Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  On the Lorentz Group representation

+ 3 like - 0 dislike
1738 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 (145 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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverflo$\varnothing$
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...