Showing an operator is essentially self-adjoint 
I have a question about checking if an operator is essentially self-adjoint. Given the operator $$H=-\frac{1}{2}\partial^2_{r}-\frac{1}{r}\partial_r$$
with domain $C^{\infty}_0((0,\infty))$ (i.e. smooth functions with compact support on $(0,\infty)$) on the Hilbert space $L^2([0,\infty))$ with the scalar product $$<\psi,\phi>=\int_{0}^{\infty}\overline{\psi(r)}\phi(r)r^2dr,$$
I wish to find if the operator is essentially self-adjoint. It is straight forward to check if the operatory is symmetric; I just applied integration by parts a couple of times then used the fact that we have compact support to make the boundary terms disappear.
When it comes to checking the operator to see if it is essentially self-adjoint, I'm unsure exactly what to check. My first thought was to see if $ker(H^*\pm i)=\{0\}$, but I got some weird solutions and wasn't sure about the domain of $H^*$. Any help pointing me in the right direction would be appreciated!
 A: Consider the formal differential operator (i.e., has no specific domain):
$$
    L= -\frac{1}{2r^{2}}\partial_{r}r^{2}\partial_{r}
     = -\frac{1}{2}\partial_{r}^{2}-\frac{1}{r}\partial_{r}.
$$
The solutions of $L\psi=0$ are easily derived:
$$
                 \psi = C\frac{1}{r}+D.
$$
Neither of these solutions is square-integrable at $\infty$ with respect to the weighted inner product you have defined. There are no possible conditions at $\infty$. However, both solutions are square integrable at $0$, which means that there are two endpoint conditions at $r=0$, and, therefore, one endpoint condition must be imposed in order to obtain a selfadjoint operator. But setting both to $0$ will not lead to a selfadjoint operator, and the domain you have chosen is one where both have been set to $0$. So, no, your operator is not essentially selfadjoint. 
You obtain an essentially selfadjoint operator $H_{1}$ by adding to the domain of $H$ any $C^{\infty}$ function which vanishes identically for large $r$ and which is identically $1$ near $0$. This is the correct operator to do Physics. The alternative is to add some $C^{\infty}$ function which vanishes identically for large $r$ and which equals a linear combination of $1/r$ and $1$ near $0$ to the domain instead, which is not desirable if $1/r$ has a non-zero multiplier.
Added: I have a little more time now to explain further. Let $L_{0}$ be $L$ acting on the domain $C_{0}^{\infty}(0,\infty)$. That is $\mathcal{D}(L_{0})=C_{0}^{\infty}(0,\infty)$ consists of infinitely differentiable functions which vanish outside a compact subset of $(0,\infty)$. Then $L_{0}$ is densely-defined and symmetric, which means it is also closable to an operator $L_{min}$ on some domain $\mathcal{D}(L_{min})$. I have labeled this 'min' because one normally refers to $L_{min}$ as the minimal operator. What I'll show you is that $L_{min}^{\star} \ne L_{min}$ which means that $L_{0}$ cannot be essentially selfadjoint. The operator $L_{min}$ is a closed linear operator, and it remains symmetric on its domain.
The standard tool for dealing with operators of this type on the weighted space $L^{2}_{r^{2}}[0,\infty)$ with inner product $(f,g)_{r^{2}}=\int_{0}^{\infty}r^{2}f(r)\overline{g(r)}\,dr$ is the Lagrange identity
$$
 r^{2}\{ (Lf)\overline{g}-f(L\overline{g})\} 
    = \frac{d}{dr}\{ r^{2}(f\overline{g}'-f'\overline{g})\}
$$
Let $L_{max}$ be $L$ defined on the domain $\mathcal{D}(L_{max})$ of twice absolutely continuous functions $f \in L^{2}_{r^{2}}[0,\infty)$ with $Lf \in L^{2}_{r^{2}}[0,\infty)$. Then
$$
   \left.(L_{max}f,g)_{r^{2}}-(f,L_{max}g)_{r^{2}} = r^{2}(f\overline{g}'-f'\overline{g})\right.|_{0}^{\infty}.
$$
The evaluation limits on the right are guaranteed to exist for all $f,g \in\mathcal{D}(L_{max})$ because the integrals on the left are absolutely convergent. Using this identity, one can show that
$$
    (L_{min}f,g)=(f,L_{max}g),\;\;\; f\in\mathcal{D}(L_{min}),g\in\mathcal{D}(L_{max}).
$$
This is done in two steps: First show the above for $f\in\mathcal{D}(L_{0})$, and then use the fact that the graph of $L_{0}$ is dense in the graph of $L_{min}$. In the language of graph inclusion, the above yields
$$
                            L_{min} \preceq L_{max} \preceq L_{min}^{\star}
$$
That is to say that every $f \in \mathcal{D}(L_{max})$ is also in $\mathcal{D}(L_{min}^{\star})$ and $L_{min}^{\star}f=L_{max}f$. Actually, one can show that $L_{min}^{\star}=L_{max}$, but that fact is not so easy to prove, and it is not needed here.
In order to show that $L_{min}^{\star} \ne L_{min}$, it is enough to show that $L_{min}^{\star}$ is not symmetric on its domain. And this I'll do by showing that $L_{max}$ is not symmetric on its domain. To do this, let $k$ be a $C^{\infty}$ function on $[0,\infty)$ which is identically $1$ near $0$ and identically $0$ for large $r$. Then $l(r)=\frac{k(r)}{r}$ is in $\mathcal{D}(L_{max})$: it is easy to check that $l \in L^{2}_{r^{2}}$, and $Ll =0$ for all $r$ near $0$, and for all large $r$. So $Ll \in L^{2}_{r^{2}}$, which means $l \in \mathcal{D}(L_{max})$. Similarly, $k\in \mathcal{D}(L_{max})$. However,
$$
         (Lk,l)_{r^{2}}-(k,Ll)_{r^{2}}=r^{2}(kl'-k'l)|_{0}^{\infty}=-r^{2}(kl'-k'l)|_{0}=-r^{2}l'|_{r=0}=1.
$$
Therefore $(L_{max}k,l)\ne (k,L_{max}l)$.
