2\otimes2\otimes2\otimes2.$$ If I go through the usual steps to construct such a sum, there is no way I get the expected result.
On the other hand, if I write $(2,2)=1\oplus 3$ and "distribute $\oplus$ over $\otimes$" as if they were actual products and sums, I get the result given by Srednicki, but I feel there is something wrong about that. Maybe I feel it's wrong just because there is something I'm missing or that I don't understand.
If this "distribute $\oplus$ over $\otimes$" is the right thing to do, I would really appreciate someone to explain why is that. If it's not the right thing, then I'd be glad if someone told me how should I deal with "group theoretic relations" like these, or where could I find some literature about this subject.
For example, on page 218, Srednicki writes $$(2,1)\otimes(1,2)\otimes(2,2)=(1,1)\oplus...$$
If my approach is the right one, then, as $$(2,1)\otimes(1,2)=1\oplus3=(2,2),$$ the full answer is $$(2,1)\otimes(1,2)\otimes(2,2)=(2,2)\otimes(2,2)=(1,1)\oplus(1,3)\oplus(3,1)\oplus(3,3).$$ Is this it?
This post imported from StackExchange Physics at 2015-07-03 22:40 (UTC), posted by SE-user Taylor