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I have a $\mathfrak{su}(3)$ Lie algebra spanned by 8 generators, $$\left\{J_1,J_2,J_3,J_4,J_5,J_6,J_7,J_8 \right\}$$

Now, I can choose infinitely many $\mathfrak{su}(2)$ sub-algebras composed of three operators (does not have to be generators but also linear combinations). For every such sub-algebra I can define Casimir operator. I want to know if Casimir operators from different sub-algebras commute with each other or not? Maybe there is some theorem I don't know about?

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  • $\begingroup$ Related: physics.stackexchange.com/q/64929/2451 $\endgroup$ – Qmechanic Dec 12 '14 at 13:57
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    $\begingroup$ Why don't you just test if they commute? You know all the commutation relations. (Spoiler: As $\mathfrak{su}(3)$ is semisimple, it has only one quadratic Casimir, meaning they won't.) $\endgroup$ – ACuriousMind Dec 12 '14 at 13:58
  • $\begingroup$ possible duplicate of Definition of Casimir operator and its properties $\endgroup$ – Brandon Enright Dec 13 '14 at 0:21
  • $\begingroup$ Perhaps it is a better question than ACuriousMind gave it credit for? The OP is probably impressed that the quadratic Casimirs of the I,V, and U-spin subalgebras are proportional to diag(1,1,0), diag(1,0,1), and diag(0,1,1) in the triplet, 3: He does not care for SU(3) Casimirs. $\endgroup$ – Cosmas Zachos Mar 7 '16 at 17:12
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A good question, especially, as I indicated in my comment above, the Casimirs of U-spin, I-spin and V-spin, all diagonal in the Gell-Mann basis do commute. But I will provide below a simple counterexample, where they don't. By the way, the title of the question confuses people into thinking about SU(3) Casimirs, not SU(2) ones, which you are asking about. You might have retitled it to "SU(2) Casimirs in SU(3)".

So here are two SU(2)s I am considering, in the Gell-Mann basis. The first one is just I-spin, comprised of $\vec{I}=(\lambda_1,\lambda_2,\lambda_3)$, and the 2nd one the formally equivalent one, $U \vec{I} U^{-1}$, where, for simplicity, take $U\equiv \exp(i\theta \lambda_4)$. The corresponding quadratic SU(2) Casimirs are, un-normalized, C=diag(1,1,0), idempotent, and $UCU^{-1}$, respectively.

You are asking if $[C,UCU^{-1} ]$vanishes or not. It suffices to check that the term linear in θ fails to vanish (the zeroth order term vanishing, of course). That is, you are inspecting $$ i\theta (C\lambda_4 C -C\lambda_4 -\lambda_4 C+C\lambda_4C). $$ Since $C\lambda_4 C=0$, and the other two terms in the parenthesis add up to $-\lambda_4$, the commutator fails to vanish, $=-i\theta \lambda_4$ plus higher orders in θ, and the answer to your question is: not.

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