1
$\begingroup$

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?

$\endgroup$
0
1
$\begingroup$

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 $$L^2(\mathbb R^+ \times S^2) \simeq L^2(\mathbb R^+) \otimes L^2(S^2).$$

For further references, have a look at section II.4 in Vol. 1. of Reed, Simon's Methods of modern mathematical physics.

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$ – Gold Feb 13 '16 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.