I am looking into the decays of quarkonia, specifically charmonium and bottomonium bound states. The decay widths depend on the quantum numbers $n^{2S+1}L_J$ of the states and the type of force mediating the decay. For example, electromagnetic transitions are related to the fine structure constant $\alpha$, while hadronic transitions to the strong coupling constant $\alpha_s$ from QCD. One can therefore relate the experimentaly found decay widths to the strong coupling constant, which depends on the mass of the state.

The end goal is to produce a rough plot of $\alpha_s$, calculated from the decay widths, versus mass of the quarkonium state. I would like to be able to compare my plot to an analytical expression. As quarkonia are produced in $e^- e^+$ collisions, the mass of the state created depends on the centre of mass energy in the collision, $s$, through (in natural units): $$\sqrt{s} = 2m_q \,.$$ The problem is, I can only find analytical expressions in terms of $Q^2=-q^2$, the negative of 4-momentum transfer (link): $$\alpha_s(Q^2) = \frac{12\pi}{ \left(33-n_f \right)\ln \left(\frac{Q^2}{\Lambda^2}\right)} , $$ where $n_f$ - number of quark flavours and $\Lambda$ - QCD scale (about $250 \, MeV/c$).

Hence, I have a way of relating the quarkonium mass to $s$, but not to $Q^2$. Is there a way to relate $Q^2$ to $\sqrt{s}$ in $e^- e^+$ collisions? Or how else could I check if my values agree with theory?

  • $\begingroup$ You give no link for the formula $\endgroup$
    – anna v
    Commented Mar 10, 2018 at 6:59
  • $\begingroup$ Sorry, this was my first question ever. The formula for $\alpha_s$ as I state it is given in the textbook Particles and Nuclei by Povh et al., but also in a paper Heavy-Quark Systems by Kwong, Rosner, Quigg (1993). $\endgroup$
    – dzejkob
    Commented Mar 10, 2018 at 11:55
  • $\begingroup$ The explicit asymptotic freedom formula you give is for space like momentum transfers (scattering), whereas the onium decay couplings pertain to timelike gluons---always more than two! The two are related, and "similar" as a starting point, but review articles on onia often outline the systematics you envision... $\endgroup$ Commented Mar 10, 2018 at 14:31
  • $\begingroup$ However, if I recall right, the onium decays depend quite sensitively on the overlap wave functions at the origin, highly sensitive to s , more s than the coupling... $\endgroup$ Commented Mar 10, 2018 at 14:36

1 Answer 1


In the copy of the paper here describing the parameters of the formula you quote

.... nf is the number of fermion flavors with mass below Q,

it treats Q as an energy scale, as happens with the running coupling constant plots. In fig 4 it shows that timelike and space like couplings approach each other. In figure 8 there is comparison with data which extends up to the center of mass energies in experiments up to the time of publication.

In a sense you are trying to see the running of $α_s$ over the width of the resonance, so you could use the formula as is, imo.

I will try to alert Cosmas Zachos to your question, in case I am making wrong assumptions.

  • $\begingroup$ Looks reasonable to me. The more modern state of the art incorporates all such effects in more detail, but the picture is less simple. $\endgroup$ Commented Mar 10, 2018 at 17:28
  • $\begingroup$ And it looks like fig 8a of your Prosperi et al. ref actually carries out part of the OP's project... $\endgroup$ Commented Mar 10, 2018 at 17:46
  • $\begingroup$ @CosmasZachos yes, except they are points, not scans of the width $\endgroup$
    – anna v
    Commented Mar 10, 2018 at 18:09
  • $\begingroup$ Thank you both. Indeed, Fig. 8 is precisely what I'd like to obtain, of course very roughly and with only a few data points. The method I'm following is the one described in the paper linked by @annav As a side question, how do I find out the decay widths/branching ratios for hadronic decays into 2 gluons? The data given here specifies a hadronic channel for the $J/ \psi$ meson, but not for any other charmonium state. The paper linked above states the number without explanation. $\endgroup$
    – dzejkob
    Commented Mar 10, 2018 at 18:19
  • $\begingroup$ maybe this has references arxiv.org/abs/1001.0848 $\endgroup$
    – anna v
    Commented Mar 10, 2018 at 18:52

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