Why are $E > 0$ particle waves influenced by a Delta-function well potential?

Question

One of the simple examples of potential wells we learn about in Quantum Physics is the Dirac delta function well

$$V(x) = -\alpha\delta(x)$$

and we learn that this function has a single bound state, for particles that have an energy lower than 0 ($E < 0$), and that particles with $E > 0$ will scatter, with a certain probabilities to be refected or transmitted.

Why are these particles whose energy is above $E = 0$ impacted by this well at all? In other cases like the infinite square well and the harmonic oscillator, $V = \infty$ everywhere outside of the well, so there it is clear that any particle is influenced (and indeed, we only have bound states for those examples). And in the case of the free particle, we have $V = 0$, there is no reflection at all, and wave packets happily carry on their pre-existing path(s).

Why is there the possibility of reflecting off of potentials like the delta function well (and similarly the finite square well)? I expect that I am bringing some kind of 'classical intuition' to the table that does not work in this quantum-mechanical situation, so if there is any other 'proper' kind of intuitive way of thinking about it, I think that would be very helpful to quench the wrong intuition here.

Answer 1

Why wouldn't they be influenced by the potential? In the case of the infinite potential, no wave function can have finite amplitude outside the potential well, as it will immediately infer infinite energy. To draw a parallel with your question, you would need that all the wave functions will have zeros at the point of the delta function in order for them not to be influenced by it, but that is not the case. Plane waves are spread throughout the entire axis with equal amplitude.

For square potential or a delta-like potential, the potential can affect the wave function just like any other finite potential. The infinite potential is the exception, not the rule. Note that a delta-function potential is not "infinite" because its measure is finite, therefore $\int dx \psi(x)\delta(x) < \infty$ for finite $\psi(x)$, in contrast to the case of $V=\infty$ for an infinite potential well.