Why do we have only 4 SU(2) total spin and isospin 3/2 baryons?
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In the context of SU(2), we find that for a quark triplet qqqqqq there are 4 wavefunctions of total Isospin 3/23/2, ranging for I3I3 going from −3/2−3/2 to 3/23/2. Furthermore, these wavefunctions are completely symmetric with respect to the exchange of any two quarks.
Since spins also obey the same laws, we also have 4 totally symmetric wavefunctions for the addition of three 1/21/2 spins.
Now, knowing the total wavefunction of a bound qqqqqq state must be totally antisymmetric, we find that for L=0L=0 ground state baryons, ϕflavourχspinϕflavourχspin must be totally symmetric.
Knowing that, it seems to me we would be able to find 16 different combinations that are totally symmetric. My book (Modern Particle Physics, Thomson) however seems to tell me otherwise, implying there are only 4 such baryons (Δ−,Δ0,Δ+,Δ++Δ−,Δ0,Δ+,Δ++). Is this there a reason or there indeed exists 12 more baryons of total isospin and spin 3/2 ?
Also, is there a systematic way of combining mixed symmetry wavefunctions to yield total symmetric ones ? Again, in the Thomson, he just happens to see that some combination of mixed symmetry wavefunctions yields a totally symmetric one.
Since spins also obey the same laws, we also have 4 totally symmetric wavefunctions for the addition of three 1/21/2 spins.
Now, knowing the total wavefunction of a bound qqqqqq state must be totally antisymmetric, we find that for L=0L=0 ground state baryons, ϕflavourχspinϕflavourχspin must be totally symmetric.
Knowing that, it seems to me we would be able to find 16 different combinations that are totally symmetric. My book (Modern Particle Physics, Thomson) however seems to tell me otherwise, implying there are only 4 such baryons (Δ−,Δ0,Δ+,Δ++Δ−,Δ0,Δ+,Δ++). Is this there a reason or there indeed exists 12 more baryons of total isospin and spin 3/2 ?
Also, is there a systematic way of combining mixed symmetry wavefunctions to yield total symmetric ones ? Again, in the Thomson, he just happens to see that some combination of mixed symmetry wavefunctions yields a totally symmetric one.
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Hey dear here is the answer
Instead of answering your question directly let me detour into a few details about isospin. Isospin is a global SU(2)SU(2) symmetry rotating between the ``up-down'' quark content. Three important SU(2)SU(2) multiplets are the singlet, doublet (fundamental) and the triplet representations given by. There are two orthogonal doublets given by,
u≡∣∣∣12,12⟩=(10)d≡∣∣∣12,−12⟩=(01)u≡|12,12⟩=(10)d≡|12,−12⟩=(01)
From the studies of spin angular momentum, which is also an SU(2)SU(2) symmetry, we know that two doublets combine to form a triplet and a singlet. In particular,
|u⟩|d⟩=12–√(|1,0⟩+|0,0⟩)|u⟩|d⟩=12(|1,0⟩+|0,0⟩)
(I'm not sure about the sign of the expression above)
Since the product of these states form a linear combination of two states with different total isospin it isn't an eigenstate of the total isospin operator, even though it is an eigenstate of the IzIz, the isospin along the zz axis.
The intuition here is identical to the reason a spin up and spin down particle aren't an eigenstate of the total spin operator.
Hope its help you
Instead of answering your question directly let me detour into a few details about isospin. Isospin is a global SU(2)SU(2) symmetry rotating between the ``up-down'' quark content. Three important SU(2)SU(2) multiplets are the singlet, doublet (fundamental) and the triplet representations given by. There are two orthogonal doublets given by,
u≡∣∣∣12,12⟩=(10)d≡∣∣∣12,−12⟩=(01)u≡|12,12⟩=(10)d≡|12,−12⟩=(01)
From the studies of spin angular momentum, which is also an SU(2)SU(2) symmetry, we know that two doublets combine to form a triplet and a singlet. In particular,
|u⟩|d⟩=12–√(|1,0⟩+|0,0⟩)|u⟩|d⟩=12(|1,0⟩+|0,0⟩)
(I'm not sure about the sign of the expression above)
Since the product of these states form a linear combination of two states with different total isospin it isn't an eigenstate of the total isospin operator, even though it is an eigenstate of the IzIz, the isospin along the zz axis.
The intuition here is identical to the reason a spin up and spin down particle aren't an eigenstate of the total spin operator.
Hope its help you
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