I understand that in 3+1 dimensions according to classical physics atoms should be unstable however atoms are stable in 3+1 dimensions because the behavior of atoms is governed by quantum physics instead of classical physics.

I have read before that atoms cannot exist in 4+1 dimensions but I'm not sure if this is taking quantum mechanics into account. Are atoms possible in 4+1 dimensions when quantum physics is taken into account or are they still unstable even after taking quantum mechanics into account?


1 Answer 1


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:

    1. $Z\leq 0$: The Hamiltonian (2) has no bound states, i.e. the hydrogen atom is ionized.

    2. $Z>1$: The Hamiltonian (2) is unbounded from below, i.e. the hydrogen atom is unstable.

    3. $0<Z\leq 1$: 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. $\Box$


  1. M. Bures & P. Siegl, Annals of Physics 354 (2015) 316, arXiv:1409.8530.


$^1$ 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 $\cong \mathbb{R}^d$.

  • 2
    $\begingroup$ Just a side note. You can roughly understand the instability as follows. The kinetic energy scales as $1/r^2$ so for any potential that goes like $1/r^2$ or faster towards zero, the wave function does not push back hard enough when it is compressed and the particle falls to the origin. Such potentials are called singular potentials. For a pedagogical treatment see: A.M. Essin & D.J. Griffiths, "Quantum mechanics of the $1/x^2$ potential", Am. J. Phys. 74 (2006) 109. $\endgroup$
    – Praan
    Dec 5, 2015 at 4:24
  • $\begingroup$ Shouldn't the uncertainty principle eliminate the instability? An electron falling into the nucleus would be confined in a very small space, which would require it to have higher momentum, thus energy; at some point, I would expect that effect to balance out the attractive potential, producing a stable energy valley. $\endgroup$ May 1 at 23:54
  • $\begingroup$ Hi Logan R. Kearsley. Thanks for the feedback. No, HUP does not block the instability. One may e.g. check the above Gaussian test/trial wavefunction satisfies HUP. $\endgroup$
    – Qmechanic
    May 2 at 7:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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