Highest and Lowest $SU(3)_F$ states

Question

For the finite dimensional $(p,q)$-irreducible representation of $SU(3)_F$, we can label the states as $\mid T_3,Y\rangle$. Where $T_3$ is the third component of isospin and $Y$ is the hypercharge. How can I obtain the highest and Lowest weight states? I mean what are the $T_3$ and $Y$ values?

Answer 1

You'd best read up on the classic texts by Gasiorowicz Elementary particle physics ISBN-13: 978-0471292876, Ch 17; or Greiner & Mueller QM Symmetries Ch 7, etc, or else WP and also.

The rep D(p,q) built out of p triplets and q anti triplets will have a (provably unique!) maximum $T_3$ state $|M\rangle$ s.t. $T_+|M\rangle=0$ with obvious maximum eigenvalue $$T_3=(p+q)/2$$ given the triplet D(1,0) on the $T_3$ , Y plane

and the antitriplet being its reflection around the origin. The iso-axis should be 1/3 below the isodoublet and 2/3 above the (strange) isosinglet here, so Y=1/3 for $|M\rangle$ and -2/3 for its minimum. We now exploit triplet and antitriplet additivity.

So, on this plane, $|M\rangle$ is the easternmost state. It can be shown its hypercharge is $$ Y=(p-q)/3 , $$ so, $$ |M\rangle= | T_3=(p+q)/2,Y=(p-q)/3 \rangle. $$

To get to the lowest Y state, apply the V-spin southwestward shift $V_- ^p$, so that $V_-^{p+1}| M\rangle=0$.

The resulting minimum Y expression in p,q is messier, $$ Y_{min}= -(q+2p)/2 . $$

So you may work out how the baryon decuplet D(3,0) has minimum Y=-2, the $\Omega^-$, and maximum $T_3=3/2$. For the pseudoscalar or baryon octet D(1,1), minimum Y=-1, and maximum $T_3=1$. For their paradigmatic D(7,3), below, maximum $T_3=5$ and minimum Y=-5-2/3= -17/3, etc.

You can get the maximum Y in complete analogy, by taking q steps to the northwest, via U-spin, so $U_+^q$. These states are not unique, i.e. there is no guarantee they are an isosinglet.

Recall all multiplets are east west symmetric, (maximum and minimum isospin equal in magnitude), but not necessarily north-south symmetric!