# Angular Momentum Operators Non-Degenerate

Typically one writes simultaneous eigenstates of the angular momentum operators $J_3$ and $J^2$ as $|j,m\rangle$, where

$$J^2|j,m\rangle = \hbar^2 j(j+1)|j,m\rangle$$ $$J_3 |j,m\rangle = \hbar m|j,m\rangle$$

There seems to be an implicit assumption that the eigenvalues of these operators are non-degenerate. I can't immediately see how this is obvious. Could someone point me in the direction of a reference, or clarify it in an answer? Apologies if I've missed something trivial!

The degeneracy or non-degeneracy of these states depends on the problem's hamiltonian as well as on the system's Hilbert space, so this question doesn't really have an answer. However, the angular momentum states will typically be degenerate in the sense that multiple (linearly independent) states will have the same angular momentum characteristics.

For a simple case, consider a single particle in 3D with a spherically symmetric potential. Then you can decompose the wavefunction into radial and angular parts and the latter can always be assumed to have well-defined total and $z$-component angular momenta: you can always write $$\Psi(\mathbf{r})=\psi(r)Y_{lm}(\theta,\phi).$$ However, you still need to deal with the radial wavefunction, and that will typically will have an infinity (either discrete or discrete + continuum) of energy eigenstates. In that sense the angular momentum states are "degenerate" though of course the energies can depend on $l$.

On a more general, representation theoretic sense, this is still true. If you have some system of particles in 3D then you can always decompose the total system Hilbert space into a direct sum of subspaces with well-defined $J^2$, within which the $J_3$ eigenstates are a good basis. That much is the theorem. However, this doesn't say anything about how many such subrepresentations there will be, what their total angular momentum can be, or even whether it's a good idea to make such a decomposition in the first place (which it won't if the system has other, stronger symmetries!).

What that means in practice is that you need to add a third quantum "number" to your states to get uniquely defined states. This is usually done by notations of the form $$|\alpha,j,m\rangle$$ where $\alpha$ stands for "all the other quantum numbers of the problem" and therefore will generally be an ordered tuple of numbers. (In the hydrogen atom, for example, it suffices to take $\alpha=n$, the principal quantum number.) This index $\alpha$ then tells you which of the many $J^2=\hbar^2j(j+1)$ representations the state belongs to. To see this notation in action see e.g. these notes on the Wigner-Eckart theorem.

Edit: a word on ladder operators.

Angular momentum ladder operators are linear combinations of angular momentum components ($J_\pm=J_1\pm i J_2$) and since representations are invariant under the action of $\mathbf{J}$, that means the action of $J_\pm$ on a state with well-defined $\alpha$ and $j$ will take it to a state with the same $\alpha$ and $j$ (i.e. in the same subrepresentation).

What this means is that you can define the ladder operators without worrying about what subrepresentation they act on - since their action is the same on all - and then restrict your attention to a fixed subrepresentation with no consequence. When you consider superpositions of states from different representations (like you would if you have an arbitrary radial wavefunction, for instance), the ladder operators work like they should on the different $|\alpha,j,m\rangle$ states, and by linearity this is enough to see how they should behave.

The take-home message is that the angular momentum algebra works fine no matter how many representations you have. If you want to find that out, though, then you do need to worry about exactly how your system looks like.

• Ah right - that makes sense and ties in with other quantum mechanics I've done. With reference to TMS's answer, this surely means that the ladder operators 'miss out' some eigenstates. I guess when you first introduce the ladder operators it's okay to make the simplifying assumption of non-degeneracy. Later on when you add further degrees of freedom this ceases to be true. But so long as you know how those degrees of freedom behave you are alright. So really I should be adjusting my notional Hilbert space according to how much information I have, right? Many thanks! Commented Dec 14, 2012 at 17:05
• But how can I see that $J_{\pm}$ leaves the $\alpha$ unchanged? I can see it if its corresponding operator commutes with $J_{1}$ and $J_{2}$, but it doesn't need to be, does it? Commented Jan 5, 2015 at 19:48
• I agree with @guillefix above -- it's not immediately clear to me that $J_{\pm}$ leave the $\alpha$ unchanged. That the other operators in the CSCO (whose eigenvalues are represented by $\alpha$) commute with $J_{\pm}$ is sufficient to guarantee this, but how are we assured of this in general? Can you comment on this if you get the chance? I can also ask a new question if you prefer.
– EE18
Commented Feb 6, 2023 at 16:41

Good for you that you noticed that, this is not obvious from the equations you wrote above, but this is assumed during the derivation and introducing the Ladder operators, check there derivation in any standard text book, more precisely this happens when they derives that: $$\hat{T}_{+}\,\psi_{m}=\alpha\left(m\right)\,\psi_{m+1}$$ Where $\alpha$ is some function of $m$, one derives it by non-degeneracy assumption.

If you still can't get it, I will explain it for you in more detail once I have time.

• Yes that's the point at which the assumption becomes important - thanks for clarifying! Commented Dec 14, 2012 at 17:01
• I am stuck at this point and I find Edward Hughes's post is not satisfactory.. Why does the simultaneous eigen-problem of $J^2$ and $J_z$ have no degeneracy regarding $J_z$-eigenvalues? Commented Oct 17, 2015 at 22:40
• @Guldam, note that the so-called "Casimir Operator" $J^2$ commutes with all of $J_x,J_y,J_z$, thus the non-degeneracy assumption of $J_z$ implies non-degeneracy of $J^2$.