# Schroedinger operator in 2 dimensions with singular potential

Consider the Schroedinger operator

$$H = -\Delta + \frac{c}{\vert x \vert^2}$$ in two dimensions with $$c >0$$

This operator has a self-adjoint realization, since it is a positive symmetric operator on $$C_c^{\infty}( \mathbb R^2 \setminus \{0\}).$$ I would like to understand three things:

0.) What are possible domains $$D(H)$$ on which $$H$$ is self-adjoint?

1.) Does there exist a domain $$D(H)$$ of $$H$$ on which $$-\Delta$$ and $$H$$ are both self-adjoint.

2.) On these domains $$D(H)$$, can we expect $$\Vert \frac{f}{\vert x \vert^2}\Vert_{L^2} \le \alpha_1 \Vert f \Vert_{L^2} + \alpha_2 \Vert \Delta f \Vert_{L^2} \text{ for some }\alpha_i \ge 0?$$

I would appreciate insights on any of these questions. My main interest is in small $$c>0.$$

One way to think about it may be to follow Example 4 in Reed-Simon II on page 160.

There the authors prove, going via a 1D reduction, that if

$$c \ge 1,$$ then $$H$$ is essentially self-adjoint (there is a unique self-adjoint extension) on $$C_c^{\infty}(\mathbb R^2 \setminus \{0\})$$ and if

$$c \in [ \frac{1}{4},1),$$ then $$H$$ is not essentially self-adjoint on $$C_c^{\infty}(\mathbb R^2 \setminus \{0\}).$$

The case $$c \in (0,\frac{1}{4})$$ seems to be missing, but I think one quickly sees by analysing the argument that it is the same case as the last one above (not essentially self-adjoint).

In a separate question Iosif Pinelis has shown that the quadratic form generated by the potential is not relatively form bounded with respect to the Dirichlet form.

The reason I asked this question was that I suspected that $$H$$ was self-adjoint on the Dirichlet domain of $$-\Delta$$, i.e. $$H_0^1(\mathbb R^2\setminus \{0\}) \cap H^2(\mathbb R^2\setminus \{0\})$$, but I am now not so sure about my conjecture anymore.

• This physics literature has a number of papers about finding the correct self-adjoint realization of this operator, e.g. starting from here: pubs.aip.org/aapt/ajp/article/42/11/960/1045509/…
– Buzz
Commented Feb 26 at 4:49
• See also the more recent papers by Derezinski. Note that this operator is usually called the "Bessel operator" which might help when doing a literature search. The radial part is equivalent to the generator of a Bessel process which might also help with intuition. Commented Feb 26 at 9:40

Taking advantage of the spherical symmetry to decompose this into a sum of one-dimensional problems sounds like the right approach. I will probably just be redoing what Reed-Simon had in mind here.

The Laplacian is given in polar coordinates by $$\Delta=\frac{1}{r}\frac{\partial}{\partial r} r\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial\theta^2} .$$ We identify $$L^2(\mathbb R^2)=L^2((0,\infty); rdr)\otimes L^2(S^1)$$ and decompose $$L^2(S^1)=\bigoplus L(e^{in\theta})$$ and also get rid of the weight in the radial component with the help of the unitary map $$U:L^2(r\, dr)\to L^2(0,\infty)$$, $$(Uf)(r)=r^{1/2}f(r)$$, then $$H$$ becomes (is unitarily equivalent to, to be precise) an orthogonal sum $$H=\bigoplus H_n$$, with each $$H_n$$ acting in $$L^2(0,\infty)$$ and $$H_n f = -f''+\frac{n^2-1/4+c}{r^2}f .$$ This operator is essentially self-adjoint if and only if $$n^2-1/4+c\ge 3/4$$ or $$c\ge 1-n^2$$. See for example the corresponding discussion in my answer here.

So $$H$$ on $$C_0^{\infty}(\mathbb R^2\setminus\{ 0\})$$ itself is essentially self-adjoint if and only if this happens for all $$n\in\mathbb Z$$, which is indeed equivalent to $$c\ge 1$$. For $$0, $$H_n$$ is still essentially self-adjoint on $$C_0^{\infty}(0,\infty)$$ for $$|n|\ge 1$$, while the closure of $$H_0$$ on this domain is symmetric with deficiency index $$(1,1)$$. So in this case there is a one-parameter family of boundary conditions, to be imposed on the elements of $$D(H_0^*)$$, and this gives all self-adjoint realizations.

As for (1), we can just set $$c=0$$ in this analysis and everything is still correct, and we are now discussing the self-adjoint extensions of $$-\Delta$$ on $$C_0^{\infty}(\mathbb R^2\setminus \{ 0\})$$. Since there is nothing to choose in the higher harmonics, the question is if $$-d^2/dr^2+k/r^2$$ can ever be given the same domain as $$-d^2/dr^2-1/(4r^2)$$ for $$-1/4. The discussion here strongly suggests that the answer is no, though it's not completely conclusive since we fixed specific boundary conditions there.

Added later (with one detail corrected later still): This seems quite clear actually: When $$c=0$$, the possible domains $$D(H_0)$$ can be obtained by putting an arbitrary solution of $$-f''-f/(4r^2)=0$$ into $$D(H_0)$$. This is an Euler equation with solution basis $$f_1=r^{1/2}$$, $$f_2=r^{1/2}\log r$$. Note also that when we undo the unitary transformation from above, these correspond to $$U^*f_j=1, \log r$$. On the other hand, $$f$$ can never be in $$D(H_0)$$ when $$c>0$$ because then $$-f''+(c-1/4)f/r^2= cf/r^2\notin L^2$$. So we never have $$D(H)=D(\Delta)$$.

For the record ("what is the correction?"), in an earlier version I put $$f_1=r^{1/2}$$ into $$D(H_0)$$ and claimed that this corresponds to $$D_0=H_0^1\cap H^2$$, which you suggested as a possible domain. This was nonsense of course (I mixed up my signs in the exponent when applying $$U^*$$), since in that case we put $$g=1$$ (near $$r=0$$) into $$D(\Delta)$$ and we simply obtain the plain Laplacian on $$H^2(\mathbb R^2)$$. Any other choice, namely put $$g=b+\log r$$ into $$D(\Delta)$$, is certainly closer to what you had in mind, but it's not really the domain $$D_0$$. In fact, $$H_0^1$$ on $$\mathbb R^2\setminus \{0\}$$ is simply $$H^1$$ since the function $$h(r)=\log a^2/\log r^2$$ gets us from $$h(0)=0$$ to $$h(a)=1$$ and $$\|h\|_{H^1(r as $$a\to 0$$.

Finally, (2) is now moot, but it would be a consequence of (1) since if $$D(S)\supseteq D(T)$$ for closed operators, then $$S$$ is $$T$$-bounded (= Theorem 5.9 in Weidmann, Linear operators in Hilbert spaces).

• So you are saying the Dirichlet domain is indeed an admissible domain? If I read your answer correctly, then you are saying that we just have to know the self-adjoint extensions of the Laplacian and we will have a complete description for $c<1$? Commented Feb 25 at 21:12
• Thanks Christian, I think I still stumble over the same point: You say "we can just set 𝑐=0 in this analysis and everything is still correct". Are you suggesting that the conditions are in some sense independent of setting $c=0$ as long as $c<1$? This is not obvious to me. Commented Feb 25 at 21:45
• @AntónioBorgesSantos: No, I'm just saying that we can decompose $-\Delta=\bigoplus H_n$ and then extend this, and since the $H_n$, $n\not= 0$ are already essentially self-adjoint, it's only about finding a domain for $H_0$. This works for any $c\in\mathbb R$, including $c=0$. Commented Feb 25 at 21:49
• ah, but you think the domains will be different for $c \neq 0$ from the ones you can obtain for $c=0.$ So both (1) and (2) are wrong. Commented Feb 25 at 21:50