Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

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.

share|improve this answer
    
you're right, I forgot the ^3 - it's corrected now. I'll look at the references! –  Emilio Pisanty Apr 19 '12 at 13:45
add comment
up vote 2 down vote accepted

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 $_2F1\rightarrow_1F_1$ confluence by letting the length $\lambda=\frac{1}{k}$ go to infinity. Thus at zero energy, $$ F_l(r) =\lim_{\lambda\rightarrow\infty} {}_1F_1\left(l+1+i\lambda;2(l+1);\frac{-2r}{i\lambda}\right) =_0F_1\left(;2(l+1);-2r\right) =\frac{(2l+1)!}{2^\frac{2l+1}{2}}r^{-\frac{2l+1}{2}}J_{2l+1}(\sqrt{8r}). $$ (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 $2l+1\ll r\ll \lambda$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.