I understand that if a particle approaches a finite potential barrier of height $V_0$ with energy $E < V_0$, there is still a finite probability of finding the particle on the other side of the barrier due to quantum tunneling.

My question is, since the wavefunction is nonzero inside the barrier region, is it possible to actually make a position measurement and locate a particle inside the barrier?

I mean, if we can say that "there is a nonzero probability that the particle is inside the barrier", surely this would suggest we can do so?

If not, why not? Am I understanding the whole wavefunction/probability distribution thing right?

  • $\begingroup$ Possibly related: physics.stackexchange.com/q/9842/2451 $\endgroup$ – Qmechanic Jun 17 '11 at 3:14
  • 3
    $\begingroup$ @Josh Chen: Is the question that you would like to have an example of a realistic experimental setup of a quantum system + a position measuring device, that could be used to detect a point particle in a classically forbidden tunneling region? $\endgroup$ – Qmechanic Jun 17 '11 at 9:40
  • $\begingroup$ Well, that would be very helpful, if such a thing is physically possible. Because we're learning this stuff in 2nd year undergrad physics now, and my lecturer says that due to the uncertainty principle we can never actually physically detect a particle inside the barrier. In his words: "The wave-like nature of the particle penetrates into the barrier, but not the particle-like properties". $\endgroup$ – Josh Chen Jun 17 '11 at 10:31
  • 3
    $\begingroup$ If the particle can be found inside the barrier, how is energy conserved? (Assuming the wave function is an energy eigenfunction with energy E < V_0, finding the particle inside the barrier would violate conservation of energy, no?) $\endgroup$ – user1247 Jun 17 '11 at 11:55
  • 1
    $\begingroup$ @user1247 : The energy is E, it is the same everywhere and it is conserved. It is just the classical notion of kinetic energy which is not applicable when E < U. $\endgroup$ – Vladimir Kalitvianski Jun 17 '11 at 19:02

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.

  • 1
    $\begingroup$ Yes, though it's important to note that this is essentially a classical phenomenon. $\endgroup$ – Emilio Pisanty Mar 5 '13 at 15:34

If you have been following particle physics, you will know that a proton is composed of three quarks that are never able to come out of the nucleon formation, i.e. the potential well in a quantum mechanical picture.

How do we know they exist if not by measurement inside that potential well?

Our tool is scattering. When one scatters an electron against a proton, in the potential barrier formulation, the trajectory of the electron after scatter reflects the wave function of the particles inside the nucleus because the electron wave function penetrates, i.e. has a solution, inside the potential barrier.

The first experiments decided that there are partons inside the nucleus, because the scattering showed a hard core. Eventually these were studied in various experiments and the Standard Model of physics was established.

So you are correct in your surmises.

Edit: from thinking on the answer of Peter Morgan I will add that Xrays are an even simpler example. Even simple xrays of one's hand. The skin is a potential barrier and the xray photons wave function is scattered within, conveying information when they come out by their scattering angle and density, on the bones etc.

  • 1
    $\begingroup$ This. The terminology is that the tunneling particle is "off-shell" when it is "in" the forbidden region, but can be knocked "on-shell" by the appropriate scattering event. Use a high resolution detector stack and you can observe the "negative" net energy of the target particle. $\endgroup$ – dmckee Jun 17 '11 at 16:25
  • $\begingroup$ Dear @anna v: I get the impression that the answer (v2) is mostly discussing particles inside a (classically allowed) potential well, rather than inside a (classically forbidden) potential wall, as Josh Chen seems to be asking about. $\endgroup$ – Qmechanic Jun 17 '11 at 18:37
  • $\begingroup$ Dear @Qmecanic when the particle is outside the potential well, the inside is "forbidden" classically. An xray photon goes "through" the barriers. If you are saying that the energy sets it higher than the barrier, still it is a scattering exploring the inside of the barrier. There is a wave function that describes the whole, xray-skin-bone that is modified by the contents inside the barrier and allows measurement by the interference pattern of the beam. Any potential well is "forbidden" classically, imo. The E<V0 is really irrelevant, because classically the E>V0 would be destructive anyhow. $\endgroup$ – anna v Jun 18 '11 at 4:08
  • $\begingroup$ Dear @anna v: I agree with your comment for its actual physics content. In terms of semantics, however, it should be stressed that the traditional definition of a classically forbidden region in quantum mechanics just refers to the region with $E < V_0$. In particular, the traditional definition does not depend on the initial position the particle. $\endgroup$ – Qmechanic Jul 14 '11 at 15:33
  • $\begingroup$ This is nonsense. There is no potential barrier for an electron scattering from a proton. The entire interior of the proton is classically allowed. $\endgroup$ – Ben Crowell May 21 '18 at 2:38

Yep, it is indeed possible for a measurement of the particle's position to reveal that it is inside the barrier, because the wavefunction there is nonzero - or more precisely, because the quantity

$$\int_\text{barrier} \langle\psi|x\rangle\langle x|\psi\rangle\mathrm{d}^n x$$

which represents the probability of the particle being measured within the barrier, is nonzero. Keep in mind that a "measurement" can be any interaction with another particle, it doesn't have to be performed by an actual measuring device. So if the particle interacts with the barrier itself, that counts. In that case you could have something like the particle tunneling partway through the barrier and possibly getting stuck in the middle (but you wouldn't see it because your measuring device can't get in there).

  • $\begingroup$ I've always found this sort of idea surprising. Shouldn't we distinguish a recorded event in a detector, the statistics of which can be checked against a QM model, from an unrecorded "interaction with another particle", which in principle cannot be checked? If the "interaction with another particle" causes (in a QM sense) recorded events, then we can check statistics against the whole model. I'm happy enough to call these something like "measurement events" and "indirect measurements", but it seems best to make some distinction. Indirect measurements are not exactly experimental physics. $\endgroup$ – Peter Morgan Jun 17 '11 at 15:42
  • 1
    $\begingroup$ @Peter In an exclusive (or nearly so) elastic (or quasi-elastic) measurements we can reconstruct the energy and momentum of the target particle in classically forbidden region of the phase space. $\endgroup$ – dmckee Jun 17 '11 at 16:31
  • $\begingroup$ @dmckee It's the "nearly so" that's trouble. Sure, if you assume that there's a one or few particle Schrodinger equation that describes the recorded statistics of events, then enough data (but not too much or too accurate) fixes what the potential and the state must have been, and, to that precision, whatever we might like to measure, even if we don't or can't. But that doesn't extend to the details, when QFT is necessary, and we have to accommodate, e.g., infinite numbers (or a continuum) of Feynman diagrams. Such concerns being awkward, we're careful about going there. $\endgroup$ – Peter Morgan Jun 17 '11 at 17:10

To be close to second year undergrad physics, consider a non-relativistic electron with energy $E$, bounded to a double well potential $V({\bf r})$ with a classically forbidden tunneling region with potential energy $V_0$ in between, i.e., $E<V_0$. (Let us assume for simplicity that the full potential profile $V({\bf r})\leq V_0$, i.e., $V_0$ is a global maximum for the profile.)

An example of a double well
(source: orst.edu)

Figure 1: An example of a double well.

As I read the question, Josh Chen is not disputing that an electron prepared in one well can reappear in the other well. Instead, the question is that since the integral of the square of the wave function over the classically forbidden tunneling region

$$ \int_{\{{\bf r}\in\mathbb{R}^3\mid V({\bf r}) > E\}} d^{3}r \ |\Psi({\bf r},t)|^2~>~0,$$

is strictly non-zero, does that actually mean experimentally that there is a non-negative probability to find the electron inside the classically forbidden tunneling region as the Born rule tells us, and how would one measure that probability, at least in principle?

Yes, the Born rule holds also in this situation. To measure the position of the electron, we will here use a photon with wave length $\lambda$ and of energy


We will assume that the energies involved

$$|E|, |V|, E_{\lambda} ~\ll~ E_0=m_0 c^2$$

are much smaller than the rest energy $E_0$ of the electron, so we can treat the electron using non-relativistic quantum mechanics.

The electron's wave function $\Psi({\bf r},t)$ is exponentially decaying in the classically forbidden tunneling region with a characteristic tunneling penetration depth

$$\delta ~\sim~ \frac{h}{\sqrt{2m_0(V-E)}}~=~\frac{hc}{\sqrt{2E_0(V-E)}}.$$

(Since we are not really interested in the possibility that the electron could reach the other well, let us for simplicity assume that the electron penetration depth $\delta<\Delta$ is smaller than the separation $\Delta$ of the two wells, i.e. we are effectively studying a single well.) To use the photon as a 'microscope' in order to claim that we have detected the electron inside the classically forbidden tunneling region, the 'microscope' should have a resolution better than the electron penetration depth. In other words,

$$\lambda \ll \delta \qquad \Leftrightarrow \qquad E_{\lambda} \gg \sqrt{E_0(V_{0}-E)} $$ $$ \Rightarrow \qquad \frac{E_{\lambda}}{E_0} \gg \sqrt{\frac{V_{0}-E}{E_0}} > \frac{V_{0}-E}{E_0} \qquad \Rightarrow \qquad E+ E_{\lambda}\gg V_{0},$$

i.e., the photon could knock the electron completely out of the well profile, so that the electron continues to spatial infinity. In principle, the incoming photon could be aimed at the classically forbidden tunneling region, and we could have prepared detectors in a $4\pi$ solid angle to capture and measure energy and momentum of all outgoing particles (the electron plus photons), and then calculate backwards to determine that a scattering event must have taken place inside the classically forbidden tunneling region. The missing energy between the incoming and the outgoing particles will be equal to the classically forbidden energy $E-V_{0}<0$.

On the other hand, if we had used soft photons with energy $E_{\lambda}<V_{0}-E$, the above inequalities get reversed, and the resolution will be too poor to determine whether the electron is inside or outside the classically forbidden tunneling region, cf. the uncertainty principle.


If one can observe individual events on the other side of the potential barrier from the source, but only when the source is turned on, then one might say that something must have caused the events. On the macroscopic scale, it must have been the source because it's only when the source is turned on that there are events. One might say, further, on the microscopic scale, that whatever it was that caused the events must have come through the space between the source and the apparatus in which the events happened, it can't have just jumped the gap. In this sense, we might say that detecting a particle on the other side of the barrier shows that a particle must have passed through the barrier, so this is a detection of a particle in the barrier. We know, however, that thinking in terms of something like classical particles is problematic, so that it's best not to think of there being a particle that travels from one place to another.

I liked Anna v's Answer enough to upvote it, but I point out that it is only by a specific kind of semi-classical inference that we can say that an event on the other side of the lab (which we say is caused by a scattered particle) means that there was another particle that caused the scattering. In this case, where there is a potential barrier, whatever material apparatus implements the potential barrier will get in the way of being able to do such a scattering experiment in any clean way. It would only be by subtracting the background caused by the material apparatus that we could tell that there were events that were caused by the "particles that classically couldn't be there because of the potential barrier". The significance of the potential barrier for the interpretation of the recorded events in terms of particles is that there is more background. [There's never no background to subtract, but often the background can be made small enough that we don't have to pay much attention to it. A common failing of interpretations, IMO, is to ignore the background in principle.]

Here we come to the statistical character of QM. Because we have to subtract the background, we cannot say that any individual event is caused by an individual particle scattering another particle. In that sense I think we cannot say that we can "locate [an individual] particle inside the barrier" by scattering (or by other interactions inside the barrier), because any individual event may have been caused by the background. Where there is more background, as there must be when we are trying to detect inside a potential well, we are less able to attribute individual events to identifiable individual particles.

Nonetheless, we can establish that a given wave function is a very effective tool for predicting just how many detection events we would record if we put a given type of detection apparatus at any given point, and if we ensure that there is very little background. Then, we can extrapolate to say what we would observe if we put that type of apparatus in a place where we in fact cannot put it. That's not "physical" in the sense that I think you mean it, but it's mathematically reasonable just up to the point that we think about actually implementing the measurement, and it lets us imagine how we would engineer new experiments in a useful way. I take this to be the basis of David Zaslavsky's Answer (which is more-or-less OK, but I did not upvote it because I think it does not enough address your question).

IMO, however, without explication here, it's good to start to think in terms of the quantum field being the intermediary when modelling an experiment, instead of particles, even when you're doing 2nd year undergraduate Physics.

  • $\begingroup$ Scattering experiments are very carefully done in vacuum, in order to have as clean as possible interactions. Or in bubble chambers where the tracks of the input scattering and output scattered can be observed. Of course there are efficiencies for the detector equipment but these are modelled and taken into account. When one measures structure functions one is doing "an Xray" of the nucleon target. $\endgroup$ – anna v Jun 17 '11 at 15:02
  • $\begingroup$ @Anna Right, we do experiments in vacuum when possible, but if there's a potential well as well as the quantum field that we want to measure, that's not a vacuum. "taken into account" means that any individual recorded event may be attributable to the background, not to the particles that are of principal interest to a particular experiment. But at the statistical level we may be able to say something very detailed and specific, such as that 41.2%, say, of the scattering events can be attributed to the quantum field of interest. Structure functions are statistical. $\endgroup$ – Peter Morgan Jun 17 '11 at 15:21

Frédéric's answer is the best one. I myself was going to write something like scattering light in the forbidden region by admixtures or photo-reactions in the forbidden region. Note, to cause a photo-reaction, you have to have sufficient energy. In a quasi-stationary case the energy is $E=\hbar \omega$. So it is the same as outside the barrier. No energy conservation violation occurs. You just observe a particle in a classically forbidden region, but quantum mechanically it is not forbidden. That is why the probability is not zero.


A barrier region can be thought as just a different mean, and like Fredderic said if the wave interacts somehow with something in this mean it is a empirical manifestation of the wave. This interaction can be interpreted as the particle detection, or as the manifestation of the presence of the quantum system inside the region. whatever you call it particle or not.


Actuality as i understand it the problem is that as you approach the particle in any way to measure it IT wont be there, further more just looking at it will change the nature of the experiment.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.