shereen1 said:
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.
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 .
Now what quantum number the particles of SU(3) multiplet share.
Thank you
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.
