How can quantum tunnelling lead to spontaneous decay?

I thought I would indicate how radiactive decay can be argued to occur from quantum tunneling. This derives in a very “back of the envelope” way the phenomenological equation for radioactive decay. From there I can argue some about the role of observing radioactive decay.

Quantum tunneling can be see with the square potential barrier. A square potential barrier with potential energy $V$ admits wave functions of the form $$exp\left(i\frac{\sqrt{2mE}x}{\hbar}\right),~exp\left(i\frac{\sqrt{2m(E - V)}x}{\hbar}\right)$$ outside and inside the square barrier. With the wave inside the barrier there is the wave vector $$k = \frac{\sqrt{2m(E – V}}{\hbar},$$ which is imaginary valued for $E < V$. The particle is tunneling through the barrier and is a decreasing function $exp(-kx)$. This holds in the region for the barrier $0 < x < d$ of width $d$. The larger $k$ is the more quickly the function exponentially decays and the probability for tunneling out is reduced. Also if the width of the potential barrier is increased there is more decrease in the function from the left to right, so again the probability for tunneling is reduced. This is computed as a transmission coefficient.

To do this problem in its fullest we have to match boundary conditions at the barrier. We might further approximate the $L^2/2mr^2$ term with an infinite barrier to the left of the square barrier. This gets into a fair amount of algebra, nothing difficult but lengthy and a bit tedious. For the square barrier we have the transmission amplitude $$T = \frac{4k_{out}k_{in}e^{-id(k_{out} - k_{in})}}{(k_{out} + k_{in})^2 + (k_{out}^2 + k_{in}^2)sin(dk_{out})}$$ and the modulus square of this is the probability for tunneling $$P_T = \left(1 + \frac{V^2sinh^2(k_{in}d)}{4E(V – E)}\right)^{-1},$$ called the transmission coefficient.

This is not entirely satisfactory. The problem is that this is for the long term stationary case, which is based on the time independent Schroedinger equation, for a square barrier. However, we can think of our alpha particle in a nuclear well moving back and forth, so that with each time it reaches the barrier it has this transmission probability of tunneling through. This transmission function is not an exact fit for this problem, but we can make use of it. The probability for the alpha particle remaining in the well is $P = 1 – P_T$ with each recurrence or orbit/oscillation of the alpha particle. I am now going to assume that the transmission probability is rather small for each oscillation or orbit. Therefore for $N$ orbits of the alpha particle $P_N \simeq 1 – NP_T$. We may then approximate this with the Taylor rule for the exponential function so that with time $t = N/\omega$, $\omega$ the oscillation frequency $\omega \simeq \hbar k_{out}/d$ that $$P(t) \simeq exp\left(-P_T\frac{\hbar k_{out}t}{d}\right)$$ which is the phenomenological rule for radioactive decay.

This question is largely about the role of measurement in radioactive decay. I think it is clear that radioactive decay has been going on long before people were measuring it. The whole enterprise of dating rocks and fossils relies upon that fact radioactive decay has been going on for millions and even billions of years. The occurrence of a radioactive decay event is a form of state reduction. In the above argument with probabilities estimated from a quantum amplitude there is an implicit assumption that with each time the alpha particle orbits there is some coupling to a set of states in the environment that can induce decoherence of the wave function. The actual decay event is then a decoherence or in the language of Fermi's golden rule a massive sum (slash and burn on quantum states) to estimate spontaneous emission of bosons.