Upsilon mesons are bound states of b-quark and anti-b-quark, of $J^{PC} = 1^{– –}$. First three states are located below $B\bar{B}$ threshold so they decay either via OZI-suppressed strong transitions to light hadrons, or via electromagnetic transitions (dilepton decays, etc). Notably, they can also decay to lower bottomonium states, for example, $\Upsilon(2S)$ can decay to $\Upsilon(1S) \pi \pi$.

There is apparently nothing which is different for these states, except that with their mass getting larger, more decay channels open up. Therefore, naively, I would expect that the first state, $\Upsilon(1S)$, should be the narrowest of all them (in other words, should have the longest lifetime), while the width of the $\Upsilon(2S)$ should be larger, and that of the $\Upsilon(3S)$ larger again.

This assumption, however, seems to contradict to the experimental data from the PDG:

Particle Mass (MeV) Full width (keV)
$\Upsilon(1S)$ 9460.30±0.26 54.02±1.25
$\Upsilon(2S)$ 10023.26±0.31 31.98±2.63
$\Upsilon(3S)$ 10355.2±0.5 20.32±1.85

So, the higher-mass states actually get narrower, which looks counter-intuitive! This controversy is absent in the charmonium system:

Particle Mass (MeV) Full width (keV)
$J/\psi(1S)$ 3096.900±0.006 92.9±2.8
$\psi(2S)$ 3686.10±0.06 294±8

where the second state is wider that the first state, as expected. (The third one is above the $D\bar{D}$ threshold so useless for this discussion.)

Question: What is the reason for such inverted behavior of the Upsilon system?

My hypothesis was falling back to the OZI rule, predicting that higher-mass mesons would annihilate into gluons of higher energy, which end up having lower running coupling constant $\alpha_s$ (the general trend that Upsilons are narrower the Psi's could point in this direction, too). This would however not explain why the effect is absent in the charmonium system; and why the presence of decays to lower $b\bar{b}$ resonances does not compensate for that effect.

[I tried to search for papers on the topic, but found only discussions of the dielectron decay width (not the full one), or of the higher states above the $B\bar{B}$ threshold.]


1 Answer 1


I think it is misleading to look at the total width, because the breakdown into partial widths is quite different.

One thing to compare is the $3g$ width (basically the total hadronic width). According to PDG $$ \Gamma(\psi(2s),3g)=299 keV\cdot 10\% = 30 keV $$ $$ \Gamma(J/\psi(1s),3g)=93 keV\cdot 66\% = 60 keV $$ The same pattern is seen in Upsilon states $$ \Gamma(\Upsilon(3s),3g)=20 keV\cdot 36\% = 7.2 keV $$ $$ \Gamma(\Upsilon(2s),3g)=32 keV\cdot 59\% = 19.2 keV $$ $$ \Gamma(\Upsilon(1s),3g)=54 keV\cdot ?\% = ? $$ I'm not sure why PDG does not attempt to give a $3g$ width for the $1s$ Upsilon. The results seem consistent with the perturbative estimate $\Gamma\sim \alpha_s^3 |\psi_n(0)|^2$ which we expect to be smaller for larger $n$ because the wave function at the origin is smaller.

What increases the total $\psi(2s)$ width is a large branching ratio into $J/\psi (\pi\pi)$. This is a different kind of decay, governed by non-perturbative QCD multipole transition. There is some theory for these transitions, see, for example, here. I'm not an expert, but it makes sense to me that the corresponding matrix elements would be larger in the $J/\psi$ compared to the $\Upsilon$.

  • $\begingroup$ Thanks! Could you please elaborate on the "we expect to be smaller for larger 𝑛 because the wave function at the origin is smaller" argument? $\endgroup$
    – Martino
    Commented Feb 6, 2021 at 10:02
  • $\begingroup$ Higher radial wave functions are more extended, so the wave function at the origin is smaller. For the Coulomb problem $\psi_n(0)|^2$ can be found in QM text books. $\endgroup$
    – Thomas
    Commented Feb 6, 2021 at 16:55

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.