ndung200790 said:
Then,what is the speciality of 3 color ''charges'' and isospin ''charges''?
Do not confuse the index space (= degrees of freedom or the dimensions of the smallest representation space) of a symmetry group with the (Noether’s) symmetry charges. When we deal with SU_{c}(3) symmetry, we take a single quark to be a fermion field taking values in a 3-dimensional vector space, i.e., a vector in 3-dimensional “colour” space;
q = \left( \begin{array}{c}r \\ b \\ g \\ \end{array}\right) \in \{3\}
and anti-quark as vector in the (inequivalent) conjugate space (the “anti-colour” space);
<br />
q^{\dagger} = \left( r^{\dagger} \ b^{\dagger} \ g^{\dagger} \right) \in \{\bar{3}\}<br />
Now let us form the tensor product,
<br />
q^{\dagger}\otimes q = \left( \begin{array}{ccc}r^{\dagger}r & r^{\dagger}b & r^{\dagger}g \\ b^{\dagger}r & b^{\dagger}b & b^{\dagger}g \\ g^{\dagger}r & g^{\dagger}b & g^{\dagger}g \end{array} \right),<br />
and reduce it into invariant subspaces (irreducible tensors). First notice that the trace
<br />
Tr\{q^{\dagger}\otimes q \} = r^{\dagger}r + b^{\dagger}b + g^{\dagger}g \equiv q^{\dagger}_{i}q^{i},<br />
defines an invariant (scalar) product in the colour space, i.e., colour singlet.
So, let us decompose the above tensor product in the following way
<br />
q^{\dagger} \otimes q = \left( q^{\dagger} \otimes q - \frac{1}{3}(q^{\dagger}_{i}q^{i})I_{3\times 3}\right) + \frac{1}{3}(q^{\dagger}_{i}q^{i})I_{3 \times 3}<br />
The first object on the right hand side is a traceless 3\times 3 matrix therefore having 8 independent elements. We can think of it as a vector living in 8-dimensional (irreducible) subspace \{8\};
<br />
\{8\} = \left( \begin{array}{ccc}\{8\}^{1}_{1} & r^{\dagger}b & r^{\dagger}g \\ b^{\dagger}r & \{8\}^{2}_{2} & b^{\dagger}g \\ g^{\dagger}r & g^{\dagger}b & \{8\}^{3}_{3} \end{array}\right), \ \ (1)<br />
where
<br />
\{8\}^{1}_{1} = \frac{1}{3}( 2 r^{\dagger}r - b^{\dagger}b - g^{\dagger}g),<br />
<br />
\{8\}^{2}_{2} = \frac{1}{3}( - r^{\dagger}r + 2 b^{\dagger}b - g^{\dagger}g),<br />
and
<br />
\{8\}^{3}_{3} = \frac{1}{3}( - r^{\dagger}r - b^{\dagger}b + 2 g^{\dagger}g).<br />
Now, using the anti-commutation relations
<br />
\{ r^{\dagger}, r \} = \{ b^{\dagger} , b \} = \{ g^{\dagger} , g \} = 1,<br />
one can show that the \{8\} transform irreducibly among themselves in the so-called adjoint representation
<br />
[\{8\}^{i}_{j} , \{8\}^{k}_{l} ] = \delta^{k}_{j}\{8\}^{i}_{l} - \delta^{i}_{l}\{8\}^{k}_{j}<br />
This completes the proof of the following identity
<br />
\{\bar{3}\} \otimes \{3\} = \{8\} \oplus \{1\}<br />
Now, any traceless 3 \times 3 matrix can be expanded in terms of the Gell-Mann matrices;
<br />
\{8\} = J^{a}_{0}\lambda^{a} = \left( \begin{array}{ccc}(J^{3}_{0} + \frac{J^{8}_{0}}{\sqrt{3}}) & (J^{1}_{0} + i J^{2}_{0}) & (J^{4}_{0} + i J^{5}_{0}) \\ (J^{1}_{0} - i J^{2}_{0}) & ( - J^{3}_{0} + \frac{J^{8}_{0}}{\sqrt{3}}) & ( J^{6}_{0} + i J^{7}_{0}) \\ (J^{4}_{0} - i J^{5}_{0}) & (J^{6}_{0} - i J^{7}_{0}) & ( - \frac{2}{\sqrt{3}}J^{8}_{0}) \end{array} \right). \ \ (2)<br />
Comparing Eq(1) with Eq(2), we can solve for the 8 numbers J^{a}_{0} (the meaning of the lower index {0} will become clear very soon);
<br />
J^{a}_{0} = q^{\dagger}_{i} ( \frac{\lambda^{a}}{2})_{ij}q_{j} \ \ \ (3)<br />
where i ,j = 1,2,3, that is q_{1} = r , q^{\dagger}_{2} = b^{\dagger}, q_{3} = g , …. In order to treat the quark fields as Dirac spinors, we insert ( \gamma_{0})^{2} = 1 in Eq(3);
<br />
J^{a}_{0}(x) = \bar{q}_{i}(x) \gamma_{0} ( \frac{\lambda^{a}}{2})_{ij}q_{j}(x)<br />
Now it is easy to show that the integral
Q^{a} = \int d^{3}x \ J^{a}_{0}(x),
(1) satisfies the Lie algebra of SU(3);
[Q^{a},Q^{b}] = i f^{abc}Q^{c}
(2) generates the correct SU(3) infinitesimal transformations on the quark fields;
[iQ^{a}, q_{i}(x)] = \delta^{a}q_{i}(x)
Thus, we conclude that J^{a}_{0}(x) is the time component of the Noether current associated with the GLOBAL SU_{c}(3) symmetry of the “FREE” quark Lagrangian. The remarkable fact here is that we derived the form of the current using only group theory and the (fermionic) anti-commutation relations, i.e., we made no reference to the dynamics; the form of the Lagrangian and the equations of motion played no role in our derivation. However, in order to show that the current is conserved (if it is) we need the form of the Lagrangian (or the Hamiltonian).
Sam