Why can't $uus$ and $uds$ quark content have $I = 1$ Isospin?

Question

Well, I know the isospin singlet with $I=0$ and isospin triplet with $I = 1$. But for $I = 0$, why it must be $uds$ quark content rather than other assembly like $uus$ or $dds$? $ud$ can have $I = 0$ and $I = 1$, but $uu$ and $dd$ can only have $I = 1$, so why?

Answer 1

can have =0 and =1, but and can only have =1, so why?

I am not quite sure what you are really asking, or where your logical snag lies. The obvious answer is interchange symmetry.

Ignoring the s, which is an isosinglet, the antisymmetric combination of two different light quarks, ${ud-du \over \sqrt{2}}$, cannot have its isospin raised, so it is an isosinglet.

But the symmetric combination, ${ud+du \over \sqrt{2}}$, can have it raised to uu or lowered to dd, so it is part of an isotriplet. We assign the former to the Λ baryon, and the latter to the Σ⁰.

Note the respective isosymmetry in the full octet baryon strange wavefunctions, $$ |\Lambda \uparrow\rangle = \frac{1}{\sqrt{12}} \Bigl ( |( u\downarrow d\uparrow -d\downarrow u\uparrow +d\uparrow u\downarrow -u\uparrow d\downarrow ) ~ s\uparrow\rangle \\ + | u\downarrow s\uparrow d\uparrow -d\downarrow s\uparrow u\uparrow +d\uparrow s\uparrow u\downarrow -u\uparrow s\uparrow d\downarrow \rangle \\ + |s\uparrow ( u\downarrow d\uparrow -d\downarrow u\uparrow +d\uparrow u\downarrow -u\uparrow d\downarrow)\rangle \Bigr ),\\ |\Sigma^0 \uparrow\rangle = \frac{1}{6} \Bigl ( 2|( u\uparrow d\uparrow +d\uparrow u\uparrow)~ s\downarrow\rangle +2|s\downarrow ( u\uparrow d\uparrow +d\uparrow u\uparrow) \rangle \\ -|( u\downarrow d\uparrow +d\downarrow u\uparrow +d\uparrow u\downarrow +u\uparrow d\downarrow ) ~ s\uparrow\rangle -|s\uparrow ( u\downarrow d\uparrow +d\downarrow u\uparrow +d\uparrow u\downarrow +u\uparrow d\downarrow ) \rangle \\ +2 | u\uparrow s\downarrow d\uparrow +d\uparrow s\downarrow u\uparrow \rangle - | u\downarrow s\uparrow d\uparrow +d\downarrow s\uparrow u\uparrow +d\uparrow s\uparrow u\downarrow +u\uparrow s\uparrow d\downarrow \rangle \Bigr ). $$

These wavefunctions have mixed flavor and spin symmetries, but an iso-cum-spin interchange is always symmetric (as in synchronized swimming!), so that, given the inevitable color antisymmetrizantion (suppressed here), all quarks are antisymmetrized w.r.t. any interchange, as fermions should always be.

You can now appreciate how, in contrast to the above, in the baryon decuplet (spin 3/2) hence always spin symmetric, there can be no isosinglet: no Λ* !! Quarks being fermions prevent that.