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

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  • $\begingroup$ Do you mean in this in the conceptual or mathematical sense? $\endgroup$ – xish Dec 14 '13 at 4:04
  • $\begingroup$ I guess both would be delightful. $\endgroup$ – user24082 Dec 14 '13 at 4:05
  • $\begingroup$ 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. $\endgroup$ – Geoffrey Dec 14 '13 at 4:22
  • $\begingroup$ I'm asking why ONLY the states with zero angular momentum are symmetric. $\endgroup$ – user24082 Dec 14 '13 at 5:05
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    $\begingroup$ 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. $\endgroup$ – Slaviks Dec 14 '13 at 5:56
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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 :enter image description here

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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 $\phi$.

Conceptually speaking, a spherically symmetric state gives the electron the option to be in orbit around any axis. In other words, it orbits around no axis.

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