This question asked how to prove the completeness of the spherical harmonics in the sense that $\{Y_{Lm}\}$ spans the set of square-integrable functions on the sphere. I'm looking for an elementary proof of what I think should be a much simpler theorem. Suppose that we're interested in the set of all functions on the sphere that have the same $L$, i.e., the set of all complex-valued functions that are eigenfunctions of the Laplacian with eigenvalue $-L(L+1)$. I'm looking for the most elementary possible proof that for this fixed $L$, the set $\{Y_{Lm}\}$ for varying $m$ is a basis. That is, I want an elementary proof that the multiplicity is $2L+1$. Ideally I would like this to be at a freshman physics level, intelligible even to people who haven't had linear algebra. (So the language would actually be less sophisticated that in my statement of the problem above.) I would be perfectly happy with a proof restricted to $L=1$.
The closest I've been able to come is an approach that is kind of complicated, and I'm not sure it's even right. First we reduce the problem to real-valued functions by forming linear combinations of $Y_{Lm}$ and $Y_{L,-m}$. Next we argue that if the hypothesis were false, we would be able to witness its failure with a state $W$ of definite $m$ (i.e., one whose azimuthal variation is $e^{im}$) that is independent of $Y_{Lm}$. Now form linear combinations of $W$ with $Y_{Lm}$ such that the linear combination $X$ vanishes at some $\theta$. This can be done at almost all $\theta$. Finally, tune the choice of $\theta$ so that for this particular value of $\theta$, not just $X$ but $\partial X/\partial \theta$ vanishes as well. Now argue that by uniqueness of solutions to 2nd order differential equations, $X$ vanishes identically, which is a contradiction. I'm not sure that the last part even works, since the standard uniqueness theorems may not work for a space with this topology.
Is there a simpler way to go about this?
I also thought about physical arguments involving state counting. If the $L=1$ state has multiplicity 3, then coupling two spin-1 states gives 9 states, and this makes sense because you get spin-0, spin-1, and spin-2 couplings, for a total of 1+3+5=9 states. But all this really seems to do is prove that if the multiplicity of spin $L$ is $2L+1$, we get self-consistency — it doesn't seem to show that $2L+1$ is necessary.
Maybe there is some sort of nice argument saying that if we can couple two states of spin 1 to make spin 0, then this spin 0 state must be unique?
Best Answer
The spherical harmonics are $Y_l^m(\theta,\varphi) = C_{lm} P_l^m(\cos\theta) e^{im\varphi},$ where $C_{lm}$ are some constants for normalization and $l = 0, 1, 2, \ldots$ and $m = -l, \ldots, +l.$
The completeness of $\mathcal B = \{ Y_l^m(\theta,\varphi) \}$ depends on the completenesses of $\mathcal B_\theta = \{ P_l^m(\cos\theta) \}$ and $\mathcal B_\varphi = \{ e^{im\varphi} \}.$ If $H_1, H_2$ are two Hilbert spaces with bases $\mathcal B_1, \mathcal B_2$ then $H_1 \otimes H_2$ has basis $\mathcal B_1 \otimes \mathcal B_2 = \{ b1 \otimes b_2 \mid b_1 \in \mathcal B_1, \, b_2 \in \mathcal B_2 \}$.
The completenesses of $\mathcal B_\theta$ and $\mathcal B_\varphi$ follow from Sturm-Liouville theory.
Here we however have a small complication. The physicality of the solutions require that $|m| \leq l.$ Therefore we don't use the full $\mathcal B_\varphi$ for a given $l$ so we actually don't use a complete basis in $\varphi$. But if we took the whole $\mathcal B_\varphi$ we would have to set the coefficients for $P_l^m$ to $0$ when $|m| > l$ to get physical (bounded) solutions.