# Decomposing an SU(3) product in irreps

1. Feb 7, 2015

### ChrisVer

I am trying to work out with Young graphs the tensor product of:
$\bar{3} \otimes \bar{3}$

The problem is that I end up with:

$\bar{3} \otimes \bar{3} = 15 \oplus 6 \oplus 3 \oplus 3$

Is that correct? It doesn't seem correct at all (dimensionally speaking I should have taken something like $\bar{6} \oplus 3$ - like baring the $3 \otimes 3 =6 \oplus \bar{3}$)...
In fact I am unable to understand the rule that says:
looking from the right-to-left in rows and from the top-to-bottom collumns, the number of the $b$s (in this case) must be less or equal to the number of $a$'s.
For example that's not the case for any of my graphs execpt for the $15$.

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Last edited: Feb 7, 2015
2. Feb 7, 2015

### Orodruin

Staff Emeritus
Remember that in SU(N), a column with N boxes corresponds to a fully anti-symmetrized object and is therefore a singlet. This means you can remove such columns. You can remove your 15 because it has a "b" in the top right and the same goes for one of your 3s.

3. Feb 7, 2015

### ChrisVer

I am not understanding why to drop it away though... Why would I drop them? I mean for the 15:
the first row has 1 a and 1 b... But the 2nd collumn has only one b, is that's why it's dropped?
If that's correct then the other 3 should also be dropped, because in the 1st collumn I have only one b..

4. Feb 7, 2015

### Orodruin

Staff Emeritus
You start at the right of the first row and go left, then continue with the second row to the right and go left (and continue from the right of the third row if you have SU(N) with N > 3, etc). You delete the tableaux if at any point you have counted more bs than as (you do not reset the count after changing rows).

5. Feb 7, 2015

### ChrisVer

Hmm...I think I get it....
When you say at any point you mean precisely at any point (like counting from right to left, I count the first box as b, so b>a and it's over, no matter if next to b I have 1 or more a's), right?

6. Feb 7, 2015

### Orodruin

Staff Emeritus
Yes, at any point, not waiting for the full row to be counted.

7. Feb 7, 2015

### ChrisVer

OK, I got it, thanks

8. Feb 7, 2015

### ChrisVer

Oh yes, I still have one question. Is it possible to understand whether you have a bar or unbarred irrep in the docomposition?
For example the $6$ or $\bar{6}$? or any kind of $N$ or $\bar{N}$

9. Feb 12, 2015

### samalkhaiat

2
Yes, once you fix your convention for the fundamental rep. there will be no confusion with higher dimensional irreps. The irreps $D^{(p,q)}$ are carried by traceless symmetric tensor $T^{ ( i_{1} \cdots i_{p} ) }_{( j_{1} \cdots j_{q} ) }$. In $SU(3)$ this corresponds to a two-row Young tableau with $(p + q)$ boxes in the upper row and $(q)$ boxes in the lower one. The dimension of this representation space (i.e, the number of independent components in T) is given by $$\mbox{dim} ( D^{( p , q )} ) = (1/2) ( p +1 ) ( q + 1 ) ( p + q + 2 ) .$$ To distinguish between different irreps, we use the two Casmir operators of the group $$G^{( 3 )} D^{ ( p , q ) } = \frac{1}{9} ( p - q ) ( 2p + q + 3 ) ( p + 2q + 3 ) D^{ ( p , q ) } ,$$ $$F^{ ( 2 ) } D^{ ( p , q ) } = \left( \frac{1}{3} ( p^{2} + p q + q^{2} ) + p + q \right) D^{ ( p , q ) } .$$
Notice that $$G^{ ( 3 ) } D^{ ( p , q ) } = - G^{ ( 3 ) } D^{ ( q , p ) } = - G^{ ( 3 )} \bar{ D }^{ ( p , q ) }$$ So, the convention is this, the eigenvalue of $G^{(3)}$ is $\geq 0$ for $D^{( p , q )}$ and $\leq 0$ for $\bar{ D }^{ ( p , q )}$. Notice also that the notation $[ n ]$ for irrep is not always safe to use. This is because there may be more than just two in-equivalent irreps of the same dimension. For example $D^{ ( 4 , 0 ) } \sim T^{( i j k l )}$, $D^{ ( 0 , 4 ) } \sim T_{( i j k l )}$, $D^{ ( 2 , 1 ) } \sim T^{ ( i j ) }_{ k }$ and $D^{ ( 1 , 2 ) } \sim T^{ k }_{( i j )}$ all have the same dimension $[15]$.

Last edited: Feb 12, 2015