Consider the potential $V(x)=\frac{2}{x^2}$ and let $\frac{\hbar^2}{2m}=1$ for convenience. Now consider the function $\psi(x)=\delta(x)$. According to Griffiths (electrodynamics book) problem 1.45(a), $x\delta'(x)=-\delta(x)$. I'm$$x\delta'(x)=-\delta(x)\tag{1}.$$ I'm not sure if I can do this but if I write
$$\delta'(x)=-\frac{\delta(x)}{x},$$$$\delta'(x)=-\frac{\delta(x)}{x},\tag{2}$$
$$\frac{d^2}{dx^2}\psi(x)=-\frac{d}{dx}\left[\frac{\delta(x)}{x}\right]=\frac{2\delta(x)}{x^2}.$$$$\frac{d^2}{dx^2}\psi(x)=-\frac{d}{dx}\left[\frac{\delta(x)}{x}\right]=\frac{2\delta(x)}{x^2}.\tag{3}$$
The Schrodinger equation now looks like
\begin{align} &-\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E \psi(x)\\ &-\frac{2\delta(x)}{x^2}+\frac{2}{x^2}\delta(x)=0 \delta(x). \end{align}\begin{align} &-\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E \psi(x)\tag{4}\\ &-\frac{2\delta(x)}{x^2}+\frac{2}{x^2}\delta(x)=0 \delta(x).\tag{5} \end{align}
So it looks like $\delta(x)$ is an eigenstate with eigenvalue zero. But this goes against my intuition and is probably wrong but I'm not sure where the fault lies. Is it the derivative of the delta function? Can energy (eigenvalue of Hamiltonian) be zero? Potential is maximum at $0$, so how can the probability be maximum at $0$?
