I was given a problem where I have a spherically symmetric potential (the exact form is not relevant to this question, I think - but anyway it is 0 for $r\in[a,b]$ and $\infty$ everywhere else) and I was asked to find the ground state energy. Note: I am actually able to solve the problem, but I first assumed $\ell=0$. Now that I think about it, I was just making the assumption blindly and I would like a justification for it.

Is it always true that the ground state of any spherically symmetric potential function has zero orbital angular momentum? [The case of hydrogen atom seems to fall out from algebra, and a physical explanation that works in general would be way nicer]

  • $\begingroup$ related: physics.stackexchange.com/q/69550 (If the ground state has nonzero angular momentum, then it's degenerate.) $\endgroup$
    – user4552
    Nov 15, 2014 at 1:32

2 Answers 2


Here's a paper with a proof that the ground state must be l=0 for spherically symmetric potentials for a single particle, assuming there's a bound state. Abstract:

The variational principle is used to show that the ground-state wave function of a one-body Schrödinger equation with a real potential is real, does not change sign, and is nondegenerate. As a consequence, if the Hamiltonian is invariant under rotations and parity transformations, the ground state must have positive parity and zero angular momentum.


  1. It can be proven that the ground state must be real and non-negative everywhere, assuming a bound state exists and assuming zero spin. The argument in the paper is abstract, but the idea is that, after making the wavefunction into a product of amplitude and phase factors which may vary with position, after calculating the potential and kinetic energy parts of the expectation value $<H>$, the only part of the energy which depends on the phase is a kinetic energy term which is minimized by making the phase constant. If the phase is constant, the wavefunction can be assumed to be real and can't change sign.

  2. From there, it can be proved that the ground state is unique. Basically, if it weren't, then there would be a second ground state wavefunction (which also can't change sign), but they can't be orthogonal since you can't integrate the product of two real non-sign-changing wavefunctions and get zero.

The spherical symmetry of the Hamiltonian demands that if there's an $l \neq 0$ state with an energy, there must also be a $-l$ state with the same energy, so the only possibility is to have $l=0$.

(Having said that, the paper cites this paper, saying that if we have a collection of two particles with strong spin-orbit coupling between them, the overall Hamiltonian can have spherical symmetry, but the system will have spontaneous symmetry breaking and end up with $l \neq 0$.)

  • $\begingroup$ Very nice, exactly what I was hoping to look for! :) $\endgroup$
    – suncup224
    Nov 16, 2014 at 0:22
  • 2
    $\begingroup$ Re the final paragraph -- the deuterium nucleus is an example. $\endgroup$
    – user4552
    Nov 16, 2014 at 4:13
  • $\begingroup$ Just to emphasize, the single-particle condition is absolutely crucial. As detailed in Ben's link, any atom other than a full-shell group is an example of a degenerate ground state in a symmetric potential; they break this answer's argument by having fermionic symmetry, which forbids the particles flocking to the $s$ state. $\endgroup$ Sep 18, 2017 at 18:15

Consider (at least perturbative) contribution of the additional effective "potential" in the radial equation when $l\ne 0$, analyze its sign and judge correspondingly.

The physical explanation is simple - it is an additional kinetic energy of a rotational motion.

  • 1
    $\begingroup$ For the physical explanation, it is not clear apriori that I can remove the rotational motion and end up with a "valid" solution...or am I asking too much of a physical expalnation? $\endgroup$
    – suncup224
    Nov 14, 2014 at 23:30
  • $\begingroup$ @suncup224: Yes, it is clear a priori - there may be purely radial solution. Equations for the angular part of Laplasian and for the radial part are separated. Whether $l=0$ will be the ground state or not is a question to be analyzed with different means. $\endgroup$ Nov 15, 2014 at 6:50

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