This is not a duplicate, I am not asking whether the proton is a stable particle, or why it is. I am asking about the definition of stability/decay whether it is boolean or not.

I have read this question:

How can beta plus decay be possible?

where John Rennie says:

There isn't a potential barrier in beta decay, whether it's beta minus or beta plus decay. In both cases the decay is slow simply because the transition probability is so slow.

and where Emilio Pisanty says in a comment:

Stability is a boolean variable, in QM as well as anywhere. If the system is in an unstable state, then it's in an unstable state, period.

There must be a misunderstanding, because beta minus decay happens, because in QM, it is all about probabilities. In this case, it happens slowly. But the proton rich nucleus is unstable.

Now if the proton rich nucleus is unstable, and it is all about just probabilities as per QM, then everything, every quantum system (composite) is unstable. Period.

One would say that there cannot exist a single quantum system, that would be stable infinitely. Even a stable atom (or a stable proton) will decay eventually. Since even the protons inside it have an average lifetime of 2.1*10^29 years.


The free proton (a proton not bound to nucleons or electrons) is a stable particle that has not been observed to break down spontaneously to other particles. The spontaneous decay of free protons has never been observed, and protons are therefore considered stable particles according to the Standard Model. However, some grand unified theories (GUTs) of particle physics predict that proton decay should take place with lifetimes between 10^31 to 10^36 years and experimental searches have established lower bounds on the mean lifetime of a proton for various assumed decay products.[22][23][24] Experiments at the Super-Kamiokande detector in Japan gave lower limits for proton mean lifetime of 6.6×10^33 years for decay to an antimuon and a neutral pion, and 8.2×1033 years for decay to a positron and a neutral pion.[25] Another experiment at the Sudbury Neutrino Observatory in Canada searched for gamma rays resulting from residual nuclei resulting from the decay of a proton from oxygen-16. This experiment was designed to detect decay to any product, and established a lower limit to a proton lifetime of 2.1×10^29 years.

Now the contradictions is where one says that anything that is a composite particle will eventually decay, does have a mean lifetime (even free protons according to GUT but not SM), but stability is a boolean variable.

If stability is a boolean, and everything (composite) does have a mean lifetime, then basically the definition of stability is the same as the definition of elementary vs non-elementary particles.

But here comes the contradiction, there are even elementary particles that are not stable, and some are stable.

Like the electron, is defined as a stable particle, but has a mean lifetime of 6.6*10^26 years.

There are elementary particles that spontaneously decay into less massive particles. An example is the muon, with a mean lifetime of 2.2×10^−6 seconds, which decays into an electron, a muon neutrino and an electron antineutrino. The electron, on the other hand, is thought to be stable on theoretical grounds: the electron is the least massive particle with non-zero electric charge, so its decay would violate charge conservation.[77] The experimental lower bound for the electron's mean lifetime is 6.6×10^28 years, at a 90% confidence level.


So the muon is unstable, because we have observed muons to decay, but the electron is stable, because we have never observed one to decay, but the electron does have a mean lifetime, 6.6*10^28 years.

Even as per the SM, the contradiction is there. Do we only say that stability is boolean, because we have never observed the electron and proton (free) to decay, but we give them a mean lifetime?

As per QM, it is all about probabilities, and nothing lasts forever. Even the stable particles ( electron and the proton) will have a mean lifetime.

Does QM probabilities win or does the SM stability (boolean) definition win?


  1. Is stability a boolean, that is, are we defining stability as the particles that have never been observed to decay (electron and free proton), and are we defining unstable the particles that have been already observed to decay?

  2. Or are we saying that QM is all about probabilities, and even the stable particles (electron, proton) do have a mean lifetime and will eventually decay?

  • $\begingroup$ You are misreading things. Eg, that quote about electron stability says that we believe that the electron is stable. "The experimental lower bound for the electron's mean lifetime is 6.6×10^28 years, at a 90% confidence level." means that experiments have demonstrated that if the electron is unstable then its mean lifetime is probably greater than $6.6\times19^{28}$ years, and we can be 90% confident in that figure. $\endgroup$ – PM 2Ring Jul 4 at 12:39
  • $\begingroup$ @PM2Ring "if the electron is unstable". Is this unsurety part of the SM, or is this just a probability as per QM? $\endgroup$ – Árpád Szendrei Jul 4 at 12:48
  • $\begingroup$ In the Standard Model, the electron is assumed to be completely stable. But it's just a model, and even though it's a very good model, and it's predictions have been tested to high precision, in physics experiment & observation always trumps theory. $\endgroup$ – PM 2Ring Jul 4 at 13:04
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    $\begingroup$ Mathematicians prove theorems, but physics isn't mathematics. We can prove that free neutrons are unstable, by watching them decay. We can't prove that the electron is stable, but we can do experiments that fail to detect electron decay. To prove that the electron never decays, you'd have to watch every electron in the universe for an infinite timespan, which is obviously impossible. $\endgroup$ – PM 2Ring Jul 4 at 13:05
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    $\begingroup$ You're attaching too much weight to common phrases. "It's all probabilities" isn't a rigorous statement, it's an intuitive way to summarize some results in quantum mechanics. It does not mean that everything in quantum mechanics fluctuates randomly. For example, in standard quantum mechanics the energy is always exactly conserved. $\endgroup$ – knzhou Jul 4 at 17:01

Now if the proton rich nucleus is unstable, and it is all about just probabilities as per QM, then everything, every quantum system (composite) is unstable. Period.

That is a truly amazing exercise in non sequitur. To be frank, logic seems to have completely deserted your post in its entirety $-$ nothing seems to be logically connected to what comes before or after.

So, let's start with the dictionary.

Boolean variable: Any variable, from the domain of Boolean algebra, having one of only two values,

i.e. a variable that can (only!) be "true" or false".

As applied to the stability of a system, saying "stability is a boolean variable" means that for any given system, it is either stable or unstable, period. There's no probabilities involved.

To be crystal clear: this is completely consistent with the existence of systems which are stable. It is also consistent with systems which are unstable. It just means that any state of any system needs to be either stable or unstable, i.e. there's no middle ground.

As applied specifically to nuclei which undergo radioactive decay (of any kind), saying "stability is a boolean variable" means that if a given nucleus is in a state that can undergo radioactive decay, then it must undergo radioactive decay. (In this form, it is a particular subform of the Gell-Mann totalitarian principle.) To be crystal clear:

  • The time at which this decay happens is probabilistic $-$ we do not know (and cannot know) when it will decay.
  • The fact that the system will decay is not probabilistic $-$ it is written in stone. If you initialize the system in the unstable state, there will always be a time at which it decays.

An additional misunderstanding in your question, which is basic enough that you should know better:

The experimental lower bound for the electron's mean lifetime is 6.6×10^28 years, at a 90% confidence level.

[...] but the electron does have a mean lifetime, 6.6*10^28 years.

No it doesn't. Lower bounds are not the same thing as values. The result you've quoted means that lifetimes shorter than $6.6\times 10^{28}$ years are ruled out, but it does not give a value for the electron lifetime, nor does it even begin to imply that said lifetime exists or is finite. If the electron has a finite lifetime, the lower bound implies that it cannot be shorter than $6.6\times 10^{28}$ yr, but the lower bound is perfectly consistent with electrons being stable.

This has nothing to do with QM or the Standard Model. Infinity cannot be accessed by physical experiments, regardless of the arena. Even if every bit of the conceptual framework indicates that a given parameter will be infinite, we are finite beings and all we can do is to perform finite experiments which provide increasingly large (but always finite) lower bounds.

This is an elementary bit of language. If its usage is not clear, then you should go back and take a long, detailed take at introductory maths textbooks before carrying on with QM.

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    $\begingroup$ This answer is correct but it seems a bit hostile, to be honest. I don't think there's any need for that. $\endgroup$ – Javier Jul 4 at 20:29
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    $\begingroup$ @Javier I think there's a need for a minimal amount of critical thought before posting questions here, particularly for experienced site members. If that bothers others then I would seriously question why they think that critical thought is optional. $\endgroup$ – Emilio Pisanty Jul 5 at 1:56
  • $\begingroup$ I agree, but given how confusing QM can be, I think we could forgive OP in this instance for asking naive questions. What seems obvious to you is not necessarily so to others. $\endgroup$ – Javier Jul 5 at 1:58
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    $\begingroup$ @Javier That might be true for a first-time poster, but that is not the case with OP. I am astounded by the disregard for physics, and this site, shown by OP in this question - and even more so by the presence of folks who stand up to defend this objectively-awful question despite the harm it does to this site. This question represents an abuse of this site's mechanisms, there's no other word for it $\endgroup$ – Emilio Pisanty Jul 5 at 2:10

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