Consulting the list of pseudoscalar mesons, we found that for charm and bottom quarks there are two mesons with quark content given by:

$$\eta_c=c\bar{c},\qquad \eta_b = b\bar{b}$$

on the other hand, for light quarks, we only have:

$$\pi^0=\frac{u\bar{u}-d\bar{d}}{\sqrt{2}}, \qquad \eta' \sim \frac{u\bar{u}+d\bar{d}+s\bar{s}}{\sqrt{3}}, \qquad \eta \sim \frac{u\bar{u}+d\bar{d}-2s\bar{s}}{\sqrt{6}}$$

It is not entirely clear to me why different pseudoscalar mesons such as $$\eta_u = u\bar{u}, \qquad \eta_d = d\bar{d}, \quad \text{or} \quad \eta_s = s\bar{s}$$ do not exist instead of the above.

My attempt: I understand that the fact that the masses of the two lightest quarks are practically the same ($m_u \approx m_d$) leads to the fact that $\eta_u$ and $\eta_d$ do not exist separately, and, instead, we have a superposition like $\pi^0=(u\bar{u}-d\bar{d})/\sqrt{2}$. In the same vein, I assume that $m_s$ is not so different from $m_u \approx m_d$, and because of these we have mesons like $\eta$ and $\eta'$ but here we have combinations with $+$ and $-$ (in the case of $\pi^0 = (u\bar{u}-d\bar{d})/\sqrt{2}$, we only found a combination with $-$. What are the reasons for the no existence of the missed combinations?

  • $\begingroup$ ...but you have learned about ω-φ mixing in vector mesons, etc... $\endgroup$ Commented Jun 5, 2023 at 20:37
  • 1
    $\begingroup$ It's basically an advanced QCD topic mystery: Godfrey, S & Isgur, N (1985) "Mesons in a relativized quark model with chromodynamics" Physical Review D32 (1), 189. $\endgroup$ Commented Jun 5, 2023 at 21:17
  • 1
    $\begingroup$ Linked. $\endgroup$ Commented Jun 5, 2023 at 21:26

1 Answer 1


You have a slight normalisation mistake between $\eta$ and $\eta^\prime$, (this has since been edited out of the question), but that is a small issue. By combining them, you can get $\eta_s$ and $\dfrac{u\bar u+d\bar d}{\sqrt2}$, and so, actually, you have already spanned the vector space of possibilities.

It is just about computational convenience. Much less computational effort if we keep track of these variables than if we worked with the conceptually cleaner version you want.

  • 3
    $\begingroup$ He is asking why it is these states that are mass eigenstates. They are not a physically irrelevant convenient basis! $\endgroup$ Commented Jun 5, 2023 at 20:39
  • $\begingroup$ I would not agree with the characterisation that anyone is saying that they are "a physically irrelevant basis". A basis just needs to span the vector space, and oftentimes it is useful to have something trivial to state correctly. Hence why we deal with LS basis for H atom, as opposed to J basis, even though the latter is the one that commutes with the Hamiltonian. No part of the question also asked about mass eigenstates. I am totally lost over where you are reading that from. I answered his "What are the reasons for the no existence of the missed combinations?" directly. $\endgroup$ Commented Jun 5, 2023 at 20:45
  • $\begingroup$ Of course redundancy of states is to be avoided, but: The heart of his question is why these combinations are the physical mass eigenstates for pseudo scalars -- and he ought to extend it to: why are they different for vectors, etc... It is a deep mystery, Feldmann, T. (2000) "Quark structure of pseudoscalar mesons", International Journal of Modern Physics A15 (02), 159-207, and just counting states to prevent over counting leaves the OP puzzled, I gather... His title asks about the missing states in lieu of the extant states! $\endgroup$ Commented Jun 5, 2023 at 21:05
  • 1
    $\begingroup$ I am not denying that your version of the question is more physically meaningful, but I am definitely not reading that from the way he posed the question. I think he is less likely to be thinking about your question, and more likely to just be confused over why it was not defined in the simple way. $\endgroup$ Commented Jun 5, 2023 at 22:09
  • $\begingroup$ I would let him clarify if he is asking a question on elementary linear algebra instead of "why mesons mix like that". His preamble on charmonium and bottomonium suggest the latter, but there is no point in arguing, if he is happy with linear algebra. $\endgroup$ Commented Jun 5, 2023 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.