Where does particle borrow energy from to tunnel?

Question

Where does particle borrow energy from to tunnel? It is implied that particle can borrow energy and leaped over to the other side wherever that is, the shorter the gap the more energy it borrows my question is where does a particle borrow its energy from and what criteria allows it to do the borrowing, last but not least how does the particle return the borrowed energy? Is it linked to quantum foam?

Answer 1

As mentioned in the link you provided, it is due to Heisenberg's uncertainty relation $-$ during the short-lasting tunneling, the particle may temporarily borrow some energy from the potential of the barrier, so sometimes it can jump over it.

Well, the energy and time may be depicted as a sort of Fourier transform pair (see Fourier transform) because the energy level is associated with quantum oscillations (see Planck relation):

$\qquad E = \hbar \omega$

which ends up as a planewave factor ($e^{-i \omega t}$) for all solutions associated with $E$ of the time-dependent Schrödinger's equation with non-varying potentials:

$\qquad \Psi_E(x,t) = \Psi_E(x,t=0)\ e^{-i \omega t} = \Psi_E(x,t=0)\ e^{-i \frac E \hbar t}$

where $\psi_E$ solutions for particular energy level. Of course, summing up / integrating over all energies gives the final wavefunction for any initial wavefunction profile (group envelope) $\Psi_E(x,0)$ arbitrarily taken.

The energy is just a useful parameter which is well defined during a longer period of time (when used to denote a specific frequency). If you deal with short time intervals and try to localize the particle in time, the energy parameter is not a discrete value anymore, but is spread over a range of frequencies. When the wavefunction in time domain starts to interact with the potential of the barrier, the wavefunction gets deformed and at that moment some of the planewave components will be over the barrier's height ! So, these planewave, Fourier components may temporarily exist and come from complex interactions with the barrier's potential (which is, btw, constant, so uncorrelated with the waveforms of the incoming particles). In case the particle tunneled, they vanish and thus return the "borrowed" energy to the barrier when they cross it in order to preserve momenta of the system.

Since the energy and time form a Fourier transform pair, they are mutually constrained by Uncertainty principle in global. For example, if the tunneling period lasts $1\ \mathrm{fs}$, the temporal energy one may observe is spread at least $\hbar/{(2\Delta t)} \simeq 10^{-34}\ \mathrm{Js}/{(2\ \mathrm{fs})} \simeq 10^{-19}\ \mathrm{J}\ [=\mathrm{C} \cdot \mathrm{V}] \simeq 1\ \mathrm{eV}$ which gives us the lower limit of spreading.

The same mechanism lies behind virtual particles $-$ a quantum field may be shortly excited without violating the law of conservation of energy in long terms, so even if there is no permanent energy around, one may encounter a myriad of virtual particles as the observation time gets shorter.

Answer 2

One should always remember that quantum mechanics predicts probabilities and not energy distributions . The energy a particle will have is an eigenvalue of the energy operator operating on the wavefunction, but the probability of finding a particle at (x,y,z) at time t is given by the complex conjugate square of the wavefunction which is the solution of the quantum mechanical equations with the boundary values imposed.

It is worth reading this link, and generally searching hyperphysics for such questions. "Hyperphysics tunneling" brings up this:

According to classical physics, a particle of energy E less than the height U0 of a barrier could not penetrate - the region inside the barrier is classically forbidden. But the wavefunction associated with a free particle must be continuous at the barrier and will show an exponential decay inside the barrier. The wavefunction must also be continuous on the far side of the barrier, so there is a finite probability that the particle will tunnel through the barrier.

So the energy level on which the particle resides is the same inside and outside the barrier. It is the probability of finding it outside that is smaller than the probability inside.

Answer 3

You mean quantum tunneling?

The particle doesn't really "borrow" energy, actually a particle will have a higher probability of tunneling through a barrier if it has a high kinetic energy. A naive analogy would be that the more energetic the bullet is, the higher the probability it has of piercing through a wall, that is, tunneling.

What makes quantum tunneling (as opposed as classical tunneling, i.e. a bullet trying to go through a wall) special is that there is always a finite probability of a particle tunneling through a barrier regardless of the "thickness" of the barrier. This differs from the classical world where there are situations where a barrier is too thick for a particle to go through. For example, in the classical world if a human runs against a wall of steel, the probability of it going through the wall is zero. While in the quantum world, a human has a non-zero probability of going through that wall of steel.

Answer 4

The particle doesn't borrow energy. Your idea that it gains energy is a guess about what is happening during the experiment, and it is wrong.

You are thinking of a particle in a tunnelling experiment as being like a ball rolling up a hill that is too high for the ball to get over the top with the amount of energy you have given it. The ball rolls part way up the hill, gaining energy in the process, and then rolls back down again. If you write down the ball's equation of motion in the classical approximation, then the ball's vertical position depends on where it is on the hill, and so its gravitational potential energy depends on its position on the hill.

The particle in the tunnelling experiment isn't doing that. It doesn't interact with the potential in such a way as to gain energy. A rough document about this specific issue is available here

https://conjecturesandrefutations.files.wordpress.com/2014/11/tunnelling41.pdf.

It's probably not a good idea to cite this since it hasn't been published and has some spelling errors. The maths in the document is based largely on this paper:

http://scitation.aip.org/content/aip/journal/jap/33/12/10.1063/1.1702424.

A detailed discussion of tunnelling that addresses some misconceptions, but not this specific error is here

http://www.sitemaker.umich.edu/herbert.winful/files/physics_reports_review_article__2006_.pdf.

Answer 5

On a quantum level, particles don't really have "momentum". They're waves. The way the Schrodinger equation works, they move faster if they have a shorter wavelength. So we defined momentum based on the wavelength. Kinetic energy also is part of the whole conservation of energy thing, so we have a very good reason to define it how we did, but again, it's based on the wavelength. But kinetic energy doesn't have to be positive. If it is, you get a sine wave, and if it's not, you get an exponent. The end result is that particles don't propagate very far if they have negative kinetic energy. But they can still get somewhere. So we call it quantum tunneling.

They don't literally borrow energy. They just have negative kinetic energy for a bit. But sometimes it's easier to think of particles as little billiard balls bouncing around like Newton would predict, and somehow getting extra energy to get past those hills. So sometimes people say the particles borrow energy.