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!


1 Answer 1


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 $(L_{max}k,l)\ne (k,L_{max}l)$.

  • $\begingroup$ Thank you for your response! Is it just enough to say that none of the solutions are square-integrable at $\infty$? It seems like it would be, because no matter what extension you try you're not going to get a square-integrable function, so there's no self adjoint extension possible. Is that right? $\endgroup$ Oct 28, 2014 at 7:35
  • $\begingroup$ Because none of the solutions of $L\psi=0$ are square-integrable at $\infty$ with respect to the weighted inner-product, then the equation is forced to be the limit point case at $\infty$, which means there are no endpoint conditions which can be specified at $\infty$. However, because both independent solutions of $L\psi=0$ are square-integrable at $0$ in the weighted norm, then there are two continuous endpoint boundary functionals at $0$, and that means one endpoint condition must be imposed on the unrestricted domain or, equivalently, a function must be added to your domain to get s.a.. $\endgroup$ Oct 28, 2014 at 7:50
  • $\begingroup$ @TinaBelcher : The condition needed at $r=0$ is somewhat artificial as it arises because you start with a domain of functions which vanish near $0$. So the adjoint domain includes all twice absolutely continuous $f \in L^{2}_{r^{2}}$ for which $Lf \in L^{2}_{r^{2}}$. That means the adjoint domain includes functions which are $1/r$ identically near $0$, as well as functions which are identically constant near $r=0$. If you had started with function that could be constant near $0$, then the adjoint domain would not have included things which are $1/r$ near $0$; that would be essentially s.a.. $\endgroup$ Oct 28, 2014 at 7:57
  • $\begingroup$ @TinaBelcher : After giving you the intuition, I supplied all of the details to show you how this machinery works. Look for the "Added" section in the answer. I show you directly that the closure of your operator cannot be selfadjoint because the adjoint of your operator is not symmetric. $\endgroup$ Oct 28, 2014 at 16:22
  • $\begingroup$ Wow, thank you very much! It's all very clear and I understand it. That's fantastic. $\endgroup$ Oct 28, 2014 at 17:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .