# How to construct a complete set in $L^2(\mathbb{R}^3)$ starting with the Spherical Harmonics?

The Spherical Harmonics form a complete set of functions on the sphere $S^2$, so that any function of $f: S^2\to \mathbb{R}$ can be written uniquely as

$$f(\theta,\phi)=\sum_{l=0}^\infty \sum_{m=-l}^{l}a_{lm}Y^m_l(\theta,\phi).$$

If we use spherical coordinates in $\mathbb{R}^3$, those functions $\{Y^m_l\}$ are eigenfunctions of the angular momentum operators $L^2$ and $L_z$ which do not depend the radial coordinate $r$. Now since the eigenvalue equations

$$L^2Y^m_l=l(l+1)\hbar^2Y^m_l \\ L_zY^m_l = m\hbar Y^m_l$$

have a unique solution to within a constant factor and since if we search for general solutions $\psi_{l,m} \in L^2(\mathbb{R}^3)$ which depend on the $r$ coordinate, the $r$ coordinate behaves just like a parameter, we know that $\psi_{l,m}$ must be a constant times $Y^m_l$. Now, because $r$ is just a parameter this constant is really a function of $r$, that is, we have

$$\psi_{l,m}(r,\theta,\phi)=f(r)Y^m_l(\theta,\phi).$$

This is the general form of the solutions to that eigenvalue equations. Now, how does one build, from that, a complete set of functions in $L^2(\mathbb{R}^3)$?

In the Quantum Mechanics book I'm reading, the author introduces another index $k$ and calls the functions of the complete set $\psi_{k,l,m}$ so that they have the form

$$\psi_{k,l,m}(r,\theta,\phi)=R_{k,l,m}(r)Y^m_l(\theta,\phi),$$

because we want them to be eigenfunctions of $L^2$ and $L_z$. Then one proves that $R_{k,l,m}=R_{k,l,m\pm 1}$. But why does one need another index? Why don't we just pick the functions $\psi_{l,m}$ and write them in the form $\psi_{l,m}(r,\theta,\phi)=f_{l,m}(r)Y^m_l(\theta,\phi)$?

In summary, how starting from the complete set of functions on the sphere $\{Y^m_l\}$ does one build a complete set of functions in $\mathbb{R}^3$?

EDIT: I think I've found a way to do this. It is basically as presented in one Quantum Mechanics book: we consider instead of the whole $L^2(\mathbb{R}^3)$, the subspaces $\mathcal{E}(l,m)$. Those subspsaces are characterized by the fact that if $\psi\in \mathcal{E}(l,m)$ then $L^2\psi=l(l+1)\hbar^2\psi$ and $L_z\psi = m\hbar\psi$.

In that case, we see that if $\psi\in \mathcal{E}(l,m)$ then by virtue of everything I said above, $\psi$ is necessarily of the form $\psi(r,\theta,\phi)=f(r)Y^m_l(\theta,\phi)$. Thus to identify $\psi$ we just need to give $f$. More than $Y^m_l$ are orthonormal with respect to the angular variables, so that if $\psi,\varphi\in \mathcal{E}(l,m)$ we have

$$(\psi,\varphi)=\int_0^\infty f^\ast(r)g(r)r^2dr,$$

in that case it seems we can identify $\mathcal{E}(l,m)$ with $L^2(\mathbb{R}_+)$ with inner product with weight $\rho(r)=r^2$ under the association $\psi\mapsto f$. Thus to get a basis of $\mathcal{E}(l,m)$ we must pick a basis of $L^2(\mathbb{R}_+)$, say $\{R_{k,l,m}\}$, where the índices $l,m$ just indicate that we are picking those to use as basis of the specific space $\mathcal{E}(l,m)$. Thus we have the basis $\{\psi_{k,l,m}\}$ on $\mathcal{E}(l,m)$ given by

$$\varphi_{k,l,m}(r,\theta,\phi)=R_{k,l,m}(r)Y^m_l(\theta,\phi)$$

After that we decompose $L^2(\mathbb{R}^3)$ as the direct sum

$$L^2(\mathbb{R}^3)=\bigoplus_{l=0}^\infty \bigoplus_{m=-l}^l \mathcal{E}(l,m)$$

Is that the idea?

The basic idea here is that we'd like to decompose every function in $g \in L^2(\mathbb R)$ into a 'radial' part $f$ an angular part $Y$. The first obstacle here is that functions from $\mathbb R^3$ aren't the same as functions from $[0,\infty) \times S^2$. Even if we identify them with each other (via polar coordinates), then it isn't clear that every funtion can be written in the desired way $g(r,\theta,\phi) = f(r)Y(\theta,\phi)$.
The way to solve this is the following: Introduce $\otimes$, the so-called tensor product of Hilbert spaces. For instance, $L^2(\mathbb R^+) \otimes L^2(S^2)$ is such a tensor product, and acutally a Hilbert space itself. It can be shown that this Hilbert space is isomorphic to $L^2(\mathbb R^+ \times S^2)$, so $$L^2(\mathbb R^+ \times S^2) \simeq L^2(\mathbb R^+) \otimes L^2(S^2).$$
• Thanks for the answer @Roland. Indeed, one of the things that are lately causing me more trouble is that idea of "radial" and "angular" parts of functions. The Quantum Mechanics book I'm studying talks a lot about these but I haven't seem any way to properly define this idea. I'll take a look at the reference you suggested, but could you elaborate a little more on the matter? How the tensor product allows us to decompose functions into radial an angular parts and how this allows one to build a complete set in $L^2(\mathbb{R}^3)$ from the Spherical Harmonics? Thanks very much again! – Gold Feb 13 '16 at 20:09