First of all, note that different authors disagree on what should be the Coulomb potential $V$ in $d$ spatial$^1$ dimensions. We will assume that it satisfies Gauss's law, i.e.
$$ V(r)~\propto~\left\{\begin{array}{rcl} r^{2-d} &\text{for}& d~\neq~ 2, \\
\ln(r)&\text{for}& d~=~2. \end{array}\right.\tag{1}$$
We will here only discuss the quantum mechanical hydrogen atom with $d\geq 3$. Let us normalize the Hamiltonian as
$$ H~=~-\frac{\hbar^2}{2m}\Delta - e_d^2r^{2-d} .\tag{2}$$
For a rigorous discussion of unbounded operators, domains and self-adjoint extensions, etc., see e.g. Ref. 1 and references therein. Let us here summarize the results:
The hydrogen atom in three spatial dimensions $d=3$ is stable and has bound states.
Four spatial dimensions $d=4$ is an interesting border case, where the Coulomb potential and the centrifugal potential have the same $1/r^2$ behaviour. If we define a dimensionless constant
$$ Z~:=~\frac{2me_{d=4}^2}{\hbar^2},\tag{3}$$
then there are three cases, cf. e.g. this Phys.SE post:
$Z\leq 0$: The Hamiltonian (2) has no bound states, i.e. the hydrogen atom is ionized.
$Z>1$: The Hamiltonian (2) is unbounded from below, i.e. the hydrogen atom is unstable.
$0<Z\leq 1$: It is possible to define asymptotic boundary conditions (ABCs) at $r=0$ / selfadjoint extensions of the Hamiltonian, such that the spectrum is bounded from below. Some of these extensions have bound states, others have not.
In more than four spatial dimensions $d>4$, the hydrogen atom is unstable. Roughly speaking, for $d>4$ the Coulomb potential (1) dominates the $1/r^2$ centrifugal potential at sufficiently small radius $r$ close to the nucleus.
The instability can be rigorously proven via e.g. the variational method, cf. the following theorem.
Theorem. An attractive singular power law potential
$$ V(r)~\propto~ -r^{-n}, \qquad
n~>~2,\tag{4} $$
has a spectrum that is unbounded from below, i.e. it has no ground state and is unstable.
Proof of theorem: Consider a normalized Gaussian test/trial wavefunction
$$\begin{align}\psi(r)~=~&Ne^{-\frac{r^2}{2L^2}}
~=~Ne^{-\frac{x^2+y^2+z^2}{2L^2}}, \cr
\int d^dr~|\psi(r)|^2
~=~&\langle\psi|\psi \rangle~=~1,\end{align}\tag{5}$$
where $N,L>0$ are two constants. For dimensional reasons, the constant $L$ must have dimension of length, and the normalization constant $N$ must scale as
$$N ~\propto~ L^{-\frac{d}{2}}.\tag{6}$$
The expectation value
$\langle\psi| K|\psi \rangle$ of the kinetic energy operator
$K=-\frac{\hbar^2}{2m}\Delta$ must scale as
$$ 0~\leq~\langle\psi| K|\psi \rangle ~\propto~ L^{-2},\tag{7}$$
essentially because the Laplacian $\Delta=\vec{\nabla}^2$ contains two position derivatives.
The expectation value $\langle\psi| V|\psi \rangle$ of the potential (4) must scale as
$$ 0~\geq~\langle\psi|V|\psi \rangle ~\propto~ - L^{-n}\tag{8}$$
for similar reasons. Thus by choosing $L\to 0^{+}$ smaller and smaller, the negative potential energy $\langle\psi| V|\psi \rangle\leq 0$ beats the positive kinetic energy $\langle\psi| K|\psi \rangle\geq 0$, so that the average energy $\langle\psi| H|\psi \rangle$ becomes more and more negative,
$$\begin{align} \langle\psi| H|\psi \rangle ~=~&\langle\psi| K|\psi \rangle + \langle\psi| V|\psi \rangle\cr
~\to~& -\infty
\quad\text{for}\quad L\to 0^{+}.\end{align} \tag{9}$$
Hence, the spectrum is unbounded from below. $\Box$
References:
- M. Bures & P. Siegl, Annals of Physics 354 (2015) 316, arXiv:1409.8530.
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$^1$ One may show that for small compact dimensions much smaller than the characteristic size of the hydrogen atom (as predicted by e.g. string theory), such dimensions get averaged over and can effectively not be felt by the hydrogen atom. In other words, one effectively only has to consider large spatial dimensions $\cong \mathbb{R}^d$.
