- #1

lalo_u

Gold Member

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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.

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##.

$$q_iq_j^*=\left(q_iq^j-\frac{1}{2}\delta_j^iq_kq^k\right)+\frac{1}{2}\delta_j^iq_kq^k.$$

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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