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:



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 $\psi_{lm}(r,\theta,\phi)$.

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

1 Answer 1


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 $L_z$.

  • $\begingroup$ 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. $\endgroup$
    – Salmon
    Jan 19, 2022 at 16:34
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 19, 2022 at 16:42
  • $\begingroup$ Is this related to the method of separating variables used to solve PDEs since only derivatives with respect to angles appear in the system? $\endgroup$
    – Salmon
    Jan 19, 2022 at 16:56
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    $\begingroup$ 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 $r$. $\endgroup$ Jan 19, 2022 at 17:01
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    $\begingroup$ 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. $\endgroup$ Jan 19, 2022 at 17:26

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