1

  1. The necessary condition for $\int\limits_{-\infty}^{+\infty}|\psi(x)|^2dx$ to be integrable is that $\psi(x)\rightarrow 0$ as $x\rightarrow\pm\infty$. But this is not the sufficient condition. For example, $\delta(x)$ vanishes as $x\rightarrow \pm \infty$, but it is not square-integrable. What is the sufficient criterion for $\int\limits_{-\infty}^{+\infty}|\psi(x)|^2dx$ to converge?

  2. In the previous question $\psi(x)$ was an arbitrary function. If now $\psi(x,t)$ be an arbitrary solution (not necessarily a stationary state) of the time-dependent Schrodinger equation what is the sufficiency condition for $\int\limits_{-\infty}^{+\infty}|\psi(x,t)|^2dx$ to be finite?

  3. Does this conditions remain valid in three-dimensions?

EDIT: Can we have a continuous function $\psi(x)$ which does not go to zero as $x\rightarrow \pm\infty$ and yet the integral $\int\limits_{-\infty}^{+\infty}|\psi(x)|^2dx$ converges?

SRS
  • 26,333
  • 5
    This is a math question. Consider Math SE – ClassicStyle Apr 04 '16 at 14:09
  • 2
    Also, see this. As it turns out, going to zero isn't even necessary http://physics.stackexchange.com/q/75527/ – ClassicStyle Apr 04 '16 at 14:17
  • Yeah, going to zero is not necessary. A function that is $e^{-x^2}$ for $x\in \mathbb{R}\setminus \mathbb{Z}$ and $1$ for $x\in \mathbb{Z}$ is perfectly square-integrable wrt Lebesgue measure. – yuggib Apr 04 '16 at 14:21
  • @TylerHG If this is a mathematical question then various theorems we prove in quantum mechanics also belong to mathematics. – SRS Apr 04 '16 at 14:24
  • Concerning 2, a sufficient condition for a solution of the Schrödinger equation to be in $L^2$, is that it is in $L^2$ at some $t\in\mathbb{R}$. However this is sort of tautological: in fact we define the solution of Schrödinger's equation to be unitary, i.e. to preserve the $L^2$ norm. In principle, there can be non-unitary solutions of the Schrödinger equations in some function space. – yuggib Apr 04 '16 at 14:34
  • 2
    Qmechanic gives a smooth counterexample to the condition being necessary here. I'm not sure what your question is - in general, there is no better characterization of square-integrable functions other than them being, well, square-integrable. Also, $\delta(x)$ doesn't "vanish" as $x\to \pm\infty$, that doesn't make sense to begin with - the Dirac delta is not a function. – ACuriousMind Apr 04 '16 at 14:50
  • 1
    Also, consider replacing "the necessary condition" by "a necessary condition". There never is a unique necessary (or sufficient) condition. – Antoine Apr 04 '16 at 14:51

0 Answers0