# If a particle and its antiparticle annihilate upon contact, how do they form bound states?

I am reading about some old discoveries in particle physics and early collider experiments from Perkin's Introduction to High Energy Physics. However, I didn't get the answers to all my questions.

If two beams of electrons and positrons are collided head-on, the collision can produce various quark-antiquark meson states called quarkonium. For example, $$e^-e^+$$ annihilations have produced various $$c\bar{c}$$ and $$b\bar{b}$$ meson bound states came to be collectively known as charmonium and bottomonium respectively. The most stable charmonimum is $$J/\psi$$ and bottomonium is $$\Upsilon$$. Toponium states do not exist since top quarks decay too fast to form mesons. The also exists bound states of $$e^+e^-$$ and $$\mu^+\mu^-$$, respectively called positronium and muonium.

If particles and antiparticles annihilate each other how can there be a bound state of them in the first place?

• Have you read up on positronium and muonium? NR bound states ensure the UP provides the usual QM repulsive pressure for the formation of such "atoms"; but their constituents decay weakly at vastly different rates. – Cosmas Zachos Oct 26 '19 at 13:55
• @CosmasZachos Thanks for the link on muonium and for the comment on its stability.. Any idea about how one makes a NR bound state of $e^-e^+$ who constituents are stable? When $e^-e^+$ annihilates and when it makes a bound state? – mithusengupta123 Oct 26 '19 at 14:21
• The article on positronium details that. Production mechanisms require a QED calculation, I suspect... The decay to 3 photons is detailed in WP and refs therein. – Cosmas Zachos Oct 26 '19 at 14:51