According to Wikipedia, a harmonic function is one which satisfies: $$ \nabla^2 f = 0 $$ The spherical harmonics (also according to Wikipedia) satisfy the relation $$ \nabla^2 Y_l^m(\theta,\phi) = -\frac{l(l+1)}{r^2} Y_l^m(\theta,\phi) $$ which is 0 only if $l = 0$. So by this definition, are the spherical harmonic functions only harmonic when $l,m = 0$?

As a related question, if I change coordinates to some $\alpha(\theta,\phi), \beta(\theta,\phi)$ and I want to find the orthogonal functions similar to the spherical harmonics in the new $\alpha,\beta$ basis, how would I do that? Would I need to solve Laplace's equation $$ \nabla^2_{\alpha,\beta} f(\alpha,\beta) = 0 $$ where $\nabla^2_{\alpha,\beta}$ is the Laplacian expressed in the new coordinate system?

  • $\begingroup$ why did u ask $m=0$? isn't $l=0$ enough? $\endgroup$ Jun 25, 2020 at 7:07
  • $\begingroup$ @mathworker21 $|m| \leq l$, so $l = 0$ implies $m=0$. So not necessary to say $m=0$, but no loss of generality either. $\endgroup$
    – tparker
    Jan 13, 2022 at 4:06

2 Answers 2


There is more than one Laplacian $\Delta=\nabla^2$ in play here: the ambient Euclidean one, and the invariant Laplacian on the sphere. The "spherical harmonics" extended in the natural way as polynomials, are harmonic for the ambient Euclidean space. They are eigenfunctions for the Laplacian on the sphere, with (generally) non-zero eigenvalues for that operator.

Yes, there is some connection between the two, which can be found in many places...


The spherical harmonics are restrictions to the unit sphere of harmonic functions on $\mathbb R^3$. There are no non-constant harmonic functions on the sphere itself.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .