Neutral sigma-star baryon, neutral sigma baryon and lambda baryon have masses 1387 MeV, 1192 MeV, 1116 MeV respectively, whereas each of these 3 particles are composed of 1 up, 1 down and 1 strange quark. Neutral sigma-star baryon (spin 3/2) is more massive than neutral sigma baryon (spin 1/2) because of hyperfine splitting but why neutral sigma baryon is more massive than lambda baryon, which is also spin 1/2 particle like neutral sigma baryon?
[Physics] Why is neutral sigma baryon more massive than lambda baryon
baryonsisospin-symmetrystandard-model
Related Solutions
These two Figures are excerpt of a page in : A Modern Introduction to Particle Physics, by Fayyazuddin & Riazuddin, 2nd Edition 2000.
The first shows the well-known octet of the mixed antisymmetric tensor $\:\boldsymbol{8}\:$ while the second shows the octet of the mixed symmetric tensor $\:\boldsymbol{8'}\:$. I don't know what particles, if any, are represented by the latter octet.
See also my answer here : Symmetry in terms of matrices. Therein octet $\:\boldsymbol{8}\:$ is produced by the mixed antisymmetric tensor $\:Y_{ijk}\:$, see equations (B.25) and (B.35), while octet $\:\boldsymbol{8'}\:$ is produced by the mixed symmetric tensor $\:X_{ijk}\:$, see equations (B.24) and (B.37).
The Figures below are excerpts from : A Modern Introduction to Particle Physics-Volume 1: Quantum Field Theory and Particles, by Y.Nagashima, Edition 2010.
In 'QUARKS AND LEPTONS: An Introductory Course in Modern Particle Physics', F.Halzen-A.Martin, Edition 1984, we meet the following concerning the spin-up proton:
We take the mixed-antisymmetric and mixed-symmetric states
\begin{align}
\mathrm p_{_A} & =\sqrt{\tfrac12}\left(\mathrm u \mathrm d- \mathrm d\mathrm u \right)\mathrm u \:\: \left\{\in \boldsymbol{8}_{_{MA}}\equiv \boldsymbol{8}\right\}
\tag{2.60}\\
\mathrm p_{_S} & =\sqrt{\tfrac16}\bigl[\left(\mathrm u \mathrm d+ \mathrm d\mathrm u \right)\mathrm u-2\mathrm u \mathrm u \mathrm d \bigr]\:\: \left\{\in \boldsymbol{8}_{_{MS}}\equiv \boldsymbol{8'}\right\}
\tag{2.62}
\end{align}
The state $\:\mathrm p_{_A} \:$ is the first member of octet $\:\boldsymbol{8}\:$ shown in the first Figure above while the state $\:\mathrm p_{_S} \:$ is the first member of octet $\:\boldsymbol{8'}\:$ shown in the second Figure.
These are multiplets produced in $\:\rm SU(2)-$isospin and by analogy the $\:\rm SU(2)-$spin antisymmetric, symmetric multiplets are produced by replacing in (2.60), (2.62)
\begin{align}
\mathrm u & \quad \Longrightarrow \quad \uparrow
\nonumber\\
\mathrm d & \quad \Longrightarrow \quad \downarrow
\tag{01}
\end{align}
so
\begin{align}
\chi\left(M_A\right) & =\sqrt{\tfrac12}\left(\uparrow \downarrow\uparrow -\downarrow\uparrow\uparrow \right)
\nonumber\\
\chi\left(M_S\right) & =\sqrt{\tfrac16}\left(\uparrow \downarrow\uparrow +\downarrow\uparrow\uparrow -2\uparrow\uparrow \downarrow \right)
\tag{2.65}
\end{align}
Next the spin-up proton is derived from
\begin{equation}
\vert \mathrm p\!\uparrow\, \rangle=\sqrt{\tfrac12}\bigl[\mathrm p_{_A}\chi\left(M_A\right)+\mathrm p_{_S}\chi\left(M_S\right) \bigr]
\tag{02}
\end{equation}
where
\begin{align}
\mathrm p_{_A}\chi\left(M_A\right) & \simeq \left(\mathrm u \mathrm d \mathrm u-\mathrm d \mathrm u \mathrm u\right)\left(\uparrow \downarrow\uparrow -\downarrow\uparrow\uparrow \right)
\nonumber\\
& =\bigl(\mathrm u\!\!\uparrow\!\mathrm d\!\!\downarrow\!\mathrm u\!\!\uparrow -\mathrm u\!\!\downarrow\!\mathrm d\!\!\uparrow\!\mathrm u\!\!\uparrow -\mathrm d\!\!\uparrow\!\mathrm u\!\!\downarrow\!\mathrm u\!\!\uparrow +\mathrm d\!\!\downarrow\!\mathrm u\!\!\uparrow\!\mathrm u\!\!\uparrow \bigr)
\tag{03}
\end{align}
and
\begin{align}
\mathrm p_{_S}\chi\left(M_S\right) & \simeq\left(\mathrm u \mathrm d\mathrm u + \mathrm d\mathrm u \mathrm u-2\mathrm u \mathrm u \mathrm d \right)\left(\uparrow \downarrow\uparrow +\downarrow\uparrow\uparrow -2\uparrow\uparrow \downarrow \right)
\nonumber\\
& =\mathrm u\!\!\uparrow\!\mathrm d\!\!\downarrow\!\mathrm u\!\!\uparrow + \mathrm u\!\!\downarrow\!\mathrm d\!\!\uparrow\!\mathrm u\!\!\uparrow-2 \mathrm u\!\!\uparrow\!\mathrm d\!\!\uparrow\!\mathrm u\!\!\downarrow +\mathrm d\!\!\uparrow\!\mathrm u\!\!\downarrow\!\mathrm u\!\!\uparrow +\mathrm d\!\!\downarrow\!\mathrm u\!\!\uparrow\!\mathrm u\!\!\uparrow +\cdots
\tag{04}
\end{align}
and finally
\begin{align} \vert \mathrm p\!\uparrow\, \rangle & =\sqrt{\tfrac{1}{18}}\bigl[\mathrm u \mathrm u \mathrm d \left(\uparrow \downarrow\uparrow +\downarrow\uparrow\uparrow -2\uparrow\uparrow \downarrow \right)+\mathrm u \mathrm d \mathrm u \left(\uparrow\uparrow\downarrow +\downarrow\uparrow\uparrow -2\uparrow\downarrow \uparrow \right)+\mathrm d \mathrm u \mathrm u \left(\uparrow\downarrow \uparrow+\uparrow\uparrow\downarrow -2\downarrow \uparrow\uparrow \right)\bigr] \nonumber\\ &=\sqrt{\tfrac{1}{18}}\bigl[\mathrm u\!\!\uparrow\!\mathrm u\!\!\downarrow\!\mathrm d\!\!\uparrow +\mathrm u\!\!\downarrow\!\mathrm u\!\!\uparrow\!\mathrm d\!\!\uparrow -2\mathrm u\!\!\uparrow\!\mathrm u\!\!\uparrow\!\mathrm d\!\!\downarrow +\text{permutations}\bigr] \tag{2.71} \end{align}
From above notes we conclude that the real baryon $\:(1/2)^{+}\:$ octet is a combination of the two octets with mixed symmetry $\:\boldsymbol{8},\boldsymbol{8'}$.
The $\Lambda^0$ and $\Sigma^0$ have respectively I=0 and I=1. The source of your confusion may come from the fact that the electromagnetism doesn't respect the isospin symmetry. In E.M. processes, only $I_3$, the third component, is conserved. The total isospin $I$ is not. A trivial example is the neutral pion decay $\pi^0 \to \gamma \gamma$. The pion is a member of an isospin triplet $(\pi^+,\pi^0,\pi^-)$ (thus $I=1$ and $I_3=0$) while photons cannot carry any isospin numbers (since not made of $u$ or $d$ quarks).
Edit (to answer a comment): how do we know that $\Sigma^0$ has $I=1$?
There are several experimental facts. The $\Sigma^+,~\Sigma^0,~\Sigma^-$ have almost the same mass (about 1190 MeV). So historically, they were considered in the same multiplet of isospin, reasonably a triplet as for $\pi^+,~\pi^0,~\pi^-$. In addition, the reaction $K^- + p \to \Sigma^0$ is seen (strong interaction). Knowing that $p$ has $I=1/2,~I_3=+1/2$ and $K^-$ has $I=1/2,~I_3=-1/2$, the initial state $K^-+p$ can have $I=0$ or $I=1$ (sum of 2 Isospin 1/2). If you admit that $\Sigma^0$ is a member of a multiplet (previous argument), you're forced to conclude that $I=1$. I'm sure that many other reactions can be found in the literature justifying $I=1$.
In the modern quark languages, $\Sigma^+ \equiv uud,~\Sigma^0\equiv uds,~\Sigma^-\equiv dds$. With the group theory, assuming the (approximate) SU(3) flavor symmetry, $I=1$ comes "naturally".
Best Answer
If one considers the baryon octet
the lambda 0 is a singlet in isospin and the sigma zero is in a triplet.
The explanation of the mass difference
Also there exists the Gell Mann- Okubo phenomenological formula which gives the mass according to the isospin.