Wolfram Alpha:
Search by keyword:
Astronomy
Chemistry
Classical Physics
Climate Change
Cosmology
Finance and Accounting
General Relativity
Lagrangian and Hamiltonian Mechanics
Macroeconomics
Mathematics
Microeconomics
Particle Physics
Probability and Statistics
Programming and Computer Science
Quantum Field Theory
Quantum Mechanics
Semiconductor Reliability
Solid State Electronics
Special Relativity
Statistical Mechanics
String Theory
Superconductivity
Supersymmetry (SUSY) and Grand Unified Theory (GUT)
test
The Standard Model
Topology
Units, Constants and Useful Formulas
Special Unitary Groups and the Standard Model  Part 1
Special Unitary Groups and the Standard Model  Part 2
Special Unitary Groups and the Standard Model  Part 3

MultiParticle States

_
In constructing the 3 and the 3 we made use of
the raising and lowering operators I_{±}, U_{±} and V_{±}
which acted on single particles. We now need to
consider how we handle multiparticle states.
Tensor Methods

Let particle 1 live in a vector space, V, with an
operator T. Let particle 2 live in a vector space,
W, with an operator S. The 2 particles form a
system. We represent the state of the 2 particle
system as v ⊗ w in a new vector space V ⊗ W.
Will will use the angular momentum operators, J,
as our example. Our claim is that the ‘sum’ of
angular momenta is an angular momentum in the
tensor product:
J_{i} ≡ J_{i}^{(1)} ⊗ 1 + 1 ⊗ J_{i}^{(2)}
This acts on v ⊗ w and satisfies [J_{i},J_{j}] = ihε_{ijk}J_{k}
Certainly the sum operator, as defined above, is
an operator on v ⊗ w. We just need to check
that the commutator holds.
[J_{i},J_{j}] = [J_{i}^{(1)} ⊗ 1 + 1 ⊗ J_{i}^{(2)},J_{j}^{(1)} ⊗ 1 + 1 ⊗ J_{j}^{(2)}]
= [J_{i}^{(1)} ⊗ 1,J_{j}^{(1)} ⊗ 1] + [1 ⊗ J_{i}^{(2)},1 ⊗ J_{j}^{(2)}]
= [J_{i}^{(1)},J_{j}^{(1)}] ⊗ 1 + 1 ⊗ [J_{i}^{(2)},J_{j}^{(2)}]
= ihε_{ijk}(J_{k}^{(1)} ⊗ 1 + 1 ⊗ J_{k}^{(2)})
= ihε_{ijk}J_{k}
Therefore, we can write:
J_{i}v ⊗ w> = (J_{i}^{(1)} ⊗ 1 + 1 ⊗ J_{i}^{(2)})(v ⊗ w)>
Lets look at the individual terms with the
understanding that each operator on the right
hand side acts on the appropriate factor in
the tensor product.:
(J_{i}^{(1)} ⊗ 1)(v ⊗ w) = J_{i}^{(1)}v ⊗ w
and,
(1 ⊗ J_{i}^{(2)})(v ⊗ w) = v ⊗ J_{i}^{(2)}w
In ket form we get:
J_{i}v ⊗ w> = J_{i}^{(1)}v> ⊗ w + v ⊗ J_{i}^{(2)}w>
The eigenvalue for this equation is:
λ_{i}v ⊗ w> = λ_{i}^{(1)}v> ⊗ w + v ⊗ λ_{i}^{(2)}w>
Where we have replaced J_{i}v ⊗ w> = λ_{i}v ⊗ w>,
J_{i}^{(1)}v> = λ_{i}^{(1)}v> and J_{i}^{(2)}w> = λ_{i}^{(2)}w>.
This becomes:
λ_{i}v ⊗ w> = (λ_{i}^{(1)} + λ_{i}^{(2)})v ⊗ w>
and,
(J_{i}^{(1)} ⊗ J_{i}^{(2)})(v ⊗ w)> = J_{i}^{(1)}v> ⊗ J_{i}^{(2)}w>
= λ_{i}^{(1)}v> ⊗ λ_{i}^{(2)}w>
= λ_{i}^{(1)}λ_{i}^{(2)}(v ⊗ w)>
Therefore, the eigenvalues ADD in the tensor
product representation.
Let us now apply this to the Cartan generators.
We can write:
H^{1'}ψ_{R1} ⊗ ψ_{R2}> = H^{1}ψ_{R1}> ⊗ ψ_{R2}> + ψ_{R1}> ⊗ H^{1}ψ_{R2}>
and,
H^{2'}ψ_{R1} ⊗ ψ_{R2}> = H^{2}ψ_{R1}> ⊗ ψ_{R2}> + ψ_{R1}> ⊗ H^{2}ψ_{R2}>
So the weight of ψ_{R1} ⊗ ψ_{R2}> is found by adding
the weights of the individual representations,
ψ_{R1} and ψ_{R2}.
2 ⊗ 2 Decomposition

The weights for the 2 are:
(1/2,1/2√3) ... u
(1/2,1/2√3) ... d
2 ⊗ 2 is spanned by:
u ⊗ u = (1,1/√3)
d ⊗ d = (1,1/√3)
u ⊗ d = (0,1/√3)
d ⊗ d = (0,1/√3)
This looks like:
_
2 ⊗ 2 Decomposition

_
The weights for the 2 and 2 are:
(1/2,1/2√3) ... u
(1/2,1/2√3) ... d
_
(1/2,1/2√3) ... u
_
(1/2,1/2√3) ... d
_
2 ⊗ 2 is spanned by:
_
u ⊗ u = (0,0)
_
d ⊗ d = (0,0)
_
d ⊗ u = (1,0)
_
u ⊗ d = (1,0)
This looks like:
_
3 ⊗ 3 Decomposition

_
The weights for the 3 are:
_
(1/2,1/2√3) ... u
_
(1/2,1/2√3) ... d
_
(0,1/√3) ... s
_
3 ⊗ 3 is spanned by:
_
u ⊗ s = (1/2,1/2√3) + (0,1/√3) = (1/2,√3/2)
_
u ⊗ d = (1/2,1/2√3) + (1/2,1/2√3) = (1,0)
_
d ⊗ s = (1/2,1/2√3) + (0,1/√3) = (1/2,√3/2)
_
u ⊗ u = (1/2,1/2√3) + (1/2,1/2√3) = (0,0)
_
d ⊗ d = (1/2,1/2√3) + (1/2,1/2√3) = (0,0)
_
s ⊗ s = (1/2,1/2√3) + (1/2,1/2√3) = (0,0)
_
d ⊗ u = (1/2,1/2√3) + (1/2,1/2√3) = (1,0)
_
s ⊗ u = (0,1/√3) + (1/2,1/2√3) = (1/2,√3/2)
_
s ⊗ d = (0,1/√3) + (1/2,1/2√3) = (1/2,√3/2)
This looks like:
3 ⊗ 3 Decomposition

The weights for the 3 are:
(1/2,1/2√3) ... u
(1/2,1/2√3) ... d
(0,1/√3) ... s
3 ⊗ 3 is spanned by:
u ⊗ u = (1,1/√3)
d ⊗ d = (1,1/√3)
s ⊗ s = (0,2/√3)
u ⊗ d = (0,1/√3)
d ⊗ u = (0,1/√3)
u ⊗ s = (1/2,1/2√3)
s ⊗ u = (1/2,1/2√3)
d ⊗ s = (1/2,1/2√3)
s ⊗ d = (1/2,1/2√3)
This looks like:
6 ⊗ 3 Decomposition

Now that we know the '6' we can easily compute
the 6 ⊗ 3 as:
d ⊗ d ⊗ u = (1,1/√3) + (1/2,1/2√3) = (1/2,√3/2)
u ⊗ d ⊗ u = (0,1/√3) + (1/2,1/2√3) = (1/2,√3/2)
d ⊗ u ⊗ u = (0,1/√3) + (1/2,1/2√3) = (1/2,√3/2)
u ⊗ u ⊗ u = (1,1/√3) + (1/2,1/2√3) = (3/2,√3/2)
d ⊗ s ⊗ u = (1/2,1/2√3) + (1/2,1/2√3) = (0,0)
s ⊗ d ⊗ u = (1/2,1/2√3) + (1/2,1/2√3) = (0,0)
u ⊗ s ⊗ u = (1/2,1/2√3) + (1/2,1/2√3) = (1,0)
s ⊗ u ⊗ u = (1/2,1/2√3) + (1/2,1/2√3) = (1,0)
s ⊗ s ⊗ u = (0,2/√3) + (1/2,1/2√3) = (1/2,√3/2)
d ⊗ d ⊗ d = (1,1/√3) + (1/2,1/2√3) = (3/2,√3/2)
u ⊗ d ⊗ d = (0,1/√3) + (1/2,1/2√3) = (1/2,√3/2)
d ⊗ u ⊗ d = (0,1/√3) + (1/2,1/2√3) = (1/2,√3/2)
u ⊗ u ⊗ d = (1,1/√3) + (1/2,1/2√3) = (1/2,√3/2)
d ⊗ s ⊗ d = (1/2,1/2√3) + (1/2,1/2√3) = (1,0)
s ⊗ d ⊗ d = (1/2,1/2√3) + (1/2,1/2√3) = (1,0)
u ⊗ s ⊗ d = (1/2,1/2√3) + (1/2,1/2√3) = (0,0)
s ⊗ u ⊗ d = (1/2,1/2√3) + (1/2,1/2√3) = (0,0)
s ⊗ s ⊗ d = (0,2/√3) + (1/2,1/2√3) = (1/2,√3/2)
d ⊗ d ⊗ s = (1,1/√3) + (0,1/√3) = (1,0)
u ⊗ d ⊗ s = (0,1/√3) + (0,1/√3) = (0,0)
d ⊗ u ⊗ s = (0,1/√3) + (0,1/√3) = (0,0)
u ⊗ u ⊗ s = (1,1/√3) + (0,1/√3) = (1,0)
d ⊗ s ⊗ s = (1/2,1/2√3) + (0,1/√3) = (1/2,√3/2)
s ⊗ d ⊗ s = (1/2,1/2√3) + (0,1/√3) = (1/2,√3/2)
u ⊗ s ⊗ s = (1/2,1/2√3) + (0,1/√3) = (1/2,√3/2)
s ⊗ u ⊗ s = (1/2,1/2√3) + (0,1/√3) = (1/2,√3/2)
s ⊗ s ⊗ s = (0,2/√3) + (0,1/√3) = (0,√3)
This looks like:
Invariant Tensors

We want Q^{i*}Q^{i} to be invariant under a transformation,
U.
Q^{i} > Q'^{i} = UQ^{i}
Q^{i*} > Q'^{i*} = U^{*}Q^{i*}
Q^{i†}Q^{i} > Q'^{i†}Q'^{i} = Q^{i†}U^{†}UQ^{i}
Therefore, the requirement is: U^{†}U = I
We write (Q^{i})^{*} as Q_{i} and get:
Q^{i} > Q'^{i} = U^{i}_{j}Q^{j}
Q_{i} > Q'_{i} = Q_{j}(U^{†})^{j}_{i}
Q^{i}Q_{i} > Q'^{i}Q'_{i} = Q_{j}(U^{†})^{j}_{i}U^{i}_{k}Q^{k}
= Q_{j}δ^{j}_{k}Q^{k}
= Q
Now consider the following mixed tensor:
Q^{ij}_{k} > Q'^{ij}_{k} = U^{i}_{l}U^{j}_{m}(U^{†})^{n}_{k}Q^{lm}_{n}
So the upper indeces transform with U and the lower
indeces transform with U^{†}.
Multiply both sides by δ^{k}_{j} to get:
δ^{k}_{j}Q^{ij}_{k} > δ^{k}_{j}Q'^{ij}_{k} = δ^{k}_{j}U^{i}_{l}U^{j}_{m}(U^{†})^{n}_{j}Q^{lm}_{n}
= Q^{i}
Therefore, δ^{k}_{j}Q^{ij}_{k} represents the trace of Q^{ij}_{k}.
δ^{k}_{j} is invariant because:
δ^{k}_{j} = δ^{i}_{l}U^{l}_{i}(U^{†})^{k}_{j}
Such that,
δ^{k}_{j}Q^{ij}_{k} = δ^{i}_{l}U^{l}_{i}(U^{†})^{k}_{j}Q^{ij}_{k}
= Q^{i} as before.
Note that δ is the Kroneker Delta which is nothing
more than the unit matrix in N dimensions.
The LeviCivita Tensor

There 2 other invariant tensors in SU(3)  the
completely antisymmetric tensors ε^{ijk} and ε_{ijk}.
These are the LeviCivita symbols in tensor form.
The LC symbol is not a tensor and does not change
under coordinate transformations.
In order to make it a tensor we need to look at
Leibniz formula for determinants.
Leibnitz Formula

The Leibnitz formula expresses the determinant
of a square matrix in terms of permutations of
the matrix elements. The formula is:
det M = ΣM_{1α}M_{2β} ... M_{nζ}
^{n!}
Proof by example:
 a b c  aei  afh
 d e f  = + bfg  bdi
 g h i  + cdh  ceg
Term Row Column #Perm Sign
    
aei 123 123 0 +
afh 123 132 1 
bfg 123 231 2 +
bdi 123 213 1 
cdh 123 312 2 +
ceg 123 321 1 
#Perm is the number of permutations (flips) it
takes to get back to the row index with even = +
and odd = . Therefore,
det M = M_{11}M_{22}M_{33}  M_{11}M_{23}M_{32} + M_{12}M_{23}M_{31}  M_{12}M_{21}M_{33}
+ M_{13}M_{21}M_{32}  M_{13}M_{22}M_{31}
For SU(3):
det M = ΣM^{μ1}_{μ1'}M^{μ2}_{μ2'}M^{μ3}_{μ3'}
^{n!}
Multiplying both sides by ε_{ijk} gives:
ε_{μ1'}_{μ2'}_{μ3'}det(M) = ε_{μ1}_{μ2}_{μ3}(∂x^{μ1}/∂x^{μ1'})(∂x^{μ2}/∂x^{μ2'})(∂x^{μ3}/∂x^{μ3'})
M is the Jacobean matrix. Therefore, we can write
det(M) as det(∂x^{μ}/∂x^{μ'}). This gives:
ε_{μ1'}_{μ2'}_{μ3'}det(∂x^{μ}/∂x^{μ'}) = ε_{μ1}_{μ2}_{μ3}(∂x^{μ1}/∂x^{μ1'})(∂x^{μ2}/∂x^{μ2'})(∂x^{μ3}/∂x^{μ3'})
Or,
ε_{μ1'}_{μ2'}_{μ3'} = det(∂x^{μ'}/∂x^{μ})ε_{μ1}_{μ2}_{μ3}(∂x^{μ1}/∂x^{μ1'})(∂x^{μ2}/∂x^{μ2'})(∂x^{μ3}/∂x^{μ3'})
This is close to the tensor transformation law,
except for the determinant out front. Objects
which transform in this way are known as TENSOR
DENSITIES. It is a quantity whose transformation
law under change of basis involves the determinant
of the transformation matrix (as opposed to a tensor,
whose transformation law does not involve the
determinant). Now consider,
g_{μ'ν'} = (∂x^{μ1}/∂x^{μ1'})(∂x^{μ2}/∂x^{μ2'})g_{μν}
The RHS can be viewed as the product of 3 matrices.
Taking the determinant of both sides of the equation,
and using the fact that the determinant of a matrix
product is the product of the determinants, this
becomes:
det(g_{μ'ν'}) = det(∂x^{μ1}/∂x^{μ1'})det(∂x^{μ2}/∂x^{μ2'})det(g_{μν})
= (det(∂x^{μ}/∂x^{μ'}))^{2}det(g_{μν})
Therefore,
√{det(g_{μ'ν'})/det(g_{μν})} = det(∂x^{μ}/∂x^{μ'})
Or,
√{det(g_{μν})/det(g_{μ'ν'})} = det(∂x^{μ'}/∂x^{μ})
ε_{μ1'}_{μ2'}_{μ3'} = √{det(g_{μν})/det(g_{μ'ν'})}ε_{μ1}_{μ2}_{μ3}(∂x^{μ1}/∂x^{μ1'})(∂x^{μ2}/∂x^{μ2'})(∂x^{μ3}/∂x^{μ3'})
In Euclidean space det(g_{μ'ν'}) = 1 leaving:
ε_{μ1'}_{μ2'}_{μ3'} = √{det(g_{μν})}ε_{μ1}_{μ2}_{μ3}(∂x^{μ1}/∂x^{μ1'})(∂x^{μ2}/∂x^{μ2'})(∂x^{μ3}/∂x^{μ3'})
Switching notation we get:
ε_{ijk} = √gε_{ijk}
^ ^
 
tensor symbol
Likewise,
ε^{ijk} = (1/√g)ε^{ijk}
Where g = det(g_{μν}) = 1 for Euclidean space.
The symbol now transforms according to the tensor
transformation law. Since this is a real tensor,
we can use ε_{ij} and ε^{ij} to raise and lower indices
in SU(2) and ε_{ijk} and ε^{ijk} to raise and lower indices
in SU(3).
We will now prove the invariance of ε.
ε_{μ1'}_{μ2'}_{μ3'}det(U) = (U^{μ1}_{μ1'}U^{μ2}_{μ2'}U^{μ3}_{μ3'})ε_{μ1}_{μ2}_{μ3}
Bet det(U) = 1. Therefore,
ε_{μ1'}_{μ2'}_{μ3'} = (U^{μ1}_{μ1'}U^{μ2}_{μ2'}U^{μ3}_{μ3'})ε_{μ1}_{μ2}_{μ3}
Therefore, the antisymmetric tensor, ε_{ijk}, is
invariant when transformed by the U's. This is
also true for ε^{ijk}.
LeviCivita Symbol and Kronecker Delta Relationship

In 3D, the LeviCivita symbol is related to the
Kronecker delta in the following manner:
_{ }  δ^{i}_{k} δ^{i}_{m} δ^{i}_{n} 
ε^{ijl}ε_{kmn} =  δ^{j}_{k} δ^{j}_{m} δ^{j}_{n} 
_{ }  δ^{l}_{k} δ^{l}_{m} δ^{l}_{n} 
or,
ε^{ijl}ε_{kmn} = δ^{i}_{k}(δ^{j}_{m}δ^{l}_{n}  δ^{j}_{n}δ^{l}_{m})  δ^{i}_{m}(δ^{j}_{k}δ^{l}_{n}  δ^{j}_{n}δ^{l}_{k})
+ δ^{i}_{n}(δ^{j}_{k}δ^{l}_{m}  δ^{j}_{m}δ^{l}_{k})
For the special case:
ε^{ijl}ε_{lmn} = δ^{i}_{m}δ^{j}_{n}  δ^{j}_{n}δ^{i}_{m}
Symmetric and Antisymmetric Tensors

Just as we do with wavefunctions in QM, we can
decompose any tensor, Q^{ij}, into symmetric and
antisymmetric parts as follows:
Q^{{ij}} = (1/2)(Q^{ij} + Q^{ji}) ∴ Q^{ij} = Q^{ji}
Q^{[ij]} = (1/2)(Q^{ij}  Q^{ji}) ∴ Q^{ij} = Q^{ji}
Where we have used the notation {} to indicate that
the tensor is symmetric in i and j and [] to indicate
that the tensor is antisymmetric in i and j. (note
the analogy with the anticommutator and the commutator).
Therefore,
Q^{ij} = Q^{{ij}} + Q^{[ij]}
= (1/2)(Q^{ij} + Q^{ji}) + (1/2)(Q^{ij}  Q^{ji})
= Q^{ij}
Example:
Q^{ij} = Q^{{ij}} + Q^{[ij]}
     
 7 10 3   7 7 6   0 3 3 
 4 1 2  =  7 1 1  +  3 0 3 
 9 4 5   6 1 5   3 3 0 
     
Also, note that:
Q^{[ij]}Q^{{ij}} = 0
Proof:
Q^{[ij]}Q^{{ij}} = Q^{ji}Q^{ij}
= Q^{ji}Q^{ji}
= Q^{ij}Q^{ij} (after swapping i <> j)
Therefore,
Q^{[ij]}Q{^{ij}} = 0
Notice that a completely antisymmetric tensor
cannot have diagonal elements, because we have
the equation Q_{ii...} = Q_{ii...} ∴ Q_{ii...} = 0
Note that in the case of mixed tensors, the upper
indices and lower indices are to be permuted
separately. It doesn't make sense to interchange
an upper index and a lower index since they are
are different entities.
A tensor with one upper and one lower index is
considered to be symmetric.
Trivially, all scalars and vectors are totally
antisymmetric (as well as being totally symmetric).
Separating the Trace

When a tensor has both upper and lower indeces it
can be contracted to obtain the trace which can then
be subtracted out. The process is illustrated in
the following examples.
Example:
~
Q^{i}_{j} = Q^{i}_{j}  (1/3)δ^{i}_{j}Q^{k}_{k}
     
 0 10 3   7 10 3   1 0 0 
 4 0 2  =  4 1 2   (11/3) 0 1 0 
 9 4 0   9 4 5   0 0 1 
     
Where ~ means traceless (i.e. Tr = 0).
This was found in the following way:
~
Q^{i}_{j} = Q^{i}_{j}  αδ^{i}_{j}Q^{k}_{k}
If we contract i with j we can write this as:
~
Q^{i}_{i} = Tr(Q^{i}_{j}) = 0 = Q^{i}_{i}  αδ^{i}_{i}Q^{k}_{k}
But δ^{i}_{i} = 3. Therefore,
0 = Q^{i}_{i}  3αQ^{k}_{k}
Therefore, α = 1/3 and Q^{k}_{k} = 11.
We can generalize this result to the following:
~
Q^{ij}_{kl} = Q^{ij}_{kl}  β(δ^{i}_{k}Q^{j}_{l} + δ^{j}_{k}Q^{i}_{l} + δ^{i}_{l}Q^{j}_{k} + δ^{j}_{i}Q^{i}_{k})
where Q^{j}_{l} = Q^{j}_{l}Q^{n}_{n} etc.
Contract k with i:
~
Q^{ij}_{il} = 0 = Q^{ij}_{il}  β(δ^{i}_{i}Q^{j}_{l} + δ^{j}_{i}Q^{i}_{l} + δ^{i}_{l}Q^{j}_{i} + δ^{j}_{i}Q^{i}_{i})
= Q^{ij}_{il}  β(δ^{i}_{i}Q^{j}_{l} + δ^{j}_{i}Q^{i}_{l} + δ^{i}_{l}Q^{j}_{i})
= Q^{ij}_{il}  β(δ^{i}_{i} + δ^{j}_{i} + δ^{i}_{l})Q^{j}_{l}
= Q^{ij}_{il}  β(3 + 1 + 1)Q^{j}_{l}
= Q^{ij}_{il}  β5Q^{j}_{l}
∴ β = 1/5
Likewise,
~
S^{ij}_{k} = Q^{ij}_{k}  β(δ^{i}_{k}Q^{j} + δ^{j}_{k}Q^{i})
Contract k with i:
S^{ij}_{i} = 0 = Q^{ij}_{i}  β(δ^{i}_{k}Q^{j} + δ^{j}_{k}Q^{i})
= Q^{ij}_{i}  β(δ^{i}_{i}Q^{j} + δ^{j}_{k}Q^{i})
= Q^{ij}_{i}  β(3 + 1)Q^{j}
∴ β = 1/4
Note, that number of traces that can be subtracted
equals the number of possible contractions. For
example:
Q^{i}_{j} has 1 trace.
Q^{ij}_{k} has 2 traces (k with i, k with j)
Q^{ijk}_{l} has 3 traces (l with i, l with j, l with k)
Q^{ij}_{kl} has 4 traces (l with i, l with j, k with i
and k with j).
etc. etc.

Digression:
We stated that contraction on a pair of indices that
are either both contravariant or both covariant is
not possible. However, in differential geometry the
metric tensor can be used used to raise or lower one
of the indices, as needed, i.e.
T^{k}_{l} = g_{jk}T^{ij}
Such that contraction is possible. This operation is
known as METRIC CONTRACTION.

Dimensions of Symmetric Treceless Tensors

The dimensions of a completely symmetric tensor
is given by:
D(n,m) = (1/2)(n + 1)(m + 1)(n + m + 2)
The formula comprehends the traceless condition
ΣQ^{ijk}_{lmk} = 0 for tensors with upper and lower indeces
that can be contracted.
Proof:
Assume the tensor is completely symmetric with
n upper indeces and m lower indeces. For the
upper indeces:
= (d + n  1)!/n!(d  1)!
= (2 + n)!/2!n!
= Γ(n + 3)/2Γ(n + 1)
= (1/2)(n + 1)(n + 2)
And for the lower indeces:
= (d + m  1)!/m!(d  1)!
= (2 + m)!/2!m!
= Γ(m + 3)/2Γ(m + 1)
= (1/2)(m + 1)(m + 2)
The total is then:
= (1/4)(n + 1)(n + 2)(m + 1)(m + 2)
Contraction of 1 upper and 1 lower index reduces
the number of independent components by:
= (1/4)nm(n + 1)(m + 1)
= # of components in EACH trace.
Examples:
Q^{{ij}} (not contractable):
 
 ^{11} ^{ } ^{ } 
 ^{21} ^{22} ^{ }  = 6
 ^{31} ^{32} ^{33} 
 
Q^{i}_{j} (contractable):
 
 ^{1}_{1} ^{1}_{2} ^{1}_{3} 
 ^{2}_{1} ^{2}_{2} ^{2}_{3} 
 ^{3}_{1} ^{3}_{2} ^{3}_{3} 
 
_{3}
Tr = ΣS^{k}_{k} = Q^{1}_{1} + Q^{2}_{2} + Q^{3}_{3} = 0
^{k=1}
Now since the RHS are just scalars we can write:
Q^{3}_{3} = (Q^{1}_{1} + Q^{2}_{2})
So Q^{3}_{3} is not independent and we can ignore it.
As a consequence D(1,1) = 9  1 = 8
 
 ^{1}_{1} ^{1}_{2} ^{1}_{3} 
 ^{2}_{1} ^{2}_{2} ^{2}_{3} 
 ^{3}_{1} ^{3}_{2} ^{3}_{3} 
 
Q^{{ij}}_{k} (contractable):
     
 ^{11}_{1} ^{ }_{ } ^{ }_{ }   ^{11}_{2} ^{ }_{ } ^{ }_{ }   ^{11}_{3} ^{ }_{ } ^{ }_{ } 
 ^{21}_{1} ^{22}_{1} ^{ }_{ }   ^{21}_{2} ^{22}_{2} ^{ }_{ }   ^{21}_{3} ^{22}_{3} ^{ }_{ } 
 ^{31}_{1} ^{32}_{1} ^{33}_{1}   ^{31}_{2} ^{32}_{2} ^{33}_{2}   ^{31}_{3} ^{32}_{3} ^{33}_{3} 
     
Contract i and j with k and sum to get the 2
traces:
_{3}
Tr = ΣS^{ik}_{k} = Q^{11}_{1} + Q^{22}_{2} + Q^{33}_{3}
^{k=1}
= Q^{1} + Q^{2} + Q^{3}
and,
_{3}
Tr = ΣS^{kj}_{k} = Q^{11}_{1} + Q^{22}_{2} + Q^{33}_{3}
^{k=1}
= Q^{1} + Q^{2} + Q^{3}
     
 ^{11}_{1} ^{ }_{ } ^{ }_{ }   ^{11}_{2} ^{ }_{ } ^{ }_{ }   ^{11}_{3} ^{ }_{ } ^{ }_{ } 
 ^{21}_{1} ^{22}_{1} ^{ }_{ }   ^{21}_{2} ^{22}_{2} ^{ }_{ }   ^{21}_{3} ^{22}_{3} ^{ }_{ } 
 ^{31}_{1} ^{32}_{1} ^{33}_{1}   ^{31}_{2} ^{32}_{2} ^{33}_{2}   ^{31}_{3} ^{32}_{3} ^{33}_{3} 
     
So the 2 traces are the same and removing them
only reduces the number of independent components
by 3. Note that we are now dealing with vectors
so we cannot say.
Q^{3} = (Q^{1} + Q^{2})
As this requires Q^{1} + Q^{2} + Q^{3} = 0 which can only
be satisfied if Q^{1} = Q^{2} = Q^{3} = 0.
Summarizing:
(0,0) = 1
(1,0) = (0,1) = 3
(2,0) = (0,2) = 6
(1,1) = 8
(2,1) = (1,2) = 15
(2,2) = 27
(3,0) = 10
(4,0) = 15
Dimensions of Antisymmetric Tensors

The dimensions of a completely antisymmetric tensor
is given by:
D_{[]} = d!/r!(d  r)!
Where d is the dimension of the space and r is
the tensor rank. Therefore, for d = 3 we get:
Q: 1
Q^{i} = Q_{i}: 3
Q^{[ij]} = Q_{[ij]}: 3
Q^{[ijk]} = Q_{[ijk]}: 1
Using the example from before we can see that:
     
 7 10 3   7 7 6   0 3 3 
 4 1 2  =  7 1 1  +  3 0 3 
 9 4 5   6 1 5   3 3 0 
     
Q^{ij} = Q^{{ij}} + Q^{[ij]}
Dim(Q^{ij}) = 9
Dim(Q^{{ij}}) = (1/2)(n + 1)(m + 1)(n + n + 2) = 6
Dim(Q^{[ij]}) = d!/r!(d  r)! = 3
Alternative Formulas for Dimensions

By plugging in numbers it is easy to show that for
the following specific cases of SU(N) we can also
write:
Q^{i} = N
Antisymmetric: Q^{[ij]} = d!/r!(d  r)! ≡ N(N  1)/2
Symmetric: Q^{{ij}} = (n + 1)(m + 1)(n + m + 2)/2 ≡ N(N + 1)/2
Q^{{i}}_{{j}} = (n + 1)(m + 1)(n + m + 2)/2 ≡ N^{2}  1
Mixed: Q^{[ij]}_{k} = N(N  2)(N + 1)/2
Tensors Components as States

It should be easy to see that there is a correspondence
between the states of the 3, 6, 8, 10 etc. and the
number of independent components of traceless symmetric
tensors.
Q^{i} has 3 independent elements. If we make
the substitution 1 = u, 2 = d, 3 = s we get:
2 1
d u
3
s
Q^{{ij}} has 6 independent elements.
22 12 11
dd ud uu
du
23 13
ds us
sd su
33
ss
Q^{{ijk}} has 10 independent elements.
     
 ^{111} ^{ } ^{ }   ^{ } ^{ } ^{ }   ^{ } ^{ } ^{ } 
 ^{211} ^{221} ^{ }   ^{ } ^{222} ^{ }   ^{ } ^{ } ^{ } 
 ^{311} ^{321} ^{331}   ^{ } ^{322} ^{332}   ^{ } ^{ } ^{333} 
     
222 221 211 111 (I_{3} = 3/2)
ddd ddu duu uuu
322 321 311 (I_{3} = 1)
sdd sud suu
332 331 (I_{3} = 1/2)
ssd ssu
333 (I_{3} = 0)
sss
ClebschGordan Decomposition for SU(3)

The ClebschGordan decomposition for SU(3) is the
decomposition of the tensor product of 2 representations
into a direct sum of irreducible representations.
The basic process is as follows:
1). Split the tensor into the sum of symmetric and
antisymmetric parts.
2). When contraction is possible, subtract out all
traces and then add them back as separate terms.
Removing all the traces is necessary otherwise
the tansor is reducible and can be contracted
into smaller dimensional representation. IN
OTHER WORDS, THE TENSOR IS IRREDUCIBLE IF A
LOWER RANK TENSOR CANNOT BE FOUND.
3). Use the LeviCivita tensor to symmetrize the
antisymmetric parts, i.e.
Q^{k} = ε_{ijk}Q^{ij}
This now transforms like a tensor with a single
index which is symmetric.
Note: Symmetrization does not alter the underlying
antisymmetry and is performed only for the purpose
of determining the CG decomposition.
ClebschGordan Series

3 ⊗ 3

u^{i} ⊗ v^{j} = Q^{{ij}} + Q^{[ij]}
Q^{{ij}} = (1/2)(u^{i}v^{j} + u^{j}v^{i})
Q^{[ij]} = (1/2)(u^{i}v^{j}  u^{j}v^{i})
We can write the 2nd terms as (1/2)ε^{ijl}(ε_{lmn}u^{m}v^{n})
Proof:
(u^{i}v^{j}  u^{j}v^{i}) = (δ^{i}_{m}δ^{j}_{n}  δ^{j}_{n}δ^{i}_{m})u^{m}v^{n}
= ε^{ijl}ε_{lmn}u^{m}v^{n}
The antisymmetric term is automatically traceless
since u^{i}v^{i} = u^{i}v^{i}. Therefore,
Q^{ij} = (1/2((u^{i}v^{j} + u^{j}v^{i}) + (1/2)ε^{ijl}(ε_{lmn}u^{m}v^{n})
^{ } = (1/2)(u^{i}v^{j} + u^{j}v^{i}) + (1/2)ε^{ijl}(uv_{l})
The first term is the '6' (2,0). The second
_
term is the '3' (0,1).
_
3 ⊗ 3

u^{i} ⊗ v_{j} = Q^{i}_{j}
~
Q^{i}_{j} = u^{i}v_{j}  βδ^{i}_{j}u^{k}v_{k}
Contract i with j and demand that Tr(S^{i}_{j}) = 0:
~
Q^{i}_{i} = 0 = u^{i}v_{i}  βδ^{i}_{i}u^{k}v_{k}
= u^{i}v_{i}  β3u^{k}v_{v}
∴ β = 1/3
Therefore,
u^{i}v_{j} = (u^{i}v_{j}  (1/3)δ^{i}_{j}u^{k}v_{k}) + (1/3)δ^{i}_{j}u^{k}v_{k}
The term inside the () is a traceless symmetric
tensor and the second term is antisymmetric. Note
the scalar terms can be regarded as being totally
symmetric or totally antisymmetric.
The first term is the '8' (1,1). The second
term is the '1' (0,0).

Digression:
The product of the fundamental representation with
its complex conjugate is always reducible into a
sum that contains at least the singlet state. We
can see this in the following manner by considering
the Kronecker δ in tensor form:
_
δ_{i}^{j} > (R ⊗ R)δ_{i}^{j}
It is convenient to introduce upper and lower indices
to write:
(T_{a}(C))^{i}_{j} = (T_{a}(R))_{i}^{j}
So that complex conjugation is achieved by changing
the lower indices to upper ones, and vice versa.
Also, transposition is achieved by exchanging rows
and columns. Thus, (T_{a}(R))^{j}_{i} is the transpose of
(T_{a}(C))_{i}^{j}.
We can now write the infinitessimal generators
as:
δ_{i}^{j} > (1 + iθT_{a}(R))_{i}^{k}(1 + iθT_{a}(C))^{j}_{l}δ_{k}^{l}
_{ } > (1 + iθT_{a}(R))_{i}^{k}(1  iθT_{a}(R))_{j}^{l}δ_{k}^{l}
_{ } > (1 + iθT_{a}(R))_{i}^{k}(1  iθT_{a}(R))_{l}^{j}δ_{k}^{l}
_{ } > δ_{i}^{j} + O(θ^{2})
_
This means that R ⊗ R must contain the singlet
representation. Now, the dimension of the tensor
product is the product of the dimensions of R and
its conjugate. Therefore, the nonsinglet states
must number N^{2}  1. Thus,
_
R ⊗ R = 1 ⊕ (N^{2}  1)
For SU(3) N^{2}  1 = 8 which is the dimension of
_
the adjoint representation. So the 3 ⊗ 3 is
reducible into a sum that contains at least the
singlet and the adjoint representations. i.e.
_  
3 ⊗ 3 = 1 ⊕ 8 =  [1x1] 0 
_{ }  0 [8x8] 
_{ }  

3 ⊗ 8

u^{i} ⊗ v^{j}_{k} = Q^{{ij}}_{k} + Q^{[ij]}_{k}
~
Q^{{ij}}_{k} = (1/2)(u^{i}v^{j}_{k} + u^{j}v^{i}_{k}  α(δ^{i}_{k}u^{l}v^{j}_{l} + δ^{j}_{k}u^{l}v^{i}_{l}))
~
Q^{[ij]}_{k} = (1/2)(u^{i}v^{j}_{k}  u^{j}v^{i}_{k}  β(δ^{i}_{k}u^{l}v^{j}_{l}  δ^{j}_{k}u^{l}v^{i}_{l}))
~
Consider Q^{{ij}}_{k}. Contract i with k to get:
~
Q^{{ij}}_{i} = 0 = (1/2)(u^{i}v^{j}_{i} + u^{j}v^{i}_{i}  α(δ^{i}_{i}u^{l}v^{j}_{l} + δ^{j}_{k}u^{l}v^{i}_{l}))
= (1/2)(u^{i}v^{j}_{i}  α(δ^{i}_{i} + δ^{j}_{k})u^{l}v^{j}_{l})
= (1/2)(u^{i}v^{j}_{i}  α(3 + 1)u^{l}v^{j}_{l}) since δ^{i}_{i} = 3
Therefore, α = 1/4. Rewriting we get:
~
Q^{{ij}}_{k} = (1/2)(u^{i}v^{j}_{k} + u^{j}v^{i}_{k})  (1/8)δ^{i}_{k}u^{l}v^{j}_{l}  (1/8)δ^{j}_{k}u^{l}v^{i}_{l}
~
Now consider Q^{[ij]}_{k}. Contract i with k to get:
Q^{[ij]}_{i} = 0 = (1/2)(u^{i}v^{j}_{i}  β(δ^{i}_{i}u^{l}v^{j}_{l}  δ^{j}_{k}u^{l}v^{i}_{l}))
= (1/2)(u^{i}v^{j}_{i}  β(δ^{i}_{i}  δ^{j}_{k})u^{l}v^{j}_{l})
= (1/2)(u^{i}v^{j}_{i}  β(3  1)u^{l}v^{j}_{l})
Therefore, β = 1/2. Rewriting we get:
~
Q^{[ij]}_{k} = (1/2)(u^{i}v^{j}_{k}  u^{j}v^{i}_{k})  (1/4)δ^{i}_{k}u^{l}v^{j}_{l} + (1/4)δ^{j}_{k}u^{l}v^{i}_{l})
So that,
~ ~
Q^{{ij}}_{k} + Q^{[ij]}_{k} = (1/2)(u^{i}v^{j}_{k} + u^{j}v^{i}_{k})  (1/8)δ^{i}_{k}u^{l}v^{j}_{l}  (1/8)δ^{j}_{k}u^{l}v^{i}_{l}
+ (1/2)(u^{i}v^{j}_{k}  u^{j}v^{i}_{k})  (1/4)δ^{i}_{k}u^{l}v^{j}_{l} + (1/4)δ^{j}_{k}u^{l}v^{i}_{l})
= u^{i}v^{j}_{k}  (3/8))δ^{i}_{k}u^{l}v^{j}_{l} + (1/8)δ^{j}_{k}u^{l}v^{i}_{l}
Need to add (3/8)δ^{i}_{k}u^{l}v^{j}_{l}  (1/8)δ^{j}_{k}u^{l}v^{i}_{l} to make
both sides balance. Therefore:
~
Q^{ij}_{k} = (1/2)(u^{i}v^{j}_{k} + u^{j}v^{i}_{k})  (1/8)δ^{i}_{k}u^{l}v^{j}_{l}  (1/8)δ^{j}_{k}u^{l}v^{i}_{l}
+ (1/2)(u^{i}v^{j}_{k}  u^{j}v^{i}_{k})  (1/4)δ^{i}_{k}u^{l}v^{j}_{l} + (1/4)δ^{j}_{k}u^{l}v^{i}_{l})
+ (3/8)δ^{i}_{k}u^{l}v^{j}_{l}  (1/8)δ^{j}_{k}u^{l}v^{i}_{l}
The blue term is:
(1/2)ε^{ijl}ε_{lmn}u^{m}v^{n}_{k}
The red term is:
(1/4)δ^{i}_{k}u^{l}v^{j}_{l} + (1/4)δ^{j}_{k}u^{l}v^{i}_{l}) = (1/4)u^{i}v^{j}_{k} + (1/4)u^{j}v^{i}_{k}
Which can be written as:
(1/4)ε^{ijl}ε_{kmn}u^{m}v^{n}_{l}
Proof:
We use the identity:
 δ^{i}_{k} δ^{i}_{m} δ^{i}_{n} 
 δ^{j}_{k} δ^{j}_{m} δ^{j}_{n}  = ε^{ijl}ε_{kmn}
 δ^{l}_{k} δ^{l}_{m} δ^{l}_{n} 
(1/4)ε^{ijl}ε_{kmn}u^{m}v^{n}_{l} =
(1/4)(δ^{i}_{k}(δ^{j}_{m}δ^{l}_{n}  δ^{j}_{n}δ^{l}_{m})u^{m}v^{n}_{l}
 δ^{i}_{m}(δ^{j}_{k}δ^{l}_{n}  δ^{j}_{n}δ^{l}_{k})u^{m}v^{n}_{l} = δ^{i}_{m}(u^{m}v^{j}_{k}  u^{m}v^{j}_{k})
+ δ^{i}_{n}(δ^{j}_{k}δ^{l}_{m}  δ^{j}_{m}δ^{l}_{k})u^{m}v^{n}_{l} = δ^{i}_{n}(u^{m}v^{n}_{k}  u^{j}v^{n}_{k}))
The 2nd and 3rd terms are 0 so we get:
(1/4)ε^{ijl}ε_{kmn}u^{m}v^{n}_{l} = (1/4)(δ^{i}_{k}(δ^{j}_{m}δ^{l}_{n}  δ^{j}_{n}δ^{l}_{m})u^{m}v^{n}_{l}
= (1/4)δ^{i}_{k}(u^{j}v^{l}_{l}  u^{l}v^{j}_{l})
= (1/4)(u^{j}v^{i}_{k}  u^{i}v^{j}_{k})
= (1/4)u^{i}v^{j}_{k} + (1/4)u^{j}v^{i}_{k}
Finally, we get:
~
Q^{ij}_{k} = (1/2)(u^{i}v^{j}_{k} + u^{j}v^{i}_{k})  (1/8)δ^{i}_{k}u^{l}v^{j}_{l}  (1/8)δ^{j}_{k}u^{l}v^{i}_{l}
+ (1/4)(ε^{ijl}(ε_{lmn}u^{m}v^{n}_{k} + ε_{kmn}u^{m}v^{n}_{l}))
+ (1/8)(3δ^{i}_{k}u^{l}v^{j}_{l}  δ^{j}_{k}u^{l}v^{i}_{l})
The first term is the '15' (2,1). The second term
_
is the '6' (0,2) and the third terms is the '3' (1,0).
6 ⊗ 3

u^{i} ⊗ v^{jk} = Q^{{ijk}} + Q^{[ij]k} + Q^{[ik]j}
Q^{{ijk}} = (u^{i}v^{jk} + u^{j}v^{ki} + u^{k}v^{ij})
Q^{[ij]k} = (u^{i}v^{jk}  u^{j}v^{ik})
Q^{[ik]j} = (u^{i}v^{kj}  u^{k}v^{ij})
u^{i} ⊗ v^{jk} = Q^{{ijk}} + (u^{i}v^{jk}  u^{j}v^{ik})
+ (u^{i}v^{kj}  u^{k}v^{ij})
As before, we can write:
(u^{i}v^{jk}  u^{j}v^{ik}) = ε^{ijl}(ε_{lmn}u^{m}v^{nk})
and,
(u^{i}v^{kj}  u^{k}v^{ij}) = ε^{ikl}(ε_{lmn}u^{m}v^{nj})
u^{i} ⊗ v^{jk} = (u^{i}v^{jk} + u^{j}v^{ki} + u^{k}v^{ij}) + ε^{ijl}(ε_{lmn}u^{m}v^{nk})
+ ε^{ikl}(ε_{lmn}u^{m}v^{nj})
= (u^{i}v^{jk} + u^{j}v^{ki} + u^{k}v^{ij}) + ε^{ijl}(uv^{k}_{l})
+ ε^{ikl}(uv^{j}_{l})
The first term is the '10' (3,0). The second term
is the '8' (1,1). Note that the second term is
traceless because it is antisymmetric (A^{ij}_{k} = A^{ji}_{k}
and A^{ij}_{k} = A^{ki}_{j}).
Note that the nonseparability of the trace in
the 6 ⊗ 3 prevents the decomposition into
9 ⊕ 8 ⊕ 1.
Young's Tableaux

We can also represent tensor products and irreps using
YOUNG'S TABLEAUX. This an alternative (and simpler)
way of looking at tensor decompositions.
A Young tableau is a diagram of leftjustified rows
of boxes where any row is not longer than the row on
top of it and each column is no longer than the one
to the left of it. There is 1 box for each particle.
The rows represent symmetric tensors, and the columns
represent antisymmetric tensors. The remaining
combinations represent mixed states. A general tableau
will be of mixed symmetry. A tableau can consist of
any number of boxes provided that no more than N boxes
occur in any column (for SU(N))
We will restrict the discussion to SU(2) and SU(3).
In these cases the maximum column height is 2 and
3 respectively. Let:
n = # of boxes beyond the 2nd row.
m = # of boxes beyond the 3rd row.
We can then describe the box combinations in tensor
notation (n,m) as follows (tensors are denoted by
Q to differentiate them from group generators.
SU(2):
Q^{i} = (1,0):

 

Q^{{ij)} = (2,0):

  

Q = (0,0):

 

 

Note: This is not (0,1) because SU(2) can only
have a maximum of 2 rows.
SU(3):
Q^{i} = (1,0):

 

Q^{{ij}} = (2,0):

  

Q^{{ijk}} = (3,0):

   

Q = (0,0):

 

 

 

Q^{i}_{j} = (1,1):

  

 

Q^{{ij}}_{k} = (2,1):

   

 

Relation to Dynkin Labels

Note that the numbers in the brackets are
identical to the Dynkin labels for the highest
weights described earlier, i.e. for SU(3),
Rep. Dynkin Q Highest weight
(n,m) (ν)

3 (1,0) Q^{i} (1/2,1/2√3)

3 (0,1) Q_{i} (1/2,1/2√3)
8 (1,1) Q^{i}_{j} (1,0)
Dimensions

1. Write down the quotient of 2 copies of the tableau.
2. In the numerator, fill the boxes as follows:

 N N+1N+2

N1 N N+1

N2N1

3. In the denominator, the number to be entered in a
given box is called the HOOK LENGTH. It is the number
of boxes to the right of the given box PLUS the number
of boxes below the given box PLUS 1 for itself.
Therefore, the abpve would look like:

 5  4  2 

 4  3  1 

 2  1 

4. The dimension can be found by evaluating the quotient
by calculating the product of their respective entries
and dividing. For example, for SU(3) we get:
3.4.5.2.3.4.1.2/5.4.2.4.3.1.2.1 = 3
For SU(2) we have:
 
 2  ÷  1 
 
D = 2/1 = 2
 
 2  3  ÷  2  1 
 
D = 2.3/2.1 = 3
 
 2   2 
 ÷ 
 1   1 
 
D = 2.2/2.1 = 1
For SU(3) we have:
 
 3  ÷  1 
 
D = 3/1 = 3
 
 3  4  ÷  2  1 
 
D = 3.4/2.1 = 6
 
 3  4  5  ÷  3  2  1 
 
D = 3.4.5/3.2.1 = 10
 
 3   3 
 
 2  ÷  2 
 
 1   1 
 ...
D = 3.2.1/3.2.1 = 1
 
 3  4  ÷  3  1 
 
 2   1 
 
D = 3.4.2/3.1.1 = 8
 
 3  4  5   4  2  1 
 ÷ 
 2   1 
 
D = 3.4.5.2/4.2.1.1 = 15
Number of States

In terms of multiplets, the integers n and m also represent
the number of steps across the top and bottom levels of the
multiplet diagram. This is illustated below,
The number of states associated with each combination is
given by:
SU(2): (n + 1) (top diagram)
SU(3): (n + 1)(m + 1)(n + m + 2)/2 (bottom 2 diagrams)
For example, (3,0) represents (4)(1)(3 + 2)/2 = 10 states.
Conjugate Representation

The complex conjugate of a given irrep is represented
by a tableaux obtained by switching any column of k
boxes with a column of (N  k) boxes. For example,
for SU(3) N = 3.
 
    >    
 
   

SU(2):
Any irrep is self conjugate.
(0,1) = '2':

 

(0,2) = '3':

  

(0,0) = '1':

 

 

SU(3):
_
(0,1) = '3':

 

 

With dimension:
 
 3   2 
 ÷ 
 2   1 
 
D = 3.2/2/1 = 3
_
(0,2) = '6':

  

  

With dimension:
 
 3  4   3  2 
 ÷ 
 2  3   2  1 
 
D = 3.4.2.3/3.2.2.1 = 6
Tensor Product of Multiplets

The following recipe tells us how to find the multiplets
that occur in coupling 2 multiplets together. To couple
together more than 2 multiplets, first couple 2, then
couple a third with each of the multiplets obtained from
the first 2, etc.
Draw the Young diagrams for the 2 multiplets, but in one
of the diagrams replace the boxes in the first row with
a's and the boxes in the second row with b’s, etc. The
unlettered diagram forms the upper left hand corner of all
the enlarged diagrams constructed below.
Add the a’s from the lettered diagram to the right hand
ends of the rows of the unlettered diagram to form all
possible legitimate Young diagrams that have no more
than one a per column. In general, there will be several
distinct diagrams, and all the a’s appear in each diagram
Use the b’s to further enlarge the diagrams already
obtained, subject to the same rules. Then throw away
any diagram in which the full sequence of letters formed by
reading right to left in the first row, then the second row,
etc., is not admissible. Note that the height of the left
most column cannot exceed N (3 in this case). As an example
we consider the somewhat complex 8 ⊗ 8 product to
illustrate the process.
This gives:
__
8 ⊗ 8 = 27 ⊕ 10 ⊕ 10 ⊕ 8 ⊕ 8 ⊕ 1
2 ⊗ 2 Decomposition

(1,0) ⊗ (1,0) = (2,0) ⊕ (0,0)
2 ⊗ 2 = 3 ⊕ 1
3 ⊗ 3 Decomposition

T^{i} ⊗ T^{j} = T^{ij}
(1,0) ⊗ (1,0) = (2,0) ⊕ (0,1)
_
3 ⊗ 3 = 6 ⊕ 3
_
3 ⊗ 3 Decomposition

T^{i} ⊗ T_{j} = T^{i}_{j}
(1,0) ⊗ (0,1) = (1,1) ⊕ (0,0)
_
3 ⊗ 3 = 8 ⊕ 1
3 ⊗ 3 ⊗ 3 Decomposition

T^{i} ⊗ T^{j} ⊗ T^{k} = T^{ijk}
(1,0) ⊗ (1,0) ⊗ (1,0) = (0,0) ⊕ (3,0) ⊕ (1,1) ⊕ (1,1)
3 ⊗ 3 ⊗ 3 = 1 ⊕ 10 ⊕ 8 ⊕ 8
We saw from before that this the decomposition:
3 ⊗ 3 ⊗ 3
_
= (3 ⊕ 6) ⊗ 3
_
= (3 ⊗ 3) ⊕ (6 ⊗ 3)
= 1 ⊕ 8 ⊕ 8 ⊕ 10
6 ⊗ 3 = 10 ⊕ 8:
This gives the baryon octet and the baryon
decuplet.
_
3 ⊗ 3 = 1 ⊕ 8:
This gives the meson octet and the antisymmetric
singlet.
Boxes as Particles

In the sense of u > d > s as increasing values:
 Rows cannot decrease.
 Columns must increase.
The various admissible combinations of u, d and s
_
for the 3 ⊗ 3 composition for mesons.
 
 u  u  = p  u  u  = Σ^{+}
 
 d   s 
 
 
 u  d  = n  u  d  = Γ^{0}
 
 d   s 
 
 
 u  s  = Σ^{0}  u  s  = Ξ^{0}
 
 d   s 
 
 
 d  d  = Σ^{}  d  s  = Ξ^{}
 
 s   s 
 

 u 

 d  = singlet

 s 

Flavor Symmetry

It was shown before that the values of (n,m) (the
Dynkin labels) correspond to the highest weight state
in the representation. We can therefore construct
all states in (n,m) by acting on the tensor with
lowering operators.
Orthonormality of States

Two states are orthogonal if there is 0 probability to
measure one of the states when the system is prepared
in the other one. For states to be orthogonal their
dot product has to be 0. Therefore, ψ_{1} and ψ_{2} are
orthogonal if:
<ψ_{1}ψ_{2}> = 0
The rules are:
<ud> = <du> = 0 etc.
_ _
<uuuu> = 1 etc.
_ _
<uudd> = 0 etc.
<uddu> = 0 etc.
States also need to be normalized such that:
<ψ_{1}ψ_{1}> = 1
Deriving the Isospin Wavefunctions

Of particular interest is the isospin operator, I_{±}.
When we consider multiple particle states we need
to take into account how their individual isospins
combine. Now isospin combines in the same way as
angular momentum, therefore we need to include the
CLEBSCHGORDAN coefficients in the calculations.
I_{±} is then written as:
I_{±}I,I_{3}> = √[I(I + 1)  I_{3}(I_{3} ± 1)]I,I_{3} ± 1>
= C_{±}I,I_{3} ± 1>
Then,
C_{±}I,I_{3} ± 1> = I_{±}ψ>
= ψ_{±}
Where ψ is the combination of quarks being raised
or lowered.
I,I_{3}> is a notation that labels states in terms of
total isospin and the third component of isospin.
This is directly analagous to the way states of
angular momentum are specified, i.e.
eigenvalue of total isospineigenvalue of z component>
For example 1,0> corresponds to the combination
of 2 isospins ud + du>.
The ClebschGordan calculation automatically ensures
the orthonormality of the states. Therefore,
I_{±}ψ> = Cψ_{±}
Such that,
<Cψ_{±}Cψ_{±}> = 1
For example,
_ _ _ _
<C(uu  dd)C(uu  dd)> = C^{2}(1  0  0 + 1) = 2C^{2}
∴ C = √(1/2)
Therefore, the correct state is:
_ _
√(1/2)(uu  dd)
In essence the CG are just the normalization
coefficients for composite states.
Using the raising and lowering operators to find the
coefficients can be very tedious. Fortunately, these
coefficients can also be found directly from published
tables (see the separate note entitled ClebschGordan
coefficients).
From before we had:
T_{±}v ⊗ w> = T_{±}v>w + vT_{±}w>
Let's return to the reaising and lowering operators
again. They operate on quarks and antiquarks as follows:
I_{+}d> = u>
_ _
I_{+}u> = d>
I_{}u> = d>
_ _
I_{}d> = u>
U_{+}s> = d>
_ _
U_{+}d> = s>
U_{}d> = s>
_ _
U_{}s> = d>
V_{+}s> = u>
_ _
V_{+}u> = s>
V_{}u> = s>
_ _
V_{}s> = u>
All other combinations give 0.
_
The 2 ⊗ 2

_
The ordering and the minus sign in the 2 ensures
that antiquarks and quarks transform in the
same way. To see this we start with defining
the antiquark doublet as:
 
q* =  d^{*} 
 u^{*} 
 
This transforms as:
   
 d'*  = U* d* 
 u'*   u* 
   
 
Now mutiply both sides by the matrix  0 1 .
 1 0 
 
       
 0 1  d^{*}'  = U* 0 1  d^{*} 
 1 0  u^{*}'   1 0  u^{*} 
       
Which gives:
   
 u'*  = U* u* 
 d'*   d* 
   
This compares to the quark doublet which transforms
as:
   
 u'  = U u 
 d'   d 
   
So far so good. However, let's stick with the
previous form. Now multiply both sides by the
 
inverse of  0 1  to get:
 1 0 
 
       
 d'^{*}  =  0 1 U* 0 1  d^{*} 
 u'^{*}   1 0   1 0  u^{*} 
       
In pecial Unitary Groups and the Standard Model  Part 1
we stated that a 2 x 2 unitary matrix can be written as:
 
U =  a b 
 b* a* 
 
Therefore,
 
U* =  a* b* 
 b a 
 
Substituting this into the previous equation
gives:
         
 d'^{*}  =  0 1  a* b*  0 1  d^{*} 
 u'^{*}   1 0  b a  1 0  u^{*} 
         
Which gives:
     
 d^{*}'  =  a b  d^{*} 
 u^{*}'   b* a*  u^{*} 
     
Which is just:
   
 d^{*}'  = U d^{*} 
 u^{*}'   u^{*} 
   
So the quark and antiquark doublets as defined
transform in the same way.
Note: This is a special property of SU(2). There
is no analogous representation for SU(3)!
The quark/antiquark I_{3}'s are:
_
u = d = 1/2
_
d = u = 1/2
Therefore,
_
1,1> = ud
_
1,1> = du
_ _ _
I_{}ud> = I_{}u>d  uI_{}d>
_ _
= uu  dd
and,
I_{}1,1> = √[1(1 + 1)  1(1  1)]1,0>
= √21,0>
Therefore,
_ _
1,0> = √(1/2)(uu  dd)
The second state needs to be orthogonal to this
state. This state is:
_ _
ψ_{2} = √(1/2)(uu + dd)
Check:
_ _ _ _
<ψ_{1}ψ_{2}> = <uu  dduu + dd>
_ _ _ _ _ _ _ _
= <uuuu> + <uudd>  <dduu>  <dddd>
= 1  0  0  1
= 0
The singlet state is ALWAYS a deadend from the point of
view of the raising and lowering operators. In other
words for the state, ψ to be a singlet it must satisfy:
I_{±}ψ> = 0
The only (0,0) state that satifies this is ψ_{2}.
_
The Mesons, 3 ⊗ 3

Quark isospins can combine to give a total of 3/2 or 1/2
This can be achieved as follows:
_
I_{3} for s, s = 0
_
1,1> = ud
_
1,1> = du
_ _
1,1/2> = us, ds
_ _
1,1/2> = sd, su
Starting from the outer states we can reach the
center in 6 ways.
_ _ _
I_{+}du> = √(1/2)(uu>  dd>)
_ _ _
I_{}ud> = √(1/2)(dd>  uu>)
_ _ _
V_{+}su> = √(1/2)(uu>  ss>)
_ _ _
V_{}us> = √(1/2)(ss>  uu>)
_ _ _
U_{+}sd> = √(1/2)(dd>  ss>)
_ _ _
U_{}ds> = √(1/2)(ss>  dd>)
_ _
We have already found the state ψ_{1} = √(1/2)(uu>  dd>)
_
at (0,0) from the 2 ⊗ 2 analysis. However, there
are 2 more states at (0,0) that we need to find.
Whatever these states are, they must be a linear
combination of:
_ _ _ _ _
uu>  dd>, uu>  ss> and dd>  ss>
The second state can be obtained by taking the
linear combination of the other two states which
is orthogonal to ψ_{1}.
_ _ _ _
ψ_{2} = uu>  ss> + dd>  ss>
_ _ _
= uu> + dd>  2ss>
Orthogonality check:
_ _ _ _ _
<ψ_{2}ψ_{1}> = <uu + dd  2ssuu  dd>
_ _ _ _ _ _ _ _
= <uuuu>  <uudd> + <dduu>  <dddd>
_ _ _ _
 2<ssuu> + 2<ssdd>
= 1  0 + 0  1  0 + 0
= 0
Therefore, ψ_{1} and ψ_{2} are orthogonal.
_ _ _ _ _ _
<uu + dd  2ssuu + dd  2ss>
= 1 + 0  0 + 0 + 1 + 0  0  0 + 4
= 6
The normalized state is:
_ _ _ _ _ _
<C(uu + dd  2ss)C(uu + dd  2ss)> = 6C^{2}
Therefore,
_ _ _
ψ_{2} = √(1/6)(uu + dd  2ss)
Likewise, the final state can be obtained by
requiring it to be orthonormal to both ψ_{1} and ψ_{2}.
_ _ _
ψ_{3} = √(1/3)(uu + dd + ss)
This is the singlet state. Again, the singlet state
is ALWAYS a deadend from the point of view of the
raising and lowering operators. In other words for
the state, ψ to be a singlet it must satisfy:
I_{±}ψ> = 0 and U_{±}ψ> = 0 and V_{±}ψ> = 0
The only (0,0) state that satifies this is ψ_{3}.
The complete list for the mesons is:
_
1,1> = ud
_ _
0,0> = √(1/2)(uu  dd)
_
1,1> = du
_ _ _
0,0> = √(1/6)(uu + dd  2ss)
_ _
0,0> = √(1/2)(uu + dd)
_ _ _
0,0> = √(1/3)(uu + dd + ss)
_
1/2,1/2> = us
_
1/2,1/2> = ds
_
1/2,1/2> = sd
_
1/2,1/2> = su
The Baryons

Quark isospins can combine to give a total of 3/2 or 1/2
This can be achieved as follows:
3/2,3/2> = uuu>
3/2,1/2> = positive combinations duu>, udu> and uud>
3/2,1/2> = positive conbinations of udd>, dud> and
ddu>
3/2,3/2> = ddd>
1/2,1/2> = positive and negative combinations of duu>,
udu> and uud>
1/2,1/2> = positive and negative combinations of udd>,
dud> and ddu>
We start with the 2 ⊗ 2 = 3 ⊕ 1 where the '2' is:
 
2 =  u 
 d 
 
The LHS is the same as the top row of the '6'. We
can now form a 'parial' tensor product consisting
of this and the top row of the '3'. We get:
(3 ⊕ 1) ⊗ 2 = (3 ⊗ 2) ⊕ (1 ⊗ 2)
= 4 ⊕ 2 ⊕ 2
3 ⊗ 2 is:
   
 uu   u 
 ud + du _{ }⊗  
 dd   d 
   
Which gives the combinations uuu, uud, udu, duu,
ddu and ddd. We can now use I_{±} operator to
find the intermediate states as follws:
I_{+}ddd> = I_{+}d>dd + dI_{}dd>
= udd + d[I_{+}d>d + dI_{+}d>]
= udd + d[ud + du]
= udd + dud + ddu
and,
I_{+}3/2,3/2> = √[3/2(3/2 + 1)  (3/2)(3/2 + 1)]3/2,1/2>
= √33/2,1/2>
Therefore,
√33/2,1/2> = udd + dud + ddu
or,
3/2,1/2> = √(1/3)(udd + dud + ddu)
We now raise this state.
√(1/3)I_{+}udd + dud + ddu>
I_{+}udd> = I_{+}u>dd + uI_{+}dd>
= 0 + u[I_{+}d>d + dI_{+}d>]
= u[ud + du]
= uud + udu
I_{+}dud> = I_{+}d>ud + dI_{+}ud>
= uud + d[I_{+}u>d + uI_{+}d>]
= uud + d[0 + uu
= uud + duu
I_{+}ddu> = I_{+}d>du + dI_{+}du>
= udu + d[I_{+}d>u + dI_{}u>]
= udu + d[uu + 0]
= udu + duu
Therefore,
√(1/3)I_{+}udd + dud + ddu> = 2√(1/3)(uud + udu + duu)
and,
I_{+}3/2,1/2> = √[3/2(3/2 + 1)  (1/2)(1/2 + 1)]3/2,1/2>
= 23/2,1/2>
Therefore,
23/2,1/2> = 2√(1/3)(uud + udu + duu)
or,
3/2,1/2> = √(1/3)(uud + udu + duu)
Finally, after repeating the procedure again we
get:
I_{+}udd + dud + ddu> = uuu
The 4 states, ddd, udd + dud + ddu, uud + udu + duu
and uuu found above form a symmetric quadruplet
called the '4'.
1 ⊗ 2 is:
 
   u 
 ud  du _{ }⊗  
   d 
 
Which gives the states udu, udd, duu and dud.
These are paired as:
udu  duu = (ud  du)u and udd  dud = (ud  du)d
as required and form the top row of one octet.
So we have found the '4' and one of the '2's. The
remaining '2' can be found by making use of the
orthonormality property. The states are:
√(1/6)(udd + dud  2ddu)
and,
√(1/6)(2uud  udu  duu)
We can use the raising and lowering operators I_{±},
U_{±} and V_{±} to move around the '10' and '6' because
they have the same equilateral triangular structure
as the '3' and '8'.
The complete list for the '10_{S}' is:
3/2,3/2> = uuu: uuu
3/2,1/2> = uud: √(1/3)(uud + udu + duu)
3/2,1/2> = ddu: √(1/3)(ddu + dud + udd)
3/2,3/2> = ddd: ddd
uus: √(1/3)(uus + usu + suu)
uds: √(1/6)(uds + usd + dsu + dus + sud + sdu)
dds: √(1/3)(dds + dsd + sdd)
ssu: √(1/3)(ssu + sus + uss)
ssd: √(1/3)(ssd + sds + dss)
sss: sss
These are symmetric for all quark interchanges.
Again we have used the orthonormality condition to
determine the coefficients. For example,
<uusuus> =
<(uus + usu + suu)(uus + usu + suu)> = 3
Mixed Symmetry Octet, 8_{MS}

√(1/6)I_{}2uud  udu  duu>
2I_{}uud> = 2(I_{}u>ud + uI_{}ud>)
= 2(dud + u[I_{}u>d + uI_{}d>])
= 2dud + 2udd
I_{}udu> = I_{}u>du + uI_{}du>
= ddu + u[I_{}d>u + dI_{}u>]
= ddu + udd
I_{}duu> = I_{}d>uu + dI_{}uu>
= d[I_{}u>u + uI_{}u>]
= d[du + ud]
= ddu + dud
√(1/6)I_{}2uud  udu  duu> = √(1/6)(2ddu + dud + udd)
and,
I_{}1/2,1/2> = √[1/2(1/2 + 1)  (1/2)(1/2 = 1)]1/2,1/2>
= 11/2,1/2>
Therefore,
1/2,1/2> = √(1/6)(2ddu + dud + udd)
and so on.
The complete list for the '8_{MS}' is:
1/2,1/2> = uud: √(1/6)(2uud  udu  duu)
1/2,1/2> = ddu: √(1/6)(2ddu + dud + udd)
uus: √(1/6)(2uus  usu  suu)
0,0: √(1/12)(2uds  usd  dsu + 2dus  sud  sdu)
0,0: (1/2)(usd + sud  sdu  dsu)
dds: √(1/6)(2dds  dsd  sdd)
ssu: √(1/6)(sus + uss  2ssu)
ssd: √(1/6)(sds + dss  2ssd)
These are all symmetric for q_{1} <> q_{2}.
This looks like:
Mixed Symmetry Octet, 8_{MA}

I_{}udu  duu>
I_{}udu> = I_{}u>du + uI_{}du>
= ddu + u[I_{}d>u + dI_{}u>]
= ddu + udd
I_{}duu> = I_{}d>uu + dI_{}uu>
= d[I_{}u>u + uI_{}u>]
= d[du + ud]
= ddu + dud
I_{}udu  duu> = udd  dud
and so on.
The complete list for the '8_{MA}' is:
1/2,1/2> = ddu: √(1/2)(udd  dud)
1/2,1/2> = uud: √(1/2)(udu  duu)
uus: √(1/2)(usu  suu)
0,0: (1/2)(usd + dsu  sdu  sud)
0,0: √(1/12)(2uds  dsu  sud  2dus  sdu + usd)
dds: √(1/2)(dsd  sdd)
ssu: √(1/2)(uss  sus)
ssd: √(1/2)(dss  sds)
These are all antisymmetric for q_{1} <> q_{2}.
Totally Antisymmetric Singlet, 1_{A}

The singlet state is:
ψ = √(1/6)(uds  usd + dsu  dus + sud  sdu)
As before, this can be verified by checking that:
I_{±}ψ> = 0 and U_{±}ψ> = 0 and V_{±}ψ> = 0
This state is antisymmetric for all interchanges:
This looks like:
Again, we can use the raising and lowering operators
to move around the 8. For example:
If we put the '10_{S}', '8_{MS}', '8_{MA}' and '1_{A}' together we get:
The 3 ⊗ 3 ⊗ 3 Approach

Constructing Baryon states is a fairly tedious
process. We can also obtain the same results if
we use:
3 ⊗ 3 ⊗ 3 = 3 ⊗ (3 ⊗ 3)
_
Now 3 ⊗ 3 = 6 ⊕ 3. This looks like:
So we can write:
_
3 ⊗ 3 ⊗ 3 = 3 ⊗ (6 ⊕ 3)
_
= 3 ⊗ 6 ⊕ 3 ⊗ 3
Now 3 ⊗ 6 looks like:
We can treat each side of the '6' as a '3' as
we did before and use the U_{±} and V_{±} operators
to get the different states. For example,
consider:
   
 dd   d 
 ds + sd _{ }⊗  
 ss   s 
   
Which gives the combinations ddd, dds, dsd,
sdd, ssd and sss.
Likewise,
 
   d 
 ds  sd _{ }⊗  
   s 
 
We can then use U_{}ddd> to generate the '4'
and, using orthogonality, the corresponding '2'
for that particular side. If we do this for
all 3 sides of the '6' and then add them together
we have:
3 ⊗ 2 ⊕ 3 ⊗ 2 ⊕ 3 ⊗ 2 = 3 ⊗ (2 ⊕ 2 ⊕ 2)
= 3 ⊗ 6
So we have performed the tensor product giving a
total of 18 states.
_
3 ⊗ 3 = 8 ⊕ 1 yields the second mixed symmetry
'2' ad the singlet.
This gives:
dsd  ssd
and so on.
Combining Spin

Spin is not the same as isospin. However, both
share exactly the same mathematics. Therefore,
we can take any of the isospin wavefunctions
derived previously and create a corresponding spin
wavefunction. For the mesons we have,
Spin 1 (symmetric) Triplet:
1,1> = ↑↑
1,0> = √(1/2)(↑↓ + ↓↑)
1,1> = ↓↓
Spin 0 (antisymmetric) Singlet:
0,0> = √(1/2)(↑↓  ↓↑)
This is often written as:
1/2 ⊗ 1/2 = 1 ⊕ 0
Isospin 1 (symmetric) Triplet:
_
1,1> = ud
_ _
1,0> = √(1/2)(uu  dd)
_
1,1> = du
Isospin 0 (antisymmetric) Singlet:
_ _
0,0> = √(1/2)(uu + dd)
ud> ⊗ ↑↑
_
Note: The reason the signs are flipped is that d
isospin transforms in the opposite way under
transformations.
We can now create a composite wavefunction as:
_
π^{+} = ud ⊗ √(1/2)(↑↓  ↓↑) with S = 0
_ _
= √(1/2)(u↑d↓  u↓d↑)
For the baryons we have:
For the '10_{S}' with S = 3/2 ∴ J = 3/2:
3/2,3/2> = ↑↑↑: ↑↑↑
3/2,1/2> = ↑↑↓: √(1/3)(↑↑↓ + ↑↓↑ + ↓↑↑)
3/2,1/2> = ↓↓↑: √(1/3)(↓↓↑ + ↓↑↓ + ↑↓↓)
3/2,3/2> = ↓↓↓: ↓↓↓
For the '8_{MA}' S = 1/2 ∴ J = 1/2:
1/2,1/2> = ↓↓↑: √(1/2)(↑↓↓  ↓↑↓)
1/2,1/2> = ↑↑↓: √(1/2)(↑↓↑  ↓↑↑)
For the '8_{MS}'S = 1/2 ∴ J = 1/2:
1/2,1/2> = ↑↑↓: √(1/6)(2↑↑↓  ↑↓↑  ↓↑↑)
1/2,1/2> = ↓↓↑: √(1/6)(2↓↓↑ + ↓↑↓ + ↑↓↓)
We can now create a composite wavefunction as:
ψ_{Flavor} ⊗ ψ_{Spin}.
For example,
ψ_{Flavor} ⊗ ψ_{Spin} = √(1/6)(2uud  udu  duu)√(1/6)(2↑↑↓  ↑↓↑  ↓↑↑)
Color Symmetry

The shrewd observer will note that there is a
problem when we consider flavor states such as
uuu> and ddd> in the Baron decuplet because
they appear to violate the Pauli Exclusion
Principle that 2 or more identical particles
cannot occupy the same quantum state. In other
words, it is impossible to construct a totally
antisymmetric wavefunction function for the state.
This led researchers to believe that an additional
quantum number had to exist. Enter color!
Exactly analagous to labelling u, d, s flavor
states by I_{3} and Y we can assign color quantum
numbers I_{3}^{C} and Y^{C} to color states.
     
 1   0   0 
r =  0  g =  1  b =  0 
 0   0   1 
     
_
The 3 and the 3 then look like:
Flavor symmetry is not exact because the quark
masses are not identical. On the other hand
color symmetry is is exact.
Only combinations of particles that have a COLOR
SINGLET state can be observed in nature. Color
singlets are colourless combinations that obey:
1. Their quantum numbers, I_{3}^{C} = Y^{C} = 0
2. They are invariant under SU(3) colour
transformations.
3. The raising and lowering operatos I_{±}, U_{±} and
V_{±} all yield 0.
The implication of this is that we can never observe
a free quark (which would carry a color charge).
Consider:
_
3 ⊗ 3 has the completely symmetric singlet:
_ _ _
√(1/3)(rr + gg + bb).
Proof:
_ _ _ _ _
I_{+}rr + gg + ss> = rg + rg = 0
_
Therefore, qq objects are colorless and can
exist as free particles (mesons).
Likewise, 3 ⊗ 3 ⊗ 3 has completely antisymmetric
singlet:
√(1/6)(rgb  grb + gbr  bgr + brg  rbg).
Proof:
I_{+}rgb  grb + gbr  bgr + brg  rbg>
= rrb  rbr + rbr  rrb + brr  brr
= 0
Therefore, qqq objects are colorless and can
exist as free particles (baryons).
However, the 3 ⊗ 3 = 6 ⊗ 3 has no color singlet
therefore the qq combination does not appear in
nature.
Quark Properties

_ _ _
u d s u d s
     
Baryon Number, B 1/3 1/3 1/3 1/3 1/3 1/3
Strangeness, S 0 0 1 0 0 1
Hypercharge, Y 1/3 1/3 2/3 1/3 1/3 2/3
Charge, Q 2/3 1/3 1/3 2/3 1/3 1/3
I_{3} 1/2 1/2 0 1/2 1/2 0
Spin, J ±1/2 ±1/2 ±1/2 ±1/2 ±1/2 ±1/2
Intrinsic
Parity, P +1 +1 +1 1 1 1
These relationships are encapsulated on the following
formulas:
Y = S + B
= 2(Q  I_{3})
The Total Wavefunction

ψ_{Total} = ψ_{Spin} ⊗ ψ_{Spacial} ⊗ ψ_{Flavor} ⊗ ψ_{Color}
(ψ_{Spacial} aka ψ_{Orbital})
There are 2 types of mesons, PSEUDOSCALAR and
VECTOR. Pseudoscalar mesons have antialigned
spins resulting in spin 0. Vector mesons have
aligned spins resulting in spin 1. The list is:
State Pseudoscalar Vector
  
1,1> π^{+} ρ^{+}
1,0> π^{0} ρ^{0}
1,1> π^{} ρ^{}
The Mesons are bosons so it is required that the
total wavefunction is symmetric.
For the ground state L = 0.
_
P(ud) = P_{u}P_{antid}(1)^{0} = 1 (see note on Parity)
Therefore, the spatial wavefunction is antisymmetric.
Now,
_ _ _
ψ_{Color} = √(1/3)(rr + gg + bb)
is totally symmetric. Therefore, ψ_{Spacial} ⊗ ψ_{Color}
is totally antisymmetric. This means that the
combination ψ_{Flavor} ⊗ ψ_{Spin} needs to be be totally
antisymmetric for the particles to be bosons.
Consider:
ψ_{Flavor} ⊗ ψ_{Spin} = ud ⊗ √(1/2)(↑↓  ↓↑)
= √(1/2)(u↑d↓  u↓d↑)
= √(1/2)(u↓d↑  u↑d↓)
This is antisymmetric under the interchange of any 2
quarks.
ψ_{Total} is now symmetric. This is, in fact, the
spin 0 wavefunction for the π^{+} pseudoscalar meson.
The ρ^{+} meson is considered to be an excited state of
the π^{+}.
For the excited state L = 1 we get.
_
P(ud) = P_{u}P_{antid}(1)^{1} = 1
_
ρ^{+} = ud ⊗ ↑↓ with S = 1
= u↑d↓
So now ψ_{Spatial}, ψ_{Flavor}, ψ_{Spin} and ψ_{Color} are all
symmetric so ψ_{Total} is symmetric;
The Baryons are fermions so it is required that
the total wavefunction is antisymmetric (Pauli
Exclusion Principle).
P(qqq) = P_{q} P_{q} P_{q}(1)^{L} = 1 for L = 0
Therefore, the spatial wavefunction is symmetric.
Now,
ψ_{Color} = √(1/6)(rgb  grb + gbr  bgr + brg  rbg)
is totally antisymmetric. Therefore, ψ_{Spacial} ⊗ ψ_{Color}
is totally antisymmetric. This means that the
combination ψ_{Flavor} ⊗ ψ_{Spin} needs to be be totally
symmetric for the particles to be fermions. This
can be achieved by constructing the normalized
linear combination:
ψ_{Flavor} ⊗ ψ_{Spin} = √(1/2)ψ_{MS}χ_{MS} + √(1/2)ψ_{MA}χ_{MA}
Where ψ_{MS} is from the 8_{MS} and ψ_{MA} is the corresponding
state from the 8_{MA}.
Consider the state ψ = uud> constructed as follows:
ψ_{Flavor} ⊗ ψ_{Spin} = √(1/2)[√(1/6)(2uud  udu  duu)
⊗ √(1/6)(2↑↑↓  ↑↓↑  ↓↑↑)]
+ √(1/2)[√(1/2)(udu  duu)
⊗ √(1/2)(↑↓↑  ↓↑↑)]
= √(1/18)(2u↑u↑d↓  u↑u↓d↑  u↓u↑d↑
+ 2u↑d↓u↑  u↓d↑u↑  u↑d↑u↓
+ 2d↓u↑u↑  d↑u↓u↑  d↑u↑u↓)
This is symmetric under the interchange of any 2
quarks.
ψ_{Total} is now antisymmetric. This is, in fact, the
spin up wavefunction for the proton, p↑>.
The Particle Zoo

_
In summary, the 3 ⊗ 3 gives the following particles:
The 3 ⊗ 3 ⊗ 3 gives the following particles:
Triality, τ

Triality is defined as:
τ = (n  m) mod 3 (the remainder of (n  m)/3)
D R τ
  
(0,0) 0
(1,0) 3 1
_
(0,1) 3 2
(1,1) 8 0
(2,0) 6 2
_
(0,2) 6 1
(3,0) 10 0
Irreps with triality 0 correspond to observable
particles. Those with triality ≠ 0 correspond
to particles that have not been observed in nature.