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To my understanding if I have a finite barrier with potential $V(x)>E$, then to the left of the barrier, the wavefunction can be represented as two exponentials:

$$\psi= e^{(ik_{left} x)} + e^{-(ik_{left} x)}$$

Where the negative exponent represents a wave travelling in the opposite direction. Firstly is this correct, or me making things up?

So in the barrier, as $V>E$, $k$ is imaginary so we get a solution of the form:

$$\psi = e^{-k x}$$

i.e. exponential decay.

However I can't understand why it shouldn't be reflected upon leaving the finite barrier. If this is so, then it would produce a similar wave but of negative exponent to the incident wave, i.e. a wave which is exponentially growing from left to right? Is this correct? Is this allowed?

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In general, both the positive and negative real exponentials are needed, but there are special cases where only one is needed.

If the barrier is "one-sided", that is it looks like a step function, then the region of space where the classical energy would be negative extends to infinity. One of those exponentials will approach zero as distance increases into the step, but the other exponential will go to infinity as distance increases into the step. Such a function is not normalizable, doesn't correspond to any physical state, and cannot be a wave function. Only one exponential is needed in the wave function.

If the barrier is "two-sided", that is, it looks like a brick or a top hat, then the space of negative classical energy is bounded. Neither the positive nor the negative exponential will go to infinity in that limited real estate. For that case you must do what you suggest, and use both the positive and negative real exponentials. Try it.

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  • $\begingroup$ Oh I understand now, no infinities reached. However I'm having difficulty in understanding the actual physical interpretation of the matter wave within the barrier. Surely as probability = $\psi^2$, as $\psi$ is non-zero within the barrier there is a probability that the particle can be detected within the barrier. Is this not violating conservation of energy laws? $\endgroup$
    – AlexLipp
    Mar 20, 2014 at 23:43
  • $\begingroup$ I think that if an interaction occurs within the barrier, you have to take the energy of the particle as being negative, and the other entities in the interaction have to make up the energy. I'm not certain about that; someone may correct me. Something akin to this happens in semiconductors near interfaces where the energy bands are bent forming a barrier. The evenescent tail of an electron wave function can be involved in electronic transitions such as optical absorption. This is called "tunneling-assisted absorption" or the Franz-Keldysh effect. $\endgroup$
    – garyp
    Mar 21, 2014 at 1:50

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