Two questions in spectral theory: the spectrum of the Fourier transform and the Hamiltonian of the hydrogen atom. I have the following two questions:


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 $\sigma$? 
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.
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
$$
\begin{align}
   (-\Delta -\frac{1}{\|x\|})f &amp; = -\Delta f -\frac{1}{r}f \\
             &amp; = -\frac{1}{r^{2}}\frac{d}{dr}r^{2}\frac{df}{dr}-\frac{1}{r}f.
\end{align}
$$
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:
$$
\begin{align}
    (-\Delta-\frac{1}{r})e^{-r/2}
      &amp; = -\frac{1}{r^{2}}\frac{d}{dr}(r^{2}(-\frac{1}{2})e^{-r/2})-\frac{1}{r}e^{-r/2} \\
      &amp; = -\frac{1}{4}e^{-r/2}+\frac{1}{r}e^{-r/2}-\frac{1}{r}e^{-r/2}=-\frac{1}{4}e^{-r/2}.
\end{align}
$$
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.
The answer to the first question is simply the following: take $f(z)=z^4$ in the spectral mapping theorem and use $\mathfrak{F}^4=1$.