# Do particle and antiparticle annihilate when they meet?

Do particle and antiparticle annihilate when they meet? As we know, an electron and a positron will annihilate when they meet. However, many quarks and antiquarks do not annihilate, but coexit as mesons. For example, a neutral pion $\pi^{0}$ is made up of $u, \overline{u}, d, \overline{d}$; a meson $\eta$ is made up of $u, \overline{u}, d, \overline{d}, s, \overline{s}$; a meson $\eta_{c}$ is made up of $c, \overline{c}$; a meson $\phi$ is made up of $s, \overline{s}$; and a meson $\Upsilon$ is made up of $b, \overline{b}$ . Why do these quark-antiquark pairs not annihilate?

• Those are bound states. – Turgon Apr 7 '17 at 6:11
• @Turgon- Why do particle-antiparticle not annihilate in bound states? – Shen Apr 7 '17 at 8:32

Mesons are unstable and decay , either through the weak decay of a quark in the charged cases, or because quark + antiquark annihilate in the neutral through the strong interaction. The strong interaction allows also for other decays, with no annihilations by the ability of gluons to generate a quark antiquark pair and create new resonances or particles ( example phi to K+K-) if the energy is enough and quantum numbers conservation allow it .

The pi0 can only decay electromagnetically, quark annihilating on antiquark and giving two photons, because there is nothing composed of quarks it can decay to through the strong interaction. Example the phi, which has a branching ratio to s s_bar and can decay to rho pi, i.e. the s annihilates on the s_bar, (it also goes into K+K-). The decay is very fast because it is a strong interaction one.

In the interactions, they form a temporary bound state . Similar to the positronium bound state, but much more complicated because of the strong interaction. Given the appropriate conditions a bound state exists temporarily, but there is a probability of annihilating from an overlap of the wavefunctions. When looking at the Y decays (page 147) there are so many channels open by quantum numbers and energy so some decays carry the b b_bar through, but the dominant involve an annihilation.

BTW within a proton, quark antiquark pair are continuously created and annihilated, for example.

• a google search for feynman diagrams for Ypsilon decays is enlightenting google.gr/…: – anna v Apr 11 '17 at 4:19

The meson you describe are made up of a quark and its antiquarks and are called quarkonia. Of course, all mesons are unstable, but they do not necessarily decay via annihilation.

When you say "annihilate", you probably think of a process $$q\bar q \longrightarrow \text{single intermediate particle}\longrightarrow \text{decay products}\,.$$

However, this process is forbidden fpor QCD, where the intermediate particle would be a gluon, because the meson is colour-neutral. Hence, only electroweak decays are possible (via $\gamma$ or $Z$). Depending on the state, there can be additional conserved quantities (spin, parity) that prohibit a direct annihilation of this type -- for example, the $c\bar c$ state $\eta_c$ has $J=0$, so it cannot decay into an intermediate vector boson. And finally, even when annihilation is possible, it might not be the dominant decay mode, as direct electroweak annihilation competes with QCD loop processes or weak decays of heavy quarks.

(Note that generally, meson properties are quite hard to compute because of the nonperturbative nature of QCD at low energies.)