The radioactive decay is caused by the instability of the nucleus. It is a random process in which an unstable nucleus loses energy by emitting radiation and goes into a lower and more stable energy state. Are quantum fluctuations responsible for triggering the decay?

  • $\begingroup$ Please define what you mean by "quantum fluctuation". $\endgroup$
    – ACuriousMind
    Oct 1, 2017 at 10:57
  • $\begingroup$ The value of a quantum field in any point in space is always jiitering around a bit. When I say quantum fluctuation, I am referring to this kind of fluctuation. We can't precisely determine the value of the field and precisely how it is changing. $\endgroup$
    – Nemo
    Oct 3, 2017 at 8:49
  • $\begingroup$ The value isn't "jittering around" - like most quantum observables, it simply isn't well-defined until you measure it, just like the position of an electron in an atom is not jittering around, but undetermined. It is indeed this indeterminacy that can be seen as responsible for decays - this is the tunnelling John Rennie's answer refers to - but calling it "quantum fluctuation" obscures the issue more than it clarifies it. $\endgroup$
    – ACuriousMind
    Oct 3, 2017 at 10:45

1 Answer 1


Quantum fluctuations don't exist, or at least not in the way typically described in popular science articles. For more on this see Are vacuum fluctuations really happening all the time?

Radioactive decays fall roughly into two classes: those that occur by tunnelling and those that don't.

Some decays require a large reorganisation of the nucleus. These would typically be fission decays, as in uranium, or alpha decays. In these cases there is usually a large potential barrier to the decay and the kinetics of the decay are best described by calculating the tunnelling probability through the barrier.

With other decays, such as beta and gamma emission, there is no barrier and in this case the kinetics are best described by a Fermi's golden rule calculation. (For beta decay we'd use the Fermi calculation but this is still essentially a golden rule calculation.)

In neither case are quantum fluctuations involved.


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