# Construct components of tensor operator [closed]

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?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ZeroTheHero, Emilio Pisanty, sammy gerbil, JMac, AccidentalFourierTransform
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – WhatIAm Aug 21 '18 at 2:34
• Could you point to this source please? – Rourke Sekelsky Aug 21 '18 at 3:38
• 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. – WhatIAm Aug 21 '18 at 3:47

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 $[J^{-(s=2)}]_{m(+1)}$.