In $\rm SU(2)$, taking up quark and down quark as a doublet we can easily apply the isospin ladder operators to write the combination of 2 quark or 3 quark (baryon) systems. In $\rm SU(3)$ quark model, to get light pseudoscalar mesons, we need to combine a triplet and antitriplet to form an octet and singlet. But how to explicitly write down the states?

E.g. the singlet state is $$|\eta’\rangle = \frac{|u \bar u\rangle + |d\bar d\rangle + |s \bar s\rangle }{\sqrt{3}}$$ It can be verified that this is indeed a singlet by operation of $\hat{T_{\pm}}|\eta‘ \rangle=0$, where $\hat{T}_{\pm}$ are the isospin ladder operators. From the condition that it should be a $Y=0,T_3=0$ state, we can find that the terms are linear combination of $|u \bar u\rangle ,|d\bar d\rangle$ and $|s \bar s\rangle$.How to find the coefficients?

In $\rm SU(2)$ the singlet state could be found by allowing orthogonality with the triplet. So the problem becomes evaluating the quark compositions for all the octet states, so that we can find the singlet by orthogonality. The quark composition at the vertices of the meson hexagon in the eightfoldway weight diagram of the pseudoscalar mesons are easy, but how to get those at the center?

My approach: By applying ladder operators we get 6 linearly dependent states since there are 6 ladder operators $T_{\pm},U_{\pm},V_{\pm}$, but we should get 2 states, because we already got 6 at the vertices of the hexagon, to complete octet we need 2 more.

In general how to obtain all the quark composition of flavour states in the nonet systematically, and how to do the same for vector mesons preferably without invoking QCD?


In point of fact the 3 central members of octets (+singlet $\leadsto$ nonets) are not the ideal states you find in the pseudo scalars, as QCD effects weird mixings: a very different question. But the pseudoscalars are ideal and easy and the ladder method you have in mind of course works.

You got the six outside pseudoscalars, so let us focus on the $|\pi^+\rangle = |u\bar{d}\rangle$ and $|K^+\rangle=|u\bar{s}\rangle$. Application of $T_-$ on $|\pi^+\rangle$ yields the neutral member of the isotriplet,
$$|\pi^0 \rangle = \frac{|u\bar{u} \rangle- |d\bar{d}\rangle}{\sqrt{2}},$$ which you may likewise lower to the third isotriplet member $|\pi^-\rangle = |d\bar{u}\rangle$.

Now, there are two more combinations with the same quark content orthogonal to that $|\pi^0 \rangle$: both isosinglets, $$|\eta'\rangle = \frac{|u \bar u\rangle + |d\bar d\rangle + |s \bar s\rangle }{\sqrt{3}}\\ |\eta\rangle = {\frac{|u\bar{u}\rangle + |d\bar{d}\rangle - 2|s\bar{s}\rangle}{\sqrt{6}}} , $$ corresponding to the traceful SU(3) singlet I, and traceless $\lambda_8$, respectively.

You are asking how to determine the relative coefficients of their summands. Both are annihilated by $T_+$; but only one is annihilated by $V_+$, which sends an s to a u, and the converse for their conjugates with a minus sign, $$ V_+|\eta'\rangle=0, \qquad V_+|\eta\rangle=|K^+\rangle . $$

So you can see the η' is a Τ,U,V singlet, i.e. an SU(3) singlet, as stated, and the η, the state orthogonal to the other two, is an isosinglet, but still firmly in the octet: it connects to four outer states of the octet by suitable raising and lowering operators, as illustrated. That's why it corresponds to the traceless Gell-Mann matrix mentioned. Convince yourself these are the only coefficient arrangements with these properties.

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  • $\begingroup$ Yes, I am convinced that these are unique. Thanks for the details but you wrote down the states and then verified. How did the first person guess the coefficients which u wrote for eta? $\endgroup$ – Manas Dogra Oct 4 '20 at 19:30
  • $\begingroup$ Well, as I said, given $\pi^0$ and $\eta'$, the unique linear combination of the same constituents orthogonal to both is the $\eta$. It's linear algebra. $\endgroup$ – Cosmas Zachos Oct 4 '20 at 19:36
  • $\begingroup$ I am so sorry to confuse between eta and eta primed. That above comment would have been--- If i get eta primed,i will get eta by orthogonality with eta primed and neutral pion and normalization condition.To get eta prime's coefficients----I need three coefficients so 3 equations...one is normalization,another is orthogonality with $|\pi_0 \rangle$,what's the other one? $\endgroup$ – Manas Dogra Oct 4 '20 at 19:40
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    $\begingroup$ As indicated, you get $\pi^0$ by $T_- \pi^+$. You get $\eta'$ as the unique SU(3) singlet annihilated by all 8 of its generators. You get $\eta$ as the state consisting of their 3 pieces but orthogonal to these two. $\endgroup$ – Cosmas Zachos Oct 4 '20 at 19:40
  • $\begingroup$ Now its clear. Thank you. Just don't believe I got a chance to talk with you...great fan. $\endgroup$ – Manas Dogra Oct 4 '20 at 19:40

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