2
$\begingroup$

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!

Improve this question
$\endgroup$
5
  • $\begingroup$ Are you sure of your definition? It seems to me that if $\psi(x)$ is discontinuous at $x = 0$ then $\psi'(0)$ is not defined. If the delta prime potential you're referring to is to be understood as the derivative of the delta potential in the sense of tempered distributions, then I think your boundary condition should look more like $\psi'(0^+) - \psi'(0^-) = \psi'(0) = \frac{\psi'(0^-) + \psi'(0^+)}{2}$ (the $\psi'(0)$ here is not technically defined but I think you should take the average of the left and right limit). $\endgroup$
    – QuantumApple
    Commented Aug 26, 2020 at 9:11
  • $\begingroup$ @QuantumApple as i said, $\psi '(0)$ is not defined in the usual sense of differentiability - the limit in the usual sense doesn't have to be defined (if it were, then $\psi$ would be continuous). I only need the one sided derivatives to exist and to be equal, and I define them to be $\psi '(0)$ just to make notation simple (but yeah, it's an abuse of notation - it's basically the expression you wrote). I'll change the post to clarify $\endgroup$
    – GSofer
    Commented Aug 26, 2020 at 9:15
  • $\begingroup$ Just to be sure, is $\delta'$ the derivative of the Dirac delta function in the sense of tempered distributions? In that case I disagree with the boundary condition you've written. $\endgroup$
    – QuantumApple
    Commented Aug 26, 2020 at 9:24
  • $\begingroup$ I don't see how $\psi(x)$ can be discontinuous at 0. That would mean the probability of detecting the particle at that point has two values (or rather, is undefined). Anyway, the answer for the "delta-prime" potential seems to be derived in this SE post. $\endgroup$
    – Philip
    Commented Aug 26, 2020 at 10:02
  • 1
    $\begingroup$ Duplicate/cross-post: math.stackexchange.com/q/3804047/168433 $\endgroup$
    – md2perpe
    Commented Aug 28, 2020 at 15:40

1 Answer 1

Reset to default
1
$\begingroup$

Here you are talking about Schrödinger equation with measure valued potentials, or a distribution of order zero. We covered theory for such equations in 2 papers. You can look at

  1. Spectral theory for systems of ordinary differential equations with distributional coefficients

  2. Sign-changing points of solutions of homogeneous Sturm–Liouville equations with measure-valued coefficients

You can contact me, I will be glad to help you!

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you! I have actually learnt a lot about this theory since the post was written. The correct way (or at least one of them) to understand the proper boundary conditions as the coupling parameter goes to infinity is by considering the quadratic form associated with the Schrodinger operator. I will mark your answer as correct so that future reader may refer to your papers. $\endgroup$
    – GSofer
    Commented Nov 20, 2022 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.