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There is no tunneling in the case of infinite potential barrier, but there is when we have a finite well. In the classical analog, in the first case we have a particle bouncing between to infinitely rigid impenetrable walls and there is no tunneling, same as the quantum case. But if we have a finite barrier, means we have walls of finite rigidity, say made of cork or something. Then the particle would just break through some of the cork and it's probability of being found further in the cork wall will decay steadily.

I can understand discrete energy levels being a new thing, because they behave like a wave that's confined and not like particles confined, but why tunneling?

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Marek's answer is a good one. One additional comment: there definitely is tunneling in the behavior of classical waves. Set up a light wave in glass so that it should totally internally reflect off of a face of the glass. Put another piece of glass near the first one, and some of the light will get through. This is precisely the same phenomenon as quantum tunneling, and it occurs even when the electric field is treated purely classically. So rather than saying that tunneling is quantum not classical, I'd say it's associated with waves. –  Ted Bunn Mar 18 '11 at 19:31

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Not a good analogy. Here's why.

Your cork wall system is very complicated. That wall isn't just a potential with finite height. Rather, it contains holes where the potential is zero (or near zero), so that the particle can just propagate further and it contains full woody parts where the potential is infinite and the particle bounces back. So it is a very complicated many particle system which you can reduce to one parameter (like penetration depth, possibly depending on energy). But the resulting system will be probabilistic and very different from just a single particle with finite height potential. Because in this simple system the particle will bounce back with 100% probability if its energy is below the potential and pass with 100% probability if its energy is bigger.

Now, if you take this simple system (with just one degree of freedom) and investigate it from the point of view of quantum mechanics you'll discover that there's non-zero probability that particle will pass through the potential even if it doesn't have sufficient energy to pass it.

Why is this strange? Well, the particle has a total energy $E = T + V$. This means that its kinetic energy is ${1\over 2}mv^2 = E - V$. As you can see, this equation makes no sense classically if the quantity on the r.h.s. is negative because you'll obtain imaginary velocity. But in quantum mechanics this still makes sense because the particle is also a wave and that imaginary factor means that this wave will exponentially decay, rather than just propagate freely.

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Marek's answer is rather too non-statistical classical. Your cork analogy looks to me at first sight not statistical in a way that gets close enough to quantum theory. As an aside, it's generally better to make your models slightly more abstract-idealized mathematical — note that Marek has replied with a model that is both much more abstract and much more idealized than your model.

If you're willing to make the potential well have expected depth $E$ with some finite variance because the apparatus that constrains the classical particle is noisy, however, then the analogy is not so bad, if the potential well is then sometimes shallow enough for the classical particle to escape. There is noise in quantum theory —insofar as one carefully thinks in terms of quantum fluctuations and is careful to distinguish them from thermal fluctuations—, so it doesn't seem unreasonable to introduce noise into classical models.

You can alternatively surround the classical particle with some kind of noisy environment, in which case the energy of the particle will fluctuate. If the fluctuations are enough that the energy of the classical particle is sometimes more than the depth of the potential well, then again the particle can escape from the potential well with some finite probability. If you haven't previously looked at Nelson models, get hold of his "Quantum fluctuations / by Edward Nelson, Princeton University Press, c1985." I don't know of and couldn't find a web resource that gives an adequate description of this approach, however, even though it's arguably closer to being acceptable to Physicists than de Broglie-Bohm type approaches.

Personally, your intuition at this level is OK, but it's not in this context that the really difficult issues arise, and also not in the double-slit kind of context. Kochen-Specker and the violation of Bell inequalities are much more challenging. In response to those challenges I have found it much more productive to characterize the differences between classical random fields and quantum fields, both of which are intrinsically statistical, instead of trying to think in terms of classical particles in comparison with quantum particles. The random field context is close enough to the quantum field context that discrete energy levels are not a novelty. I presume you're not getting my Answers to your Questions, but your Questions are very familiar to me.

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There is no qualitative difference between the quantum and the classical case:

A quantum particle has a small probability of crossing the barrier because of its random kinetic energy.

So does a Brownian particle in a classical barrier potential.

  • In both cases, as the barrier gets higher, one needs more and more kinetic energy to cross the barrier, and the probability of getting this much kinetic energy through random collisions (or whatever else the -classical or quantum - random process is) gets smaller and smaller.
  • In both cases, crossing a narrow barrier takes much less time than crossing a wide barrier, as the particles crossing most likely will have very little kinetic energy left when on top of the barrier.
  • In both cases, only an infinitely high barrier cannot be crossed at all.
  • Edit2: In both cases, given a fixed barrier width, one can make the mean crossing time arbitrarily large by making the barrier high enough. And given a fixed barrier height, one can make the mean crossing time arbitrarily small by making the barrier narrow enough.

    Thus tunneling is a misnomer, as it is actually a process of climbing across a barrier, and has nothing to do with going through a tunnel.

    Thus qualitatively, the classical and the quantum situation behave identical. Of course, a quantitative treatment reveals the usual differences between a classical and a quantum treatment of the same phenomenon.

    Edit3: Therefore it is purely a matter of taste whether one considers the process of crossing a barrier as a process of tunneling through a barrier or of climbing across a barrier, and both figures of speech are applicable in the classical case and in the quantum case.

    Edit1: More quantitatively, for a Brownian particle with unit diffusion constant in a potential $V(x)$ with the values infinite for $x\le -1$, $U$ for $x\in[0,1]$, and zero otherwise, the mean escape time from the square well is $\frac{L^2+1}{2}+Le^{U}$ (I used formula (5.2.60) from Gardiner, Handbook of stochastic methods, 1985, with $x=a=-1$, $b=L+\epsilon$). Taking $L=0$, we see that the mean time to reach the barrier is $\frac{1}{2}$; the mean time needed to cross the barrier once reached is therefore $T=\frac{L^2}{2}+Le^{U}$. At given $L>0$, we can make $T$ arbitrarily large by sufficiently increasing $U$. And at given $U$, we can make $T$ arbitrarily small by sufficently decreasing $L$.

    Something similar holds in the quantum case, although the usual treatment of tunneling is not in terms of a mean crossing time. http://galileo.phys.virginia.edu/classes/751.mf1i.fall02/OneDimSchr.htm says that the probability of tunneling of a free quantum particle through a square barrier of width $L$ and height $U$ is $p=4k^2a^2/[(k^2+a^2)^2\sinh^2(aL)+4k^2a^2]$, which decays like $e^{-2aL}$ (Here $a$, roughly proportional to the square root of $V_0$ depends on the energy of the state considered). I don't know how to translate that into a mean crossing time, as it is a probability rather than a rate. Maybe someone else can help out. However, as the tunneling probability decays with increasing $L$, the mean crossing time will also grow strictly monotonically with $L$ and with $U$, and will vanish when $L=0$, leading to the qualitative picture described above.

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    If it only has to do with climbing, why does the tunneling probability decrease exponentially with barrier width? –  user2963 Mar 13 '12 at 17:20
        
    Can you post a reference for the calculations, so that I can check and explain it for myself? –  Arnold Neumaier Mar 13 '12 at 17:26
        
    search "Tunneling through a Square Barrier " in galileo.phys.virginia.edu/classes/751.mf1i.fall02/… –  pcr Mar 13 '12 at 17:35
        
    See for example the last equation on this page, the transmission coefficient for a finite barrier: en.wikipedia.org/wiki/Quantum_tunnelling –  user2963 Mar 13 '12 at 17:40
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    I removed the disqualifying phrase ''misnomer'' and replaced it by saying it is a matter of taste how to regard it. But I still don't understand why the tunneling metaphor should apply in the quantum case but not in the classical case. @zephyr: What counts classically is an integral over the barrier width involving the potential, hence a very high barrier is counterbalanced by a very narrow integration interval. If you find this counterintuitive, you should instead acknowledge that a classical particle can also tunnel! –  Arnold Neumaier Mar 16 '12 at 17:11

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