# Why does $\ell=0$ correspond to spherically symmetric solutions for the spherical harmonics?

In quantum mechanics why do states with $\ell=0$ in the Hydrogen atom correspond to spherically symmetric spherical harmonics?

• Do you mean in this in the conceptual or mathematical sense? – xish Dec 14 '13 at 4:04
• I guess both would be delightful. – user24082 Dec 14 '13 at 4:05
• Are you asking why $Y_0^0(\theta,\phi)$ is spherically symmetric, or why the spherical harmonics are a part of the solution, or why the state with zero angular momentum is spherically symmetric? Actually, the answer to any of those questions might very well be: because the math says so. – Geoffrey Dec 14 '13 at 4:22
• I'm asking why ONLY the states with zero angular momentum are symmetric. – user24082 Dec 14 '13 at 5:05
• Having a non-zero angular momentum means it has to point somewhere, hence not all directions in space are equivalent, hence lack of spherical symmetry. – Slaviks Dec 14 '13 at 5:56

One way to understand it is to recognize that for the spherical harmonic $|l,m\rangle$ with $l=0$ (and obviously $m=0$), we have $\hat L_i|0,0\rangle=0$, where $\hat L_i$ is the angular momentum operator in the direction $i=x,y,z$. It is obvious for $\hat L_z$, which eigenvalue is $m=0$, and can be verified for the other two.
Then, the rotation operator $\hat R(\theta)$ around a direction $\vec n$ with angle $\theta$ is given by $$\hat R(\theta)=\exp(i\theta \,\vec n . \vec{\hat L} )$$ from which we clearly see that the state $|0,0\rangle$ is invariant for all rotations : $\hat R(\theta)|0,0\rangle=|0,0\rangle$ and is thus spherically symmetric.
In this formulation, you see that it is the only state like that. You can also show that the state $|l,0\rangle$ is axially symmetric (along $z$), etc. See for instance this nice picture :
Suppose that there existed a spherically symmetrical wavefunction $\psi({\bf r})=f(r)$ for which $l\neq0$. This cannot be, for if we calculate $\langle \psi | L^2 | \psi \rangle$ we will always get zero, as each term in $L^2$ has derivatives with respect to $\theta$ and $\phi$.