Proof that eigenvalues of the Coulomb hamiltonian are not nonnegative Let's consider the Coulomb Hamiltonian
$$
-\Delta - \frac{1}{|x|}$$
in $\mathbb{R}^3$.
It is known that eigenvalues of the Coulomb Hamiltonian are negative.
I know it has negative eigenvalues, but I can't prove it hasn't nonnegative eigenvalues.
That is, I want to prove the following.
If smooth function $u$ satisfies the following
$$\int_{\mathbb{R}^3
} |u(x)|^2 dx + \int_{\mathbb{R}^3
} |\nabla u(x)|^2 dx ＜ \infty$$ and
$$-\Delta u - \frac{u}{|x|} = \lambda u$$ for some $\lambda \geq 0$, then $u=0$.
Any advice would be appreciated.
 A: Let $V(x) = -\frac{1}{|x|}$ be the potential and $H = -\Delta + V$. We know that $H$ is a self-adjoint operator on $H^2(\mathbb{R}^3)$. Let $U(t)\psi(x) = e^{-\frac{3t}{2}}\psi(e^{-t}x)$. $U$ is a strongly continuous unitary group (dilation).
Now let $\lambda$ be an eigenvalue of $H$. Since $H$ is self-adjoint and $U$  a unitary group, we have $\lambda \in \mathbb{R}$ and
we observe that $\langle \psi, [U(t),H] \psi \rangle = \langle U(-t)\psi, H \psi \rangle + \langle H\psi, U(t)\psi \rangle = \langle U(-t)\psi, \lambda \psi \rangle + \langle \lambda\psi, U(t)\psi \rangle= 0$
Observe that the potential $V$ satisfies $U(-t)VU(t) = e^{-t}V$. So, $ 0 = \lim_{t \to 0} \langle \psi, \frac{1}{t}[U(t),H] \psi \rangle = \lim_{t \to 0} \langle U(-t)\psi, \frac{1}{t}(H - U(-t)HU(t)) \psi \rangle = \lim_{t \to 0} \langle \psi, \frac{1-e^{-2t}}{t} (-\Delta \psi)\rangle + \langle \psi, \frac{1-e^{-t}}{t} V\psi \rangle = \langle \psi, (-2\Delta + V) \psi \rangle = \langle \psi, -\Delta \psi + H\psi \rangle = \langle \psi, -\Delta \psi + \lambda\psi \rangle$
The calculation shows that, if $||\psi|| = 1$:
$-\lambda = \langle \psi, -\Delta \psi \rangle > 0$. So, the point spectrum is contained in $(-\infty, 0 )$.
A: As I recall, there's a theorem that bound states with potentials that go to zero at infinity necessarily have negative total energy, otherwise the wave function is not normalizable. This implies eigenvalues of the energy will be non-positive. The remaining question is whether the ground state has 0 energy. Waves in a bounded region of space give rise to interference which tends to limit solutions to the associated wave equation, so bound states imply discrete Eigenvalues.
Now a Coulomb potential has scattering states, i.e. unbounded: Coulomb Scattering. These have a continuous spectrum of positive eigenvalues.
