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While solving the quantum mechanical case of potential barrier meaning - $$\text{E} <\text{V} $$ The transmission coefficient is nonzero. My problem is what is happening with particle motion

  • Has particle really crossed the barrier in measurement term or it is just error in measurement at that scale which reflect as transmission coefficient.

  • Resnick and Eisenberg book on quantum physics compares it with optical tunneling, i can't understand that one as well.

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  • $\begingroup$ It's not really a classical 'particle' at this point. Tunnelling is a purely quantum phenomenon. $\endgroup$ – Avantgarde Jul 16 at 11:13
  • $\begingroup$ Have a look at hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html and keep in mind that quantum mechanics is about prpbababilities, in this case the probability for a particle to be outside the barrier $\endgroup$ – anna v Jul 16 at 11:56
  • $\begingroup$ @Avantgarde Light can tunnel though a barrier in a classical description. So tunneling is more appropriately a wave phenomenon. $\endgroup$ – my2cts Jul 16 at 12:56
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/11188/2451 and links therein. $\endgroup$ – Qmechanic Jul 16 at 13:07
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Quantum tunneling is a QM phenomenon where the subatomic particle passes through the potential barrier.

In quantum mechanics, the rectangular (or, at times, square) potential barrier is a standard one-dimensional problem that demonstrates the phenomena of wave-mechanical tunneling (also called "quantum tunneling") and wave-mechanical reflection. The problem consists of solving the one-dimensional time-independent Schrödinger equation for a particle encountering a rectangular potential energy barrier. It is usually assumed, as here, that a free particle impinges on the barrier from the left.

https://en.wikipedia.org/wiki/Rectangular_potential_barrier

Now there is a classical interpretation.

Although classically a particle behaving as a point mass would be reflected, a particle actually behaving as a matter wave has a non-zero probability of penetrating the barrier and continuing its travel as a wave on the other side.

This means that as per QM, the particle behaving as a wave can actually exist at some probability on the other side of the barrier.

In classical wave-physics, this effect is known as evanescent wave coupling. The likelihood that the particle will pass through the barrier is given by the transmission coefficient, whereas the likelihood that it is reflected is given by the reflection coefficient. Schrödinger's wave-equation allows these coefficients to be calculated.

So there is a way to do this in the classical way, using evanescent wave coupling.

Especially in optics, evanescent-wave coupling refers to the coupling between two waves due to physical overlap of what would otherwise be described as the evanescent fields corresponding to the propagating waves.[7] One classical example is frustrated total internal reflection in which the evanescent field very close (see graph) to the surface of a dense medium at which a wave normally undergoes total internal reflection overlaps another dense medium in the vicinity. This disrupts the totality of the reflection, diverting some power into the second medium.

https://en.wikipedia.org/wiki/Evanescent_field#Evanescent-wave_coupling

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  • $\begingroup$ Thanks all the writer...i got answer $\endgroup$ – Ashutosh Pandey Jul 16 at 12:54
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In quantum tunneling, the particle has an evanescent state when it is in the barrier. Light can also have an evanescent state in a Bragg stack or some other situations. Since both the light and a particle treated using non-relativistic quantum mechanics are in evanescent states some of the physics is similar in these two cases and you could call them both examples of tunneling. For a detailed review of tunneling for both light and quantum mechanical particles see:

http://winful.engin.umich.edu/wp-content/uploads/sites/376/2018/01/physics_reports_review_article__2006_.pdf

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  • $\begingroup$ Guess you mean: the state is evanescent inside the barrier, but it become a propagating state again when it has passed through the barrier. $\endgroup$ – flippiefanus Jul 16 at 12:36
  • $\begingroup$ I have edited the answer to clarify that issue. $\endgroup$ – alanf Jul 16 at 13:24

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