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When giving fundamental reason for radioactivity, books or references most often state that this is due to the fact that nucleus is not in a stable enough state thus it goes back to the stability.

This is not a fundamental enough reason to me. Fundamental would need to involve the level of elementarly particle physics.

What is the fundamental reason, at particle level, of the radioactivity ?

Is it just due to the fact that Standard Model elementary particles couple with other particles, from Standard Model Feynman rules, and thus the radioactivity is only these reactions that occur, with the rates that just follow the cross-sections of the relevant particle level Feynman diagrams ?

If so, why books never explain it in this way, which looks much more "natural" than the explanations with "stability".

Mathieu Krisztian
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2 Answers2

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Radioactivity is an exothermic reaction: the decay products have kinetic energy. An unstable state is such that the "left-hand side" has more energy than the right-hand side, so there is enough kinetic energy for the products to fly apart.

You may think of "stability" as the low elevation in an energy graph, and "instability" a relative high elevation. If you are at the top of the hill, you are guaranteed to roll/slide downhill to a low point, which you reach with extra kinetic energy. But the converse does not happen, unless you get a kick by a bump that imparts the requisite kinetic energy to you.

How fast your decay will go will depend on 1) the absolute square of the decay matrix element, a fundamental, quantum quantity, as WP details; but also, 2) the phase space: the kinematic distribution of momenta and energies (relativistically invariant, of course) which you might (very loosely) analogize to the steepness off the hill.

The fundamental physics (computed, e.g., in QFT) is in the matrix element, and contributes to how fast you roll down (decay rate). But the same matrix element is totally powerless in the face of unfavorable phase-space: stability.

So, the weak interaction involved in β-decay, say, of a free neutron, involves virtually the same weak matrix element as the frustrated decay of a bound neutron in a nucleus; it does not decay, because its energy level is lowered below threshold by (strong) nuclear binding: it has been stabilized, analogous to the top of the hill sinking to a small sinkhole.

(Actually, a line-reversed matrix element is involved in inverse β-decay a scattering reaction involving the same particles.)

So, stability, is the first line of defense against decay, namely, unavailability of excess energy. However, if you do have the excess energy to decay to some products, as in the sought-after baryon decay, you will never decay if the fundamental theory does not support a non-vanishing matrix element. If that element is small, as in GUT nucleon decay, the decay will be seriously slow. If it vanishes, the decay won't go.

Cosmas Zachos
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The object is in a superposition of "nucleus" and "reduced nucleus plus decay products" states. There is a potential barrier between the "nucleus" states and the "reduced nucleus plus decay products" states, so if the object starts measured in a " nucleus" state, the probability of finding it on the far side of the potential barrier is low. Iff the total configuration energy of the system in a decayed state $\le$ the energy in a non decayed state, the probability is nonzero, so if you observe for long enough you will eventually measure the system in a decayed state.

g s
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