# Why isn't the delta-function-potential solution the same as the infinite square well solution?

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

1. Why can we not equate these two situations - the double delta function scenario and the infinite square well
2. 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.
3. 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.

1. Why would you equate the two situations? They are completely different. In the case of the infinite square well, the particle is restricted to be in a finite region of space, because the potential is infinite outside that region. In the case of the attractive double-Dirac potential, the particle is allowed to be anywhere, because the potential is finite (in fact bounded above) everywhere. The particle is more likely to be found "at" the positions of the delta functions, because the potential is "smaller" there than everywhere else ($$-\infty < 0$$, after all), but it still is allowed to be anywhere.
3. Again, your misconceptions have to do with the fact that you are thinking about a time-dependent problem, and the solutions above are the stationary states, i.e., the ones that don't evolve in time other than trivially. (That is, recall that in going from the TDSE to the TISE, one assumes that the solutions have the form $$e^{-iEt/\hbar}\psi(x)$$, and the trivial exponential time-dependence has no physical consequences in the case of the isolated system.) The fact that the solutions have a certain symmetry comes from the fact that the system itself is symmetric under the parity operation.