I Why is a pair of an electron and a positron unstable?

We know that an electron and a positron will annihilate and emit gamma ray. But the electron and positron possess initial kinetic energy meaning that it is difficult for them to really collide in each other. Just like earth is not dropping into the Sun even with the gravitational pull. So I am curious how scientists can annihilate the two. And what is the scale of their distance such that annihilation occurs?
 

Drakkith

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But the electron and positron possess initial kinetic energy meaning that it is difficult for them to really collide in each other. Just like earth is not dropping into the Sun even with the gravitational pull.
Their attraction results in an acceleration of both charges, which causes the emission of electromagnetic radiation. This robs energy from both of them and causes them to very quickly spiral into each other.

As for the distance scale where annihilation occurs, I admit I don't know.
 
Their attraction results in an acceleration of both charges, which causes the emission of electromagnetic radiation. This robs energy from both of them and causes them to very quickly spiral into each other.

As for the distance scale where annihilation occurs, I admit I don't know.
We know that electrons are not emitting EM waves while "orbiting" around a proton, as long as they stay at the steady state (the orbital). Is it possible that an electron-positron system also has such a steady state so that they don't annihilate?
 

Drakkith

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2018 Award
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We know that electrons are not emitting EM waves while "orbiting" around a proton, as long as they stay at the steady state (the orbital). Is it possible that an electron-positron system also has such a steady state so that they don't annihilate?
To my knowledge, there is not, but I don't know the details of why. Perhaps someone else here can provide a more in-depth explanation.
 
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We know that electrons are not emitting EM waves while "orbiting" around a proton, as long as they stay at the steady state (the orbital). Is it possible that an electron-positron system also has such a steady state so that they don't annihilate?
There is only one such steady state, and that is 1s.
In which electron does not orbit around the proton, but oscillates through the proton.
The density of electron at the proton, though finite, is nonzero at the cusp of 1s orbital, and smaller but still nonzero at the cusp of all higher s orbitals.
p, d and higher angular momentum orbitals possess nodes that go through proton and thus have zero density at proton.

There are many other nuclei besides positron that can annihilate an electron. Starting with Be-7, most common K-40 etc. While annihilation of 1s electrons is most common, electrons can also be annihilated from 2s and higher orbitals, as well as molecular and crystal orbitals of s character at that nucleus. But not from p, d, etc. orbitals or molecular or crystal orbitals of p, d etc. character at that nucleus.
 
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They can annihilate directly if they come close enough (not exactly a well-defined thing in quantum mechanics, but the classical analogy is not too bad here). The probability (the cross section) decreases with increasing energy. They can also form positronium first, decay to its ground state and then annihilate from there.
 
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They can annihilate directly if they come close enough (not exactly a well-defined thing in quantum mechanics, but the classical analogy is not too bad here). The probability (the cross section) decreases with increasing energy. They can also form positronium first, decay to its ground state and then annihilate from there.
Or decay to any higher s orbital and annihilate from there. Or to a molecular or crystal orbital with s character at positron.
Does an unbound orbit have a well-defined orbital angular momentum (s, p etc.)?
 
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Not impossible but very unlikely.
Not in case of 2s orbital.
The radiative decay of 2s positronium into 1s is strongly (though not completely) forbidden, just as 2s hydrogen and for the same reasons.
Whereas annnihilation from 2s state is slower than from 1s, but not drastically so.
And states that can decay to either 2s and 1s (such as 3p) have no strong reason to prefer 1s over 2s.
 
https://arxiv.org/abs/hep-ph/9911410
The lifetime of a positronium "atom" is calculated with quantum electrodynamics. The amplitude for the decay is some kind of a path integral.

Since it is not a standard particle collision experiment, but involves a bound state, the positronium, people probably use some extra assumptions in the calculations.
 

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