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I have the following two questions:

  1. The Fourier transform defines a unitary (provided that it is normalized properly) map $\hat{\cdot}:L^2(\mathbf{R})\rightarrow L^2(\mathbf{R})$. I figured out its point spectrum, which is very easy; is it possible to determine the whole spectrum of this operator? I know $\sigma_p(\hat{\cdot})=\mu_4(\mathbf{C})$ ($4$-th roots of unity) already; is it the case $\sigma(\hat{\cdot})=\mu_4(\mathbf{C})$ also, maybe because $\sigma_p$ might be dense in $\sigma$?

  2. Is there is a concise way (not taking more than two pages say) way to see that the closure $H$ of the Hamiltonian operator of the hydrogen atom (defined on $C^\infty_0$), viz. $$-\frac{1}{2}\Delta-\frac{1}{\|x\|},$$ has domain $H^2(\mathbf{R}^3)$, is self-adjoint, and determine the spectrum.

Remarks on 2: I found a reasonably short proof of self-adjointness in Reed/Simon's Methods of modern mathematical physics, vol. II, Thm. X.15.

I look forward to your answers.

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    $\begingroup$ If $\mathscr{F}$ is the Fourier transform, then $\mathscr{F}^{4}=I$. The spectrum is $\sigma(\mathscr{F})=\{1,i,-1,-i\}$. mathoverflow.net/questions/12045/… $\endgroup$ – DisintegratingByParts Jul 24 '14 at 14:25
  • $\begingroup$ @T.A.E.: this is known to me already, these are just the eigenvalues (point spectrum). How come that $\sigma(\mathscr{F})=\sigma_p(\mathscr{F})$. $\endgroup$ – HAMMER Jul 24 '14 at 14:43
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    $\begingroup$ These questions, while in the same field, seem sufficiently different to me that I don't see why they're together. $\endgroup$ – Semiclassical Jul 24 '14 at 15:16
  • $\begingroup$ The eigenspaces with different eigenvalues are closed and orthogonal and the direct sum of these is everything. So, that's the spectral decomposition of $\mathscr{F}$. Whenever you have an annihilating polynomial $p$ with distinct roots for a linear operator $T$ on a complex linear space (not even necessarily complete) the space decomposes in the direct sum of the eigenspaces associated with the polynomial. The projections are polynomials in $T$ which means they're bounded if $T$ is bounded. math.stackexchange.com/questions/845016/… $\endgroup$ – DisintegratingByParts Jul 24 '14 at 15:56
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    $\begingroup$ That's not the spectrum. There are bound states. $\endgroup$ – DisintegratingByParts Jul 24 '14 at 18:25
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The spectrum is not as stated. The spectrum of the Hamiltonian for the non-relativistic Hydrogen atom has eigenvalues corresponding to the bound states of Hydrogen, and these are negative. The positive spectrum corresponds to unbound states, and is a continuous spectrum. In spherical coordinates, for a function which depends on the radius $r$ only, one has $$ \begin{align} (-\Delta -\frac{1}{\|x\|})f & = -\Delta f -\frac{1}{r}f \\ & = -\frac{1}{r^{2}}\frac{d}{dr}r^{2}\frac{df}{dr}-\frac{1}{r}f. \end{align} $$ There is a purely radial eigenfunction of this operator which corresponds to the ground state of the Hydrogen atom. This eigenfunction has the form $f(r)=e^{-r/2}$ in this case where the physical constants are missing: $$ \begin{align} (-\Delta-\frac{1}{r})e^{-r/2} & = -\frac{1}{r^{2}}\frac{d}{dr}(r^{2}(-\frac{1}{2})e^{-r/2})-\frac{1}{r}e^{-r/2} \\ & = -\frac{1}{4}e^{-r/2}+\frac{1}{r}e^{-r/2}-\frac{1}{r}e^{-r/2}=-\frac{1}{4}e^{-r/2}. \end{align} $$ Therefore $-1/4 \in \sigma_{P}(-\Delta-\frac{1}{|x|})$. There are lots of other negative eigenvalues and eigenfunctions for this operator.

I already answered the first part of your question in the comment section. The spectrum of the Fourier transform is $\{ 1, i, -1, -i \}$.

Final Note: I see that you changed the question again. This was a complete and valid answer to the original question, and the answer is not a trivial one. Before that, the last paragraph above was a complete answer to the question. HAMMER, if you don't like the answer, it's best to ask a separate question, rather than invalidate a perfectly good answer that someone has put thought and effort into. You obviously learned something from my answers, and it's not polite to invalidate that in such a way that it makes my answer look completely irrelevant to the topic.

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  • $\begingroup$ One thing this doesn't address (which is reasonable given the direction of the problem) is the structure/degeneracy of the bound states. The typical discussion of the origin of the seemingly 'accidental' degeneracies is in terms of a hidden $SO(4)$ symmetry group. (A less orthodox explanation is the concept of superintegrability. (Section 3.4 discusses the hydrogen atom case.)) $\endgroup$ – Semiclassical Jul 24 '14 at 20:12
  • $\begingroup$ I find it odd to see an answer to an entirely different question, which I find impolite also. (I asked for a proof that the spectrum is $[0,\infty)$, and I clearly thought that this is true. Instead you should have corrected me in a comment, said what the spectrum is and proved it. This would have been much more appropriate.) $\endgroup$ – HAMMER Jul 25 '14 at 12:46
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    $\begingroup$ Showing that your conjecture is false IS an answer. $\endgroup$ – DisintegratingByParts Jul 25 '14 at 14:53
  • $\begingroup$ Noooooooooooot. $\endgroup$ – HAMMER Jul 26 '14 at 15:20
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    $\begingroup$ You originally asked us to show that the spectrum of the Hamiltonian for the Hydrogen atom has spectrum $[0,\infty)$, and I told you in comments that was false, and I hinted for you to look at the bound states of Hydrogen. You questioned me on that, and seemed to feel that the spectrum of the operator was somehow different than the classical Physics spectrum. So I produced an eigenfunction with eigenvalue $\lambda < 0$ associated with the ground state of Hydroen, thereby proving that it would be impossible to show the thing you asked us to show. I think that was a very good answer. $\endgroup$ – DisintegratingByParts Jul 26 '14 at 16:48
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The answer to the first question is simply the following: take $f(z)=z^4$ in the spectral mapping theorem and use $\mathfrak{F}^4=1$.

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  • $\begingroup$ What does the spectral mapping theorem have to do with it? $\endgroup$ – DisintegratingByParts Jul 25 '14 at 0:35
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    $\begingroup$ Well $f(\sigma(\mathscr{F}))=\sigma(\mathscr{F}^4)=\sigma(1)=\{1\}$ implies $\sigma(\mathscr{F})\subset\mu_4$? $\endgroup$ – HAMMER Jul 25 '14 at 12:43

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