Why does charmonium (and phi mesons) not decay via quark and antiquark annihilation?

The decay of heavy quark/antiquark pairs (say $$c\bar{c}$$, $$s\bar{s}$$) is supposedly 'suppressed because of the Zweig/OZI rule', see for instance Phi meson.

And they certainly have a longer lifetime than expected. However, the Zweig suppression only comes into play because

• a) We expect these mesons to decay into other mesons

• b) For the low mass states (e.g. the ground state of charmonium) it is kinematically not possible for it to decay into other mesons with a c quark, hence why you would not get J/psi decaying into D mesons via the weak interaction. Instead it would decay into pions by the Zweig rule.

My question is: why do we not consider simply the annihilation of the quark and antiquark, either to a gluon and subsequent quark pair production (say to an up, antiup which would be kinematically favourable and conserve angular momentum and parity) or the same via a photon to leptons or quarks?

EDIT: A great response explained that annihilation to a gluon is impossible because it does not conserve colour. However I have just seen Feynman diagrams on google:

which has this exact process occurring! And it also has a free gluon (in the final diagram) which is not possible for a non-colourless gluon, i.e. one that actual interacts/exists?

• As for the diagrams in the edit, the difference to the quark antiquark pair in the phi or charmonium, is that they can carry any color to balance the color of the exchange, It is when they are bound into a colorless hadron that the color counting is constraining. see hyperphysics.phy-astr.gsu.edu/hbase/Forces/feyns.html – anna v May 30 at 12:59
• According to the Zweig-rule, the decay of J/Psi into three pions is suppressed?! – Ben Aug 19 at 13:04

That is more or less what we do, except...

Annihilation through a photon is electromagnetic rather than strong, so this is a suppression.

Annihilation through a single gluon can't happen because the gluon has a colour+anticolour charge, whereas the initial meson is colourless. The gluon has to have a charge like red-antiblue and you can't make that from a red-antired quark-antiquark pair.

If you annihilate through two gluons you can make the colour charges balance - but not the charge conjugation property. This is odd (C=-1) for the gluon, and therefore even (C=+1) for two gluons.

So the annihilation has to take place through 3 gluons. That means the decay rate is proportional to $$\alpha_s^6$$. Even though $$\alpha_s$$ is not as small as $$\alpha_{em}$$, the extra powers are still enough to suppress the decay - particularly for high mass particles ($$\Upsilon$$ and $$\psi$$ rather than $$\phi$$) where the running of $$\alpha_s$$ makes it even smaller.

• Thank you very much for your clear reply. I was wondering if you could explain the images I have found on google which seem to show this quark annihilation to gluon process, within the context of your answer? Thank you very much! – Meep May 30 at 10:46
• These show the production of $t$ and $\overline t$ quarks which have different colours. (Could be any quark flavour, doesn't have to be $t$.) These diagrams are part of a bigger picture which will include the hadronisation of the quarks produced, involving lots of gluon interactions such that they end up as colourless mesons or baryons. These interactions occur at low energy scales so $\alpha_s$ is large. – RogerJBarlow May 30 at 10:52
• The gluons in the pictures on the bottom line likewise are just part of a picture: the full scenario would show two colourless protons as $r-g-b$ coloured quarks interacting and producing coloured gluons, two of which are the ones shown – RogerJBarlow May 30 at 10:56
• Annihilation through a single gluon can't happen because the gluon has a colour+anticolour charge, whereas the initial meson is colourless. Isn't this also forbidden simply by conservation of energy-momentum? If you annihilate through two gluons you can make the colour charges balance - but not the parity+angular momentum. Could you explain more about why this is? The two gluons, each with intrinsic spin 1, can couple to make spin 0, 1, or 2 (or higher values as well, incuding orbital angular momentum). If the parity is negative, then why can't the gluons be in a $+\otimes -$ state? – Ben Crowell May 30 at 12:27
• These are virtual processes so the laws of conservation of energy and momentum can be violated in parts of the picture provided they hold overall: the intermediate gluon(or gluons) is off shell. – RogerJBarlow May 30 at 12:45