How can $\Lambda^0$ and $\Sigma^0$ both have $uds$ quark content? Title says it all: How can $\Lambda^0$ and $\Sigma^0$ both have $uds$ quark content? Doesn't this make them the same baryon?
 A: *

*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.

A: To start 3 quarks u,d,s form an irreducible representation for the 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 a 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:
$$
p_A = (ud-du)u/\sqrt2
$$
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-symmetric state is
$$
\psi_A = [(ud-du)s+(su-us)d+(ds-sd)u]/\sqrt6
$$
$\Lambda$ 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.
A: 
how can $\Lambda^0$ and $\Sigma^0$ both have $uds$ quark content? Doesn't this make them the same baryon?

Well, the basic difference which started their group classification is that $\Lambda$ has a mass of 1115.683±0.006 and $\Sigma$=1189.37±0.07 in MeV. The difference is very  significant within the measurement errors, so after they were detected, they had obviously to belong to a different in energy  bound state of quarks. Then the group structure enters, which fits that the $\Sigma$ is a triplet at that mass, thus Isospin 1, whereas $\Lambda$ is a singlet because no charged modes were found at that mass.
Then the weak SU(3) structure conveniently allowed for the singlet and triplet in representations of the quarks. The same quarks can be at different energy levels which may be stable to strong decays due to quantum number conservation laws.
_____________-
Aside: somehow I find that the present generation of physicists answering questions on this site first grab the theoretical aspect and usually stay there, whereas it is from the measurements/data that the theoretical structure developed.
A: 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: I will give a simple picture here as I do not know much theory. The baryons consists of quarks which have spin 1/2. The spins can add up and the quarks can go around each other to have orbital angular momentum. The consequence is that given the same uds quarks, you can still have many states.
If you look at Table III in this paper: https://arxiv.org/abs/1508.07233 You will find how $\Sigma_c$s and $\Lambda_c$s are different. Basically the two light quarks in $\Sigma_*$ form a spin triplet while for $\Lambda_*$ the two light quarks form a spin singlet.
