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Are proton, antiproton, electron, positron the only subatomic particles that can freely exist (i.e. I don't want particles that only exist in bound state as constituents such as quarks) and don't decay, i.e. are stable?

What about muons?

Are there other particles/hadrons that can exist freely (i.e. not in some bound states) and don't decay?

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    $\begingroup$ Electron, positron, all neutrinos, photons, gravitons, and possibly the lightest supersymmetric particle (LSP) if the R-parity is broken are exactly stable. However, protons and antiprotons aren't "subatomic particles". Proton is really a hydrogen nucleus, so you should also include hundreds of stable nuclides (isotopes) as well. In GUT theories, proton (and nuclei) are long-lived but ultimately unstable. They decay. Similarly, there is a risk that there exists a tiny decay of heavier neutrino flavors to lighter ones and photons/gravitons. $\endgroup$
    – Luboš Motl
    Dec 19, 2011 at 21:57
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    $\begingroup$ With the condition "observed subatomic particles", there are just electron, positron, all neutrinos ($\nu_e$, $\nu_\mu$, $\nu_\tau$, $\bar \nu_e$, $\bar \nu_\mu$, $\bar \nu_\tau$), photon? $\endgroup$
    – qazwsx
    Dec 19, 2011 at 23:28
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    $\begingroup$ It depends on the meaning of "observed". If one allows the level of complexity in observation that is needed to "observe" quarks, they are also observed. You must mean "stable observed". $\endgroup$
    – anna v
    Dec 20, 2011 at 4:45

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No. At a minimum neutrinos and anti-neutrinos are also stable{*} and exist in unbound systems.

Additionally the electromagnetic carrier boson (i.e. the photon) can exist for arbitrarily long times in the reference frames of massive objects (it's proper time is necessarily zero). The same could be said for the graviton if we had experimental confirmation of it's existence.

Further, many beyond-the-standard-model candidate theories feature (anti-)proton decay, though the current experimental limits require this to be a very slow process indeed.

{*} We have to be a little careful about what we mean here. The pure mass states $\nu_i$ are stable in free propagation, however neutrinos are created and destroyed in flavor states. So we have time-dependent superpositions of mass-states, in which the sum of the lepton flavor numbers is conserved.

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  • $\begingroup$ So, is what people usually mean by $\nu_e$, $\nu_\mu$, $\nu_\tau$, $\bar \nu_e$, $\bar \nu_\mu$, $\bar \nu_\tau$ the stable mass states? $\endgroup$
    – qazwsx
    Dec 19, 2011 at 20:33
  • $\begingroup$ The states labeled $e$, $\mu$, and $\tau$ are the flavor states. We typically label the mass states $1$, $2$, and $3$. The flavor states are eigenstates of the weak interaction, the mass states are eigenstates of the free Hamiltonian. Because they are created/destroyed in flavor states but propagate under the free Hamiltonian they mix; but they don't decay for any reasonable meaning of that word. $\endgroup$
    – dmckee --- ex-moderator kitten
    Dec 19, 2011 at 20:50
  • $\begingroup$ Ok, so what people usually mean by $\nu_e$ etc. are flavor states which are created and destroyed frequently. So they are not "stable", i.e. persistently existing. $\endgroup$
    – qazwsx
    Dec 19, 2011 at 21:08
  • $\begingroup$ It is incorrect (or at least not entirely correct) to think of these states being "created" and "destroyed" during mixing. A neutrino continues to exist, it's just that it's flavor content is a function of time. There is no decay mode for $\nu \to \mathrm{other stuff}$ and the global lepton number remains conserved. $\endgroup$
    – dmckee --- ex-moderator kitten
    Dec 19, 2011 at 21:17
  • $\begingroup$ Right, @dmckee. User: "decay" is linked to the function $\exp(-t)$, oscillations (caused by mixing) are linked to the function $\sin(t)$. These are different functions. Only the first one converges to zero, so only the first one deserves to be called "decay" and only the first one is linked to an instability. Incidentally, one may prepare any superposition of flavor states of neutrinos as an energy eigenstate. It won't be a momentum eigenstate at the same moment, however. $\endgroup$
    – Luboš Motl
    Dec 19, 2011 at 21:53
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In addition to the other answers, the lightest magnetic monopole is stable by magnetic charge conservation.

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    $\begingroup$ If it exists... $\endgroup$
    – Anixx
    Dec 31, 2011 at 13:58
  • $\begingroup$ One can be more certain of the monopole, both existence and stability, than the neutralino. $\endgroup$
    – Ron Maimon
    Dec 31, 2011 at 16:27
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The lightest super symmetric neutrinos are theorised to be stable, and as such represent a candidate for dark matter.

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    $\begingroup$ These new supersymmetric particles (and top dark matter candidates) are called "neutralinos", not "neutrinos", they're completely different particles although both of them are neutral spin-1/2 fermions. $\endgroup$
    – Luboš Motl
    Dec 19, 2011 at 21:54

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