# How can $\Lambda^0$ and $\Sigma^0$ both have $uds$ quark content?

The isospin is different. $I=0$ for the $\Lambda^0$ and $I=1$ for the $\Sigma^{0}$. This makes the $\Lambda^0$ an isospin singlet state but the $\Sigma^0$ is part of an isospin triplet.

There are quite few other examples e.g. compare a proton (uud with $I=1/2$) with a $\Delta^{+}$ (uud with $I=3/2$).

A similar question is the following.

*How can $\pi^0$ and $\eta$ in the $SU(3)_F$ meson octet both have the same $SU(3)_F$ flavor content?*One could answer that this is because $\pi^0$ is part of an isospin triplet of pions with $I=1$, while $\eta$ is an isospin singlet with $I=0$. Or one may point out that their explicit ket linear combinations of quark flavors are orthogonal, $\pi^0 = (u\bar{u} - d\bar{d})/\sqrt{2},$ and $\eta = (u\bar{u} + d\bar{d} - 2s\bar{s})/\sqrt{6},$ respectively. See also e.g. this Phys.SE post. The $J=0$ spin state $(\uparrow\downarrow+\downarrow\uparrow)/\sqrt{2}$ factorizes.

Now back to OP's question.

*How can $\Lambda^0$ and $\Sigma^0$ in the $SU(3)_F$ baryon octet both have $uds$ quark content?*Rob Jeffries has already correctly answered that they have different isospin. Alternatively, one may point out that their ket linear combinations of quark flavors are different. Naively, one would expect $\Lambda^0 = (ud + du)s/\sqrt{2}$ and $\Sigma^0 = (ud - du)s/\sqrt{2}$. However the last sentence is not completely correct, since the $J=1/2$ spin state $\uparrow\uparrow\downarrow$ of the three quarks does not factorize. It turns out that the explicit ket linear combinations of quark flavor and spin of $\Lambda^0_{\uparrow}$ and $\Sigma^0_{\uparrow}$ contain 12 and 18 terms, respectively, cf. Ref. 1.

References:

- W. Greiner & B. Müller,
*Quantum Mechanics: Symmetries;*Exercise 8.15.

To start 3 quarks u,d,s form an irreducible representation for SU(3) group.
Now $\mathbf{3\times 3\times 3} = \mathbf{10+8+8+1}$ for the SU(3) group. The first irreducible representation **10** has 10 purely symmetric states. The next representation **8** has 8 mixed symmetric states. The next representation **8** has 8 mixed anti-symmetric states. The last **1** is purely anti-symmetric state.

There exists six ladder operators which take u,d,s quarks to (d,s),(u,s),(u,d) and back to those quark states. $$I_-u=d,\qquad I_+d=u$$ $$V_-u=s,\qquad V_+s=u$$ $$U_-d=s,\qquad U_+s=d$$ The above figure shows the quark content of those 8 quarks. Now, anti- symmetric proton state in 1,2 indices is: $$ p_A = (ud-du)/\sqrt2 $$ By using ladder operator $U$ it gives $$ U_- p_A= \Sigma^+ = (us-su)u/\sqrt2 $$ Using ladder operator $I$ and normalizing it gives $$ I_- \Sigma^+ = \Sigma^0 =[(ds-sd)u+(us-su)d]/2 \, . $$ The totally anti-symetric state $$ \psi_A = [(ud-du)s+(su-us)d+(ds-sd)u]/\sqrt6 $$ $\Lambda$ is orthogonal to both $\psi_A$ and $\Sigma^0$. This gives $$ \Lambda = [2(ud-du)s+(us-su)d+(sd-ds)u]/\sqrt(12) $$ Here, we observe that although $\Lambda,\Sigma^0$ have same quark content, the flavor wave function of two baryons are different from each other. Also, $I=1,I_3 =0$ for $\Sigma^0$ and $I=0,I_3 =0$ for $\Lambda$. Hence, both are different.