radioactivity - How can quantum tunnelling lead to spontaneous decay?

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

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.