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cyleung_2001

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Can the phenomenon of quantum tunneling be explained? Any theory that can account for it? Or is it just empirical? Would someone kindly answer my questions?

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- Thread starter cyleung_2001
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cyleung_2001

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Suppose you have a potential that is continuous everywhere (for simplicity. It need not be continuous, just so long as it remains finite for any finite point). Classically, in a conservative system, if the energy of your particle is less than the potential energy, the particle will never be found there. It's a "classically forbidden" area.

Now, in quantum mechanics, for various mathematical as well as physical reasons, the wave function for a particle must be continuous for such potentials, and therefore you get a particle that "leaks" into the classically forbidden area. In fact, if you have an INFINITE potential of finite length separating two areas (experimentally this would be approximated by, for example, an enormous electric field that is confined spacially), there is a nonzero chance that the particle will pop through that potential and end up on the other side. In the case of free particles with a potential, the decay is exponential, so as the distances grow large the probability drops, but tunneling is perfectly accounted for in conventional quantum mechanics.

- #3

seratend

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MalleusScientiarum said:In fact, if you have an INFINITE potential of finite length separating two areas ... there is a nonzero chance that the particle will pop through that potential and end up on the other side.

Are you sure? really sure?

Seratend.

- #4

Doc Al

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For anMalleusScientiarum said:In fact, if you have an INFINITE potential of finite length separating two areas (experimentally this would be approximated by, for example, an enormous electric field that is confined spacially), there is a nonzero chance that the particle will pop through that potential and end up on the other side.

Absolutely.... but tunneling is perfectly accounted for in conventional quantum mechanics.

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selfAdjoint

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Believe it.

- #6

cyleung_2001

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- #7

Oh yes...you're right...because the wave function goes as [tex]\sim e^{-V}[/tex] more or less. My bad.

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