What is the relation between irreducible representation of $SO(3)$ and spherical harmonics?

Question

I just studied the rep theory of $SU(2)$ (now I know its the double cover of $SO(3)$ so I guess their reps are highly similar) and I also know spherical harmonics (I will use SH for short) are the complete basis of function $S^2\rightarrow C$.

There indeed exists some materials on this topic, but most are out of purely algebraic approach. So I can't grasp the logic and intuition behind this fact.

$SO(3)$ is the group used to describe the 3D rotation, which I can almost definitely claim that, the domain of definition of SH is "$S^2$", is not a coincidence. Intuitively I guess if $G$ is a group and can act on some space $M$, then the action can somehow induce a rep of $G$ whose rep space is $M$, then the rep on $M$ can somehow form a basis of the space of function $M \rightarrow C$.

I don't know how to make my statement formally. I wonder whether the previous description can be constructed systematically, or just some confusing hugwash.

Answer 1

At fixed $l$, the $2l+1$ spherical harmonics:

$$ Y_l^m(\theta, \phi) \ \ \ m \in [-l, -l+1, \cdots, l-1, l] $$

are a $2l+1$ dimensional irreducible$^1$ representation of $SO(3)$, with Casimir invariant $l(l+1)$. [1] means they are closed under rotations, so for some arbitrary rotation, $R_{{\bf r} \rightarrow {\bf r'}}$:

$$ Y_l^m(\theta', \phi')= \sum_{m'=-l}^l \big[D^{(l)}_{mm'}(R)\big]^*Y_l^{m'}(\theta, \phi) $$

That is: dipoles rotate into other dipoles, quadrupoles rotate into other quadrupoles, and so on.

I'm not sure how it relates to representation theory, but one thing that makes the $Y_l^m(\theta,\phi)$ useful in physics is that they are eigenfunctions under $z$-rotations, with eigenvalue $\exp(im\phi)$:

$$ R_{\phi}\big[Y_l^m(\theta, \phi)\big] = e^{im\phi} Y_l^m(\theta, \phi)$$

and also something that is not obvious (to me) from pictures of atomic orbitals:

$$ \sum_{m=-l}^l ||Y_l^m(\theta, \phi)||^2 \propto 1 $$

that is, a closed shell is isotropic, it has spherical symmetry.

Answer 2

You could do worse than simply going to your text, e.g., Modern Quantum Mechanics (3rd Ed) by J J Sakurai & Jim Napolitano, and reviewing the definition, $$\langle \hat {\mathbf n}|\ell, m\rangle= Y^m_\ell (\theta, \phi)=Y^m_\ell (\hat {\mathbf n}). \tag{3.232}$$

_{Recall that $\langle \hat {\mathbf n}|L_z= -i\partial_\phi \langle \hat {\mathbf n}|~; \quad \langle \hat {\mathbf n}|{\mathbf L}^2= -\frac{1}{\sin^2\theta}\left (\sin\!\theta ~\partial_\theta ~ \sin\!\theta ~\partial_\theta +\partial_\phi^2 \right)\langle \hat {\mathbf n}|$. The representation logic is then self-evident from $L_z|\ell, m\rangle = m|\ell, m\rangle $ and ${\mathbf L}^2|\ell, m\rangle= \ell(\ell+1)|\ell, m\rangle $.}