Redshift Academy


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
------------------------------------------------------

Multi-Particle 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 multi-particle 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_ijkJ_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_ijkJ_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> = H¹|ψ_R1> ⊗ |ψ_R2> + |ψ_R1> ⊗ H¹|ψ_R2> 

and,

H^2'|ψ_R1 ⊗ ψ_R2> = H²|ψ_R1> ⊗ |ψ_R2> + |ψ_R1> ⊗ H²|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*Qⁱ to be invariant under a transformation, 
U.

Qⁱ -> Q'ⁱ = UQⁱ

Q^i* -> Q'^i* = U^*Q^i*

Q^i†Qⁱ -> Q'^i†Q'ⁱ = Q^i†U^†UQⁱ

Therefore, the requirement is:  U^†U = I

We write (Qⁱ)^* as Q_i and get:

Qⁱ -> Q'ⁱ = Uⁱ_jQ^j

Q_i -> Q'_i = Q_j(U^†)^j_i

QⁱQ_i -> Q'ⁱQ'_i = Q_j(U^†)^j_iUⁱ_kQ^k

              = Q_jδ^j_kQ^k

              = Q

Now consider the following mixed tensor:

Q^ij_k -> Q'^ij_k = Uⁱ_lU^j_m(U^†)ⁿ_kQ^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_jQ^ij_k -> δ^k_jQ'^ij_k = δ^k_jUⁱ_lU^j_m(U^†)ⁿ_jQ^lm_n

                  = Qⁱ

Therefore, δ^k_jQ^ij_k represents the trace of Q^ij_k.

δ^k_j is invariant because:

δ^k_j = δⁱ_lU^l_i(U^†)^k_j

Such that,

δ^k_jQ^ij_k = δⁱ_lU^l_i(U^†)^k_jQ^ij_k

       = Qⁱ as before.

Note that δ is the Kroneker Delta which is nothing 
more than the unit matrix in N dimensions.

The Levi-Civita Tensor
----------------------

There 2 other invariant tensors in SU(3) - the 
completely antisymmetric tensors ε^ijk and ε_ijk.
These are the Levi-Civita symbols in tensor form.
The L-C 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_1αM_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 = M₁₁M₂₂M₃₃ - M₁₁M₂₃M₃₂ + M₁₂M₂₃M₃₁ - M₁₂M₂₁M₃₃

        + M₁₃M₂₁M₃₂ - M₁₃M₂₂M₃₁

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'))²det(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 = √|g|ε_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. 

Levi-Civita Symbol and Kronecker Delta Relationship
---------------------------------------------------

In 3D, the Levi-Civita symbol is related to the 
Kronecker delta in the following manner:

     | δⁱ_k δⁱ_m δⁱ_n |
ε^ijlε_kmn = | δ^j_k δ^j_m δ^j_n |
     | δ^l_k δ^l_m δ^l_n |

or,

ε^ijlε_kmn = δⁱ_k(δ^j_mδ^l_n - δ^j_nδ^l_m) - δⁱ_m(δ^j_kδ^l_n - δ^j_nδ^l_k) 

           + δⁱ_n(δ^j_kδ^l_m - δ^j_mδ^l_k)

For the special case:

ε^ijlε_lmn = δⁱ_mδ^j_n - δ^j_nδⁱ_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 anti-commutator 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^jiQ^ij

       = -Q^jiQ^ji

       = -Q^ijQ^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:

   ~
   Qⁱ_j       =    Qⁱ_j     -       (1/3)δⁱ_jQ^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:

~
Qⁱ_j = Qⁱ_j - αδⁱ_jQ^k_k 

If we contract i with j we can write this as:

~
Qⁱ_i = Tr(Qⁱ_j) = 0 = Qⁱ_i - αδⁱ_iQ^k_k

But δⁱ_i = 3.  Therefore,

0 = Qⁱ_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(δⁱ_kQ^j_l + δ^j_kQⁱ_l + δⁱ_lQ^j_k + δ^j_iQⁱ_k)

where Q^j_l = Q^j_lQⁿ_n etc.

Contract k with i:

~
Q^ij_il = 0 = Q^ij_il - β(δⁱ_iQ^j_l + δ^j_iQⁱ_l + δⁱ_lQ^j_i + δ^j_iQⁱ_i)

          = Q^ij_il - β(δⁱ_iQ^j_l + δ^j_iQⁱ_l + δⁱ_lQ^j_i)

          = Q^ij_il - β(δⁱ_i + δ^j_i + δⁱ_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(δⁱ_kQ^j + δ^j_kQⁱ)

Contract k with i:

S^ij_i = 0 = Q^ij_i - β(δⁱ_kQ^j + δ^j_kQⁱ)

         = Q^ij_i - β(δⁱ_iQ^j + δ^j_kQⁱ)

         = 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:

Qⁱ_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_jkT^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 |
 -       -

Qⁱ_j (contractable):

 -       -
| ¹₁ ¹₂ ¹₃ |  
| ²₁ ²₂ ²₃ |  
| ³₁ ³₂ ³₃ |  
 -       -
     ₃
Tr = ΣS^k_k = Q¹₁ + Q²₂ + Q³₃ = 0
    ^k=1

Now since the RHS are just scalars we can write:

Q³₃ = -(Q¹₁ + Q²₂)

So Q³₃ is not independent and we can ignore it.
As a consequence D(1,1) = 9 - 1 = 8

 -       -
| ¹₁ ¹₂ ¹₃ |  
| ²₁ ²₂ ²₃ |  
| ³₁ ³₂ ³₃ |  
 -       -

Q^{ij}_k (contractable):

 -          -    -          -    -          -
| ¹¹₁   |  | ¹¹₂   |  | ¹¹₃   |
| ²¹₁ ²²₁  |  | ²¹₂ ²²₂  |  | ²¹₃ ²²₃  |
| ³¹₁ ³²₁ ³³₁ |  | ³¹₂ ³²₂ ³³₂ |  | ³¹₃ ³²₃ ³³₃ |
 -          -    -          -    -          -

Contract i and j with k and sum to get the 2 
traces:

     ₃
Tr = ΣS^ik_k = Q¹¹₁ + Q²²₂ + Q³³₃
    ^k=1

   = Q¹ + Q² + Q³

and,

     ₃
Tr = ΣS^kj_k = Q¹¹₁ + Q²²₂ + Q³³₃
    ^k=1

   = Q¹ + Q² + Q³

 -          -    -          -    -          -
| ¹¹₁   |  | ¹¹₂   |  | ¹¹₃   |
| ²¹₁ ²²₁  |  | ²¹₂ ²²₂  |  | ²¹₃ ²²₃  |
| ³¹₁ ³²₁ ³³₁ |  | ³¹₂ ³²₂ ³³₂ |  | ³¹₃ ³²₃ ³³₃ |
 -          -    -          -    -          -

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.
 
Q³ = -(Q¹ + Q²)

As this requires Q¹ + Q² + Q³ = 0 which can only 
be satisfied if Q¹ = Q² = Q³ = 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

Qⁱ = 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:

Qⁱ = 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 ≡ N² - 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.

Qⁱ 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   (I₃ = 3/2)
  ddd   ddu   duu   uuu   

     322   321   311      (I₃ = 1)
     sdd   sud   suu

        332   331         (I₃ = 1/2)
        ssd   ssu
  
           333            (I₃ = 0)
           sss

Clebsch-Gordan Decomposition for SU(3)
--------------------------------------

The Clebsch-Gordan 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 Levi-Civita tensor to symmetrize the
     antisymmetric parts, i.e.

     Q^k = ε_ijkQ^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 C-G decomposition.

Clebsch-Gordan Series
---------------------

3 ⊗ 3
------

uⁱ ⊗ v^j = Q^{ij} + Q^[ij]

Q^{ij} = (1/2)(uⁱv^j + u^jvⁱ) 

Q^[ij] = (1/2)(uⁱv^j - u^jvⁱ)

We can write the 2nd terms as (1/2)ε^ijl(ε_lmnu^mvⁿ)

Proof:

(uⁱv^j - u^jvⁱ) = (δⁱ_mδ^j_n - δ^j_nδⁱ_m)u^mvⁿ

              = ε^ijlε_lmnu^mvⁿ

The antisymmetric term is automatically traceless
since uⁱvⁱ = -uⁱvⁱ.  Therefore,

Q^ij = (1/2((uⁱv^j + u^jvⁱ) + (1/2)ε^ijl(ε_lmnu^mvⁿ)

  = (1/2)(uⁱv^j + u^jvⁱ) + (1/2)ε^ijl(uv_l)
                                                   
The first term is the '6' (2,0).  The second 
             _
term is the '3' (0,1). 
    _
3 ⊗ 3
------

uⁱ ⊗ v_j = Qⁱ_j
                 
~
Qⁱ_j = uⁱv_j - βδⁱ_ju^kv_k

Contract i with j and demand that Tr(Sⁱ_j) = 0:

~
Qⁱ_i = 0 = uⁱv_i - βδⁱ_iu^kv_k

        = uⁱv_i - β3u^kv_v

∴ β = 1/3

Therefore,

uⁱv_j = (uⁱv_j - (1/3)δⁱ_ju^kv_k) + (1/3)δⁱ_ju^kv_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_lδ_k^l

  -> (1 + iθT_a(R))_i^k(1 - iθT_a(R))_j^lδ_k^l

  -> (1 + iθT_a(R))_i^k(1 - iθT_a(R))_l^jδ_k^l

  ->  δ_i^j + O(θ²)
                    _
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 non-singlet states 
must number N² - 1.  Thus,   
    _
R ⊗ R = 1 ⊕ (N² - 1)

For SU(3) N² - 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
------

uⁱ ⊗ v^j_k = Q^{ij}_k + Q^[ij]_k

~
Q^{ij}_k = (1/2)(uⁱv^j_k + u^jvⁱ_k - α(δⁱ_ku^lv^j_l + δ^j_ku^lvⁱ_l))

~
Q^[ij]_k = (1/2)(uⁱv^j_k - u^jvⁱ_k - β(δⁱ_ku^lv^j_l - δ^j_ku^lvⁱ_l))

         ~
Consider Q^{ij}_k.  Contract i with k to get:

~
Q^{ij}_i = 0 = (1/2)(uⁱv^j_i + u^jvⁱ_i - α(δⁱ_iu^lv^j_l + δ^j_ku^lvⁱ_l))

      = (1/2)(uⁱv^j_i - α(δⁱ_i + δ^j_k)u^lv^j_l)

      = (1/2)(uⁱv^j_i - α(3 + 1)u^lv^j_l) since δⁱ_i = 3

Therefore, α = 1/4.  Rewriting we get:

~
Q^{ij}_k = (1/2)(uⁱv^j_k + u^jvⁱ_k) - (1/8)δⁱ_ku^lv^j_l - (1/8)δ^j_ku^lvⁱ_l

             ~
Now consider Q^[ij]_k.  Contract i with k to get:

Q^[ij]_i = 0 = (1/2)(uⁱv^j_i - β(δⁱ_iu^lv^j_l - δ^j_ku^lvⁱ_l))

      = (1/2)(uⁱv^j_i - β(δⁱ_i - δ^j_k)u^lv^j_l)

      = (1/2)(uⁱv^j_i - β(3 - 1)u^lv^j_l)

Therefore, β = 1/2.  Rewriting we get:

~
Q^[ij]_k = (1/2)(uⁱv^j_k - u^jvⁱ_k) - (1/4)δⁱ_ku^lv^j_l + (1/4)δ^j_ku^lvⁱ_l)

So that,

~       ~
Q^{ij}_k + Q^[ij]_k = (1/2)(uⁱv^j_k + u^jvⁱ_k) - (1/8)δⁱ_ku^lv^j_l - (1/8)δ^j_ku^lvⁱ_l

                 + (1/2)(uⁱv^j_k - u^jvⁱ_k) - (1/4)δⁱ_ku^lv^j_l + (1/4)δ^j_ku^lvⁱ_l)

               = uⁱv^j_k - (3/8))δⁱ_ku^lv^j_l + (1/8)δ^j_ku^lvⁱ_l

Need to add (3/8)δⁱ_ku^lv^j_l - (1/8)δ^j_ku^lvⁱ_l to make 
both sides balance.  Therefore:

~
Q^ij_k = (1/2)(uⁱv^j_k + u^jvⁱ_k) - (1/8)δⁱ_ku^lv^j_l - (1/8)δ^j_ku^lvⁱ_l

      + (1/2)(uⁱv^j_k - u^jvⁱ_k) - (1/4)δⁱ_ku^lv^j_l + (1/4)δ^j_ku^lvⁱ_l)

      + (3/8)δⁱ_ku^lv^j_l - (1/8)δ^j_ku^lvⁱ_l

The blue term is:

(1/2)ε^ijlε_lmnu^mvⁿ_k

The red term is:

(1/4)δⁱ_ku^lv^j_l + (1/4)δ^j_ku^lvⁱ_l) = (1/4)uⁱv^j_k + (1/4)u^jvⁱ_k

Which can be written as:

(1/4)ε^ijlε_kmnu^mvⁿ_l

Proof:

We use the identity:

| δⁱ_k δⁱ_m δⁱ_n |
| δ^j_k δ^j_m δ^j_n | = ε^ijlε_kmn
| δ^l_k δ^l_m δ^l_n |

(1/4)ε^ijlε_kmnu^mvⁿ_l = 

(1/4)(δⁱ_k(δ^j_mδ^l_n - δ^j_nδ^l_m)u^mvⁿ_l 

      - δⁱ_m(δ^j_kδ^l_n - δ^j_nδ^l_k)u^mvⁿ_l = δⁱ_m(u^mv^j_k - u^mv^j_k) 

      + δⁱ_n(δ^j_kδ^l_m - δ^j_mδ^l_k)u^mvⁿ_l = δⁱ_n(u^mvⁿ_k - u^jvⁿ_k))

The 2nd and 3rd terms are 0 so we get:

(1/4)ε^ijlε_kmnu^mvⁿ_l = (1/4)(δⁱ_k(δ^j_mδ^l_n - δ^j_nδ^l_m)u^mvⁿ_l

                  = (1/4)δⁱ_k(u^jv^l_l - u^lv^j_l) 

                  = (1/4)(u^jvⁱ_k - uⁱv^j_k)

                  = -(1/4)uⁱv^j_k + (1/4)u^jvⁱ_k 

Finally, we get:

~
Q^ij_k = (1/2)(uⁱv^j_k + u^jvⁱ_k) - (1/8)δⁱ_ku^lv^j_l - (1/8)δ^j_ku^lvⁱ_l

      + (1/4)(ε^ijl(ε_lmnu^mvⁿ_k + ε_kmnu^mvⁿ_l))

      + (1/8)(3δⁱ_ku^lv^j_l - δ^j_ku^lvⁱ_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
------

uⁱ ⊗ v^jk = Q^{ijk} + Q^[ij]k + Q^[ik]j

Q^{ijk} = (uⁱv^jk + u^jv^ki + u^kv^ij)

Q^[ij]k = (uⁱv^jk - u^jv^ik) 

Q^[ik]j = (uⁱv^kj - u^kv^ij)

uⁱ ⊗ v^jk = Q^{ijk} + (uⁱv^jk - u^jv^ik) 

           + (uⁱv^kj - u^kv^ij)

As before, we can write:

(uⁱv^jk - u^jv^ik) = ε^ijl(ε_lmnu^mv^nk)

and,

(uⁱv^kj - u^kv^ij) = ε^ikl(ε_lmnu^mv^nj)

uⁱ ⊗ v^jk = (uⁱv^jk + u^jv^ki + u^kv^ij) + ε^ijl(ε_lmnu^mv^nk)

           + ε^ikl(ε_lmnu^mv^nj)

         = (uⁱv^jk + u^jv^ki + u^kv^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 non-separability 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 left-justified 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):

Qⁱ = (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):

Qⁱ = (1,0):

 ---
|   | 
 ---

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

 -------     
|   |   |  
 -------  

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

 -----------     
|   |   |   |  
 -----------    

Q = (0,0):

 ---
|   |
|---|
|   |  
|---|
|   |
 ---

Qⁱ_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)    Qⁱ     (1/2,1/2√3) 
 -
 3    (0,1)    Q_i     (1/2,-1/2√3)

 8    (1,1)    Qⁱ_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+1|N+2|   
    |---|---|---|    
    |N-1| N |N+1|  
    |---|---|---    
    |N-2|N-1|       
     -------

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 
--------------------



Tⁱ ⊗ T^j = T^ij

(1,0) ⊗ (1,0) = (2,0) ⊕ (0,1)
             _
3 ⊗ 3 = 6 ⊕ 3
    _
3 ⊗ 3 Decomposition
--------------------



Tⁱ ⊗ T_j = Tⁱ_j

(1,0) ⊗ (0,1) = (1,1) ⊕ (0,0)
    _
3 ⊗ 3 = 8 ⊕ 1

3 ⊗ 3 ⊗ 3 Decomposition
------------------------



Tⁱ ⊗ 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 anti-symmetric 
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 |  = Γ⁰
 ---|---         |---|---
| d |            | s |
 ---              ---

 -------           -------
| u | s |  = Σ⁰   | u | s |  = Ξ⁰
 ---|---          |---|---
| 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, ψ₁ and ψ₂ are 
orthogonal if:

<ψ₁|ψ₂> = 0 

The rules are:

<u|d> = <d|u> = 0 etc.
  _  _               
<uu|uu> = 1 etc.   
  _  _
<uu|dd> = 0 etc.

<ud|du> = 0 etc.

States also need to be normalized such that:

<ψ₁|ψ₁> = 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
CLEBSCH-GORDAN coefficients in the calculations.  
I_± is then written as:

I_±|I,I₃> = √[I(I + 1) - I₃(I₃ ± 1)]|I,I₃ ± 1>

         = C_±|I,I₃ ± 1>

Then, 

C_±|I,I₃ ± 1> = I_±|ψ>

             = ψ_±

Where ψ is the combination of quarks being raised 
or lowered.

|I,I₃> 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 isospin|eigenvalue of z component>

For example |1,0> corresponds to the combination 
of 2 isospins |ud + du>.

The Clebsch-Gordan 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)> = C²(1 - 0 - 0 + 1) = 2C²

∴ C = √(1/2)

Therefore, the correct state is:
        _    _
√(1/2)(uu - dd)

In essence the C-G 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 Clebsch-Gordan 
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 anti-quarks 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 anti-quarks and quarks transform in the 
same way.  To see this we start with defining 
the anti-quark 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 anti-quark 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/anti-quark I₃'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>

        = √2|1,0>

Therefore,
                _    _
|1,0> = √(1/2)(uu - dd)

The second state needs to be orthogonal to this
state.  This state is:
             _    _
ψ₂ = √(1/2)(uu + dd)

Check:
            _    _  _    _
<ψ₁|ψ₂> = <uu - dd|uu + dd> 
            _  _      _  _      _  _      _  _
        = <uu|uu> + <uu|dd> - <dd|uu> - <dd|dd>

        = 1 - 0 - 0 - 1

        = 0

The singlet state is ALWAYS a dead-end 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 ψ₂.
            _
The Mesons, 3 ⊗ 3
------------------

Quark isospins can combine to give a total of 3/2 or 1/2 
This can be achieved as follows:
          _
I₃ 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 ψ₁ = √(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 ψ₁.  
       _      _      _      _
ψ₂ = |uu> - |ss> + |dd> - |ss>
       _      _       _
   = |uu> + |dd> - 2|ss>

Orthogonality check:
            _    _     _  _    _
<ψ₂|ψ₁> = <uu + dd - 2ss|uu - dd>
            _  _      _  _      _  _      _  _      
        = <uu|uu> - <uu|dd> + <dd|uu> - <dd|dd> 
               _  _       _  _
          - 2<ss|uu> + 2<ss|dd>

        = 1 - 0 + 0 - 1 - 0 + 0

        = 0

Therefore, ψ₁ and ψ₂ are orthogonal.
  _    _     _  _    _     _
<uu + dd - 2ss|uu + dd - 2ss>

     = 1 + 0 - 0 + 0 + 1 + 0 - 0 - 0 + 4 

     = 6

The normalized state is:
    _    _     _     _    _     _
<C(uu + dd - 2ss)|C(uu + dd - 2ss)> = 6C²

Therefore,
             _    _     _
ψ₂ = √(1/6)(uu + dd - 2ss)

Likewise, the final state can be obtained by 
requiring it to be orthonormal to both ψ₁ and ψ₂.
             _    _    _
ψ₃ = √(1/3)(uu + dd + ss)

This is the singlet state.  Again, the singlet state 
is ALWAYS a dead-end 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 ψ₃.

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>

             = √3|3/2,-1/2>

Therefore,

√3|3/2,-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>

             = 2|3/2,1/2>

Therefore,

2|3/2,1/2> = 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,

<uus|uus> = 

       <(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>

             = 1|1/2,-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 q₁ <-> q₂.

This looks like:



Mixed Symmetry Octet, 8_MA 
-------------------------

I_-|udu - duu> 

I_-|udu> = I_-|u>du + uI_-|du>

        = ddu + u[I_-|d>u + d|I_-|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 q₁ <-> q₂.

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 I₃ and Y we can assign color quantum 
numbers I₃^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, I₃^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

I₃                 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 - I₃)

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>        π⁰           ρ⁰
  
       |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_uP_anti-d(-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_uP_anti-d(-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 wave-function is anti-symmetric (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_MSχ_MS + √(1/2)ψ_MAχ_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.