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$\Sigma^0$ and $\Lambda^0$ consist of $uds$. $\Sigma^+_c$ and $\Lambda^+_c$ consist of $udc$. $\Sigma^0_b$ and $\Lambda^0_b$ consist of $udb$. There is probably something similar for $udt$. They have the same spin and differ in isospin. $\Sigma$ has a greater mass than mass of $\Lambda$.

In a similar question How can $\Lambda^0$ and $\Sigma^0$ both have $uds$ quark content?, the accepted answer refers to isospin. In another question What is the difference between $\Lambda$ and $\Sigma$ baryons?, the answer https://physics.stackexchange.com/a/248668/327171 also refers to isospin. This question has a some another formulation and accent. What is the difference in the interaction between quarks in $\Sigma$ and $\Lambda$ baryons?

Qmechanic
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Imyaf
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1 Answers1

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The answer is still iso-spin.

For example what's the difference between a iso-singlet deuteron:

$$ d = \frac 1 {\sqrt 2} \big(|pn\rangle - |np\rangle\big) $$

and the non-existent iso-triplet containing helium-2, the scalar-deuteron, and the di-neutron:

$$ ^2{\rm He} = |pp\rangle $$ $$ d^0 = \frac 1 {\sqrt 2} \big(|pn\rangle + |np\rangle\big) $$ $$ n^2 = |nn\rangle $$

The only difference is the iso-spin. (and the spin, because identical fermions need to be antisymmetric, so the deuteron is spin-1, and the hypothetical iso-triplet would be spin-0).

Now writing out the flavor times spin state for baryons can be tedious, I'm not going to do it, but it is a good exercise.

It is similar to what is shown with the deuteron, just with nucleons replaced by quarks $(p, n) \rightarrow (u, d)$, and then symmetrized over 3 quarks.

Here's an example of what that will look like:

Proton spin/flavor wavefunction

Though you will note that for quarks, unlike nucleons, the isospin (or flavor) times spin state will be symmetric, since you still have the color wave function, which has to be an $SU(3)$ singlet:

$$ \chi_{color}= \frac 1 {\sqrt 6} \big(|rgb\rangle + |gbr\rangle + |brg\rangle - |bgr\rangle - |grb\rangle - |rbg\rangle\big) $$

JEB
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