This question has been bugging me for some time.

Take the hamiltonian for the hydrogen atom: $$\hat{H}=-\frac{1}{2}\nabla^2-\frac{1}{r},$$ acting on (a domain contained in) $L^2(\mathbb{R}^3)$. It is standard fact that this is an unbounded operator which has a countable infinity of eigenvalues, all of which are negative and which accumulate around 0, and has a continuous spectrum on the whole of $(0,\infty)$. Physically, the former are bound states which correspond to elliptic Keplerian orbits in the classical problem, and the latter are unbound states and correspond to hyperbolic orbits. I also know that the spectrum of all unbounded operators is a closed set, so that 0 is definitely in $\sigma\left(\hat{H}\right)$.

My question is then: to what part of the spectrum does 0 belong to (i.e. point, continuous, residual)? What are the corresponding eigenfunctions? What kind of degeneracy does it have? (I would expect it to admit a common eigenvector with any $l,m$ angular momentum numbers, but I'm far from sure.) How do the eigenfunctions correspond to the nearby bound and unbound states?


2 Answers 2


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 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.

  • $\begingroup$ you're right, I forgot the ^3 - it's corrected now. I'll look at the references! $\endgroup$ Apr 19, 2012 at 13:45
  • $\begingroup$ I'm unsure why I didn't accept this answer ::checks notes:: ten years ago, but I'm happy to correct this now. $\endgroup$ Apr 22, 2022 at 15:03

I looked in Anatoly's references, and Quantum mechanics for mathematicians by Leon A. Takhtajan does have the calculation of the continuum wavefunctions, though it does not do the $k=0$ case.

The eigenfunction $f_l$ at energy $E=\frac{1}{2}k^2$ and angular momentum $l$ must satisfy the eigenvalue equation $$ f_l''+\left(\frac{2}{r}-\frac{l(l+1)}{r^2}+k^2\right)f_l=0. $$ After the obligatory asymptotics factorization of $f_l(r)=r^{l+1}e^{ikr}F_l(r)$, this equation reads $$ F_l''+\left(\frac{2(l+1)}{r}-2ik\right)F_l'+\left(\frac{2}{r}-\frac{2ik(l+1)}{r}\right)F_l=0, $$ and Takhtajan gives the solution as a confluent hypergeometric function, $F_l(r)={}_1F_1\left(l+1+\frac{i}{k};2(l+1);2ikr\right)$, under $F_l(0)=1$.

From this the $k=0$ case can be recovered as a limit in the same spirit as the ${}_2F_1\rightarrow{}_1F_1$ confluence by letting the length $\lambda=\frac{1}{k}$ go to infinity. Thus at zero energy, \begin{align} F_l(r) & =\lim_{\lambda\rightarrow\infty} {}_1F_1\left(l+1+i\lambda;2(l+1);\frac{-2r}{i\lambda}\right) \\ & ={}_0F_1\left(;2(l+1);-2r\right) \\ & =\frac{(2l+1)!}{2^\frac{2l+1}{2}}r^{-\frac{2l+1}{2}}J_{2l+1}(\sqrt{8r}). \end{align} (This still obeys $F_l(0)=1$.) Alternatively, the Bessel function solution can be obtained directly by the appropriate transformations or by plugging the $k=0$ equation into Mathematica.

In the asymptotic regime, $r\gg 2l+1$, one then gets $$ f_l(r) =\frac{(2l+1)!}{2^l\sqrt{2\pi \sqrt{2}}}r^{\frac{1}{4}} \cos\left(\sqrt{8r}-(2l+1)\frac{\pi}{2}-\frac{\pi}{4}\right) $$ if my maths is right. However, in real cases this can only happen if $2l+1\ll r\ll \lambda$.


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