I'm working through, developing the formalism for an octonion generalization of angular momentum operators, but the whole time I'm thinking that this has probably been done elsewhere. My searches have come up with some hints that this is already out there, but nothing solid.

Working this through is fine, if a little slow. But if this development is already out there I'd love to get a lead so I can avoid reinventing formalism.

Quaternions have these commutation relations:

For $ i,j,k \in \{1,2,3\} $

$$ [e_i, e_j]=2 \epsilon_{ijk} e_k$$ And the usual angular momentum operators have a similar commutation relation. $$[J_i, J_j]=i \hbar \epsilon_{ijk} J_k$$

I'm interested in generalizing the usual angular momentum operator formalism from the usual 3 to include 7 operators which have the same commutation rules as the octonions.

Octonions have these commutation relations:

For $ i,j,k \in \{1,2,3,4,5,6,7\} $ $$ [e_i, e_j]=2 \epsilon_{ijk} e_k$$ We use this to define a generalization of the angular momentum operators. $$[J_i, J_j]=i \hbar \epsilon_{ijk} J_k$$

In the ordinary case of spin operators, with spin-1/2, there is a multiplication rule that matches the quaternion multiplication rule $$S_1 S_2=\frac{i \hbar}{2} S_3$$ But with spin-1, there isn't $$S_1 S_2 \neq \frac{i \hbar}{2} S_3$$ So when we manipulate these octonion $J_i$ operators, it's critical to remember that while they have the same commutation relation as the octonions, these don't nescessarily have the same multiplication rules.

I'm using the number convention where $$e_0 \equiv 1 $$ $$e_1 \equiv i, \enspace e_2 \equiv j, \enspace e_3 \equiv k $$ $$e_4 \equiv l $$ $$e_5 \equiv il=m, \enspace e_6 \equiv jl=n, \enspace e_7 \equiv kl=p $$ Which is consistent with the multiplication table found in the Wikipedia article on Octonions.

There are 3 triples (quaternion loops) that include each octonion. The three loops that include $e_3$ are 123, 473 and 653. $$e_1 e_2=e_3, \enspace e_4 e_7=e_3, \enspace e_6 e_5=e_3$$

If we define vector operators that use these loops we get the following: $$ \vec A \equiv (J_1, J_2, J_3),\enspace \vec B \equiv (J_4, J_7, J_3) ,\enspace \vec C \equiv (J_6, J_5, J_3)$$ then $$A^2=J_1^2+J_2^2+J_3^2, \enspace B^2=J_4^2+J_7^2+J_3^2, \enspace C^2=J_6^2+J_5^2+J_3^2$$ and we get three pair of ladder operators $$A_\pm \equiv J_1 \pm iJ_2, \enspace B_\pm \equiv J_4 \pm iJ_7, \enspace C_\pm \equiv J_6 \pm iJ_5, \enspace $$

And some of the new identities are pretty simple.

$$[A_+, B_+]=-2 \hbar C_-, \enspace [A_-, B_-]=2 \hbar C_+$$ $$[A_+, B_-]=0, \enspace [A_-, B_+]=0$$

If we define $$\vec J \equiv (J_1, J_2, J_3, J_4, J_5, J_6, J_7)$$ then $$J^2 = J_1^2+J_2^2+J_3^2+J_4^2+J_5^2+J_6^2+J_7^2$$ It's also easy to show that $$J^2 =A^2+B^2+C^2-2 J_3^2$$ and $$[J_3, J^2]=0, \enspace [J_3^2, J^2]=0 $$
Using the definitions $$\{X, Y\} \equiv XY+YX$$ and $$\{X, Y, Z\} \equiv X\{Y, Z\}+\{Y, Z\}X$$ It can be shown that $$[A^2, B^2]=[B^2, C^2]=[C^2, A^2]=i \hbar (\{ J_1, J_4, J_5 \}+\{ J_1, J_7, J_6 \}+\{ J_2, J_4, J_6 \}-\{ J_2, J_7, J_5 \})$$ which, then, gives us $$[A^2, J^2]=[A^2, A^2+B^2+C^2-2 J_3^2]=[A^2, B^2]+[A^2, C^2]=0$$
This gives us the first main result, that we have three main compatible observables: $$J_3, A^2, J^2$$
It seems like this formalism would have been developed somewhere else.

Has anyone seen a development like what I'm trying to do (probably something better)?

  • $\begingroup$ I'm closing this question as non-mainstream. Our Q&A system does not do well with open-ended "researchy" questions like "does anyone have any general advice on how to proceed with this development", and we also explicitly don't do peer review. Comments are also not the correct place to debate such things. $\endgroup$
    – ACuriousMind
    Dec 30, 2019 at 19:30