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In comments to a Phys.SE question, it has been written:

'Tunneling' is perfectly real, even in classical physics. [...] For sufficiently large temperatures this can put the system above a hump in its potential energy.

and

the only difference between the classical case and the quantum mechanical one is that classical physics is a random walk in real time, while QM is a random walk in imaginary time.

I understand that in a system of particles with finite temperature some particles can overcome a potential barrier. That's how I interpret the first statement. I don't understand the business of "random walk in imaginary time". Can someone explain?

Update

What I was originally looking for was 1.) classical system that can transport mass through a forbidden region and 2.) explanation of "random walk in imaginary time". So far, I don't see anything for question 1.), but I think I'll grok 2.) if I invest some time and energy.

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    $\begingroup$ The Brownian motion of a particle subjected to random forces can be described by the Fokker-Planck equation, which is very similar to the Schroedinger equation of the free particle after you transform the time variable by multiplying it with an imaginary constant t -> it. That's called a Wick rotation and relates statistical mechanics to (single particle) quantum mechanics. If you look at the path integral solution for the Schroedinger equation, you will find that one can interpret the probability of finding a particle in a new position as an integral over all possible paths (random walks). $\endgroup$ – CuriousOne Sep 12 '14 at 19:18
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    $\begingroup$ The quotes in your question are of course perfectly wrong... or at least absolutely sloppy and then obviously dangerous. You should try checking the literature about instantons [ en.wikipedia.org/wiki/Instanton ] to understand the trick of Wick rotation @CuriousOne is talking about, and also the difference between tunnelling and classical rolling of the particle in the bottom of a double-well potential. Also a look at this article arxiv.org/abs/cond-mat/0602288 and other paper from Dykman might be helpful. $\endgroup$ – FraSchelle Sep 17 '14 at 5:47
  • $\begingroup$ The quotes in my question denote the fact that I'm copying the phrase verbatim from another source. What's wrong with that? $\endgroup$ – garyp Sep 17 '14 at 12:30
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Frustrated total internal reflection is an optical phenomenon. It's such a close analogue to quantum tunneling that I sometimes even explain it to people as "quantum tunneling for photons". But you can calculate everything about it using classical Maxwell's equations.

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    $\begingroup$ Ok, that's a wave phenomenon. The electric field in the gap is perfectly well defined, and there's no problem with conservation of energy. I don't think. Also, I don't buy "photon tunneling". But it suggest that we look in the realm of the Hamilton-Jacob equation for analogs. $\endgroup$ – garyp Sep 12 '14 at 19:32
  • $\begingroup$ @garyp: Photons tunnel just fine. Total internal reflection is a perfectly suitable phenomenon for that. Of course it can be explained as a wave phenomenon. To the best of my knowledge almost everything that photons in do IN OPTICS in quantity can be explained as a wave phenomenon. Lower the intensity and install a photomultiplier tube, and you will recover the things the optical waves do with particle counts. $\endgroup$ – CuriousOne Sep 12 '14 at 19:42
  • $\begingroup$ @CuriousOne Well photons do not exist as particles. They are excitations of the EM mode. So I see what you are saying, but I think the interpretations of photons as particles in this discussion does not help me understand "tunneling in classical mechanics" $\endgroup$ – garyp Sep 12 '14 at 19:53
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    $\begingroup$ @garyp: Photons are just as much particles as any other elementary particles. They all have both particle and wave properties, just different fields are involved. $\endgroup$ – Jan Hudec Sep 12 '14 at 20:43
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    $\begingroup$ @garyp you say "photons are not particles, they are excitations of the EM mode". But the same is true for electrons: they are not particles, they (and positrons) are excitations of the corresponding Dirac field. There's not much difference other than electrons are charged massive fermions and photons are uncharged massless bosons. $\endgroup$ – Ruslan Nov 24 '14 at 13:00
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There are two ways to see the analogy between the "quantum diffusion" and classical diffusion. The first one, I think the easier one is comparing the Schrödinger equation with the diffusion equation: $$i \partial_t \psi = -\sum \partial_{xx} \psi$$ (forgetting all the $\hbar,m$ factors) When you transform $t \to -i \tau$ you get the usual diffusion equation $$\partial_\tau \psi= \sum \partial_{xx} \psi$$ The diffusion equation is precisely the one which governs the probability distribution of a random walk so we can say the quantum propagation is a random walk (or rather diffusion) in complex time. But I don't really think this gives a clue why tunneling should happen with both.


The second way is through the path integral. It was derived even before Feynman or Dirac in 1923 by Wiener that we can formulate the random walk transition probability $P(x_0,x_1,T)$ from $x_0$ to $x_1$ in time $T$ as $$P(x_0,x_1,T) = \int \exp(-S_0[x(t)]/\xi) \mathcal{D}[x(t)],\; S_0[x(t)] = \int_0^T \frac{1}{2} \dot{x}^2 d\tau$$ With $x(0)=x_0$ and $x(T)=x_1$ and $\xi$ a parameter of the random walk. When we substitute $t \to -i\tau$ we will have three $i$ in the action - two in the time derivative and one in the action. In the end we get the Feynman path integral $$\int \exp(iS_0[x(t)]/\xi) \mathcal{D}[x(t)]$$ But same as the wave-function, the propagator isn't the same object as the transition probability - it has to be squared. Once again - quantum propagation is kinda like a random walk in complex time (but not entirely).


Now to a classical analogue of tunneling. Suppose you have a small charged particle in a fluid and an electrostatic potential blocking it from entering and area with a potential barrier $V$. I say it is charged so that the quasineutral fluid is not affected and we can focus only on the particle but a general case would work similarly. The particle has an average kinetic energy $3 k_B T/2<V$ from the molecules bouncing off from it. But even though there seems not to be enough energy in the surroundings, there is a non-zero probability it will cross the barrier because the particles can "bounce it through" even though with some difficulty.

This is analogous to the quantum particle which can never be found to have an energy larger than $V$, yet you find it on the other side. Maybe not surprisingly, this bouncing of a particle in a fluid is modeled as a random walk with a diffusion equation.

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  • $\begingroup$ The particle gets across the barrier because by happenstance a series of collisions occur which give it enough energy to get over. I'm not disputing what you are all saying, I'm trying to understand it. It's news to me. I'm not clear on the tunneling analogy. Can we say that for the quantum particle that in the time it would take for the tunneling to occur ($\Delta t$) that there is enough uncertainty in the particle's energy ($\Delta E$) that it might be found to have enough energy to get over the barrier? $\endgroup$ – garyp Sep 12 '14 at 21:38
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Consider the path integral:

$$\int Dx \exp\left(i\int\left(\frac{m\dot{x}^2}{2}-V\right)dt\right)$$

You can consider paths in "imaginary time" by performing a Wick rotation $t\to i\tau$

$$\int Dx \exp\left(-i^2\int\left(\frac{m}{2}\left[\frac{1}{i}\frac{dx}{d\tau}\right]^2-V\right)d\tau\right)$$

$$\int Dx \exp\left(\int\left(\frac{m}{2}\left[\frac{1}{i}\frac{dx}{d\tau}\right]^2+V\right)d\tau\right)$$

So we're now considering a classical action $S[x]=\int\left(\frac{m}{2}\left[\frac{1}{i}\frac{dx}{d\tau}\right]^2+V\right)d\tau$.

The potential has been inverted. Where we had, for example, a double well potential, we now have a double "hump" potential. There is a classical trajectory associated with going from one hump to the other i.e. it rolls down the hill into the centre, and then up the other hill coming to a stop at the top. But this path is in imaginary time (the instanton solution).

The path integral lets us calculate the probability of the particle travelling from one position to another over some, here imaginary, time. a particle will move around space like a random walk with these probabilities.

Can somebody perhaps be more formal/give a more in depth explanation? This is a little vague I realise...

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  • $\begingroup$ Hmm. I'll need more to chew on. Maybe I need a book. $\endgroup$ – garyp Sep 12 '14 at 19:34
  • $\begingroup$ As an experimentalist I can't comment on the details, but ifca.unican.es/users/wio/Index_archivos/Talks/StEtienne.pdf seems to give a nice introduction to path integrals for stochastic equations. So do a great many textbooks, of course. Hagen Kleinert has made a cottage industry out of using path integrals for almost everything that seems to move in some random way, see e.g. his book "Path Integrals in Quantum Mechanics, Statistics, and Polymer Physics, and Financial Markets, Third Edition". I remember the first edition covering the basics well. $\endgroup$ – CuriousOne Sep 12 '14 at 19:35
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As for "a classical analog to quantum mechanical tunneling", theoretically one can jump over a classical barrier having lesser kinetic energy than the potential energy one's mass would have at the top of the barrier. In fact, in the course of a high jump, one can bend over the barrier in such a way that one's center of gravity will be outside of the body and pass under the barrier. See, e.g., A. Cohn, M. Rabinowitz, Classical Tunneling, Int'l Journ. Theor. Phys., v. 29, #3, 1990, p. 215.

If you prefer a strictly one-dimensional classical analog, you can imagine a train of masses connected with springs, which passes a one-dimensional potential barrier - again, if the thickness of the barrier is less than the length of the train, the latter does not have to have the kinetic energy equal to or greater than that of its total mass at the top of the barrier. However, the high jumper example seems more graphic.

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  • $\begingroup$ But in none of those cases does the system pass through a forbidden energy state. $\endgroup$ – garyp Sep 13 '14 at 2:28
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    $\begingroup$ @garyp: Because of the uncertainty relation for time and energy, I don't think we can be sure a quantum system passes through a forbidden energy state. $\endgroup$ – akhmeteli Sep 13 '14 at 3:28
  • $\begingroup$ Yeah. See my comment to the answer by @Void $\endgroup$ – garyp Sep 13 '14 at 10:57
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The last part of @Void's answer is spot on. The magic words are "Kramer's escape rate problem". See: https://home.icts.res.in/~abhi/notes/kram.pdf

This phenomenon occurs when a classical particle is both confined to propagate in a local potential energy minimum and placed in an ambient thermal bath. The particle can "tunnel" out of the potential energy well, despite not having enough deterministic kinetic energy of its own, due to the thermal fluctuations of the bath. As in the case of quantum tunneling, the probability that the particle escape its potential energy well is proportional to the negative exponential of the energy "height" of the barrier. The fact that this phenomenon is not called "thermal tunneling" is just historical accident/Copenhagen propaganda.

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Evanescent waves are the mechanism beind both quantum tunneling and frustrated total internal reflection in @SteveB's answer. Evanescent waves and frustrated total internal reflection are not limited to light, but can occur in any phenomena governed by the wave equation, including sound and water waves.

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  • $\begingroup$ In evanescent wave phenomena, no mass crosses the forbidden region, in contrast to tunneling. $\endgroup$ – garyp Sep 13 '14 at 2:32
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    $\begingroup$ Quantum tunneling is an evanescent wave of the Schroedinger equation, so in that case it does. Classical evanescent waves transfer energy across the forbidden zone so there is not that much conceptual difference. $\endgroup$ – Daniel Mahler Sep 13 '14 at 2:57
  • $\begingroup$ Take a look at my update to my question. $\endgroup$ – garyp Sep 13 '14 at 3:01
  • $\begingroup$ Not sure how the update applies to this. Quantum tunneling is mathematically an evanescent wave. EWs also have classical realizations as totally frustrated internal reflection. The update does not really help specify what it is you are looking for or why EWs and TFIR is not a classical analog of quantum tunneling $\endgroup$ – Daniel Mahler Sep 13 '14 at 3:06
  • $\begingroup$ Yeah, unfortunately I was thinking out loud. In the meantime, I've replaced that update with a clarification of the original question. $\endgroup$ – garyp Sep 13 '14 at 3:18
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In the case of complex velocity, there is not only a tunnel effect, but also heat transfer from a cold to a hot body. This is due to the fact that the real part of the velocity square may become negative $(Re V+iImV)^2=(ReV)^2-(ImV)^2+2iReV ImV$. The imaginary member of the complex number is compensated by the complex conjugate member. This is the effect of complex speed. This corresponds to the turbulent flow regime. $$\frac{\partial T}{\partial t}+V_k\frac{\partial T}{\partial x_k}=\chi \Delta T+\frac{\nu}{2c_p}(\frac{\partial V_i}{\partial x_k}+\frac{\partial V_k}{\partial x_i})^2$$

For a large imaginary part of the velocity, the real part of the last term may be negative. At real speed, he describes heating of the body by changing the speed. At complex speed, it describes the cooling flow. It may happen that the de-negative part of the velocity square will cool the cold environment. This is Ranque-Hilsch effect. The existing thermodynamics describes only the real velocities of the medium; in the case of a complex velocity, deviations from the laws of thermodynamics are possible

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