I recently came across the concept of a $\delta'(x)$ (delta prime) potential, which is basically a potential which imposes the boundary condition:
- $\frac{\partial\psi}{\partial x}$ is 'continuous' at $0$, in the sense that $\frac{\partial\psi}{\partial x}|_{0^+}=\frac{\partial\psi}{\partial x}|_{0^-}=$:$\psi'(0)$, meaning that both one sided limits exist and are equal (although $\psi$ doesn't have to be differentiable in the usual sense).
- The double sided limits of $\psi (0)$ exist and we have - $$\psi(0^+)-\psi(0^-)=\psi'(0)$$
Note that $\psi$ doesn't have to be continuous at $0$.
I wish to look at the potential of the form $-\sigma\delta'(x)$ (where $\sigma\in\mathbb R$), which is defined just as before, only this time we have the condition: $$\psi(0^+)-\psi(0^-)=-\sigma\psi'(0)$$
This gives me a certain boundary condition for any choice of $\sigma$. My question is - what is the condition as $\sigma\rightarrow \infty$? In a sense - what happens if I take $\sigma=\infty$? Is there a 'proper' way to describe this boundary condition?
It seems to me that for me to take this limit, I must have in the corresponding boundary condition that $\psi'(0)=0$. But what I'm interested in is - does $\psi$ now have to be continuous at $0$? We now have something of the form: $$\psi(0^+)-\psi(0^-)=-\infty\cdot 0$$ Naively, we can't estimate the RHS and claim if $\psi$ is continuous at $0$. But maybe anyone here knows of a 'proper' way to take this limit (via distribution theory or something) so that we can know if maybe $\psi$ also needs to be continuous at $0$ for some reason?
I'd be happy to hear all kinds of answers - intuitions, formal proofs (this is probably best), and even references to papers/books which do something similar to what I described.
Thanks in advance!
