$$\Lambda^+_c$$ and $$\Sigma_c^+$$ are both made up of the same quarks, and have the same I_3, but have different isospin quantum numbers. How is the isospin quantum number determined? Also is the isospin quantum number indicated by the greek letter used?
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3 Answers
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This answer is talking about the simple lamda and sigma; the logic is the same.
The isospin quantum number is determined by the experiments where the lamda and the sigma were discovered. The three same mass states of sigma+ sigma- sigma0 allowed to identify them as I_vector=1_vector. The lamda is a neutral singlet with a different unique mass, therefore it has I_vector =0_vector. These are the triplet and singlet representations respectively of the group SU(2).
mass of lamda (1115.683±0.006) mass of Sigma0 (1,192.642 ± 0.024),mass of Sigma+ (1,189.37 ± 0.07), mass of Sigma- (1,197.449 ± 0.030)
The difference in the Sigma masses is of the same order as the difference in the proton neutron masses, and is due to the different quark content and interactions within the baryons. To first order they can be considered to have the same mass for strong interactions.
The mass of the Lamda is significantly different than that of the Sigma triplet, which allowed the identification of isospin zero as a singlet.
After the quark model was elaborated, it was noted that the isospin projection was related to the up and down quark content of particles. The relation is
By this, the value of I3 of the nucleons proton (symbol p) and neutron (symbol n) is determined by their quark composition, uud for the proton and udd for the neutron
The fact of the different masses for Lamda and Sigma zero shows that they are different "bound states" of the strong interaction.
The charmed Sigma is a triplet, and triplets can be in Ispin SU(2) I_vector=1 . The charmed Lamda is a singlet, and thus I=0 is its representation.
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exactly like for a composition of two spin 1/2, you get (0,1) but you have two states composed by the same single states and have the same $S_z$ but have different total spin. I forgot the details about these particles in question, but you should find it somewhere written as a quantum states.
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From Wikipedia (excluding the anti-quarks for now to keep it simple):
In the modern formulation, isospin (I) is defined as a vector quantity in which up and down quarks have a value of $I = 1/2$, with the 3rd-component ($I_3$) being 1/2 for up quarks, and -1/2 for down quarks, while all other quarks have $I=0$. In general, for hadrons, therefore \begin{eqnarray} I_3=\frac {1}{2}(n_u-n_d) \end{eqnarray} where $n_u$ and $n_d$ are the numbers of up and down quarks respectively.
Or otherwise it can be stated in form of $|{I, I_3}>$ that, $u =|{1/2, 1/2}>$, $d=|{1/2, -1/2}>$ and other quarks as $|{0, 0}>$.
Thus for the hyperon with quark content $uds$, it can have its composite isospin $I = |I_u - I_d|, I_u + I_d$. i.e. 0, 1. Thus, it can be expressed as two states: $|{0,0}>$ and $|{1,0}>$.
This is similar to your example, $\Lambda^+_c$ and $\Sigma_c^+$, which have quark content $udc$ as the $strange$ has been replaced with a $charm$. Then their $| I, I_3>$ should be $|1/2 - 1/2, 1/2-1/2>$ and $|1/2+1/2, 1/2-1/2>$, $i.e.$ $|0,0>$ and $|1,0>$
