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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?

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  • $\begingroup$ 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. $\endgroup$ – Cosmas Zachos Oct 26 '19 at 13:55
  • $\begingroup$ @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? $\endgroup$ – mithusengupta123 Oct 26 '19 at 14:21
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    $\begingroup$ 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. $\endgroup$ – Cosmas Zachos Oct 26 '19 at 14:51
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A bound state is possible for awhile because the particle and antiparticle are separated in space and not significantly “in contact”. You can think of them as orbiting each other; their kinetic energy and angular momentum keep them apart. But, just as in a hydrogen atom, they are really described by wavefunctions. Eventually they annihilate because their wave functions do overlap a bit and there is a small probability that the particle and antiparticle are at the same point at the same time. So the bound state has a probability to decay and has a finite average lifetime.

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    $\begingroup$ I'm not sure, but might be slightly better to say that the wavefunction of the system displays a non-zero probability density that the two particles exist at the same point. $\endgroup$ – garyp Oct 26 '19 at 18:10

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