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On the Lorentz Group representation

+ 3 like - 0 dislike
154 views

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
asked Jul 2, 2015 in Mathematics by Taylor (15 points) [ no revision ]
retagged Jul 3, 2015
Related: physics.stackexchange.com/q/149455/2451

This post imported from StackExchange Physics at 2015-07-03 22:40 (UTC), posted by SE-user Qmechanic
Perhaps I should add that I have no problems when dealing with individual representations (if that makes sense): I understand why (1,1) is a scalar representation, its properties and so on. Also, I know how to tensor representations (e.g. $(3,2)=4\oplus 1$, where the isomorphism is given by, for example, the Clebsch—Gordan coefficients). My problem is about dealing with expressions like $(m_1,n_1)\otimes(m_2,n_2)\otimes...=(a,b)\oplus(c,d)\oplus...$; I feel like there are many possible results, all related by isomorphisms (is this right?).

This post imported from StackExchange Physics at 2015-07-03 22:40 (UTC), posted by SE-user Taylor
Also, I know I could go with the usual $j_1+j_2,j_1+j_2-1,...,|j_1-j_2|$, buy that feels extremely inefficient when adding several momenta, so that I know there must be a better way...

This post imported from StackExchange Physics at 2015-07-03 22:40 (UTC), posted by SE-user Taylor
Your notation makes no sense to me. The scalar rep should be $(0,0)$, not $(1,1)$, as it is the zero spin representation. Why are there equations where some rep has two labels, like $(2,1)$, and another only one?! Finally, what is your actual question (besides `"Is this correct")?

This post imported from StackExchange Physics at 2015-07-03 22:40 (UTC), posted by SE-user ACuriousMind
Im following Srednicki's notation, where a representation is written as $(2n+1,2n'+1)$, so the scalar ($n=0, n'=0$) is $(1,1)$. Also, as I understand it, $(a,b)=a\otimes b$ (perhaps I should write $\cong$ instead of =, as it is an isomorphism, right?. So $(2,1)=2\otimes1$ and so on (that's why I sometimes write $(a,b)$ and sometimes individual numbers. Finally, my question is related to how to prove the relation given by Srednicki, and how to calculate analogous expressions. Than everybody for your time :)

This post imported from StackExchange Physics at 2015-07-03 22:40 (UTC), posted by SE-user Taylor

 $$(2,2)\otimes(2,2) = (2\otimes 2,2\otimes 2)=(1\oplus 3,1\oplus 3)=(1,1)\oplus(1,3)\oplus(3,1)\oplus(3,3)$$ 

1 Answer

+ 1 like - 0 dislike

There is a subtle difference between saying $(2,2)$ and $2\otimes 2$. In the latter case we are thinking of both reps as transforming under the same element of the group $SU(2)$. In the former case we are thinking of $(2,2)$ as transforming under the Lorentz group, which contains two distinct copies of $SU(2)$. Call one copy the $L$ copy and the other the $R$ copy. Then the four basis vectors of $(2,2)$ are $0_L 0_R, 0_L 1_R,\dots$ etc. These four basis vectors do not separate out into $1\oplus 3$ since I can choose elements of the Lorentz group that only rotate one of the two representations.

So think of $(2,2)\otimes(2,2) = (2\otimes 2,2\otimes 2),$ which has basis vectors like e.g. $0_{L1} 1_{R1} \otimes 0_{L2} 1_{R2}$, so then I can apply addition of angular momenta between the two Ls and Rs. Then $(1\oplus 3,1\oplus 3)$ means you take all of the basis vectors of $(1\oplus 3)_L$ and tensor product with all the basis vectors of $(1\oplus 3)_R$. So it does distribute.

So when you write $(2,1)\otimes (1,2)$ think of it like $$(2,1)\otimes (1,2)=(2\otimes 1,1\otimes 2)= (2,2)$$

This post imported from StackExchange Physics at 2015-07-03 22:40 (UTC), posted by SE-user octonion
answered Jul 2, 2015 by octonion (140 points) [ no revision ]
Well that was really helpful :) So when dealing with larger expressions like $(1,2)\otimes(2,3)\otimes(1,3)$, we have to actually add the three momenta, right? I mean, we have $=(1\otimes2\otimes1,2\otimes3\otimes3)$, and we have to calculate $1\otimes2\otimes1$ and so on.

This post imported from StackExchange Physics at 2015-07-03 22:40 (UTC), posted by SE-user Taylor
Yeah, you got it.

This post imported from StackExchange Physics at 2015-07-03 22:40 (UTC), posted by SE-user octonion

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