If, say, a particle with energy $E<V_0$, approaches a finite potential barrier with height $V_0$, and happens to tunnel through, where would the particle be during the time period when it is to the left of the potential barrier and to the right of the potential barrier? Surely there must be a finite amount of time for it to travel through to the other side, unless it simply teleports there? If it travels through with energy less than $V_0$, however, doesn't that mean it cannot enter in the region of the potential barrier?
Quantum Mechanics – Determining the Position of a Particle During a Tunneling Event
quantum mechanicsquantum-tunneling
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This model is something that I've seen a lot of students (including myself) struggle with, and a big part of it is that calling it the "Coulomb barrier" is quite over-simplified.
Going back to fundamental physics principles, the Coulomb force is repulsive as you say, so I'll write this as $F\propto 1/r^2$, where $F$ is the force and $r$ is the distance from the center of the nucleus and the positive sign indicates that the force is in the positive $r$ direction, away from the nucleus. The potential, which is what's shown on the graph, is the negative of the integral of the field (which is proportional to the force), this results in $E\propto 1/r$. For an analog, imagine lifting a weight where "up" is the positive z-axis. The path involves an increase in z-coordinate and the force from the field (gravitational) is negative. The negative integral of the field over distance is the potential.
The missing part is why they "chop off" this $1/r$ function at some radius, presumably the radius of the nucleus. Well, this allows us to infer some things about the other force present, which is the nuclear force. It would appear that the force acts weakly beyond the radius of the nucleus, since the $1/r$ form is unaltered beyond that point. It also absolutely must have a potential that increases faster than $1/r$ as $r \rightarrow 0$. Both of these would be loosely satisfied if the nuclear potential was $-1/r^n$ (note the negative) where $n>1$. The alpha particle, while in the nucleus, by the way, has kinetic energy which is the reason its energy is higher than the hypothetically lowest energy level possible for it. This has to do with the fact that quantum mechanics only allows certain (quantized) energy levels.
Just For an example of how this is possible, I'll say $E_{nuclear}=-1/r^4$ (I am not saying this is how the force acts, it's just for utility).
$$E(r) = E_{nuclear}(r) + E_{Coulomb}(r)$$ $$E(r) = 1/r - 1/r^4$$
$$F(r) = -\frac{d}{dr} \left( E_{nuclear}(r) + E_{Coulomb}(r) \right) = F_{nuclear}(r) + F_{Coulomb}(r) $$ $$F(r) = 1/r^2 - 1/r^5$$
My intent is that this answer allows you to explicitly answer to yourself what the potential and force contributions are, and give some very rudimentary example for how the "edge" can appear. Beyond that I hope it's clear how someone may take the shape of the above curve and lump it into a "wall".
Yes you appear to "understand the whole wavefunction / probability distribution thing right", and it is possible to measure a probability of presence in the forbidden region, but this probability is usually small.
I don't know whether any experiment has shown it with "real particle" like atoms, electrons or neutron, but it has been definitely shown with photons. The "classically forbidden region" in the particle pictures corresponds to evanescent waves, and there have been experiment where evanescent waves have been used to excite the fluorescence of atoms. Each photon emitted by an atom in the evanescent wave can then be seen as a position measurement of a photon in the evanescent wave region.
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Isn't the whole point here that one cannot say where the particle IS exactly? One can only calculate the probabilities of it being at one place.
Tunneling means the probability of it being inside the barrier isn't zero (since we want the probability distrubition to be continuous). There is always penetration of the wave function into the barrier.
IMHO tunneling means the penetration goes deep enough to actually reach the other side, so the wave function of the particle is propagated further on that side too, meaning there's a chance the particle went through. During the passing the particle has had a chance of being inside the barrier.
I don't know if it's correct to say that when the particle has passed, it has been inside the barrier, but that just because the notion of the particle actually being somewhere is somewhat wrong.