In this pdf


talking about the particle in a central field problem, on page 3, the following sentence is written:

$H=\frac{-1}{r^2}\frac{\partial }{\partial r}r^2\frac{\partial }{\partial r}+\frac{L^2}{r^2}+V(r)$ hence, all of the angular dependence of $H$ is contained in $L^2$ and we immediately conclude that:

[$H$,$L^2$]=[$H,L_z$]=$0$ which means that the eigenfunctions of $H$ are also angular momentum eigenfunctions!

I can't understand this last sentence.

All I know from theory is that "If two operators do commute , they have a complete set of simultaneous eigenfunctions."

So I agree that $H,L^2,L_z$ have a complete set of simultaneous eigenfunctions, but I also know that not all the eigenstates of one operator are also eigenstates of the other operators i.e, there might be eigenstates of the Hamiltonian that are not simultaneous eigenstates of the three operators.

So what is the meaning of that sentence?

Maybe it refers to the fact that I can express any eigenstates of the Hamiltonian as a linear combination of simultaneous eigenvectors (since they form a complete set) and so:

$$L^2|H\rangle=L^2\sum_{i} c_i|H,L^2,L_z\rangle~?$$


1 Answer 1


Yes this is not clearly written. $H$ and $L^2$ have a common set but this does not mean that every eigenstate of $H$ is also an eigenstate of $L^2$. The issue comes down to repeated eigenvalues of one or both of the operators.

A simple example would be a combination of two hydrogen atom states with the same energy but different $\ell$ values: clearly this is still an eigenstate of $H$ but by construction not of $L^2$. Alternatively, two hydrogen wavefunctions with different energies but the same $\ell$ are eigenstates of $L^2$ but not of $H$.

The same holds for $L_z$.

Thus: if they commute, there exists a common set, but in a commuting pair the eigenstates of one need not be eigenstates of the other.

  • $\begingroup$ so do you think that the sentence refers to the last equation I've written? $\endgroup$
    – Salmon
    Jan 29, 2022 at 22:08
  • $\begingroup$ your notation is funny…. not sure about $\vert H\rangle$ since $H$ is an operator. Maybe you mean $\vert \epsilon_i\ell_i m_i\rangle$. $\endgroup$ Jan 29, 2022 at 22:10
  • $\begingroup$ Sorry, by $|H\rangle$ I mean an eigenstate of $H$, so $L^2|\epsilon\rangle=L^2\sum_{i} c_i|\epsilon_i,l_i,m_i\rangle~?$ $\endgroup$
    – Salmon
    Jan 29, 2022 at 22:13
  • 1
    $\begingroup$ yes something like that. $\endgroup$ Jan 29, 2022 at 22:17

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