# Predicting Decay Rates via the Standard Model

Question 7584 illustrated a procedure to forecast the decay rates of isotopes with known long average lifetimes. Lifetimes of the many U isotopes vary from micoseconds to gigayears. F has only one stable isotope while Sn has 10. Can Standard Model principles be used to predict the stability of isotopes and the average lifetimes for unstable isotopes, or can this only be done by measurement?

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## 2 Answers

The nuclear forces are a complex amalgam of primarily Quantum Chromodynamics forces and electromagnetic ones, but to deal with the diagramatic way of calculating in Quantum Field Theory, is not possible. The weak force responsible for beta decays should also be in the calculations.Too many diagrams and too convoluted.

Quantum mechanical models with a potential well to estimate the collective forces are used for this. Nuclear physics has been using various models successfully, like the shell model. to predict energy levels in nuclei.

Ways of estimating lifetimes are taught in nuclear engineering, for example:

Course Outcomes: Students must be able to... calculate the consequences of radioactive growth and decay and nuclear reactions. calculate estimates of nuclear masses and energetics based on empirical data and nuclear models. calculate estimates of the lifetimes of nuclear states that are unstable to alpha-,beta- and gamma decay and internal conversion based on the theory of simple nuclear models. use nuclear models to predict low-energy level structure and level energies.

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In principle, the Standard Model contains the answer to all these questions about half-lives. In practice, it is impossible to calculate at this point - as far as I know.

However, it is very easy to explain why the lifetimes span such an exponentially huge interval of possible time scales. Here's why.

An alpha-decay - and only alpha-decaying nuclei with "really long-lived" half-lives will be explained by the simple argument below - may be approximated as the confinement and quantum tunneling of an alpha particle.

Imagine that a big nucleus $M$ decays to a smaller one $N$ and an alpha-particle: $$M \to N + \alpha$$ In this context, $M$ may be viewed as a bound state of $N$ and $\alpha$. The alpha-particle is confined by the attractive potential of the nucleus $N$. However, we know that this bound state isn't quite stable - it decays. It follows that a free $N$ plus a free $\alpha$ must have a lower total mass/energy.

Consequently, the bound state of $N$ and $\alpha$, and that's what we called $M$, must be metastable and the alpha-particle is confined by a potential wall. Classically, it would stay there forever. Quantum mechanically, there is quantum tunneling. There's a nonzero probability that the alpha-particle tunnels through the potential wall and gets out. However, the rate of this event is exponentially small because the wave function, having an imaginary momentum inside the potential wall (a negative kinetic energy needed to go through the wall), exponentially decreases in the wall.

The probability or decay rate $\Gamma$ is then comparable to $$\Gamma = dP/dt \sim \exp (-V/V_0) \times (10^{-24}{\rm seconds})^{-1}$$ where $V$ is the total "thickness" of the potential created by the nucleus $N$ inside the bound state $M$, and $V_0$ is some typical QCD-scale unit for such thickness. To get the right dimensions, I also included $10^{-24}$ seconds as the typical "QCD time scale". However, as you can see, the half-lives of the isotopes will be exponentially larger. Relatively moderate changes of $V/V_0$ - from 20 to 100 - may change the lifetime from a fraction of a second to billions of years.

This qualitative pattern is surely well understood. In fact, approximate estimates of the lifetimes can be done by models in nuclear physics, and those models can be at least approximately be showed compatible with the Standard Model. But no one has calculated the half-life of uranium-235 via a bound state of many quarks and gluons - as far as I know. Also, however, there is not a single glimpse of a demonstrable discrepancy.

Well, I was cheating a bit: even beta-decay may lead to incredibly long lifetimes. An example is potassium-40 which can decay by any three types of the beta-decay - and its lifetime is still over a billion of years. Microscopically, the beta-decay is a bit different than the alpha-decay because it requires a real transformation of the types of particles - not just a different state of protons' and neutrons' motion and position.

At the same moment, beta-decay is somewhat simpler because its "microscopic process" is just a single weak interaction vertex with a W-boson - and the rest of the nucleus is less changed (only 1 nucleon is added/killed, instead of 4). In some sense, the beta-decays combine the quantum tunneling above with the elementary W-boson-induced weak interaction and they may have very diverse lifetimes, too.

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This is a rather small gain in reputation compared to Your standards, Lubos, I wonder why? (tongue in cheek) :=( – Georg Apr 1 '11 at 11:27
You know, people suck compared to what they normally do. ;-) – Luboš Motl May 20 '11 at 15:08
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