All of the baryons in the spin-1/2 octet except the $\Lambda$ isospin singlet have spin-3/2 excited states in the baryon decuplet. What is it that prevents the existence of a $\Lambda^{*}$ baryon with $I=0$, $s=-1$, and $S=3/2$?
[Physics] Why is there no baryon isospin singlet with spin 3/2
baryonsisospin-symmetryparticle-physicsquarks
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 $\:\boldsymbol{8},\boldsymbol{8'}$.
To write the wavefunction of a baryon, you write it as a direct product of the different parts of the wavefunction (just as you would for any other particle):
\begin{equation} \left| \psi \right\rangle = \left| \mbox{spatial} \right\rangle \otimes \left| \mbox{spin} \right\rangle \otimes \left| \mbox{Isospin} \right\rangle \otimes \left| \mbox{color} \right\rangle \end{equation}
Furthermore, the difference between a proton and $ \Delta ^+ $ is that they have different spins and total isospin. The proton is a spin $ 1/2 $ and total isospin $ 1/2 $ object while the $ \Delta ^+ $ is a spin $ 3/2 $ and total isospin $ 3/2 $ object.
Best Answer
First, the neutron $n$ and the proton $p$ don't have "excited counterparts" in the decuplet, either, do they?
Now, the two multiplets are completely different. One has eight $SU(3)_{\rm flavor}$ components, the other has ten. So it's clearly invalid to call one group "excitations" of the other.
Because the quark content (charge and strangeness) is the same, the names $\Sigma^\pm$, $\Sigma^0$, $\Xi^-$, and $\Xi^0$ from the octuplet are reused with asterisks to represent some components of the decuplet that happen to have the same quark content.
But it's a coincidence that they have the same quark content: they still transform as entirely different representations of the flavor group.
The quark content $uds$ has two baryons in the octuplet, namely $\Sigma^0$ and $\Lambda$. It is no coincidence that there are two of them. It's simply because in the diagram, the 8 components are written according to their weights i.e. eigenvalues of the $U(1)^2$ maximum commuting subgroup of $SU(3)$. And because there are unavoidably 6 components along the hexagon, the remaining 2 must be at the center. Because the 8-dimensional representation is the adjoint, those correspond to nothing else than the two generators of the $U(1)^2$ "Cartan torus" in the Lie algebra.
On the other hand, the decuplet doesn't have any degeneracies of this sort. The 10 components are organized into rows $1+2+3+4$ and there is no doubling anywhere. The component with the $uds$ quark content is just called $\Sigma^{*0}$. It could perhaps be called $\Lambda^*$ as well – because it's really neither. But it's mathematically guaranteed that there is only one component of the decuplet with the $uds$ quark content. It's guaranteed by group theory, by the decomposition of representation of $SU(3)$ under the $U(1)^2$ subgroup (the weights of the representations).
One may imagine what the representations look like in terms of tensors. The 8-dimensional representation is the adjoint with $3^2-1$ components. It's like the tensor $T_{a\bar b}$ except that we have three quarks, not one quark and one antiquark. Even with three quarks, it's possible, we may replace the antiindex $\bar b$ by an antisymmetric $cd$ pair. So this representation is a tensor $T_{acd}$ which is $cd$-antisymmetric but whose $acd$-antisymmetrization (that's the trace) vanishes.
On the other hand, the decuplet is a symmetric tensor $T_{acd}$. You can see that the symmetric tensor admits all combinations of $u,d,s$, and when you specify how many $u$, how many $d$, how many $s$, the component is determined by the symmetry of the tensor. The three corners $uuu,ddd,sss$ are allowed.
On the other hand, in the mixed symmetry tensor, $uuu,ddd,sss$ are forbidden due to the antisymmetry in the $cd$ indices. On the other hand, there are two independent components $T_{uds}$ and $T_{dsu}$. The remaining four permutations are determined by the vanishing trace and the $cd$ antisymmetry.