Skip to main content
Physics

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

Required fields*

4
  • $\begingroup$ Since $\psi (x)$ is not differentiable at $x = 0$, does that mean that $\left< P^2 \right>$ is valid only for $x \ne 0$ $\endgroup$ Commented Jun 27, 2022 at 12:01
  • $\begingroup$ What I meant was, “Is the value of $\left< P^2 \right>$ only for $x \ne 0$?” $\endgroup$ Commented Jun 27, 2022 at 12:11
  • $\begingroup$ Hi Atom. Thanks for the feedback. The expectation value $\langle P^2\rangle=\int_{\mathbb{R}}\!dx\ldots$ integrates over the whole $x$-axis; the origin is not excluded. $\endgroup$ Commented Jun 27, 2022 at 12:25
  • $\begingroup$ I think I got it. $\psi’’(x)$ is differentiable at $x = 0$ but not $\psi’(x)$, right? $\endgroup$ Commented Jun 28, 2022 at 7:31