In Griffiths, there is a worked-through derivation for the solutions to the wave function for a delta-square potential case and for the infinite square well case. The infinite square well solution takes on the form of sine. There is also a problem in Griffiths discussed here: Bound States in a Double Delta Function Potential, which shows that the solutions in this double delta function potential case take on the exponential form. They are visualized here:
I know that the sum of exponentials can be equated to sine, so it could be such that the "inside" of these delta-potentials is the same as the infinite square well case. However, the "outside" region of the potentials is non-zero. My questions about this are
- Why can we not equate these two situations - the double delta function scenario and the infinite square well
- Why isn't it relevant where the particle starts? My intuition is that there should be a rather small probability of the particle being outside of the delta barriers if the particle starts between them, but the solutions are actually symmetric about them.
- Why is the probability not smaller on one side of the potential, and why do the solutions reach an extrema at the value of the potential? I would expect them to not have a very large chance of being there because of the infinite potential, and even if they were, I would expect the amount that tunnels through to be very low, so I would expect the probability to at least decrease substantially on one side.