Let $V := \operatorname{span}{(J_1, J_2, J_3)}$ be a Lie algebra over the complex numbers such that

  1. $J_1$, $J_2$, and $J_3$ are essentially self-adjoint operators on some Hilbert space $\mathcal{H}$.
  2. The Lie bracket is given by the commutator.
  3. We have that $[J_i, J_j] = \mathrm{i}\epsilon_{ijk}J_k$.

Let $J^2 := \sum_i J_i \circ J_i$. Then is it true that

  1. $J_i$'s have a pure point spectrum.
  2. The common eigenvectors of $J^2$ and $J_3$ come as families $\psi_{j(j+1), m}$, where $m = -j, -j+1, \ldots, j-1, +j$. The eigenvalue $j(j+1)$ is associated to $J^2$ and $m$ is associated to $J_3$ with $j \in \mathbb{N}_0/2$.

I ask this question here as the above conclusions hold for finite dimensional Hilbert spaces, where all linear operators are bounded. We find the above analysis done in almost all introductory QM textbooks, for spin operators, using ladder operators. I want to ask if we can do the same analysis for orbital angular momentum operators.

  • $\begingroup$ I find that in many books, the authors explicitly find the eigenvalues and eigenvectors of the orbital angular momentum operators, and show that they satisfy the two conditions. $\endgroup$ Commented May 17 at 20:00
  • $\begingroup$ One another way of finding the eigenvalue-eigenvector relations for the orbital angular momentum operators seems by showing that the operators have pure point spectrum explicitly (or perhaps by showing that they are elliptic differential operators on S^2 which have pure point spectrum? i don't know), and then using the fact that two (strongly) commuting self-adjoint operators on a Hilbert space have a common eigenbasis. $\endgroup$ Commented May 17 at 20:05
  • $\begingroup$ Although this is a purely mathematical question (as phrased), I believe this should be on-topic here for its relevance to angular momentum operators in QM. $\endgroup$ Commented May 17 at 20:07
  • $\begingroup$ @ValterMoretti Thank you. I've removed commuting from 1. $\endgroup$ Commented May 17 at 20:15
  • $\begingroup$ You are asking something about spherical harmonics, but it is not evident what... $\endgroup$ Commented May 17 at 20:22

1 Answer 1


With the only assumption that $J^2$ is essentially selfadjoint in a dense invariant common domain where the symmetric operators $J_k$ satisfy the usual commutation relation, the thesis is essentially true: These operators are essentially selfadjoint as well and the closures are the generators of a unique strongly continuous rep of $SU(2)$ in the considered Hilbert space. That is consequence of the Nelson theorem. In turn, as the Lie group is compact, the Hilbert space is the Hilbert direct sum of finite dim irreducible reps of $SU(2)$ due to the celebrated Peter-Weyl theorem. Since these irreducible representations are finite dimensional, the standard argument applies to each of them, giving rise to the usual spectra in each invariant subspace. The total spectrum is nothing but the union of these spectra with a suitable multiplicity. In summary, the spectra of the $J_k$ always are pure point spectra with the standard values.

The spectrum of the $J_j$ in the whole Hilbert space is not necessarily bounded. It depends on the spectrum of $J^2$: in each irreducible subrepresentation it is constant, but the number of these constants (of the form $j(j+1)$ with $j\in \mathbb{N}$) may be arbitrarily large.

  • $\begingroup$ I guess it should be that "the spectrum of J^2 is bounded from below" instead of J_i's. I will change that. $\endgroup$ Commented May 17 at 20:48
  • 3
    $\begingroup$ Indeed, $\sigma(J^2) \geq 0$ necessarily, but $\sigma(J_k)$ is not necessarily bounded below (or above). $\endgroup$ Commented May 17 at 21:01

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