# How do we obtain radial part and angular part with spherical harmonics?

In a website (sorry, I can't find it anymore) I've read this when talking about spherical harmonics:

if we want to determine simultaneous eigenfuntions of $$L^2$$ and $$L_z$$ we have to solve this system:

$$L^2|l,m\rangle=\hbar^2l(l+1)|l,m\rangle$$ $$L_z|l,m\rangle=\hbar m|l,m\rangle$$

in Schrodinger's representation:

$$-(\frac{1}{sin(\theta)}\frac{\partial}{\partial{\theta}}(sin(\theta)\frac{\partial}{\partial{\theta}})+\frac{1}{sin^2(\theta)}\frac{\partial^2}{\partial{\phi^2}})\psi_{lm}(r,\theta,\phi)=l(l+1)\psi_{lm}(r,\theta,\phi)$$

$$-i\frac{\partial}{\partial{\phi}}\psi_{lm}(r,\theta,\phi)=m\psi_{lm}(r,\theta,\phi)$$

we can note that operators $$L^2$$ and $$L_z$$ don't contain variable $$r$$ so the $$\psi_{lm}(r,\theta,\phi)$$ must be of the form $$\psi_{lm}(r,\theta,\phi)=f(r)Y_{lm}(\theta,\phi)$$.

Question: Why if operators $$L^2$$ and $$L_z$$ don't contain variable $$r$$ the $$\psi_{lm}(r,\theta,\phi)$$ must be of the form $$\psi_{lm}(r,\theta,\phi)=f(r)Y_{lm}(\theta,\phi)$$? I think it is related to the fact that $$L^2$$ and $$L_z$$ have a complete set of simultaneous eigenfunctions but I don't understand why this leads to the separated form of the $$\psi_{lm}(r,\theta,\phi)$$.

• The decomposition $\psi=f(r) Y_m^l (\theta,\phi)$ is only adequate for central potentials. Jan 19, 2022 at 15:25

There are several possible more or less elaborated answers, I will discuss here a version which may be of interest, I hope, because it can be extended to other cases. It happens because this approach relies upon general (though elementary) abstract results of Hilbert space theory.

Let $$d\Omega$$ be the unique rotationally invariant (Borel) measure on the two-dimensional sphere of unit radius $$S^2$$ such that $$\int_{S^2} 1 d\Omega = 4\pi$$.

We have an Hilbert space isomorphism, $$L^2(\mathbb{R}^3, dx^3) \equiv L^2(S^2, d\Omega)\otimes L^2([0,+\infty), r^2dr)\tag{1}$$ where the isomorphism is the unique linear and bounded extension of the map $$L^2(S^2, d\Omega)\otimes L^2([0,+\infty), r^2dr) \ni Y\otimes f \mapsto Y \cdot f \in L^2(\mathbb{R}^3, d^3x)$$ where $$(Y \cdot f)(\theta, \phi, r) := Y(\theta,\phi) f(r)\:.$$

That is a special case of the general result $$L^2(X, dx) \otimes L^2(Y, dy)\equiv L^2(X\times Y, dx\otimes dy) \tag{2}$$ where the isomorphism is again the unique linear and bounded extension of the map $$L^2(X,dx)\otimes L^2(Y,dy) \ni f\otimes g \mapsto f \cdot g \in L^2(X\times Y, dx\otimes dy)$$ In the case (1), $$\mathbb{R}^3 = S^2 \times [0,+\infty)$$ and $$d^3x= d\Omega \otimes r^2 dr$$ as is well known passing from Cartesian to polar coordinates.

Let us focus attention on (1). If one writes down the explicit expression of the three selfadjoint angular momentum operators $$L_x,L_y,L_z$$ in $$L^2(\mathbb{R}^3, dx^3)\equiv L^2(S^2, d\Omega)\otimes L^2([0,+\infty), r^2dr)$$ he/she discovers the following interesting fact. $$L_k = {\cal L}_k \otimes I_{L^2([0,+\infty), r^2dr)}\:,\tag{3}$$ for some operators $${\cal L}_k$$ which are differential operators on the dense subspace of smooth (complex valued) functions in $$L^2(S^2, d\Omega)$$. For instance $${\cal L}_z = - i \hbar \frac{\partial}{\partial \phi}$$ and similar identities are valid for the other components, where only derivatives with respect to the angles $$\phi$$ and $$\theta$$ take place: these are the variables used to define wavefunctions in $$L^2(S^2, d\Omega)$$.

From the general theory of strongly continuous unitary representations of compact topological groups, in this case $$SU(2)$$, as the three $$-i{\cal L}_k$$ define a representation of its Lie algebra, one can construct a Hilbert basis of simultaneous eigenvectors of $${\cal L}^2$$ and $${\cal L}_z$$ in $$L^2(S^2, d\Omega)$$. This basis is the very well known family of spherical harmonics $$Y^\ell_m(\theta,\phi).$$

Notice that the coordinate $$r$$ does not appear just because we are working in a Hilbert space where it does not exist!

Let us conclude. As a general result related with (2), if $$L^2(X\times Y, dx\otimes dy)$$ is a Hilbert space and $$\{Y_a\}_{a\in A} \subset L^2(X, dx)$$ is a Hilbert basis whereas $$\{R_b\}_{b\in B} \subset L^2(Y, dy)$$ is another Hilbert basis, then $$\{Y_a \otimes R_b\}_{(a,b) \in A\times B} \subset L^2(X, dx) \otimes L^2(Y, dy)\equiv L^2(X\times Y, dx\otimes dy)$$ is a Hilbert basis as well for $$L^2(X\times Y, dx\otimes dy)$$.

Specializing the result to (1), we see that a Hilbert basis of $$L^2(\mathbb{R}^3, d^3x)$$ is made of the products $$Y^\ell_m(\theta, \phi) f_n(r)$$ where $$\{f_n\}_{n \in N}$$ is any given Hilbert basis of $$L^2([0,+\infty). r^2dr)$$.

Notice that, in view of (3), $$L^2 (Y^\ell_m f_n) = ({\cal L}^2 \otimes I) Y^\ell_m \otimes f_n = \hbar^2 \ell(\ell+1) (Y^\ell_m f_n)$$ and $$L_z (Y^\ell_m f_n) = ({\cal L}_z \otimes I) Y^\ell_m \otimes f_n = \hbar m (Y^\ell_m f_n)$$

More generally, with the same argument, $$L^2 (Y^\ell_m f) = \hbar^2 \ell(\ell+1) (Y^\ell_m f)$$ and $$L_z (Y^\ell_m f) = \hbar m (Y^\ell_m f)$$ for every $$f\in L^2([0,+\infty), r^2dr)$$.

In summary, the functions $$Y^\ell_m(\theta,\phi)f_n(r)$$ form a symultaneous Hilbert basis of eigenvectors of $$L^2$$ and $$L_z$$ nomatter the choice of the Hilbert basis $$\{f_n\}_{n\in N}$$ in $$L^2([0,+\infty), r^2 dr)$$.

With a slightly more elaborated argument, it is possible to choice the functions $$f_n$$ also depending on $$\ell$$ and possibly $$m$$, changing the original functions $$f_n$$ in every eigenspace of $$L^2$$ and $$L_z$$.

• Thanks, but isn't there an easier way to answer the question? It's my first course in quantum mechanics and I'm not able to follow the things you wrote. Jan 19, 2022 at 16:34
• Sorry, I wanted to present a general argument. The easiest way is to check that it works and to use this specific result as a general paradigm for other cases: when an operator does not depend on a variable $z$, it is better to try to find its eigenfunctions as products $f(...)g(z)$, where $...$ does not contain $z$. The reason why it works always is the one I wrote. Jan 19, 2022 at 16:42
• Is this related to the method of separating variables used to solve PDEs since only derivatives with respect to angles appear in the system? Jan 19, 2022 at 16:56
• Yes it is related to it, but not completely. In that case the operator can be decomposed as a sum of operators referred to different variables. Here the variable $r$ appears nowhere. Eigenfunctions of the only variables $\theta$ and $\phi$ exist, but they cannot belong to the whole Hilbert space because the radial part of the integration diverges. The easiest way to fix the problem is to multiply the eigenfunctions with a cutoff function of the only variable $r$. Jan 19, 2022 at 17:01
• It is different if you are looking for the eigenfunctions of the angular momentum, but it becomes the standard problem with separation of variables if you next deals with a Hamiltonian with spherical symmetry. Jan 19, 2022 at 17:26