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Special Unitary Groups and the Standard Model  Part 2
Special Unitary Groups and the Standard Model  Part 3
Special Unitary Groups and the Standard Model  Part 1

The Fundamental (Defining) Representation

The special unitary group of degree N, SU(N), is
the LIE GROUP where the elements are N x N unitary
matrices with determinant = +1. The group operation
is that of matrix multiplication. Lie groups are
best studied in terms of their Lie algebra involving
the generators that create the group elements.
Consider a general complex transformation in 2
dimensions, x' = Ux which, in matrix form, reads:
   
 x'  = U x 
 y'   y 
   
     
 x'  =  a b  x 
 y'   c d  y 
     
This has 2N^{2} = 8 free parameters. To satisfy
the condition x^{2} + y^{2} = x'^{2} + y'^{2} requires
that the matrix be unitary. Thus, U^{†}U = I.
       
 a* c*  a b  =  aa* + cc* a*b + c*d  =  1 0 
 b* d*  c d   b*a + d*c b*b + d*d   0 1 
       
Thus,
aa* + cc* = 1
a*b + c*d = 0
b*a + d*c = 0
b*b + d*d = 1
These four conditions means that the original 8
free parameters are reduced to 4. In other words,
if the matrix is unitary it is characterized by
N^{2} = 4 parameters. If, in addition to these
conditions, we require that the determinant of the
matrix = 1, the matrix must have the form:
     
 x'  =  a b  x 
 y'   b* a*  y 
     
It is easy to see that U^{†}U = I.
       
 a* b  a b  =  aa* + bb* a*b  ba*  =  1 0 
 b* a  b* a*   b*a  ab* b*b + aa*   0 1 
       
So,
aa* + bb* = 1 ... 1.
a*b  ba* = 0 ... 2.
b*a  ab* = 0 ... 3.
b*b + aa* = 1 ... 4.
1. and 4. are the same so now there are only 3
free parameters which we can generalize to N^{2}  1.
This formula holds for all unitary groups.
The implication of this is that there are only
3 matrices that we can be constructed that meet
these requirements. These are the 3 SKEW
HERMITIAN (antiHermitian) Pauli matrices iσ_{1},
iσ_{2} and iσ_{3}.
Aside: A skew Hermitian (antiHermitian) matrix
is one that has the property that A^{†} = A. This
is accomplished by multiplying a Hermitian matrix
by i. Thus, for example:
 
iσ_{1} =  0 i 
^{ }  i 0 
 
   
 0 i ^{†} =  0 i  = iσ_{1}
 i 0 ^{ }  i 0 
   
 
 0 i  a = 0, a* = 0, b = i, b* = i
 i 0 
 
 
 0 1  a = 0, a* = 0, b = 1, b* = 1
 1 0 
 
 
 i 0  a = i, a* = i, b = 0, b* = 0
 0 i 
 
If we break U down into its real and imaginary
components we get:
 
U =  a_{R} + ia_{I} b_{R} + ib_{I} 
 b_{R} + ib_{I} a_{R}  ia_{I} 
 
_{ }   _{ }   _{ }   _{ }  
= a_{R} 1 0  + ia_{I} 1 0  + b_{R} 0 1  + ib_{I} 0 1 
_{ }  0 1 _{ }  0 1 _{ }  1 0 _{ }  1 0 
_{ }   _{ }   _{ }   _{ }  
_{ }   _{ }   _{ }   _{ }  
= a_{R} 1 0  + ia_{I} 1 0  + ib_{R} 0 i  + ib_{I} 0 1 
_{ }  0 1 _{ }  0 1 _{ }  i 0 _{ }  1 0 
_{ }   _{ }   _{ }   _{ }  
= a_{R}I + a_{I}iσ_{3} + b_{R}iσ_{2} + b_{I}iσ_{1}
So any element can be represented as a linear
combination of I and the Pauli matrices.
It is worthwhile pointing out that this is just
the Lie algebra of SU(2). We are in a vector
space where σ_{0}, iσ_{1}, iσ_{2} and iσ_{3} are the bases
which produce, through exponentiation, the special
unitary matrices that are the elements of the
group SU(2).
In terms of infnitesimal generators we can write:
U = exp(iθ_{i}σ_{i}/2)
The generators satisfy the following commutation
rule:
[T_{i},T_{j}] = iε^{ijk}T_{k}
Where, T_{i} = σ_{i}/2 and ε^{ijk} is the LEVICIVITA symbol.
Hermitian Generators

The requirement U^{†}U = I requires that the generators
be Hermitian. To see this we write:
U = exp(iθ_{i}σ_{i}/2) = I + iθT (dropping the factor of 2)
and,
U^{†} = I  iθT^{†}
Thus,
UU^{†} = (I + iθT)(I  iθT^{†})  i^{2}θ^{2}TT^{†} = I
= I + iθ(T  T^{†}) = I
This implies that (T  T^{†}) = 0. For this to be true,
T has to be Hermitian.
For example,
   
σ_{2}σ_{2}^{†} =  0 i  0 i  = I
^{ }  i 0  i 0 
   
σ_{2}  σ_{2}^{†} = 0
What about the determinant of T? We can write for U:
     
 1 0  +  iθT_{11} iθT_{12}  =  1 + iθT_{11} iθT_{12} 
 0 1   iθT_{21} iθT_{22}   iθT_{21} 1 + iθT_{22} 
     
det of RHS = (1 + iθT_{11})(1 + iθT_{22})  (iθT_{12}iθT_{21})
= 1 + iθ(T_{11} + T_{22})
Therefore, to get det = 1, T_{11} = T_{22}.
In other words, the trace of the generators has be 0.
Equivalently, we can also see the from the Jacobi Formula:
1 = detU = det(exp(iθG)) = exp(iθTr(T)) ∴ Tr(T) = 0.
This is clearly the case for the σ's.
Note: SU(1) ≡ U(1) since T is I (unit matrix) and
therefore is just a phase rotation. Therefore,
U = exp(iθ)
For SU(3) we replace the σ's with the 8 GELLMANN
matrices. Thus,
U = exp(iθ_{i}λ_{i}/2)
 
_{ }  0 1 0 
λ_{1} =  1 0 0 
_{ }  0 0 0 
 
 
_{ }  0 i 0 
λ_{2} =  i 0 0 
_{ }  0 0 0 
 
 
_{ }  1 0 0 
λ_{3} =  0 1 0 
_{ }  0 0 0 
 
 
_{ }  0 0 1 
λ_{4} =  0 0 0 
_{ }  1 0 0 
 
 
_{ }  0 0 i 
λ_{5} =  0 0 0 
_{ }  i 0 0 
 
 
_{ }  0 0 0 
λ_{6} =  0 0 1 
_{ }  0 1 0 
 
 
_{ }  0 0 0 
λ_{7} =  0 0 i 
_{ }  0 i 0 
 
 
_{ }  1 0 0 
λ_{8} = 1/√3  0 1 0 
_{ }  0 0 2 
 
The GellMann matrices generalize the Pauli matrices
for SU(2) that are indicated in orange. Like, the σ's
the matrices are both traceless and Hermitian.
The generators satisfy the following commutation rule:
[T_{i},T_{j}] = if_{ijk}T_{k}
Where, T_{i} = λ_{i}/2 and f_{ijk} are the structure constants.
Structure Constants

The structure constants are found from the fundamental
representation as:
f^{ijk} = 2iTr([T_{i},T_{j}]T_{k})
Proof:
Consider:
[T_{a},T_{b}]T_{d}
This is equivalent to:
if^{abc}T_{c}T_{d}
Therefore,
[T_{a},T_{b}]T_{d} = if^{abc}T_{c}T_{d}
If we now take the trace of this we get:
Tr([T_{a},T_{b}]T_{d}) = if^{abc}Tr(T_{c}T_{d})
But from before, Tr(T_{c}T_{d}) = (1/2)δ_{bc}. Therefore,
Tr([T_{a},T_{b}]T_{d}) = if^{abc}(1/2)δ_{bc}
Or,
f^{abc} = 2iTr([T_{a},T_{b}]T_{d})
Example: f^{458}
f^{458} = 2iTr[[T_{4},T_{5}]T_{8}]
^{ } = 2iTr(T_{4}T_{5}T_{8}  T_{5}T_{4}T_{8})
   
^{ }  i/8√3 0 0   i/8√3 0 0 
^{ } = 2iTr( 0 0 0    0 0 0 
^{ }  0 0 i/4√3   0 0 i/4√3 
   
 
^{ }  i/4√3 0 0 
^{ } = 2iTr( 0 0 0 )
^{ }  0 0 i/2√3
 
^{ } = (2i)(3i/4√3)
^{ } = √3/2
f ijk
 
1 123
1/2 147, 246, 257, 345
1/2 156, 367
(√3)/2 458, 678
All others permutations are 0.
The structure constants obey the following rules:
f^{abc} = f^{bca} = f^{cab} = f^{bac} = f^{acb} = f^{cba}
Thus, f^{123} = f^{312} etc.
Structure Constants as a Tensor

Consider the easier case of the LeviCivita symbol
in 3D, ε_{ijk}.
_{ } { +1 if ijk is (1,2,3), (2,3,1) or (3,1,2)
ε_{ijk} = { 1 if ijk is (3,2,1), (1,3,2) or (2,1,3)
_{ } { 0 if i = j or j = k or k = i
We can represent this as a 3D matrix as follows.
 
 0 1 0 
 1 0 0  ij3
 0 0 0 
 
 
 0 0 1 
 0 0 0  ij2
 1 0 0 
 
 
 0 0 0 
 0 0 1  ij1
 0 1 0 
 
Thus, ε_{123} = 1 and ε_{321} = 1
In general, the dimension of the f tensor is:
(N^{2}  1) x (N^{2}  1)
And there would be (N^{2}  1) of them.
Trace Orthonormality

The GM matrices also obey the trace orthonormality
relation:
Tr(λ_{a}λ_{b}) = 2δ_{ab}
Consider Tr(λ_{1}λ_{1}) and Tr(λ_{8}λ_{8})
     
 0 1 0  0 1 0   1 0 0 
Tr( 1 0 0  1 0 0 ) = Tr( 0 1 0 ) = 2
 0 0 0  0 0 0   0 0 0 
     
     
 1 0 0  1 0 0   1 0 0 
Tr((1/3) 0 1 0  0 1 0 ) = Tr((1/3) 0 1 0 ) = 2
 0 0 2  0 0 2   2 0 4 
     
Or, in terms of T_{i}:
Tr(T_{a}T_{b}) = (1/2)δ_{ab}
     
 0 1/2 0  0 1/2 0   1/4 0 0 
Tr( 1/2 0 0  1/2 0 0 ) = Tr( 0 1/4 0 ) = 1/2
 0 0 0  0 0 0   0 0 0 
     
NOTE: WE USE BOTH λ AND T IN THIS DISCUSSION SO
BE MINDFUL OF FACTORS OF 2 IN SOME OF THE RESULTS.
SU(2) Subgroups

The following form SU(2) subgroups of SU(3):
T_{1}, T_{2} and T_{3}
T_{4}, T_{5} and (1/2)(√3T_{8} + T_{3})
T_{6}, T_{7} and (1/2)(√3T_{8}  T_{3})
Proof:
[T_{4},T_{5}] = if^{453}T_{3} + if^{458}T_{8}
[T_{4},T_{5}] = (i/2)λ_{3}/2 + (i√3/2)λ_{8}/2
   
_{ }  i/4 0 0   i/4 0 0 
_{ } =  0 1/4 0  +  0 i/4 0 
_{ }  0 0 0   0 0 i/2 
   
 
_{ }  i/2 0 0 
_{ } =  0 0 0 
_{ }  0 0 i/2 
 
[T_{6},T_{7}] = (i/2)λ_{3}/2 + (i√3/2)λ_{8}/2
   
_{ }  i/4 0 0   i/4 0 0 
_{ } =  0 i/4 0  +  0 i/4 0 
_{ }  0 0 0   0 0 i/2 
   
 
_{ }  0 0 0 
_{ } =  0 i/2 0 
_{ }  0 0 i/2 
 
Casimir Operators

The Casimir operator is an operator that commutes
all other generators. As such it is the center
of the Lie algebra. Thus,
_
[C_{2}(R),T_{a}(R)] = 0
_
The QUADRATIC CASIMIR OPERATOR, C_{2}(R), operator is:
Where (R) denotes the particular representation.
_
C_{2}(R) = ΣT_{a}(R)T_{a}(R)
^{n}
Note: For SU(3) there is also a cubic CO given by:
_
C_{2}(R) = Σ_{ijk}f^{ijk}λ_{i}λ_{j}λ_{k}/8
However, we can ignore it for our purposes.
Casimir Invariant

_
Since C_{2}(R) commutes with all operators we can use
SCHUR'S LEMMA that says:
If D is an irreducible representations of a group
G, A is a matrix and [D(g),A] = 0 then A = kI where
k is some number and I is the identity matrix.
In our case we can write:

C_{2}(R) = C_{2}(R)I
Where C_{2} is the QUADRATIC CASIMIR INVARIANT. The
invariant also follows the relationship:
dim R * C_{2}(R) = dim T * T(R)
Where,
dim R is the dimension of the representation.
dim T is the number of generators.
T(R) is the INDEX of the representation given by:
T(R)δ^{ab} = Tr(T_{a}T_{b}) (δ^{aa} = δ^{bb} = ... = 1, 0 otherwise)
For the fundamental representation of SU(2), and
using σ_{2} we get:
Index:
     
 0 i/2  0 i/2  =  1/4 0  ∴ T(F) = 1/2
 i/2 0  i/2 0   0 1/4 
     
Commutator:
       
 1/4 0  0 i/2    0 i/2  1/4 0  = 0
 0 1/4  i/2 0   i/2 0  0 1/4 
       
Casimir Invariant:
2 * C_{2}(F) = 3 * 1/2
∴ C_{2}(F) = 3/4
Casimir Operator:
_
C_{2}(F) = Σ_{i}(σ_{i}/2)^{2}
_{ } = (σ_{1}/2)^{2} + (σ_{2}/2)^{2} + (σ_{3}/2)^{2}
_{ }  
_{ } =  3/4 0 
_{ }  0 3/4 
_{ }  
_{ }  
_{ } = 3/4 1 0 
_{ }  0 1 
_{ }  
_
Which is in agreement with C_{2}(F) = C_{2}(F)I
For the fundamental representation of SU(3) and using
λ_{1} we get:
Index:
     
 0 1/2 0  0 1/2 0   1/4 0 0 
 1/2 0 0  1/2 0 0  =  0 1/4 0  ∴ T(F) = 1/2
 0 0 0  0 0 0   0 0 0 
     
Commutator:
       
 1/4 0 0  0 1/2 0   0 1/2 0  1/4 0 0 
 0 1/4 0  1/2 0 0    1/2 0 0  0 1/4 0  = 0
 0 0 0  0 0 0   0 0 0  0 0 0 
       
Caimir Invariant:
3 * C_{2}(F) = 8 * 1/2
∴ C_{2}(F) = 4/3
Casimir Operator:
_
C_{2}(F) = Σ_{i}(λ_{i}/2)^{2}
_{ } = (λ_{1}/2)^{2} + ... (λ_{8}/2)^{2}
_{ }  
_{ }  4/3 0 0 
_{ } =  0 4/3 0 
_{ }  0 0 4/3 
_{ }  
_{ }  
_{ }  1 0 0 
_{ } = (4/3) 0 1 0 
_{ }  0 0 1 
_{ }  
_
Which is in agreement with C_{2}(F) = C_{2}(F)I
Raising and Lowering Operators

For SU(3) we can define the operators:
I_{±} = (1/2)(λ_{1} ± iλ_{2}) gives d <> u
V_{±} = (1/2)(λ_{4} ± iλ_{5}) gives s <> u
U_{±} = (1/2)(λ_{6} ± iλ_{7}) gives s <> d
Therefore,
   
^{ }  0 1 0 ^{ }  0 0 0 
I_{+} =  0 0 0  I_{} =  1 0 0 
^{ }  0 0 0 ^{ }  0 0 0 
   
   
^{ }  0 0 1 ^{ }  0 0 0 
V_{+} =  0 0 0  V_{} =  0 0 0 
^{ }  0 0 0 ^{ }  1 0 0 
   
   
^{ }  0 0 0 ^{ }  0 0 0 
U_{+} =  0 0 1  U_{} =  0 0 0 
^{ }  0 0 0 ^{ }  0 1 0 
   
The operators follow the commutation relations:
[I_{+},I_{}] = λ_{3}
[V_{+},V_{}] = (√3/2)λ_{8} + (1/2)λ_{3}
[U_{+},U_{}] = (√3/2)λ_{8}  (1/2)λ_{3}
[I_{+},V_{}] = U_{} [I_{+},U_{+}] = V_{+}
[U_{+},V_{}] = I_{} [I_{},U_{}] = V_{}
[I_{},V_{+}] = U_{+} [U_{},V_{+}] = I_{+}
and every other commutation of I,V and U = 0
These relationships are emphasized in the following
diagrams. The left hand case is not allowed and so
the commutator is 0. The right hand case is [I_{+},U_{+}]
producing the root vector, V_{+}.
These operators act on the column vectors:
     
 1   0   0 
u =  0  d =  1  s =  0 
 0   0   1 
     
Proof for d > u:
     
^{ }  0 1 0  0   1 
I_{+} =  0 0 0  1  =  0 
^{ }  0 0 0  0   0 
     
Because I_{+} and I_{} are associated with d > u and
u > d associated with neutrons and protons, they
are referred to as the isospin or Ispin operators.

Digression:
Isospin

The proton and neutron have very similar masses
and the nuclear force is found to be approximately
chargeindependent. In 1932, Heisenberg proposed
that if you could switch off the electric charge
of the proton there would be no way to distinguish
between a proton and neutron.
He proposed that the neutron and proton should be
considered as two states of a single entity  the
nucleon.
   
p =  1  n =  0 
 0   1 
   
The 2 states are analogous to the spin up and spin
down states of a spin 1/2 particle. He referred
to this as isospin. Isospin has nothing to do
with spin but shares the same mathematics.
The neutron and proton form an isospin doublet with
total isospin I = 1/2 and third component I_{3} = ±1/2
We can extend this idea to the u and d quarks since
they have similar masses:
   
u =  1  d =  0 
 0   1 
   

These operators are the operators that make the
transitions between the weights.
Cartan Generators

The Cartan subalgebra of a Lie algebra is the
largest subset of elements of the algebra that
are traceless, Hermitian and can be diagonalized
simultaneously. In other words, it is the largest
subset of elements of the algebra that all commute
with one another. The RANK of a Lie algebra, r,
is the number of elements contained in its Cartan
subalgebra. It is conventional to denote these
elements as H^{i}. For SU(3) the Cartan generators
are:
 
^{ }  1 0 0 
H^{1} = (1/2) 0 1 0  = λ_{3}/2 = I_{3}
^{ }  0 0 0 
 
 
_{ }  1 0 0 
H^{2} = (1/2√3) 0 1 0  = λ_{8}/2 = Y/2 = I_{8}
_{ }  0 0 2 
 
It is easy to show that [H^{1},H^{2}] = 0 (orthogonal)
and H^{i} = (H^{i})^{†}.
Hence SU(3) has a rank of 2.
These matrices share the same set of linearly
independent eigenvectors but not necessarily
the same eigenvalues. As such they can be
simultaneously diagonalized.
Proof:
       
 1/2 0 0   1   0   0 
 0 1/2 0  > 1/2 0 ; 0 0 ; 1/2 1 
 0 0 0   0   1   0 
       
       
 1/2√3 0 0   1   0   0 
 0 1/2√3 0  > 1/2√3 0 ; 1/√3 0 ; 1/2√3 1 
 0 0 1/√3   0   1   0 
       
SU(2) has 1 Cartan generator so it has a rank of
1. The Cartan generator is σ_{3}/2.
 
H =  1/2 0 
 0 1/2 
 

Digression:
Simultaneous Diagonalization

2 n x n matrices A and B are simultaneusy diagonalizable if
they commute with one another and both can be diagonalized
using the same invertible matrix S such that:
S^{1}AS = M and S^{1}BS = P
Where M and P are both diagonal matrices. The use of the
same invertible matrix S implies that simultaneously
diagonalizable matrices share a common basis of
linearly independent eigenvectors.
Example:
   
A =  2 2  and B =  1 2 
 2 2   2 1 
   
     
eigenvectors/eigenvalues of  2 2  = 0 1  and 4 1 
 2 2   1   1 
     
     
eigenvectors of  1 2  = 1 1  and 3 1 
 2 1   1   1 
     
   
 1 1 ^{1} =  1/2 1/2 
 1 1 ^{ }  1/2 1/2 
   
       
 1/2 1/2  2 2  1 1  =  4 0  = C
 1/2 1/2  2 2  1 1   0 0 
       
       
 1/2 1/2  1 2  1 1  =  3 0  = D
 1/2 1/2  2 1  1 1   0 1 
       
[A,B] = [C,D] = 0

Weight Diagrams

The Weights of a representation contain the
eigenvalues of the elements of the Cartan
subalgebra. The eigenvalues of the Cartan
generators act on the basis of the fundamental
representation give the weight vectors of the
algebra.
For SU(3) we get:
H^{1}u> = h^{1}u> gives:
     
 1/2 0 0  1   1 
 0 1/2 0  0  = (1/2) 0 
 0 0 0  0   0 
     
H^{2}u> = h^{2}u> gives:
     
 1/2√3 0 0  1   1 
 0 1/2√3 0  0  = (1/2√3) 0 
 0 0 1/√3  0   0 
     
∴ w^{1} = (h^{1},h^{2}) = (1/2,1/2√3)
H^{1}d> = h^{1}d> gives:
     
 1/2 0 0  0   0 
 0 1/2 0  1  = (1/2) 1 
 0 0 0  0   0 
     
H^{2}d> = h^{2}d> gives:
     
 1/2√3 0 0  0   0 
 0 1/2√3 0  1  = (1/2√3) 1 
 0 0 1/√3  0   0 
     
∴ w^{2} = (h^{1},h^{2}) = (1/2,1/2√3)
H^{1}s> = h^{1}s> gives:
     
 1/2 0 0  0   0 
 0 1/2 0  0  = (0) 0 
 0 0 0  1   1 
     
H^{2}s> = h^{2}s> gives:
     
 1/2√3 0 0  0   0 
 0 1/2√3 0  0  = (1/√3) 0 
 0 0 1/√3  1   1 
     
∴ w^{3} = (h^{1},h^{2}) = (0,1/√3)
Diagramatically, the weight space for SU(3) in the
fundamental representation looks like:
Q = I_{3} + Y/2
Normalize to λ_{8} ∴ Y => λ_{8}/√3
Q = λ_{3}/2 + λ_{8}/2√3
The coordinates are the ISOSPINs, I_{3} and √3/2 times
the HYPERCHARGEs, Y. Thus we can redraw the
above diagram in terms of I_{3} and Y as follows:
Therefore, H^{1} generates the isospin and H^{2} generates
the hypercharge.
Finally, using Q = (2I_{3} + Y)/2
= I_{3} + Y/2
   
 1/2 0 0   1/2√3 0 0 
=  0 1/2 0  + (1/√3) 0 1/2√3 0 
 0 0 0   0 0 1/√3 
   
   
 1/2 0 0   1/6 0 0 
=  0 1/2 0  +  0 1/6 0 
 0 0 0   0 0 1/3 
   
 
 2/3 0 0 
=  0 1/3 0 
 0 0 1/3
 
This is referred to as the CHARGE MATRIX.
There following relationship is also used in the
literature:
Y = S + B
Where B is the Baryon number and S is the Strangeness
quantum number = (n_{S}  n_{S'}) where n_{S'} is the number
of anti strange quarks.
The above representation is referred to as the '3'.
Note: For SU(2) there is only 1 Cartan generator.
This generates the WEAK ISOSPIN. In order to get
the WEAK HYPERCHARGE it is necessary to consider
the SU(2)_{L} ⊗ U(1)_{Y} gauge group and the electroweak
model developed by GLASHOWWEINBERGSALAM. Thus,
gauge group = SU(2)_{L} ⊗ U(1)_{Y}
^ ^
 
weak weak
isospin, I_{3} hypercharge, Y_{W}
The weak isospin part is:
     
 1/2 0  1  = (1/2) 1 
 0 1/2  0   0 
     
∴ w^{1} = (1/2,?)
     
 1/2 0  0  = (1/2) 0 
 0 1/2  1   1 
     
∴ w^{2} = (1/2,?)
To get the weak hypercharge we use the formula:
Y_{W} = 2(Q  I_{3})
Therefore, for the u and d quarks we get:
_{ }    
Y_{W} = 2( 2/3 0    1/2 0 )
_{ }  0 1/3   0 1/2 
_{ }    
_{ }  
_{ } = 2 1/6 0 
_{ }  0 1/6 
_{ }  
_{ }  
_{ } =  1/3 0 
_{ }  0 1/3 
_{ }  
_{ }  
_{ } = (1/3) 1 0 
_{ }  0 1 
_{ }  
_{ } = (1/3)I
Therefore, the eigenvalue is:
_{ }    
Y_{W} 1  = (1/3) 1 
_{ }  0   0 
_{ }    
Y_{W} as the generator of U(1)_{L} is just a phase
rotation.
For the left handed leptons, e_{L} and ν_{L} we get:
_{ }    
Y_{W} = 2( 1 0    1/2 0 )
_{ }  0 0   0 1/2 
_{ }    
_{ }  
_{ } = 2 1/2 0 
_{ }  0 1/2 
_{ }  
_{ }  
_{ } =  1 0 
_{ }  0 1 
_{ }  
_{ }    
Y_{W} 1  = 1 1 
_{ }  0   0 
_{ }    
The Complex Conjugate (AntiFundamental) Representation

In the complex conjugate representation:
_
U = exp(iε_{i}T_{a}(C))
The generators of the group are given by T_{a}(C) = (T_{a}(F))^{†}.
Thus,
_
U = exp(iε_{i}T_{a}(F)^{†})
or
_{ } __{ } _
ψ > exp(iε_{i}T_{a}(F)^{†})ψ and ψ = exp(iε_{i}T^{a}(F)^{†})ψ
For SU(2) the fundamental and conjugate representation
turn out to be equivalent and SU(2) is referred to as
a real group. This is not true for the SU(3) and the
representations are inequivalent.
The generators follow the same commutations rules as
for the fundamental case:
[T_{a}(C),T_{b}(C)] = if^{abc}T_{c}(C)
Proof:
[T_{1}(C),T_{2}(C)] = if^{123}T_{3}(C)
 
_{ }  0 1/2 0 
T_{1}(C) = λ_{1}^{†}/2 =  1/2 0 0 
_{ }  0 0 0 
 
 
_{ }  0 i/2 0 
T_{3}(C) = λ_{3}^{†}/2 =  i/2 0 0 
_{ }  0 0 0 
 
       
 0 1/2 0  0 i/2 0   0 i/2 0  0 1/2 0 
 1/2 0 0  i/2 0 0    i/2 0 0  1/2 0 0 
 0 0 0  0 0 0   0 0 0  0 0 0 
       
 
 i/2 0 0 
=  0 i/2 0 
 0 0 0 
 
= if^{123}(λ_{3})^{†}/2
The weight for SU(3) are:
_ _ _
H^{1}u> = h^{1}u> gives:
     
 1/2 0 0  1   1 
 0 1/2 0  0  = (1/2) 0 
 0 0 0  0   0 
     
_ _ _
H^{2}u> = h^{2}u> gives:
     
 1/2√3 0 0  1   1 
 0 1/2√3 0  0  = (1/2√3) 0 
 0 0 1/√3  0   0 
     
∴ w^{1} = (h^{1},h^{2}) = (1/2,1/2√3)
_ _ _
H^{1}d> = h^{1}d> gives:
     
 1/2 0 0  0   0 
 0 1/2 0  1  = (1/2) 1 
 0 0 0  0   0 
     
_ _ _
H^{2}d> = h^{2}d> gives:
     
 1/2√3 0 0  0   0 
 0 1/2√3 0  1  = (1/2√3) 1 
 0 0 1/√3  0   0 
     
∴ w^{2} = (h^{1},h^{2}) = (1/2,1/2√3)
_ _ _
H^{1}s> = h^{1}s> gives:
     
 1/2 0 0  0   0 
 0 1/2 0  0  = (0) 0 
 0 0 0  1   1 
     
_ _ _
H^{2}s> = h^{2}s> gives:
     
 1/2√3 0 0  0   0 
 0 1/2√3 0  0  = (1/√3) 0 
 0 0 1/√3  1   1 
     
∴ w^{3} = (h^{1},h^{2}) = (0,1/√3)
Diagramatically, the weight space for SU(3) in the
antifundamental representation looks like:
When acting on antiquarks we need to use:
I_{+} > I_{+}^{*} = I_{}
I_{} > I_{}^{*} = I_{+}
etc. etc.
This is because the generators in the antifundamental
case obey
_
T(C) = T(F)^{†}
For example,
 
_{ }  0 0 0 
I_{} =  1 0 0 
_{ }  0 0 0 
 
 
_{ }  0 1 0 
I_{}^{T} =  0 0 0  = I_{+}
_{ }  0 0 0 
 
_ _
I_{+}d> = u>
The complete list of raising and lowering operators
for the antiquarks is:
_ _
I_{+}u> = d>
_ _
V_{+}u> = s>
_ _
U_{+}d> = s>
_ _
I_{}d> = u>
_ _
V_{}s> = u>
_ _
U_{}s> = d>