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  1. I didn't think that it would be possible for a wave function to get through the delta function because there is no "leakage" of the wave function through an infinite potential barrier. I can understand why a particle could get through a non infinite barrier because of this exponential leakage but I don't understand how a particle could get through an infinite delta potential. My thinking right now is that maybe there is some imaginary (as apposed to real) leakage, similar to how a quadratic that doesn't cross the $x$ axis in the real plane may cross the $x$ axis in the imaginary plane. Is this the reason?

  2. I tried to solve for the reflection and transmission coefficients by solving for the two solutions outside the delta function and you obviously get 4 different exponentials with imaginary exponents. I know that these two functions have to touch so you equate them at $x=0$ and get $A+B=C+D$. Then I did the finding the change in slope trick and I get another relation between these 4 coefficients. The problem is that this gives me 2 equations and 4 coefficients to solve for, so I'm not sure what to do now (also a k value so 5 variables I guess). For the bound solution the coefficients were much nicer and it was easy to solve, but I'm not sure how to solve this one. My initial thinking is that I can get rid of the D coefficient since I only want the transmitted wave to be right-moving, but that only cuts me down to 3 variables when I need either 2 or another equation.

Qmechanic
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Shrodinger 2016
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    You can solve or look up the solution for the rectangular well (or barrier). What happens to the solution when you vary the parameters in such a way that the sequence of barriers approximates a delta function? – CuriousOne Jul 14 '16 at 20:03
  • See e.g. Wikipedia. – Qmechanic Jul 14 '16 at 20:09
  • I'm not exactly sure what the question here is. What about the standard solution to the $\delta$-function potential is unclear to you? – ACuriousMind Jul 14 '16 at 21:18
  • I don't understand why a particle can get through a delta potential. – Shrodinger 2016 Jul 14 '16 at 21:21
  • The Dirac potential is just a mathematical device, it's not literally a spike up to infinity. There is nothing in the real world that is infinite, potentials do fade away given enough distance. –  Jul 14 '16 at 21:41
  • Ok but our math shows that it is possible if it did exist. Why? – Shrodinger 2016 Jul 14 '16 at 21:47
  • Math is just our best way of describing the world, because we can't imagine what, say an electron or a black hole, really look like. Electrical engineers use complex numbers to work out problems but they don't worry too much that the square root of minus 1 is involved, as long as it gives them the right answer. –  Jul 14 '16 at 22:07
  • Under the WKB approximation, the relevant quantity for tunneling is $L \sqrt{V}$ where $L$ is the length of the barrier. So one would expect a very thin barrier to have little effect, and this overpowers the fact that $V$ diverges. – knzhou Jul 14 '16 at 22:50
  • @Shrodinger2016: Please don't deface your post. – Qmechanic Aug 04 '16 at 05:31

1 Answers1

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Even with a delta function potential, continuity of the wave function is still required. (Please see comment from ACuriousMind below on this).

The derivative of the wavefunction is obviously not continuous, however. You can find the discontinuity by integration about the delta function from +s to -s, where s is a small parameter.

You then let s go to 0 and check the behaviour of the derivative. You don't state the wavefunction(s) you are using, but here is a plot of a typical one.

enter image description here

You can find the actual calculations here:

Delta Function Wikipedia

  • Strictly speaking, continuity of the wavefunction is not required - the space of wavefunctions is the space of (equivalence classes of) square-integrable functions, which includes non-continuous functions. See discussion of this very potential here, and Qmechanic's answer here for why the stationary states tend to be nice in spite of the fact that there is no general constraint on the smoothness of a wavefunction. – ACuriousMind Jul 14 '16 at 21:23
  • @ACuriousMind thanks very much, I self study, so this site reduces my chances of taking what's written in a textbook as always correct. –  Jul 14 '16 at 21:37