I am trying to understand the topic of the title. If I consider a central Hamiltonian, so an Hamiltonian of the form $H=\frac{p^2}{2m}+V(r)$ what are the logical steps that lead me to the known result?

What I think I've understood is:

1. We notice that a central Hamiltonian commutes with operators $L^2$ and $L_z$ so there exist an orthonormal basis formed by their simultaneous eigenstates and since it is convenient work with states that give us immediately the eigenvalues of three operators, that form a basis and that are also eigenstates of $H$, so we can know the time evolution of every state expressed as a linear combination of the simultaneous eigenstates we search for these.

P.S.: we could also look just for the eigenstates of the Hamiltonian with TISE but the simultaneous eigenstates of three operators are even more convenient.

2. We want to know these simultaneous eigenstates (or eigenfunctions) of $H,L^2,L_z$, how do we do this? With the time-independent Schroedinger equation so we write the equation: $H\psi(r,\theta,\phi)=E\psi(r,\theta,\phi)$ in spherical coordinates (because it's easier to resolve).

Now comes the problem and the question: Why do we assume that those simultaneous eigenstates are, in Schroedinger representation, equal to $R(r)\Theta(\theta)\Phi(\phi)$?

Is this what we assume? I know that expressing the TISE in spherical coordinates leads to a PDE that can be separated into 3 ODE assuming a solution of the form $R(r)\Theta(\theta)\Phi(\phi)$ but what comes first? Do we search the total eigenfunctions in a factorized form beacuese we want the PDE to become an ODE or because in a factorized form it's equal to the simultaneous eigenstates that we're searching?

My problem is I am not understanding the order of logical steps, starting from the central Hamiltonian how do we arrive to search the solution of the TISE as $\psi(r,\theta,\phi)=R(r)\Theta(\theta)\Phi(\phi)$.

Please, if you can, explain the steps a physicist does when in front of this Hamiltonian like $H_{central} \rightarrow H,L^2,L_z$ $commute$ $\rightarrow$ there exist a basis...

  • $\begingroup$ @CosmasZachos Yes I did, spherical harmonics are the product of $\Theta(\theta)\Phi(\phi)$ in particular the $\Theta(\theta)$ is Legendre polynomials and $\Phi(\phi)=\frac{1}{\sqrt{2\pi}}e^{im\phi}$. $\endgroup$
    – Salmon
    Commented May 16, 2022 at 11:20
  • $\begingroup$ True, but the Hamiltonian is a function of r and $L^2$, not $L_z$, which you use to organize solutions. As a rule, the factorization of the wf is in terms of sums of Ys. So your generic solution is a sum of the form you wrote. $\endgroup$ Commented May 16, 2022 at 13:53

1 Answer 1


The point is that, as you doubtlessly covered in class, the homogeneous TISE PDE is separable in spherical coordinates, $$ \left ( \nabla^2 + \frac{2m}{\hbar^2}(E-V(r))\right ) \psi =0, \hbox {where} \\ \nabla^2 = \frac{1}{r^2} \partial_r ( r^2\partial_r) -\frac{L^2}{\hbar^2r^2}, \\ -L^2/\hbar^2 \equiv \frac{1}{\sin \theta} \partial_\theta (\sin\theta ~ \partial_\theta) +\frac{1}{\sin^2\theta } \partial^2_\phi . $$ You have already studied, presumably, the eigenspectrum of $L^2$ and its degeneracies, quantified by m, the eigenvalues of $L_z$, so $L^2 Y_l^m (\theta, \phi)= \hbar^2 l(l+1) Y_l^m$.

The TISE then separates to two sides with operators dependent on completely different variables, $$ {L^2\over \hbar^2} \psi = \left (\partial_r (r^2 \partial_r) + \frac{2mr^2}{\hbar^2}(E-V(r)) \right )\psi, $$ so a factorized $\psi= R(r) Y_l^m(\theta, \phi)$ will provide a solution as long as the eigenvalues of both sides match (identify), $$ {L^2 Y_l^m\over \hbar^2 Y_l^m}=l(l+1) ={ \left (\partial_r (r^2 \partial_r) + \frac{2mr^2}{\hbar^2}(E-V(r)) \right ) R(r) \over R(r)}. $$

Indeed, this matching provides the standard radial eigenvalue equation, upon rearrangement specifying E and the R(r) eigenfunctions (which depend on l, in general).

NB. In the above, the eigenfunctions of $L^2$ are really $\sum_m Y^m_l$, since m does not appear in the equation (ugh! bad notation for the mass there!). It is a convenience index to parametrize the degeneracy of the eigenfunctions of $L^2$ ! So the solutions need a specific l, but not a specific m.

  • 2
    $\begingroup$ 1. In this, (inferior, since PDE, not matrix mechanics) method, all solutions must solve the TISE so you should not miss any. The completeness issue is one of the spectra of Hermitian hamiltonians. I showed to you the separable equation cannot have non-factorized solutions! 2. This is correct. It's easier. For the Kepler potential, you may use parabolic coordinates, which is not as easy to the uninitiated. You might not wish to pay too much attention to your book, unless it is Sakurai & Napolitano or Landau & Lifschitz, of course. Of course! $\endgroup$ Commented May 16, 2022 at 21:29
  • 2
    $\begingroup$ A bit. BUT, as I insisted, m is a bit of a canard of convenience in organizing your solutions (and physics). Any linear combination of spherical harmonics of different m s with the same l will be a solution as well. The commutation of $L_z$ with $L^2$ is a consequence of spherical symmetry. $\endgroup$ Commented May 16, 2022 at 21:43
  • 2
    $\begingroup$ Nono... I was sloppy. Any $R_l(r)\sum_m c_m Y_l^m (\theta,\phi)$ in the above will be a solution with the same energy. The above eigenvalue equation does not depend on m. This is a degeneracy index due to spherical symmetry, but the mathematical problem does not care about it. $\endgroup$ Commented May 16, 2022 at 22:01
  • 1
    $\begingroup$ No, there is no explicit $L_z$ in your equation. This is something you introduced out of a hat to organize/resolve the eigenfunctions of $L^2$. The linear combination above solves the TISE, but is not an eigenfunction of $L_z$, something you simply chose to consider to expedite the systematics. $\endgroup$ Commented May 16, 2022 at 22:11
  • 1
    $\begingroup$ Yes to both. In point of fact, H commutes with $L^2$, $L_z$, $L_x$, $L_y$, which is what spherical symmetry means. But, of course, the 4 operators I wrote don't all commute among themselves... $\endgroup$ Commented May 16, 2022 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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