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
5,082 questions , 2,232 unanswered
5,352 answers , 22,786 comments
1,470 users with positive rep
820 active unimported users
More ...

  Supermultiplet dimensions from Young Tableaus

+ 3 like - 0 dislike
822 views

In John Terning's book, on pages 14 and 15, there are lists of $\mathcal{N} = 2$ and $\mathcal{N} = 4$ supermultiplets, labeled in terms of the dimensions of the corresponding R-symmetry $d_R$ and spin-symmetry $2j+1$. I want to figure out a way to get all these numbers by Young Tableaus in a systematic way.

Of course, the $\mathcal{N}=2$ case is relatively straightforward. Its clear that the numbers in the labels for $\mathcal{N}=4$ can separately be recovered using

$4_{R} \otimes 4_{R} = 10_R \oplus 6_R$

for $SU(4)_R$ and tensor products of this basic identity with $4_R$. Of course, one can also do the same thing for the $SU(2)$ spin symmetry

$2_{SU(2)} \otimes 2_{SU(2)} = 3_{SU(2)} + 1_{SU(2)}$

But if one writes $(\textbf{R}, 2j+1)$ as the label, then how does one justify

$(\textbf{4}_R, 2)\otimes(\textbf{4}_R, 2) = (\textbf{10}_R,1) \oplus (\textbf{6}_R,3)$

or

$(\textbf{4}_R, 2)\otimes((\textbf{10}_R,1) \oplus (\textbf{6}_R,3)) = (\bar{\textbf{20}}, 2) + (\bar{\textbf{4}},4)$

What I'm asking is: how do you get this particular grouping?

This post imported from StackExchange Physics at 2015-06-02 11:45 (UTC), posted by SE-user leastaction
asked Jun 2, 2015 in Theoretical Physics by leastaction (425 points) [ no revision ]

1 Answer

+ 4 like - 0 dislike

The way to justify this is to realize that when you write the label as  \((\textbf{R}, 2j+1)\) you are embedding the direct product of those groups into a larger group i.e.  \(SU(N) \otimes SU(M) \in SU(NM)\). Where the tensor indices of the larger group can be thought of as an ordered pair of indices \((a,\dot\alpha)\), where \(a=1,...,N\) and \(\dot\alpha = 1,...,M\) for this case we have N=4 and M=2.

In Terning's book we can see that each charge carries an order pair (which is embedded in the larger SU(NM) group) and has to be completely antisymmetric which every other charge. So if we have n charges we want to represent that as a completely antisymmetric rank n tensor living in the SU(NM) group. The way to represent this with Young's Tableaus is to think of it as n boxes arranged vertically. The trick is the embedded group has to transform the same way as the representation in the larger group when permutating the ordered indices.

For example. \((\textbf{4}_R, 2)\otimes(\textbf{4}_R, 2) = (\textbf{10}_R,1) \oplus (\textbf{6}_R,3)\)

This represents a completely antisymmetric rank 2 tensor in SU(8). The first thing to do when trying to come up with possible representations is to write down all the ways 2 (note this number is the same as rank of the embedding group) boxes can be arranged in SU(4) and SU(2), which happens to be the same 2 horizontal and 2 vertical or the 10 and 6 for SU(4) and the 3 and 1 for SU(2) respectively. (I don't know how to draw boxes here...). Where the boxes arranged horizontally are completely symmetric and the ones arranged vertically are completely antisymmetric. Since we need the permutation of ordered pair to be antisymmetric, the only way to do this is to have the symmetric rep in one group be ordered with the antisymmetric of the other rep, for permutations of the symmetric rep give you a plus while the antisymmetric give you a minus resulting in an overall minus.

The next rep is a little more complicated. \((\textbf{4}_R, 2)\otimes(\textbf{4}_R, 2)\otimes(\textbf{4}_R, 2) = (\bar{\textbf{20}}, 2) \oplus (\bar{\textbf{4}},4)\)

Here we have a rank 3 tensor. The way to arrange 3 boxes in SU(4) result in \(\bar 4, \bar{20''}, \bar{20} \) check out his appendix B.4 to align the box notation with those numbers. While the SU(2) only gives the 2 and 4 rep (note the 2 rep is drawn with 3 boxes the same way as the \bar{20}!!) The \((\bar{\textbf{4}},4)\) follows the same logic as previous, the \bar{4} is completely antisymmetric while the 4 is completely symmetric. The \((\bar{\textbf{20}}, 2)\) is more complicated, and I wasn't able to find a completely generic way to show it, but essentially when you take the tensor product of the two reps in the permutation group (S_n, n being the 3 in this case) you find that they can decomposed into a completely antisymmetric tensor (and actually a completely symmetric  tensor).

Georgi explains all of this in gory detail this in his book sections 15.2 and 1.21-24.

Long story short the decomposition of the representations should transform under permutations the same way as the rank n tensor embedded in the higher group. The left hand side is completely antisymmetric so too should the right hand side.

(Sorry the previous edits were made under some errors and I didn't know how to delete so I just redid it)

answered Jun 15, 2015 by Peter Anderson (205 points) [ revision history ]
edited Jun 19, 2015 by Peter Anderson

Thank you for the detailed reply @PeterAnderson! As it turns out, there is a more heuristic method which I managed to figure out, which is purely tableau-based. It is described in the comments of the original post on Physics.SE, from which this post was derived.

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$\hbar$ysic$\varnothing$Overflow
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
...