Delta baryon quartet?

  • #1
lalo_u
Gold Member
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0

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##.
 
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Answers and Replies

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

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

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

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

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

Okay, let us now look at [itex]\Delta_{ijk} \in [4][/itex]. To obtain a 3-index tensor we can multiply our symmetric tensor [itex]S_{ij} \in [3][/itex] by the vector [itex]q_{k}[/itex]. However, the resultant tensor [itex]S_{ij}q_{k} \in [3] \otimes [2][/itex] is reducible because it is not symmetric with respect to all indices [itex](ijk)[/itex]. 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 [itex]S_{ij}q_{k}[/itex]:

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