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In Shankar exercise 5.4.2 we consider the reflection and transmission of a wavefunction through a delta potential $V(x) = aV_0\delta(x)$. I understand that to mean that the potential is zero everywhere except for the origin, where it is infinite.

Intuitively, I thought that an incoming quanton should not transmit through the potential, but I was ready to write myself off as trying to force a classical interpretation on the problem. Quantum tunneling exists, after all. So there will be a transmitted wave.

However, I have trouble reconciling this with my understanding of the infinite square well. In the infinite square well, the quanton's wavefunctions are resonant modes of the potential barriers (which are infinite). Thus, the solutions to the infinite square well - which have periodic boundary conditions - must require that the wave not be able to transmit through the infinite potential. Otherwise, the quanton is not trapped and we cannot have resonant modes.

These two situations indicate conflicting pieces of information. The former implies that particles can transmit through a delta potential, while the latter case implies the opposite. Is there a contradiction here, or have I made a mistake in my understanding of the concepts of QM?

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  • $\begingroup$ For infinite square well, the potential is infinite everywhere outside the well, not just at the boundaries of the well. $\endgroup$ – K_inverse Oct 24 '18 at 3:48
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The two situations are distinct. For the infinite square, you have a finite region of space where the potential is finite, and for the rest of space the potential is infinite. Regions where the potential is infinite are forbidden, even in quantum mechanics, the wave function must vanish in those regions.

But that is not the only problem. Even if you had 2 infinite potential wells, separated by a finite distance, the solutions would not "communicate", they would be independent, so no transmission could occur.

The difference for the delta potential is that even though the two regions of space, $x<0$ and $x>0$, are separated by an infinite potential at the origin, $V=\delta(x)$, the width of this potential is $0$, and that is what makes it possible for transmission to occur.

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