Delta Baryon Quartet: Constructing from SU(2) Symmetry

[lalo_u]

TL;DR 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.

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 $SU(3)$.