# Quarks and spin states

vertices
According to my notes "any pair of similar quarks must be in identical spin states". What is the reason for this?

humanino
Hi
According to my notes "any pair of similar quarks must be in identical spin states". What is the reason for this?
Very unclear. I do not expect the quark spin states in your left leg to be anyhow related to my grandmother's right arm quark spin states. So, which pairs are you talking about ?

vertices
for example |uu>, |ss>, |dd>... the spins must be pointing in the same direction.

even in hadrons, you can't have for example, the state |uus> having spins pointing |up-down-up> (the spins of u quarks have to be pointing in the same direction).

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vertices

due to colour confinement, the colour charge of a given state has to be 0 (a singlet). Therefore, the colour wavefunction is antisymettric.

Now mesons have to be symetric under exchange because they are bosons, right? So the spin-space wavefunction of |quark,anti-quark> has to be antisymetric too for the overall wavefunction to be symetric, right?

humanino
Quarks are fermions. Two identical quarks are undistinguishable inside a hadron. From the spin-statistics theorem, the "wavefunction" of a hadron in terms of quark and gluon degrees of freedom (assuming we can construct such a thing, although it should exist obviously in principle) must be anti-symmetric under the exchange of two quarks. It is one thing that is postulated, but for which people believe there should be a rigorous demonstration, that all hadrons occurring in Nature as free states must be color singlets. The color part of the wavefunction is therefore antisymmetric. So the rest of the wavefunction, in the space of flavor times spin times position for instance, should be symmetric. Note that the ground state, with the space part being obviously symmetric, has spin times flavor symmetric as well. So for two identical flavor, you get the symmetric spin state you were asking about.

vertices
Quarks are fermions. Two identical quarks are undistinguishable inside a hadron. From the spin-statistics theorem, the "wavefunction" of a hadron in terms of quark and gluon degrees of freedom (assuming we can construct such a thing, although it should exist obviously in principle) must be anti-symmetric under the exchange of two quarks. It is one thing that is postulated, but for which people believe there should be a rigorous demonstration, that all hadrons occurring in Nature as free states must be color singlets. The color part of the wavefunction is therefore antisymmetric. So the rest of the wavefunction, in the space of flavor times spin times position for instance, should be symmetric. Note that the ground state, with the space part being obviously symmetric, has spin times flavor symmetric as well. So for two identical flavor, you get the symmetric spin state you were asking about.

i see. It just gets a little bit confusing when you consider composite systems of quarks, because ofcourse quarks are fermions but a system of quarks in a given state, may not be - thus a composite state may or may not possesses exchange symettry.

many that humanin