Delta Baryon Quartet: Constructing from SU(2) Symmetry

[lalo_u]

TL;DR

I'm looking for a tensor representation for a $SU(2)$ quartet in order to build a $\Delta$ baryon states from quark multiplications. I know that it is possible for $\pi$ triplet and it is shown in the thread.

In $SU(2)$ symmetry, we can define a triplet as $2\otimes 2^*=3\oplus 1$ with a tensor representation like this:
$$q_iq_j^*=\left(q_iq^j-\frac{1}{2}\delta_j^iq_kq^k\right)+\frac{1}{2}\delta_j^iq_kq^k.$$
The upper index denotes an anti-doublet and the traceless part in parentheses represents a triplet.
For instance, let $q=\begin{pmatrix}u\\d\end{pmatrix}$ be the doublet and $\bar{q}=\begin{pmatrix}\bar{u}&\bar{d}\end{pmatrix}$ the anti-doublet. If we build the $\pi$ meson triplet $T$ from the tensor defined above,
$$
T_1^1=u\bar{u}-\frac{1}{2}\left(u\bar{u}+d\bar{d}\right)=\frac{1}{2}\left(u\bar{u}-d\bar{d}\right)=\frac{\pi^0}{\sqrt{2}};\\
T_1^2=u\bar{d}=\pi^+;\;T_1^2=d\bar{u}=\pi^-;\\
T_2^2=d\bar{d}-\frac{1}{2}\left(u\bar{u}+d\bar{d}\right)=-\frac{1}{2}\left(u\bar{u}-d\bar{d}\right)=-\frac{\pi^0}{\sqrt{2}}.
$$
We obtain $T=\begin{pmatrix}\frac{\pi^0}{\sqrt{2}}&\pi^+\\ \pi^-&-\frac{\pi^0}{\sqrt{2}}\end{pmatrix}$ as usual for $SU(2)$ triplets like the vector bosons in SM.

I'm trying to do the same for the $\Delta$ baryons if possible. The problem is that it is a quartet and i need to express it as a tensor. I did not find any reference in order to define a $SU(2)$ quartet in a tensor representation and I could not do it on my own as in the previous case.
Any help please?

Hint: Recall that the $\Delta$ baryons are constructed from three quarks: $\Delta^{++}=uuu,\,\Delta^+=uud,\,\Delta^0=udd,\,\Delta^-=ddd$. Besides, the quartet representation comes from $2\otimes 2\otimes 2=2\oplus 2\oplus 4$.

 

[samalkhaiat]

Summary:: I'm looking for a tensor representation for a $SU(2)$ quartet in order to build a $\Delta$ baryon states from quark multiplications. I know that it is possible for $\pi$ triplet and it is shown in the thread.$\Delta^{++}=uuu,\,\Delta^+=uud,\,\Delta^0=udd,\,\Delta^-=ddd$. Besides, the quartet representation comes from $2\otimes 2\otimes 2=2\oplus 2\oplus 4$.

Notations: q_{i} \in [2], \ q_{1} = u, \ q_{2} = d, S_{ij} \equiv q_{(i}q_{j)} = q_{i}q_{j} + q_{j}q_{i}, q_{[i}q_{j]} = q_{i}q_{j} - q_{j}q_{i}. Now consider the tensor q_{i}q_{j} \in [2] \otimes [2] and decompose it into irreducible parts (i.e., symmetric and anti-symmetric combinations) q_{i}q_{j} = \frac{1}{2} q_{(i}q_{j)} + \frac{1}{2} q_{[i}q_{j]}, or q_{i}q_{j} = \frac{1}{2} S_{ij} + \frac{1}{2} \epsilon_{ij}\left(\epsilon^{kl}q_{k}q_{l} \right) . This is nothing but the tensor form of the Clebsch-Gordan series [2] \otimes [2] = [3] \oplus [1] . We now make the following observation about the irreducible representations of SU(2):

1) The 1-dimensional irrep [1] is carried by the scalar or 0-rank tensor S = \epsilon^{ij}q_{i}q_{j}.

2) The 2-dimensional (fundamental or defining) irrep [2] is carried by the vector or rank-1 tensor q_{i}.

3) The 3-dimensional irrep [3] is carried by the symmetric rank-2 tensor S_{ij}.

Thus, it is clear from the above that the 4-dimensional irrep [4] must be carried by totally symmetric rank-3 tensor, call it \Delta_{ijk} = \Delta_{(ijk)}. Indeed, any irrep [n] of SU(2) is carried by totally symmetric tensor of rank (n-1).

Okay, let us now look at \Delta_{ijk} \in [4]. To obtain a 3-index tensor we can multiply our symmetric tensor S_{ij} \in [3] by the vector q_{k}. However, the resultant tensor S_{ij}q_{k} \in [3] \otimes [2] is reducible because it is not symmetric with respect to all indices (ijk). So, we need to decompose it into irreducible parts (i.e., subtract all invariant subspaces). We do that by the following trick of rewriting the tensor S_{ij}q_{k}:

\begin{align*}S_{ij}q_{k} &= \frac{1}{3} \left( S_{ij}q_{k} + S_{jk}q_{i} + S_{ki}q_{j}\right) \\ & + \frac{1}{3}\left( S_{ij}q_{k} - S_{jk}q_{i}\right) \\ & + \frac{1}{3} \left( S_{ij}q_{k} - S_{ki}q_{j}\right) . \end{align*} The first line is what you are after, i.e., the totally symmetric tensor \Delta_{(ijk)} which carries the irrep [4], as it has only 4 independent components. The second line is anti-symmetric under i \leftrightarrow k and, therefore, can be written as \epsilon_{ik}Q_{j} where Q_{j} \equiv S_{jl}q^{l}. Similarly, the third line can be written as \epsilon_{kj}Q_{i}. So we find S_{ij}q_{k} = \Delta_{(ijk)} + \epsilon_{k(i} \hat{Q}_{j)}, \ \ \ \ (1) where \hat{Q}_{i} \equiv - \frac{1}{3} S_{ik}q^{k} \in [2], and \Delta_{(ijk)} = \frac{1}{3} \left( S_{ij}q_{k} + S_{jk}q_{i} + S_{ki}q_{j}\right) . \ \ \ \ (2) Equation (1) is nothing but the tensor form of the Clebsch-Gordan series [3] \otimes [2] = [4] \oplus [2] , and Eq(2) gives you the four Iso-spin (\frac{3}{2}) states in [4]: \Delta^{+2} = \Delta_{(111)}, \ \Delta^{+} = \sqrt{3}\Delta_{(112)}, \ \Delta^{0} = \sqrt{3} \Delta_{(122)} and \Delta^{-} = \Delta_{(222)}. You will also see these states appear naturally in the irrep [10] of SU(3).