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I'm reading Georgi's textbook on Lie Algebras and have been struggling with this question for quite awhile. The entirety of Chapter 4 (Tensor Operators) has been much more difficult than anything I've seen following it in this book, or maybe just much less intuitive.

The operator $(r_{+1})^2$ satisfies $[J^+,(r_{+1})^2] = 0$. It is therefore the $O_{+2}$ component of a spin 2 tensor operator. Construct the other components $O_m$. (There is more to the question, but I'll only attempt that once I figure out this first part).

My Attempt: In an example prior to this Georgi lists a couple dozen commutation relations for some arbitrary operators $a_{\pm 1},b_{\pm 1},a_0,b_0,c_0$. This includes $[J^+,a_{+1}] = 0$, implying that the $a_{+1}$ operator produces the highest weight state. He takes $A_{+1} = a_{+1}$ as the highest weight "state," and then applies the lowering operator such that $A_0 = [J^-,A_{+1}] = \frac{1}{2}( a_0 + b_0 + c_0)$; this is just from the commutation relations given, so it's most likely arbitrarily made up. He applies $A_{-1} = [J^-,A_0] = 2a_{-1} + b_{-1}$ again to get to the lowest state, which is the lowest because $[J^-,a_{-1}] = [J^-,b_{-1}] = 0$.

So, I decided to apply the definition of a tensor operator, that is $[J_a, O_l^s] = O_m^s [J_a^s]_{ml}$. In our case the $J_a = J^-$ (with $s=2$), $O_l^s = O_{+2}^2 = (r_{+1})^2$, giving me $[J^-,O_{+2}^2] = 2O_{+1}$. But also there is an equation for tensor products, which gives $[J^-,(r_{+1})(r_{+1})] = \sqrt{6} \left( r_0r_{+1} + r_{+1}r_0 \right)$. This tells me that $O_{+1} = \sqrt{ \frac{3}{2}} \left( r_0r_{+1} + r_{+1}r_0 \right)$. Is this correct?

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closed as off-topic by ZeroTheHero, Emilio Pisanty, sammy gerbil, JMac, AccidentalFourierTransform Sep 1 '18 at 2:20

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  • $\begingroup$ You seem to be essentially on the right track. I've noticed a number of sources have $O_{+1} = \frac {1}{ \sqrt{2}}(r_0 r_{+1} + r_{+1 }r_0 )$. However when I just did it myself I got a $\sqrt{\frac{3}{2}}(r_0 r_{+1} + r_{+1 }r_0 )$ like you did. I think we are at least right up to a scaling factor. $\endgroup$ – WhatIAm Aug 21 '18 at 2:34
  • $\begingroup$ Could you point to this source please? $\endgroup$ – Rourke Sekelsky Aug 21 '18 at 3:38
  • $\begingroup$ info.phys.unm.edu/~ideutsch/Classes/Phys531F11/… or en.wikipedia.org/wiki/… note that this theory is generalized for the product of two different vector operators, so in this case v=w=a=b=r. $\endgroup$ – WhatIAm Aug 21 '18 at 3:47
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This is the correct way to do this, but the normalization is off due to the $r$s being spin 1 operators. That is, the commutator $[J^-,r_{+1}r_{+1}] = \sqrt{2} \left( r_0r_{+1} + r_{+1}r_0 \right)$ as we use $[J^{-(s=1)}]_{m(+1)}$ as the matrix component instead of $[J^{-(s=2)}]_{m(+1)}$.

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