# How many ''charges'' are there in SU(2) and SU(3) symmetry?

1. Jan 1, 2012

### ndung200790

How many conserved observations(''charges'') are there in SU(2) and SU(3) symmetries?I know that U(1) has only one charge that is electric charge.
Thank you very much for your kind helping.

2. Jan 1, 2012

### DrDu

The number of independent charges is equal to the number of generators of the group, i.e. N^2-1. However, not all of these charges commute among themselves.

3. Jan 1, 2012

### ndung200790

Then,what is the speciality of 3 color ''charges'' and isospin ''charges''?

4. Jan 1, 2012

### tom.stoer

There is a nice qm example, namely the N-dim. harmonic oscillator which has an U(N) = U(1)*SU(N) symmetry. The generators are

$$T^a;\;a=0\ldots N^2-1;\;T^0 = 1$$

The charges are

$$Q^a = \sum_{ik}\,a_i^\dagger (T^a)_{ik} a_k$$

Both the matrices T and the charges Q generate the same algebra

$$[T^a,T^b] = if^{abc}T^c$$
$$[Q^a,Q^b] = if^{abc}Q^c$$

Based on the algebra of the T's the algebra of the Q's can be derived.

The charge for a=0 is related to the Hamiltonian itself

$$H = Q^0 + \frac{N}{2}$$

The charges are conserved, i.e.

$$[Q^a,H] = [Q^a,Q^0] = 0$$

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

Hm, I don't know, they are realized in nature ;-)

SU(2) isospin is a special case of the SU(N) flavor symmetry which is not an exakt symmetry due to the different quark masses; SU(3) color is a local gauge symmetry, i.e. it is related with unphysical degrees of freedom; the charges of SU(3) color are special as all physical states must be color singulets:

$$Q^a |\text{phys}\rangle = 0$$

Last edited: Jan 1, 2012
5. Jan 1, 2012

### ndung200790

So,physical observations relate only with global symmetry?

6. Jan 1, 2012

### ndung200790

And,SU(2) in weak interaction and SU(3) in strong interaction is considered as local or global symmetries?Because we are considering physical observations.

7. Jan 2, 2012

### tom.stoer

Gobal flavor symmetry was observed rather directly via the hadron multiplets: http://en.wikipedia.org/wiki/Eightfold_Way_(physics [Broken])

Local color symmetry was constructed as a non-abelian gauge symmetry (Yang-Mills theory) + quantization (Veltman, 't Hooft, Gross, Politzer, Wilczek, ...). There are direct indications for SU(Ncolor) symmetry (requirement for an additonial quantum number in the |?++> = |uuu>, asymptotic freedom, color confinement). But the indications for Ncolor=3 are more indirect.

1) in the naive quark model one works with three constituent quarks; in order to get the correct symmetry for the hadrons one has to introduce an additional 'color' degree of freedom and b/c of the three constituent quarks Ncolor=3 seems to be natural for color, too (in order to get color singulets which means that the new degree of freedom is not observed macroscopically).

2) in the renormalization group equation for the string coupling constant there is a pre-factor (11 Ncolor - 2 N flavor).

So there are experimental tests both for the non-abelian gauge symmetry in general (confinement, asymptotic freedom) as well as for Ncolor

Last edited by a moderator: May 5, 2017
8. Jan 2, 2012

### samalkhaiat

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);
$$q^{\dagger} = \left( r^{\dagger} \ b^{\dagger} \ g^{\dagger} \right) \in \{\bar{3}\}$$
Now let us form the tensor product,
$$q^{\dagger}\otimes q = \left( \begin{array}{ccc}r^{\dagger}r & r^{\dagger}b & r^{\dagger}g \\ b^{\dagger}r & b^{\dagger}b & b^{\dagger}g \\ g^{\dagger}r & g^{\dagger}b & g^{\dagger}g \end{array} \right),$$
and reduce it into invariant subspaces (irreducible tensors). First notice that the trace
$$Tr\{q^{\dagger}\otimes q \} = r^{\dagger}r + b^{\dagger}b + g^{\dagger}g \equiv q^{\dagger}_{i}q^{i},$$
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
$$q^{\dagger} \otimes q = \left( q^{\dagger} \otimes q - \frac{1}{3}(q^{\dagger}_{i}q^{i})I_{3\times 3}\right) + \frac{1}{3}(q^{\dagger}_{i}q^{i})I_{3 \times 3}$$
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 $\{8\}$;
$$\{8\} = \left( \begin{array}{ccc}\{8\}^{1}_{1} & r^{\dagger}b & r^{\dagger}g \\ b^{\dagger}r & \{8\}^{2}_{2} & b^{\dagger}g \\ g^{\dagger}r & g^{\dagger}b & \{8\}^{3}_{3} \end{array}\right), \ \ (1)$$
where
$$\{8\}^{1}_{1} = \frac{1}{3}( 2 r^{\dagger}r - b^{\dagger}b - g^{\dagger}g),$$
$$\{8\}^{2}_{2} = \frac{1}{3}( - r^{\dagger}r + 2 b^{\dagger}b - g^{\dagger}g),$$
and
$$\{8\}^{3}_{3} = \frac{1}{3}( - r^{\dagger}r - b^{\dagger}b + 2 g^{\dagger}g).$$
Now, using the anti-commutation relations

$$\{ r^{\dagger}, r \} = \{ b^{\dagger} , b \} = \{ g^{\dagger} , g \} = 1,$$
one can show that the $\{8\}$ transform irreducibly among themselves in the so-called adjoint representation
$$[\{8\}^{i}_{j} , \{8\}^{k}_{l} ] = \delta^{k}_{j}\{8\}^{i}_{l} - \delta^{i}_{l}\{8\}^{k}_{j}$$
This completes the proof of the following identity
$$\{\bar{3}\} \otimes \{3\} = \{8\} \oplus \{1\}$$
Now, any traceless $3 \times 3$ matrix can be expanded in terms of the Gell-Mann matrices;
$$\{8\} = J^{a}_{0}\lambda^{a} = \left( \begin{array}{ccc}(J^{3}_{0} + \frac{J^{8}_{0}}{\sqrt{3}}) & (J^{1}_{0} + i J^{2}_{0}) & (J^{4}_{0} + i J^{5}_{0}) \\ (J^{1}_{0} - i J^{2}_{0}) & ( - J^{3}_{0} + \frac{J^{8}_{0}}{\sqrt{3}}) & ( J^{6}_{0} + i J^{7}_{0}) \\ (J^{4}_{0} - i J^{5}_{0}) & (J^{6}_{0} - i J^{7}_{0}) & ( - \frac{2}{\sqrt{3}}J^{8}_{0}) \end{array} \right). \ \ (2)$$
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);
$$J^{a}_{0} = q^{\dagger}_{i} ( \frac{\lambda^{a}}{2})_{ij}q_{j} \ \ \ (3)$$
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);
$$J^{a}_{0}(x) = \bar{q}_{i}(x) \gamma_{0} ( \frac{\lambda^{a}}{2})_{ij}q_{j}(x)$$
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

Last edited: Jan 2, 2012
9. Jan 3, 2012

### tom.stoer

@Sam, thx for the excellent explanation

@ndung200790, two remarks:
1) this construction differes from mine (harmonic oscillator toy model), nevertheless, the algebraic context is identical; the relation is basically λa/2 = Ta
2) due to local gauge invariance the color-currents and the charges in QCD will have both a quark- and a gluon content which has not yet been discussed.