# Are proton, antiproton, electron, positron the only observed subatomic particles that can freely exist and don't decay, i.e. are stable?

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?

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

-
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. – Luboš Motl Dec 19 '11 at 21:57
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? – Problemania Dec 19 '11 at 23:28
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". – anna v Dec 20 '11 at 4:45

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.

-
 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? – Problemania Dec 19 '11 at 20:33 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. – dmckee♦ Dec 19 '11 at 20:50 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. – Problemania Dec 19 '11 at 21:08 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. – dmckee♦ Dec 19 '11 at 21:17 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. – Luboš Motl Dec 19 '11 at 21:53