Quick question on Baryon spin states

[unscientific]

Things I don't understand:

• What do they mean by "two spin-1/2 doublets and a spin-3/2 quadruplet"?
• Why do they use the two flavours "+2/3e and -1/3e" ?

 

[mfb]

What do they mean by "two spin-1/2 doublets and a spin-3/2 quadruplet"?

Do you understand the easier case of 2 constituents?

Why do they use the two flavours "+2/3e and -1/3e" ?

The observed charges go from -1 to +2. The only option to get this from all combinations of two types of quarks is +2/3e and -1/3e as charges. Other numbers just do not work.

 

[unscientific]

Do you understand the easier case of 2 constituents?
The observed charges go from -1 to +2. The only option to get this from all combinations of two types of quarks is +2/3e and -1/3e as charges. Other numbers just do not work.

Just referenced Griffiths book and now I understand. Thanks!

 

[nikkkom]

Why spin 1/2 uuu and ddd states don't exist?

 

[Vanadium 50]

Why spin 1/2 uuu and ddd states don't exist?

Pauli exclusion.

 

[ChrisVer]

color : antisymmetric
so spin*flavor = symmetric.
The uuu and ddd belong to the 10-dim irreducible representation of 3*3*3... it is totally symmetric.
However the 2*2*2 = 4 + 2 + 2 for the three spin 1/2 combination, the only totally symmetric irrep is the 4-dimensional one (J=3/2) ... The other two 2-dimensitonal have mixed symmetry.
http://www.physics.umd.edu/courses/Phys741/xji/chapter3.pdf

 

[nikkkom]

Pauli exclusion.

"Naively", in baryon every quark has a different color charge, thus even in uuu or ddd there aren't two quarks with all identical quantum numbers, so Pauli exclusion should not apply here.

 

[ChrisVer]

"Naively", in baryon every quark has a different color charge, thus even in uuu or ddd there aren't two quarks with all identical quantum numbers, so Pauli exclusion should not apply here.

As I already said- The color will be antisymmetric, The rest will have to be combined into a symmetric part, so that the whole wavefunction will be totally antisymmetric - that is (not naively saying) Pauli's exclusion principle.

 

[Vanadium 50]

"Naively", in baryon every quark has a different color charge, thus even in uuu or ddd there aren't two quarks with all identical quantum numbers, so Pauli exclusion should not apply here.

Of course it does. You need a total antisymmetric wavefunction. Color is antisymmetric, so the rest has to be symmetric. uuu and ddd are flavor symmetric, so they also have to be spin-symmetruic,

 

[unscientific]

Of course it does. You need a total antisymmetric wavefunction. Color is antisymmetric, so the rest has to be symmetric. uuu and ddd are flavor symmetric, so they also have to be spin-symmetruic,

Why is colour anti-symmetric for uuu, ddd and sss for $J^P = \frac{1}{2}^{+}$ ?

Is it because for $J^P = \frac{3}{2}^{+}$ we already set colour to be anti-symmetric, so the same colour for $J^P = \frac{1}{2}^{+}$ must be anti-symmetric?

 

[Vanadium 50]

Color is antisymmetric because particles don't carry naked color.

 

[ChrisVer]

Because the singlet (color neutrality) you can achieve from SU(3) corresponds to a totally antisymmetric projection.

 

[unscientific]

Color is antisymmetric because particles don't carry naked color.

Because the singlet (color neutrality) you can achieve from SU(3) corresponds to a totally antisymmetric projection.

Yes, isolated quarks have never been observed, due to quark confinement. But uuu, ddd and sss for $J^P = \frac{1}{2}^{+}$ are not isolated quarks. Since a quark carries colour while an anti-quark carries anti-colour, shouldn't uuu, ddd and sss carry colour? My question is why is the colour anti-symmetric. I'm guessing it's because we already assign it to be anti-symmetric in $J^P = \frac{3}{2}^{+}$.

 

[ChrisVer]

Yes, isolated quarks have never been observed, due to quark confinement. But uuu, ddd and sss for $J^P = \frac{1}{2}^{+}$ are not isolated quarks. Since a quark carries colour while an anti-quark carries anti-colour, shouldn't uuu, ddd and sss carry colour? My question is why is the colour anti-symmetric. I'm guessing it's because we already assign it to be anti-symmetric in $J^P = \frac{3}{2}^{+}$.

Confinement has little to do with that. Every hadron is a color singlet/color neutral because of confinement. That's what counts. So even if you want to combine 3 quarks together, you have to do that so that you'll obtain a singlet in the end. You can see that doing that with 2 quarks for example, it's not possible. Doing that with a quark+antiquark in SU(3)_flavor it's possible, and that's why you get the mesons. But here you have SU(3)_color, so you must always look for singlets when you write down quark bound states.

In particular they belong to the the 1-dimensional representation of SU(3)_color , coming from the $3 \otimes 3 \otimes 3 = 10 \oplus 8 \oplus 8 \oplus 1$ which is antisymmetric... To illustrate the multiplication imagine you have 1 quark which can have (R,G,B) a 2nd quark with (R,G,B) and a third quark with (RGB) and you want to combine them... Their combination you see gives you the singlet, that's the meaning of the above multiplication and its decomposition into irreducible representations which tells you that there exists a combination that transforms trivially under SU(3)_color transformations...If you do the calculation of this with Young Diagrams, you will see that the singlet corresponds to the antisymmetry, you will have to write it in some form like $\epsilon^{abc} C_{1a} C_{2b} C_{3c}$ with $C_{1,2,3~ a}$ the color of 1,2,3 quarks . So it's not what you assign for any spin/parity state. The color is always antisymmetric so that you won't get colorful bound states that are not observable in nature.

Singlet= transforms trivially , so it's equivalent to being neutral.