Why is hydrogen stable but lepton pairs aren't?

[Hetware]

I probably knew the textbook answer to this at one time, but I don't recall. Feynman states in the beginning of Vol II of the FLP that electrons don't collide into the nucleus due to the uncertainty principle. But an electron positron pair will mutually annihilate.

I suspect the answer has something to do with Bose-Einstein and/or Fermi-Dirac statistics, conservation of quantum numbers, etc.

 

[Jano L.]

I have wondered about this too some time ago. From the standpoint of quantum theory of fields, as I understood it, the electron and the positron are described by the same field and the formalism leads to annihilation easily. The proton, however, is described by different field than the electron and the annihilation is not so easy. So, the charge of positron and proton is the same, but otherwise they are represented by different mathematical quantities and this makes the distinction.

 

[jtbell]

Actually, atomic electrons do "collide" with the nucleus if the wave function is nonzero at the origin. In fact, for the ground state (1s orbital) the wave function is maximum at the nucleus!

If energy considerations allow it, this leads to the nuclear decay mode called "electon capture" which has similar results to beta+ decay.

It doesn't happen with hydrogen because you have to supply extra energy in order to make the reaction p + e --> n + neutrino "go".

 

[Hetware]

I have wondered about this too some time ago. From the standpoint of quantum theory of fields, as I understood it, the electron and the positron are described by the same field and the formalism leads to annihilation easily. The proton, however, is described by different field than the electron and the annihilation is not so easy. So, the charge of positron and proton is the same, but otherwise they are represented by different mathematical quantities and this makes the distinction.

I hate those kinds of "explanations"! It always seems like: "it works that way, because that's the way it works." Not to say that yours is not the answer I asked for.

I guess one might argue that the pair annihilation takes place in a fuzzy enough space-time region as to not violate the uncertainty principle. Regarding the electron and the proton, the reason they don't annihilate is that certain quantum numbers wouldn't preserved. Such answers always frustrate.
https://www.youtube.com/watch?v=NHx00XG6-jU
https://www.youtube.com/watch?v=lr8sVailoLw

 

[Hetware]

Actually, atomic electrons do "collide" with the nucleus if the wave function is nonzero at the origin. In fact, for the ground state (1s orbital) the wave function is maximum at the nucleus!

If energy considerations allow it, this leads to the nuclear decay mode called "electon capture" which has similar results to beta+ decay.

It doesn't happen with hydrogen because you have to supply extra energy in order to make the reaction p + e --> n + neutrino "go".

Thanks. I didn't know that.

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/radact2.html#c3

So a neutrino is the soul of a dying electron?

 

[Jano L.]

I hate those kinds of "explanations"! It always seems like: "it works that way, because that's the way it works." Not to say that yours is not the answer I asked for.

I have no idea which explanation you asked for, but I tried to give you one I know. Actually I am not very much satisfied by the above explanation either. However, I do not believe it is of the kind you indicated. There is some non-trivial mathematical difference in the description of proton and positron in QFT and this is connected to the different behaviour of the couples proton-electron and positron-electron.

Of course, no explanation is ever perfect, one can always go on step further and ask why.