Neumann function in 2D Schrödinger equation in polar coordinates

Question

In the question 2D Schrodinger equation in polar coordinates, @Qmechanic has shown that $\mathrm{Y}_0(r)\sim \ln r$ should be omitted since it's energy expectation value will diverge:

$$ \def\dd{\mathrm{d}} \def\bm{\boldsymbol} \int_0^{a} \psi^*\nabla^2\psi\, \dd^2 \bm{x} =\int_0^{a} -\frac{\ln r}{r^2} r \, \dd r \, \dd\theta \to\infty $$

However, if we calculate the Neumann function instead of logarithm, it is actually not divergency:

$$ \def\Y{\mathrm{Y}} \nabla^2_{r,\theta}\Y_0(r)=-\Y_0(r) \implies\int_0^{a} \psi^*\nabla^2\psi\, \dd^2 \bm{x} =\int_0^{a} -\Y_0^2(r) \, r \, \dd r \, \dd\theta <\infty $$

and the function itself is normalizable similarly.

So should we omit this solution? How can we explain the infinity at the origin?

Answer 1

The Neumann function satisfies $\nabla^2Y_0=-Y_0$ only for $r>0.$ If one was to include the origin in the solution, this functional relation is changed.

Indeed, since asymptotically towards the origin $Y_0(r)\asymp\frac{2}{\pi}\ln r$ we observe that:

$$\int\limits_{r\leq \epsilon} dA[ \nabla^2Y_0+ Y_0]=\int\limits_{r\leq \epsilon} dA [\nabla^2(\frac{2}{\pi}\ln r)+ \mathcal{O}(1)]=\int\limits_{r=\epsilon} \textbf{$\nabla$}(\frac{2}{\pi}\ln r)\cdot d\textbf{l}+\mathcal{O}(\epsilon^2)=\int_{r=\epsilon}\frac{2}{\pi r} r d\theta+\mathcal{O}(\epsilon^2)=4+\mathcal{O}(\epsilon^2)$$ and thus we conclude that actually as we let $\epsilon\rightarrow 0$:

$$\nabla^2Y_0(r)+ Y_0(r)=4\frac{\delta(r)}{r}$$

and now as we substitute into the kinetic energy expression,

$$\int[Y^*_0\nabla^2Y_0]rdrd\theta=-\int Y_0^2 rdrd\theta+4\int Y_0(r)\delta(r)drd\theta$$

we see that while the first integral converges, the second diverges, reconciling the discrepancy.