# What triggers a nuclear decay?

I understand that the decay of a nucleus is a random event and one cannot predict exactly when it will happen for a particular nucleus. What I would like to know is what triggers this event to happen?

The surprising answer is that nothing triggers it. In quantum mechanics all we can talk about is the probabilities of various events happening: whether they actually happen in a given period is truly random. There is no secret mechanism which we could find which controls whether an event happens or not.

Well, there are, really, three-and-a-half possibilities:

• the first possibility is that there is no secret mechanism, things really are random as I said above;
• the second possibility is that there is some secret mechanism, but the rules of the game say that we could never observe such a mechanism, even in principle;
• the third possibility is that there's not really a mechanism, but somehow there's just a huge list saying what happens when for every event, which list we could also never discover, even in principle;
• the final half possibility are that the experiments which show that one of the preceding three possibilities must be true are incorrect.

I think most physicists think that the first possibility is true, but there are significant minorities who are unhappy with it in various ways. Well, perhaps everyone is unhappy with it, but there is a significant minority who are so unhappy with it that they devote a lot of effort into investigating other options. Einstein, famously, was one of this minority.

## Bell's theorem

The core thing here is a famous result called Bell's theorem. What Bell concluded was that

In a theory in which parameters are added to quantum mechanics to determine the results of individual measurements, without changing the statistical predictions, there must be a mechanism whereby the setting of one measuring device can influence the reading of another instrument, however remote. Moreover, the signal involved must propagate instantaneously, so that such a theory could not be Lorentz invariant.

What this means is that if we want to explain the predictions of quantum mechanics by what I've called a 'secret mechanism' above, and which physicists call a 'hidden variable', then this mechanism must allow the instantaneous transmission of information between objects, however far separated they are. He then says that 'such a theory could not be Lorentz invariant': how bad is that?

It's bad. What it means is that such a theory is not compatible with special relativity, a theory which has been extremely well tested. In particular it means that if we had access to this secret mechanism, then assuming all the tests we've done of special relativity are not just wrong, we could build a time machine. In particular we could build a time machine which will send information into our own past. And this isn't just some kind of thorectical 'if we could make a black hole we could do this' thing: we could actually build such a thing for a reasonable amount of money (I don't know how much, but let's say for less money than was spent on the Apollo programme). This is very bad, to put it mildly.

So if we aren't willing to accept that this leads to the four options above, which I'll present in a different order.

• Perhaps the experiments which show that the predictions of quantum mechanics are correct are wrong, and we're off the hook. These experiments are really hard to do without loopholes, but all of the experiments done so far have been compatible with what quantum mechanics predicts. I think it's a safe assumption that the predictions made by quantum mechanics are in fact correct.
• Perhaps there is a secret mechanism, but life is arranged so that it can never be observed, even in principle. That's a horrible option I think, and in particular if there is this secret mechanism which can never be observed why not just assume there isn't? Science has to do with things that can be observed (even if such observation is very hard and perhaps beyond our abilities for the forseeable future), not with things that can't, even in principle.
• Perhaps there isn't a mechanism but things still are not random: everything is just predetermined, and in particular the results of the experiments we do and the choices we make while doing those experiments is all predetermined, so the experiments are meaningless. This is called 'superdeterminism' and, again, it's kind of uninteresting: if it's true then we can't know it is because, well, everything we do is predetermined.
• Perhaps there isn't a mechanism and things really are random and all we can know is various probabilities.

The last of these is, I think, the standard view, and it's the view that gives rise to my initial statement: nothing triggers the decays, whether or not a decay happens is truly random and all we can know is the probability that it will happen in a given interval.

## Entanglement and randomness

Bell's theorem is usually understood to refer to a phenonemon in quantum mechanics called entanglement: this is where measurements on two physically separated objects are correlated, and it turns out that they are correlated in such a way that the awkward options above are the only valid explanations (indeed Bell's theorem itself is the bit of maths that shows that these are the only options).

The reason this matters for atomic decay is that the theory which controls atomic decay is the same: quantum mechanics. So although Bell's theorem deals with entanglement, the theory that predicts entanglement is also the theory that controls atomic decay: if there are hidden variables behind atomic decay which mean it's not random they will be the same hidden variables that Bell's theorem show have such awkward properties if they exist.

It's also the case, I think, that atomic decay should produce particles which are entangled and which therefore should, in principle, be amenable to using as candidates in tests of quantum mechanics. I am not an expert on this so this is somewhat speculative on my part, but in beta decay the results are an electron and an antineutrino (or a positron and a neutrino). Both of these have spin, and I presume that their spins must be entangled (or entangled with each other and the spin of the nucleus which decayed). So in principle you could use these things in tests of entanglement. This is very much in principle because neutrinos are absurdly hard to detect.

Although such an experiment would be extremely hard to do, it would rule out the possibility that there is somehow some completely other, hitherto unknown, theory which controls atomic decay and which does allow prediction of when it happens. I think there are a lot of other reasons why this possibility is implausible: quantum mechanics works superbly well for one thing and we see no trace of any other theory which might apply for another, but such an experiment would conclusively show that it is what governs decay.

Note that Einstein was dead by the time Bell published his theorem: we don't know what he would have said if he had known about it.

• Wouldn't there be another possibility where there is some trigger but we just haven't identified it yet? – Mike Aug 19 '19 at 23:02
• @Mike: that's the 'all the experiments which validate the predictions of QM are wrong' case. QM predicts, testably, that such a mechanism ('trigger') must agree with Bell's theorem. So yes, this is a possibility, and that possibility is that QM is wrong. – tfb Aug 20 '19 at 10:06
• Doesn't this go too far? Op just asked about randomness, but here the discussion veers into the fact that entangled quantum states exist. These things seem orthogonal. The randomness question would be valid in a universe governed by purely classical probabilities and local phenomena. The entanglement question arises because "probabilities" switch to complex-valued, aka "the wave function", which is really a lot more interesting than a real-valued probability distribution. – David Tonhofer Aug 20 '19 at 15:11
• @DavidTonhofer: that's a good question. I think I have a good answer to it, and I'm going to add it to the answer: if you want to check it in a few minutes & comment again if you think I'm wrong or confused? – tfb Aug 20 '19 at 15:30
• @IronGremlin: I don't understand what you're asking, but if it can be formulated as a question then it's probably better to ask it as such where people better at this than me will see it! – tfb Aug 21 '19 at 9:04

If you look at the actual equations governing quantum mechanics, there is no randomness at all. The nucleus starts out in a state where it hasn't decayed. Over time, it evolves into a mixture of the undecayed state and the decayed state. It's like Schrodinger's cat. Gradually the mixture shifts more and more toward decay.

If an observer watches the nucleus to see whether it's decayed yet, then the observer also becomes a mixture of a state in which they have seen it decay and a state in which they haven't. For practical reasons, it's not possible to observe wave interference effects between human-scale objects, so we can't detect interference between the different states of the observer.

Therefore each state of the observer is cut off from the others, and they can't detect each other, it seems to the observer, in that particular state, as if something random has happened. It's then natural for that observer to stop thinking about the other possibilities which, to them, might as well not exist. If they stop keeping track of those other possibilities, they are doing something called the Copenhagen interpretation of quantum mechanics. The Copenhagen interpretation is an optional add-on to quantum-mechanics.

• That should be "superposition", not "mixture", right? Because superposition is the complex-linear combination of two basis state vectors (both states at the same time for real!), but mixture is the real-valued classical probability mix of two state vectors (one or the other but we don't know which). – David Tonhofer Aug 20 '19 at 14:52
• @DavidTonhofer: Yes, I'm just writing in nontechnical language, because of my perception of the level of the OP's knowledge. – Ben Crowell Aug 25 '19 at 0:44
• Not wanting to get into interpretations of quantum mechanics but ... just found at the arxiv: Interpretations of quantum theory: A map of madness – David Tonhofer Aug 25 '19 at 17:30

I am surprised that no one has discussed vacuum fluctuations as the trigger for these spontaneous nuclear decays. This question is just the nuclear physics analog of this question: An explanation of the Spontaneous Emission which applies to atomic physics. That spontaneous decays in atomic physics are triggered by vacuum fluctuations of the E&M field is the essence of Wigner-Weisskopf Theory.

Unlike atomic physics spontaneous nuclear decays can involve the strong and weak interactions as well as E&M. Vacuum fluctuations occur for these interactions as well. Physical manifestations of vacuum fluctuations include the Casimir force and the Lamb shift among other phenomena. A strong hint that these fluctuations are the trigger for nuclear decays is the fact that some decay half-lives have been observed to undergo small changes when the decaying nucleus is subjected to an altered (or polarized) environmemnt (see the discussion on "Changing Decay Rates" here https://en.wikipedia.org/wiki/Radioactive_decay#Theoretical_basis_of_decay_phenomena )

It is true that many theoretical treatments of vacuum fluctuations are based on perturbative calculations. This leads some to question whether vacuum fluctuations are not just an artifact of an approximation scheme, hence lacking physical reality. The success of some nonperturbative treatments for certain toy QFT models argues against that view (see: https://ncatlab.org/nlab/show/non-perturbative+quantum+field+theory).

• But that just shifts the problem to "vacuum fluctuations" (whatever these are ... and they are "ontologically suspect" aren't they) – David Tonhofer Aug 25 '19 at 17:24
• @DavidTonhofer I will edit my answer to address your comment. – Lewis Miller Aug 26 '19 at 2:32
• I think vacuum fluctuations fall with the definition of the wavefunction for a system , and just are within the general quantum mechanical probability frame. Experiments showing effects on decays (again statistically, not per decay) actually show that different boundary conditions affect the probability for the reaction. – anna v Aug 26 '19 at 3:30
• The link in the last sentence does not mention the words "vacuum" or "fluctuations". – fqq Aug 28 '19 at 12:31

This is a simpler statement:

Take the probability curve for throwing a dice and coming up with six. If the dice is true, not deliberately weighted, there is a probability of $$1/6$$ and there is no way for the player to know when juggling the dice what the outcome would be. If the dice were weighted so that some numbers came up more often, again it is only statistically that this will be seen, not at the individual throw.

Quantum mechanical wave functions ($$Ψ$$) ,with the boundary conditions of the problem, in this case the decay of a nucleus, weight the statistical distribution of quantum mechanical observations according to $$Ψ^*Ψ$$ in an analogous fashion as with a weighted dice.

The argument on how the dice behavior can be in principle calculated exactly in classical mechanics when all factors are taken into account, including the way of the throw, does not hold by axiomatic definition in quantum mechanics. Some physicists are still trying to find underlying deterministic mathematics , for quantum mechanics, but at the moment not successfully .

In any case, even for classical probabilities it is not possible to give an exact prediction of a probable path. even though in principle it is there when all the boundary values are taken in, because of the complexity. That is why probabilities were invented.

There cannot be a "trigger" for 6 coming up, even for weighted dice. More so for quantu mechanical cases, axiomatically at the moment, but even if a deterministic underlying theory is ever established , no "trigger" can exist because of the complexity.

The short answer is that systems tend to minimise their energy over time. The electrons around an atom will spontaneously drop to the lowest possible energy orbitals. Carbon spontaneously combines with oxygen to form carbon dioxide. Radioactive nuclei lower their energy by decaying.

A ball on the crest of a hill will not spontaneously roll down the hill; it needs some impetus to get itself moving. As soon as you push it, it will quickly roll down, because the position "ball in the valley" has less energy than "ball on top of the hill". But you need to get it moving. With very small things, like nuclei, you don't even need to do that - the nucleus doesn't have a well-defined position or momentum; it's as if the rock in our example could spontaneously move a few inches to the side, or gain a bit of velocity. The rock wouldn't stay on top of the hill for very long.

• This seems like a somewhat misleading classical analogy. Randomness in quantum mechanics doesn't have anything to do with equilibrium, and the Schrodinger equation doesn't say anything about things going to lower energy levels. The Schrodinger equation is symmetric under time-reversal, and it describes absorption of energy on an equal footing with emission. – Ben Crowell Aug 19 '19 at 22:48
• This Answer simply begs the question. The OP asks what triggers the decay of a nucleus at one time and not another, and you offer the analogy of a stone that rolls down a hill-- because it spontaneously moved an inch from the peak and o began to roll. Well, then, what triggered that initial movement at one time and not another? – Beta Aug 19 '19 at 23:22
• @BenCrowell The classical analogy is also symmetric; and the reason it (almost never) happens in the opposite direction is essentially the same as with the quantum case - though obviously, it's a lot easier for a stable nuclei to absorb e.g. a photon and an antineutrino of just the right energy than for a rock to absorb all of the energy released as it was rolling down the hill; it would be so unlikely for a lock to spontaneously roll back uphill that we can't really expect it would ever happen once in the whole universe. – Luaan Aug 20 '19 at 6:58
• @Beta No, I contrast the classical macroscopic phenomenon with what actually happens on the level of e.g. individual nuclei. The crucial distinction is the uncertainty in position and momentum - a rock doesn't spontaneously start rolling down the hill, but if you scaled it small enough, it would (for a narrow enough peak, of course). Of course there's no real macroscopic analogue (though classical waves do share the uncertainty, it's not quite the same kind). – Luaan Aug 20 '19 at 7:01