6
$\begingroup$

I am going through the notes on QFT by Srednicki.

When describing fermions, from the very beginning 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 (half) 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 angular 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 first glance, 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?

$\endgroup$

closed as off-topic by ACuriousMind, Kyle Kanos, Kyle Oman, yuggib, Qmechanic Jul 3 '15 at 12:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Oman, yuggib, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Related: physics.stackexchange.com/q/149455/2451 $\endgroup$ – Qmechanic Jul 2 '15 at 20:28
  • 1
    $\begingroup$ 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")? $\endgroup$ – ACuriousMind Jul 2 '15 at 21:11
  • 1
    $\begingroup$ 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 :) $\endgroup$ – AccidentalFourierTransform Jul 2 '15 at 21:35
  • 1
    $\begingroup$ This question (v2) seems like an archetype of a math problem encountered in many areas of physics, e.g. QCD, and which the community consistently wants to not migrate to Math.SE, cf. this meta post. $\endgroup$ – Qmechanic Jul 3 '15 at 12:23
  • 1
    $\begingroup$ However, reviewers are voting to migrate it to Math.SE. I close this question as a homework-like question, partly to prevent it from being migrated to Math.SE. $\endgroup$ – Qmechanic Jul 3 '15 at 12:25
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – AccidentalFourierTransform Jul 3 '15 at 7:21
  • $\begingroup$ Yeah, you got it. $\endgroup$ – octonion Jul 3 '15 at 16:53

Not the answer you're looking for? Browse other questions tagged or ask your own question.