How can quantum tunnelling lead to spontaneous decay?

Question

I have never understood what measuring process (if any) is supposed to be continuously polling the quantum state of an unstable bound system subjected to decay via quantum tunnelling. The reason I reckon some kind of polling process should exist in the first place is the following:

According to the QM postulates, the unitary evolution of such a system should by definition keep it reversible, so it is only when measured that a decay can be observed or not. But this would make the decay rate dependent on the measurement rate, while we well know that the decay probability is constant, and the decay deemed "spontaneaous".

What am I missing?

Edit:

From the discussion in the comment section I gather I have not been clear about what I am asking here exactly. Let me try to reformulate the question.

It is about how quantum tunnelling is supposed to explain the exponential dynamics of a decay process. I am not asking about the Zeno effect; quite the opposite, actually: why, in the absence of any measurement, do we have an exponential decay at all? I just do not understand at what point in the unitary evolution of the unstable system the tunnelling effect is spontaneously happening.

The polling process I imagined is just a way to ask "why does the tunnelling effect manifest iself ?" because I cannot see how it can manifest without a measurement. Please do not infer that I am making up my pet theory here. I am only looking for a way to picture the situation, which at this time I don't get at all.

Answer 1

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.

Answer 2

I am adding a few points that Lawrence's answer didn't yet address, but should help clarify the nature of the quantum tunneling process:

According to the QM postulates, the unitary evolution of such a system should by definition keep it reversible, so it is only when measured that a decay can be observed or not. But this would make the decay rate dependent on the measurement rate, while we well know that the decay probability is constant, and the decay deemed "spontaneaous". What am I missing?

The reversibility of the unitary evolution does not preclude the irreversibility of single tunneling events. This is because "reversibility" means different things in the two contexts:

1) A unitary evolution $\hat U$ is said to be time-reversible if whenever $|\psi(0)\rangle \rightarrow |\psi(t)\rangle = {\hat U}(t)|\psi(0)\rangle$ is a valid dynamics, then for $|{\bar \psi}(0)\rangle = {\hat T} |\psi(t_0)\rangle$ at some time $t_0$ and with ${\hat T}$ the antilinear time-reversal operator, $ |{\bar \psi}(0)\rangle \rightarrow |{\bar \psi}(t)\rangle = {\hat U}(t)|{\bar \psi}(0)\rangle \equiv {\hat T} |\psi(t_0-t)\rangle$ gives the exact time-reversed dynamics. This implies that $[{\hat T}, {\hat U}] = 0$, and similarly for the generator of ${\hat U}$, the Hamiltonian ${\hat H}$. For a spin-0 particle the time-reversal operator ${\hat T}$ amounts to complex conjugation in the position representation, but for spin-1/2 and higher it also includes a unitary component acting on the spin degrees of freedom.

2) A tunneling event is irreversible simply because once the particle tunnels through the barrier, the probability that it will tunnel back at some later time is practically null. The actual tunneling corresponds to the $|\psi(0)\rangle \rightarrow |\psi(t)\rangle = {\hat U}(t)|\psi(0)\rangle$ evolution. But let $D$ denote the spatial domain "inside" the barrier, and let $$ P_D(t) = \int_D{dV\;|\langle {\bf x} |\psi(t)\rangle|^2} $$ be the probability to localize the particle within $D$. In general it is assumed that the particle is initially localized entirely within D, such that $P_D(0) = \int_D{dV\;|\langle {\bf x} |\psi(0)\rangle|^2} = 1$. The statement that the probability of reverse tunneling is null then amounts to saying that $P_D(t)$ vanishes asymptotically, $$ \lim_{t \rightarrow \infty}{P_D(t)} = \lim_{t \rightarrow \infty}{\int_D{dV\;|\langle {\bf x} |\psi(t)\rangle|^2}} \;\rightarrow \;0 $$ The fact that the tunneling evolution ${\hat U}(t)$ is still time reversible means that if we now take as initial state the time reversed of the state at some asymptotically large time $t_0 \rightarrow \infty$, $|{\bar \psi}(0)\rangle = {\hat T} |\psi(t_0)\rangle$, when the particle is definitely localized outside of $D$, $\int_D{dV\;|\langle {\bf x} |{\bar \psi}(0)\rangle|^2} \rightarrow 0$, then the evolution of $|{\bar \psi}\rangle$ will be the exact time-reversed of the original one for $|\psi\rangle$, and will asymptotically drive the particle back into domain $D$ through reverse tunneling, $$ \lim_{t \rightarrow \infty}{\int_D{dV\;|\langle {\bf x} |{\bar \psi}(t)\rangle|^2}} \;\rightarrow \;1 $$ In the sense of an individual event, this reverse tunneling into $D$ is just as irreversible as the original tunneling out of $D$, despite the fact that ${\hat U}(t)$ is a time-reversible evolution.

If you prefer, think of an analogy with a free particle wave-packet of well-defined average momentum ${\bf p}$: it always travels in the direction of $\bf p$ and statistically speaking never turns around, despite inherent zitterbewegung and dispersion. But the free particle evolution is time-reversible, and the time-reversed of this wave packet will travel in the $-\bf p$ direction, etc.

[…] why, in the absence of any measurement, do we have an exponential decay at all? I just do not understand at what point in the unitary evolution of the unstable system the tunneling effect is spontaneously happening.

It is not so much at what point in time tunneling takes place, since this is essentially a statistical process, but what is the probability that at time $t$ the particle is located outside the inner domain $D$. To determine this probability we do not really need to follow the dynamics of the same system through multiple queries. In general we cannot even do this without altering the entire dynamics. What we must do is to query ensembles of identically prepared systems at different times $t$, ideally using a different ensemble for each such query. In principle, this would ensure that the measurement procedure does not interfere with the dynamics up to that point in time, and that the measured probability is indeed reliable.

Exponential decay then means that the fraction of particles detected outside domain $D$ in each such query diminishes exponentially with the time elapsed since the preparation of the respective ensemble. Note that the average tunneling time is statistically well defined, although we do not need to consider an average over exact individual tunneling times for each of the particles in an ensemble.

(From a comment to Lawrence's answer) to get a neat exponential law, it seems to me that some quasi-continuous interaction is needed. I have not seen that point discussed.

Actually it has been extensively researched in relation to a multitude of distinct fields, from nuclear physics to chemical kinetics and biophysics. See these "Lectures on Dissipative Tunneling", and google the Leggett-Caldeira model.