According to Kepler's First Law, the orbit of a planet is an ellipse round the sun with the sun at one focus. There's an inherent asymmetry in this. Instead of the sun being in the dead center, its shifted over a little bit.

In the hydrogen atom, all the orbitals of the electron are symmetric about the proton at the dead center. Why is there no similar asymmetry?

You can convert the function for the position of a classical simple harmonic oscillator with respect to time to a space dependent probability distribution where the probability is higher at the classic turning points where the velocity is at its lowest. The ground state of the quantum harmonic oscillator has a higher probability exactly between the classical turning points. The quantum solutions more closely match the classical probability at higher quantum numbers.

I was thinking the classic "lopsidedness" of gravity could be recovered at higher quantum numbers for a Coulomb like potential Schrodinger Equation. But higher principle quantum numbers just enlarge the orbitals, they all remain symmetric about the proton. So that can't give you higher probabilities on an ellipse. The technique to recover classical behavior that works for the harmonic oscillator fails for Coulomb like potentials.

Are there circumstances where any asymmetry appears in the probability of the electron, in particular, concentrations of probability along an ellipse?

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    $\begingroup$ I would think it is the same reason why orbitals live in 3D, but orbits live in 2D. $\endgroup$
    – DanDan0101
    Jun 1 at 22:51
  • $\begingroup$ The Kepler problem is usually discussed in center of mass coordinates using reduced masses. We are giving up on analyzing the motion of the "central body" and accept that the simplified problem does not conserve momentum. In the simplified derivation of the hydrogen spectrum and orbitals the mass of the proton is assumed to be infinite. In reality it is 2000 times the mass of the electron, so the proton wave function doesn't matter enough to worry about it for the simplified problem, but you are correct... it is there and matters for precision spectroscopy. $\endgroup$ Jun 1 at 23:02
  • $\begingroup$ In the case of H atom, if the proton is not at the center, a dipole moment appears and this state cannot be the groundstate of a single atom. However, when two H atoms interact at a large distance a dipole moment appears and you obtain this asymmetric state in every atom. $\endgroup$
    – M06-2x
    Jun 1 at 23:40
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    $\begingroup$ Are you asking how the vanishing L state, s, could connect to a parabola? Have you worked out the LRL invariants? $\endgroup$ Jun 2 at 1:54

2 Answers 2


First things first. You know that for the Coulomb potential there are additional invariants, the Laplace-Runge-Lenz vector beyond angular momentum, nonvanishing for elliptical orbits, classically, but vanishing for circular ones. The "lopsidedness" you are talking about amounts to the ellipse's eccentricity, a dimensionless quantity $|{\mathbf A} /me^2|$. Classically, the elliptical orbit is tantamount to angular momentum conservation in conjunction with LRL invariance.

Quantizing, classical trajectories have a very uncomfortable connection to both Wigner functions and wavefucntions. In QM, this eccentricity quantity presents as $$ \frac{\hbar \sqrt{2}}{e^2 \sqrt{m}} \sqrt{|H(L^2/\hbar^2+1)|} \qquad \leadsto \qquad \frac{\hbar \sqrt{2}}{e^2 \sqrt{m}} \sqrt{|E|( l(l+1)+1)}~~. $$

You then see this eccentricity analog never vanishes in QM, even for s waves, spherical states! This angular momentum zero-point-shift is well known in deformation quantization, and informs the proper correspondence principle, however counter-intuitive. It has been thoroughly investigated in phase space by Dahl and collaborators, J P Dahl & M Springborg, Mol Phys 47 1001 (􏰑1982)􏰀.

So, formally, the classical eccentricity survives in QM and is crucial for its spectrum. Recall it is the basis of the original modern quantization of the hydrogen atom, W Pauli Jr, "Über das Wasserstoffspektrum vom Standpunkt der neuen Quantenmechanik" Zeitschrift für Physik 36 no. 5 (1926): 336-363.

For large n, and for the largest $l=n-1$ for that n, the square root tends to 1.


In general, you should simply stop thinking of quantum theory in terms of classical ideas.

However, this particular issue has a simple answer.

There is an analogous problem in scattering, that when a beam is incident upon an atom, say, the scattered wavefunction is spherically symmetric, despite the fact that when we actually observe the scattered particle, the correct system wavefunction is with the scattered particle in a specific momentum direction, and the atom recoils in the opposite direction to conserve momentum.

The standard interpretation, silly as it is using collapse terminology, is that the wavefunction before measurement conserves the spherical symmetry, but upon measurement, it collapses to a final state that conserves momentum but not spherical symmetry. If you accept Many Worlds Interpretation, then the system wavefunction spherically symmetric in the multiverse, but each universe branch sees a specific direction.

This gives us a simple explanation for the problem of yours. The Keplerian ellipse is easily expressed in terms of orbital elements. The parts of them that are related to energy, e.g. eccentricity and semimajor axis, we can imagine as being fixed, but the others that are merely the fixing the orientation of the ellipse, those angles really do not matter, despite them being fixed for any single trajectory. If you allow them unrestricted variation, that the ensemble has the angles uniformly distributed over all possible values, then the resultant wavefunction ought to be spherically symmetric despite each ellipse being not.

Note that if you have a tiny perturbation anywhere, even just temporary ones, say, from an energy uncertainty, then the same kind of statistical thermodynamics argument that motivates the ergodic hypothesis will also explain why we ought to expect that the system should be uniformly distributed over all possible angles, i.e. spherically symmetric.

Let us include a historical tangent. Bohr-Sommerfeld quantisation in Old Quantum Theory went quite far into considering ellipses and relativistic corrections, so that when Schrödinger originally wrote down his wave equation, he knew that Klein-Gordon equation had the relativistic corrections in the wrong direction from the experiments that agrees with Bohr-Sommerfeld analysis for the H atom. This is why he went a step down and published just the non-relativistic wave equation.

But you should really pay lesser and lesser attention to these things, because the correct worldview is the quantum one, not the classical one. There is no way to get the ellipse ideas to work out, because the correct counting of quantum states comes from quantum theory, where zero orbital angular momentum is explicitly possible (and ground state needs it). You really should treat the Bohr-Sommerfeld model as a lucky fluke and nothing more.

  • $\begingroup$ I think I follow you. The actual trajectory of planet isn't simply an ellipse, but an ellipse whose semi-major axis essentially sweeps out through 360 degrees over time so the classical case IS spherically symmetric over time. Does that sweeping out happen in a two body system? $\endgroup$
    – R. Romero
    Jun 2 at 17:19
  • $\begingroup$ You are getting the correct idea, but in the case of the planets the conservation laws make it such that the orbital plane will not be affected unless some other thing disturbs things. In the ideal classical two-body system, it will not do this sweeping out, but in quantum, it is thought that it will do so. $\endgroup$ Jun 2 at 17:24
  • $\begingroup$ @R.Romero The semi-major axis of a single planet does not show any precession in Newtonian mechanics. In general relativity it is just one of many effects, another one of which is the emission of gravitational waves which lead to an instability of the system. That's why we are observing black hole mergers. That kind of thing does, of course, not exist in quantum mechanics. $\endgroup$ Jun 2 at 22:54

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