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I am wondering about electron-positron annihilation. At https://www.princeton.edu/~romalis/PHYS312/Positron.pdf, I read:

During this slowing down process, which is similar to what would be experienced by any charged particle, annihilation of the positron is unlikely. When a positron has become “thermalized” at the thermal energy level of the lattice (about 0.025 eV at room temperature), the probability of its annihilation becomes large.

Why is that so?

Potentially related question: There seem to be two modes of electron-positron annihilation, a free one and another one via positronium formation (compare https://sundoc.bibliothek.uni-halle.de/diss-online/08/08H048/t2.pdf). Positronium formation is said to be unlikely at high kinetic energies as well.

Again, why is that so?

Bonus question: The rate of free annihilation is independent of the kinetic energy of the positronium, as the cross section is inversely proportional to the positron velocity ($\sigma \propto 1/v$), so that the velocity cancels out when the cross section is multiplied with the number of possible interactions per unit time (which is $\propto v * n_e$) - again, compare https://sundoc.bibliothek.uni-halle.de/diss-online/08/08H048/t2.pdf.

Why is the cross-section of free annihilation inversely proportional to the velocity?

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  • $\begingroup$ The first link is not working. The site wants one clear question , not a series of questions. for the positronium :Remember that positronium is the bound state of electron positron, so there will be a specific energy state that is stable, before annihilation takes over, and the energiy will be low. $\endgroup$
    – anna v
    Commented Jun 9, 2022 at 19:16
  • $\begingroup$ @annav the first link is working great for me. Could you tell me what error message you are getting so I can double-check? $\endgroup$
    – bers
    Commented Jun 10, 2022 at 20:20
  • $\begingroup$ I am getting "Sorry, the page you are looking for is not available. " $\endgroup$
    – anna v
    Commented Jun 11, 2022 at 2:47
  • $\begingroup$ @annav it works for me in Chrome and using wget and curl. I don't have another URL to the same file, unfortunately. $\endgroup$
    – bers
    Commented Jun 11, 2022 at 15:59
  • $\begingroup$ it worked in chrome for me too. $\endgroup$
    – anna v
    Commented Jun 11, 2022 at 16:11

1 Answer 1

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at the thermal energy level of the lattice (about 0.025 eV at room temperature), the probability of its annihilation becomes large.

you ask

Why is that so?

See here the calculation of electron positron scattering, where the average crossection falls with the square of the energy. At low energy the probability of scattering is large and then annihilation is inevitable to have larger probability, also because of the quantum probability of the positron meeting an almost free electron in the fermi level of the lattice an bind into the positronium. (The lifetime of the positronium is very small, it annihilates into two photons.)

You ask:

Why is the cross-section of free annihilation inversely proportional to the velocity?

Connected with the calculation in the link above, the energy is in the denominator, and the kinetic energy is part of it ($1/2mv^2$).

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  • $\begingroup$ Thanks! One follow-up: According to eq. (1.6) from the second link in my question, $\sigma \propto 1/v$. According to eq. (7.335) from the link in your answer, $\sigma \propto 1/E^2$ which implies $\sigma \propto 1/v^4$ if I am not mistaken. Are these different cross-sections? $\endgroup$
    – bers
    Commented Jun 11, 2022 at 18:05
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    $\begingroup$ I would say different estimates, it is not my specialty and it could be the difference between a truly free electron positron scattering ( the link I give) and one where the conduction band electrons are bound to the whole lattice (your link) $\endgroup$
    – anna v
    Commented Jun 11, 2022 at 19:07

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