2 edited body

First of all, the Hamiltonian in question is defined on $L^2(\mathbb R^3)$, not on $L^2(\mathbb R)$. This is important because in the one-dimensional case the potential vould would have a non-integrable singularity which complicates things seriously. On $L^2(\mathbb R^3)$, the operator, defined as a closure from $C_0^\infty$, is selfadjoint. This is proved, for example, in the book by

T. Kato, Perturbation theory for linear operators, Springer, 1966.

Thus the residual spectrum is impossible. A rigorous calculation of eigenvalues and eigenfunctions can be found in the books

L. D. Faddeev and O. A. Yakubovskii, Lectures on quantum mechanics for mathematics students. American Mathematical Society, 2009;

L. A. Takhtajan, Quantum mechanics for Mathematicians, American Mathematical Society, 2008.

The point 0 is an accumulation point of negative eigenvalues and the limit point of continuous spectrum, thus it belongs to the essential spectrum.

1

First of all, the Hamiltonian in question is defined on $L^2(\mathbb R^3)$, not on $L^2(\mathbb R)$. This is important because in the one-dimensional case the potential vould have a non-integrable singularity which complicates things seriously. On $L^2(\mathbb R^3)$, the operator, defined as a closure from $C_0^\infty$, is selfadjoint. This is proved, for example, in the book by

T. Kato, Perturbation theory for linear operators, Springer, 1966.

Thus the residual spectrum is impossible. A rigorous calculation of eigenvalues and eigenfunctions can be found in the books

L. D. Faddeev and O. A. Yakubovskii, Lectures on quantum mechanics for mathematics students. American Mathematical Society, 2009;

L. A. Takhtajan, Quantum mechanics for Mathematicians, American Mathematical Society, 2008.

The point 0 is an accumulation point of negative eigenvalues and the limit point of continuous spectrum, thus it belongs to the essential spectrum.