# How to Derive Baryon Octet Matrix?

In every textbook on particle physics that I've read, I encounter the following matrix when reading about matrix of wave functions for the baryon octet:

$$\left[\begin{array}{ccc} \frac{\Sigma^0}{\sqrt{2}} + \frac{\Lambda}{\sqrt{6}} & \Sigma^+ & P \\ \Sigma^- & -\frac{\Sigma^0}{\sqrt{2}} + \frac{\Lambda}{\sqrt{6}} & N \\ \Xi^- & \Xi^0 & -\frac{2\Lambda}{\sqrt{6}} \end{array}\right],$$

but thus far, no author has bothered to explain where the entries of this matrix come from. I understand that the baryons transform as an $SU(3)$ flavour octet, and I understand how the octet decomposes under restriction to the isospin $SU(2)$ subgroup, but I have no idea where the above matrix comes from. How does one derive it?

The baryon octet comes out of the triple tensor product of the triplet, $$\boldsymbol{3}\otimes\boldsymbol{3}\otimes\boldsymbol{3}=\boldsymbol{8}\oplus\dotsm\,,$$ (see also here) where you can think of the triplet as $\boldsymbol{3}=(u,d,s)^T$. The octet is of mixed symmetry - one way to explicity write out the quarks content is to first constructing the antitriplet in $\boldsymbol{3}\otimes\boldsymbol{3}=\bar{\boldsymbol{3}}\oplus \boldsymbol{6}$, which is the antisymmetric combination ($\bar{\boldsymbol{3}}^i=\frac{1}{2}\epsilon^{ijk}\boldsymbol{3}_j\boldsymbol{3}_k$): $$\bar{\boldsymbol{3}}=\left([ds],[su],[ud]\right)\,.$$ Here $[ds]=\frac{1}{2} (ds-sd)$. Now the tensor product becomes $$\bar{\boldsymbol{3}}\otimes\boldsymbol{3} =\begin{pmatrix}u\\d\\s\end{pmatrix}\otimes \left([ds],[su],[ud]\right)\big|_\text{traceless} = \left.\begin{pmatrix}u[ds] & u[su] & u[ud]\\d[ds]& d[su] & d[ud]\\s[ds] & s[su] & s[ud]\end{pmatrix}\right|_\text{traceless}\,,$$ whence you can read off the quark content of the entries. The remaining tricky part is the precise identification of the diagonals so that the octet is traceless and the wavefunctions are properly normalised.
The symmetry of the full state is really more involved (flavour, colour, spin, spatial), and needs to be antisymmtric overall since the quarks are fermions. The colour state is antisymmetric as well (i.e. colourless according to $SU(3)_\text{colour}$). Hence the remaining state has to be symmetric in flavour$\times$spin$\times$spatial part. Ground states have orbital angular momentum $L=0$, so the spatial part is symmetric as well. Thus, flavour$\times$spin has to be symmetric.
The trace of the matrix above corresponds to the totally antisymmetric flavour combination of $uds$, so the spin state would have to be totally antisymmetric as well, but this is not possible - hence, the trace (singlet) correspond to an excited state (of zero isospin, i.e. some higher $\Lambda$). Now all that remains is to identify the other two combinations on the diagonal. Here, rather than going through the details, we can use the factr that the isospin $SU(2)$ acts on the first two rows/columns in this formulation: Hence, the $(33)$ entry has $I=0$, while the $(11)$ and $(22)$ parts are combinations of $I=0$ (singlet -- unit matrix in $(12)$-space) and $I=1$ (triplet -- diagonal $\sim\sigma^3$). The isosinglet is called $\Lambda$, the triplet is $\Sigma^0$ (well, the neutral part of the triplet with $\Sigma^\pm$). Overall tracelessness then imposes the diagonal $$\begin{pmatrix} a\Sigma^0 +b\Lambda &&\\&-a\Sigma^0 +b \Lambda&\\&&-2b\Lambda\end{pmatrix}\,,$$ and the coefficients $a$ and $b$ are finally fixed by requiring the normalisation $$\operatorname{tr} B^\dagger B =1$$ for each state.
• @Toffomat Also, isn't the andtisymmetric combination $ds - sd$, rather than $ds + sd$? Apr 24 '17 at 16:27
• @Kristoll: Of, course, antisymmetry is with $-$, I've corrected that. As for your other comment: I'll try to write something later today if I have time. Apr 25 '17 at 8:05