# A SU(3) multiplet

1. Mar 3, 2016

### shereen1

Dear All
I just have a question. We say that the SU(2) doublet have the same value of isospin but the particles of this multiplet differs by I3. Now what quantum number the particles of SU(3) multiplet share.
Thank you

2. Mar 3, 2016

### ChrisVer

The SU(3) of color triplet has the colors R,G,B for the quarks...

3. Mar 4, 2016

### samalkhaiat

This is because $SU(2)$ is a rank-1 group: Only the representation matrix of $J_{3}$ is diagonal. This allows us to picture each irreducible representation, also called multiplet or the manifold of states,
$$|j , m = -j\rangle , |j , m = -j +1\rangle , \cdots , |j , m = j-1\rangle , |j , m = +j\rangle ,$$
by one-dimensional graph (line) with equally spaced points $(j,m)$ representing the $(2j+1)$ states $| j , m \rangle$
$$J^{2}|j,m \rangle = j(j+1) |j,m \rangle , \ \ \ J_{3}|j,m \rangle = m |j,m \rangle$$
The raising and lowering operators $J_{\pm}$ move the points $(j,m)$ along the line, i.e., they connect states which have the same $j$ but different values of $m$
$$J_{\pm}|j,m \rangle = \sqrt{(j \mp m)(j \pm m + 1)} |j , m \pm 1\rangle .$$
Only the mass, spin and baryon number are the same for each $SU(3)$ multiple.
$SU(3)$ is a rank-2 group: because there are only two diagonal generators $\lambda_{3} \equiv T_{3}$ and $\lambda_{8} \equiv \frac{\sqrt{3}}{2} Y$,
$$[ T_{3} , Y ] = 0 ,$$
the states in an $SU(3)$ irreducible multiplet need two labels, the eigen-value $t_{3}$ of $T_{3}$ and $y$ of $Y$. So, we can picture the multiplet by a figure on the $t_{3}-y$ plane. We can repeat what we did with $SU(2)$ and define the operators
$$T_{\pm} = \lambda_{1} \pm i \lambda_{2},$$
which connect states on the $t_{3}$-axis or any other line parallel to $t_{3}$, i.e., $T_{+}(T_{-})$ raises (lowers) $t_{3}$ by one unit and leaves $y$ unchanged.
So for the horizontal line connecting the $u$ quark, i.e., the point $(1/2 , 1/3)$ to the $d$ quark, i.e., the point $(-1/2,1/3)$ we have
$$T_{+} = | u \rangle \langle d | , \ \ T_{-} = | d \rangle \langle u | ,$$ and $$2T_{3} = [ T_{+} , T_{-} ] = | u \rangle \langle u | - | d \rangle \langle d | .$$ So, you can say that $T_{+} = u^{\dagger}d$ destroys a $d$ quark and creates a $u$ quark. Recalling the electric charge operator
$$Q = \frac{2}{3}u^{\dagger}u - \frac{1}{3}d^{\dagger}d - \frac{1}{3}s^{\dagger}s ,$$ and using the Gell-Mann-Nishijima relation among $Q$, the diagonal I-spin generator $T_{3}$ of $SU(2)$ and the hypercharge operator $Y = 2 (Q - T_{3})$, we find
$$Y = \frac{1}{3} ( u^{\dagger}u + d^{\dagger}d - 2 s^{\dagger}s ) .$$ We can rewrite this in a form which will be useful bellow $$d^{\dagger}d - s^{\dagger}s = \frac{3}{2} Y - T_{3} . \ \ \ \ \ \ \ (1)$$
This is not the end of the story, the lines parallel to $t_{3}$ end on two diagonal lines defined by the remaining two set of raising and lowering operators (there are 3 $SU(2)$ groups living inside $SU(3)$ with old saying : I spin, you (U) spin and we (V) all spin together)
$$V_{\pm} = \lambda_{4} \pm i \lambda_{5}, \ \ \ U_{\pm} = \lambda_{6} \pm i \lambda_{7} .$$
$V_{+}$ makes $60$ degree angle with $T_{+}$. So it raises $t_{3}$ by $1/2$ unit and raises $y$ by $1$ unit; $U_{+}$ makes $120$ degree angle with $T_{+}$. So it lowers $t_{3}$ by $1/2$ unit and raises $y$ by $1$ unit, etc.
So, graphically each irreducible representation $(p,q)$ shows up as a figure with hexagonal boundary on the $t_{3}-y$ plane: 3 sides having $p$ units of length and the other 3 sides having $q$ units; when either $p = 0$ or $q = 0$, the hexagonal collapses into equilateral triangle $(p,0)$ or $(0,q)$. The most important irreducible representations are: the fundamental triplets $(0,1) = \{ \bar{3}\}$ and $(1,0) = \{3\}$; the octet (adjoint rep.) $\{8\} = (1,1)$ and the decuplet $\{10\} = (3,0)$.
Again for the fundamental representation, on the diagonal line connecting $d(-1/2,0)$ with $s(0,-2/3)$ we have
$$U_{+} = d^{\dagger} s , \ \ \ U_{-} = s^{\dagger} d .$$
So, we can define a $U_{3}$ for that diagonal line by
$$2U_{3} = [ U_{+} , U_{-} ] = d^{\dagger}d - s^{\dagger}s .$$
Now, if we use Eq(1) we get
$$2U_{3} = \frac{3}{2}Y - T_{3} = 2 Y - Q .$$ From this we conclude that the electric charge $Q$ commutes with $Y , T_{3}$ and $U_{3}$.
The last diagonal line in the equilateral triangle connects $u(1/2,1/3)$ to $s(0,-2/3)$. Again, we do the same thing. We write down the ladder operators
$$V_{+} = u^{\dagger}s , \ \ \ \ V_{-} = s^{\dagger}u ,$$ and define a $V_{3}$ for that line by $2V_{3} = [ V_{+} , V_{-}]$.
So, along each of the $T, U$ and the $V$ lines, the irreducible representations of $SU(3)$ decompose into representation of the corresponding $SU(2) \times U(1)$. In particular, along the horizontal lines we have multiplets of $SU(2)_{T} \times U(1)_{Y}$. For example, in the adjoint representation $(p = 1 , q = 1) = \{8\}$, we have: for the $y = 1$ line
$$K^{0}/n = (-1/2,1) , \ \ \ K^{+}/p = (1/2,1) \ : (\vec{1/2}, 1) \in SU(2)_{T} \times U(1)_{Y} .$$
For the line $y = 0$, we get the following states (or points)
$$\pi^{-}/ \Sigma^{-} = ( -1,0) , \ \ \pi^{0}/ \Sigma^{0} = (0,0) , \ \ \pi^{+}/ \Sigma^{+} = (1,0) : ( \vec{1},0) \in SU(2)_{T} \times U(1)_{Y} .$$
On this line, there are also the states $\eta^{0} / \Lambda^{0}= (0,0)$ corresponding to $(\vec{T} = \vec{0} , Y= 0) \in SU(2)_{T} \times U(1)_{Y}$.
And finally for the $y = -1$ we have
$$K^{-}/ \Xi^{-} = (-1/2 , -1) , \ \ \ \bar{K}^{0}/ \Xi^{0} = (1/2 , -1) \ : (\vec{1/2} , -1) \in SU(2)_{T} \times U(1)_{Y} .$$
If you put these points on the $t_{3}-y$ plane you obtain a hexagonal. Of course, the same states and picture can be obtained using the U-Spin and the V-Spin operators $(U_{\pm}, V_{\pm})$ going diagonally in the $t_{3}-y$ plane.