Why Are There Only 2 uds Combinations in the Baryon Octet?

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Homework Statement

Construct the Baryon Octet. (J^p)=1/2 +

Specifically I am puzzled as to why there are only 2 uds combinations.

Homework Equations

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The Attempt at a Solution

Ok so I know to make an overall spin state of 1/2 I have a antisymetric spin state and so in the octet there can't be uuu, ddd or sss because anti symmetrical color * anti symm spin would be symmetrical overall which is not allowed.
But my question is why is there only 2 variations of the uds combination.
Why can't each of the u/d/s have oposite spin to the other 2. That would make 3 possibilities.

 

[mark726]

Hello, thank you for your question. The Baryon Octet is a classification of particles that consists of eight baryons made up of three quarks each. These baryons are composed of up (u), down (d), and strange (s) quarks. The spin state of the baryons in the Baryon Octet is 1/2, meaning that each baryon has a total spin of 1/2.

To answer your question, there are only two variations of the uds combination because of the Pauli exclusion principle. This principle states that no two identical fermions (particles with half-integer spin) can occupy the same quantum state simultaneously. In other words, the three quarks in a baryon must have different quantum numbers, including spin. Since the up and down quarks have the same spin of 1/2, they cannot have opposite spins in the same baryon. Therefore, the only possible combinations for the u, d, and s quarks in the Baryon Octet are uud and uds.

I hope this helps clarify your question. If you have any further questions, please don't hesitate to ask.