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What is the precise statement of the OZI Rule?

I've heard that a diagram is OZI suppressed if it can be "cut in two by cutting only gluon lines", but I don't really understand.

For example, consider the decay $\phi \to K^+ K^-$, which is supposedly not OZI suppressed. This diagram is from Griffiths:

phi meson decays into a pair of pions

Well, it seems to me that I can cut this diagram in half. I just have to insert the scissors in between the u and s quark on the bottom left, snip both the gluon lines, then exit between the u and s quark on the top right.

Or are we not allowed to separate two quarks in the same hadron?

What about a diagram where one of the quarks in a hadron emits a gluon and that gluon then decays to say a $\pi^0$? Would that be OZI suppressed?

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  • $\begingroup$ shouldn't particles, that move leftwards, be anti-particles (the lower strange-quark here)? $\endgroup$
    – Ben
    Aug 19, 2019 at 20:16

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A better way to phrase the rule is that a diagram is OZI-suppressed if you can arrange it so that there's some time at which the quarks annihilate and all of the energy/momentum is carried by gluons. Graphically, this is equivalent to being able to draw a vertical line (or curve), the "cut," such that all the initial state particles are on the left side of the cut, all the final state particles are on the right side, and the cut goes through only gluon lines. Hopefully you can see why the diagram in your question doesn't fit these criteria. (i.e. no, you're not allowed to separate two quarks in the same hadron, but it's more than that.)

The reason for this OZI rule, by the way, is that the interactions of high-energy gluons are relatively weak (because of the running of the strong coupling). So if the gluons are stuck carrying all the energy, their probability of interaction will be relatively low. But as long as there are quarks around, so that some of the energy is tied up in the quarks' masses, the gluons can have low energy and thus retain a high interaction strength.

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  • $\begingroup$ So the second example I gave, where one of the quarks in a hadron emits a gluon and that gluon decays to $\pi^0$, is that also not OZI suppressed? Because the quarks lines in the original hadron are continuous all the way across $\endgroup$
    – Brian Bi
    Oct 22, 2013 at 7:01
  • $\begingroup$ Yep, that sounds right. $\endgroup$
    – David Z
    Oct 22, 2013 at 7:03
  • $\begingroup$ But, in that case, isn't the gluon forced to carry enough energy to produce the pion? $\endgroup$
    – Brian Bi
    Oct 22, 2013 at 7:13
  • $\begingroup$ Yes, so that particular Feynman diagram might have a relatively low contribution. But there are probably going to be other Feynman diagrams for the same process ($h\to h\pi$) where the pion is produced in a different way, that doesn't require a high energy gluon. The OZI rule was kind of invented for the case of an entire process, where energy has to "pass through" the gluons in all the contributing Feynman diagrams. (And besides, the pion mass isn't really that high anyway.) $\endgroup$
    – David Z
    Oct 22, 2013 at 7:22

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