Derivation of the Angular Component of Spherical Harmonics

Question

Every derivation of spherical harmonics seems to tell me that $e^{im\phi}$ is the most obvious solution in the world to $\frac{\partial^2 f}{\partial \phi^2} = -m^2f$. But what about $Ae^{im\phi} + Be^{-im\phi}$ ? Isn't that a more general, more "correct" solution? Whether it's spherical harmonics or particle on a ring, I always run into $e^{im\phi}$ and $e^{-im\phi}$ as separate solutions. But why, when the general form involves writing them as a linear combination?

I think at the heart of it is their orthogonality, and in physics the subspace formed by the degeneracy of $m$ and $-m$, as well as the connection to the angular momentum operator $L_z$. But I can't really unpack it into a strictly mathematical explanation for why in none of the derivations they are ever presented initially as a linear combination. I feel like the derivation should at least begin there, before separating the two terms into distinct solutions. Or am I missing something, and is there a good reason why they are treated as separate from the start?

Answer 1

Maybe the "extra seasoning" that you need is that the solutions must be periodic:

$$f(\phi) = f(2 \pi m + \phi).$$

This is what forces $m$ to be an integer, and then leads to the general solution

$$f(\phi) = Ae^{im \phi} + Be^{-im \phi}.$$

Answer 2

You're correct. And this is a good question.

The general solution is $A_m e^{im\phi}+B_me^{-im\phi}$ where $m\in\mathbb{Z}$. (Note that there is some amount of redundancy in this notation since just the meanings of $A_m$ and $B_m$ are interchanged for $m<0$ compared to $m>0$.) As usual, we require some sort of boundary condition to select the desired constants $A_m, B_m$.

So consider general solutions of the form $\tilde{Y}_{\ell}^m = N_{\ell m} P_\ell^m [A_m e^{im\phi}+B_me^{-im\phi}]$, where $N_{\ell m}$ is some normalization and $P_\ell^m$ are the associated Legendre functions.

Now, we know that $-\ell \le m \le \ell$ and that there exists an operator $L_+$, which when acting on $\tilde Y_\ell^\ell$ gives zero. ($L_+$ is the 'raising operator.' I'm not quoting sources or formulas here since the material is standard.) When you work-out the details, using the form of $L_+$ you'll find that you cannot construct eigenfunctions of $L^2$ and $L_z$ (say) for functions which have both $A_m$ and $B_m$ non-zero.