# Is it true that an isolated fundamental particle does not decay?

Is it true that an isolated fundamental/elementary particle does not decay?

It seems logical to me.

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What exactly is a "fundamental particle" to you? – ungerade Oct 22 '12 at 13:13

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Fundamental = elementary – mick Oct 22 '12 at 13:22
Concerning the edit made , i assumed it ( fundamental explained now by the then missing link ) was well known on this site to all. – mick Oct 22 '12 at 13:25

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– ungerade Oct 22 '12 at 13:34
re: 2 comments up: (some) people here will get in all sorts of arguments about what it means for a particle to be fundamental. But the definition on Wikipedia that you linked to is the commonly accepted one. – David Zaslavsky♦ Oct 22 '12 at 14:10
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## 3 Answers

No, because not all the fundamental particles (in the sense of the standard model) are stable. In particular, electrons and photons are stable; muons and tau leptons are not, and will decay into lighter leptons, e.g. $\mu^-\to e^- \bar\nu_e \nu_\mu$.

Neutrinos are kind of funny because, while they are prevented from decaying in the traditional sense by conservation of energy, momentum, and lepton family number, they do oscillate - so if you start with an electron neutrino, for example, it will turn into something that can be observed as any flavor of neutrino, and in this way you can find yourself measuring something like $\nu_e\to\nu_\tau$. But then you can just as well measure $\nu_\tau\to\nu_e$, and if you have two (actually three) particles which can all decay into each other, does it even make sense to call it a decay?

Quarks try to be funny but actually wind up just being annoying, because they are never found in isolation so nobody is exactly sure how an isolated quark would behave if you could put one in a universe by itself. That being said, heavy quarks (charm, bottom, theoretically top) are routinely observed to decay in high-energy collisions, where asymptotic freedom presumably applies, so it's not much of a stretch to imagine that an isolated heavy quark would decay into lighter quarks plus a pion or leptons.

Likewise, the weak gauge bosons decay all the time in collisions, so if you managed to create a universe that contained only a $W^\pm$ or $Z^0$ and nothing else, it would presumably decay quickly into a lepton and antineutrino or into some combination of hadrons. Same goes for the Higgs, except with different possible decay products.

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For now the best and accepted answer , although a bit above me for now. maybe some wiki-ing. – mick Oct 22 '12 at 13:59
However what about diffusion of light ? – mick Oct 22 '12 at 14:00
@mick what about it? Diffusion isn't really decay. Also I could edit in some Wikipedia links or more information if you let me know what parts are above you. – David Zaslavsky♦ Oct 22 '12 at 14:01
Im confused about diffusion then. If a wave diffuses , how about the particle interpretation then ? Isnt it particle decay in the particle interpretation ?? – mick Oct 22 '12 at 14:20
No, that's just the probability amplitude for measuring the particle in a particular position spreading out (assuming we're talking about diffusion in position space). – David Zaslavsky♦ Oct 22 '12 at 14:23
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It's only true if you take "isolated" to mean "isolated for an infinitely long time", in which case, an unstable particle would never form either. An isolated particle is exactly stable, this is the S-matrix state, and the exactly stable particles in our world are the electron/positron, the neutrino, the photon, the graviton and some dark matter.

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What about decays in waves or virtual particles ? or diffusion of light ? – mick Oct 22 '12 at 13:23
The proton is, to experimental evidence, exactly stable, but not fundamental. (I know you know this) – Jerry Schirmer Oct 22 '12 at 13:57
Im no expert Jerry. – mick Oct 22 '12 at 14:19
@JerrySchirmer: The proton can't be exactly stable because of proton decay in the standard model, and gravitational proton decay (really neutron decay) in the formation of black holes from neutron stars and their subsequent evaporation. The photon is exactly stable and not fundamental (meaning it is a mixture of SU(2) and U(1) gauge bosons), and I thought you said "photon" not "proton" at first. The S-matrix definition of "fundamental" is "exactly stable", though, so the photon is S-matrix fundamental, although field theoretically not fundamental. The proton is not fundamental in either view. – Ron Maimon Oct 22 '12 at 18:11
@RonMaimon: there is no proton decay in the standard model.  The formation of neutron stars involves electron capture and are not proton decay, but an interaction with electrons.  Most GUTs predict proton decay, but this has never been observed. – Jerry Schirmer Oct 22 '12 at 21:04

Presumably if it was fundemental there would be nothing for it to decay into ?

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there is no conservation law of amount of particles or waves as far as I know. – mick Oct 22 '12 at 13:22
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