When a wave function in QM potential well problems interact with a potential barrier with height more than the energy of the wave, the amplitude of the wave doesn't immediately falls off to zero, rather it decays exponentially, allowing phenomenon like quantum tunneling. In optics, the electromagnetic waves behave likewise to give the frustrated total internal reflection. How do we explain this behavior physically? I know Schrödinger's equation and Maxwell's equations require the wave to decay exponentially, but is there something else to it?
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1$\begingroup$ It is a good question to which I have no other answer than that is how QM rolls. $\endgroup$– my2ctsJul 25, 2019 at 18:19
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$\begingroup$ In English, only capitalize the first word of a sentence, and names of persons or places. $\endgroup$– DanielSankJul 25, 2019 at 23:05
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$\begingroup$ You seem to give the answer in your own question: because it is like a wave, and waves suffer that kind of "frustrated total reflection". $\endgroup$– FGSUZJul 26, 2019 at 0:23
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2 Answers
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Physically, the Schrödinger's equation expresses conservation of energy. The term $d^2\Psi/dx^2$ measures the curvature of the wavefunction, which is an indirect indication of its wavelength, and therefore its momentum and kinetic energy. At a "kink" where the second derivative blows up, the kinetic energy is $\pm\infty$, which isn't possible without violating conservation of energy.
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1$\begingroup$ I am still a bit confused how this answers my question. $\endgroup$ Jul 25, 2019 at 18:32
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$\begingroup$ @HarshdeepSingh Falling immediately to $0$ requires an infinite second derivative? $\endgroup$ Jul 25, 2019 at 23:45
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Why should it?
Think of this as a boundary condition problem. You solve the PDE, Schrodinger's equation, which is a wave equation. The solutions must have continuous first and zeroth derivations, zeroth being a fancy way of saying that the function is continuous.
This is what allows some solution beyond the barrier. This is typical of all wave equations. In QM we are attributing this "wave" to the probabilistic measure of locating a particle at some location in space (in the Copenhagen interpretation). So if you are trying to make sense of how the wave function should behave using your knowledge of how classical particles are expected to behave you will always wind up confused. What you should do is solve the wave equation, apply boundary conditions, and then ask what behavior does this predict relative to particle observation. In your example the non vanishing wave function beyond the barrier is observed as tunneling and penetration.
As Ben Crowell mentioned in his answer, if V0 --> infinity then the wave function does in fact vanish at the boundary and everywhere beyond it.
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$\begingroup$ All of this doesn't justify that wave function and its derivative must be continuous. $\endgroup$– JhorJul 26, 2019 at 7:06