Existence of a function $\psi$ satisfying $\lim_{r\to 0}r\psi\left(r\right)=\lim_{r\to \infty}r\psi\left(r\right)=0$

Question

I stumbled across an exercise in quantum mechanics concerning the self-adjointess of the radial momentum operator. In the exercise, the statement was made that this operator is self-adjoint for wave functions that satisfy

$$\lim_{r\to0}r\psi\left(r\right)=\lim_{r\to \infty}r\psi\left(r\right)=0$$

I was wondering, from a pure mathematical point of view, what an example of such a function $\psi\left(r\right)$ would be.

Answer 1

How about $\psi(r) = \frac{1}{\sqrt{2\pi}\; r^2} \exp(-1/r)$?

It satisfies both of your limits, and also is normalized in the sense that $$ \int_0^\infty dr\, 4\pi r\; {\left|\psi(r)\right|}^2 = 1\, . $$

Something like $\frac{8^{1/4}}{{\pi}^{3/4}} \exp(-r^2)$ also works in this regard. Come to think of it, any function that is a polynomial in $r$ times $\exp(-a r^2)$ will work, where $a>0$. I believe this class would include the radial part of the eigenstates of the hydrogen atom.

Answer 2

maybe $\varphi(r)=0$? that's a trivial example.

you can also take $\varphi(r)=2^{1+\varepsilon}$ for $r<1/2$ and $\varphi(r)=1/r^{1+\varepsilon}$ for $r\geq 1/2$ for every $\varepsilon>0$.

In particular, everything that is bounded in an interval around zero and tends to zero faster than $1/r$ as $r$ goes to infinity will work.

Answer 3

$1/(a^2+r^2)$ will work for $a>0$; the normalisation constant is an exercise.