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In a first (or second) course on quantum mechanics, everyone learns how to solve the time-independent Schrödinger equation for the energy eigenstates of the hydrogen atom: $$ \left(-\frac{\hbar^2}{2\mu} \nabla^2 - \frac{e^2}{4\pi\epsilon_0 r} \right)|\psi\rangle = E|\psi\rangle .$$ The usual procedure is to perform separation of variables to obtain a radial equation and an angular equation, which are solved separately for the radial wavefunction $R_{n\ell}(r)$ and the spherical harmonics $Y_\ell^m(\theta, \phi)$. By combining these factors, we obtain the hydrogenic stationary-state wavefunctions $$ \psi_{n\ell m}(r, \theta, \phi) = R_{n\ell}(r) Y_\ell^m(\theta, \phi). $$ It seems to me that nothing about this problem (which is essentially the quantum-mechanical analogue of the classical two-body problem) is inherently three-dimensional, so we can just as well consider the corresponding problem in $d = 1, 2, 4, 5, \dots$ spatial dimensions. In fact, by transforming into hyperspherical coordinates \begin{align} x_1 &= r \cos(\phi_1) \\ x_2 &= r \sin(\phi_1) \cos(\phi_2) \\ x_3 &= r \sin(\phi_1) \sin(\phi_2) \cos(\phi_3) \\ &\vdots\\ x_{d-1} &= r \sin(\phi_1) \cdots \sin(\phi_{d-2}) \cos(\phi_{d-1}) \\ x_d &= r \sin(\phi_1) \cdots \sin(\phi_{d-2}) \sin(\phi_{d-1}) \,. \end{align} the same procedure employed in the 3D case should carry over (with a considerable increase in algebraic complexity). Has this problem been solved before? If so, what is known about the solution? In particular,

  • How many quantum numbers are required to describe hydrogenic stationary states in $d$ spatial dimensions? Do these quantum numbers have a clear physical interpretation, like $n, \ell, m$?
  • What is the $n$-dimensional analogue of the Bohr formula $E_n = E_1/n^2$? Do energies continue to depend on a single (principal) quantum number? As the dimension tends to infinity, is the energy spectrum discrete or continuous?
  • What is the $d$-dimensional analogue of the spherical harmonics $Y_\ell^m(\theta, \phi)$? Can these functions be described as eigenfunctions of $d$-dimensional angular momentum operators, analogous to $L^2$ and $L_z$? If so, what are the eigenvalues?
  • Is there a reasonable closed form for the $d$-dimensional hydrogenic stationary-state wavefunctions? If not, is there a reasonable (asymptotic) approximation formula?
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    $\begingroup$ Comments to the post (v5): 1. Note that the physically relevant Coulomb potential $V$ in $d$ spatial dimensions is not just $\propto 1/r$ but instead $\propto 1/r^{d-2}$ (in order to ensure that the electric field satisfies Gauss's law), cf. this related Phys.SE post. 2. Note in particular that the quantum mechanical hydrogen atom is unstable for $d>4$ (and with a sufficiently strong EM coupling constant also for $d=4$). $\endgroup$ – Qmechanic May 15 '16 at 10:19
  • $\begingroup$ How far off would it be to say this is a good hand wavy justification on why we are three dimensional beings? I don't think this would be an acceptable question to the site (as there might be a very simple explanation to why this question I just asked is stupid) so I'll just ask it in this comment to lower humiliation... $\endgroup$ – Vendetta May 19 '16 at 12:49
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A nice overview of the problem is given in arXiv:1205.3740. I'll summarise the most important points here.

Let $d$ be the number of space dimensions. Then the Laplace operator is given by $$ \Delta=\frac{\partial^2}{\partial r^2}+\frac{d-1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\Delta_S\tag{1} $$ where $\Delta_S$ is the Laplace operator on the $d-1$ sphere.

The Coulomb potential is given by the solution of $$ -\Delta V=e\delta(\boldsymbol r)\tag{2} $$ which is solved by $$ V(r)=\frac{2(d/2-1)!}{(d-2)\pi^{(d-2)/2}}\frac{e}{r^{d-2}}\tag{3} $$

One way to show the expression above is to consider the Gauss Law in $d$ dimensions, that is, $E(r)=e/\int_r\mathrm ds$ where the area of the $d-1$ sphere can be found here.

With this, the Schrödinger equation reads $$ \left[\frac{1}{2m}(-\Delta)+V(r)-E\right]\psi(\boldsymbol r)=0\tag{4} $$

The problem has spherical symmetry so that, as usual, we may write $$ \psi(r,\boldsymbol \theta)=\frac{1}{r^{(d-1)/2}}u(r)Y_\ell(\boldsymbol\theta)\tag{5} $$ where $\boldsymbol \theta=(\theta,\phi_1,\phi_2,\cdots,\phi_{d-2})$ and the power $(d-1)/2$ of $r$ is chosen to remove the linear term in $(1)$. The superspherical harmonics ($\sim$ Gegenbauer polynomials) are the generalisation of the usual spherical harmonics to $d$ dimensions: $$ \Delta_S Y_\ell+\ell(\ell+d-2)Y_\ell=0\tag{6} $$

Using this form for $\psi$, the Schrödinger equation becomes $$ u''(r)+2m[E-V_\ell(r)]u(r)=0\tag{7} $$ where the effective potential is $$ V_\ell=V(r)+\frac{1}{2m}\frac{\ell_d(\ell_d+1)}{r^2}\tag{8} $$ with $\ell_d=\ell+(d-3)/2$.

Now, this eigenvalue problem has no known analytical solution, so that we must resort to numerical methods. You can find the numerical values of the energies on the arxiv article. An important point that is addressed in that article is that there are no negative eigenvalues for $d\ge 4$, that is, there are no bound states in more than three dimensions. But for $d\ge 5$ there are stable orbits, with positive energy, with well behaved wave functions.

To answer to some of your questions:

  • In general you need $d$ quantum numbers for $d$ dimensions, modulo degeneracies. In the case of the hydrogen atom in 3D, there are the spherical symmetry and the accidental symmetry$^1$, so that you only need one quantum number. In $d$ dimensions the spherical symmetry remains, but I think that the accidental does not, which would mean that you need $d-1$ quantum numbers for $d\neq 3$. One would need to check whether the Runge-Lenz vector is conserved in $d$ dimensions or not (left as an exercise). If this vector is actually conserved, then the energies would depend on $d-2$ quantum numbers.

  • As there is no analytical solution to the radial Schrödinger equation, we don't know. In the case of $d=3$ dimensions, the Bohr-Sommerfeld quantisation rule turns out to be exact. We could check what this scheme predicts for $E_n$ (though we couldn't know whether it would be exact or not: we must compare to the numerical results).

  • These are well-known for mathematicians. You can read about them on the wikipedia article.

  • In closed form, no. I don't know of an asymptotic formula, but it should be rather easy to derive from the radial Schrödinger equation, where the centrifugal term $r^{-2}$ dominates for $r\to\infty$, so that we can neglect the Coulomb term.


$^1$ In spherical symmetric systems, the potential depends on neither $\theta$ nor $\phi$. In these systems energies don't depend on $m$, the azimuthal quantum number. Therefore, in general for radial potentials the energies depend on two quantum numbers, $n,\ell$. In the specific case of $V=1/r$, there is another symmetry that is kind of unexpected (or at least it is not very intuitive geometrically). When $V=1/r$ the rotational symmetry $SO(3)$ is enlarged into a $SO(4)$ symmetry, and this new symmetry is known as accidental symmetry. This symmetry makes the energy levels independent of $\ell$, that is, $E=E_n$. Note that this symmetry is not present in the rest of the atomic table, that is, multielectronic atoms, which makes the energy levels depend on the angular momentum (and thus, Hund's rules).

One may illustrate the above as \begin{aligned} \text{If $V=V(r,\theta,\phi)$} &\longrightarrow E=E_{n\ell m}\\ \text{If $V=V(r)$}&\longrightarrow E=E_{n\ell}\\ \text{If $V=1/r$}&\longrightarrow E=E_{n} \end{aligned} The first line is the general result for 3D systems; the second line is the result of spherical symmetry; and the third line is the result of the accidental symmetry. If you want to read more about this symmetry, you'll find some nice references in Why are hydrogen energy levels degenerate in $\ell$ and $m$? and/or here.

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  • $\begingroup$ Thanks for this great answer! A couple of questions: (1) Does equation 3 have the right constants? For $d = 3$ I get $V(r) = \frac{3e}{2r}$, not $\frac{e^2}{r}$ as expected. (2) What do you mean by "accidental symmetry" in your first bullet point? I thought the two relevant symmetries were the rotational symmetries in $\theta$ and $\phi$. $\endgroup$ – David Zhang May 16 '16 at 3:15
  • $\begingroup$ @DavidZhang 1) whoops youre right: I meant $\Gamma(d/2)$ instead of $(d/2)!$. I should have written $(d/2-1)!$. 2) I'll extend my answer. $\endgroup$ – AccidentalFourierTransform May 16 '16 at 11:16

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