Ground state of Spherical symmetric potential always have $\ell=0$? 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]
 A: 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.

Essentially, 


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*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. 

*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 $l \neq 0$.)
A: 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.
