3
$\begingroup$

In $3$-dimensional space, any given irreducible representation of the rotation group has a basis whose states are uniquely labeled by the eigenvalues $m$ of a single observable $J_z$, which is one of the components of the angular momentum. The notation $|j,m\rangle$ is often used, where $j$ specifies the irreducible representation. The other two components $J_x$ and $J_y$ don't commute with $J_z$.

In $N$-dimensional space, with $N=2k$ or $N=2k+1$, angular momentum has $\binom{N}{2}$ linearly independent components. We can choose up to $k$ components that commute with each other. But when $N> 3$, specifying the eigenvalues of these $k$ commuting observables often does not uniquely specify a single state within a given irreducible representation.$^\dagger$ Therefore, to uniquely label the states in a basis, we need at least one additional observable that commutes with the $k$ commuting generators. A Casimir won't work, because Casimirs are invariant under rotations: they can't distinguish between states within an irreducible representation.

Question: What are these extra observable(s) that we need to uniquely label the basis states in an irreducible representation of $SO(N)$ when $N>3$?

Example: For $N=5$, let $J_{jk}$ denote the generator of rotations in the $j$-$k$ plane. Then $J_{12}$ and $J_{34}$ commute with each other, but we need at least one more observable that commutes with these. The combination $J_{15}^2+J_{25}^2+J_{35}^2+J_{45}^2$ is a candidate: it commutes with $J_{12}$ and $J_{34}$, and it's not invariant under rotations. But is this the only extra observable we need? What's the general pattern for arbitrary $N$?


$^\dagger$ In the language of Lie-algebra representation theory, this is because an irreducible representations can have some weights with multiplicity $>1$. For examples of irreducible representations of $SO(5)$ having weights with multiplicity $>1$, see https://arxiv.org/abs/1511.02015.

$\endgroup$
1
$\begingroup$

Looks like you already specified your CSCOs, unless I am missing something. For even N, your k commuting su(2)~so(3) $j_z$s will completely characterize your N-vector by their k $j_z$s; while for odd N, you may extend your construction, which is manifestly adequate for complete characterization of any 5-vector: $$J_{12} \leadsto J_{12}^2; ~~ J_{34} \leadsto J_{34}^2; ~~J_{15}^2+J_{25}^2+J_{35}^2+J_{45}^2;$$ it should then work for any tensoring of such.

For generic N=2k+1, then, the manifestly commuting set $$ J_{12} ; J_{34}; ...; J_{2k-1,2k}; ~~J_{1, 2k+1}^2+J_{2, 2k+1}^2+ ...+ J_{2k,2k+1}^2 $$ appears adequate, no?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for the answer. I can see that these are SCOs (sets of commuting observables), but it's not clear to me that they are CSCOs (complete sets of commuting observables). How would one demonstrate that the $k$ commuting generators are sufficient when $N$ is even, and that those together with the single extra observable $\sum_{n<N} J_{nN}^2$ are sufficient when $N$ is odd? $\endgroup$ – Chiral Anomaly Aug 25 at 23:04
  • 1
    $\begingroup$ Well, for the N vector rep, you just work out the antisymmetric NxN matrices... you do see that, no? The k guys and their symmetric squares net 2k, and the extra symmetric one the odd one out. The tensor products of the vector reps will preserve this structure, I think, in the relevant subspaces. I am unsure about the spinor reps. Simple examples might illuminate the structure... $\endgroup$ – Cosmas Zachos Aug 26 at 0:05
  • $\begingroup$ Thank you for your patient comment! It seems clear to me now. By considering both the $k$ generators and their squares, we get $2k$ linearly independent and mutually commuting matrices in the defining (vector) representation. In the odd-$N$ case, we get one more using $\sum_{n<N} J_{nN}^2$. Altogether, we have $N$ mutually commuting matrices in the defining representation, so this must be a complete set in any representation. Going to the covering group (spin group) doesn't add any new dimensions to the group, so the same set of matrices must still be complete in the spinor case. $\endgroup$ – Chiral Anomaly Aug 26 at 0:34
  • $\begingroup$ Yes, something like that. Francesco Iachello's nifty 2006 Springer classic Lie Algebras and Applications has a lot in Ch.Sec 5.3: 5.3.1, 5.3.2, but it pays to come prepared.... $\endgroup$ – Cosmas Zachos Aug 26 at 0:37
  • 1
    $\begingroup$ Minor correction to my previous comment, in case anybody's watching: I meant to write "...so the same set of operators must still be complete in the spinor case." The matrices that represent those operators are different in different representations, by definition. $\endgroup$ – Chiral Anomaly Aug 26 at 0:50
0
$\begingroup$

The hidden $SO(4)$ symmetry of the $1/r$ central potential is a very instructive example. I'll attempt to construct an answer to your question motivated by this example:

The conservation of the Runge-Lenz vector $\mathbf{M}$ around a $1/r$ potential leads to a hidden $SO(4)$ symmetry (see Section 4.1 of Sakurai's Modern Quantum Mechanics for an excellent discussion). Re-scaling to give it units of angular momentum, $\mathbf{M} \rightarrow \mathbf{N}$ you can define new operators from $\mathbf{N}$ and the $SO(3)$ physical angular momentum $\mathbf{L}$: $$\mathbf{I} = (\mathbf{L} + \mathbf{M})/2, $$ $$\mathbf{K} = (\mathbf{L} - \mathbf{M})/2, $$ which satisfy independent commutation relations $$ [I_i, I_j] = i \hbar \epsilon_{ijk} I_k,$$ $$ [K_i, K_j] = i \hbar \epsilon_{ijk} K_k,$$ $$ [I_i, K_j] = 0.$$ From this it might appear that there is a redundancy, and a state might be labelled with the eigenvalues of, say, $I_1$, $K_1$, $\mathbf{I}^2$, and $\mathbf{K}^2$, but it is easy to check (see Sakurai) that $\mathbf{I}^2 - \mathbf{K}^2 = \mathbf{L} \cdot \mathbf{N} = 0$, which enforces that the eigenvalues of $\mathbf{I}^2$ and $\mathbf{K}^2$ must be equal. Therefore, you have yourself a full set of 3 physical observables to label your quantum state $\lvert m_I, m_K, i\rangle$ or $\lvert m_I, m_K, k\rangle$ .

For a general group $SO(N)$, I posit without proof that it should be sufficient to replicate the above process by decomposing the space into independent (commuting) subspaces of the "angular momentum", and to identify relationships between the eigenvalues of these operators to eliminate redundancies.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, this is interesting. But from what I understand about the LRL vector, it's defined only within a particular class of models. In particular, it seems to rely on the existence of a position operator, and I don't see a clear way to generalize it for an arbitrary model with $SO(4)$ symmetry (or $SO(N)$ symmetry), like model of quantum fields, for example. Still, for models where the LRL vector is defined, I'd be interested in knowing why the set of observables you listed is expected to be sufficient for uniquely labeling the basis states in an $SO(4)$ irrep. That's not obvious to me. $\endgroup$ – Chiral Anomaly Aug 25 at 22:56
  • $\begingroup$ I don't believe that the existence of the LRL vector is actually necessary, I merely used it as an example to motivate the construction of the operators $\mathbf{I}$ and $\mathbf{K}$ from "more physical" obvservables. In any case, it seems to me from the example of $SO(4)$ that one should always be able to construct mutually-commuting sub-groups of the larger group of rotations. $\endgroup$ – Matthew O'Brien Aug 25 at 23:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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